Blog | EGMO 2018http://www.egmo2018.org/blog/2018-02-14T00:38:47+00:00BlogThinking Out Loud – EGMO 2014 Problem 22018-02-06T08:36:24+00:002018-02-06T10:37:16+00:00Camilla Casamento Tumeohttp://www.egmo2018.org/blog/author/camilla/http://www.egmo2018.org/blog/EGMO2014-P2/<p> <script type="text/x-mathjax-config">// <![CDATA[ MathJax.Hub.Config({ tex2jax: {inlineMath: [['$','$'], ['\$','\$']]} }); // ]]></script> </p> <blockquote> <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> </blockquote> <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/author/camilla/feeds/atom/#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>My EGMO – Camilla2017-12-05T08:30:00+00:002018-02-14T00:38:47+00:00Camilla Casamento Tumeohttp://www.egmo2018.org/blog/author/camilla/http://www.egmo2018.org/blog/my-egmo-cct/<p>An interview with/Un'intervista con <strong>Camilla Casamento Tumeo</strong></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/author/camilla/feeds/atom/#ita" style="text-decoration: underline;">Vai alla versione italiana</a></span></p> <p><strong>When did you first take part in a Mathematical contest?</strong></p> <p><span>When I first took part in a maths contest I was 13 years old. I was attending the second year of middle school, and I absolutely did not want to participate. That was probably because the idea of a contest gave me some anxiety. My father is a maths teacher, and he has always pushed me to take challenges. I usually listen to his good advice, but in that situation I was hardly persuadable: he had to promise me he would come with me to my favourite band's concert — and he actually did it! — to convince me to take part in the competition. Surprisingly, I really had fun and I achieved a good result. Since then, I never missed a maths contest I could take part in.</span></p> <p><strong>What has the EGMO experience given to you?</strong></p> <p><span>When I was in high school, I was introduced to the world of Mathematical Olympiads, which is a sort of parallel Universe, with its own dynamics and passed on traditions! EGMO has been one of the greatest experiences of those years. Having the chance to take part in an international competition gave me a strong motivation to do my best. I really learned a lot and at the same time I had great fun. I will always keep the memory of the people I met, the games, the opening ceremonies, the trips, the strange perception of time during the contest. EGMO is not only an occasion to learn more mathematics; it is a comprehensive experience that will enrich you under many points of view.</span></p> <p><span><img alt="Clara, Camilla, Federica and Giulia at EGMO 2014 in Antalya" src="http://www.egmo2018.org/static/media/uploads/post/egmo3.jpg" style="padding-top: 10px;" width="100%"/><em>Clara, Camilla, Federica and Giulia at EGMO 2014 in Antalya.</em></span></p> <p><strong>Are you still in contact with the world of Mathematical Olympiads?</strong></p> <p><span>I remember the feeling I perceived while I was doing the last high school mathematical contest of my life. It was quite a bittersweet sensation, as when you are reading the last page of a book that has kept you company for a while. I could not suppress a bit of melancholy, but I was grateful for all the happiness and the satisfactions I had during the high school years. Anyway, that was not a real ending: I now cooperate with the organisers of Italian Mathematical Olympiad, and this also gives me much. Inventing problems is fun — though sometimes even more difficult than solving them! — and it is a good excuse to keep in touch with all the amazing people I have met when I was a contestant. It is definitely worth it!</span></p> <p><strong>Is there some advice you would like to give to the contestants of this edition of EGMO?