points equally distant from a point on the altitude

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points equally distant from a point on the altitude

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Source: MEMO 2016 T5

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#1 h Aug 25, 2016, 2:22 PM • 3 Y

Y by pisgood, Adventure10, Mango247

Let $ABC$ be an acute triangle for which $AB \neq AC$ , and let $O$ be its circumcenter. Line $AO$ meets the circumcircle of $ABC$ again in $D$ , and the line $BC$ in $E$ . The circumcircle of $CDE$ meets the line $CA$ again in $P$ . The lines $PE$ and $AB$ intersect in $Q$ . Line passing through $O$ parallel to the line $PE$ intersects the $A$ -altitude of $ABC$ in $F$ .

Prove that $FP = FQ$ .

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#2 h Aug 25, 2016, 2:32 PM • 3 Y

Y by buratinogigle, pisgood, Adventure10

Since $\angle AFO=\angle AEB=180^\circ-\angle DPA$ so $F$ is the circumcenter of $\triangle ADP$ . Other hand, $\angle EPD=\angle ECD=\angle EAB$ so $A,P,D,Q$ are concyclic implies $F$ is the circumcenter of $\triangle APQ$ i.e $FP=FQ$

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#3 h Aug 29, 2016, 2:55 AM • 2 Y

Y by pisgood, Adventure10

$APDQ$ cyclic + $PE \perp AD \implies$ $F$ is the center of $APD$

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#4 h Sep 5, 2016, 6:09 PM • 2 Y

Y by Adventure10, Mango247

Assume wlog that $AB<AC$ . We first observe that $\angle QAD=\angle BAD=\angle BCD=\angle ECD=\angle EPD=\angle QPD$ and so $A,P,D,Q$ are concyclic. Secondly, $\angle EPC=180^{\circ}-\angle EDC=180^{\circ}-\angle ABC$ yields that $B,P,C,Q$ lie on a circle. Let $F^*$ denote the circumcenter of the triangle $APQ$ . Clearly, $F^*$ lies on the $A$ altitude in $ABC$ and $F^*P=F^*Q$ . Thus, we only need to show that $OF^* \parallel PQ$ . Notice that $\angle AOF^*=90^{\circ}$ since $OF^*$ is the perpendicular bisector of $AD$ . Now, $\angle F^*AO=\angle (OF^*,BC)=\angle B-\angle C=\angle PEC$ which yields that $\angle (OF^*,BC)=\angle (PQ,BC)$ and so $OF^* \parallel PQ$ . The result follows.

This post has been edited 1 time. Last edited by anantmudgal09, Sep 5, 2016, 6:15 PM
Reason: Latex

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#5 h Sep 15, 2016, 4:23 AM • 2 Y

Y by Adventure10, Mango247

Note that by angle chase $APDQ$ and $BPCQ$ are cyclic. Now $FO \perp AD$ and bisects it $\Longrightarrow$ circumcenter of $APDQ \in OF$ and since $AF, AE$ are isogonal in $\triangle APQ$ circumcenter lies on $AF$ and thus the circumcenter actualy is $F$ and thus $FP = FQ$ holds.
Does T5 mean a hard problem? I don't think that this is hard or even a medium problem.

This post has been edited 1 time. Last edited by WizardMath, Sep 15, 2016, 5:43 AM

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#7 h Jan 12, 2022, 9:15 AM

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Ig this solution is bit different.
$\angle EPC=180°-\angle EDC=180°-\angle ABC=\angle QBC$
Hence, $QBPC$ is a cyclic quadrilateral.
By PoP,
$PE.EQ=BE.EC=AE.ED$
$\implies PE.EQ=AE.ED \implies \text{APDQ is cyclic quadrilateral}$
Also simlpe angle chasing gives $F$ is circumcenter of $\Delta APD$ Hence $F$ is the circumcenter of $(APDQ)$ . Result immediate.

This post has been edited 2 times. Last edited by sanyalarnab, Jan 12, 2022, 9:16 AM

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