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Source: ELMO Shortlist 2024 G5

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1595 posts

#1 h Jun 22, 2024, 3:54 PM • 1 Y

Y by Rounak_iitr

Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$ . Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ . Let $P$ and $Q$ be points on the circumcircles of triangles $AOB$ and $AOC$ , respectively, such that $A$ , $P$ , and $Q$ are collinear. Prove that if the circumcircle of triangle $OPQ$ is tangent to $\omega$ at $T$ , then $\angle BTD=\angle CAP$ .

Tiger Zhang

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#2 h Jun 22, 2024, 4:26 PM • 7 Y

Y by VicKmath7, Rounak_iitr, CT17, ihatemath123, Funcshun840, ehuseyinyigit, NonoPL

My problem!

$[asy] size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.7; pen bluefill,pinkfill,turquoisedraw,bluedraw,purpledraw,pinkdraw,graydraw; bluefill = RGB(204,233,255); pinkfill = RGB(255,204,255); turquoisedraw = RGB(0,170,153); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); graydraw = RGB(119,119,119); pair A,B,C,M,N,O,T,P,Q,D; A = dir(85); B = dir(195); C = dir(345); O = circumcenter(A,B,C); M = (A+B)/2; N = (A+C)/2; path c,d,e,f; c = circumcircle(A,B,C); T = intersectionpoints(c,11M-10N--N)[0]; d = circle((O+T)/2,distance(O,T)/2); e = circumcircle(A,B,O); f = circumcircle(A,C,O); P = intersectionpoints(d,e)[0]; Q = intersectionpoints(d,f)[0]; D = foot(A,B,C); filldraw(c,bluefill,bluedraw); fill(d,pinkfill); filldraw(circumcircle(A,M,N),pinkfill,pinkdraw); draw(d,pinkdraw); draw(e,purpledraw); draw(f,purpledraw); draw(A--B--C--cycle,bluedraw); draw(A--P,turquoisedraw); draw(N--T,turquoisedraw); draw(A--D,graydraw); dot("$A$",A,1.9*dir(85)); dot("$B$",B,dir(225)); dot("$C$",C,dir(315)); dot("$M$",M,dir(140)); dot("$N$",N,dir(30)); dot("$O$",O,1.9*dir(270)); dot("$T$",T,dir(150)); dot("$P$",P,dir(265)); dot("$Q$",Q,dir(10)); dot("$D$",D,dir(270)); [/asy]$

Let $\measuredangle$ denote directed angles modulo $180^\circ$ .

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ . The crux of the problem is the following claim.

Claim: $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$ .

Proof: Notice that the circumcircle of $AMN$ has diameter $\overline{AO}$ , so it is tangent to $\omega$ . Thus, the circumcircles of $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$ . Also, we have $\measuredangle OTP=\measuredangle OQP=\measuredangle OCA=\measuredangle NAO$ , so $N$ and $P$ are symmetric about the perpendicular bisector of $\overline{AT}$ . Similarly, $M$ and $Q$ are symmetric. $\square$

Since $A$ , $P$ , and $Q$ are collinear, $M$ , $N$ , and $T$ are collinear. Since $\overline{MN}$ is the perpendicular bisector of $\overline{AD}$ , we have $TA=TD$ . Let $\overline{AP}$ and $\overline{MN}$ intersect at $X$ . By symmetry, $XAT$ is isosceles. Thus, we have
$\measuredangle BTD=\measuredangle BTA+\measuredangle ATD=\measuredangle BCA+2\measuredangle ATX=\measuredangle MNA+\measuredangle AXN=\measuredangle PAC$ and we are done. $\square$

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1524 posts

#3 h Jun 22, 2024, 6:50 PM • 2 Y

Y by IYxMT, ehuseyinyigit

Nice problem, here is a solution found while watching Portugal vs Turkey Match lol.
Let the tangency point be $T$ , let $M,N$ midpoints of $AC, AB$ respectively and let $(OPQ) \cap (ANMO)=G$ , note that they are circles with diameters $TO,AO$ respectively and since $TO=AO$ and $\angle TGO=90$ we get $G$ midpoint of $TA$ , so these circles are symetric w.t.t. $GO$ . Now $\angle APO=\angle ABO=\angle NAO=\angle NMO$ which shows that arcs $QO,NO$ have the same lenght, in the same way arcs $PO,MO$ have the same lenght and thus $(N,Q), (M,P)$ are symetric w.r.t. $GO$ , now this means $T,M,N$ colinear. To finish note that $\angle GTP=\angle AQG=\angle TNG=\angle NTB$ (last one is just midbase), therefore $\angle NTD=\angle ATN=\angle BTP$ which means $\angle BTD=\angle PTN=\angle CAP$ as desired thus we are done :cool:

:cool:

.

