Blog | EGMO 2018http://www.egmo2018.org/blog/2018-02-27T11:47:19+00:00BlogThinking Out Loud – EGMO 2017 Problem 12018-02-27T11:47:08+00:002018-02-27T11:47:19+00:00Samuele Mongodihttp://www.egmo2018.org/blog/author/Sam/http://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 $\widehat{ABC} &gt; \widehat{CDA}$. 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> </blockquote> <p> <script type="text/x-mathjax-config">// <![CDATA[ MathJax.Hub.Config({ tex2jax: {inlineMath: [['$','$'], ['\$','\$']]} }); // ]]></script> </p> <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> </div> <p><attachment aria-label="figureEgmo17-A1.pdf" class="Apple-web-attachment" id="&lt;275A27ED-4277-4B08-9419-A2F22DCAF7B0@Home&gt;" role="button" title="figureEgmo17-A1.pdf" type="application/pdf"></attachment></p>