19/5/2018: JOM 2013 G6

by navi_09220114, May 18, 2018, 5:24 PM

Promoting: This is Junior Olympiad of Malaysia 2013 Shortlist problem.

Unfortunately this tradition had stopped since 2017. I hope this will revive soon. Anyway...

10. Consider a triangle $ABC$ . Points $P,Q$ lie on $AB,AC$ respectively such that the four points $B, C, P, Q$ are
concyclic. The reflection of $P$ across $AC$ is $P'$ , and the reflection of $Q$ across $AB$ is $Q'$ . The circumcircles $\omega_1$ and $\omega_2$ of $(APP')$ and $(AQQ')$ respectively, intersect again at $S$ distinct from $A$ . Furthermore, $BS$ intersects $\omega_1$ again at $X$ , and $CS$ intersects $\omega_2$ again at $Y$ .

Prove that the four points $P, Q, X, Y$ lie on a circle.

Solution: By Radical Axis Theorem, it suffices to prove $AS, PX, QY$ are concurrent.

Step 1: $AS\perp BC$ .

Proof: Redefine $S=PQ'\cap QP'$ instead, then $\angle P'SP=180-\angle SPQ-\angle SQP=180-(180-2\angle C)-(180-2\angle B)=180-2\angle A=180-\angle P'AP=180-\angle Q'AQ$ . So $S$ lies on $(APP')$ and $(AQQ')$ as desired.

Now let $AU, AV$ be diameters of $(APP'), (AQQ')$ , then $U, V$ lies on $AB, AC$ , and $U, S, V$ colinear. Then $\angle AUV=\angle SQC=\angle AQP=\angle ABC\Rightarrow UV\parallel BC\Rightarrow AS\perp BC$ .

Now here comes the key idea: $AS, PV, QU$ are altitudes in $\triangle AUV$ so they are concurrent. So it suffices to show that $(A, V; S, X)=(A, U; S, Y)$ . Indeed if it is true, by projecting from $P, Q$ respectively, if $H$ is the orthocenter of $\triangle AUV$ , then $(A, H; S, PX\cap AS)=(A, H; S, QY\cap AS)$ implying $PX, QY, AS$ are concurrent.

Let $AS\cap BC=T$ , tangent to $S$ of $(APSUX)$ intersect $BC$ at $M$ , and tangent to $S$ of $(AQSVY)$ intersect $BC$ at $N$ , then $(SA, SV; SS, SX)=(T,\infty;B, M)=\frac{BT}{TM}$ . Likewise for the other side, it suffices to prove that $\frac{BT}{TM}=\frac{CT}{TN}$

Step 2: $TB\cdot TN=TC\cdot TM=TS\cdot TA$ .

Proof: This is because $\angle SMT=\angle VSM=\angle SAC\Rightarrow SAMC$ is cyclic. Likewise $SANB$ is cyclic, hence the equality. $\blacksquare$

This post has been edited 3 times. Last edited by navi_09220114, May 18, 2018, 5:28 PM

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You can just angle chase all the way

Observe $\angle PSQ=\angle ASP+\angle ASQ=180^{\circ}-2\angle A$ hence $S=\overline{PQ'} \cap \overline{P'Q}$ . Thus, $\angle AXB=\angle BPS=\angle APQ=\angle ACB$ hence $X \in \odot(ABC)$ . Likewise $Y \in \odot(ABC)$ . Now $\angle AXP=\angle AYQ=90^{\circ}-\angle A$ . Let $H_B, H_C \in \odot(ABC)$ with $\overline{BH_B} \perp \overline{AC}$ and $\overline{CH_C} \perp \overline{AB}$ . Then clearly, $P \in \overline{H_BX}$ and $Q \in \overline{H_CY}$ . Since $\overline{H_BH_C} \parallel \overline{PQ}$ ; by Reim's Theorem $PQXY$ is cyclic.

by anantmudgal09, May 22, 2018, 10:22 PM

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oh dear reflection of orthocenters!! That would have describe the lines PX and QY very well!!

Even the original solution provided by Justin is quite long and uses brokard's theorem, i never thought this can be angle chased right away even after i know X and Y are on (ABC)!

Thank you for your great idea today

This post has been edited 1 time. Last edited by navi_09220114, May 23, 2018, 4:17 AM

by navi_09220114, May 23, 2018, 4:16 AM

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Hi IMO god, CANBANKAN
by 799786, Aug 6, 2022, 5:28 PM
Hi geo god
by CANBANKAN, May 7, 2022, 4:35 AM
Updates please !
by Kayak, Aug 6, 2018, 6:47 AM
Wow Navi, so this is what you have been up to lately. Interesting!
by SeanGee, May 27, 2018, 3:12 PM
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by claserken, May 18, 2018, 6:43 PM
Contrib please
by WizardMath, Mar 16, 2018, 8:38 PM
Contrib?
by anantmudgal09, Mar 15, 2018, 7:35 AM
Nice blog! Second shout lol
by WizardMath, Mar 9, 2018, 10:41 AM
first shout >
by doitsudoitsu, Mar 5, 2018, 12:20 PM

9 shouts

lol

19/5/2018: JOM 2013 G6

by navi_09220114, May 18, 2018, 5:24 PM

2 Comments