Junior Balkan Mathematical Olympiad 2024- P2

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Lukaluce

268 posts

Lukaluce

#1 h Jun 27, 2024, 11:05 AM • 3 Y

Y by Rounak_iitr, ItsBesi, farhad.fritl

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

This post has been edited 2 times. Last edited by Lukaluce, Jun 28, 2024, 12:38 PM

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Gryphos

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Gryphos

#2 h Jun 27, 2024, 11:35 AM • 1 Y

Y by WallyWalrus

Solution

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P2nisic

406 posts

P2nisic

#3 h Jun 27, 2024, 12:06 PM • 1 Y

Y by WallyWalrus

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

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Assassino9931

#4 h Jun 27, 2024, 12:11 PM • 3 Y

Y by Strudan_Borisov, Jalil_Huseynov, ehuseyinyigit

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.

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giannis2006

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giannis2006

#5 h Jun 27, 2024, 12:17 PM • 1 Y

Y by Assassino9931

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

64 posts

Z4ADies

#6 h Jun 27, 2024, 12:29 PM • 1 Y

Y by ehuseyinyigit

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 $RQ \cap BC =P$ .

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Marinchoo

407 posts

Marinchoo

#7 h Jun 27, 2024, 12:50 PM • 2 Y

Y by Strudan_Borisov, ehuseyinyigit

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.

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sami1618

909 posts

sami1618

#8 h Jun 27, 2024, 1:07 PM • 1 Y

Y by ehuseyinyigit

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

2 posts

Alex9100

#9 h Jun 27, 2024, 1:15 PM • 1 Y

Y by Alex_9100

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.

This post has been edited 2 times. Last edited by Alex9100, Jun 27, 2024, 1:22 PM
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cursed_tangent1434

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cursed_tangent1434

#10 h Jun 27, 2024, 1:26 PM • 1 Y

Y by Rounak_iitr

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.

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trigadd123

134 posts

trigadd123

#11 h Jun 27, 2024, 3:41 PM • 1 Y

Y by ehuseyinyigit

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.

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hukilau17

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hukilau17

#12 h Jun 27, 2024, 4:46 PM • 1 Y

Y by L13832

complex bash

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ItsBesi

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ItsBesi

#13 h Jun 27, 2024, 5:35 PM • 1 Y

Y by ehuseyinyigit

Storage

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g0USinsane777

48 posts

g0USinsane777

#14 h Jun 27, 2024, 9:04 PM

Y by

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.

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X.Luser

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X.Luser

#15 h Jul 6, 2024, 1:01 PM • 1 Y

Y by anirbanbz

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

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atdaotlohbh

186 posts

atdaotlohbh

#16 h Jul 9, 2024, 8:43 PM

Y by

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

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Assassino9931

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Assassino9931

#17 h Jul 11, 2024, 7:00 AM

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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).

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Alex009

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Alex009

#18 h Aug 4, 2024, 9:53 PM

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$JDAER$ 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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Primeniyazidayi

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Primeniyazidayi

#19 h Apr 20, 2025, 12:09 PM

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.........

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