Kvant Math Problem 771
Let $O$ be the common point which is simultaneously the circumcenter of $\triangle ABC$ and the incenter of $\triangle ABK$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m38s
Source on kvant.digital
Problem
In triangle $ABC$, the bisector $AK$ is drawn. It is known that the centers of the circles: the incircle of triangle $ABK$ and the circumcircle of triangle $ABC$ coincide. Find the angles of triangle $ABC$.
Exploration
Let $O$ be the common point which is simultaneously the circumcenter of $\triangle ABC$ and the incenter of $\triangle ABK$. Since $AK$ is the bisector of $\angle A$, the point $K$ lies on $BC$.
Because $O$ is the incenter of $\triangle ABK$, it lies on the bisector of $\angle BAK$. Since $AK$ bisects $\angle A$ of $\triangle ABC$, we have
$$\angle BAK=\frac A2.$$
Hence
$$\angle BAO=\frac12\angle BAK=\frac A4.$$
Because $O$ is the circumcenter of $\triangle ABC$, the angle between chord $AB$ and radius $AO$ equals
$$90^\circ-\angle ACB=90^\circ-C.$$
Therefore
$$\angle BAO=90^\circ-C.$$
Combining the two expressions for $\angle BAO$ gives
$$C=90^\circ-\frac A4.$$
A symmetric argument at vertex $B$ of triangle $ABK$ should give another relation. Since $O$ is the incenter of $\triangle ABK$,
$$\angle ABO=\frac12\angle ABK.$$
Because $K\in BC$,
$$\angle ABK=\angle ABC=B.$$
On the other hand, from the circumcenter property,
$$\angle ABO=90^\circ-C.$$
Thus
$$\frac B2=90^\circ-C.$$
Combining this with the previous equation yields
$$\frac B2=\frac A4,$$
hence
$$A=2B.$$
Using
$$A+B+C=180^\circ$$
and
$$C=90^\circ-\frac A4,$$
gives a unique solution. Computing,
$$2B+B+90^\circ-\frac B2=180^\circ,$$
so
$$\frac52 B=90^\circ, \qquad B=36^\circ.$$
Hence
$$A=72^\circ,\qquad C=72^\circ.$$
The potentially dangerous step is the identification of $\angle BAO$ and $\angle ABO$ with $90^\circ-C$; this must be justified carefully from the geometry of a circumcenter.
Problem Understanding
We are given a triangle $ABC$ with angle bisector $AK$, where $K\in BC$. The circumcenter of $\triangle ABC$ coincides with the incenter of $\triangle ABK$. The task is to determine all possible angles of $\triangle ABC$.
This is a Type A problem. We must determine all angle triples satisfying the condition and prove that no other triangle can satisfy it.
The core difficulty is translating the coincidence of the two centers into angle relations. The circumcenter gives relations involving radii and chords of $\triangle ABC$, while the incenter gives bisector relations in $\triangle ABK$. Their combination should determine the angles uniquely.
The expected answer is
$$(A,B,C)=(72^\circ,36^\circ,72^\circ).$$
The reason is that the common center lies on angle bisectors of $\triangle ABK$ and simultaneously on perpendicular bisectors associated with $\triangle ABC$, producing two independent linear relations among the angles.
Proof Architecture
The first lemma is that
$$\angle BAO=90^\circ-C.$$
This follows because $OA=OB$, so $\triangle AOB$ is isosceles, and the central angle subtending chord $AB$ equals $2C$.
The second lemma is that
$$\angle BAO=\frac A4.$$
This follows because $O$ is the incenter of $\triangle ABK$, hence lies on the bisector of $\angle BAK$, and $\angle BAK=\frac A2$.
Combining the first two lemmas yields
$$C=90^\circ-\frac A4.$$
The third lemma is that
$$\angle ABO=\frac B2.$$
This follows from the incenter property in $\triangle ABK$ and the equality $\angle ABK=B$.
The fourth lemma is that
$$\angle ABO=90^\circ-C.$$
This is another application of the circumcenter property to the isosceles triangle $AOB$.
Combining the third and fourth lemmas yields
$$\frac B2=90^\circ-C.$$
The hardest step is deriving the equalities
$$\angle BAO=\angle ABO=90^\circ-C$$
from the circumcenter condition without introducing an incorrect angle-chord relation.
