equal segments concerning circumcircle

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Source: IGO Elementary 2016 2

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#1 h Jul 22, 2018, 11:49 AM • 1 Y

Y by Adventure10

Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$ . Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$ , such that $CX = CY = AB$ . (The points $A$ and $Y$ lie on different sides of the line $BC$ ). The line $XY$ intersects $\omega$ for the second time in point $P$ . Show that $PB = PC$ .

by Iman Maghsoudi

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RaduAndreiLecoiu

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RaduAndreiLecoiu

#2 h Jul 22, 2018, 12:48 PM • 2 Y

Y by Adventure10, Mango247

Redefine $P$ the midpoint of arc $BAC$ and $YP \cap AC = \{Z\}$
Hence arcs $BP$ and $CP$ are equal so $\angle CYZ = \angle CBP = \angle BUP$
Notice that $YB \parallel AC$ so $\angle BYZ = \angle YZC = \angle CYZ$ so $CY = CZ = CX$ so $X \equiv Z$ and we are done.

This post has been edited 1 time. Last edited by RaduAndreiLecoiu, Jul 22, 2018, 12:48 PM

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#3 h Jul 23, 2018, 6:36 AM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$ . Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$ , such that $CX = CY = AB$ . (The points $A$ and $Y$ lie on different sides of the line $BC$ ). The line $XY$ intersects $\omega$ for the second time in point $P$ . Show that $PB = PC$ .

by Iman Maghsoudi

$\angle PBC=\angle XYC=\angle YXC=(\smile AP+\smile CY)/2=(\smile AP+\smile AB)/2=\smile BP/2=\angle BCP.$ As I know, it is the problem from Iranian geometric olimpiad.

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zuss77

#4 h Jul 12, 2020, 1:28 PM

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Let $Q$ be another point on $\omega$ so that $CQ=AB$ . With $CX=CY=CQ$ by Fact 5 $X$ - incenter of $AYQ$ , so $AP=PQ$ . $P$ - midpoint of the arc $BAC$ .

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FriIzi

#6 h Mar 22, 2023, 9:20 AM

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$\angle XYC=\angle YXC=\alpha$
$\angle ABP=\angle ACP=\beta$
$\angle PBC=\alpha$
if $\angle BCA=\alpha-\beta$ we are done
note that $ABYC$ is isosceles trepazoid
$\angle XCY= 180-2\alpha$
$\implies$
$\angle YBC=\alpha-\beta$
and $\triangle BYC$ has two known angle and we can use them for find the third
so, $\angle BCY=180-3\alpha-\beta$
$\implies$
$\angle BCA=\alpha-\beta$
as we desired

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