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Source: Indonesian Mathematical Olympiad 2014 Day 2 Problem 6

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#1 h Sep 4, 2014, 3:07 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle. Suppose $D$ is on $BC$ such that $AD$ bisects $\angle BAC$ . Suppose $M$ is on $AB$ such that $\angle MDA = \angle ABC$ , and $N$ is on $AC$ such that $\angle NDA = \angle ACB$ . If $AD$ and $MN$ intersect on $P$ , prove that $AD^3 = AB \cdot AC \cdot AP$ .

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#2 h Sep 4, 2014, 5:22 PM • 2 Y

Y by Adventure10, Mango247

Since $\angle ADN+\angle ADM=\angle ACB+\angle ABC=180^{\circ}-\angle MAN,$ then $AMDN$ is cyclic, being $D$ the midpoint of its circumcircle arc $MN.$ Hence $\angle ADN=\angle AMN$ and $\angle DAN=\angle DAM$ gives $\triangle ADN \sim \triangle AMP$ $\Longrightarrow$ $AM \cdot AN=AP \cdot AD \ (1).$ But $\angle MDA=\angle ABC$ $\Longrightarrow$ $AD^2=AM \cdot AB$ and similarly $AD^2=AN \cdot AC,$ hence $AD^4=AB \cdot AC \cdot AM \cdot AN \ (2).$ Combining $(1)$ and $(2)$ yields $AD^3=AB \cdot AC \cdot AP.$

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#3 h Sep 23, 2020, 12:15 AM

Y by

Lemma 1. $BMNC$ is cyclic.
Proof. Since $AD$ bisects $\angle BAC$ , $\triangle{AMD} \sim \triangle{ADB}$ and $\triangle{ADN} \sim \triangle{ACD}$ . Then
$AD^2=AM \cdot AB$ $(1)$
$AD^2=AN \cdot AC$
Then, $AM \cdot AB = AN \cdot AC$ . So, we proved.

In other words, by Lemma 1,
$\angle{ACD}=\angle{AMP}$ $\Rightarrow$ $\triangle AMP \sim \triangle ACD$ $\Rightarrow$ $AD \cdot AM = AP \cdot AC$ $(2)$
Then, multiplying $(1)$ and $(2)$ :
$AD^3=AP \cdot AB \cdot AC$ . So we are done! :showoff:

:showoff:

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#4 h Jul 22, 2021, 10:37 AM

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this was also 2000 Brazil Cono Sur TST 2.3

This post has been edited 1 time. Last edited by parmenides51, Jul 22, 2021, 10:37 AM

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