AoPS - Problem

Problem

Source: JBMO 2024

Tags: geometry, circumcircle, collinear, Balkan, JBMO

Lukaluce

27.06.2024 14:05

Let $ABC$ be a triangle such that $AB < AC$ . Let the excircle opposite to A be tangent to the lines $AB, AC$ , and $BC$ at points $D, E$ , and $F$ , respectively, and let $J$ be its centre. Let $P$ be a point on the side $BC$ . The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$ . Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$ . Prove that the points $P, Q$ , and $R$ are collinear. (The excircle of a triangle $ABC$ opposite to $A$ is the circle that is tangent to the line segment $BC$ , to the ray $AB$ beyond $B$ , and to the ray $AC$ beyond $C$ .) Proposed by Bozhidar Dimitrov, Bulgaria

Gryphos

27.06.2024 14:35

By Thales,

$D,E,R$ lie on the circle

$k$ with diameter

$AJ$ . We now claim that

$Q$ lies on

$k$ as well. Indeed, because of the cyclic quadrilaterals

$BDQP$ ,

$CPQE$ we have

$\angle EQD = \angle EQP + \angle PQD = \angle ACB + \angle CBA = 180^\circ - \angle BAC = 180^\circ - \angle DAE$ . Now since

$AR \parallel BC$ , by angle chasing we get

$\angle RQD = 180^\circ - \angle DAR = \angle CBA = \angle PQD$ , implying that

$P,Q,R$ are collinear.

P2nisic

27.06.2024 15:06

Lukaluce wrote: Let $ABC$ be a triangle such that $AB < AC$ . Let the excircle opposite to A be tangent to the lines $AB, AC$ , and $BC$ at points $D, E$ , and $F$ , respectively, and let $J$ be its centre. Let $P$ be a point on the side $BC$ . The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$ . Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$ . Prove that the points $P, Q$ , and $R$ are collinear. (The excircle of a triangle $ABC$ opposite to $A$ is the circle that is tangent to the line segment $BC$ , to the ray $AB$ beyond $B$ , and to the ray $AC$ beyond $C$ .) $angle DQE=\angle DQP+\angle PQE=\angle ABC+\angle ACB=180-\angle DAE$ so $D,Q,E,A$ are concyclic. More over $\angle ADI_A=\angle ARI_A=\angle AEI_A=90$ so $D,A,R,E,I_A,Q$ are concyclic. No w we have that: $\angle DQP=\angle ABC=\angle DI_AR=\angle DQR$ So $Q,P,R$ are collinear

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Assassino9931

27.06.2024 15:11

Following last year, we have another high class JBMO Geometry problem by Bozhidar Dimitrov (Strudan_Borisov) from Bulgaria; or as several leaders said: bomba. The original submission also has a harder version, which hides the point $R$ and requires to show that $PQ$ passes through a fixed point as $P$ varies - very beautiful, but of course not suitable for juniors. I only think it should have been P1, but oh well.

giannis2006

27.06.2024 15:17

From the cyclic quadrilaterals, we get that: $\angle DQE = \angle DQP + \angle EQP = \angle ABC + \angle ACB = 180 - \angle BAC = 180 - DAE$ , hence $Q$ lies on $(ADE)$ . Also, $\angle ARJ = \angle AEJ = \angle ADJ = 90$ , so $A,R,E,J,D$ are concyclic in a circle with diameter $AJ$ and hence $A,R,E,J,Q,D$ are all concyclic. Now we have that: $AR \perp FJ => AR \parallel BC$ . From the cyclic quadrilaterals and using that $AR \parallel BC$ : $\angle RQE = \angle RAE = \angle RAC = \angle ACB = 180 - \angle PCE = \angle PQE$ and $P,Q,R$ are collinear, as needed. Note that the same holds for any point $J$ , not just the excenter, where $D,E,F$ are the projections of $J$ in the sides of the triangle.

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Z4ADies

27.06.2024 15:29

Just using angle chasing.... $\angle QDB=\angle QPC= 180- \angle CEQ$ so, $DAEQ$ is cyclic. Also, $JDAE$ cyclic $\implies$ $JEADQ$ is cyclic.Also we know $AREJ$ cyclic.So,by written some angles we deduce $RQ \cap BC =P$ .

