Blog | EGMO 2018http://www.egmo2018.org/blog/BlogenEGMOEGMO CountriesEnglishFlorenceGender IssuesIMOItalianoMy EgmoOlimpiadiProblem solvingWomen in Maths beyond StereotypesTue, 27 Feb 2018 11:47:08 +0000Thinking Out Loud – EGMO 2017 Problem 1http://www.egmo2018.org/blog/EGMO2017-P1/<blockquote>
<p><strong>EGMO 2017, Problem 1</strong>. Let \(ABCD\) be a convex quadrilateral with \(\widehat{DAB} = \widehat{BCD} = 90^\circ\) and \(\widehat{ABC} > \widehat{CDA}\). Let \(Q\) and \(R\) be points on segments \(BC\) and \(CD\), respectively, such that line \(QR\) intersects lines \(AB\) and \(AD\) at points \(P\) and \(S\), respectively. It is given that \(PQ = RS\). Let the midpoint of \(BD\) be \(M\) and the midpoint of \(QR\) be \(N\). Prove that the points \(M, N, A\) and \(C\) lie on a circle.</p>
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<div dir="ltr">Ok, so, problem one. Should be easy. Oh. Geometry. Ok, well maybe it's... oh, it's not a triangle.
<div>A quadrilateral...</div>
<div>Ok ok ok, like the Guide says, <a href="https://youtu.be/tOltAxX_KKA">don't panic</a>!</div>
<div>Let's see... we have a quadrilateral with some right angles, a lot of equal segments and we want a cyclic quadrilateral.</div>
<div></div>
<div>Ah, I said "cyclic"... well, the quadrilateral we start with, $ABCD$ that is, has two opposite right angles, by hypothesis, which is the most famous instance of a cyclic quadrilateral. So, good news: we start with $ABCD$ cyclic and we want $AMNC$ cyclic. </div>
<div></div>
<div><span>A drawing could help...</span></div>
<div><span><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2017-1/egmo2017.1-1.jpeg" width="70%"/></span></div>
<div></div>
<div>But $M$ and $N$ are not really defined in a good way that relates to the angles of $ABCD$... actually, $N$ is defined in a very peculiar way... oh, wait, $N$ is the midpoint or $QR$, but extending $QR$ to meet $AD$ and $AB$ we obtain $S$ and $P$ such that $SR=PQ$, so, to sum it up, $N$ is the midpoint of $PS$ and, as we already knew, $M$ is the midpoint of $BD$.</div>
<div></div>
<div>Aaaaand we have right angles all around! Look at the picture! $PS$ is the hypothenuse of the right-angled triangle $APS$, $BD$ is the hypothenuse in two right triangles, $ABD$ and $CBD$. Oh, wait, $RQ$ is the hypothenuse of $CRQ$. We have a pattern here...</div>
<div></div>
<div>The midpoint of the hypothenuse is the circumcenter of the right triangle: this fact allows us to move a lot of angles around!</div>
<div><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2017-1/egmo2017.1-2.jpeg" width="70%"/></div>
<div></div>
<div>So, let's get some stuff done with $AMNC$... we want to show that it is cyclic. This can be obtained in various ways, but we notice that we have two right-angled triangles with circumcenter $M$, which have the right angle in $A$ and $C$ and two right-angled triangles with circumcenter $N$, which have the right angle in $A$ and $C$. So, it seems reasonable to look at the angles $\widehat{MAN}$ and $\widehat{MCN}$.</div>
<div></div>
<div>For example: let us consider $\widehat{MAN}=\widehat{MAB}-\<wbr/>widehat{PAN}$. In the triangle $BAD$, $MA=MB$, so $\widehat{MAB}=\widehat{ABD}$; in the triangle $PAS$, $NA=NP$, so $\widehat{PAN}=\widehat{APS}$.</div>
<div><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2017-1/egmo2017.1-3.jpeg" width="70%"/></div>
<div>We end up with</div>
<div>$$\widehat{MAN}=\widehat{ABD}-<wbr/>\widehat{APS}$$</div>
<div>and, if we extend $BD$ and $PS$ and let them intersect each other in $E$, we finally obtain $\widehat{MAN}=\widehat{BEP}$ (by viewing $\widehat{ABD}$ as an external angle of $BEP$).</div>
<div></div>
<div>Now, we know what to do: $\widehat{MCN}=\widehat{NCQ}-\<wbr/>widehat{MCB}=\widehat{RQC}-\<wbr/>widehat{DBC}=\widehat{BEQ}$ (by viewing $\widehat{RQC}$ as an external angle of $BEQ$).</div>
<div></div>
<div>This finishes the proof: as $\widehat{MAN}=\widehat{MCN}$, the quadrilateral $AMNC$ is cyclic.</div>
<div>We are done and happy with our job... uhm wait, what was that annoying hypothesis about angles for? Oh, an inequality... that triggers a pavlovian reaction: I have to look for configuration issues (typical, with cyclic quadrilaterals) .... probably, the inequality guarantees that there is only one configuration possible.</div>
<div>Once we make sure of this last thing, the problem is really over and we can happily go on to problem 2, which won't be geometry...</div>
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<p><attachment aria-label="figureEgmo17-A1.pdf" class="Apple-web-attachment" id="<275A27ED-4277-4B08-9419-A2F22DCAF7B0@Home>" role="button" title="figureEgmo17-A1.pdf" type="application/pdf"></attachment></p>Samuele MongodiTue, 27 Feb 2018 11:47:08 +0000http://www.egmo2018.org/blog/EGMO2017-P1/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2014 Problem 3http://www.egmo2018.org/blog/EGMO2014-P3/<p>
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<p>It is a truth universally acknowledged that mathematicians like to state the obvious. As further confirmation of this well-established fact, it was recently pointed out to me that my 'thinking out loud' posts cannot rival with Marcello's. Now, what is one supposed to do in the face of this self-evident truth? Certainly not get offended – how can you get offended by something you agree with? So, maybe panic? Stop making these posts, to everyone's relief? After some deliberation, I decided that what the comment <em>really</em> meant was that I should strive to be more creative in my writing. I cannot guarentee that I succeeded, but it is a fact that I ended up composing an <em>even longer</em> post than usual. So read ahead at your own peril, you've been warned: long-winded ramblings follow.</p>
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<p><strong>EGMO 2014, Problem 3</strong>. We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ for any positive integers $a, b$ satisfying $a + b = n$.</p>
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<p>Now that's an ugly statement if I ever saw one! When a problem is this hideous, it often means that it's been built around a solution, instead of the other way around. And this should make it easier: I suspect that the requirement of having precisely $k$ prime factors might be there just to obfuscate what the problem is really about. Can I solve it with just one prime factor, that is, $\omega(n)=1$?</p>
<p>Let's see: I need to take $n=p^r$ for some prime $p$ and a positive integer $r$. Also I want $d(n)=r+1$ <em>not</em> to divide something, so maybe taking $r+1=q$ to be a large prime could be a good idea. Let's do just that. As for $p$, I don't see any good reason why one choice should be better than another, except that 2 (while one might argue that deep down $2$ is not <em>really</em> a prime number – come on, it's <em>even</em>!) tends to have stranger properties than other primes, so let's take $p=2$. If this doesn't work I'll try $p=3$, but I refuse to believe that I need to take $p=32003$.</p>
<p>So suppose $n=2^{q-1}$, $a+b=n=2^{q-1}$, and $d(n)=q$ divides $d(a^2+b^2)$. Now, where does $d(a^2+b^2)$ take its prime factor $q$ from? From some prime power $p^{q-1}$, or rather, $p^{kq-1}$. But this is a huge number: let's find out how big it really is. We have $a^2+b^2 \leq n^2 = 2^{2q-2}$, so $p^{kq-1} \leq 4^{q-1}$. This definitely cannot happen for $p \geq 5$. What happens for $p=2,3$? Assuming that $q$ is not too small (...to be made precise later), $p^{kq-1} \leq 4^{q-1}$ implies $k=1$. But is it possible that $3^{q-1} \mid a^2+b^2$? A well-known lemma (ok, I must confess – I'm a number theorist by trade. All the same, this problem is still not the most appealing one I've come across in my life) says that in this case $3^{(q-1)/2}$ divides both $a$ and $b$ – and if $(q-1)/2$ is not an integer, then probably I need to round it up. (Do I? Let's see: the basic lemma is that if $3 \mid a^2+b^2$, then $3 \mid a$ – because $3$ is $3 \bmod 4$. So for $q=1$... ehr, ok, I was mistaken: $a,b$ are divisible by $3^{\lceil q/2 \rceil}$. Well whatever, they are divisible by some large power of $3$). So anyway, if $3^{q-1}$ divides $a^2+b^2$, then $a,b$ are divisible by something like $3^{q/2}$. In this case $a+b \geq 2 \cdot 3^{q/2}$, which I was hoping to be larger than $2^{q-1}$, but that doesn't seem to be the case. Oh but wait, I'm being really obtuse: if $3^{\text{something}}$ divides both $a$ and $b$, then $3$ divides $a+b=n=2^{q-1}$, which is more than a little absurd. Ok, so the only possibility is $a^2+b^2 = 2^{kq-1} \times \text{something}$ with $k=1$ or maybe $2$. Can $k=2$ really happen? If I write down the obvious inequalities I find\[2^{2q-2} = n^2 \geq a^2+b^2 \geq 2^{2q-1}\]so no, that does not happen. Fine. Suppose $a^2+b^2=2^{q-1} \times \text{something}$: this would be a little surprising, because the powers of 2 need to recombine in very interesting ways to ensure that the same power of 2 divides both $n$ and $a^2+b^2$ with $a+b=n$. Can I turn this philosophy into an actual statement? Write $a=2^{m} x, b=2^{n} y$ and\[a^2+b^2 = 2^{2m} x^2 + 2^{2n} y^2.\]If (say) $x < y$ then I can collect a term $2^{2m}$ and find that $a^2+b^2 = 2^{2m}(x^2+2^{2n-2m}y^2)$, so the largest power of $2$ that divides $a^2+b^2$ is $2^{2m}$. It should also be $2^{q-1}$, however, so $2m=q-1$ and $n$ is larger. What's the other condition? $a+b=n$. But then no, the largest power of $2$ in $a+b$ is going to be $2^m$ (same argument), and $2^m = 2^{(q-1)/2} < 2^{q-1}$. Ok, this seems to do it, right? Ah no, I'm forgetting one case: what happens if $m=n$? Let me see again,\[a^2+b^2 = 2^{2m} x^2 + 2^{2n} y^2 = 2^{2m} (x^2 + y^2)\]with $x$ and $y$ odd. So $x^2+y^2 \equiv 2 \pmod 4$, and $a^2+b^2$ is divisible exactly by $2^{2m+1}$. So this time $2m+1=q-1$, which however cannot happen because $q$ is odd. Ok, that was not so hard... but boy am I glad that I'm writing this down! I think I've already forgotten 80% of what I've done. I'm also slightly worried at having used this lemma $3 \mid a^2+b^2 \Rightarrow 3 \mid a, b$. Is this going to generalize nicely when I take $n$ with $k$ prime factors?</p>
<p>Ok, let's stop for a sec and think before trying the general case – what have I used so far?</p>
<ul>
<li>That $n$ is small: this seems to be crucial, so I want to make $n$ very small.</li>
<li>On the other hand, I need (maybe? Or maybe I'm just on the wrong track altogether?) a prime with a largish exponent, in order to fabricate a largish prime in $d(n)$.</li>
<li>I've also used that $n$ is not divisible by $3$. I have no clue if this is going to be necessary in general, but it might be.</li>
<li>Putting everything together, I think that I should try to take $n = 2^{q-1} p_2 p_3 \cdots p_k$ where the $p_i$'s are distinct primes that as small as possible (which might mean either $p_2=3, p_3=5, p_4=7, ...$ or $p_2=5, p_3=7, p_4=11, ...$, depending on whether I want to enforce the condition $3 \nmid n$ or not).</li>
</ul>
