Kvant Math Problem 537

Let $O$ be the center of the circumcircle of the isosceles triangle $ABC$, and let $M$ be the midpoint of $PQ$.

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Problem

A circle is tangent internally to the circumcircle of an isosceles triangle $ABC$, as well as to the equal sides $AB$ and $AC$ of this triangle at points $P$ and $Q$, respectively. Prove that the midpoint of segment $PQ$ is the center of the incircle of triangle $ABC$.

International Mathematical Olympiad (XX, 1978)

Exploration

Let $O$ be the center of the circumcircle of the isosceles triangle $ABC$, and let $M$ be the midpoint of $PQ$. Since $AB=AC$, the axis of symmetry of the triangle is the altitude from $A$ to $BC$. Any circle tangent to both $AB$ and $AC$ has its center on this axis. The given circle is tangent to the circumcircle internally, hence its center also lies on the symmetry axis.

Denote by $S$ the center of the given circle. Because $AB$ and $AC$ are symmetric with respect to the axis, the tangency points $P$ and $Q$ are symmetric. Hence $M$ lies on the symmetry axis as well.

The goal is to show that $M$ is the incenter. Since the incenter of an isosceles triangle lies on the symmetry axis, it suffices to prove that the distance from $M$ to $AB$ equals the inradius.

Let the radius of the given circle be $r$. Then $SP\perp AB$ and $SQ\perp AC$, with $SP=SQ=r$. Because $M$ is the midpoint of the chord joining the symmetric points $P,Q$, the segment $SM$ is perpendicular to $PQ$ and lies on the symmetry axis. The quadrilateral formed by $S,P,M$ suggests computing the distance from $M$ to the side $AB$.

A coordinate model seems natural. Put the symmetry axis on the $y$-axis, let the equal sides make angle $\theta$ with it. Then $P$ and $Q$ occur at the same distance from $A$ along the sides. Expressing everything through the position of $S$ may reveal a simple relation between the radius of the given circle and the distance from $M$ to the sides.

The condition of internal tangency to the circumcircle should provide the missing equation. Let the circumcircle have radius $R$. If $OS=R-r$, and both centers lie on the symmetry axis, then the vertical coordinate of $S$ is determined. After eliminating parameters, one may obtain that the distance from $M$ to either side equals $R\cos\theta/(1+\cos\theta)$, which is exactly the inradius of an isosceles triangle with circumradius $R$ and vertex angle $2\theta$.

The step most likely to conceal an error is the derivation of the position of $S$ from the tangency condition and the subsequent computation of the distance from $M$ to a side. That calculation must be carried out explicitly.

Problem Understanding

We are given an isosceles triangle $ABC$ with $AB=AC$. A circle is tangent internally to the circumcircle of $ABC$ and tangent to the sides $AB$ and $AC$ at points $P$ and $Q$. We must prove that the midpoint of $PQ$ coincides with the center of the incircle of triangle $ABC$.

This is a Type B problem.

The core difficulty is relating the geometry of the auxiliary circle, which is tangent to the circumcircle and the two equal sides, to the incenter of the original triangle. The essential task is to compute the distance from the midpoint of $PQ$ to the sides of the triangle and show that it equals the inradius.

Proof Architecture

Lemma 1. The center $S$ of the given circle lies on the axis of symmetry of the isosceles triangle, and $P,Q$ are symmetric with respect to that axis.

Sketch. A circle tangent to both sides $AB$ and $AC$ has equal distances from them, hence its center lies on the angle bisector at $A$, which is the symmetry axis.

Lemma 2. In suitable coordinates, if the circumradius is $R$ and the equal sides form angle $\theta$ with the symmetry axis, then the center of the given circle is at height $R\cos\theta$ above the circumcenter.

Sketch. Use the internal tangency condition $OS=R-r$ together with the fact that the radius equals the distance from $S$ to either side.

Lemma 3. If $M$ is the midpoint of $PQ$, then the distance from $M$ to either side equals $\dfrac{R\cos\theta}{1+\cos\theta}$.

Sketch. Compute the coordinates of $P,Q,M$ from the coordinates of $S$.

Lemma 4. The inradius of triangle $ABC$ equals $\dfrac{R\cos\theta}{1+\cos\theta}$.

Sketch. Use the standard relation $r_{!in}=4R\sin\frac A2\sin\frac B2\sin\frac C2$ with $A=2\theta$ and $B=C=\frac{\pi}{2}-\theta$.

The most delicate part is Lemma 3, where the coordinates of $M$ and its distance to a side must be computed correctly.

Solution

Let $O$ be the center of the circumcircle of triangle $ABC$, and let $S$ be the center of the circle tangent to $AB$, $AC$, and internally tangent to the circumcircle. Let $r$ be the radius of this circle.

Choose coordinates so that

$$O=(0,0),$$

and the axis of symmetry of the isosceles triangle is the $y$-axis. Let the circumradius be $R$.

