Let \(ABC\) be an acute triangle with circumcircle \(\omega\). Let \(P\) be a variable point on the arc \(BC\) of \(\omega\) not containing \(A\). Squares \(BPDE\) and \(PCFG\) are constructed such that \(A\), \(D\), \(E\) lie on the same side of line \(BP\) and \(A\), \(F\), \(G\) lie on the same side of line \(CP\). Let \(H\) be the intersection of lines \(DE\) and \(FG\). Show that as \(P\) varies, \(H\) lies on a fixed circle. Proposed by Karthik Vedula
Problem
Source: ELMO Shortlist 2023 G5
Tags: Elmo, geometry
29.06.2023 04:51
By straightforward complex bash, $\overline h=\frac{4i}{b-c}+\frac{\frac{b-c-2i(b+c)}{b-c}}p$ which obviously lies on a circle.
29.06.2023 23:02
We rephrase the problem as follows: Quote: Let $B,C$ be different points on circle $\omega$. Let $P$ be a variable point on the arc $BC$ of $\omega$ not containing $A$. Let $H$ be the point such that $HX=PB, HY=PC,$ with $X,Y$ being the projections of $H$ on $PB,PC$ respectively, and $H,P$ lie on different sides of line $BC$. Then, prove that as $P$ varies, $H$ lies on a fixed circle. Let $P_1,P_2$ be the midpoints of minor and major arcs $BC$, respectively. Let $P$ coincide with $P_1,P_2$ (ignore the condition that $P$ lies on minor arc $BC$) and assume that $H_1,H_2$ are the coressponding points. Note that points $H_1,H_2$ are fixed. We claim that $H$ lies on the circle of diameter $H_1H_2$, which will evidently solve the problem. We present the following Claim. Claim: $HH_1 \parallel PP_1$. Proof: Note that $HX/HY=PB/PC,$ and so $H$ lies on the $P-$ symmedian of triangle $PBC$. Let $HP,H_1P_1$ intersect at point $X$. Moroever, let $\angle BP_2C=x$ and $M$ be the midpoint of $BC$. Then, we compute: $HX/HP=\sin \angle HPC=\sin \angle MPC,$ and so $HP=PB/\sin \angle MPC, \,\,\, (1)$. Moreover, triangles $PBX$ and $PMC$ are similar, hence $PX/PM=BX/MC=BX/BM=1/\sin \angle MXB=1/\cos x,$ hence $PX=PM/\cos x, \,\,\, (2)$. Now, in a similar manner we obtain that $H_1P_1=P_1B/\sin \angle MP_1C, \,\,\, (3)$ and $P_1X=P_1M/\cos x, \,\,\, (4)$. Now, note that $\dfrac{PX}{PH} \cdot \dfrac{P_1H_1}{P_1X}=\dfrac{PM}{P_1M} \cdot \dfrac{P_1H_1}{PH}=\dfrac{PM}{P_1M} \cdot \dfrac{P_1B}{PB} \cdot \dfrac{\sin \angle MPC}{\sin \angle MP_1C}$ However, $PM/PB \cdot \sin \angle MPC=\sin \angle PBC/\sin \angle PMC \cdot \sin \angle MPC=\sin \angle PBC \cdot MC/PC=MC/2R,$ and similarly $P_1M/P_1B \cdot \sin \angle MP_1C=MC/2R,$ and so the above expression equals $1$, hence $PX/PH=P_1X/P_1H_1,$ and so the desired result follows from Thales' theorem $\blacksquare$ Now, to the problem, in a similar manner to the above Claim we obtain that $HH_2 \parallel PP_2,$ and so $\angle H_1HH_2=\angle P_1PP_2=90^\circ$, hence $H$ belongs on the circle with diameter $H_1H_2,$ as desired.
30.06.2023 08:46
Sol:- Construct square $BCRS$ such that $A,R,S$ lie on same side of $BC$.Let the tangent at $B$ to $(ABC)$ meet $RS$ at $L$ and the tangent at $C$ meet $RS$ at $M$.Clearly $L,M$ are fixed points. Step 1 $DE$ passes through $L$ and $FG$ passes through $M$. Proof By spiral similarity at $B$ sending square $BPDE$ to $BCRS$ we know that $\Delta BES \sim \Delta BPC$.So $\measuredangle BLS=\measuredangle LBC=\measuredangle BPC=\measuredangle BES$ so $LSEB$ is cyclic.So $\angle BEL=90^\circ=\angle BED \implies D,L,E$ are collinear. Similarly $FG$ passes through $M$ Step 2 $H$ lies on fixed circle. Proof $L,M$ are fixed points. Due to parallel lines $\measuredangle LHM=\measuredangle BPC=\measuredangle BAC$ which is a fixed angle.Since $\measuredangle LHM$ is a fixed angle subtended on fixed segment $LM$ , $(LHM)$ is a fixed circle.
30.06.2023 16:07
Really nice problem Claim $:$ As $P$ varies all $ED$ lines are concurrent. Proof $:$ Assume $P$ and $P'$ on arc $BC$ and assume $BPDE$ and $BP'D'E'$ are squares. Let $DE$ and $D'E'$ meet at $X$. Note that $\angle BEX = \angle BE'X = \angle 90 \implies BEE'X$ is cyclic so $\angle BXE = \angle BE'E = \angle BP'P$ and since $BP || DE$ we have $\angle 180 - \angle PBX = \angle BXE = \angle BP'P \implies BX$ is tangent to $ABC$. so as $P$ varies $ED$ lines all concur at $X$. we have the same case with all $FG$ lines so assume they concur at some point $Y$. Now Note that $PDGH$ is cyclic so $\angle XHY = \angle 180 - \angle GPD$ so we have to prove $\angle GPD$ is constant which is true since $\angle GPD = \angle BPD+\angle GPC-\angle BPC = 180 - \angle BPC = \angle BAC$