<p>Ok, let's get started. Have I already mentioned that I don't like this problem? I don't like this problem. So, $n=2^{q-1} p_2 p_3 \cdots p_k$, $a+b=n$, and $d(a^2+b^2)$ is a multiple of $q$. Can this really happen? It means that $a^2+b^2$ (of order roughly $n^2$...) has a prime power factor of the form $r^{kq-1}$ with $r$ prime. Now, let's say something which is a bit false but philosophically very true: if $q$ is ginormous, then $n^2$ looks very much like $2^{2q-2}$. Ok, not really, but let's say that it's sort of true on a logarithmic scale, which is what I care about at the moment (because I'm looking at exponents). So anyway, mathematics, not philosophy:\[a^2+b^2 \leq n^2 \leq 2^{2q-2} (p_2\cdots p_k)^2,\]and on the other hand $r^{kq-1}$ divides $a^2+b^2$, so\begin{equation}\tag{1}\label{eq:ineq}r^{kq-1} \leq 2^{2q-2} (p_2\cdots p_k)^2\end{equation}If $r$ is at least 5, well, then we obtain an inequality of the form\[5^{q-1} \leq 4^{q-1} (p_2\cdots p_k)^2\]which (since the product $p_2 \cdots p_k$ is fixed... ok, not yet, but it will be fixed eventually) is impossible when $q$ is large. So $r$ is either 2 or 3. This is good, because it's the same conclusion we had in the $\omega(n)=1$ case. I know how to rule out 3: it suffices to take $(n,3)=1$. So we're saying $a^2+b^2=2^{kq-1} (\text{something})$, and \eqref{eq:ineq} gives $k \leq 2$. That IS unfortunate: before we had $k=1$ automatically. Ugh, this means one extra case. Fine, let's do it. Actually no, first let's check that the case $k=1$ generalizes. And by the way I've just realized that I'm such an idiot, I'm using $k$ both for $\omega(n)$ and for the exponent of $r$ in $a^2+b^2$. I'll need to fix this when I write an actual solution. Anyway: suppose $a+b=n=2^{q-1}(p_2\ldots p_k)$ and $a^2+b^2=2^{q-1}(\text{something})$. If $v_2(a) \neq v_2(b)$<a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#foot1">*</a> we have $q-1=v_2(a^2+b^2)=2 \min\{v_2(a),v_2(b)\}$ and $v_2(a+b)=\min\{v_2(a),v_2(b)\}$, which leads to a contradiction, just like before. What if $v_2(a)=v_2(b)$? Then $q-1=v_2(a^2+b^2)=2v_2(a)+1$, which is impossible because $q$ is odd. Good, this also works just as before, and leaves me with the case $a+b=n=2^{q-1}(p_2\ldots p_k)$ and $a^2+b^2=2^{2q-1}(\text{something})$. This $2$-adic valuation business is working so well that I think I should probably try using it on the missing case as well. I'll need to deal with the same two cases as above:</p>
<ul>
<li>$v_2(a) \neq v_2(b)$: then $2q-1=v_2(a^2+b^2)=2\min\{v_2(a),v_2(b)\}$. This is odd = even, so nope!</li>
<li>$v_2(a)=v_2(b)$: then $2q-1 = v_2(a^2+b^2) = 2v_2(a)+1$. This is not a contradiction per se, it just gives $v_2(a)=v_2(b)=q-1$. Hopefully this will contradict $a+b=n$, otherwise I'll throw the laptop out of the window in a fit of frustration... So: is it possible that $a+b=n$ with $v_2(a)=v_2(b)=v_2(n)=q-1$? If I write\[a+b = 2^{q-1}x + 2^{q-1} y = 2^{q-1}(x+y) = n...\](at this point I've stared at this equation for more than I care to admit, until...) I AM A TOTAL MORON. $x$ and $y$ are odd, so $x+y$ is even, and $v_2(a+b)=v_2(2^{q-1}(x+y))$ is at least $q$, contradiction.</li>
</ul>
<p>And with this we're finally done. So, to summarise (and write a decent proof in which we try to hide all the ugliness above)</p>
<ul>
<li>We want to construct such an $n$ with $\omega(n)=k$. Take $n=2^{q-1} (p_2 \cdots p_k)$, where $p_2,\ldots,p_k$ are the smallest $k-1$ primes larger than $3$, and $q$ is a prime so large that $5^{q-1} > 4^{q-1} (p_2\ldots p_k)^2$.</li>
<li>In this situation, $d(n)$ is divisible by $q$. We want to show that $a+b=n$ implies $d(n) \nmid d(a^2+b^2)$, so it suffices to show that $q \nmid d(a^2+b^2)$.</li>
<li>Suppose $a+b=n$ are such that $q \mid d(a^2+b^2)$. Then there exist a prime $r$ and an integer (which is most definitely not called $k$, but say $e$) such that $r^{eq-1} \mid a^2+b^2$. Since $a^2+b^2 \leq n^2 = 4^{q-1} (p_2 \ldots p_k)^2 < 5^{q-1}$, this implies that we are in one of the following three situations: $r=2, e=1$; $r=2, e=2$; $r=3, e=1$.</li>
<li>if $r=3$, then $3 \mid a^2+b^2$, which implies $3 \mid a, b$ (in a competition I would write down a proof for this lemma), which in turn implies $3 \mid a+b = n$, contradiction.</li>
<li>if $r=2$, let $A=v_2(a), B=v_2(b)$ [notice that these were called respectively $m$ and $n$ above. Using $n$ here is maybe not such a great idea...]. Then $v_2(a^2+b^2)=eq-1$, and there are two cases:
<ul>
<li>$A \neq B$: then $v_2(a^2+b^2)=2 \min\{A,B\}$, which is even, so $e=1$. But then $q-1=v_2(n)=v_2(a+b) = \min\{A,B\} = \frac{1}{2} v_2(a^2+b^2)=\frac{q-1}{2}$, contradiction.</li>
<li>$A=B$: then $eq-1=v_2(a^2+b^2)=2A+1$, so $e=2$ for parity reasons. But then $q-1=v_2(n) = v_2(a+b) \geq A+1=q$, contradiction.</li>
</ul>
</li>
<li>Since we reach a contradiction in every case, this shows that there are no pairs of positive integers $(a,b)$ such that $a+b=n$ and $d(n) \mid (a^2+b^2)$, as required.</li>
</ul>
<h5><strong>Closing remarks</strong></h5>
<p>$p$-adic valuations are your friends! Especially in a problem like this, where (almost) the only relevant property of $n$ and $a^2+b^2$ is their prime factorisation, you should definitely think in terms of valuations. And if you are not familiar with the notation $v_p(\cdot)$ and its basic properties (one above all: $v_p(x+y) \geq \min\{v_p(x), v_p(y)\}$, with equality whenever $v_p(x) \neq v_p(y)$), you may want to read up on it (one of many possible starting points: <a href="http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/olympiad-number-theory.pdf">http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/olympiad-number-theory.pdf</a>).</p>
<p>A final word of wisdom: powerful as they are, <strong>congruences</strong> (or $p$-adic valuations, in this case) <strong>can't do everything!</strong> As one of the organisers of our national Mathematical Olympiad likes to say, any good number theory problem involves both congruences and inequalities. This should be apparent in the problem we just solved!</p>
<p><a id="foot1"></a></p>
<p><strong>*</strong>here $v_2(x)$ is the exponent of $2$ in the factorisation of $x$: it's called the<em> $2$-adic valuation of $x$.</em></p>Davide LombardoTue, 13 Feb 2018 12:47:37 +0000http://www.egmo2018.org/blog/EGMO2014-P3/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2014 Problem 2http://www.egmo2018.org/blog/EGMO2014-P2/<p>
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<p><strong>EGMO 2014, problem 2.</strong> Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB=BC=CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ are collinear.</p>
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<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/tol.egmo2014.2.jpeg" style="float: left; padding-right: 10px;" width="45%"/>It is nice to sit in front of my pc and write this solution, remembering the first day of contest at my first EGMO, in Antalya, 2014.</p>
<p>Needless to say, a large well-drawn figure is the first thing you need, when facing a geometry problem. It is usually a good idea to initially make a quick sketch of the figure in order to understand what adjustments can be taken to make the drawing clearer, and then draw the accurate final figure.</p>
<p>Now, let us analyze the first hypothesis: since $DB=BC=CE$, $CBE$ and $BCD$ are isosceles triangles. In particular $\angle BCD = \angle BDC$ and $\angle CEB=\angle CBE$. Along with highlighting equal angles with the same colour on the figure<strong><a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#fig" style="text-decoration: none;">*</a></strong>, if it is easy as in this case, it may be a good idea to write the measures of the angles explicitly. Using the standard notation we have $\angle BCD = \angle BDC = 90 - \frac{\beta}{2}$ and $\angle CEB=\angle CBE = 90-\frac{\gamma}{2}$.</p>
<p>Since $I$ is the incentre of $ABC$ we have $\angle IBC=\frac{\beta}{2}$ and $\angle BCI = \frac{\gamma}{2}$. Now we see that having explicitly written the measures of the angles has proved to be useful: the presence of a $\frac{\gamma}{2}$ and a $90-\frac{\gamma}{2}$ reveals the existence of a right angle: defining $X = CI \cap BE$ and $Y=CD \cap BI$, we have $\angle CXB=90$ and similarly $\angle BYC=90$. We now look at the figure, and try to use all of the other hypotheses. $H$ is an orthocentre and we have just discovered two right angles, so let's work on this! Being $H$ the orthocentre of $DEF$, we have $DH \perp FE$ . We have just noticed that $FE \perp CI$, so lines $DH$ and $CI$ are parallel. Analogously, lines $EH$ and $BI$ are parallel. So we have now found the parallelogram $HSIT$, where $S = HE \cap CI$ and $T= BI \cap DH$. At this point of the contest I was happy because of the discovery of a parallelogram in my figure, but I then remained stuck for a while. I was looking at my well-drawn figure, but had some difficulties at finding an input for probably more than 20 minutes: I thus looked for some comfort in the cookies that I had found on my table at the beginning of the contest, and then decided to try to add something to my figure. When you are stuck, sometimes, it is worth it to work up the courage and define a new point or prolong a segment in order to find something new!</p>
<p>We would like to use that $M$ is the midpoint of arc $BC$. Since $M$ lies on the circumcircle of $ABC$, let's try to prolong the two couples of parallel sides of $HSIT$, and give a name to the intersections of this lines with the circumcircle. We define $J, R, Q, P, N, L$, as in the figure below, and also name the arcs: $CL =a, LN=b, NP=c, PM=d, MA=e, AQ=f, QR=g, RJ=h, JB=i$. Now that we have named them – and the figure has become a bit more threatening! – we should try to write some relationships among them. We are ready to use the information that $M$ is the midpoint of arc $BC$:</p>
<p>\begin{equation}a+b+c+d=e+f+g+h+i\tag{1}\end{equation}</p>
<p>Now, let's look for other relationships; $BI$ is the bisector of $\angle ABC$, then $N$ is the midpoint of the arc $AC$ not containing $B$, which leads to:</p>
<p>$$a+b=c+d+e\tag{2}$$</p>
<p>Analogously $R$ is the midpoint of arc $AB$:</p>
<p>$$f+g=h+i\tag{3}$$</p>
<p>Then, we can deduce another identity since the couple of lines $DH$ and $IC$ are parallel, and a couple of parallel lines cuts two equal arcs on a circle, so:</p>
<p>$$g=a\tag{4}$$</p>
<p>and analogously</p>
<p>$$c=i\tag{5}$$</p>
<p>Since we have written all of these equalities, we would like now to extract new information from them. Let us substitute (2) into (1), obtaining $2c+2d=f+g+h+i$. Using (3) we can rewrite the last expression as $2c+2d=2h+2i$, and using (5) this becomes $2c+2d=2h+2c \Rightarrow d=h$. This is good news, so it deserves to be highlighted in the figure. A dear friend of mine, Federica, who went to EGMO two years later, had given to me a set of felt-tip colours some weeks before I left for Antalya, saying "May these help you at EGMO when you will face the geometry problem!". They had actually brought me luck! I used the red felt-tip-pen to colour the two arcs on my figure.</p>
<p>Now, we have just used that a couple of parallel lines cuts two equal arcs on a circle, so we can now use this information in reverse: $RM$ and $JP$ are parallel. Let us explore the equations for another while: substituting $c+d=a+b-e$ in (1), we get $2a+2b-e=e+f+g+h+i$. Using again (3) this becomes $2a+2b=2e+2f+2g$, but since (4) holds, this becomes $b=e+f$. $e$ and $f$ are contiguous arcs on the circle, so we have just found another couple of congruent arcs, $QM$ and $NL$, that I highlighted in blue in the figure. We thus have another couple of parallel lines: $MN$ and $QL$. Now, since $MN$ is parallel to $HT$ and $HSIT$ is a parallelogram, we have $MN$ parallel to $RC$. Similarly, we have $MR$ parallel to $BN$. So, another parallelogram comes out in our figure: $RINM$.</p>
<p>We have two parallelograms, the small one and the big one, which share vertex $H$. To conclude that $I, H, M$ are collinear, we would like to show that there is an homothety of centre $H$ that maps $HSIT$ to $MRIN$ which is equivalent to showing that these two parallelograms are similar. This holds if and only if $\frac{IT}{IN}=\frac{IS}{IR}$. It is natural to use the chord theorem $IN \cdot IB=IR \cdot IC$, so we can rewrite the preceding equation, equivalent to the thesis, as $IS \cdot IC=IB \cdot IT$. This is what we would like to show to conclude the proof. This equation holds if and only if $SBCT$ is a cyclic quadrilateral. I remember I became a bit worried at this point, since I did not have much time left. Showing that a quadrilateral is cyclic can be done in many different ways, but just have a look at the figure: since $XBCY$ is clearly cyclic, we already have $\angle XBY= \angle XCY$. To prove that $BCST$ is cyclic it would be enough to show that $\angle SBT = \angle SCT$, which, after the last observation, is equivalent to proving that $\angle SBE = \angle DCT$. This seems a good way to proceed. Since $BCE$ is isosceles and $BE \perp CX$, then $CX$ is the axes of segment $BE$, so, since $S$ lies on $CX$, $SBE$ is also an isosceles triangle, and the same holds for $TDC$. In particular $\angle SBE = \angle SEB$ and $\angle TCD = \angle TDC$. The thesis is now equivalent to showing that $\angle SEB = \angle TDC$. This follows easily: let us define $O=DH \cap FE$ and $Z=HE \cap DF$. The quadrilateral $DZOE$ is cyclic since $\angle DZE= \angle DOE = 90$, because $H$ is the circumcentre of $FED$ and $O$ and $Z$ are the feet of two altitudes in this triangle. But then we have $\angle SEB = \angle ZEO =\angle ZDO = \angle CDT$, as we wanted to prove.</p>
