Olympiad problems

the 6th XMO, 5

As shown in the figure, $\odot O$ is the circumcircle of $\vartriangle ABC$, $\odot J$ is inscribed in $\odot O$ and is tangent to $AB$, $AC$ at points $D$ and E respectively, line segment $FG$ and $\odot O$ are tangent to point $A$, and $AF =AG=AD$, the circumscribed circle of $\vartriangle AFB$ intersects $\odot J$ at point $S$. Prove that the circumscribed circle of $\vartriangle ASG$ is tangent to $\odot J$. [img]https://cdn.artofproblemsolving.com/attachments/a/a/62d44e071ea9903ebdd68b43943ba1d93b4138.png[/img]

2020 Princeton University Math Competition, B3

Tags: geometry , geometric inequality , tangent circle

Let $ABC$ be a triangle and let the points $D, E$ be on the rays $AB$, $AC$ such that $BCED$ is cyclic. Prove that the following two statements are equivalent: $\bullet$ There is a point $X$ on the circumcircle of $ABC$ such that $BDX$, $CEX$ are tangent to each other. $\bullet$ $AB \cdot AD \le 4R^2$, where $R$ is the radius of the circumcircle of $ABC$.

2025 Sharygin Geometry Olympiad, 15

Tags: trigonometry , geometry , tangent circle

A point $C$ lies on the bisector of an acute angle with vertex $S$. Let $P$, $Q$ be the projections of $C$ to the sidelines of the angle. The circle centered at $C$ with radius $PQ$ meets the sidelines at points $A$ and $B$ such that $SA\ne SB$. Prove that the circle with center $A$ touching $SB$ and the circle with center $B$ touching $SA$ are tangent. Proposed by: A.Zaslavsky

1953 Moscow Mathematical Olympiad, 237

Tags: concyclic , orthogonal , circles , geometry , tangent circle

Three circles are pair-wise tangent to each other. Prove that the circle passing through the three tangent points is perpendicular to each of the initial three circles.

Indonesia MO Shortlist - geometry, g4

Tags: geometry , tangent circle

Given that two circles $\sigma_1$ and $\sigma_2$ internally tangent at $N$ so that $\sigma_2$ is inside $\sigma_1$. The points $Q$ and $R$ lies at $\sigma_1$ and $\sigma_2$, respectively, such that $N,R,Q$ are collinear. A line through $Q$ intersects $\sigma_2$ at $S$ and intersects $\sigma_1$ at $O$. The line through $N$ and $S$ intersects $\sigma_1$ at $P$. Prove that $$\frac{PQ^3}{PN^2} = \frac{PS \cdot RS}{NS}.$$