Marinchoo

27.06.2024 15:50

Kudos to Bozhidar Dimitrov for the cute problem! Here's a quick solution: Introduce $Q'$ as the second intersection of $\overline{RP}$ and the cyclic $ARDJE$ . Then $\angle DQP=\angle RAD=\beta=\angle ABP$ hence $BPQD$ is cyclic. Similarly, $CPQE$ is cyclic as well, so $Q=Q'$ , as desired.

sami1618

27.06.2024 16:07

Notice that $ARDJE$ is cyclic. Define $Q'$ as the second intersection of $PR$ and $(ARDJE)$ . Then $\angle PQ'D=180^{\circ}-\angle BAR=\angle ABC$ $\angle PQ'E=\angle RAC=\angle ACB$ Thus $BPQ'D$ and $CPQ'E$ are concyclic, so $Q=Q'$ , finishing the problem.

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Alex9100

27.06.2024 16:15

Since, $\angle ARJ = \angle JDA= \angle AEJ= 90 A$ , $R$ , $E$ , $J$ and $D$ lie on a circle. Now, $JF \perp$ both $BC$ and $AR \implies BC\parallel AR$ $\implies \angle RAB + \angle ABC = 180= \angle RAB + \angle PQD$ $\implies Q \in (AERJD)$ $\implies \angle FPR = \angle PRA =180-\angle ADQ=\angle BPQ$ $\implies P$ , $Q$ and $R$ lie on a single line.

cursed_tangent1434

27.06.2024 16:26

Surprisingly very easy. I think the original submission (showing $PQ$ passes through a fixed point) wouldn't have been too hard for the Problem 2 spot. We first prove the following key claim. Claim : Points $A$ , $D$ , $E$ , $J$ , $R$ and $Q$ lie on the same circle. Proof : It is immediate that $ADJE$ is cyclic, and that $R$ lies on this circle as well, due to the right angles. Further, $\measuredangle DQE = \measuredangle DQP + \measuredangle PQE = \measuredangle ABC + \measuredangle BCA = \measuredangle BAC = \measuredangle DAE$ so $Q$ also lies on this circle, finishing the proof of the claim. Now, let $P' = \overline{QR} \cap \overline{BC}$ . We simply note that since $AR \parallel BC$ (since $JF \perp BC$ quite clearly), we have $\measuredangle DQP' = \measuredangle DQR = \measuredangle DAR = \measuredangle CBA = \measuredangle P'BA$ so $P'$ lies on $(DBQ)$ . Thus, $P'=P$ and it immediately follows that points $P, Q$ , and $R$ are collinear, as desired.

trigadd123

27.06.2024 18:41

Something more general holds: if $E$ and $F$ are fixed points on $AB$ and $AC$ respectively and $D$ varies on $BC$ , then the radical axis of $(BDE)$ and $(CDF)$ passes through a fixed point. For simplicity, assume $D, E$ and $F$ lie on the sides of $\triangle ABC$ (the same argument works in general). Let $X$ be the point on $(AEF)$ such that $AX\parallel BC$ , which is clearly fixed. We claim that $X$ is the desired point. If $(BDE)$ and $(CDF)$ meet a second time at $K,$ then $K$ lies on $(AEF)$ by Miquel's theorem. The collinearity follows because $\angle XKF=\angle XAF=\angle ACB=180^{\circ}-\angle FKD.$ In fact, an even further generalization can be found in post #9 here.

hukilau17

27.06.2024 19:46

Identify the

$A$ -excircle with the unit circle, and let the other tangent from

$P$ to the

$A$ -excircle touch it at

$G$ . (If

$P=F$ , take

$G=F$ as well.) Then

$\begin{align*} |d|=|e|=|f|=|g|&=1 \\ j &= 0 \\ a &= \frac{2de}{d+e} \\ b &= \frac{2df}{d+f} \\ c &= \frac{2ef}{e+f} \\ p &= \frac{2fg}{f+g} \\ r = \frac{a+f+(-f) - \overline{a}f(-f)}2 &= \frac{de+f^2}{d+e} \end{align*}$ Now we find the coordinate of

