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Source: ELMO Shortlist 2023 G5

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#1 h Jun 29, 2023, 1:39 AM • 3 Y

Y by GeoKing, ImSh95, Funcshun840

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

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DottedCaculator

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DottedCaculator

#2 h Jun 29, 2023, 1:51 AM • 3 Y

Y by on_gale, ImSh95, Funcshun840

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.

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555 posts

#3 h Jun 29, 2023, 8:02 PM • 1 Y

Y by ImSh95

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.

This post has been edited 1 time. Last edited by Orestis_Lignos, Jun 29, 2023, 8:19 PM

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#4 h Jun 30, 2023, 5:46 AM • 1 Y

Y by ImSh95

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 $M$

https://cdn.discordapp.com/attachments/961202142809563140/1124211109608558612/image.png

https://cdn.discordapp.com/attachments/961202142809563140/1124212085560197140/image.png

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.

https://cdn.discordapp.com/attachments/961202142809563140/1124213976536981505/image.png

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Jun 30, 2023, 1:07 PM • 2 Y

Y by ImSh95, Funcshun840

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 $\angle GPD = \angle BPD+\angle GPC-\angle BPC = 180 - \angle BPC = \angle BAC$

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#6 h Mar 27, 2025, 2:01 AM

Y by

Let $K,L$ be ppints such that $CBKL$ is a square and $KL,A$ are on the same side w.r.t. $BC$ , also consider tangents from $B,C$ to $\omega$ hit $KL$ at $I,J$ respectively, finally let $B',C'$ the diametral opposites of $B,C$ respectively.
Now redefining $I$ as $KL \cap DE$ notice that there is a spiral sim centered at $B$ taking $BEDC$ to $BKLC$ and in such we have that $IBDL$ from miquel, however from diameter we have $P,D,B'$ colinear and also $L,B',C$ colinear thus $IB'$ is diameter of $(IBDL)$ and therefore $\angle IBB'=90$ which gives that this $I$ is the original one, similarily $F,G,J$ are colinear.
Now to finish just notce that from the parallels we have $\angle IHJ=\angle BPC=180-\angle BAC$ which defines a circle through $I,J$ with this arc lenght measure thus we are done :cool:

:cool:

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