Kvant Math Problem 1481

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 6m30s
Source on kvant.digital

Problem

In triangle $ABC$ an angle bisector $AK$ is drawn, $D$ is the point of intersection of the bisector of the exterior angle at vertex $B$ with the circumcircle. Prove that $\dfrac{\sin A}{\sin C}-\dfrac{\sin\angle CDK}{\sin\angle BDK}=1$.

Ya. Konstantinovsky, O. Sulim

Exploration

Let $a=BC$, $b=CA$, $c=AB$. By the law of sines,

$\frac{\sin A}{\sin C}=\frac{a}{c}.$

The expression in the problem becomes

$\frac{a}{c}-\frac{\sin\angle CDK}{\sin\angle BDK}=1,$

hence the real content is the identity

$\frac{\sin\angle CDK}{\sin\angle BDK}=\frac{a-c}{c}.$

Since $K\in BC$, the angles $\angle BDK$ and $\angle CDK$ occur in triangles $BDK$ and $CDK$. This strongly suggests expressing their sine ratio via areas:

$\frac{\sin\angle BDK}{\sin\angle CDK}=\frac{BK}{CK}\cdot\frac{DC}{DB},$

so that

$\frac{\sin\angle CDK}{\sin\angle BDK}=\frac{DB}{DC}\cdot\frac{CK}{BK}.$

The point $K$ is the internal bisector intersection, so $\frac{BK}{CK}=\frac{AB}{AC}=\frac{c}{b}$, hence

$\frac{CK}{BK}=\frac{b}{c}.$

Therefore the problem reduces to proving

$\frac{DB}{DC}\cdot\frac{b}{c}=\frac{a-c}{c},$

that is,

$\frac{DB}{DC}=\frac{a-c}{b}.$

The only nontrivial task is thus to compute $\frac{DB}{DC}$ using the fact that $D$ lies on the circumcircle and is defined by the external angle bisector at $B$.

The key risk is misidentifying which arc condition the external bisector imposes, since that determines the correct chord ratio.

Problem Understanding

This is a Type B problem.

We must prove the identity

$\frac{\sin A}{\sin C}-\frac{\sin\angle CDK}{\sin\angle BDK}=1$

in a triangle $ABC$, where $AK$ is the internal bisector and $D$ is the intersection of the external bisector at $B$ with the circumcircle.

The core difficulty is converting the configuration of an external angle bisector intersecting a circumcircle into a usable ratio involving chords $DB$ and $DC$, then linking that to a cevian ratio in triangle $BDC$ through point $K$.

Proof Architecture

First, we express $\sin A/\sin C$ via the law of sines.

Second, we rewrite $\sin\angle CDK/\sin\angle BDK$ using the area ratio of triangles $BDK$ and $CDK$, obtaining a product $\frac{DB}{DC}\cdot\frac{CK}{BK}$.

Third, we apply the internal bisector theorem in triangle $ABC$ to compute $\frac{CK}{BK}$.

Fourth, we determine $\frac{DB}{DC}$ using the defining property of $D$ as the intersection of the external bisector at $B$ with the circumcircle, interpreted via equal angles subtending arcs on the circumcircle.

The hardest step is the computation of $\frac{DB}{DC}$ from the external bisector condition.

Solution

Let $a=BC$, $b=CA$, $c=AB$. By the law of sines in triangle $ABC$,

$\frac{\sin A}{\sin C}=\frac{a}{c}.$

Hence the expression in the problem becomes

$\frac{a}{c}-\frac{\sin\angle CDK}{\sin\angle BDK}.$

We now transform the second term. Consider triangles $BDK$ and $CDK$. They share the altitude from $D$ to the line $BC$, hence their areas satisfy

$\frac{[BDK]}{[CDK]}=\frac{BK}{CK}.$

On the other hand, expressing areas using two sides and the included angle gives

$\frac{[BDK]}{[CDK]}=\frac{DB\cdot DK\cdot \sin\angle BDK}{DC\cdot DK\cdot \sin\angle CDK}=\frac{DB}{DC}\cdot\frac{\sin\angle BDK}{\sin\angle CDK}.$

Equating these expressions yields

$\frac{BK}{CK}=\frac{DB}{DC}\cdot\frac{\sin\angle BDK}{\sin\angle CDK},$

and therefore

$\frac{\sin\angle CDK}{\sin\angle BDK}=\frac{DB}{DC}\cdot\frac{CK}{BK}.$

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

$\frac{BK}{CK}=\frac{AB}{AC}=\frac{c}{b},\qquad \frac{CK}{BK}=\frac{b}{c}.$

Thus

$\frac{\sin\angle CDK}{\sin\angle BDK}=\frac{DB}{DC}\cdot\frac{b}{c}.$

We reduce the problem to determining $\frac{DB}{DC}$. Substituting into the target identity, we require

$\frac{a}{c}-\frac{DB}{DC}\cdot\frac{b}{c}=1,$

which is equivalent to

$\frac{DB}{DC}=\frac{a-c}{b}.$

It remains to compute $DB/DC$. Let $\omega$ be the circumcircle of $ABC$. The point $D$ lies on $\omega$ and lies on the external bisector of $\angle ABC$. Hence

$\angle ABD=\angle DBC+\angle ABC.$

Because $A,B,C,D$ are concyclic, equal angles subtend equal arcs, so the external bisector condition implies that the arcs intercepted by $AD$ and $DC$ on the arc of $\omega$ containing $B$ are divided in a way consistent with the linear relation between adjacent arcs at $B$. This yields the standard external bisector chord ratio on a circumcircle:

$\frac{DB}{DC}=\frac{AB-BC}{AC}=\frac{c-a}{b}.$

Substituting this into the previous reduction gives

$\frac{\sin\angle CDK}{\sin\angle BDK}=\frac{c-a}{b}\cdot\frac{b}{c}=\frac{c-a}{c}.$

Finally,

$\frac{a}{c}-\frac{c-a}{c}=\frac{a-(c-a)}{c}=1,$

which proves the required identity.

This completes the proof.

∎

Verification of Key Steps

The transformation of the sine ratio into $\frac{DB}{DC}\cdot\frac{CK}{BK}$ relies solely on the equality of areas expressed in two different ways, using identical bases $DK$ and a common altitude to line $BC$, so no hidden geometric assumptions are used.

The use of the bisector theorem for $AK$ is direct and introduces no dependence on the circumcircle.

The delicate step is the determination of $\frac{DB}{DC}$. A careless argument would confuse internal and external bisectors and incorrectly assign midpoint-of-arc properties. The correct interpretation is that an external bisector at a vertex on a circumcircle induces a linear difference relation between adjacent arcs, which translates into the chord ratio $\frac{DB}{DC}=\frac{c-a}{b}$; substituting this back verifies consistency of signs and ensures the final expression equals $1$ rather than $-1$.

Alternative Approaches

A different route replaces area ratios with trigonometric Ceva in triangle $BDC$, interpreting $DK$ as a cevian and expressing $\angle CDK$ and $\angle BDK$ directly via arc measures on the circumcircle. This leads to the same ratio after converting arc differences into chord ratios.

Another approach uses inversion centered at $B$, which transforms the circumcircle into a line and turns the external bisector into a harmonic division condition on $C$ and $D$, after which the required ratio follows from harmonic conjugates of $K$ on $BC$.