2013 Japan Mathematical Olympiad Finals Problem 4

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2013 Japan Mathematical Olympiad Finals Problem 4

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Feb 11, 2013, 4:25 PM • 4 Y

Y by buratinogigle, Davi-8191, Adventure10, Mango247

Given an acute-angled triangle ABC, let $H$ be the orthocenter. A cirlcle passing through the points $B,\ C$ and a cirlcle with a diameter $AH$ intersect at two distinct points $X,\ Y$ . Let $D$ be the foot of the perpendicular drawn from $A$ to line $BC$ , and let $K$ be the foot of the perpendicular drawn from $D$ to line $XY$ . Show that $\angle{BKD}=\angle{CKD}$ .

This post has been edited 1 time. Last edited by Kunihiko_Chikaya, Feb 11, 2013, 4:54 PM

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#2 h Feb 11, 2013, 4:37 PM • 6 Y

Y by Kunihiko_Chikaya, buratinogigle, ts0_9, mathematiculperson, Adventure10, Mango247

Let $E \equiv BH \cap CA$ and $F \equiv CH \cap AB$ be the feet of the B- and C- altitudes. $XY, EF$ and $BC$ are pairwise radical axes of the circle $\odot(AEF)$ with diameter $\overline{AH},$ the circle $\odot(BCEF)$ with diameter $\overline{BC}$ and the circle $\odot(BCXY)$ $\Longrightarrow$ $XY,EF,BC$ concur at their radical center $R.$ From the complete quadrangle $BCEF,$ the cross ratio $(B,C,D,R)$ is harmonic. Since $DK \perp RK,$ then it follows that $KD,KR$ bisect $\angle BKD$ internally and externally.

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#3 h Feb 11, 2013, 4:54 PM • 3 Y

Y by Kunihiko_Chikaya, seyyed_khandan, Adventure10

Nice problem here is a generalization.

Let $ABC$ be a triangle and $P$ is point such that $AP\perp BC$ . $E,F$ are projection of $P$ on $CA,AB$ , respectively. $BE$ cuts $CF$ at $H$ . $AH$ cuts $BC$ at $D$ . $(K)$ is circle diameter $AP$ . $(L)$ is a circle passing through $B,C$ . $d$ is radical axis of $(K),(L)$ . $N$ is projection of $D$ on $d$ . Prove that $\angle BND=\angle CND$ .

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#4 h Feb 14, 2013, 6:29 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

buratinogigle wrote:

Nice problem here is a generalization.

Let $ABC$ be a triangle and $P$ is point such that $AP\perp BC$ . $E,F$ are projection of $P$ on $CA,AB$ , respectively. $BE$ cuts $CF$ at $H$ . $AH$ cuts $BC$ at $D$ . $(K)$ is circle diameter $AP$ . $(L)$ is a circle passing through $B,C$ . $d$ is radical axis of $(K),(L)$ . $N$ is projection of $D$ on $d$ . Prove that $\angle BND=\angle CND$ .

Let $AP \cap BC=\{U\}$ . $BUPF, CUPE, AFPE$ are cyclic, so $\widehat{CBA}=\widehat{UBF}=\widehat{APF}=\widehat{AEF}$ , so $FECB$ is cyclic, now, all you have to do is to apply the same argument as in the initial problem.

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#5 h Apr 4, 2020, 12:02 PM • 1 Y

Y by mkomisarova

Easy problem
Let $BE, CF$ be altitudes in triangle $ABC$ . It is obvious that $BCEF$ is cyclic. Note that $BC$ is radical axis of circumscribed circle $BCEF$ and circle passing through points $B, C$ , $XY$ is radical axis of circle passing through points $B, C$ and circle, whose diameter is $AH$ , $EF$ is radical axis of circle passing through points $B, C$ and circle, whose diameter is $AH$ . As a result $BC$ , $XY$ , $EF$ are concurrent at point $Z$ . Note that $(Z,D;B,C)=-1$ and $KD$ is perpendicular to $XY$ . Using well known lemma we obtain that $KD$ is bisector of $\angle BKC$ and we are done.

This post has been edited 4 times. Last edited by Kimchiks926, Apr 4, 2020, 8:08 PM
Reason: typo

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#6 h Jan 14, 2021, 11:20 PM

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Nice exercise, good for projective geometry

$\color{black}\rule{25cm}{1pt}$
Let $E$ and $F$ be the feet of the perpendiculars from $B$ and $C$ , respectively. Let $T$ be the intersection of $EF$ with $BC$ .
We apply the theorem of the radical center on $(AFHE),(BFEC)$ and $(BXYC)$ to get that $T,X$ and $Y$ are colinear.
We have that $-1=(T,D;B,C)$ and since $KD$ is perpendicular to $XY$ , we must have that $\angle BKD = \angle DKC$ .

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L567

#7 h Jan 15, 2021, 3:25 AM

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After a long time, I actually solve an olympiad geometry problem....

Let $D, E, F$ be the feet of perpendiculars and let $w$ be the circle passing through $B$ and $C$ . Using radical axes theorem on $(BCEF), (AH), (w)$ , we get that $XY, EF, BC$ are concurrent at a point $T$ . So, we get that $(T, D; B, C) = -1$ . Since $\angle TKD = 90^\circ$ , we know that $\angle BKD = \angle CKD$

This post has been edited 1 time. Last edited by L567, Jan 15, 2021, 3:25 AM

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#8 h Jul 17, 2023, 4:38 AM

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Simple outline
Let $P$ be the intersection of $XY$ and $BC$ . If we show $P, B, D, C$ are harmonic points, $\angle BKD = \angle CKD$ would be trivial.
Let's make a line that connects the perpendicular lines from $B, C$ to $AC, AB$ . Let's grab those two points $Q, R$ . If we hold 3 circles $(B, C, Q, R), (A, Q, H, R), (B, C, Y, X)$ and if we grab the 3 radical axes it intersects at one point $P$ . We know that for every triangle if we hold the intersection of $QR$ and $BC$ the point and $B, D, C$ form a harmonic point. therefore We are done.

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