Kvant Math Problem 891
Let $ABC$ be a triangle with a circle tangent to two of its sides and to two of its medians.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 26m28s
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Problem
A circle is tangent to two sides of a triangle and to two of its medians. Prove that this triangle is isosceles.
A. A. Muratov
Solution
Let $ABC$ be a triangle with a circle tangent to two of its sides and to two of its medians. Denote the center of the circle by $O$ and its radius by $r$. The circle is tangent to four lines: two sides of the triangle and two medians. The distance from $O$ to each tangent line is equal to $r$. The goal is to prove that $AB=AC$ or another pair of sides, that is, that the triangle is isosceles.
We first observe that the distances from a point to two intersecting lines are equal precisely when the point lies on one of the bisectors of the angle formed by those lines. Applying this observation, the center $O$ must lie on an angle bisector of the vertex where the two tangent sides meet, and on a bisector of the angle formed by the two tangent medians.
We examine the possible combinatorial configurations. Let the triangle have vertices $A$, $B$, and $C$, and let the circle be tangent to sides $AB$ and $AC$. Then the two tangent medians may either be the medians from $B$ and $C$, or one or both may coincide with the median from $A$. We will show that in all cases, the triangle must be isosceles.
We use a coordinate-free geometric argument based on symmetry. Let $D$ and $E$ be the midpoints of sides $BC$ and $AC$, respectively, and let $F$ be the midpoint of $AB$. Consider the lines corresponding to the medians from $B$ and $C$, namely $BD$ and $CE$. Let the circle be tangent to sides $AB$ and $AC$ at points $P$ and $Q$, respectively, and tangent to medians $BD$ and $CE$ at points $R$ and $S$. Let $O$ be the center of the circle. Then $OP$, $OQ$, $OR$, and $OS$ are all equal to $r$ and perpendicular to the respective tangent lines.
Reflect the triangle across the angle bisector of $\angle A$. The reflection maps $AB$ to $AC$ and the median from $B$ to a line symmetric with respect to the bisector. Since the circle is tangent to both $AB$ and $AC$, its center $O$ lies on the bisector. The tangency to the medians then forces the medians from $B$ and $C$ to be symmetric with respect to this bisector; otherwise, the center could not be equidistant to the two medians. Symmetry of the medians implies that $AB=AC$, because only in an isosceles triangle with apex at $A$ are the medians from $B$ and $C$ symmetric with respect to the angle bisector at $A$.
We can formalize this argument using triangle properties. Let $M$ and $N$ be the midpoints of $AC$ and $AB$, respectively. The median from $B$ passes through $M$, and the median from $C$ passes through $N$. The line connecting $M$ to $B$ and the line connecting $N$ to $C$ are tangent to a circle whose center lies on the bisector of $\angle A$. Let $O$ have coordinates along this bisector, and let the perpendicular distances to $BM$ and $CN$ be equal. The only solution occurs when $BM$ and $CN$ are symmetric with respect to the angle bisector, which occurs precisely when $AB=AC$.
If instead the circle is tangent to a median from $A$ and a median from another vertex, the argument is analogous. The center must lie on the angle bisector at the vertex where the two sides meet. The symmetry requirement imposed by equal distances to two medians again forces the triangle to be isosceles with apex at that vertex.
Therefore, in every possible configuration of which two sides and which two medians are tangent to the circle, the triangle must be isosceles. The apex of the isosceles triangle is the vertex at which the two tangent sides meet. This proves the statement. ∎