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DottedCaculator

7348 posts

DottedCaculator

#4 h Jun 22, 2024, 11:08 PM

Y by

Invert at $O$ . We have $\overline{p'}=\frac{2t-a-b}{t^2-ab}$ , so $p=\frac{t^2-ab}{2t-a-b}$ , which means $p-a=\frac{(t-a)^2}{2t-a-b}$ . Similarly, $q-a=\frac{(t-a)^2}{2t-a-c}$ , so $\frac{2t-a-b}{2t-a-c}$ is real, so $\frac{2t-a-c}{b-c}$ is real, which implies $2t-a-c=-bc\left(\frac2t-\frac1a-\frac1c\right)$ , or $2at^2-(a+c)(a+b)t+2abc=0.$
We have $d=\frac12\left(a+b+c-\frac{bc}a\right)$ . The condition $\angle BTD=\angle CAP$ is equivalent to $\frac{b-t}{d-t}\frac{c-a}{p-a}$ real, or
$\begin{align*} \frac{btca}{(d-t)\frac{(t-a)^2}{2t-a-b}}&=\frac1{\left(\overline d-\frac1t\right)\frac{\overline{(t-a)^2}}{\overline{2t-a-b}}}\\ \frac{abct(2t-a-b)}{d-t}&=\frac{a^2t^2\left(\frac2t-\frac1a-\frac1b\right)}{\overline d-\frac1t}\\ b^2c(2t-a-b)(t\overline d-1)&=t(d-t)(2ab-at-bt)\\ t^3(a+b)-t^2(ad+bd+2ab+2b^2c\overline d)+t(2abd+2b^2c+ab^2c\overline d+b^3c\overline d)-b^2c(a+b)&=0\\ t^3a(a+b)-t^2\frac12((a+b)(a^2+ab+ac-bc)+4a^2b+2b(bc+ca+ab-a^2))+t\frac12(2ab(a^2+ab+ac-bc)+4ab^2c+b(a+b)(ab+bc+ca-a^2))-ab^2c(a+b)&=0\\ t^32a-t^2(a^2+3ab+ac+bc)+tb(a^2+3ac+ab+bc)-2ab^2c&=0\\ (2at^2-(a+c)(a+b)t+2abc)(t-b)&=0. \end{align*}$

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3446 posts

#5 h Jul 29, 2024, 2:33 AM

Y by

Solved with Benjaminxiao, Zhaom.

Let $P'$ be the intersection of lines $OP$ and $AB$ , and let $Q'$ be the intersection of lines $OQ$ and $\overline{AC}$ ; in other words, $P'$ and $Q'$ are the images of $P$ and $Q$ under inversion about $\omega$ . Therefore, line $P'Q'$ is tangent to $\omega$ at $T$ .

Also, since $A$ , $P$ and $Q$ are collinear, it follows that $(AOP'Q')$ is cyclic. By Simson's theorem, the feet from $O$ onto sides $\overline{AB}$ , $\overline{AC}$ and $\overline{P'Q'}$ are collinear; in other words, $T$ lies on the $A$ -midline of $\triangle ABC$ .

Claim: We have $\triangle TPQ \sim \triangle ACB$ (with similarity factor of $\tfrac{1}{2}$ .)
Proof: We have $\angle TPQ = \angle TOQ' = 90^{\circ} - \angle OQ'P' = \angle P'AO - 90^{\circ} = \angle C,$ and likewise for $\angle TQP$ and $\angle B$ .

If we define $P_1$ and $Q_1$ as the reflections of $T$ across $P$ and $Q$ , respectively, we have that $\triangle T Q_1 P_1$ and $\triangle ABC$ are reflections (and in particular, $\omega$ is the circumcircle of $\triangle TQ_1 P_1$ ). From this, it follows by symmetry $\angle PAT = \angle (P_1 Q_1, AT) = \angle (AT, BC)$ .

