Easy right-angled triangle problem

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Easy right-angled triangle problem

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Source: SMO open 2024 Q1

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gghx

#1 h Aug 3, 2024, 2:17 AM • 1 Y

Y by GA34-261

In triangle $ABC$ , $\angle B=90^\circ$ , $AB>BC$ , and $P$ is the point such that $BP=BC$ and $\angle APB=90^\circ$ , where $P$ and $C$ lie on the same side of $AB$ . Let $Q$ be the point on $AB$ such that $AP=AQ$ , and let $M$ be the midpoint of $QC$ . Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$ .

This post has been edited 3 times. Last edited by gghx, Aug 3, 2024, 3:02 AM

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#2 h Aug 3, 2024, 2:35 AM • 2 Y

Y by centslordm, GA34-261

I believe there is an error in the statement, as $\angle B=90$ implies $AC$ is the hypotenuse of triangle $ABC$ , so $AB>AC$ cannot happen.

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gghx

#3 h Aug 3, 2024, 2:38 AM • 2 Y

Y by GeoKing, GA34-261

My apologies, it should be $AB>BC$ .

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#5 h Aug 3, 2024, 3:01 AM • 1 Y

Y by GA34-261

I think the problem statement is not true. I checked on geogebra also, it is not coming out to be true. Please check

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gghx

#6 h Aug 3, 2024, 3:03 AM • 1 Y

Y by GA34-261

g0USinsane777 wrote:

I think the problem statement is not true. I checked on geogebra also, it is not coming out to be true. Please check

I'm so sorry, it should be $BP=BC$ and not $BP=PC$ . I have checked the problem character for character already and it should be correct now.

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Ianis

#7 h Aug 3, 2024, 3:34 AM • 2 Y

Y by GA34-261, Rounak_iitr

Let $\angle PBC=2\alpha$ . Then $\angle BCP=\angle CPB=90^\circ -\alpha$ and $\angle ABP=90^\circ -2\alpha$ , so $\angle PAQ=\angle PAB=2\alpha$ , and therefore $\angle AQP=\angle QPA=90^\circ -\alpha =\angle BCP$ . Hence $BCPQ$ is cyclic and $M$ is its circumcentre, so $\angle BMP=2\angle BCP=180^\circ -2\alpha =180^\circ -\angle PAB$ , and therefore $ABMP$ is cyclic, so $AM$ is the angle bisector of $\angle PAB$ . Hence if $D$ is the midpoint of $AB$ we have $\angle MDB=2\angle MAB=\angle PAB$ , which means that $MD\parallel AP$ , as desired.

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#8 h Aug 3, 2024, 3:35 AM • 2 Y

Y by centslordm, GA34-261

Solution with coordinate bash:

Let $B=(0,0)$ , $A=(0,a)$ , $C=(c,0)$ . $P=(x,y)$ must satisfy $x^2+y^2=c^2$ for the length condition and $\frac{y-a}{x}=-\frac{x}{y}$ for the right angle condition. Solving this system results in $P=(x,y)=\bigg(\frac{c\sqrt{a^2-c^2}}{a},\frac{c^2}{a}\bigg)$
We can find that the length of $AP$ is $\sqrt{a^2-c^2}$ because of some really nice cancellations, so $Q=(0,a-\sqrt{a^2-c^2})$ . Then $M=\bigg(\frac{c}{2},\frac{a-\sqrt{a^2-c^2}}{2}\bigg)$ . We can find that the slope of $AP$ is $-\frac{\sqrt{a^2-c^2}}{c}$ , so the line in question in the problem is given by the equation $y-\frac{a-\sqrt{a^2-c^2}}{2}=-\frac{\sqrt{a^2-c^2}}{c}\bigg(x-\frac{c}{2}\bigg)$
Substituting $x=0$ results in $y=\frac{a}{2}$ , so we are done.

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#9 h May 31, 2025, 1:02 AM • 1 Y

Y by GA34-261

Lemma: $CPQB$ and $MPOQ$ are cyclic, with $O$ being the midpoint of $AB$ .

Proof: Let $\angle PAB = \alpha$ , then $\angle CPB = 90^{\circ} - (90^{\circ} - \alpha) = \alpha$ . Notice that since $\triangle CBP$ and $\triangle QAP$ are isosceles, then $\angle BCP = \angle PQA$ , so $CPQB$ is cyclic with diameter $CQ$ because $\angle B = 90^{\circ}$ .

To show that $MPOQ$ is cylic, first note that because $M$ is the center of $(CPQB)$ , $MQ = MP \iff \angle MQP = \angle MPQ$ , but this means that $\alpha = \angle CQP = \angle MPQ = \angle MOQ$ , proving the lemma. $\square$

Becuase $MPOQ$ is cylic, then $\angle MOP = \alpha \implies \angle QOP = 2\alpha$ , but since $\angle BAP = \alpha$ , then $O$ must be the center of $(BAP)$ , and the result follows. $\square$

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