19th kmo #2

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19th kmo #2

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Source: KMO round 2, problem 2

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#1 h Feb 3, 2006, 3:07 AM • 1 Y

Y by Adventure10

For triangle $ABC$ , $P$ and $Q$ satisfy $\angle BPA + \angle AQC=90^{\circ}$ . It is provided that the vertices of the triangle $BAP$ and $ACQ$ are ordered counterclockwise(or clockwise). Let the intersection of the circumcircles of the two triangles be $N$ ( $A \neq N$ , however if $A$ is the only intersection $A=N$ ), and the midpoint of segment $BC$ be $M$ . Show that the length of $MN$ does not depend on $P$ and $Q$ .

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#2 h Feb 3, 2006, 5:12 PM • 2 Y

Y by Adventure10, Mango247

lightrhee wrote:

This is the second problem of the second round of 19th Korean Mathematics Olympiad.

2.
For triangle ABC, P and Q satisfy $\angle BPA + \angle AQC=90^{\circ}$ . It is provided that the vertices of the triangle BAP and ACQ are ordered counterclockwise(or clockwise). Let the intersection of the circumcircles of the two triangles be N( $A \neq N$ , however if A is the only intersection $A=N$ ), and the midpoint of segment BC be M. Show that the length of MN does not depend on P and Q.

Under the problem's situation we have $\angle (BP {,\;} PA) + \angle(AQ {,\;} QC) = 90^\circ \Rightarrow$
$\angle(BN {,\;} NC) = \angle(BN {,\;} AN) + \angle(AN {,\;} NC)= \angle(BP {,\;} PA) +\angle(AQ {,\;} QC) = 90^\circ \Rightarrow$
in any case $\angle BNC = 90^\circ \Rightarrow$ as $MN$ is the median of a triangle with one right angle ( could you say the name of such triangle , please ?

) so $MN = \frac{1}{2} BC =$ const.

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#3 h Jan 10, 2019, 6:25 AM • 2 Y

Y by Adventure10, Mango247

Thank you so much.

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