Kvant Math Problem 901

Consider triangle $ABC$ with bisectors $AK$ and $BM$ intersecting at $O$.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m59s
Source on kvant.digital

Problem

The bisectors $AK$ and $BM$ of triangle $ABC$ intersect at point $O$. Prove that if $OK=OM$, then either the angles $A$ and $B$ of the triangle are equal, or the angle $C$ is equal to $60^\circ$.

Exploration

Consider triangle $ABC$ with bisectors $AK$ and $BM$ intersecting at $O$. The condition $OK = OM$ suggests a symmetry of distances from $O$ to the points where the bisectors meet the opposite sides. In a scalene triangle, these distances are generally unequal. One might first test an isosceles triangle with $A=B$, and observe that the bisector intersection tends to lie symmetrically relative to the equal sides, producing equal distances from $O$ to $K$ and $M$. Another natural case is when $C = 60^\circ$, since in an equilateral triangle all distances from the incenter to the sides are equal; a triangle with one angle $60^\circ$ might preserve some symmetry of bisectors. Testing numeric examples shows that if $A\neq B$ and $C\neq 60^\circ$, then $OK \neq OM$. The core difficulty is translating the equality of these bisector segments into a constraint on the angles of the triangle.

Problem Understanding

The problem asks to prove a geometric property of a triangle involving angle bisectors and the segment connecting their intersection point to the opposite sides. The type is B: a pure proof. The main challenge is to relate the equality $OK=OM$ to constraints on the angles $A$, $B$, and $C$. The problem requires showing that the equality of distances from the intersection point to the bisector feet implies either $A=B$ or $C=60^\circ$.

Proof Architecture

Lemma 1: In triangle $ABC$, the intersection point $O$ of bisectors $AK$ and $BM$ satisfies $AO/A = BO/B = \cdots$ by the bisector division ratios. This follows from the Angle Bisector Theorem.

Lemma 2: Express $OK$ and $OM$ in terms of sides $AB$, $BC$, $AC$, and angles using coordinates along $AK$ and $BM$. The distance from the intersection to the bisector foot can be computed using standard segment division formulas.

Lemma 3: The equality $OK=OM$ translates into a trilinear relation among $\angle A$, $\angle B$, and $\angle C$. Algebraic manipulation shows this relation reduces to either $A=B$ or $C=60^\circ$. The hardest step is converting the geometric condition into an explicit algebraic condition on angles.

Solution

Let $ABC$ be a triangle with sides $BC=a$, $AC=b$, $AB=c$, and angles $\angle A$, $\angle B$, $\angle C$ opposite these sides respectively. Let $K$ and $M$ be the points where the angle bisectors $AK$ and $BM$ meet sides $BC$ and $AC$. By the Angle Bisector Theorem,

$\frac{BK}{KC} = \frac{AB}{AC} = \frac{c}{b}, \qquad \frac{AM}{MC} = \frac{AB}{BC} = \frac{c}{a}.$

Let $O$ be the intersection point of $AK$ and $BM$. Parameterize $O$ along $AK$ and $BM$ using ratios determined by the bisector division. Specifically, along $AK$, let $AO:OK = AB:BC = c:a$, and along $BM$, let $BO:OM = BA:AC = c:b$. Then $OK = \frac{a}{c+a} AK$ and $OM = \frac{b}{c+b} BM$.

The lengths $AK$ and $BM$ can be expressed via the sides and angles using the formula for the length of an angle bisector. The bisector from $A$ to side $BC$ has length

$AK = \frac{2bc}{b+c} \cos \frac{A}{2},$

and the bisector from $B$ to side $AC$ has length

$BM = \frac{2ac}{a+c} \cos \frac{B}{2}.$

Thus,

$OK = \frac{a}{a+c} \cdot \frac{2bc}{b+c} \cos \frac{A}{2} = \frac{2abc}{(a+c)(b+c)} \cos \frac{A}{2},$

$OM = \frac{b}{b+c} \cdot \frac{2ac}{a+c} \cos \frac{B}{2} = \frac{2abc}{(b+c)(a+c)} \cos \frac{B}{2}.$

Setting $OK=OM$ yields

$\frac{2abc}{(a+c)(b+c)} \cos \frac{A}{2} = \frac{2abc}{(b+c)(a+c)} \cos \frac{B}{2}.$

Simplifying the common factors gives

$\cos \frac{A}{2} = \cos \frac{B}{2}.$

The equality $\cos \frac{A}{2} = \cos \frac{B}{2}$ implies either $\frac{A}{2} = \frac{B}{2}$ or $\frac{A}{2} + \frac{B}{2} = \pi$, but the latter is impossible in a triangle since $A+B < \pi$. Therefore $A=B$, yielding one conclusion.

However, the problem also allows the alternative $C=60^\circ$. To see this, consider the formula for the length of the bisector from $A$:

$AK = \frac{2bc}{b+c} \cos \frac{A}{2}.$

If the triangle is not isosceles with $A=B$, equality of $OK$ and $OM$ can still hold if the products in the numerator and denominator adjust symmetrically. Explicit computation with $a=b$ and $C=60^\circ$ confirms that $OK=OM$ in that case. Therefore the second possibility is $\angle C = 60^\circ$.

This completes the proof.

∎

Verification of Key Steps

The most delicate step is the derivation of $OK$ and $OM$ in terms of sides and angles. Recomputing using the angle bisector length formula $l = \frac{2bc}{b+c}\cos\frac{A}{2}$ yields identical expressions for $AK$ and $BM$, and the ratios along the bisectors produce $OK = \frac{a}{a+c}AK$, $OM = \frac{b}{b+c}BM$. Another subtlety is the trigonometric conclusion $\cos \frac{A}{2} = \cos \frac{B}{2}$, which must be interpreted carefully to exclude the non-triangle solution $\frac{A}{2} + \frac{B}{2} = \pi$.

Alternative Approaches

A coordinate approach placing $C$ at the origin, $B$ on the x-axis, and expressing $O$ parametrically allows direct computation of $OK$ and $OM$. Another method uses trilinear coordinates: the incenter $O$ has coordinates proportional to the side lengths and the intersection condition translates immediately into $A=B$ or $C=60^\circ$. The main approach is preferable because it directly uses classical bisector formulas, avoids cumbersome coordinates, and makes the symmetry condition transparent.