Write

$$A=(0,R).$$

Let the sides $AB$ and $AC$ make an angle $\theta$ with the $y$-axis. Their equations are

$$x=(R-y)\tan\theta, \qquad x=-(R-y)\tan\theta.$$

Since the circle with center $S$ is tangent to both sides, its center lies on the symmetry axis. Write

$$S=(0,s).$$

The distance from $S$ to either side equals the radius $r$. Using the equation

$$x-(R-y)\tan\theta=0,$$

the distance from $(0,s)$ to the side is

$$r=\frac{|(R-s)\tan\theta|}{\sqrt{1+\tan^2\theta}} =(R-s)\sin\theta.$$

Hence

$$r=(R-s)\sin\theta. \tag{1}$$

The circumcircle and the given circle are internally tangent. Since both centers lie on the $y$-axis,

$$OS=R-r,$$

that is,

$$s=R-r. \tag{2}$$

Substituting (2) into (1),

$$r=r\sin\theta+R(1-\sin\theta).$$

Therefore

$$r(1-\sin\theta)=R(1-\sin\theta),$$

and since $\sin\theta\neq1$,

$$r=\frac{R(1-\sin\theta)}{1} \Big/ (1-\sin\theta) =R-r,?$$

A cleaner elimination is obtained directly from (1) and (2):

$$r=(R-(R-r))\sin\theta=r\sin\theta.$$

This shows that $S$ lies above $O$, so we must take

$$s=r-R,$$

because the tangency point is on the extension of the line of centers. Then

$$|s|=R-r,$$

and $s<0$. Hence

$$s=r-R. \tag{3}$$

Combining (1) and (3),

$$r=(R-(r-R))\sin\theta=(2R-r)\sin\theta.$$

Thus

$$r(1+\sin\theta)=2R\sin\theta,$$

so

$$r=\frac{2R\sin\theta}{1+\sin\theta}. \tag{4}$$

From (3),

$$s=r-R =\frac{R(\sin\theta-1)}{1+\sin\theta}. \tag{5}$$

Let $P$ be the foot of the perpendicular from $S$ to $AB$. Since the line $AB$ has direction vector $(\sin\theta,-\cos\theta)$, the unit normal directed toward the interior is $(\cos\theta,\sin\theta)$. Hence

$$P=S+r(\cos\theta,\sin\theta).$$

Using symmetry,

$$Q=S+r(-\cos\theta,\sin\theta).$$

Therefore the midpoint $M$ of $PQ$ is

$$M=(0,s+r\sin\theta).$$

Substituting (4) and (5),

$$M_y = \frac{R(\sin\theta-1)}{1+\sin\theta} + \frac{2R\sin^2\theta}{1+\sin\theta} = \frac{R(2\sin^2\theta+\sin\theta-1)} {1+\sin\theta}.$$

Since

$$2\sin^2\theta+\sin\theta-1 =(1+\sin\theta)(2\sin\theta-1),$$

we obtain

$$M_y=R(2\sin\theta-1).$$

The distance from $M$ to the side $AB$ equals

$$\rho = \frac{(R-M_y)\tan\theta} {\sqrt{1+\tan^2\theta}} = (R-M_y)\sin\theta.$$

Using $M_y=R(2\sin\theta-1)$,

$$\rho = 2R(1-\sin\theta)\sin\theta. \tag{6}$$

Now compute the inradius of triangle $ABC$. Its angles are

$$A=2\theta, \qquad B=C=\frac{\pi}{2}-\theta.$$

The formula

$$r_{!in} = 4R \sin\frac A2 \sin\frac B2 \sin\frac C2$$

gives

$$r_{!in} = 4R \sin\theta \sin^2!\left(\frac{\pi/2-\theta}{2}\right).$$

Since

$$\sin^2!\left(\frac{\pi/2-\theta}{2}\right) = \frac{1-\sin\theta}{2},$$

we get

$$r_{!in} = 2R\sin\theta(1-\sin\theta).$$

Comparing with (6),

$$\rho=r_{!in}.$$

The point $M$ lies on the symmetry axis of the isosceles triangle, hence on the angle bisector at $A$. Its distances to the equal sides $AB$ and $AC$ are both $\rho=r_{!in}$. Consequently $M$ is the center of the incircle of triangle $ABC$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the position of the center $S$ relative to the circumcenter $O$. For internal tangency, the tangency point lies on the line of centers. The auxiliary circle is tangent to the sides near the vertex $A$, hence its center is on the opposite side of $O$ from $A$. This yields $s=r-R<0$. Choosing $s=R-r$ places the center between $O$ and $A$ and contradicts the geometry.

The second delicate point is the determination of $M$. Because $P$ and $Q$ are obtained from $S$ by adding the normal vectors $\pm(r\cos\theta,r\sin\theta)$, their midpoint is exactly

$$M=(0,s+r\sin\theta).$$

Any sign error in the normal vector changes the final distance computation.

The third delicate point is the inradius formula. Substituting

$$A=2\theta,\qquad B=C=\frac{\pi}{2}-\theta$$

into

$$r_{!in}=4R\sin\frac A2\sin\frac B2\sin\frac C2$$

gives

$$r_{!in} = 4R\sin\theta \cdot \frac{1-\sin\theta}{2} = 2R\sin\theta(1-\sin\theta),$$

which matches exactly the computed distance from $M$ to a side.

Alternative Approaches

A synthetic solution can be obtained by considering the homothety centered at the common tangency point of the two circles. Because the given circle is tangent to both equal sides, the contact points $P$ and $Q$ are symmetric. One then studies the polar line of the tangency chord $PQ$ with respect to the auxiliary circle and uses symmetry to show that its midpoint lies on the angle bisector and has the correct power with respect to the sides of the triangle.

Another approach uses inversion centered at $A$. The sides $AB$ and $AC$ remain fixed, while the two tangent circles transform into circles still tangent to the same pair of rays. The image configuration reduces the statement to a relation between two coaxal circles and a chord joining symmetric tangency points. The coordinate method above is preferable because every quantity can be computed directly and the equality with the inradius emerges from a short calculation.