Kvant Math Problem 128

Let the median be drawn from a vertex $A$ to the midpoint $M$ of the opposite side.

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Problem

Find the ratio of the sides of a triangle, one of whose medians is divided by the incircle into three equal parts.

Exploration

Let the median be drawn from a vertex $A$ to the midpoint $M$ of the opposite side. If the incircle cuts this median into three equal parts, then the two intersection points of the circle with the median lie at distances $\frac13 AM$ and $\frac23 AM$ from $A$.

Along any line, the midpoint of a chord is the projection of the center onto that line. Hence the projection of the incenter onto the median must lie at the midpoint of the two intersection points, that is, at $\frac12 AM$. This condition is unexpectedly strong and should translate into an algebraic relation among the side lengths.

The most delicate point is extracting that relation correctly. After introducing coordinates adapted to the median, the condition that the incenter projects to the midpoint of the median turns out to factor completely and yields either $a=2b$ or $a=2c$. A second equation comes from the fact that the half-length of the chord cut by the circle on the median equals $\frac16 AM$. Solving these two conditions determines the side ratio.

Problem Understanding

We are given a triangle with an incircle and a median. The incircle intersects that median in two points and divides the median into three equal segments. We must determine the ratio of the side lengths of the triangle.

This is a Type A problem. We must determine all possible side ratios and prove that no others occur.

The core difficulty is converting the geometric condition on the median and the incircle into equations involving the side lengths.

The answer will be that the side lengths are proportional to $5:10:13$.

Proof Architecture

Let the median be $AM$, where $M$ is the midpoint of $BC$, and let $a=BC$, $b=CA$, $c=AB$.

Choose coordinates with $A=(0,0)$ and $M=(m,0)$, where $m=AM$; represent $B$ and $C$ symmetrically about $M$.

Compute the $x$-coordinate of the incenter from barycentric coordinates. The condition that the projection of the incenter onto the median is the midpoint of the chord cut by the incircle implies $x_I=\frac m2$.

Show that the resulting equation factors as $(a-2b)(a-2c)=0$. Hence either $a=2b$ or $a=2c$.

Assume $a=2b$; the other case is symmetric. Write $t=\frac cb$ and compute $m^2$, the squared distance from the incenter to the median, and the inradius.

Use the fact that the half-length of the chord cut by the incircle on the median equals $\frac m6$. This gives an equation for $t$.

Solve the equation and obtain $t=\frac{13}{5}$, leading to the ratio $a:b:c=10:5:13$.

The hardest step is proving that $x_I=\frac m2$ forces $(a-2b)(a-2c)=0$.

Solution

Let $AM$ be the median in question, where $M$ is the midpoint of $BC$. Denote

$$a=BC,\qquad b=CA,\qquad c=AB,$$

and let

$$m=AM.$$

Choose coordinates

$$A=(0,0),\qquad M=(m,0).$$

Since $M$ is the midpoint of $BC$, we may write

$$B=(m+x,y),\qquad C=(m-x,-y).$$

Then

$$c^2=(m+x)^2+y^2, \qquad b^2=(m-x)^2+y^2,$$

hence

$$x=\frac{c^2-b^2}{4m}.$$

Let $I$ be the incenter. In barycentric coordinates,

$$I=\frac{aA+bB+cC}{a+b+c}.$$

Therefore its $x$-coordinate is

$$x_I= \frac{b(m+x)+c(m-x)}{a+b+c}.$$

The incircle intersects the median in two points dividing it into three equal parts. These points are located at distances

$$\frac m3,\qquad \frac{2m}{3}$$

from $A$. Their midpoint is at $\frac m2$.

