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Source: Bulgaria National Olympiad 2025, Day 2, Problem 4

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#1 h Apr 8, 2025, 1:53 PM • 1 Y

Y by cubres

Let $ABC$ be an acute triangle with $AB < AC$ , midpoint $M$ of side $BC$ , altitude $AD$ ( $D \in BC$ ), and orthocenter $H$ . A circle passes through points $B$ and $D$ , is tangent to line $AB$ , and intersects the circumcircle of triangle $ABC$ at a second point $Q$ . The circumcircle of triangle $QDH$ intersects line $BC$ at a second point $P$ . Prove that the lines $MH$ and $AP$ are perpendicular.

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Primeniyazidayi

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Primeniyazidayi

#2 h Apr 8, 2025, 2:31 PM • 1 Y

Y by cubres

Try to show that P,E,F are collinear where E and F are the other altitudes.Then applying Brocard on (BEFC) shows that M is the orthocenter of the triangle PAH(which also means that H is the orthocenter of PAM).
As bonus:
K,the second intersection of AP and (ABC),is the A-queue point and A,K,E,H,F are concyclic.
P,K,A and M,H,K,A' are pairwise collinear where A' is the antipode of A wrt (ABC).

This post has been edited 4 times. Last edited by Primeniyazidayi, Apr 9, 2025, 1:33 PM

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#3 h Apr 8, 2025, 2:55 PM • 1 Y

Y by cubres

Let $K$ = $(AH) \cap (ABC)$ , $X$ = $QD \cap (ABC)$ , $H' = AH \cap (ABC), A' = KM \cap (ABC)$ . A' is obviously the A antipode.

Claim) $K H D M$ is cyclic.
pf) $\angle KQD = \angle KQX = \angle BQD - \angle BQK = \angle B - \angle BQK$
$\angle DHM = \angle H'HA' = \angle KCA + \angle H'AA' =\angle C - \angle BQK + \angle B - C = \angle B - \angle BQK = \angle KQD$

Ok.

Therefore $KHDQP$ is cyclic, so $\angle HKP = \angle HDM = 90,$ also $\angle AKH = 90.$

So $\angle AKP = 180$ , so $A, K, P$ colinear. $MH \perp AK$ , so problem solved.

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292 posts

#4 h Apr 8, 2025, 4:59 PM • 1 Y

Y by cubres

Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$|a|=|b|=|c|=1$ $m=\frac{b+c}2$ $d=\frac{a^2+ab+ac-bc}{2a}$ $h=a+b+c$ Now we find the coordinate of $Q$ . Since $Q$ is on the unit circle we have $|q|=1$ , and since line $AB$ is tangent to the circumcircle of $\triangle BDQ$ , we have
$\frac{(a-b)(d-q)}{(b-d)(b-q)} \in \mathbb{R}$ $\frac{a^2+ab+ac-bc-2aq}{(a+c)(b-q)} = \frac{a^2q+2abc-abq-acq-bcq}{a(a+c)(b-q)}$ $a^3+a^2b+a^2c-abc-2a^2q = a^2q+2abc-abq-acq-bcq$ $q = \frac{a(a^2+ab+ac-3bc)}{3a^2-ab-ac-bc}$ Now we find the coordinate of $P$ . Since $P$ is on line $BC$ , we have $\overline{p} = \frac{b+c-p}{bc}$ . Since $D,H,P,Q$ are concyclic, we have
$\frac{(d-q)(h-p)}{(h-q)(d-p)} \in \mathbb{R}$ Now $\frac{(d-q)(a-b)}{(b-c)(q-b)} \in \mathbb{R}$ , and $\frac{d-p}{b-c} \in \mathbb{R}$ , so we write this as
$\frac{(q-b)(h-p)}{(a-b)(h-q)} \in \mathbb{R}$ Now
$q-b = \frac{a^3-2a^2b+a^2c+ab^2-2abc+b^2c}{3a^2-ab-ac-bc} = \frac{(a-b)^2(a+c)}{3a^2-ab-ac+bc}$ $h-q = \frac{2a^3+a^2b+a^2c-ab^2-ac^2-b^2c-bc^2}{3a^2-ab-ac-bc} = \frac{(a+b)(a+c)(2a-b-c)}{3a^2-ab-ac-bc}$ So we write this as
$\frac{(a-b)(h-p)}{(a+b)(2a-b-c)} \in \mathbb{R}$ And since $\frac{a-b}{a+b}$ is pure imaginary, we have
$\frac{a+b+c-p}{2a-b-c} \in i\mathbb{R}$ $\frac{a+b+c-p}{2a-b-c} = -\frac{\frac1a+\frac1b+\frac1c-\frac1b-\frac1c+\frac{p}{bc}}{\frac2a-\frac1b-\frac1c} = \frac{ap+bc}{ab+ac-2bc}$ $(a+b+c-p)(ab+ac-2bc) = (ap+bc)(2a-b-c)$ $p = \frac{(a+b+c)(ab+ac-2bc) - bc(2a-b-c)}{(ab+ac-2bc) + a(2a-b-c)} = \frac{a^2b+a^2c+ab^2-2abc+ac^2-b^2c-bc^2}{2(a^2-bc)}$ Then we have the vectors
$h-m = \frac{2a+b+c}2$ $a-p = \frac{2a^3-a^2b-a^2c-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} = \frac{(a-b)(a-c)(2a+b+c)}{2(a^2-bc)}$ and so
$\frac{a-p}{h-m} = \frac{(a-b)(a-c)}{a^2-bc}$ which is pure imaginary. $\blacksquare$

