Kvant Math Problem 467

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m06s
Source on kvant.digital

Problem

Points $D$ and $E$ divide the sides $AC$ and $AB$ of an equilateral triangle $ABC$ in the ratios $|AD|:|DC|=|BE|:|EA|=1:2$. The lines $BD$ and $CE$ intersect at point $O$. Prove that the angle $AOC$ is a right angle.

A. Krasnodemskaya

Exploration

The triangle is equilateral, so placing it in a coordinate system with one side horizontal allows algebraic control over all distances and angles. The ratios $AD:DC=1:2$ and $BE:EA=1:2$ indicate that $D$ and $E$ are fixed affine combinations of the vertices, suggesting that the intersection point $O$ can be found via linear equations. Since the conclusion concerns a right angle at $O$, the natural target is to prove orthogonality via a dot product condition in coordinates.

A direct geometric attempt using angle chasing is obstructed by the asymmetric placement of $D$ and $E$, so the most stable approach is to encode the configuration in Cartesian coordinates and compute $O$ explicitly. The key risk in such an approach is arithmetic consistency when solving the intersection of $BD$ and $CE$, since any error propagates into the final orthogonality check.

Problem Understanding

This is a Type B problem, since the task is to prove that $\angle AOC = 90^\circ$.

In an equilateral triangle $ABC$, points $D$ and $E$ divide $AC$ and $AB$ internally in the ratios $AD:DC=1:2$ and $BE:EA=1:2$. The lines $BD$ and $CE$ intersect at $O$, and the claim is that the segment $AO$ is perpendicular to $CO$.

The structure suggests a hidden orthogonality arising from a symmetric affine configuration inside an equilateral triangle, which is best revealed through coordinates.

Proof Architecture

The proof proceeds by placing the triangle in the Euclidean plane with explicit coordinates for $A$, $B$, and $C$. The first lemma expresses $D$ and $E$ as affine combinations of endpoints using the given ratios. The second lemma computes the parametric equations of lines $BD$ and $CE$ and determines their intersection point $O$. The third lemma verifies orthogonality by computing the dot product of vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$.

The most delicate step is the correct solution of the two linear equations defining $O$, since any computational slip invalidates the final orthogonality check.

Solution

Place the equilateral triangle in the coordinate plane by setting $A=(0,0)$, $B=(1,0)$, and $C=\left(\frac12,\frac{\sqrt3}{2}\right)$.

The point $D$ lies on $AC$ with $AD:DC=1:2$. This implies that $D$ divides $AC$ internally in the ratio $1:2$, hence

$$D=\frac{2A+C}{3}=\frac{C}{3}=\left(\frac16,\frac{\sqrt3}{6}\right).$$

The point $E$ lies on $AB$ with $BE:EA=1:2$, which is equivalent to $AE:EB=2:1$. Therefore

$$E=\frac{A+2B}{3}=\left(\frac23,0\right).$$

The line $BD$ is parametrized as

$$B+t(D-B)=\left(1,0\right)+t\left(-\frac56,\frac{\sqrt3}{6}\right)=\left(1-\frac{5t}{6},\frac{t\sqrt3}{6}\right).$$

The line $CE$ is parametrized as

$$C+s(E-C)=\left(\frac12,\frac{\sqrt3}{2}\right)+s\left(\frac16,-\frac{\sqrt3}{2}\right)=\left(\frac12+\frac{s}{6},\frac{\sqrt3}{2}-\frac{s\sqrt3}{2}\right).$$

At their intersection point $O$, the coordinates satisfy

$$1-\frac{5t}{6}=\frac12+\frac{s}{6}, \qquad \frac{t\sqrt3}{6}=\frac{\sqrt3}{2}-\frac{s\sqrt3}{2}.$$

Dividing the second equation by $\frac{\sqrt3}{6}$ yields

$$t=3-3s.$$

Substituting into the first equation gives

$$1-\frac{5(3-3s)}{6}=\frac12+\frac{s}{6}.$$

Expanding,

$$1-\frac{15}{6}+\frac{15s}{6}=\frac12+\frac{s}{6},$$

which simplifies to

$$-\frac32+\frac{5s}{2}=\frac12+\frac{s}{6}.$$

Multiplying by $6$,

$$-9+15s=3+s,$$

so $14s=12$ and hence $s=\frac67$. Substituting back,

$$t=3-3\cdot\frac67=\frac37.$$

The point $O$ is obtained from the parametrization of $BD$:

$$O=\left(1-\frac{5}{6}\cdot\frac37,\frac{3}{7}\cdot\frac{\sqrt3}{6}\right)=\left(\frac{9}{14},\frac{\sqrt3}{14}\right).$$

Now compute vectors

$$\overrightarrow{OA}=A-O=\left(-\frac{9}{14},-\frac{\sqrt3}{14}\right), \qquad \overrightarrow{OC}=C-O=\left(\frac12-\frac{9}{14},\frac{\sqrt3}{2}-\frac{\sqrt3}{14}\right).$$

These simplify to

$$\overrightarrow{OC}=\left(-\frac17,\frac{3\sqrt3}{7}\right).$$

The dot product is

$$\overrightarrow{OA}\cdot\overrightarrow{OC}=\left(-\frac{9}{14}\right)\left(-\frac17\right)+\left(-\frac{\sqrt3}{14}\right)\left(\frac{3\sqrt3}{7}\right)=\frac{9}{98}-\frac{9}{98}=0.$$

Hence $\overrightarrow{OA}\perp \overrightarrow{OC}$, which implies $\angle AOC=90^\circ$.

This completes the proof. ∎

Verification of Key Steps

The determination of $D$ and $E$ relies on converting segment ratios into affine combinations, where $AD:DC=1:2$ yields $D=(2A+C)/3$ and $BE:EA=1:2$ yields $E=(A+2B)/3$. Any reversal of these weights would move the points closer to the wrong vertices and destroy the final orthogonality.

The intersection computation depends on solving a coupled linear system in parameters $t$ and $s$. The elimination step $t=3-3s$ is essential, since it reduces the system to a single variable; any algebraic slip here propagates directly into the coordinates of $O$.

The final orthogonality check is sensitive to sign consistency in both coordinate differences. The cancellation of the dot product into zero confirms that no accumulated computational error remains.

Alternative Approaches

A synthetic approach can be developed using barycentric coordinates in $\triangle ABC$, where $D$ and $E$ are represented by fixed weight triples. In that framework, $BD$ and $CE$ are written as linear constraints in barycentric variables, and $O$ emerges as a solution of a linear system whose orthogonality condition reduces to a symmetric identity in the barycentric metric of an equilateral triangle. This method avoids coordinate geometry but replaces it with an equivalent affine computation of similar complexity, while the Cartesian approach remains more direct and transparent for explicit verification.