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Source: Mexican Math Olympiad 2015 Problem 1

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#1 h Nov 23, 2015, 9:13 PM • 4 Y

Y by mathematicsy, jhu08, Adventure10, Mango247

Let $ABC$ be an acuted-angle triangle and let $H$ be it's orthocenter. Let $PQ$ be a segment through $H$ such that $P$ lies on $AB$ and $Q$ lies on $AC$ and such that $\angle PHB= \angle CHQ$ . Finally, in the circumcircle of $\triangle ABC$ , consider $M$ such that $M$ is the mid point of the arc $BC$ that doesn't contain $A$ . Prove that $MP=MQ$

Proposed by Eduardo Velasco/Marco Figueroa

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#2 h Nov 23, 2015, 9:30 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Isn't it reaaaally easy or am I blind this evening?

$\angle {APQ}=\angle {ABH}+\angle {PHB}=\angle {ACH}+\angle {CHQ}=\angle AQP$
So triangle $APQ$ is isosceles.
Looking at triangles $PAM$ and $QAM$ , we have $AP=AQ$ , $AM$ common and $\angle {PAM}=\angle {QAM}$ , so they are congruent .
$\implies MP=MQ$ .

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suli

#3 h Nov 29, 2015, 2:37 AM • 3 Y

Y by jhu08, Adventure10, Mango247

You probably are blind for not calling $APMQ$ a kite as $A, M$ lie on perpendicular bisector of $PQ$ (just kidding

)

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eshan

#4 h Nov 29, 2015, 3:14 AM • 5 Y

Y by sarthak786, aops1221math, jhu08, Adventure10, Mango247

Okay

So here's my solution.

My Solution.

Let $\angle PHB=\angle CHQ=\theta.$

Then $\angle HPA=\angle QHA=90-A-\theta,$ as $\angle HBA=\angle HCA=90-A.$

As $P,H,Q$ are collinear, we can say that $\triangle APQ$ is isosceles.

Now, by fact 5, we can say that $AM$ bisects $\angle A.$

Now, let us denote $AM\cap PQ=X.$

As $\triangle APQ$ is isosceles, $AX\perp PQ,$ and $PM=MQ.$

Which means $AM$ is the perpendicular bisector of $PQ.$

Now, we know that the set of all points equidistant from $2$ given points is the line joining the $2$ vertices.

As $M$ lies on the perpendicular bisector of $PQ,$ $M$ must be equidistant from $P,Q.$

Which means $MP=MQ.~\Square$

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Anar24

#6 h Sep 3, 2016, 4:56 PM • 2 Y

Y by jhu08, Adventure10

can someone post a solution with complex numbers? it is interesting if there is a solution with complex baash

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#7 h Dec 28, 2018, 9:40 PM • 3 Y

Y by jhu08, Adventure10, Mango247

How is the fact that M is the midpoint of arc BC l used?

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#8 h Apr 17, 2021, 8:13 PM • 1 Y

Y by jhu08

First, let $E$ be the foot of the perpendicular from $B$ to $AC$ and $F$ be the foot of the perpendicular from $C$ to $AB$ . Let $N$ be the intersection of $PQ$ and $AM.$ Note that $\angle FHB=\angle EHC$ , so $PQ$ is just the angle bisector of $\angle FHB$ and $\angle EHC$ . So, $\triangle FHP \sim \triangle EHQ \implies \angle FPH=\angle EQH$ . Combined with $\angle AQN=\angle PAN$ and $AN=AN$ , it follows by AAS that $\triangle PAN \cong \triangle QAN$ . In particular, $AM \perp PQ$ and $PN=QN$ . Combined with $NM=NM$ , we get that $\triangle PNM \cong \triangle QNM$ , so $MP=MQ$ as desired

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L567

#9 h Apr 18, 2021, 5:27 AM • 1 Y

Y by jhu08

Simple one...

$\angle APQ = \angle PHB + \angle PBH = \angle QHC + \angle HCQ = \angle AQP$ and so $AP = AQ$ . So $APQ$ is isosceles and so its angle bisector is its perpendicular bisector. But since $AM$ is the angle bisector, $M$ must lie on the perpendicular bisector of $PQ$ and so $MP = MQ$

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#10 h Aug 4, 2021, 4:01 PM • 1 Y

Y by jhu08

Its easy becuase its P1

Claim 1: The triangle $\triangle APQ$ is isosceles.
Proof: Its known that $\angle PBH=\angle QCH$ , if we add the fact that $\angle PHB=\angle QHC$ then:
$\angle PBH+\angle PHB=\angle QCH+\angle QHC \implies \angle APQ=\angle AQP \implies \triangle APQ \; \text{isosceles}$
By Claim 1 we know that $AM$ is perpendicular bisector of $PQ$ thus $APQM$ is a kite and then $MP=MQ$ and we are done :blush:

:blush:

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#11 h Dec 29, 2021, 8:18 PM

Y by

The line $CH$ meets $AB$ at $D$ and $BH$ meets $AC$ at $E$ . Since $\angle BDC = \angle BEC = 90^{\circ}$ we get $BDEC$ is cyclic quadrilateral. Hence, $\angle PBH = \angle DBE = \angle DCE = \angle HCQ$ so that

$\angle APQ = \angle PBH + \angle PHB = \angle HCQ + \angle CHQ = \angle PQA$
Therefore, $AP = AQ$ . Finally, since $M$ lie on the internal bisector of $\angle PAQ$ , it turns out that $AM$ is perpendicular bisector of $PQ$ , thus $MP = MQ \blacksquare$

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