European Mathematical Cup 2016 problem 4 senior division

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European Mathematical Cup 2016 problem 4 senior division

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#1 h Dec 31, 2016, 2:39 PM • 4 Y

Y by anantmudgal09, MintTea, Adventure10, Mango247

Let $C_{1}$ , $C_{2}$ be circles intersecting in $X$ , $Y$ . Let $A$ , $D$ be points on $C_{1}$ and $B$ , $C$ on $C_2$ such that $A$ , $X$ , $C$ are collinear and $D$ , $X$ , $B$ are collinear. The tangent to circle $C_{1}$ at $D$ intersects $BC$ and the tangent to $C_{2}$ at $B$ in $P$ , $R$ respectively. The tangent to $C_2$ at $C$ intersects $AD$ and tangent to $C_1$ at $A$ , in $Q$ , $S$ respectively. Let $W$ be the intersection of $AD$ with the tangent to $C_{2}$ at $B$ and $Z$ the intersection of $BC$ with the tangent to $C_1$ at $A$ . Prove that the circumcircles of triangles $YWZ$ , $RSY$ and $PQY$ have two points in common, or are tangent in the same point.

Proposed by Misiakos Panagiotis

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#2 h Jan 1, 2017, 2:35 PM • 5 Y

Y by anantmudgal09, xdiegolazarox, mathlomaniac, sameer_chahar12, Adventure10

First note that $KAYC$ and $KBYD$ are cyclic by Miquel's theorem. Next, $PDYC$ is cyclic because $\angle PDY=180- \angle DXY=180- \angle PCY$ . Likewise $QDYC$ is cyclic, so $(PQY)=(DCY)$ , and likewise, $(YWZ)=(ABY)$ .

Let $\ell_a, \ell_b,\ell_c, \ell_d$ be the tangent line of $A, B, C, D$ wrt to their respective circles ( $C_1$ or $C_2$ ), and let $M=\ell_a\cap \ell_d, N=\ell_b \cap \ell_c$ . Obviously the spiral similarity centered $Y$ sends $BNC$ to $DMA$ , so $\angle (MS,SN)=\angle (AY,YC)=\angle (MY,YN)$ , so $MSNY$ is cyclic. Likewise, $MRNY$ is cyclic, so $MRSNY$ is cyclic, so $(RSY)=(MNY)$ .

Final step is to prove that $(DCY), (ABY), (MNY)$ are coaxial, which is sufficient to prove that $\frac{pow(M, (ABY))}{pow(M, (DCY))}=\frac{pow(N, (ABY))}{pow(N, (DCY))}$ $\iff \frac{DM\cdot MP}{MA\cdot MZ}=\frac{CN\cdot NQ}{NB\cdot NW}$ $\iff \frac{PM}{MZ}=\frac{QN}{NW}$ But this is because $\angle (PM,MZ)=\angle (WN,NQ)$ since $RMNS$ is cyclic, and $\angle (MP,PZ)=\angle (WQ,QN)$ because $DPQC$ is cyclic. So $\triangle MPZ\sim \triangle WQN$ , hence the ratio equality and we are done.

This post has been edited 3 times. Last edited by navi_09220114, Apr 14, 2018, 3:35 AM

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#3 h Jul 15, 2018, 5:28 PM • 2 Y

Y by Adventure10, Mango247

navi_09220114 wrote:

First note that $KAYC$ and $KBYD$ are cyclic by Miquel's theorem. Next, $PDYC$ is cyclic because $\angle PDY=180- \angle DXY=180- \angle PCY$ . Likewise $QDYC$ is cyclic, so $(PQY)=(DCY)$ , and likewise, $(YWZ)=(ABY)$ .

Let $\ell_a, \ell_b,\ell_c, \ell_d$ be the tangent line of $A, B, C, D$ wrt to their respective circles ( $C_1$ or $C_2$ ), and let $M=\ell_a\cap \ell_d, N=\ell_b \cap \ell_c$ . Obviously the spiral similarity centered $Y$ sends $BNC$ to $DMA$ , so $\angle (MS,SN)=\angle (AY,YC)=\angle (MY,YN)$ , so $MSNY$ is cyclic. Likewise, $MRNY$ is cyclic, so $MRSNY$ is cyclic, so $(RSY)=(MNY)$ .

Final step is to prove that $(DCY), (ABY), (MNY)$ are coaxial, which is sufficient to prove that $\frac{pow(M, (ABY))}{pow(M, (DCY))}=\frac{pow(N, (ABY))}{pow(N, (DCY))}$ $\iff \frac{DM\cdot MP}{MA\cdot MZ}=\frac{CN\cdot NQ}{NB\cdot NW}$ $\iff \frac{PM}{MZ}=\frac{QN}{NW}$ But this is because $\angle (PM,MZ)=\angle (WN,NQ)$ since $RMNS$ is cyclic, and $\angle (MP,PZ)=\angle (WQ,QN)$ because $DPQC$ is cyclic. So $\triangle MPZ\sim \triangle WQN$ , hence the ratio equality and we are done.