Now, we have
$\angle BTD = \angle ATD - \angle C = \angle PAT + \angle (AT, BC) - \angle C = \angle PAT + \angle (AT, AC) = \angle PAT - \angle CAT = \angle CAP.$

This post has been edited 1 time. Last edited by ihatemath123, Jul 29, 2024, 5:14 AM

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#6 h Jul 29, 2024, 3:08 AM • 1 Y

Y by ihatemath123

I regret not discovering the following version of the problem (that I found just now), which has a significantly more appealing statement.

ELMO SL 2024 G5 but better wrote:

Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$ . A line through $A$ intersects $\overline{BC}$ , the circumcircle of $AOB$ , and the circumcircle of $AOC$ at $D$ , $E \ne A$ , and $F \ne A$ , respectively. Suppose the circumcircle of $OEF$ is tangent to $\omega$ at $T$ . Prove that $\angle ATD=90^\circ$ .

Solution

This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Feb 2, 2025, 11:33 PM

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awesomeming327.

1714 posts

awesomeming327.

#7 h Feb 2, 2025, 12:12 AM

Y by

Let $M$ and $N$ be midpoints of $AB$ and $AC$ , respectively.

Claim 1: $ATNP$ and $ATMQ$ are isosceles trapezoids.

Note that $OT$ is the diameter of the circle $(OPQ)$ and $OA$ is the diameter of the circle $(OMN)$ so $\angle TPO=\angle ANO=90^\circ$
Furthermore, we have
$\measuredangle PTO=\measuredangle PQO=\measuredangle AQO=\measuredangle ACO=\measuredangle OAC=\measuredangle OAN$ Since we also have $OT=OA$ , $\triangle OTP\cong\triangle OAN$ . Similarly, $\triangle OTQ\cong \triangle OAM$ . Thus, our claim is true.

Since $A$ , $P$ , $Q$ are collinear, this means that $T$ , $M$ , $N$ are collinear. Let $MN$ intersect $(ABC)$ at $T$ and $T'$ . Then $TT'BC$ is an isosceles trapezoid. We have
$\measuredangle PTA=\measuredangle TAN=\measuredangle TAC=\measuredangle TT'C=\measuredangle BTT'=\measuredangle BTN$ This means that $\measuredangle BTP=\measuredangle NTA$ . Since $TN$ is the perpendicular bisector of $AD$ ,
$\measuredangle NTA=\measuredangle DTN$ so $\measuredangle BTD=\measuredangle PTN=\measuredangle PAN=\measuredangle PAC$ as desired.

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84 posts

akasht

#8 h Apr 30, 2025, 12:25 PM

Y by

skibidi spiral

Let $X$ and $Y$ be the inverses of $P$ and $Q$ upon inverting about $(ABC)$ . Then $A$ lies on $(OXY)$ and furthermore $XY$ is tangent to $(ABC)$ . This implies that $T$ is the foot of $O$ on $XY$ . Additionally, $A,X,B$ and $A,Y,C$ are collinear. Let $A'$ be the other intersection of $(ABC)$ with $(OXY)$ .
We now setup some points to do a spiral at $A'$ . Define

$T'$ as the reflection of $T$ about the midpoint of $XY$ .
$S, S'$ as the intersection of $AT,A'T$ with $(OXY)$ .
$O'$ as the antipode of $O$ in $(OXY)$ .

Now note that the spiral about $A'$ sending $(ABC)$ to $(OXY)$ sends $A \to O'$ , $B \to X$ , $C \to Y$ , $D \to T'$ , $T \to S$ . Also, observe that by Reim, $SS' // XY$ so $S,S'$ are symmetric about the perpendicular bisector of $XY$ . Thus $\angle BTD = \angle XST' = \angle YS'T = \angle YS'A' = \angle A'AC = \angle PAC$ as desired

This post has been edited 2 times. Last edited by akasht, Apr 30, 2025, 12:41 PM

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#9 h Apr 30, 2025, 12:41 PM

Y by

CyclicISLscelesTrapezoid wrote:

I regret not discovering the following version of the problem (that I found just now), which has a significantly more appealing statement.

ELMO SL 2024 G5 but better wrote:

Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$ . A line through $A$ intersects $\overline{BC}$ , the circumcircle of $AOB$ , and the circumcircle of $AOC$ at $D$ , $E \ne A$ , and $F \ne A$ , respectively. Suppose the circumcircle of $OEF$ is tangent to $\omega$ at $T$ . Prove that $\angle ATD=90^\circ$ .

Solution

Also it seems that if $(OEF) \cap \omega =T_1, T_2$ , then $\angle AT_1D+\angle AT_2D = 180^\circ$ .

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