$Q$ . Since

$B,D,P,Q$ are concyclic, we have

$\frac{(b-d)(p-q)}{(b-p)(d-q)} \in \mathbb{R}$ Since

$\frac{b-d}d,\frac{b-p}f \in i\mathbb{R}$ , this becomes

$\frac{d(p-q)}{f(d-q)} \in \mathbb{R}$

$\frac{d(2fg-fq-gq)}{f(f+g)(d-q)} = \frac{f(2-f\overline{q}-g\overline{q})}{(f+g)(1-d\overline{q})}$

$d(1-d\overline{q})(2fg-fq-gq) = f^2(d-q)(2-f\overline{q}-g\overline{q})$

$\overline{q} = \frac{d(2fg-fq-gq) - 2f^2(d-q)}{d^2(2fg-fq-gq) - f^2(f+g)(d-q)} = \frac{q(df+dg-2f^2) + 2df(f-g)}{q(d+f)(d-f)(f+g) - df(2dg-f^2-fg)}$ Similarly, since

$C,E,P,Q$ are concyclic, we have

$\overline{q} = \frac{q(ef+eg-2f^2) + 2ef(f-g)}{q(e+f)(e-f)(f+g) - ef(2eg-f^2-fg)}$ Setting these equal gives

$\frac{q(df+dg-2f^2) + 2df(f-g)}{q(d+f)(d-f)(f+g) - df(2dg-f^2-fg)} = \frac{q(ef+eg-2f^2) + 2ef(f-g)}{q(e+f)(e-f)(f+g) - ef(2eg-f^2-fg)}$ Clearing denominators gives

$q^2(e+f)(e-f)(f+g)(df+dg-2f^2) + (\cdots)q - 2def^2(f-g)(2ef-f^2-fg)$

$= q^2(d+f)(d-f)(f+g)(ef+eg-2f^2) + (\cdots)q - 2def^2(f-g)(2dg-f^2-fg)$ Subtracting and factoring out

$d-e$ gives

$q^2(f+g)(2df^2+2ef^2-def-deg-f^3-f^2g) + (\cdots)q + 4def^2g(f-g) = 0$ Thus by Vieta,

$pq = \frac{4def^2g(f-g)}{(f+g)(2df^2+2ef^2-def-deg-f^3-f^2g)}$ and so

$q = \frac{2def(f-g)}{2df^2+2ef^2-def-deg-f^3-f^2g}$ Now we find the vectors

$\begin{align*} p-q &= \frac{2f^2(2dfg+2efg-def-deg-f^2g-fg^2)}{(f+g)(2df^2+2ef^2-def-deg-f^3-f^2g)} \\ p-r &= \frac{2dfg+2efg-def-deg-f^3-f^2g}{(d+e)(f+g)} \end{align*}$ Then

$\frac{p-q}{p-r} = \frac{2f^2(d+e)(2dfg+2efg-def-deg-f^2g-fg^2)}{(2df^2+2ef^2-def-deg-f^3-f^2g)(2dfg+2efg-def-deg-f^3-f^2g)}$ We conjugate this, replacing each

$d,e,f,g$ with its reciprocal, and then multiply across by

$d^2e^2f^6g^2$ , to see that it equals its conjugate and is thus real.

$\blacksquare$

ItsBesi

27.06.2024 20:35

Let

$\odot(BDP)=\omega_1 , \odot(CEP)=\omega_2$

$\textbf{Claim:}$ Points

$A,D,Q$ and

$E$ -are concyclic

$\textbf{Proof:}$

$\angle DQP \stackrel{\omega_1}{=} 180-\angle DBP \equiv 180-\angle DBC=\angle ABC \equiv \angle B \implies \angle DQP=\angle B$

$\angle PQE \stackrel{\omega_2}{=} 180- \angle PCE \equiv 180 - \angle BCE = \angle ACB \equiv \angle C \implies \angle PQE=\angle C$ So

$\angle DQE=\angle DQP+ \angle PQE=\angle B+\angle C=180-\angle A \equiv 180-\angle BAC \equiv 180-\angle DAE \implies \angle DQE+\angle DAE=180 \implies$ Points

$A,D,Q$ and

$E$ -are concyclic

$\square$ . Let

$\odot(DQE)=\Omega$

$\textbf{Claim:}$

$J \in \Omega$

$\textbf{Proof:}$

$\angle JDB=90 , \angle JFB=90$ So

$\angle JDB+\angle JFB=180 \implies$ Points

$J,F,D$ and

$P$ -are concyclic

$\implies \angle FJD=180- \angle FBD \equiv 180-\angle DBC=\angle ABC \equiv B \implies \angle FJD=\angle B$ Similarly