Can you please explain slightly more with a diagram

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#4 h Mar 19, 2020, 9:07 AM • 2 Y

Y by sameer_chahar12, amar_04

Note that $Y$ is the center of spiral similarity which takes $AC$ to $DB$ , and so we have $\triangle YAD \stackrel{+}{\sim} YCB$ . This gives us $\measuredangle WAY=\measuredangle DAY=\measuredangle BCY=\measuredangle WBY \Rightarrow W \in \odot (YAB)$ Similarly, $Z \in \odot (YAB)$ , and so we have $\odot (YWZ) \equiv \odot (YAB)$ . Also, $\measuredangle QCY=\measuredangle CBY=\measuredangle ADY=\measuredangle QDY \Rightarrow Q \in \odot (CDY)$ Similarly, $P \in \odot (CDY)$ , and so we have $\odot (PQY) \equiv \odot (CDY)$ .

Now, let $T=AD \cap BC$ and $U$ (resp. $V$ ) be the point where the tangents to $C_1$ (resp. $C_2$ ) at $A$ and $D$ (resp. $B$ and $C$ ) meet. Then $\measuredangle SAT=\measuredangle ZAW=\measuredangle CBW=\measuredangle SCT \Rightarrow S \in \odot (ACT)$ But, as $Y$ is also the center of spiral similarity taking $AD$ to $CB$ , so $AYTC$ is cyclic. This means that $SAYTC$ and $BRYDT$ are cyclic pentagons (the second one following in a similar one as the first one). Thus, we have $\measuredangle SYR=\measuredangle SYT+\measuredangle TYR=\measuredangle SCT+\measuredangle TBR=\measuredangle VCB+\measuredangle CBV=\measuredangle CVB=\measuredangle SVR$ which means that $V \in \odot (RSY)$ . Similarly, $U \in \odot (RSY)$ , and so we must have $\odot (RSY) \equiv \odot (UVY)$ .

Thus, it suffices to show that $\frac{UP \cdot UD}{UZ \cdot UA}=\frac{VC \cdot VQ}{VW \cdot VB} \Leftrightarrow \frac{UP}{UZ}=\frac{VQ}{VW} \Leftrightarrow \frac{\sin \angle UZP}{\sin \angle UPZ}=\frac{\sin \angle VWQ}{\sin \angle VQW}$ But, we have $\measuredangle UZP=\measuredangle AZB=\measuredangle AWB=\measuredangle QWV \Rightarrow \sin \angle UZP=\sin \angle VWQ$ Similarly, we have $\sin \angle UPZ=\sin \angle VQW$ . Combining the above two equalities, we get the desired result. $\blacksquare$

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#6 h Jun 5, 2020, 7:51 PM • 6 Y

Y by a_simple_guy, GeoMetrix, mueller.25, Bumblebee60, MintTea, myh2910

Nice and Easy!!!

steppewolf wrote:

Let $C_{1}$ , $C_{2}$ be circles intersecting in $X$ , $Y$ . Let $A$ , $D$ be points on $C_{1}$ and $B$ , $C$ on $C_2$ such that $A$ , $X$ , $C$ are collinear and $D$ , $X$ , $B$ are collinear. The tangent to circle $C_{1}$ at $D$ intersects $BC$ and the tangent to $C_{2}$ at $B$ in $P$ , $R$ respectively. The tangent to $C_2$ at $C$ intersects $AD$ and tangent to $C_1$ at $A$ , in $Q$ , $S$ respectively. Let $W$ be the intersection of $AD$ with the tangent to $C_{2}$ at $B$ and $Z$ the intersection of $BC$ with the tangent to $C_1$ at $A$ . Prove that the circumcircles of triangles $YWZ$ , $RSY$ and $PQY$ have two points in common, or are tangent in the same point.

Proposed by Misiakos Panagiotis

Let $CQ\cap\{AZ,BW\}=\{S,T\}$ respectively.

Claim 1:- $\{C,D\}\in\odot(YPQ)$
Consider the Spiral Similarity $\sigma$ at $Y$ mapping $\triangle ADY\mapsto\triangle YCB$ . So, $\angle QCY=\angle CBY=\angle ADY$ and $\angle PDY=\angle DAY=\angle YCB$ . So, $\{C,D\}\in\odot(YPQ)$

Claim 2:- $\{S,T\}\in\odot(RSY)$
Now notice that $\sigma:AA\cap DD\mapsto CC\cap BB=T$ . Hence, $\triangle YKA\stackrel{+}{\sim}\triangle YTC$ . Hence, $\angle AKY=\angle YTS$ . Similarly $RKYT$ is a cyclic quad $\implies \{S,T\}\in\odot(RSY)$

Now by Coaxility Lemma it suffices to prove that

$\begin{align*} \frac{Pow_{\odot(YQP)}W}{Pow_{\odot(YRS)}W}=\frac{Pow_{\odot(YQP)}Z}{Pow_{\odot(YRS)}Z} &\iff \frac{WQ\cdot WD}{WR\cdot WT}=\frac{ZP\cdot ZC}{ZS\cdot ZK} \\ &\iff\frac{WD}{WR}\cdot\frac{WQ}{WT}=\frac{ZC}{ZS}\cdot\frac{ZP}{ZK} \\ &\iff\frac{\sin\angle WTQ}{\sin\angle WQT}\cdot\frac{\sin\angle WRD}{\sin\angle WDR}=\frac{\angle ZKP}{\sin\angle ZPK}\cdot\frac{\sin\angle ZSC}{\sin\angle ZCS}\cdots\cdots\cdots(\star)\end{align*}$
$(\star)$ is true due to Claim 1 and Claim 2. Hence $\{\odot(YWZ),\odot(RSY),\odot(YPQ)\}$ are coaxial. $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Jun 5, 2020, 7:51 PM

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