<p>The contest ended a few minutes after I finished writing the last lines of the proof; I had liked the problem because of the number of nice intermediate steps that had led me to the conclusion. Since I was satisfied, I was then ready to spend a beautiful rest of day in Antalya with the other girls!</p>
<p><a id="fig"></a></p>
<p><em>Here is Camilla's figure from Antalya!</em></p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/figura.jpg" width="100%"/></p>Camilla Casamento TumeoTue, 06 Feb 2018 08:36:24 +0000http://www.egmo2018.org/blog/EGMO2014-P2/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2013 Problem 1http://www.egmo2018.org/blog/EGMO2013-P1/<p>
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<p>by <strong>Marcello Mamino</strong></p>
<p><a href="http://www.egmo2018.org/blog/EGMO2013-P2">< Go to the previous post in the series</a></p>
<p>Finally problem 1. Geometry. I'm not going to like it.</p>
<blockquote>
<p><strong>EGMO 2013, problem 1.</strong> The side $\textrm{BC}$ of the triangle $\textrm{ABC}$ is extended beyond $\textrm{C}$ to $\textrm{D}$ so that $\textrm{CD} = \textrm{BC}$. The side $\textrm{CA}$ is extended beyond $\textrm{A}$ to $\textrm{E}$ so that $\textrm{AE} = 2\textrm{CA}$. Prove that if $\textrm{AD} = \textrm{BE}$, then the triangle $\textrm{ABC}$ is right-angled.</p>
</blockquote>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-1/egmo2013.1-1.jpeg" style="float: right;" width="40%"/>Yeah, sure, right-angled. See why I hate the stuff? I don't know where the right angle is. I study my figure, accurately reproduced to the right, with growing discomfort. Somehow, I decide that $\textrm{B}$ and $\textrm{C}$ look wrong, which makes $\textrm{A}$ right, right? I check by Pythagoras' theorem that if $\textrm{A}$ is right, then $\textrm{AD}=\textrm{BE}$.</p>
<p>This problem just screams for Cartesian coordinates. I need to prove that $(x,y):=\textrm{A}$ is on the circle with diameter $\textrm{BC}$. I call $\textrm{B}$ and $\textrm{C}$ respectively $(1,0)$ and $(-1,0)$, so the circle is $x^2+y^2=1$. Point $\textrm{D}$ is $(-3,0)$. Point $\textrm{E}$ is $(3x+2,3y)$. $\displaystyle (x+3)^2 + y^2 =(3x+1)^2 + (3y)^2 \;\Rightarrow\; 8 x^2 + 8 y^2 = 8 $.</p>
<p>Average score was 6.14, most contestants obtained the maximum score of 7 points, curiously 9 out of 87 got 5: those that did it with analytic geometry? Number of <em>official solutions</em>: eight. Three by synthetic geometry, plus a variant, two by trigonometry, one by vectors, one by complex numbers, and one by areal coordinates. I'm not kidding, <em>areal coordinates</em>. Seriously, it wasn't so bad, but I don't understand geometry. </p>
<p></p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-1/egmo2013.1-2.jpeg" width="100%"/></p>
<p>Alfred Tarski's quantifier elimination method guarantees that all problems of Euclidean geometry can, indeed, be solved algebraically by Cartesian coordinates. Unless you are an extremely fast computer, however, you should not be too excited about the idea of applying Tarski's method in practice: the number of operations required to solve a problem is a <em>tower of exponentials</em> as high as the text of the problem is long (when expressed in a suitable formal language which is, in general, more verbose than English). Tarski's method has been substantially improved by several people, and now computer programs can, indeed, solve geometric problems using these techniques. Yet, unfortunately, such methods are not nearly usable by mere mortals. So, is Tarski's theorem the silver bullet that kills geometry? Of course not. Can I use Tarski's theorem to justify my contempt for the stuff? Definitely.</p>Marcello MaminoTue, 30 Jan 2018 12:50:29 +0000http://www.egmo2018.org/blog/EGMO2013-P1/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2013 Problem 2http://www.egmo2018.org/blog/EGMO2013-P2/<p>
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<p>by <strong>Marcello Mamino</strong></p>
<p><span style="text-decoration: underline;"><a href="http://www.egmo2018.org/blog/EGMO2013-P3">< Go to the previous post in the series</a></span></p>
<p>Emboldened by my success with problem 3, I move with confidence up the list. To be honest, this is exactly what I did, uncountable years ago, at the IMO. Even better, I then did problem 2 in a matter of minutes. Save discovering shortly after the competition that my solution was worth precisely nought, because I misread the statement. Certainly, that was an excess of boldness. Still, at that time, we considered 7-0-7 a decent result. And I put forward, in my defence, that I could not have solved that problem anyway. Let's see if history repeats itself (it won't: I would never confess in writing a blunder so gross). It's time for combinatorics.</p>
<blockquote>
<p><strong>EGMO 2013, problem 2</strong>. Determine all integers $m$ for which the $m\times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1, 2, 3, \dotsc, 10$ in some order.</p>
</blockquote>
<p>This is a typical <em>search problem</em>, I want to spend a few words about them. Suppose for a moment that the value of $m$ is fixed, then the problem would be stated as "is there a tiling of the $m\times m$ square$\dotsc$" which question has a similar structure to "is there a disposition of eight queens on a chessboard such that they don't attack each other?", or "can a chessboard without two opposite corners be covered by dominoes?". Notably, these questions do not ask for <em>all</em> configurations, and this suggests two different approaches. If the set of solutions is small or even empty, then one needs to analyze the hypothetical solutions very carefully, and often to fully describe the set: this is the case of the dominoes. If the set of solutions is large and rather unstructured, as for the eight queens, then the best approach is to try and construct one, by trial and error, a bit of intuition, and common sense. In either case, taking the wrong strategy might spell disaster. So this type of problems often involves some kind of alternation between an experimental mode and an analytic mode, and sticking to one would be ill advice.</p>
<p>My expectation, at first sight, is that only a few values of $m$ will matter, limited by the maximum possible area of the rectangles. For each of these $m$, I have a different search problem. I expect the ones that have solutions to be easy to solve by just some trial and error. After all, problem 3 has been so easy that this one must really be a piece of cake: I have no reason to expect any difficulty with the feasible cases. For the ones that can not be solved, there will be some colouring, or parity, or modular thing going on, like in the dominoes problem. So I have my strategy: first find the admissible values of $m$ with areas, second look for a few solutions by trial and error, third figure out which values don't have a solution and look for an invariant.</p>
<p>The maximum total area of our rectangles is clearly \[ A_{\textrm{max}} \;=\; 10\times 9 + 8\times 7 + 6\times 5 + 4\times 3 + 2\times 1 \;=\; 190\] and the minimum \[ A_{\textrm{min}} \;=\; 10\times 1 + 9\times 2 + 8\times 3 + 7\times 4 + 6\times 5 \;=\; 110\] Why? Well, it just looks right, and it is also almost the <em>rearrangement inequality</em>. I will not spend time looking for a proof until I have completed the rest of the solution, but I think it likely that an argument by induction, similar to that of the rearrangement inequality, will yield these bounds. If I need it in my solution, I will find it later, or, in <em>zeitnot</em> as a chess player would say, I will just pretend that these are well known facts. Anyway, I am left with just three values of $m$ to check, $11$, $12$, and $13$, which surprises me slightly because, at first, I had the impression of a greater variety of cases.</p>
<p><img src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-1.jpeg" style="float: right;" width="30%"/>I did good progress so far, which makes me hope for a quick conclusion by just finding configurations for $m=11,12,13$. So I draw three little figures that look more or less like the one on the right: I will call this disposition of five rectangles in a square <em>windmill configuration</em>. I make sure to draw the rectangles of the figure for $m=11$ skinny and tall, and those of $m=13$ quite square-ish. I don't know whether or not there are other arrangements of the five rectangles to consider, however, for now, this one seems reasonable, so I hope to find solutions that look like this. Finally, I set out to find my configurations by writing numbers on the edges of the rectangles.</p>
<p>I spend most of the time of this blind search on the case $m=11$, somewhat less on $m=12$, I give up before even trying $m=13$. It might be bad luck, it might be that there are actually not so many solutions, but after a few minutes it becomes apparent that I am not going anywhere. Can I prune the search space somehow? I notice that the perimeter of the square, $4m$, needs to equal the sum of the dimensions of the four outer rectangles. This means that I can compute the sum of the dimensions of the inner rectangle by adding together all the sides, obtaining $1+2+\dotsb+10=55$, and then subtracting $4m$. This gives me $11,7,3$ respectively for $m=11,12,13$.</p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-2.jpeg" style="float: left; padding-right: 5px;" width="30%"/>My new observation seems more incisive for $m=13$, because, in this case, it leaves only one choice for the dimensions of the inner rectangle, namely $2\times 1$. I give it a try and, at the first attempt, with a bit of luck, I immediately hit upon a solution. (If you are curious to know how much luck did I actually use, it turns out that the likelihood was one in three. So it was not just a bit of luck, it was indeed about $1.58$ bits.) This is the configuration on the left of this paragraph. Unfortunately, my good luck does not hold for the other cases, so I set forth for more analysis.</p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-3.jpeg" style="float: right;" width="30%"/>I am still in the mood for finding solutions rather than proving that they don't exist, and I have a bad feeling about $m=12$: it's <em>even</em>. So I begin from $m=11$. I have various choices for the inner rectangle: $6\times 5,\, 7\times 4 \dotsc$ (sum$=11$). Starting from $6\times 5$, I make a list of the choices available for $a_1, a_2, a_3, a_4$ in the figure, subject to the constraint that $a_1+a_3=11-6=5$ and $a_2+a_4=11-5=6$.</p>
<p>My little table of choices now looks like this.</p>
<table width="40%">
<tbody>
<tr>
<td style="border: black;">$a_1+a_3$</td>
<td>$2+3$</td>
<td>$1+4$</td>
</tr>
<tr>
<td style="border: black;">$a_2+a_4$</td>
<td>$2+4$</td>
<td>$1+5$</td>
</tr>
</tbody>
</table>
<p></p>
