Problem 25: IMO Shortlist 2006, G3

by henderson, Jun 21, 2016, 11:39 AM

$\bf\color{red}Problem \ 25$ Let $ABCDE$ be a convex pentagon such that
$\angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle CBA = \angle DCA = \angle EDA.$ The diagonals $BD$ and $CE$ meet at $P$ . Prove that the line $AP$ bisects the side $CD$ .
$\bf\color{red}My \ solution\$ $\color{blue}\bf{1.}$ $\quad$ $\triangle AED \sim \triangle ADC$ $\Longrightarrow$ $\frac{AD}{AE}=\frac{AC}{AD}$ $\Longrightarrow$ $\boxed{AD^2=AC\cdot AE}$ $.$

$\color{blue}\bf{2.}$ $\quad$ $\triangle ADC\sim \triangle ACB$ $\Longrightarrow$ $\frac{AC}{AD}=\frac{AB}{AC}$ $\Longrightarrow$ $\boxed{AC^2=AB\cdot AD}$ $.$

$\color{blue}\bf{3.}$ $\quad$ Combining $\color{blue}\bf{1}$ and $\color{blue}\bf{2,}$ we obtain $AD^2\cdot AC^2=AC\cdot AE \cdot AB \cdot AD$ $\Longrightarrow$ $AD\cdot AC=AE\cdot AB$ $\Longrightarrow$ $\frac{AB}{AD}=\frac{AC}{AE}.$ Since $\angle BAD=\angle CAE,$ we get $\boxed{\triangle ABD\sim \triangle ACE}$ $.$

$\color{blue}\bf{4.}$ $\quad$ $\triangle ABD\sim \triangle ACE$ $\Longrightarrow$ $\angle ABD=\angle ACE$ $\Longrightarrow$ $\angle ABP=\angle ACP,$ which means that $ABCP$ is a cyclic quadrilateral. Similarly, we can conclude that $AEDP$ is also a cyclic quadrilateral.

Let $AP$ meet $DC$ at $N.$ Then, from the cyclic quadrilateral $ABCP,$ $\angle CAN=\angle CAP=\angle CBP=\angle CBA-\angle PBA=\angle NCA-\angle PCA=\angle NCP.$ Since, $\angle CNA=\angle CNP,$ we have $\triangle CAN\sim \triangle PCN,$ which means that $\frac{CN}{AN}=\frac{PN}{CN},$ so $CN^2=AN\cdot PN.$
Similarly, using the cyclic quadrilateral $AEDP,$ we can get $\triangle DAN\sim \triangle PDN,$ and so $DN^2=AN\cdot PN.$ Therefore, $CN=DN,$ or in other words, $AP$ bisects $DC.$ $\square$

This post has been edited 4 times. Last edited by henderson, Sep 27, 2016, 1:37 PM

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Epic seeing how many solutions can be there. My solution was that proving ABCD and ACDE are similar and cyclic. Then note that CD is tangent to circle ABC and AED both of which obviously pass through P. Thus AP is the radical axis. Let the intersection of AP and CD be M, the centre of (ABC) be K and that of (AED) be L. Now, CM^2 = XM^2 - XC^2 = LM^2 - LD^2 = MD^2 and the result follows. The "XM^2 - XC^2 = LM^2 - LD^2 " equality came from the radical axis definition.

by Artistic_Potato, Apr 18, 2021, 10:46 AM

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I am reading @Artistic_Potato's comment, and I see that he proved that ABCD is cyclic. But since angle ADC = angle ACB, this would mean that angle CAB is 0, which is obviously false. I don't know how to reply to his comment, which explains me making a separate comment.

by Aarush12, Dec 27, 2023, 4:26 AM

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Problem 25: IMO Shortlist 2006, G3

by henderson, Jun 21, 2016, 11:39 AM

2 Comments