ratio wanted, symmedian and isosceles triangle

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ratio wanted, symmedian and isosceles triangle

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Source: Ukraine Online MO 2020 10.6 11.5

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#1 h Sep 7, 2020, 7:46 PM

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An isosceles triangle $ABC$ is given with a base $AC, P$ is an arbitrary point on this base, $T$ is the projection of $P$ on $BC$ . In what ratio does the symmedian of $\vartriangle PBC$ derived from vertex $C$ , divides the segment $AT$ ?

(Anton Trygub)

This post has been edited 1 time. Last edited by parmenides51, Apr 3, 2021, 10:07 PM
Reason: proposer typo

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#2 h Mar 6, 2021, 7:36 PM

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Cutey.

Here, I consider $ABC$ being acute, obtuse case is similar.
Let $H$ be the reflection of $C$ over $T$ . Let $M$ the midpoint of $AH$ and let $N$ be the midpoint of $PB$ . Let $C$ -symmedian intersect $PB$ at $K$ . Let $Q$ be the intersection of $C$ -symmedian of $\triangle PCB$ and $AT$ .

We have $\triangle PCH\sim\triangle BAC$ , since $\angle PCH=\angle BAC$ and $PC=PH$ , $AB=AC$ .

Therefore, $\frac{PC}{AB}=\frac{CH}{AC}\Longleftrightarrow \frac{PC}{CH}=\frac{BC}{AC}$ and since $\angle ACH=\angle BCP$ , we have $\triangle ACH\sim \triangle BCP$ . Thus, $C$ -symmedian of $\triangle PCB$ bisects $AH$ , since $\angle ACM=\angle BCN=\angle ACK$ .

Also, since $T$ is the midpoint of $CH$ , we get that $Q$ is the centroid of $\triangle PCB$ . Therefore, we have
$\boxed{\dfrac{AQ}{QT}=\dfrac{2}{1}}$ .

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