<p>as you can see below the figure. On the line $a_2+a_4$ I didn't write $3+3$ because, of course, $a_2$ and $a_4$ need to be different. By the same principle, I can now cross out $1+5$ from the same line, because $5$ is already one of the sides of the inner rectangle. So I have established that $a_2$ and $a_4$ are $2$ and $4$, it doesn't matter in which order because $a_2$ and $a_4$ are exchanged by a rotation of the figure by $180^\circ$. Looking at the line $a_1+a_3$, I must now cross out the combination $2+3$, because $2$ is already used for $a_2$ or $a_4$. Similarly I cross out also $1+4$, because $4$ is used. So no option is left but to discard the whole case that the inner rectangle is $6\times 5$.</p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-4.jpeg" style="float: left; padding-right: 5px;" width="30%"/>The method of the little table is clearly general. I will not write here all the boring details, as anyone can turn that handle by himself; it suffices to say that, in the next case examined, $7\times 4$, all options cross out except one, which gives the solution on the left. Having found solutions for $m=11$ and $m=13$, I turn my method to $m=12$ and in short order I discard all combinations, thus excluding this case. Have I finally concluded problem $2$? Not yet, not even close$\dotsc$</p>
<p>I look back to what is missing, now, in my argument. Definitely I need to prove the inequalities, and that gives the impression of a merely technical exercise. Also I need to check whether there are other dispositions of rectangles besides the windmill configuration. This second gap might actually hold some surprise, so I decide to look into it first. Any tiling of the square needs to cover the four corners of it, and, since the side length of the square is larger than any of those of the rectangles, I can observe that two corners cannot be covered by the same rectangle. Now I want to say that the remaining rectangle is in the center. If not, it must be on one side, hence, on the opposite side, I have only two rectangles. I represent the situation like this.</p>
<p><img src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-5.jpeg" style="display: block; margin-left: auto; margin-right: auto;" width="70%"/></p>
<p>They fit like a three-stepped peg in a two-stepped hole (observing that the steps are really three and two because all the side lengths of the rectangles are different). I conclude that the fifth rectangle can not be on one of the sides, as a consequence each of the four rectangles in the corners must touch those of the adjacent corners, and I am in the windmill configuration. Or am I?</p>
<p>I have stumbled upon the problem of balancing precision and practicality in mathematics. We all agree that correctness of a mathematical proof should be a <em>purely</em> logical matter. However, in practice, one can not trace back everything to the axioms of Euclid (or, better said, their modern substitutes). If for no other reason, at least for lack of space and time. In this particular case, I have no real definition of the windmill configuration, which makes it impossible to prove anything about it. Of course, I can identify it as a configuration such that precisely four of the rectangles touch the sides of the square. However, this is hardly a definition in the sense of allowing me to <em>formally</em> prove properties like $a_1+a_3=5$. But why? One definition is never an isolated entity. What does it mean to "touch the sides"? What is a "dissection of the square"? When do two polygons "overlap"? If I go down the rabbit hole, I will end up realizing that I don't even know what a rectangle is, or a number. Everyone must accept a reasonable level of imprecision, it might even be a <em>necessity</em> for effective communication. I satisfy myself with my imperfect definition because it achieves what is important: that the reader understands it clearly, and agrees that my conclusions are correct. Nevertheless, this windmill configuration is just the tiniest bit above my level of comfort with imprecision.</p>
<p>Now that the windmills are sorted (sort of), I need to focus on the areas. Honestly, the technique that I am most eager to use on those things is called avoidance. I search the information that I managed to collect for anything that could help me bound the side of the square. My attention is attracted by the sum of the dimensions of the inner rectangle, that I know to be $11,7,3$ for $m=11,12,13$. The next term of the series $11,7,3$ is obviously $-1$. Fine, the dimensions of the inner rectangle sum to $55-4m$, hence $55-4m>0$, and $m<13.75$. I dispensed with $A_{\textrm{max}}$, but, unfortunately, I cannot find anything like this to get rid of $A_{\textrm{min}}$ as well. Clearly $m\ge 10$, because $10$ is one of the dimensions of the tiles, but my proof of the windmill configuration relies on each corner of the square belonging to a different tile, which in turn requires $m>10$. So I can not use the windmill to exclude the case $m=10$, and, since essentially all that I have proven so far is based on the windmill configuration, here I am back to square one. After considering the idea of classifying more configurations, I decide that the most expedient course of action is to just prove the inequality.</p>
<p>I want to prove \[ A_{\textrm{min}} \;=\; 10\times 1 + 9\times 2 + 8\times 3 + 7\times 4 + 6\times 5 \;=\; 110\] my impression is that I need to generalize it and then prove it by induction. I choose the following generalization.</p>
<p><em>Let $x_1<x_2<\dotsb<x_{2n}$ be positive real numbers. Partition the set $\{x_1, x_2\dotsc x_{2n}\}$ into $n$ pairs $\{a_1,b_1\},\,\{a_2,b_2\}\dotsc$ Then \[ x_1x_{2n} + x_2x_{2n-1} + \dotsb \;\le\; a_1b_1 + a_2b_2 + \dotsb \]</em> </p>
<p>Obviously I needed to replace $10$ with $n$. I decided to also replace the numbers with indeterminates because, otherwise, I might find myself embarrassed when, to make the inductive step, I remove the product $1\times 2n$ and I end up with the wrong set of numbers: $2\dotsc 2n-1$ instead of $1\dotsc 2(n-1)$.</p>
<p>The inductive step should be immediate. I assume the negation of the thesis \[ a_1b_1 + a_2b_2 + \dotsb \;<\; x_1x_{2n} + x_2x_{2n-1} + \dotsb \] If $x_1$ is paired with $x_{2n}$, I just erase the product $x_1x_{2n}$ thus contradicting the inductive hypothesis. So $x_1$ is not paired with $x_{2n}$, therefore, without loss of generality, I can assume $a_1=x_1$ and $b_2=x_{2n}$. Now it should happen that \[ a_1b_2 + a_2b_1 \;\le\; a_1b_1 + a_2b_2 \tag{$\star$} \] which, by the way, is case $n=2$. If this happens, then \[ a_1b_2 + a_2b_1 + \dotsb \;\le\; a_1b_1 + a_2b_2 + \dotsb \;<\; x_1x_{2n} + x_2x_{2n-1} + \dotsb \] and I have contradicted the hypothesis again because $a_1b_2$ is just $x_1x_{2n}$. To check ($\star$), I bring everything to the right hand side of the inequality \[ 0 \;\le\; a_1b_1 + a_2b_2 - a_1b_2 - a_2b_1 \;=\; (a_2 - a_1) (b_2 - b_1) \] and the two parentheses are positive because $a_1=x_1$ is the smallest of the indeterminates, and $b_2=x_{2n}$ is the largest.</p>
<p><em>Sonnez trompettes et clairons!</em> I am finally out of this madness.</p>
<p>I have indeed completed a proof that I am confident, if competently written, would give me full points. Writing it up, however, is no easy task (and I can tell because, indeed, I did turn my thoughts into words, at least partially, in the paragraphs above). There is the death of a thousand cuts of the tables of cases to eliminate. There is all the handwavy arguing around windmills. There is the lemma of the inequality that needs to be formulated properly. And, also, I am not satisfied by bludgeoning through the problem in this way. A quick perusal of my tables of cases for $m=12$ shows an array of all possibilities, all neatly arranged, and all carefully crossed out. Arguably, this artifact is precisely as informative as a fresh table of all cases, all neatly arranged, no one crossed out. Since I need to reconstruct an argument anyway, I might as well look for a quick one. Indeed, I spend a moment looking for a simple way to do the exclusion of cases, and I find it. But really this is not what I want to do: I want to get rid of the windmills.</p>
<p>Now, in a competition, I would probably evaluate how much time I have left. Maybe I would turn my attention to another problem, and leave the write-up of this one for later. There are two benefits in finding a much shorter proof, or one that is considerably easier to write: the first is that it takes less time to write it, the second is that the probability of introducing errors is decreased. These benefits come at the cost of time to look for a better solution, and at the risk of not finding it. The optimal strategy is not easy to divine. Fortunately for me, I am in no competition, so I just carry on thinking about this problem. (I am also known to make very bad decisions in these matters. As a student, I remember factoring during some exam a polynomial of ridiculous degree, like $10$, just because I saw that I could do it. Well, it was a polynomial in two indeterminates, and I could have as well solved for the other one, which was of degree $2$.)</p>
<p>I need a parity or colouring argument to exclude $m=12$, but I wouldn't know how to use colouring here. By parity, I know that there is an even number of rectangles of odd area. In other words, if the numbers are paired $(a_1,b_1)\dotsc(a_5,b_5)$, either all the products $a_ib_i$ are even, or there are three even ones and two odd ones (four odds can be excluded because that would require eight of the numbers to be odd). To begin with, I want to look further into the first case. Checking modulo $4$, or perhaps a larger power of two, seems to be a good idea. I notice that \[ \textrm{even} \times \textrm{odd} \;\equiv\; \textrm{even} \quad \textrm{(mod $4$)}\] or, more precisely, $(2x)\times(2y+1)=4xy+2x\equiv2x$ modulo $4$. I assume that all the products are even, namely, without loss of generality, $a_1\dotsc a_5$ are even and $b_1\dotsc b_5$ are odd. Then I can write \[ a_1b_1 +\dotsb+a_5b_5 \;\equiv\; a_1+\dotsb+a_5 \;=\; 2+4+\dotsb+10 \;=\; 30 \;\not\equiv\; 12^2 \quad \textrm{(mod $4$)}\] This leaves me with the case of two odd and three even products.</p>
<p>I attack the second case with a similar method. There are two products of the form $\textrm{odd}\times\textrm{odd}$, this leaves one mixed product and two of the form $\textrm{even}\times\textrm{even}$. Clearly the congruence modulo $4$ ignores the $\textrm{even}\times\textrm{even}$ summands. Adding the observation that \[ \textrm{odd} \times \textrm{odd} \;\equiv\; \textrm{odd} + \textrm{odd} - 1 \quad \textrm{(mod $4$)}\] I quickly compute that the mixed product $\textrm{even}\times\textrm{odd}$ must satisfy $\textrm{even}-\textrm{odd}\equiv 1$ modulo $4$. However, I can't make any more progress, and I begin to suspect that one pairing such that the sum of the products is $12^2$ might exist. Eyeballing one approximation and then improving it, I get \[ 10 \times 4 + 6 \times 2 + 8 \times 3 + 9 \times 7 + 5 \times 1 \;=\; 12^2 \] Therefore the numbers alone do not exclude $m=12$.</p>
<p><img src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-6.jpeg" style="float: left; padding-right: 5px;" width="30%"/>I want a quick way to exclude configurations that have two $\textrm{odd}\times\textrm{odd}$ rectangles, two $\textrm{even}\times\textrm{even}$ rectangles, and a mixed one, for the case $m=12$. I immediately see a dissection of the $12\times12$ square into tiles that have the right parities (even though not the right dimensions), and indeed anyone can see it, here on the left. The existence of this configuration suggests to me that colouring arguments are not likely to work, definitely not one modulo $2$. Making this tiling, I had the impression that I could not have chosen the two $\textrm{odd}\times\textrm{odd}$ rectangles with dimensions that are all different. I felt forced to stack them one above the other as in the figure. This might lead to something, so now I am attracted by the idea of studying the relative position of the odd-sided tiles.</p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-7.jpeg" style="float: right;" width="30%"/>One of the $\textrm{odd}\times\textrm{odd}$ rectangles must be in a corner of the square. No windmill here, just the good old pigeon hole principle. So the sides of the square that form that corner are divided by the tiles touching them into segments, at least one of which has odd length. Since $m=12$ is even, on each of these sides there must be another odd length segment, therefore each of the two remaining odd-sided tiles is adjacent to one of the sides of the corner. In the figure to the right, I put the corner to the bottom left, and I draw the side with the $\textrm{odd}\times\textrm{even}$ tile horizontally. I observe that, because $x\neq y$, I can draw a vertical line (dashed in the figure) that crosses one of the $\textrm{odd}\times\textrm{odd}$ tiles but not on the other. This line intersects each tile in an even length segment except one, which contradicts its length being even.</p>
