VMO 2015 problem 4

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VMO 2015 problem 4

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Source: VMO 2015

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#1 h Jul 31, 2016, 4:45 AM • 2 Y

Y by Adventure10, Mango247

Given a circumcircle $(O)$ and two fixed points $B,C$ on $(O)$ . $BC$ is not the diameter of $(O)$ . A point $A$ varies on $(O)$ such that $ABC$ is an acute triangle. $E,F$ is the foot of the altitude from $B,C$ respectively of $ABC$ . $(I)$ is a variable circumcircle going through $E$ and $F$ with center $I$ .

a) Assume that $(I)$ touches $BC$ at $D$ . Probe that $\frac{DB}{DC}=\sqrt{\frac{\cot B}{\cot C}}$ .

b) Assume $(I)$ intersects $BC$ at $M$ and $N$ . Let $H$ be the orthocenter and $P,Q$ be the intersections of $(I)$ and $(HBC)$ . The circumcircle $(K)$ going through $P,Q$ and touches $(O)$ at $T$ ( $T$ is on the same side with $A$ wrt $PQ$ ). Prove that the interior angle bisector of $\angle{MTN}$ passes through a fixed point.

This post has been edited 4 times. Last edited by quangminhltv99, Aug 3, 2016, 3:08 PM

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#2 h Jan 5, 2017, 7:02 AM • 1 Y

Y by Adventure10

The key to this problem is the intersection of $BC$ and $EF$ , which is the radical center of several circles involved in the problem.

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#3 h Apr 26, 2019, 2:58 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
a) We have: $\dfrac{BD^2}{DC^2}$ = $\dfrac{\overline{BF} . \overline{BA}}{\overline{CE} . \overline{CA}}$ = $\dfrac{BF . AB}{CE . AC}$ = $\dfrac{BC \cos \widehat{ABC} . \sin \widehat{ACB}}{BC \cos \widehat{ACB} . \sin \widehat{ABC}}$ = $\dfrac{\cot \widehat{ABC}}{\cot \widehat{ACB}}$
So: $\dfrac{BD}{DC}$ = $\sqrt{\dfrac{\cot \widehat{ABC}}{\cot \widehat{ACB}}}$
b) Let $d$ be tangent at $T$ of ( $O$ )
We have: $d$ , $PQ$ , $BC$ are radical axises of (( $O$ ); ( $TPQ$ )), (( $TPQ$ ); ( $BHC$ )), (( $BHC$ ); ( $O$ ))
Then: $d$ , $PQ$ , $BC$ concurrent
Let $S$ be the concurrent point of $d$ , $PQ$ , $BC$
So: $ST^2$ = $\overline{SB}$ . $\overline{SC}$ = $\overline{SP}$ . $\overline{SQ}$ = $\overline{SM}$ . $\overline{SN}$ or $d$ is common tangent of ( $O$ ) and ( $MTN$ )
Hence: $TM$ , $TN$ are isogonal conjugate in $\widehat{BTC}$
Therefore: the internal bisector of $\widehat{MTN}$ passes through midpoint of $\stackrel\frown{BC}$ which not containing $A$

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#4 h Apr 12, 2020, 3:13 PM • 1 Y

Y by SenatorPauline

Solution

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