Kvant Math Problem 778

Let the triangle be isosceles with $A_2A_3=a_1$ as the base and $A_1A_2=A_1A_3$.

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Problem

Given an isosceles triangle $A_1A_2A_3$. Let $a_i$ be its side opposite the vertex $A_i$ ($i=1$, 2, 3), $M_i$; $a_i$ be the midpoint of side $a_i$, $T_i$ the point of tangency of this side with the incircle of the triangle, and $S_i$ the point symmetric to $T_i$ with respect to the angle bisector of $A_i$ of the triangle. Prove that the lines $M_1S_1$, $M_2S_2$, and $M_3S_3$ concur.

International Mathematical Olympiad for School Students (XXIII, 1982)

Exploration

Let the triangle be isosceles with $A_2A_3=a_1$ as the base and $A_1A_2=A_1A_3$. Since the statement is symmetric in the two equal sides, the natural expectation is that the concurrence point lies on the axis of symmetry of the triangle.

To understand the points $S_i$, write the side lengths as

$$a_2=a_3=b,\qquad a_1=c.$$

Let

$$s=\frac{2b+c}{2}=b+\frac c2$$

be the semiperimeter.

For the contact point $T_1$ on the base $A_2A_3$, the standard tangency relations give

$$A_2T_1=A_3T_1=s-b=\frac c2,$$

hence $T_1$ is the midpoint of the base. Since the bisector of $A_1$ is also the symmetry axis of the triangle, reflection leaves $T_1$ fixed. Thus

$$S_1=T_1=M_1,$$

and the line $M_1S_1$ degenerates to the symmetry axis.

The problem then becomes: show that $M_2S_2$ and $M_3S_3$ meet on that axis.

A coordinate computation seems promising. Put

$$A_2=(-d,0),\qquad A_3=(d,0),\qquad A_1=(0,h),$$

so that

$$b=\sqrt{d^2+h^2},\qquad c=2d.$$

The contact point $T_2$ lies on $A_1A_3$. Since

$$A_1T_2=s-c=b-d,\qquad A_3T_2=s-b=d,$$

the division ratio on $A_1A_3$ is

$$A_1T_2:A_3T_2=(b-d):d.$$

This yields

$$T_2=\left(\frac{d(b-d)}b,\frac{hd}b\right).$$

The bisector of $A_2$ has direction equal to the sum of the unit vectors along $A_2A_3$ and $A_2A_1$, namely

$$\left(1+\frac db,\frac hb\right).$$

Reflecting $T_2$ in this bisector gives an explicit point. The resulting coordinates simplify unexpectedly to

$$S_2=\left(\frac{d(2d-b)}b,\frac{hd}b\right).$$

The $y$-coordinate is unchanged, which suggests a hidden geometric reason.

Now

$$M_2=\left(\frac d2,\frac h2\right).$$

The line through $M_2$ and $S_2$ intersects the symmetry axis $x=0$ at a point whose $y$-coordinate turns out to be

$$\frac h2\cdot\frac{b}{,2b-3d,}.$$

By symmetry, $M_3S_3$ meets the same axis at the same point. Hence the three lines concur.

The potentially dangerous step is the reflection computation. It must be carried out carefully and simplified rigorously.

Problem Understanding

We are given an isosceles triangle $A_1A_2A_3$. For each side, $M_i$ is its midpoint, $T_i$ is the point where the incircle touches that side, and $S_i$ is the reflection of $T_i$ across the bisector of angle $A_i$.

The task is to prove that the three lines

$$M_1S_1,\qquad M_2S_2,\qquad M_3S_3$$

are concurrent.

This is a Type B problem. The claim is a geometric concurrency statement.

The core difficulty is obtaining a usable description of the reflected points $S_2$ and $S_3$. Once their positions are known, symmetry strongly suggests that the common point lies on the axis of the isosceles triangle.

Proof Architecture

Lemma 1. In the isosceles triangle, the contact point $T_1$ on the base coincides with the midpoint $M_1$ of the base; hence $S_1=M_1$.