</strong></p> <p><span>I believe that training and elaborating good contest strategies is important. Nevertheless, the best advice I can give you is: don't get too nervous and have as much fun as you can when you are at EGMO! These are moments that you will never forget. It is natural to be concerned about your performance, especially if you have worked very hard to arrive at this point. But don't let anxiety spoil a single moment of your time, just do your best and live it peacefully. Enjoy the place, enjoy the people, enjoy all the good you can take. There are moments I have lived at EGMO that I will always carry with me. Many of these are as simple as a walk on the beach at night, in Antalya, or playing hide-and-seek in the hotel's corridors with the French girls, in Minsk. I hope that this edition of EGMO will be an amazing time for you as it has been for me in the past years. We are waiting for you! </span></p> <p></p> <p><span><img alt="Camilla, Giulia, Morena, Vittoria, Giada and Alessandra at EGMO 2015 in Minsk" src="http://www.egmo2018.org/static/media/uploads/post/egmo.jpg" style="padding-top: 10px;" width="100%"/><em>Camilla, Giulia, Morena, Vittoria, Giada and Alessandra at EGMO 2015 in Minsk.</em></span></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/author/camilla/feeds/atom/#eng" style="text-decoration: underline;">Back to English</a></span></p> <p><em><strong>Quando hai partecipato alla tua prima gara di matematica?</strong></em></p> <p><em>La prima volta che presi parte ad una gara di matematica avevo 13 anni. Frequentavo la seconda media e non avevo alcuna intenzione di partecipare, probabilmente perché la sola idea di gareggiare mi agitava. Mio padre è un insegnante di matematica e mi ha sempre spinto ad accettare nuove sfide. Di solito ascolto i suoi buoni consigli, ma quella volta gli diedi del filo da torcere: dovette promettere di accompagnarmi al concerto del mio gruppo preferito — cosa che poi fece! — per convincermi a partecipare. Un po’ a sorpresa mi divertii molto e ottenni anche un buon risultato. Da allora non ho mancato alcuna gara cui ho avuto la possibilità di partecipare.</em></p> <p><em><strong>Cosa ti hanno lasciato le EGMO?</strong></em></p> <p><em>Quando andavo alle superiori fui iniziata al mondo delle Olimpiadi di Matematica, una specie di universo parallelo, con le sue dinamiche e le sue proprie tradizioni. Le EGMO sono state una delle migliori esperienze di quegli anni. Avere l’opportunità di partecipare ad una gara internazionale mi diede una grandissima motivazione a dare il meglio di me. Imparai veramente tanto, divertendomi nel farlo. Ricorderò sempre le persone che ho incontrato, i giochi, le cerimonie d’apertura, le gite, la distorsione del tempo durante la gara. Le EGMO non sono solamente un’occasione per imparare più matematica: sono un’esperienza completa che ti arricchisce da molti punti di vista.</em></p> <p><em><strong>Sei ancora in contatto con il mondo delle Olimpiadi di Matematica?</strong></em></p> <p><em>Ricordo bene l’emozione provata durante l’ultima gara di matematica da liceale della mia vita. Una sensazione dolceamara, un po’ come leggere l’ultima pagina di un libro che ti ha tenuto compagnia per qualche tempo. Non riuscivo a sopire del tutto la nostalgia, ma ero al tempo stesso grata per tutta la felicità e le soddisfazioni avute negli anni delle superiori. In realtà non era davvero la fine: ora collaboro con gli organizzatori delle Olimpiadi di Matematica italiane e anche questo mi dà grande soddisfazione. Inventare problemi è divertente — anche se alle volte è ancor più difficile che risolverli! — ed è un’ottima scusa per rimanere in contatto con le persone meravigliose incontrate quando ero una concorrente. Ne vale assolutamente la pena.</em></p> <p><em><strong>C’è qualche consiglio particolare che vuoi dare alle concorrenti di questa edizione delle EGMO?</strong></em></p> <p><em>Credo che allenarsi e preparare buone strategie di gara sia importante. Al contempo il miglior suggerimento che mi sento di dare è il seguente: non vi agitate e cercate di divertirvi il più possibile mentre siete alle EGMO! Sono momenti che non dimenticherete mai. È inevitabile essere un po’ preoccupate del proprio rendimento, in particolare dopo tutti gli sforzi e i sacrifici fatti per arrivare fin qui. Ma non lasciate che il nervosismo rovini anche solo un momento del vostro tempo, date il vostro meglio e siate in pace. Godetevi il posto, le persone, tutto il buono che potete. Ci sono momenti vissuti alle EGMO che porterò sempre con me, alcuni apparentemente banali, come passeggiare sulla spiaggia di notte ad Antalya, o giocare a nascondino con le concorrenti francesi a Minsk. Spero che questa edizione delle EGMO sia per tutte voi un momento fantastico, come lo è stato per me negli anni passati. Vi aspettiamo! </em></p>