For every chord of a circle, the perpendicular from the center bisects the chord. Since the chord lies on the median, the projection of $I$ onto the median is the midpoint of the chord. Thus

$$x_I=\frac m2.$$

Substituting the formula for $x_I$ and the expression for $x$, we obtain

$$\frac{m(b+c)+x(b-c)}{a+b+c} = \frac m2.$$

After replacing $x$ by $\frac{c^2-b^2}{4m}$ and simplifying,

$$2m^2(b+c-a)=(b-c)^2(b+c).$$

Using the median formula

$$m^2=\frac{2b^2+2c^2-a^2}{4},$$

the left-hand side becomes

$$\frac{2b^2+2c^2-a^2}{2}(b+c-a).$$

A direct factorization yields

$$\frac12(a-2b)(a-2c)(a+b+c)=0.$$

Since $a+b+c>0$,

$$a=2b \quad\text{or}\quad a=2c.$$

The two cases are symmetric. It suffices to consider

$$a=2b.$$

Set

$$t=\frac cb.$$

Scaling does not affect ratios, so take $b=1$. Then

$$a=2,\qquad c=t.$$

The median formula gives

$$m^2=\frac{t^2-1}{2}.$$

From

$$x=\frac{t^2-1}{4m},$$

one finds

$$x=\frac{\sqrt{2t^2-2}}{4},$$

and then

$$y^2=1-(m-x)^2 =\frac{9-t^2}{8}.$$

The semiperimeter is

$$s=\frac{a+b+c}{2} =\frac{3+t}{2}.$$

The area equals

$$\Delta=my,$$

hence

$$r=\frac{\Delta}{s} =\frac{2my}{3+t}.$$

The $y$-coordinate of the incenter is

$$k=\frac{y(1-t)}{3+t}.$$

Therefore

$$r^2-k^2 = \frac{(3-t)(t-1)}{8}.$$

The quantity $\sqrt{r^2-k^2}$ is the half-length of the chord cut by the incircle on the median. Since the median is divided into three equal parts, that half-length equals

$$\frac m6.$$

Thus

$$r^2-k^2=\frac{m^2}{36}.$$

Substituting the expressions above,

$$\frac{(3-t)(t-1)}{8} = \frac{t^2-1}{72}.$$

Multiplying by $72$,

$$9(3-t)(t-1)=t^2-1.$$

This simplifies to

$$(t-1)(5t-13)=0.$$

The value $t=1$ gives $m=0$ and is impossible. Hence

$$t=\frac{13}{5}.$$

Therefore

$$a:b:c = 2:1:\frac{13}{5} = 10:5:13.$$

In the symmetric case $a=2c$, the ratio is $10:13:5$, which represents the same unordered side ratio.

Hence the side lengths are proportional to

$$\boxed{5:10:13}.$$

Verification of Key Steps

The first delicate step is the deduction $x_I=\frac m2$. The incircle intersects the median in two points at distances $\frac m3$ and $\frac{2m}{3}$ from $A$. Their midpoint is at $\frac m2$. The radius drawn to the midpoint of a chord is perpendicular to the chord, so the projection of the center onto the median must be exactly this midpoint.

The second delicate step is the factorization leading to $a=2b$ or $a=2c$. Starting from

$$2m^2(b+c-a)=(b-c)^2(b+c)$$

and substituting

$$m^2=\frac{2b^2+2c^2-a^2}{4},$$

one obtains

$$\frac12(a-2b)(a-2c)(a+b+c)=0.$$

Since $a+b+c>0$, no other possibility exists.

The third delicate step is the chord-length condition. The chord endpoints are at $\frac m3$ and $\frac{2m}{3}$, so the distance from its midpoint to either endpoint is

$$\frac12\left(\frac{2m}{3}-\frac m3\right) = \frac m6.$$

For a circle of radius $r$ whose center is at distance $k$ from the line, the half-chord length equals $\sqrt{r^2-k^2}$, giving the equation used in the proof.

Alternative Approaches

A synthetic approach is possible. Let the two intersection points of the incircle with the median be $P$ and $Q$. Since $AP=PQ=QM$, the midpoint of $PQ$ is also the midpoint of $AM$. The projection of the incenter onto the median is therefore the midpoint of the median itself. Using standard formulas for distances from the incenter to medians and angle bisectors, one can derive directly that one side equals twice another.

After obtaining $a=2b$ or $a=2c$, one may use only classical formulas for the median and the inradius, avoiding coordinates entirely. The coordinate method above is preferable because every quantity involved is expressed explicitly and the crucial factorization emerges in a transparent algebraic form.