<p>I have a new argument for the case $m=12$, and, what is better, I notice that one can clean it up substantially. Ignore the relative position of the tiles, and just call <em>horizontal</em> the direction of the odd side of the tile $\textrm{odd}\times\textrm{even}$. The intersection of a <em>vertical</em> line with any tile is a segment of even length, unless the tile is of type $\textrm{odd}\times\textrm{odd}$, in which case the segment has odd length. The horizontal lengths of the two $\textrm{odd}\times\textrm{odd}$ tiles differ, therefore one can find a vertical line that intersects just one of them, and this brings about the same contradiction as before. I am rather pleased with this argument, and I cannot fail to notice that it works for all even values of $m$, not just $12$. Beauty! I know very well that I was prompted to prove the $A_{\textrm{min}}$ inequality to exclude the single case $m=10$, so this new argument replaces also the inequality.</p>
<p>I quickly bring together my little armament. Of course $m\ge10$, and my latest advancement proves that $m$ is odd. The upper bound $m\le13$, however, still relies on $55-4m>0$, which was obtained as a consequence of those hellish windmills. At this point, I won't allow them to destroy my proof. I look into this $55-4m$, and it becomes presently apparent that the only observation required is that a tile can not have two opposite sides on the perimeter of the square. In fact, this implies that the perimeter of the square, which has length $4m$, is subdivided by the tiling into segments of distinct lengths between $1$ and $10$. Therefore $4m\le1+\dotsb+10=55$. I finally have an argument to be pleased with, read it just after this section.</p>
<p>The participants have obtained generally high scores on this exercise. In average 4.01 out of 7 points, median 5. I am confident that those of my readers that attempted the task did not fail. Personally, it took to me a marginally greater effort, compared with problem 3, to reach a first solution at the clarions. Writing that solution formally, which I didn't do, would have been quite time consuming. This problem then lingered in my mind for a few days, during which time I performed the remaining work piecemeal, often while doing something else, by mostly mental computation. Obviously I have no written record for this part, so it has been reconstructed in my narration, but I think faithfully.</p>
<p>I can't hide my satisfaction with the solution. In my opinion, it is a rather lovely argument. Comparison with the <em>official solution</em> shows that the key ideas of ours are completely different. The official solution, which argues through the windmill configuration and area inequalities, is remarkably well engineered. Even though it uses the windmill configuration, there is a visible effort to do it <em>sparingly</em>. In particular, it avoids the relations, such as our $a_1+a_3=5$, that connect the dimensions of the inner rectangle with those of the outer rectangles. These relations are, in fact, the hardest to justify formally, so I see it as an indication that its author, or authors, shared our same discomfort with windmills. The trick used in the official solution to <em>deduce</em> the $A_{\textrm{min}}$ and $A_{\textrm{max}}$ inequalities from rearrangement, instead of <em>re-proving</em> them, is really nice and worth remembering.</p>
<hr/>
<h4>Solution to problem 2</h4>
<blockquote>
<p>Determine all integers $m$ for which the $m\times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1, 2, 3, \dotsc, 10$ in some order.</p>
</blockquote>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/EGMO2013-2/egmo2013.2-8.jpeg" style="float: right;" width="30%"/></p>
<p>The values are precisely $m=11$ and $m=13$, and tilings for these values are exhibited in the figure. To exclude all other values, we begin with the observation that $m\ge10$, otherwise the tile having one side of length $10$ would not fit. Furthermore, we will prove that (a) $m\le 13$ and (b) $m$ is odd. This is enough to conclude.</p>
<p>For (a), assume $m\ge 14$. Observing that no tile can contact two opposite sides of the square, we conclude that the perimeter of the square, which has length $4m$, is subdivided by the tiling into segments of distinct lengths between $1$ and $10$. Therefore $4m\le1+\dotsb+10=55$, contradicting $m\ge 14$.</p>
<p>For (b), let $a_1\times b_1\dotsc a_5\times b_5$ be the tiles covering a square of even side length $m$. Since the area of the square must equal the total area of the tiles, either none or two of the products $a_ib_i$ are odd (four can be excluded because it would require eight of the dimensions to be odd). Considering the first case, by the pigeon hole principle, each product $a_ib_i$ contains one even factor and one odd factor: without loss of generality the $a_i$ are the even factors. Observe that \[ a_ib_i\; =\; a_i + a_i(b_i-1) \;\equiv\; a_i \quad \text{(mod $4$)}\] Therefore, computing the congruence modulo $4$ of the tiles' total area \[ a_1b_1 + \dotsb + a_5b_5 \;\equiv\; a_1 +\dotsb+ a_5 \;=\; 2 +\dotsb+10 \;\equiv\; 2 \quad \text{(mod $4$)}\] This contradicts $m^2$ being a multiple of $4$, thus excluding the first case. As a consequence, two of the products must be odd and three even. Hence two of the tiles are of type $\textrm{odd}\times\textrm{odd}$, one of type $\textrm{odd}\times\textrm{even}$, and two of type $\textrm{even}\times\textrm{even}$.</p>
<p>Call <em>horizontal</em> the direction of the odd side of the tile $\textrm{odd}\times\textrm{even}$. Any <em>vertical</em> line intersecting a tile must do it in a segment of even length, unless the tile is of type $\textrm{odd}\times\textrm{odd}$, in which case the segment is odd. The dimensions of all tiles differ, so do in particular the horizontal lengths of the two $\textrm{odd}\times\textrm{odd}$ tiles, hence we can find a vertical line intersecting one of them but not the other. This vertical line is divided into a union of segments precisely one of which has odd length, contradicting that $m$ is even. This excludes also the second case, thus establishing claim (b).</p>
<p></p>
<p><img alt="From xkcd by Randall Munroe (https://xkcd.com/556/)" src="https://imgs.xkcd.com/comics/alternative_energy_revolution.jpg" width="100%"/></p>
<p style="text-align: right;"><em>From <strong>xkcd</strong> by <strong>Randall Munroe</strong> (<span style="text-decoration: underline;"><a href="https://xkcd.com/556/">https://xkcd.com/556/</a></span>)</em></p>
<p style="text-align: right;"><a href="http://www.egmo2018.org/blog/EGMO2013-P1">Continue to the next problem ></a></p>Marcello MaminoTue, 23 Jan 2018 09:45:02 +0000http://www.egmo2018.org/blog/EGMO2013-P2/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2013 Problem 3http://www.egmo2018.org/blog/EGMO2013-P3/<p>
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<p>by <strong>Marcello Mamino</strong></p>
<div style="margin-left: 50%; padding-bottom: 10px;">
<p><em>He who would cross the Bridge of Death</em><br/><em>Must answer me</em><br/><em>These questions three</em><br/><em>Ere the other side he see.</em></p>
</div>
<p>When Alessandra sent around an e-mail looking for contributors to this series, I had already solved most of past EGMO problems. It is tempting to reconstruct solution paths retrospectively, but, doing so, one goes at risk, in all honesty, of producing unnaturally elegant solutions, and also of forgetting the most embarrassing blunders. Fortunately, I noticed that I didn't tackle yet any of the three problems of Day 1 EGMO 2013, so I set out to do them <em>from the point of view of a contestant</em>. I took this as a sort of psychological experiment, so I will describe my solution process as it happened, and not as it could or should have happened.</p>
<p>To obtain the best instructional value out of this experiment, if you are not already familiar with the problems, I suggest that you solve them before going any further. Unfortunately, it happens at times that one problem just won't give. In this predicament, if you can say on your honour that it resisted a most gallant assault, then don't hold yourself vanquished. Read, instead, the relevant section just until a new idea comes up, and then stop to try again. Good luck!</p>
<hr/>
<p>Let's do the bold thing and start from <strong>problem 3</strong>.</p>
<blockquote>
<p><strong>EGMO 2013, problem 3.</strong> Let $n$ be a positive integer.</p>
<ol style="list-style-type: lower-alpha;">
<li>Prove that there exists a set $S$ of $6n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32n^2$.</li>
<li>Prove that every set $T$ of $6n$ pairwise different positive integers contains two elements the least common multiple of which is larger than $9n^2$.</li>
</ol>
</blockquote>
<p>There are two parts, and if there is any dependency between the two, then it's going to be <strong>(b)</strong> depending on <strong>(a)</strong>, so I should better start from <strong>(a)</strong>. It wants me to build a large set $S$ of positive integers with small pairwise lcm.</p>
<p>Intuitively, elements of $S$ should have common factors, to keep the lcm low. So I suspect that $S$ might be something similar to the set of products $p_1^{e_1}\cdot p_2^{e_2}\dotsb$ with small primes $p_i$ and some balancing act to be performed on the exponents. Before I embark on something so complicated, I want to check the most trivial cases to set a baseline. For instance, what happens if I take the set of the first $6n$ numbers? Well, obviously the lcms are bounded by little less than $36n^2$: not so far off considering that I am only aiming for $32n^2$.</p>
<p>These numbers need to have some common factor, so what if instead of taking the first, say, $a$ numbers, I take the first $a$ <em>even</em> numbers? Clearly my bound on the lcms jumps from $a^2$ to $2a^2$, and that's a loss, the corresponding gain is that, now, I freed up all the odd numbers. To make a rough first attempt, I want to see what happens adding to $S$ those odd numbers that are halves of even numbers already in: clearly they cannot produce any larger lcm. My new $S$ contains $a$ even numbers and $\frac{1}{2}a$ odd numbers, total $\frac{3}{2}a$, so the square of the cardinality of $S$ has gone up by a factor of $\frac{9}{4}$ at the expense of increasing the largest lcm by a factor of $2$: that's a win.</p>
<p>I'm making some progress: even numbers are good. I want to check multiples of $3$ now. So I have the first $a$ multiples of $3$, namely all up to $3a$. In addition to these $a$ numbers, I have the non-multiples up to $\frac{1}{3}$ of $3a$, which makes a grand total of $\frac{5}{3}a$ numbers. My upper bound on the lcms is $3a^2$, and this means a loss of $3$ for a gain of $\frac{25}{9}<3$: not good.</p>
<p>I'm back to the multiples of $2$. First, I want to see how far I am from the goal. I take $a$ even numbers, namely up to $2a$, and $b$ odd numbers, namely below $2b$. I chose $a=2b$, and from $a+b=6n$ follows $b=2n$. My lcms are now bounded by more or less $2a^2=32n^2$. Surprising: it seems that this is already the solution. Or almost the solution, maybe I have ignored some $+1$ in the counts and that's where the devil is. So let's be precise. \[ S = \{ \text{even nrs.} \le 8n \} \cup \{ \text{odd nrs.} < 4n \} \]The lcm is definitely <em>no larger than</em> $32n^2$, the first set has $4n$ elements, the second $2n$. Fine: part <strong>(a)</strong> solved.</p>
