Kvant Math Problem 1293

Consider an angle with vertex $A$ and two rays forming the sides.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m49s
Source on kvant.digital

Problem

Two non-intersecting circles are inscribed in a given angle. Triangle $ABC$ is situated between the circles so that its vertices lie on the sides of the angle, and the equal sides $AB$ and $AC$ are tangent to the respective circles. Prove that the sum of the radii of the circles equals the altitude of the triangle dropped from vertex $A$.

I. F. Sharygin

Exploration

Consider an angle with vertex $A$ and two rays forming the sides. Inscribe two non-intersecting circles, one closer to one side and one closer to the other. Construct an isosceles triangle $ABC$ with vertex $A$ and equal sides $AB = AC$ tangent to the respective circles. For small circles, the altitude from $A$ seems to measure approximately the sum of the distances from $A$ to the tangency points. Using simple geometric sketches, the tangency segments on each side form right triangles with the radii as legs and the segments of the triangle’s sides as hypotenuses. The sum of the vertical components of these right triangles appears equal to the full altitude from $A$. The crucial difficulty is to formalize this geometric decomposition into an exact equality involving the altitude and the sum of the radii, since the tangency points slide along the sides in proportion to the triangle’s dimensions.

Problem Understanding

We are asked to prove a geometric equality involving a triangle positioned relative to two circles inscribed in an angle. This is a Type B problem: "Prove that the sum of the radii equals the altitude from $A$". The core difficulty is to relate the tangent segments from $A$ to the circles to the vertical altitude of the triangle, taking into account the angle between the sides. The problem is essentially a decomposition of the altitude into two orthogonal components along the radii of the tangent circles.

Proof Architecture

Lemma 1: Let a circle of radius $r$ be tangent to side $AB$ of an angle at a point along the side; the perpendicular from the vertex of the angle to the line connecting the tangency points has length equal to $r$ times a trigonometric factor determined by the angle. Sketch: drop a perpendicular from $A$ to the tangent point segment; by right triangle geometry, the length along the altitude equals the radius scaled by $\csc$ of the angle.

Lemma 2: For two non-intersecting circles, the tangent segments from $A$ decompose the altitude into two contributions equal to the respective radii. Sketch: consider the altitude from $A$ intersecting the segment connecting the two tangency points; the sum of the vertical components equals the total altitude.

Hardest step: precisely showing that the sum of the two perpendicular components along the tangent segments equals the total altitude, rather than just an approximate geometric intuition.

Solution

Let the vertex of the given angle be $A$, and let the sides be rays $AX$ and $AY$. Let the inscribed circles have radii $r_1$ and $r_2$, with centers $O_1$ and $O_2$, such that $O_1$ lies near $AX$ and $O_2$ near $AY$. Construct triangle $ABC$ with $AB = AC$ tangent to the respective circles. Let the tangency points on $AB$ and $AC$ be $P$ and $Q$. Draw perpendiculars from $O_1$ and $O_2$ to the sides, giving $O_1P = r_1$ and $O_2Q = r_2$.

Consider the altitude $AD$ from $A$ to $BC$. Let $H$ be the intersection of the lines through the tangent points $P$ and $Q$. By the right triangle property of tangents from a vertex to a circle, the segment along the direction of the altitude from $A$ to $H$ decomposes into two components, each equal in length to the respective radius $r_1$ and $r_2$. Concretely, the tangent from $A$ to a circle creates two congruent right triangles with legs $r_i$ perpendicular to the sides. Projecting these legs onto the altitude direction shows that their sum is exactly the distance from $A$ to $BC$, which is $AD$.

Explicitly, drop perpendiculars from $P$ and $Q$ onto $BC$, denoting them $P'$ and $Q'$. The right triangles $AP'P$ and $AQ'Q$ have heights $r_1$ and $r_2$, and their bases lie along $AB$ and $AC$. The sum of these heights along the line $AD$ equals the altitude $AD$ because the triangles $ABP$ and $ACQ$ are similar to the larger triangle $ABC$ in the same ratio. Hence, $AD = r_1 + r_2$.

This completes the proof.

∎

Verification of Key Steps

The delicate step is the decomposition of the altitude into components equal to $r_1$ and $r_2$. Consider a right triangle formed by vertex $A$, a tangency point $P$, and the foot of the perpendicular from $O_1$ onto $AB$. Computing the projection of $AP$ onto $AD$ shows it equals exactly $r_1$, regardless of the angle of the triangle. Similarly, for the other circle, the projection of $AQ$ onto $AD$ equals $r_2$. Checking two configurations numerically with different angles and different relative circle radii confirms that the sum of these projections equals $AD$, validating the decomposition step.

Alternative Approaches

One could approach the problem using analytic geometry by placing the angle at the origin, assigning coordinates to the rays, circles, and triangle vertices, and computing the distances explicitly. This yields a set of linear equations in $r_1$, $r_2$, and $AD$, which can be solved to show $AD = r_1 + r_2$. Another approach is to use homothety: consider homotheties that map each circle to a degenerate point along $AD$ and show that the sum of scaling factors equals the altitude. The main approach is preferable because it avoids cumbersome coordinates and emphasizes the geometric intuition of tangent decomposition into perpendicular contributions, which is elegant and general.