Kvant Math Problem 1591

Verified: yes
Verdicts: PASS + PASS
Solve time: 17m12s
Source on kvant.digital

Problem

In triangle $ABC$, the angle bisectors $BL$ and $AK$ are drawn. It turns out that $KL$ is the angle bisector of triangle $AKC$. Find the angle $BAC$.

S. I. Tokarev

Tournament of the Towns (spring, 1997)

Exploration

Let $BL$ be the internal angle bisector at $B$ meeting $AC$ at $L$, and $AK$ be the internal angle bisector at $A$ meeting $BC$ at $K$. The condition states that in triangle $AKC$, the segment $KL$ is the internal angle bisector of $\angle AKC$, so $L$ lies on $AC$ and $KL$ divides the opposite side proportionally to the adjacent sides.

This introduces three angle-bisector constraints: one in triangle $ABC$ at $A$, one at $B$, and a third in triangle $AKC$ linking $K$ and $L$. The structure suggests expressing segment ratios via side lengths and enforcing compatibility among them.

Let $AB=c$, $BC=a$, $CA=b$. The key unknown is $\angle BAC = A$. The decisive step is to express $AK/KC$ via the bisector at $A$ and match it with $AL/LC = AB/BC$ arising from the bisector at $B$. This translates the geometric condition into a trigonometric equation in $A$.

The most delicate point is ensuring the bisector condition in triangle $AKC$ is correctly converted into a ratio on $AC$ and consistently matched with the triangle $ABC$ side lengths.

Problem Understanding

This is a Type A problem: determine all possible values of $\angle BAC$.

In triangle $ABC$, two standard angle bisectors define points $K$ on $BC$ and $L$ on $AC$. An additional condition states that $KL$ is the angle bisector of triangle $AKC$ at vertex $K$.

The task is to convert this geometric concurrency condition into algebraic constraints on side ratios, which forces a unique value of $\angle BAC$. The derivation must prove both necessity and existence of the configuration.

Proof Architecture

First, express $AL/LC$ using the angle bisector theorem in triangle $ABC$, giving $AL/LC = AB/BC$.

Next, express the condition that $KL$ bisects $\angle AKC$ in triangle $AKC$, yielding $AL/LC = AK/KC$.

Then compute $AK$ and $KC$ in terms of $a,b,c$ using the angle bisector theorem at $A$ in triangle $ABC$, obtaining $KC = \frac{ab}{b+c}$ and $AK = \frac{2bc\cos(\frac{A}{2})}{b+c}$.

Equating the two expressions for $AL/LC$ produces a trigonometric equation in $A$ alone. To ensure completeness, we must also verify that a triangle with $\angle A = 120^\circ$ actually satisfies all bisector conditions.

Solution

Let $AB=c$, $BC=a$, $CA=b$, and let $\angle BAC = A$.

Since $BL$ is the internal angle bisector of $\angle B$ in triangle $ABC$, the angle bisector theorem gives

$\frac{AL}{LC} = \frac{AB}{BC} = \frac{c}{a}.$

Now consider triangle $AKC$. The condition states that $KL$ is the internal angle bisector of $\angle AKC$, hence by the angle bisector theorem in triangle $AKC$,

$\frac{AL}{LC} = \frac{AK}{KC}.$

Combining the two equal expressions for $\frac{AL}{LC}$ yields

$\frac{AK}{KC} = \frac{c}{a}.$

We now compute $AK$ and $KC$.

Since $AK$ is the internal angle bisector of $\angle A$ in triangle $ABC$, the angle bisector theorem gives

$\frac{BK}{KC} = \frac{AB}{AC} = \frac{c}{b},$

$KC = \frac{ab}{b+c}.$

The standard length formula for the angle bisector from $A$ gives

$AK = \frac{2bc\cos\left(\frac{A}{2}\right)}{b+c}.$

Therefore,

$\frac{AK}{KC} = \frac{\frac{2bc\cos\left(\frac{A}{2}\right)}{b+c}}{\frac{ab}{b+c}} = \frac{2c\cos\left(\frac{A}{2}\right)}{a}.$

Equating this with $\frac{c}{a}$ yields

$\frac{2c\cos\left(\frac{A}{2}\right)}{a} = \frac{c}{a}.$

Since $a>0$ and $c>0$, we cancel to obtain

$2\cos\left(\frac{A}{2}\right) = 1,$

$\cos\left(\frac{A}{2}\right) = \frac{1}{2}.$

Hence

$\frac{A}{2} = 60^\circ,$

which gives

$A = 120^\circ.$

To verify existence, we check that a triangle with $\angle A = 120^\circ$ can satisfy all bisector relations. By the law of cosines,

$a^2 = b^2 + c^2 - 2bc\cos A = b^2 + c^2 + bc,$

$a = \sqrt{b^2 + bc + c^2}.$

The bisector lengths then satisfy

$KC = \frac{ab}{b+c} = \frac{b \sqrt{b^2 + bc + c^2}}{b+c}, \qquad AK = \frac{2bc\cos(60^\circ)}{b+c} = \frac{bc}{b+c}.$

Then

$\frac{AK}{KC} = \frac{bc/(b+c)}{b\sqrt{b^2 + bc + c^2}/(b+c)} = \frac{c}{\sqrt{b^2 + bc + c^2}} = \frac{c}{a},$

matching the previous ratio. Similarly, $\frac{AL}{LC} = \frac{c}{a}$ as before, so $KL$ indeed bisects $\angle AKC$. This confirms that the configuration exists and satisfies all conditions.

Therefore, the unique angle satisfying the given configuration is

$\boxed{120^\circ}.$

Verification of Key Steps

The identification $\frac{AL}{LC} = \frac{AK}{KC}$ follows directly from applying the angle bisector theorem in triangle $AKC$ to $KL$, since $L$ lies on $AC$ and $K$ is the vertex of the bisected angle.

The computation of $KC$ relies on the standard angle bisector theorem in triangle $ABC$, giving $BK:KC = AB:AC$.

The formula for $AK$ is consistent with the bisector length relation derived from projecting the bisector onto the opposite side. Substitution produces a ratio depending only on $A$, and verifying the converse demonstrates that the configuration exists for $A = 120^\circ$.

Alternative Approaches

One can also approach the problem using trilinear or barycentric coordinates, where the condition that $KL$ is an angle bisector translates into a linear relation between coordinates of $K$ and $L$, leading to a homogeneous equation in side ratios that reduces to a trigonometric constraint on $\angle A$.

Another approach uses directed angles and Ceva-type relations applied twice, once in $ABC$ and once in $AKC$, eliminating segment ratios without explicit length formulas. The trigonometric bisector-length method is shorter because it reduces the condition immediately to a single equation in $\angle A$ and allows direct verification of existence.