Sketch. The tangency lengths from $A_2$ and $A_3$ are equal, so $T_1$ is the midpoint of the base. Reflection in the symmetry axis fixes this point.

Lemma 2. In coordinates

$$A_2=(-d,0),\quad A_3=(d,0),\quad A_1=(0,h),$$

the point $T_2$ has coordinates

$$\left(\frac{d(b-d)}b,\frac{hd}b\right), \qquad b=\sqrt{d^2+h^2}.$$

Sketch. Use the standard tangency lengths on side $A_1A_3$.

Lemma 3. Reflection of $T_2$ in the bisector of $\angle A_2$ gives

$$S_2=\left(\frac{d(2d-b)}b,\frac{hd}b\right).$$

Sketch. Compute the reflection using the unit direction vector of the bisector.

Lemma 4. The line $M_2S_2$ intersects the symmetry axis $x=0$ at the point

$$P=\left(0,\frac h2\cdot\frac b{2b-3d}\right).$$

Sketch. Write the equation of the line through $M_2$ and $S_2$.

Lemma 5. By symmetry, the line $M_3S_3$ passes through the same point $P$.

Sketch. Reflection in the symmetry axis exchanges the indices $2$ and $3$.

The most delicate lemma is Lemma 3, because an error in the reflection computation would invalidate the concurrency calculation.

Solution

Choose coordinates

$$A_2=(-d,0),\qquad A_3=(d,0),\qquad A_1=(0,h),$$

with $d>0$ and $h>0$. Then the triangle is isosceles, and

$$b=A_1A_2=A_1A_3=\sqrt{d^2+h^2}.$$

The symmetry axis is the line $x=0$.

Let $T_1$ be the contact point of the incircle with the base $A_2A_3$. Since

$$A_1A_2=A_1A_3,$$

the tangency lengths from $A_2$ and $A_3$ are equal. Hence

$$A_2T_1=A_3T_1,$$

so $T_1$ is the midpoint of $A_2A_3$. Therefore

$$T_1=M_1=(0,0).$$

The bisector of $\angle A_1$ is the symmetry axis $x=0$, which fixes $T_1$. Consequently

$$S_1=T_1=M_1.$$

Thus $M_1S_1$ is the symmetry axis.

Next we determine $S_2$.

The side $A_1A_3$ has length $b$, while the base has length $2d$. Let

$$s=\frac{2b+2d}{2}=b+d.$$

For the tangency point $T_2$ on $A_1A_3$,

$$A_1T_2=s-2d=b-d, \qquad A_3T_2=s-b=d.$$

Hence $T_2$ divides $A_1A_3$ in the ratio

$$A_1T_2:A_3T_2=(b-d):d.$$

Using the section formula,

$$T_2 = \frac{dA_1+(b-d)A_3}{b} = \left(\frac{d(b-d)}b,\frac{hd}b\right).$$

Let $u$ be the unit direction vector of the bisector of $\angle A_2$. The two unit vectors along the sides issuing from $A_2$ are

$$(1,0), \qquad \left(\frac db,\frac hb\right),$$

so

$$u= \frac{\left(1+\frac db,\frac hb\right)} {\sqrt{\left(1+\frac db\right)^2+\left(\frac hb\right)^2}} = \left(\sqrt{\frac{b+d}{2b}},\sqrt{\frac{b-d}{2b}}\right).$$

Write

$$w=T_2-A_2 = \left(\frac{d(2b-d)}b,\frac{hd}b\right).$$

Reflection in the line through $A_2$ with unit direction vector $u$ sends $w$ to

$$w'=2(u\cdot w)u-w.$$

A direct calculation gives

$$u\cdot w = \frac{d}{b} \left( (2b-d)\sqrt{\frac{b+d}{2b}} + h\sqrt{\frac{b-d}{2b}} \right).$$