$\angle JFC=90 , \angle JEC=90$ So

$\angle JFC+\angle JEC=180 \implies$ Points

$F,E,J$ and

$C$ -are concyclic

$\implies \angle FJE=180-\angle FCE \equiv 180-\angle BCE=\angle ACB \equiv \angle C \implies \angle FJE=\angle C$ So

$\angle DJE=\angle DJF+\angle FJE=\angle B + \angle C=\angle DQP+\angle PQE=\angle DQE \implies \angle DJE=\angle DQE \implies$ Points

$D,Q,J$ and

$E$ -are concyclic

$\iff J \in \odot (DQE) \iff J \in \Omega$

$\square$

$\textbf{Claim:}$

$R \in \Omega$

$\textbf{Proof:}$

$\angle JDA=90 , \angle JRA=90 \implies \angle JDA+\angle JRA=180 \implies$ Points

$J,D,A$ and

$R$ -are concyclic

$\iff R \in \odot(AJD) \iff R \in \Omega$

$\square$

$\textbf{Claim:}$ Points

$P , Q$ and

$R$ -are collinear

$\textbf{Proof:}$ From

$\angle ARF=\angle RFC=90$ we get that :

$AR \parallel BC$ Now we finish the problem

$\angle PQE \stackrel{\omega_2}{=} 180-\angle PCE=\angle ACB \equiv \angle C \stackrel{AR \parallel BC}{=} \angle CAR \equiv EAR \stackrel{\Omega}{=} \angle EQR \equiv RQE \implies \angle PQE=\angle RQE \implies$ Points

$P , Q$ and

$R$ -are collinear

$\blacksquare$

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g0USinsane777

28.06.2024 00:04

Claim : $A,R,E,J,Q,D$ are concyclic Proof : Clearly, the points $R,E,D$ lie on the circle with $AJ$ as diameter since they subtend $90^{\circ}$ angle at $AJ$ . Now, since $BPQD$ and $CPQE$ are cyclic, $\angle DQP = \angle B$ and $\angle EQP = \angle C$ $\implies \angle DQE = 180^{\circ} - \angle A$ Let $P'$ be the intersection of $QR$ with $BC$ Claim : $P = P'$ Proof : $\angle BPQ = 180^{\circ} - \angle BDQ = \angle ARQ = \angle ARP' = \angle CP'R = \angle BP'Q$ This means that $P$ and $P'$ lie on the same line and subtend the same angle at $BQ \implies \angle PQP' = 0 \implies \angle P = P'$ . And we are done.

X.Luser

06.07.2024 16:01

I solved it in 2 minutes it's very easy for jbmo

atdaotlohbh

09.07.2024 23:43

By Miquel Theorem $Q$ lies on $(ADJE)$ . Because it's diameter is $AJ$ , $R$ also lies on it. Now $\angle DQP=\angle ABC$ and $\angle DQR = \angle DJR=\angle DJF=\angle BAC$ , so $P$ , $Q$ and $R$ are collinear

Assassino9931

11.07.2024 10:00

trigadd123 wrote: Something more general holds: if $E$ and $F$ are fixed points on $AB$ and $AC$ respectively and $D$ varies on $BC$ , then the radical axis of $(BDE)$ and $(CDF)$ passes through a fixed point. Which, we just now realized, appears at the generalization of ISL 2018 G2 by Daniel Zhu (in post 8).

Alex009

05.08.2024 00:53

$JDAER$ is clearly cyclic and moreover $AR\parallel BC$ as they are both perpendicular to $RJ$ Claim: $Q$ lies on $(JDAER)$ Proof: $\angle ADQ=\angle BDQ=\angle QPC=180^\circ- \angle CEQ=180^\circ- \angle AEQ$ $\implies DAEQ$ cyclic $\implies JDAERQ$ cyclic. $\blacksquare$ From here we can finish with 2 ways $1$ . $\angle DQP = \angle ABP \overset{AR \parallel BC}{=} 180^\circ -\angle BAR = 180^\circ - \angle DAR = \angle DQR$ . $\blacksquare$ $2$ . $\angle EQP = \angle ACB = \angle CAR = \angle EAR =\angle EQR$ $\blacksquare$

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