Olympiad problems

2025 Taiwan Mathematics Olympiad, 5

Two fixed circles $\omega$ and $\Omega$ intersect at two distinct points $A$ and $B$. Let $C$ and $D$ be two fixed points on the circle $\omega$. Let $P$ be a moving point on $\omega$. Line $PA$ meets circle $\Omega$ again at $Q$. Prove that the second intersection $R$ of two circumcircles of triangles $QPC$ and $QBD$ always lies on a fixed circle. [i]Proposed by buratinogigle[/i]

1986 Dutch Mathematical Olympiad, 4

Tags: geometry , fixed point , fixed , fixed circle

The lines $a$ and $b$ are parallel and the point $A$ lies on $a$. One chooses one circle $\gamma$ through A tangent to $b$ at $B$. $a$ intersects $\gamma$ for the second time at $T$. The tangent line at $T$ of $\gamma$ is called $t$. Prove that independently of the choice of $\gamma$, there is a fixed point $P$ such that $BT$ passes through $P$. Prove that independently of the choice of $\gamma$, there is a fixed circle $\delta$ such that $t$ is tangent to $\delta$.

2022 All-Russian Olympiad, 3

Tags: geometry , fixed circle , perpendicular lines

An acute-angled triangle $ABC$ is fixed on a plane with largest side $BC$. Let $PQ$ be an arbitrary diameter of its circumscribed circle, and the point $P$ lies on the smaller arc $AB$, and the point $Q$ is on the smaller arc $AC$. Points $X, Y, Z$ are feet of perpendiculars dropped from point $P$ to the line $AB$, from point $Q$ to the line $AC$ and from point $A$ to line $PQ$. Prove that the center of the circumscribed circle of triangle $XYZ$ lies on a fixed circle.

Ukrainian TYM Qualifying - geometry, 2017.3

Tags: geometry , fixed circle , fixed

The altitude $AH, BT$, and $CR$ are drawn in the non isosceles triangle $ABC$. On the side $BC$ mark the point $P$; points $X$ and $Y$ are projections of $P$ on $AB$ and $AC$. Two common external tangents to the circumscribed circles of triangles $XBH$ and $HCY$ intersect at point $Q$. The lines $RT$ and $BC$ intersect at point $K$. a). Prove that the point $Q$ lies on a fixed line independent of choice$ P$. b). Prove that $KQ = QH$.