<p></p>
<p style="padding-top: 15px;">Part <strong>(b)</strong>. From part <strong>(a)</strong> I have got some intuition about the critical $S$, the largest one having all lcms below $9n^2$. It will probably contain most small numbers, because if $x\in S$ then all divisors of $x$ need to be in $S$. On the other hand the larger numbers, those of an order of magnitude larger than $n$, must be rather sparse, because they need to have lots of common factors. Then, of course, there are no elements of $S$ above $9n^2$. To begin with, I want to take a closer look at this $S$ from the sparse side, namely down from $9n^2$.</p>
<p>Note: the choice of strategy just described is the <em>key step</em> to the solution. It came to my mind in a natural and, I should say, subconscious way. It is, however, interesting to spell out in words which choices have been made. First, I decided to fix the maximum lcm to $9n^2$ and try to look for the largest $S$ that fits, with a view, obviously, of proving that this $S$ has necessarily less than $6n$ elements, thus obtaining a contradiction. A symmetric approach, which we can call direct, would have been to fix the cardinality of $S$ instead of the largest lcm, and then prove that one of the lcms is $>9n^2$. In other words, one tactic is to fit as many objects as possible in a limited space, the other is to fit a fixed number of objects in the smallest possible space. I cannot offer a rational argument to justify my preference for the first, except, possibly, this: in an attempt to figure out which mechanism limits the space, it appears natural to just fill it up until whatever it is that becomes exhausted prevents further progress, thus making itself visible. Such a general consideration is not satisfactory, and, in other circumstances, I would probably just do it the other way round without giving much thought. The second choice is to look first at the largest numbers in this critical $S$. Why? It is obvious that the <em>big end </em>of $S$ is where the hypothesis that all lcms are $\le 9n^2$ imposes the most immediate constraints.</p>
<p>So I want to look just below $9n^2$. Clearly I cannot have two numbers $x$ and $y$ <em>just below </em>$9n^2$, because their lcm would be at least $2x$ or $2y$, and this is already too much. Indeed, this argument tells me that there is at most one number between $\frac{9}{2}n^2$ and $9n^2$. I paid one single number to halve the upper bound: that's good business in my book. I call $A$ the number $9n^2$, and I draw a picture like this</p>
<p><img src="http://www.egmo2018.org/static/media/uploads/post/2013.3-1.jpeg" style="display: block; margin-left: auto; margin-right: auto;" width="60%"/></p>
<p>to signify that the interval between $\frac{1}{2}A$ and $A$ contains at most one element of $S$. Another one-element interval seems to be between $\frac{1}{3}A$ and $\frac{1}{2}A$, because two elements in there cannot differ only by a factor of $2$, thus the lcm is at least $3$ times one of them (actually even $4$ times maybe, but it's not time for fine tuning yet). My picture displays now more detail, and an obvious conjecture. <img alt="" src="http://www.egmo2018.org/static/media/uploads/post/2013.3-2.jpeg" style="display: block; margin-left: auto; margin-right: auto;" width="60%"/>Does the pattern continue? It seems so.</p>
<p>To state the last observation formally, it should happen that, for all $i\ge 1$, if $\frac{A}{i+1} < x < y \le \frac{A}{i}$, then $\operatorname{lcm}(x,y) > A$. Arguing about the common factors of $x$ and $y$ is cumbersome, so I would prefer to control the lcm somehow more algebraically. Considering that $x$ and $y$ are sandwiched in a small interval, there is an obvious upper bound on the gcd, namely the size of the interval. This calls for the well known relation connecting lcm and gcd \[ \operatorname{lcm}(x,y) = \frac{xy}{\operatorname{gcd}(x,y)} \ge\frac{xy}{y-x} > \frac{\left(\frac{A}{i+1}\right)^2}{\frac{A}{i}-\frac{A}{i+1}} = A \frac{i}{i+1} \]OK, I hoped to get $A$, but I got a little bit less (the fraction $\frac{i}{i+1}$ is quite close to $1$). Probably I just need to do my computation more carefully, because the claim itself, I already observed, wastes information. Yet it's promising enough. Before I spend more time fixing this argument, I prefer to assume the claim and see what benefit can be drawn from it. Joining the intervals, the claim tells me that there are at most $i$ elements of $S$ above $\frac{A}{i+1}$. Remembering that $A=9n^2$, it seems reasonable to try $i+1=\sqrt{A}=3n$, so $S$ has at most $3n-1$ elements above $3n$. Of course it also has at most $3n$ elements $\le 3n$, which makes $6n-1$. Here we go: the only missing step is the proof of the claim.</p>
<p>The computation above is patently very rough: I just replaced both $x$ and $y$ in the product $xy$ with their smallest possible value $\frac{A}{i+1}$, and then I replaced $y-x$ with the upper bound $\frac{A}{i}-\frac{A}{i+1}$. That's wasteful because $x$ and $y$ cannot be at the same time maximally small and maximally apart from each other. Let's try to keep $x$ and $y$ as indeterminates for a little longer \[ \operatorname{lcm}(x,y) \ge \frac{xy}{y-x} = \frac{1}{\frac{1}{x}-\frac{1}{y}} \]Now from the hypothesis $\frac{A}{i+1} < x < y \le \frac{A}{i}$, taking the reciprocals, I get \[ \frac{i}{A} \le \frac{1}{y} < \frac{1}{x} < \frac{i+1}{A} \;\;\Rightarrow \;\; \frac{1}{x}-\frac{1}{y} < \frac{1}{A} \]And, substituting in the formula above, $\operatorname{lcm}(x,y) > A$. Done.</p>
<p>It remains to collect the thoughts above and write up a formal solution, you will find it in the next section. The problem turned out to be much easier than expected, however I suspect that good luck helped me considerably. So always remember to throw gems at unicorns before contests. In hindsight, the large gap between part <strong>(a)</strong>, $32n^2$, and part <strong>(b)</strong>, $9n^2$, should have told me that there was no razor-sharp construction going on. On the other hand, I can only be grateful that it was so: this is a sort of problems where often one can make improvements by considering more and more intervals, and constructing proofs like that can be tedious.</p>
<p>Don't be put off if you encountered difficulties solving this exercise, the stats show that the best score obtained by any of the 87 participants is 3 points out of 7, and 69 persons didn't make any point at all. Comparison with the two <em>official solutions</em> shows that they are based essentially on the same mechanisms as ours. Both the official solutions, however, do part <strong>(b)</strong> through the approach that we called direct. There is a distinct smell of black magic around these solutions, which confirms, in my mind, the intuition that the approach by contradiction opposes fewer obstacles.</p>
<hr/>
<h4>Solution to problem 3</h4>
<blockquote>
<p>Let $n$ be a positive integer.</p>
<ol style="list-style-type: lower-alpha;">
<li>Prove that there exists a set $S$ of $6n$ pairwise different positiveintegers, such that the least common multiple of any two elements of $S$ is no larger than $32n^2$.</li>
<li>Prove that every set $T$ of $6n$ pairwise different positive integerscontains two elements the least common multiple of which is larger than $9n^2$.</li>
</ol>
</blockquote>
<p><strong>(a) </strong>Let $S$ contain precisely the positive even numbers $\le 8n$ and the odd numbers $<4n$, hence a total of $6n$ numbers. Let $x$ and $y$ be elements of $S$, we show that $\operatorname{lcm}(x,y)\le 32n^2$. In fact, if at least one of them is odd, then\[\operatorname{lcm}(x,y) \le xy < (8n)(4n) = 32n^2\]Otherwise\[\operatorname{lcm}(x,y) = \frac{xy}{\operatorname{gcd}(x,y)} \le\frac{xy}{2} \le \frac{(8n)^2}{2} = 32n^2\]This concludes part <strong>(a).</strong></p>
<p><strong>(b)</strong> Assume towards a contradiction that $\operatorname{lcm}(x,y) \le 9n^2$ for all $x,y\in T$: we will prove that $T$ has less than $6n$ elements. We claim that, for all $i\ge 1$, the set $T$ contains at most one element in the interval $I_i:=\left(\frac{9n^2}{i+1}, \frac{9n^2}{i}\right]$. Assuming, for the moment, the claim, $T$ has at most $3n-1$ elements in the interval $(3n,9n^2] = \bigcup_{i=1}^{3n-1} I_i$. On the other hand, all elements of $T$ must not exceed $9n^2$, and there are at most $3n$ elements $\le3n$, therefore $T$ can have at most $6n-1$ elements, hence the statement.</p>
<p>It remains to prove the claim. Let $\frac{9n^2}{i+1} < x < y \le \frac{9n^2}{i}$. Taking reciprocals we get \[ \frac{i}{9n^2} \le \frac{1}{y} < \frac{1}{x} < \frac{i+1}{9n^2} \;\; \Rightarrow \;\; \frac{1}{x}-\frac{1}{y} < \frac{1}{9n^2} \]hence \[ \operatorname{lcm}(x,y) = \frac{xy}{\operatorname{gcd}(x,y)} \ge\frac{xy}{y-x} = \frac{1}{\frac{1}{x}-\frac{1}{y}} > 9n^2\]It follows that $x$ and $y$ cannot belong both to $T$, and the claim is proven.</p>
<p style="text-align: right;"><span style="text-decoration: underline;"><a href="http://www.egmo2018.org/blog/EGMO2013-P2">Continue to the next problem ></a></span></p>Marcello MaminoTue, 16 Jan 2018 09:40:22 +0000http://www.egmo2018.org/blog/EGMO2013-P3/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2013 Problem 4http://www.egmo2018.org/blog/EGMO2013-P4/<p>
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<p>by <strong>Davide Lombardo</strong></p>
<blockquote>
<p><strong>EGMO 2013, problem 4</strong>. Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial \[P(n)=\frac{n^5+a}{b}\] takes integer values.</p>
</blockquote>
<p>The first thing I notice is that $b$ cannot be even, because the numbers $n^5+a$ and $(n+1)^5+a$ have different parity, so they can't both be divisible by an even number.</p>
<p>This being said, what is this problem really about? What we're being asked to study is really a congruence in disguise, because the condition can be rewritten as<br/>\[<br/>\begin{cases}<br/>(n-1)^5+a \equiv 0 \pmod b\\<br/>n^5+a \equiv 0 \pmod b\\<br/>(n+1)^5+a \equiv 0 \pmod b<br/>\end{cases}<br/>\]<br/>Now by the Chinese Remainder Theorem I probably only care about $b$ being the power of a prime, because if I can solve these congruences for two (powers of) primes then I can also solve them modulo the product. So let's say that $b$ is prime for now, I'll worry about powers later. Also I notice that $a$ is a bit of a red herring: if I rewrite the congruences as<br/>\[<br/>\begin{cases}<br/>a \equiv -n^5 \pmod b \\<br/>(n-1)^5-n^5 \equiv 0 \pmod b\\<br/>(n+1)^5-n^5 \equiv 0 \pmod b<br/>\end{cases}<br/>\]<br/>the truth seems to be that whenever I can find an $n$ that satisfies the last two equations, then I just choose $a$ appropriately from the first. So whatever, let's throw out the first equation altogether for now, and just work with<br/>\[<br/>\begin{cases}<br/>(n-1)^5-n^5 \equiv 0 \pmod b\\<br/>(n+1)^5-n^5 \equiv 0 \pmod b<br/>\end{cases}<br/>\]<br/>Now I'm slightly confused: it's clear that (given $b$) the first equation only has finitely many solutions, and so does the second. But how do I combine them? Oh, maybe the idea is something like "well, these equations each have some solution, but the solutions to the first and the solutions to the second are almost always all distinct modulo $b$". Or are they? Maybe there's an infinite class of $b$ for which these really have common solutions... unclear<strong><a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#footnote">*</a></strong>. While I think about how to proceed, I notice that $b$ cannot divide $n$, nor $n \pm 1$, because otherwise the congruences fail miserably for obvious reasons. I should probably rewrite the congruences again as<br/>\[<br/>(n-1)^5 \equiv n^5 \equiv (n+1)^5 \pmod b<br/>\]<br/>Now that I see them this way, I'm reminded of the fact that taking fifth powers is injective modulo 5 because of Fermat's little theorem, so $b$ cannot be 5. That is, if $b=5$ we have $(n-1)^5 \equiv n-1 \pmod 5$, $n^5 \equiv n \pmod 5$, $(n+1)^5 \equiv n+1 \pmod 5$ and therefore<br/>\[<br/>n-1 \equiv n \equiv n+1 \pmod 5,<br/>\]<br/>contradiction.