Let $A$ and $B$ be two arbitrary points on the circumference of the circle such that $AB$ is not the diameter of the circle. The tangents to the circle drawn at points $A$ and $B$ meet at $T$. Next, choose the diameter $XY$ so that the segments $AX$ and $BY$ intersect. Let this be the intersection of $Q$. Prove that the points $A, B$, and $Q$ lie on a circle with center $T$.
Problem
Source: Finland 2017, p5
Tags: circle, Circumcenter, diameter, geometry
Com10atorics
16.01.2021 16:25
Too easy. Since $XB\perp QY$ and $YA\perp QX$, $\angle BTA=180-2\angle BAT=180-2\angle BXA=2\angle BQA$ and we are done.
baldeagle123
16.01.2021 22:13
I don't think $XB \perp QY$ and $AY \perp QZ$ proves that $A, B, Q$ are concyclic because $A, Y, T$ are not collinear since $\overline{AY}$ is secant and $\overline{AT}$ is tangent edit: nvm I got confused because of my notation I solved similarly by showing that $\angle BTA = 2\angle BQA$
parmenides51
18.09.2024 15:41
Circumcenter Lemma: If for an interior point T of triangle ABC, holds that <ATB=2 < AQB and AT=TB then T is circumcenter of ABC.
Let H be the intersection of AY and BX
Let <AXB=<AYB=w
Then <TAB=<ABT=w as tangent - chord angle
So <ATB=180^o -<TAB - <ABT=180^o -2w=2(90^o-w)
And AT=TB
<XAY=90^o =<AYB because XY is diameter
So AHBQ is cyclic
<AQB=180^o -<AHB= 180^o - (90^o +w)=90^o-w
So <ATB =2 <AQB and since AT=AB by the lemma
We get that T is circumcenter of ABQ