Solution
Let $O$ be the common center, namely the circumcenter of $\triangle ABC$ and the incenter of $\triangle ABK$.
Since $O$ is the circumcenter of $\triangle ABC$, we have
$$OA=OB.$$
Hence $\triangle AOB$ is isosceles.
The central angle subtending chord $AB$ is
$$\angle AOB=2\angle ACB=2C.$$
Therefore
$$2\angle BAO+\angle AOB=180^\circ,$$
which gives
$$2\angle BAO+2C=180^\circ.$$
Thus
$$\angle BAO=90^\circ-C.$$
Since $\triangle AOB$ is isosceles,
$$\angle ABO=\angle BAO=90^\circ-C.$$
Now use the fact that $O$ is the incenter of $\triangle ABK$.
Because $AK$ is the bisector of $\angle A$ in $\triangle ABC$,
$$\angle BAK=\frac A2.$$
The incenter lies on the bisector of every angle of the triangle, so $AO$ bisects $\angle BAK$. Hence
$$\angle BAO=\frac12\angle BAK=\frac A4.$$
Comparing with the previously obtained value of $\angle BAO$,
$$\frac A4=90^\circ-C.$$
Therefore
$$C=90^\circ-\frac A4.$$
Since $K$ lies on $BC$,
$$\angle ABK=\angle ABC=B.$$
Because $BO$ bisects the angle at $B$ of $\triangle ABK$,
$$\angle ABO=\frac B2.$$
Using the circumcenter relation already established,
$$\angle ABO=90^\circ-C.$$
Hence
$$\frac B2=90^\circ-C.$$
Combining this with
$$90^\circ-C=\frac A4$$
gives
$$\frac B2=\frac A4,$$
so
$$A=2B.$$
Substituting into
$$A+B+C=180^\circ$$
and
$$C=90^\circ-\frac A4,$$
we obtain
$$A+B+90^\circ-\frac A4=180^\circ.$$
Since $A=2B$,
$$2B+B+90^\circ-\frac B2=180^\circ.$$
Thus
$$\frac52B=90^\circ,$$
and therefore
$$B=36^\circ.$$
Consequently
$$A=72^\circ, \qquad C=72^\circ.$$
These angles indeed satisfy all derived necessary relations, so the solution is unique.
$$\boxed{\angle A=\angle C=72^\circ,\ \angle B=36^\circ}$$
Verification of Key Steps
The first delicate step is the computation of $\angle BAO$. Since $O$ is the circumcenter, the central angle subtending chord $AB$ equals $2C$. The triangle $AOB$ is isosceles because $OA=OB$. Hence
$$\angle BAO=\frac{180^\circ-2C}{2}=90^\circ-C.$$
A careless argument might confuse $\angle BAO$ with the angle between a chord and a tangent. No tangent is involved here; the conclusion comes solely from the isosceles triangle $AOB$.
The second delicate step is the use of the incenter condition at vertex $A$. The angle bisected by $AO$ is not $\angle A$ of $\triangle ABC$ but $\angle BAK$ of $\triangle ABK$. Since $AK$ already bisects $\angle A$ of $\triangle ABC$,
$$\angle BAK=\frac A2,$$
and only then does
$$\angle BAO=\frac A4$$
follow.
The third delicate step is identifying
$$\angle ABK=B.$$
This is valid because $K$ lies on segment $BC$, so the rays $BK$ and $BC$ coincide. If one replaced $\angle ABK$ by $\frac B2$, the entire computation would become incorrect.
Alternative Approaches
A different route uses distances from the common center. Let $R$ be the circumradius of $\triangle ABC$. Since $O$ is also the incenter of $\triangle ABK$, the distance from $O$ to each side of $\triangle ABK$ equals the inradius of that triangle. The distance from the circumcenter of $\triangle ABC$ to line $AB$ is $R\cos C$, while the distance to line $AK$ is $R\cos(A/2)$ because $O$ lies on the bisector of $\angle A$. Equality of these distances gives
$$\cos C=\cos\frac A2.$$
A similar comparison involving side $BK$ yields
$$\cos C=\sin\frac B2.$$
These relations lead to
$$C=90^\circ-\frac A4,\qquad \frac B2=90^\circ-C,$$
and the same angle values follow.
The main approach is preferable because it uses only elementary angle chasing and the standard properties of circumcenters and incenters, avoiding any metric computations.