<br/>Actually I know more: taking 5th powers is injective modulo $b$ (a prime) whenever $5$ does not divide $b-1$. Oh, that sounds like an almost interesting fact: if $b$ is not 1 mod 5, then I also obtain $n-1 \equiv n \equiv n+1$, which can hardly happen. So $b \equiv 1 \pmod 5$, that's something. And now I'm afraid I really have to compute something, don't I? The prime $b$ divides the 2 numbers <br/>\[<br/>(n+1)^5-n^5, (n-1)^5-n^5<br/>\]<br/>so I guess I can try to run a "Euclidean algorithm-style" kind of computation to reduce the powers of $n$ involved. Let's see: $\left( (n+1)^5-n^5, (n-1)^5-n^5 \right)$ is the same as<br/>\[<br/>\left( 1 + 5 n + 10 n^2 + 10 n^3 + 5 n^4, -1 + 5 n - 10 n^2 + 10 n^3 - 5 n^4 \right)<br/>\]<br/>so clearly the very first thing I want to do is sum these expressions to get rid of the $n^4$:<br/>\[<br/>\left( 1 + 5 n + 10 n^2 + 10 n^3 + 5 n^4, 10 n + 20 n^3 \right).<br/>\]<br/>Oh, that's actually nice. I can also throw out a factor of 10 from $10n+20n^3$ (because I know that $b$ is neither 2 nor 5) and a factor of $n$ (because I know that $b$ cannot divide $n$), so I end up with having to compute<br/>\[<br/>\left( 5 n^4 + 10 n^3+10 n^2+5n+1, 2 n^2+1 \right)<br/>\]<br/>The unfortunate detail is that I cannot divide $5 n^4 + 10 n^3+10 n^2+5n+1$ by $2n^2+1$ as polynomials, because the leading terms do not divide each other. But that's not really a problem, let me double the first polynomial (which I can do for free, since $2n^2+1$ is odd anyway) and try again: I want to compute<br/>\[<br/>\left( 2(5 n^4 + 10 n^3+10 n^2+5n+1), 2 n^2+1 \right).<br/>\]<br/>Ugh, long polynomial division. Never could abide those, but fine, let's do it: after some computations on a separate sheet of paper, I find<br/>\[<br/>2(5 n^4 + 10 n^3+10 n^2+5n+1) = (2n^2+1)(5n^2+10n)+15n^2+2<br/>\]<br/>Ok, so (subtracting a multiple of the second polynomial from the first) I now have that if $b$ divides $(n+1)^5-n^5$ and $n^5-(n-1)^5$, then $b$ divides<br/>\[<br/>(15n^2+2,2n^2+1)<br/>\]<br/>and I suppose I should do one more step and compute<br/>\[<br/>\begin{aligned}<br/>(15n^2+2,2n^2+1) & = (30n^2+4,2n^2+1) \\<br/>& = (30n^2+4-15(2n^2+1),2n^2+1)\\<br/>& =(-11,2n^2+1).<br/>\end{aligned}<br/>\]<br/>But this is excellent, 11 is an actual number (as opposed to a polynomial in $n$)! So who cares about the congruences and everything else I said at the beginning: if $b$ divides both $(n+1)^5-n^5$ and $(n-1)^5-n^5$, then $(b,10n)=1$ and computing the GCD shows that $b$ divides $(11,2n^2+1)$. So $b$ is either 1 or 11. I guess when $b=1$ there is not much else to say (I can just take $a$ to be whatever I like), so I only need to deal with $b=11$.<br/>In this case I know that I have to choose $a \equiv -n^5 \pmod{11}$ (we said that at the beginning, right?), and also -- since I'm assuming that $b=11$ divides $(11,2n^2+1)$ -- I know that $2n^2+1 \equiv 0 \pmod{11}$. This is a congruence I can do: if I multiply by 5 it becomes $10n^2+5 \equiv 0 \pmod{11}$, that is $n^2 \equiv 5 \pmod{11}$. Since $4^2 \equiv 16 \equiv 5 \pmod{11}$, the only solutions are $n \equiv \pm 4 \pmod {11}$. And so $a \equiv -n^5 \equiv -(\pm 4)^5 \equiv \mp 2^{10} \equiv \mp 1 \pmod {11}$. Mmmh, I realize I haven't done a single example (in my defense, I wouldn't have found 11 by examples...), so let's check that I haven't gotten anything wrong: I'm claiming that I can take $a \equiv -1 \pmod {11}$ and $n \equiv 4 \pmod{11}$. Is it true that $3^5-1, 4^5-1$ and $5^5 - 1$ are all 0 modulo 11? Well certainly $4^5 \equiv 2^{10} \equiv 1 \pmod{11}$ by Fermat's little theorem, and $5^5 \equiv (4^2)^5 \equiv 4^{10} \equiv 1 \pmod{11}$ for the same reason; also $3^5 \equiv 9 \cdot 9 \cdot 3 \equiv 12 \equiv 1 \pmod{11}$, so yeah, it seems to work out. Let's also do $a \equiv 1 \pmod{11}$ and $n \equiv -4 \pmod{11}$: we need to check that $-5^5 \equiv -1 \pmod{11}$, $-4^5 \equiv -1 \pmod{11}$, $-3^5 \equiv -1 \pmod{11}$, and these are the same congruences as before up to a sign, so they also work out.<br/>Conclusion:<br/>\[<br/>\boxed{b=1, a \in \mathbb{Z}_{>0} \text{ or } b=11, a \equiv \pm 1 \pmod{11}}<br/>\]</p>
<p><strong>Second solution.</strong> I also sketch a less elementary (but somewhat more direct, in my opinion) solution to this problem. The idea is that from the congruences<br/>\[<br/>\begin{cases}<br/>(n+1)^5 \equiv n^5 \pmod{b}\\<br/>(n-1)^5 \equiv n^5 \pmod{b}<br/>\end{cases}<br/>\]<br/>we see that $\frac{n+1}{n}$, $\frac{n-1}{n}$ are (primitive and distinct) fifth roots of unity modulo $b$. Call $t:=1/n$; then $1+t$ and $1-t$ are fifth roots of unity, and so are $(1+t)(1-t)=1-t^2$, $(1+t)^2=1+2t+t^2$ and $(1-t)^2=1-2t+t^2$. These are five 5th roots of unity, and all of them are primitive (because none of them is equal to 1!), so 2 of them are equal modulo $b$. We can remove 1 from each of them and divide by $t \neq 0$, and we find that two of the following 5 numbers are equal modulo $b$:<br/>\[<br/>1, -1, -t, 2+t, -2+t<br/>\]<br/>Since $t \not \equiv \pm 1 \pmod b$, it's easy to see that $t \equiv \pm 3 \pmod{b}$. Since $b$ divides both $(1+t)^5-1 \equiv (1+1/n)^5-1$ and $(1-t)^5-1 \equiv (1-1/n)^5 -1$, we find that $b$ divides either $(4^5-1,(-2)^5-1)=(1023,33)=33$ or $((-2)^5-1,4^5-1)=33$. Since $b$ cannot have a factor of 3 (recall the observation in the first solution that the prime factors of $b$ are congruent to 1 modulo 5!), this gives that $b$ divides 11.</p>
<p><a id="footnote"></a></p>
<p><strong>*</strong> For the really committed readers: there is an advanced tool to answer this sort of question, called the <a href="https://en.wikipedia.org/wiki/Resultant"><em>resultant</em></a>, which shows that we only need to care about $b$ being a power of $2$, $5$ or $11$. But during a competition, and without a computer, I probably wouldn't have been able to compute resultants...</p>Davide LombardoTue, 09 Jan 2018 10:13:34 +0000http://www.egmo2018.org/blog/EGMO2013-P4/EGMOOlimpiadiProblem solvingThinking Out Loud – EGMO 2015 Problem 5http://www.egmo2018.org/blog/thinking-out-loud-egmo-2015-problem-5/<p>
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<p>by <strong>Davide Lombardo</strong></p>
<blockquote>
<p><strong>EGMO 2015, Problem 5</strong>. Let $m, n$ be positive integers with $m > 1$. Anastasia partitions the integers $1, 2, \dots , 2m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers.<br/>Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n.$</p>
</blockquote>
<p>Well, Boris' sum lies between $1+2+\cdots+m = \frac{m(m+1)}{2}$ and $(m+1)+(m+2)+\cdots+(m+m) = m^2 + \frac{m(m+1)}{2}$; so any value of $n$ outside of this interval is fine, and we need to find a partition that prevents Boris from achieving a given $n$ only for $n$ in this interval. The first idea that comes to mind is to try and force Boris to do something <em>we</em> want, independently of what he may choose (as in the best conjurer's tricks). For example, the obvious partition $(1,m+1), (2,m+2), \ldots, (m,m+m)$ has the property that – no matter what Boris chooses – we know what his choice is modulo $m$: it's going to be $1,2,\ldots,m$. So with this partition poor Boris can only obtain sums that are congruent to<br/>$$1 + \cdots + m \equiv \frac{m(m+1)}{2} \pmod{m}.$$<br/>That's pretty good already: let's say that $m$ is odd, for the moment, so that $\frac{m(m+1)}{2}$ is zero modulo $m$. Then we're saying that if $n$ is not a multiple of $m$, we can just give him this partition and he'll be done for :). So this leaves us with only a few numbers to worry about, namely the multiples of $m$ in the interval $[\frac{m+1}{2} m,\frac{3m+1}{2}m]$. Wait though, I'm being way too generous with my friend Boris here: why should I let him choose all the big numbers and achieve a super-large sum? If he wants to take $2m$ he'll have to give up on $2m-1$, and likewise with $2m-2,2m-3$ etc... in other words, what happens if I offer him the other obvious partition, namely $\{1,2\}, \{3,4\}, ..., \{2m-1,2m\}$? His sum is going to be in the interval $[1+3+\cdots+(2m-1),2+4+\cdots+2m] = [m^2,m^2+m]$. Much better! So this partition ensures that he can never realise an $n \not \in [m^2,m^2+2m]$, and the one from before rules out all the multiples of $m$. Tough luck, eh, Boris? It's not looking good for you, brother... anyway: if $n \neq m^2, m^2+m$ we've won (assuming that $m$ is odd, that is). What do we do if $n=m^2$, say? What other invariant can I come up with? I've already used "size" and "residue class modulo $m$". Can I make sure that the sum is always odd, for example? This would require me to pair together numbers of the same parity (if for even just one pair Boris can choose between numbers of opposite parity then I have no control on the parity of the sum). Can I do it, even for small values of $m$? Say $m=4$: then I should try $\{1,3\},\{2,4\}$, so the sum is indeed always odd. But this probably depends on $m$ modulo 4, so I run into further trouble if $4 \mid m$.</p>
<p>Let me check: $m=4$, $\{1,5\}, \{2,6\}, \{3,7\}, \{4,8\}$. Then the sum is... well, whatever it is, this is a partition we already considered, so it cannot give me any new information (come to think about it, why did I try $m=4$ instead of $m=3$?). No, we need to find a different way.</p>
<p>While I'm thinking about this, let me work out which numbers in the interval $[m^2,m^2+m]$ I need to worry about when $m$ is even. It's those numbers $N$ such that $N \equiv \frac{m(m+1)}{2} \pmod{m}$, where $m=2k$. Then $N \equiv k(2k+1) \equiv k \pmod{2k}$, and the only such number in the interval $[m^2,m^2+m]$ seems to be $m^2+k = m^2+\frac{m}{2}$. So this is actually 'easier', at least in the sense that there's just a single number we need to make sure Boris cannot achieve.</p>
<p>This, however, does not solve the problem of how to actually stop that pesky Boris from making his sum equal to those few numbers left. To be fair, there are so few numbers I don't know how to deal with that maybe I can afford to leave Boris some choice, though I still don't want his sum to vary too wildly, at least with respect to some property (I mean: it can vary all it likes in size, so long as its residue class modulo something is fixed, for example). Maybe I can force Boris to choose between a sum that is "large", but of the wrong parity, and one that is of the correct parity, but too small to be $m^2$? This is worth a try: I should probably pair $\{1,2m\}$, and then pair the rest of the numbers so that numbers in every pair have the same parity and roughly the same size (so that, ideally, when Boris chooses between $1$ and $2m$, he either overshoots – if he picks $2m$ – or undershoots – if he picks $1$). I realize that I'm talking about over- and under-shooting and not caring about parity, but maybe parity will intervene in some mysterious way? I don't know: let's see this in action for a smallish $m$, say $m=3$. Following the recipe above, I should take $\{1,6\}; \{2,4\}; \{3,5\}$, and the numbers to avoid are $9, 9+3=12$. Let me check... this seems to work! I'm not sure whether this is just luck, though; I feel like trying another small case. Let's say $m=5$, so that the partition is $\{1,10\}, \{2,4\}, \{3,5\}, ...$. Ah. I was going to write $\{4,6\}$, but $4$ is already taken. So I need to have $\{6,8\}$... mmh, this does not sound very promising. Anyway, here is the partition:<br/>\[<br/>\{1,10\}, \{2,4\}, \{3,5\}, \{6,8\}, \{7,9\},<br/>\]<br/>and the numbers to avoid are $25$ and $25+5=30$. Darn, I can see that tiresome Boris gloating already... $10+2+3+8+7=30$. Nope, this doesn't work.</p>
