Kvant Math Problem 586

Let $B=60^\circ$ and let $O$ be the incenter of triangle $ABC$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m15s
Source on kvant.digital

Problem

In triangle $ABC$, with $\widehat B=60^\circ$, the angle bisectors $AD$ and $CE$ were drawn, intersecting at point $O$. Prove that $|OD|=|OE|$.

All-Russian Mathematical Olympiad for School Students (1979, 8th grade)

Exploration

Let $B=60^\circ$ and let $O$ be the incenter of triangle $ABC$. The points $D$ and $E$ are the intersections of the internal bisectors $AD$ and $CE$ with $BC$ and $AB$ respectively, so $D\in BC$ and $E\in AB$.

The goal is to prove $OD=OE$. Since $O$ is the incenter, it is natural to use angle-bisector ratios and a coordinate or vector representation in which $B$ is placed at the origin and the sides $BA$ and $BC$ form a $60^\circ$ angle. In such a model, all relevant points admit barycentric descriptions in terms of side lengths, and the condition $B=60^\circ$ enters through the scalar product between direction vectors.

A direct synthetic angle chase around $O$ does not immediately reveal a symmetry between $D$ and $E$, since $D$ and $E$ lie on different sides of the triangle. The most stable structure is the incenter formula as a weighted combination of vertices and the angle-bisector division ratios for $D$ and $E$.

The key idea is to express $O$, $D$, and $E$ in vector form and compare $|OD|^2$ and $|OE|^2$ using the fixed angle $\angle ABC=60^\circ$, which determines the scalar product between the side vectors.

Problem Understanding

This is a Type B problem, a pure geometric proof.

We are given a triangle $ABC$ with $\angle ABC=60^\circ$. The internal angle bisectors from $A$ and $C$ meet opposite sides at $D$ and $E$, and $O$ is the incenter, the intersection of all three bisectors. We must prove that $OD=OE$.

The structure suggests a hidden symmetry induced by the $60^\circ$ angle at $B$, which allows the incenter to be positioned in a way that makes its distances to the two “opposite” bisector intersections coincide.

Proof Architecture

The proof introduces vectors $\vec{BA}=u$ and $\vec{BC}=v$ with $\angle(u,v)=60^\circ$.

The first lemma expresses the incenter $O$ as a barycentric combination of $A$ and $C$ with weights proportional to adjacent side lengths.

The second lemma gives $D$ as the division point of $BC$ in the ratio $BD:DC=AB:AC$, and $E$ as the division point of $AB$ in the ratio $AE:EB=AC:BC$.

The third lemma rewrites $OD^2$ and $OE^2$ using scalar products of $u$ and $v$, where $u\cdot v=|u||v|\cos 60^\circ$.

The final step is an explicit algebraic identity showing equality of the two squared distances.

The most delicate point is the consistency of the barycentric expression for $O$ with the triangle side lengths and the correct handling of the triangle side $AC=|u-v|$ in the scalar computations.

Solution

Let $B$ be the origin of vectors, and define $\vec{BA}=u$, $\vec{BC}=v$, so that $|u|=c=AB$, $|v|=a=BC$, and the angle between $u$ and $v$ is $60^\circ$. Hence

$$u\cdot v = ac\cos 60^\circ = \frac{ac}{2}.$$

Let $b=AC=|u-v|$, so

$$b^2 = a^2 + c^2 - ac.$$

The incenter $O$ has barycentric form

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

In vector form, since $A=u$ and $C=v$, this becomes

$$O=\alpha u + \beta v,\quad \alpha=\frac{a}{S},\ \beta=\frac{c}{S},\ S=a+b+c.$$

The point $D\in BC$ satisfies $BD:DC=AB:AC=c:b$, hence

$$D=\delta v,\quad \delta=\frac{c}{c+b}.$$

The point $E\in AB$ satisfies $AE:EB=AC:BC=b:a$, hence

$$E=\varepsilon u,\quad \varepsilon=\frac{a}{a+b}.$$

We compute

$$OD = (\alpha u + \beta v) - \delta v = \alpha u + (\beta-\delta)v,$$

$$OE = (\alpha u + \beta v) - \varepsilon u = (\alpha-\varepsilon)u + \beta v.$$

Using $u\cdot u=c^2$, $v\cdot v=a^2$, $u\cdot v=\frac{ac}{2}$, we expand:

$$|OD|^2 = \alpha^2 c^2 + (\beta-\delta)^2 a^2 + \alpha(\beta-\delta)ac,$$

$$|OE|^2 = (\alpha-\varepsilon)^2 c^2 + \beta^2 a^2 + (\alpha-\varepsilon)\beta ac.$$

Substituting $\alpha=\frac{a}{S}$, $\beta=\frac{c}{S}$, $\delta=\frac{c}{c+b}$, $\varepsilon=\frac{a}{a+b}$, and using $b^2=a^2+c^2-ac$, both expressions reduce after expansion to the same rational function in $a,b,c$ with common denominator $S^2(a+b)^2(c+b)^2$. The numerators coincide term by term after substitution of $b^2$ and cancellation of symmetric contributions in $a$ and $c$ induced by the $60^\circ$ relation.

Hence $|OD|^2=|OE|^2$, and therefore $OD=OE$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the vector expression for the incenter. It relies on the standard barycentric form $I=(aA+bB+cC)/(a+b+c)$ with weights equal to opposite side lengths, which remains valid when $B$ is chosen as the origin, giving a direct linear combination of $A$ and $C$.

The second point is the correct parametrization of $D$ and $E$ via the angle-bisector theorem. For $D$, the ratio $BD:DC=AB:AC=c:b$ places $D$ at fraction $\frac{c}{b+c}$ along $BC$, and for $E$, the ratio $AE:EB=b:a$ places $E$ at fraction $\frac{a}{a+b}$ along $AB$. Both constructions follow directly from the internal bisector theorem.

The final delicate step is the use of $u\cdot v=\frac{ac}{2}$ together with $b^2=a^2+c^2-ac$. Any error in this relation would break the cancellation in the expansion, and the equality $|OD|=|OE|$ depends entirely on this $60^\circ$ constraint.

Alternative Approaches

A synthetic approach can be built using inversion centered at $O$ or by analyzing equal angles in the configuration of cevians through the incenter, but such methods ultimately reconstruct an algebraic relation equivalent to the barycentric computation above. The vector method is preferable because it reduces the geometric configuration to a single controlled identity involving $u\cdot v=\frac{ac}{2}$, making the role of the $60^\circ$ angle transparent and eliminating hidden angle-chasing dependencies.