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#5 h Apr 8, 2025, 6:43 PM • 1 Y

Y by cubres

Rename $P$ to $X$ , then rename $Q$ to $P$ . Let $Q$ be the $A$ -queue point. Then $X$ is the $A$ -Ex point.

Now note that $P$ is the 11SLG4 point, because $PD\cap\Gamma$ has to be the isosceles trapezoid point. Since $Q$ lies on $(XHD)$ , let's show that $XQDP$ is cyclic. Indeed,
$\angle QXD=90^{\circ}-\angle QAD=\angle QPA'=\angle QPD,$ as desired. $\blacksquare$

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1388 posts

#6 h Apr 9, 2025, 9:37 PM • 2 Y

Y by AlexCenteno2007, cubres

It is well known that if ray $MH^\to$ intersects the circumcircle of $ABC$ at $K$ , then $\angle AKH = 90^{\circ}$ . We have $\angle KQD = \angle BQD - \angle BQK = \angle ABC - \angle BAK = 90^{\circ} - \angle BAD - \angle BAK = 90^{\circ} - \angle KAH = \angle AHK = 180^{\circ} - \angle KHD$ , so $KHDQ$ is cyclic. Hence $PKHDQ$ is cyclic, implying $\angle PKH = 180^{\circ} - \angle PDH = 90^{\circ}$ , so $A$ , $K$ and $P$ are collinear, implying $MH \perp AP$ .

This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 10:01 PM

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#7 h Apr 10, 2025, 3:54 AM • 1 Y

Y by cubres

Diagram
Solution

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1391 posts

#8 h Apr 10, 2025, 9:53 AM • 2 Y

Y by AlexCenteno2007, cubres

Trivial by angle-chasing, I am posting this just because the ratio lemma technique works nicely here.

Let $(AH) \cap (ABC)=T$ , $AT \cap BC=P$ , $R$ be such that $AR \parallel BC$ and redefine $Q=RD \cap (ABC)$ . Clearly $PTHD$ is cyclic with diameter $PH$ , $MH \perp AP$ and $(BDQ)$ touches $AB$ as $\angle BQD=\angle RCB=\angle ABC$ . We are only left to show that $P \in (DHQ)$ , so we will show that $PTDQ$ is cyclic. By Theorem 3.1 in mira74's ratio lemma handout., we have to show $f(T)f(Q)=f(P)f(D)$ (where $f(X)=\pm \frac{XB}{XC}$ ), and since $f(D)=f(Q)f(R)$ by Lemma 2.2, we need $f(T)=f(R)f(P)$ . However, $f(R)f(A)=1$ and $f(P)=f(T)f(A)$ by Lemma 2.2, which implies that $f(T)=f(R)f(P)$ .

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