<p>Going back to the idea of 'almost-forcing-Boris-to-do-something-but-not-quite', combined with 'there should be a single pair for which Boris' choice matters, and even then neither of the two choices should work (muahahaha evil laugh)', I'm eventually lead – after a significant amount of doodling, writing down partitions, noticing that they do not work, and cursing – to considering the following alternative. Let's try to fix the congruence class of the sum modulo some $M$ which is not quite $m$, but it's close enough that we can arrange most pairs to have elements that are congruent modulo $M$. In other words: let's try to force the sum to take some value modulo $M$; we'll fail, but maybe we can fail not tooo badly, so that the sum can only take <em>very few</em> values modulo $M$.</p>
<p>Natural choices are $M=m+1$ and $M=m-1$; let's try the former. The partition would then be<br/>\begin{equation}<br/>\begin{array}{cccccc}<br/>1 & 2 & 3 & \cdots & m-1 & m \\<br/>m+2 & m+3 & m+4 & \cdots & 2m & m+1;<br/>\end{array}\tag{$\star$}\label{eq:FinalPartition}<br/>\end{equation}<br/>can we finally trick Boris into making the wrong choice? Whatever he decides to do, he will end up with a sum which is either $1+2+\cdots+(m-1)+m$ or $1+2+\cdots+(m-1)+(m+1)$ modulo $m+1$. So, writing $S$ for Boris' sum, $S \equiv \frac{m(m-1)}{2} + \begin{cases} -1 \\ 0\end{cases} \pmod{m+1}$.<br/>Now the numbers we want to avoid are (let me take $m$ odd again...) $m^2+m \equiv (-1)^2-1 \equiv 0 \pmod{m+1}$ and $m^2 \equiv 1 \pmod{m+1}$, while (using that $m=2k+1$ is odd) <br/>\[<br/>S \equiv \frac{(2k+1)(2k)}{2} + \begin{cases} -1 \\ 0\end{cases} \equiv k(-1) + \begin{cases} -1 \\ 0\end{cases} \pmod{2k+2}.<br/>\]<br/>Oooh this looks very promising: it means we're done, provided that<br/>\[<br/>- k + \begin{cases} -1 \\ 0\end{cases} \not \equiv \begin{cases} 0 \\ 1\end{cases} \pmod{2k+2},<br/>\]<br/>that is, we have problems only if $k \equiv 0,-1,-2 \pmod{2k+2}$. But plainly this cannot happen, because a positive integer that is congruent to $0,-1,-2$ modulo $2k+2$ is at least equal to $2k$, so it cannot be equal to $k$. <br/>This only leaves us with the case of even $m=2k$. In this case the sum is<br/>\[<br/>S \equiv \frac{(2k)(2k-1)}{2} + \begin{cases} -1 \\ 0\end{cases} \equiv -2k+ \begin{cases} -1 \\ 0\end{cases} \pmod{2k+1}<br/>\]<br/>while the only $n$ we have to avoid is $m^2+\frac{m}{2} = 4k^2+k \equiv k(4k+1) \equiv 2k^2 \equiv -k \pmod{2k+1}$. And again we are done, provided that we do not have<br/>\[<br/>-2k + \begin{cases} -1 \\ 0\end{cases} \equiv -k \pmod{2k+1},<br/>\]<br/>that is $k \equiv -1,0 \pmod{2k+1}$, which is again impossible unless $k=0$, which however means $m=1$, that we do not need to consider. Yay, take that, Boris!</p>
<p>In conclusion,</p>
<ul>
<li>If $n \not \in [m^2,m^2+m]$, we present Boris with the partition $\{1,2\},\{3,4\},\cdots,\{2m-1,2m\}$: this forces his sum to lie in the interval $[m^2,m^2+m]$, and therefore different from $n$.</li>
<li>If $m$ is odd and $n \in [m^2,m^2+2m]$ is not $m^2,m^2+m$, then we give Boris the partition $\{1,m+1\}, \cdots, \{m,2m\}$. This forces him to end up with a sum that is congruent to $\frac{m(m+1)}{2} \pmod{m}$, which in particular is different from $n$. We use the same partition if $m$ is even, $n \in [m^2,m^2+2m]$, and $n \neq m^2+\frac{m}{2}$.</li>
<li>Finally, for the few remaining values of $n$ (namely $n=m^2+m/2$ if $m$ is even, $n=m^2, m^2+m$ if $m$ is odd), we foil Boris' evil plans by using the partition in \eqref{eq:FinalPartition}: Boris cannot obtain $n$ as sum, because this would lead to a contradiction modulo $m+1$.</li>
</ul>
<p><strong>Comments.</strong> While this problem is not <em>very</em> easy, there are many low-hanging fruits that one should reach early on in their solution attempts. Namely, the partitions $\{1,2\}, \{3,4\}, \ldots, \{2m-1,2m\}$ and $\{1,m+1\},\{2,m+2\}, \ldots, \{m,2m\}$ already rule out most of the values of $n$, and an experienced problem solver <em>should take very little time to notice that they work most of the time</em>. As for the extra idea to rule out the few remaining cases, it is less standard and certainly partially helped by doodling and making examples, but one should also keep in mind that (unless the problem is really hard!) there should be an easy description for these partitions. In other words: either we're completely off track, and the problem is done in an entirely different way<span style="text-decoration: underline;"><a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#footnote1"><sup><strong>1</strong></sup></a></span>, or <em>there has to be a way to write down the partitions we're missing</em>. So it only makes sense to try very regular partitions, obtained by following very simple patterns; and I wouldn't rule out the possibility that, among the <em>reasonable</em> partitions that one might write down, there's more than one that works for – say – $n=m^2$. As a matter of fact, more 'regular' (='easy to describe algorithmically') partitions are <em>a priori</em> better not just because there's a bias in olympiad problems towards having 'simple' solutions<span style="text-decoration: underline;"><a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#footnote2"><strong><sup>2</sup></strong></a></span>, but also because their simple description is often our only hope of <em>proving</em> something about them (in our case, congruences on the sums that can be obtained). In other words, what I'm saying is that, while you despair and attack various small cases, <em>you should not waste time trying random partitions</em>, because even if they <em>did</em> work, you would have no idea how to replicate that pattern for different values of $m$ and $n$.</p>
<p><a id="footnote1"></a></p>
<p><sup><b>1</b></sup> maybe by counting how many partitions there are in total, how many sums can be realized from any given partition, and then coming up with an incredibly clever idea</p>
<p><a id="footnote2"></a></p>
<p><sup><b>2</b></sup> yes, this is metagaming. Can you deny it's true?</p>Davide LombardoTue, 02 Jan 2018 10:27:19 +0000http://www.egmo2018.org/blog/thinking-out-loud-egmo-2015-problem-5/EGMOProblem solvingThinking Out Loudhttp://www.egmo2018.org/blog/thinking-out-loud/<p>by/di <strong>Alessandra Caraceni</strong></p>
<p><img alt="" src="http://www.egmo2018.org/static/media/uploads/post/thinking_out_loud.jpeg" width="100%"/></p>
<p><a id="eng"></a></p>
<p style="color: gray;"><img alt="" height="25" src="http://www.egmo2018.org/static/media/uploads/post/1280px-flag_of_the_united_kingdom.svg.png"/> <span> English version below / <a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#ita" style="text-decoration: underline;">Vai alla versione italiana</a></span></p>
<p>It may sound surprising to our Italian contestants today, but in 2004 Italy could only boast two gold medallists in all of the history of the IMO; our selection system was still developing into the rather articulate one we have today, and, though training resources definitely existed, we did not yet have a whole database of recorded lectures covering everything from binomials to LTE, and we were always hungry for more.</p>
<p>It was in that context that a short pdf, elegantly written, beautifully typeset, appeared one day on the Italian Olympiad website. It was called "<em>problem solving ad alta voce</em>" (roughly, problem solving out loud) and it was written by Marco Cammi, one of the two IMO gold medallists I mentioned. It was a faithful recording of Marco's thought process upon seeing the six problems of that year's Italian national olympiad: which ideas came to him first, why they worked or didn't, even how he would sometimes make a mistake and briefly wonder what was wrong with his approach, only to convince himself it had to work and go back to fix it. I still have that pdf; with its 13 pages of friendly banter and beautiful maths, I consider it a symbol of what truly got me into the olympiads and into mathematics: the sense that communication of mathematical ideas could feel incredibly natural and enjoyable, that we could lay bare the inner workings of our minds and learn from each other in a way that was so intimate, so vibrant with the will to understand and the desire to share, that we all became – I thought – automatically, authentically, irreparably friends.</p>
<p>Stream-of-consciousness-style training materials are not new nor particularly rare, but in my opinion they remain one of the most precious resources for contestants to learn how to approach problems in an efficient, mindful, ultimately successful way. This is why we'll be starting a weekly series of blog post by different authors, who will share with us their thoughts about past EGMO problems and take us on an adventure along the path that brought them to a solution.</p>
<p>For our first instalment, <span style="text-decoration: underline;"><a href="http://www.egmo2018.org/blog/thinking-out-loud-egmo-2015-problem-5/">Davide tells us about tackling <strong>Problem 5 from EGMO 2015</strong></a></span>.</p>
<p>Enjoy!</p>
<p><a id="ita"></a></p>
<hr/>
<p style="color: gray;"><img alt="" height="25" src="http://www.egmo2018.org/static/media/uploads/post/italian-flag-graphic.png"/> <span> Versione italiana a seguire / <a href="http://www.egmo2018.org/blog/category/problem-solving/feeds/rss/#eng" style="text-decoration: underline;">Back to English</a></span></p>
<p><em>L'idea potrebbe sorprendere gli olimpionici italiani del giorno d'oggi, ma nel 2004 l'Italia non aveva prodotto che due vincitori di medaglie d'oro in tutta la storia delle IMO; il sistema di selezione stava ancora evolvendo verso quello relativamente complesso che abbiamo oggi e, sebbene le risorse per allenarsi ci fossero, non avevamo a disposizione intere raccolte di lezioni incentrate sui temi più disparati, dai binomiali a LTE, ed eravamo sempre in attesa di nuovo materiale.</em></p>
<p><em>Fu in questo contesto che un breve, elegante pdf intitolato "problem solving ad alta voce" comparve un giorno sul sito delle Olimpiadi Italiane della Matematica. L'autore era Marco Cammi, uno dei due ori IMO di cui sopra, e l'opera consisteva in un fedele resoconto dei pensieri di Marco nell'approcciare i sei problemi della gara nazionale: il lettore aveva occasione di scoprire quali idee Marco avesse avuto per prime, come queste lo avessero condotto a una soluzione o non avessero portato a nulla, perfino come talvolta un errore gli desse l'impressione di aver sbagliato approccio ma ragionamenti astratti lo inducessero a tornare sui propri passi e individuare la falla. Ho ancora quel pdf; con le sue 13 pagine di chiacchiere amichevoli e splendida matematica è diventato per me un simbolo di quel che mi ha incantato della matematica e delle olimpiadi: il piacere e la naturalezza del comunicare idee matematiche, il fatto che fosse possibile portare alla luce i meccanismi al lavoro nel segreto delle nostre menti e imparare gli uni dagli altri in un modo così intimo, così pervaso dal desiderio di capire e condividere, che non si poteva che diventare – o così mi è sempre parso – automaticamente, sinceramente, irreparabilmente amici.</em></p>
<p><em>Materiali per l'allenamento olimpico in stile "flusso di coscienza" non sono nuovi né particolarmente rari, ma rimangono secondo me una delle risorse più preziose per imparare ad avvicinarsi ai problemi in maniera efficiente, consapevole e – in ultima analisi – vincente. Ecco perché stiamo per dare inizio a una serie di post in cui autori diversi condivideranno con noi quello che pensano di alcuni problemi EGMO tratti dalle gare passate, invitandoci ad avventurarci con loro alla ricerca di una soluzione.</em></p>
<p><em>Nella prima puntata, <a href="http://www.egmo2018.org/blog/thinking-out-loud-egmo-2015-problem-5/"><span style="text-decoration: underline;">Davide ci racconta come ha affrontato il <strong>problema 5 delle EGMO 2015</strong></span></a>.</em></p>
<p><em>Buona lettura!</em></p>Alessandra CaraceniTue, 02 Jan 2018 10:24:19 +0000http://www.egmo2018.org/blog/thinking-out-loud/EGMOProblem solving