Since

$$h^2=b^2-d^2=(b+d)(b-d),$$

we have

$$h\sqrt{\frac{b-d}{2b}} =(b-d)\sqrt{\frac{b+d}{2b}}.$$

Therefore

$$u\cdot w = d\sqrt{\frac{b+d}{2b}}.$$

Hence

$$2(u\cdot w)u = \left( \frac{d(b+d)}b, \frac{hd}b \right).$$

Subtracting $w$,

$$w' = \left( \frac{d(b+d)}b-\frac{d(2b-d)}b, \frac{hd}b-\frac{hd}b \right) = \left( \frac{d(2d-b)}b, 0 \right).$$

Adding the coordinates of $A_2=(-d,0)$,

$$S_2 = A_2+w' = \left( \frac{d(2d-b)}b, \frac{hd}b \right).$$

The midpoint of side $A_1A_3$ is

$$M_2=\left(\frac d2,\frac h2\right).$$

The slope of $M_2S_2$ equals

$$m= \frac{\frac{hd}b-\frac h2} {\frac{d(2d-b)}b-\frac d2} = \frac{h(2d-b)} {d(4d-3b)}.$$

Its equation is

$$y-\frac h2 = m\left(x-\frac d2\right).$$

Setting $x=0$ yields the intersection with the symmetry axis:

$$y = \frac h2-\frac d2,m = \frac h2 \left( 1-\frac{2d-b}{4d-3b} \right) = \frac h2\cdot\frac b{2b-3d}.$$

Thus

$$P= \left( 0, \frac h2\cdot\frac b{2b-3d} \right)$$

lies on $M_2S_2$.

Reflection in the symmetry axis $x=0$ exchanges the vertices $A_2$ and $A_3$, hence it exchanges the pairs $(M_2,S_2)$ and $(M_3,S_3)$. The line $M_3S_3$ is therefore the mirror image of $M_2S_2$.

Since $P$ lies on the symmetry axis, it is fixed by this reflection. Consequently $P$ also lies on $M_3S_3$.

The point $P$ lies on $M_1S_1$ because $M_1S_1$ is the symmetry axis itself. Hence

$$P\in M_1S_1\cap M_2S_2\cap M_3S_3.$$

The three lines are concurrent.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the determination of $T_2$. The tangency lengths on side $A_1A_3$ are

$$A_1T_2=s-a_1=(b+d)-2d=b-d,$$

and

$$A_3T_2=s-a_3=(b+d)-b=d.$$

Their sum is $b$, the length of $A_1A_3$, so the ratio used in the section formula is correct.

The second delicate step is the simplification

$$u\cdot w=d\sqrt{\frac{b+d}{2b}}.$$

Starting from

$$h^2=(b+d)(b-d),$$

one obtains

$$h\sqrt{\frac{b-d}{2b}} = \sqrt{\frac{(b+d)(b-d)^2}{2b}} = (b-d)\sqrt{\frac{b+d}{2b}},$$

which justifies the cancellation that makes the reflection computation manageable.

The third delicate step is the symmetry argument. Reflection in the axis $x=0$ sends $A_2$ to $A_3$, preserves the incircle, sends $T_2$ to $T_3$, and sends the bisector of $\angle A_2$ to the bisector of $\angle A_3$. Hence it sends $S_2$ to $S_3$ and $M_2$ to $M_3$. Therefore the image of the line $M_2S_2$ is exactly the line $M_3S_3$. Since every point of the symmetry axis is fixed, any point of $M_2S_2$ lying on that axis must also lie on $M_3S_3$.

Alternative Approaches

A synthetic approach is possible by introducing the symmetry axis and expressing the reflected contact points through equal angle conditions. One can show directly that the cevians $M_2S_2$ and $M_3S_3$ are symmetric and then identify their common point on the axis using ratios along the sides and Menelaus type computations in suitable auxiliary triangles. The argument avoids coordinates but requires several nontrivial ratio transformations.

Another approach uses barycentric coordinates. The contact points of the incircle have simple barycentric descriptions, and reflection across an angle bisector can be expressed algebraically. After computing the barycentric equations of the lines $M_iS_i$, the concurrence follows from a determinant calculation. The coordinate method presented above is preferable because the isosceles symmetry reduces the problem to finding a single intersection with the symmetry axis, and every computation remains elementary.