Collinearity in isosceles trapezoid

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Collinearity in isosceles trapezoid

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Source: Nordic Mathematical Contest 2017 Problem 3

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#1 h Apr 3, 2017, 5:50 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ and $N$ be the midpoints of the sides $AC$ and $AB$ , respectively, of an acute triangle $ABC$ , $AB \neq AC$ . Let $\omega_B$ be the circle centered at $M$ passing through $B$ , and let $\omega_C$ be the circle centered at $N$ passing through $C$ . Let the point $D$ be such that $ABCD$ is an isosceles trapezoid with $AD$ parallel to $BC$ . Assume that $\omega_B$ and $\omega_C$ intersect in two distinct points $P$ and $Q$ . Show that $D$ lies on the line $PQ$ .

This post has been edited 4 times. Last edited by math163, Nov 10, 2018, 8:30 AM

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#3 h Apr 3, 2017, 7:53 PM • 3 Y

Y by zuss77, Adventure10, Mango247

Let $B'$ be a point such that $BB'\parallel AC$ and $B'\in \odot (ABC)$ similarly defined $C'$ , so from $ACB'B$ is an isoseles trapezoid we get $MB$ $=$ $MB'$ $\Longrightarrow$ $BB'$ is the radical axis of $\odot (ABC)$ and $\omega_B$ similarly $CC'$ is the radical axis of $\odot (ABC)$ and $\omega_C$ $\Longrightarrow$ $E=BB'\cap CC'$ belongs in the radical axis of $\omega_B$ and $\omega_C$ .

On the other hand, so from $ABCD$ is an isosceles trapezoid and $ABEC$ is a parallelogram we get $ED\perp BC$ $\Longrightarrow$ $ED\perp MN$ $\Longrightarrow$ $ED$ is perpendicular to the union of centers of $\omega_B$ and $\omega_C$ , but $E$ belongs in the radical axis of $\omega_B$ and $\omega_C$ , hence $ED$ is the radical axis of $\omega_B$ and $\omega_C$ $\Longrightarrow$ $P$ , $Q$ and $D$ are collinear.

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#4 h Apr 4, 2017, 1:57 AM • 1 Y

Y by Adventure10

could you post all the problem? Thanks

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#5 h Apr 4, 2017, 9:44 AM • 2 Y

Y by Adventure10, Mango247

(redacted)

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GGPiku

#6 h Apr 4, 2017, 1:15 PM • 2 Y

Y by Adventure10, Mango247

Easy to see that $PQ\perp BC$ , since $PQ\perp MN$ by radical axis and $MN\parallel BC$ . Denote X the intersection of PQ and BC. Since it's on the radical axis we have $XM^2-MA^2=XN^2-NA^2$ , so $XM^2-XN^2=AM^2-AN^2$ . Now when Y is a variable point on BC, $YM^2-YN^2$ can have a fixed value k in 2 positions on BC, symmetric by the point Z where $ZM=ZN$ . WLOG, if we choose $AB<AC$ then X has an unique value, so in order to prove that $P,Q,D$ are collinear, it's enough to prove that $D', X$ coincide, where $D'$ is the projection of D on BC. So we just have to prove that $D'M^2-D'N^2=AM^2-AN^2$ which is just easy computation after we project $M,N,A,D$ onto $BC$ . Correct me if I'm wrong, and if anyone wants me to add the rest of the calculation I will, although it's pretty easy.

This post has been edited 2 times. Last edited by GGPiku, Apr 4, 2017, 1:16 PM

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