3. Given an acute triangle $A B C$ that is not isosceles. Let $K$ be the intersection of the angle bisector of $\angle A C B$ with the perpendicular bisector of side $A B$. The line $C K$ intersects the altitudes from vertices $A$ and $B$ at points, which we will denote as $P$ and $Q$ respectively. Assume that triangles $A K P$ and $B K Q$ have the same area. Determine the measure of angle $\angle A C B$.


3. Let's denote the internal angles of triangle $ABC$ in the usual way. From the congruence of the peripheral angles $ACK$ and $BCK$ in the circumcircle of triangle $ABC$, it follows that the corresponding chords $AK$ and $BK$ are congruent, so point $K$ bisects the arc $AB$ that lies opposite vertex $C$ (Fig. 3). According to the theorem of peripheral angles, the sizes of angles $AKC$ and $BKC$ are, respectively, $\beta$ and $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c9c9e610010bfde2d888g-3.jpg?height=765&width=734&top_left_y=543&top_left_x=707)

Fig. 3

Let $V_{a}, V_{b}$ be the feet of the altitudes corresponding to vertices $A, B$ of triangle $ABC$. Since $ABC$ is an acute triangle, points $V_{a}$ and $V_{b}$ are interior points of the corresponding sides. The size of angle $APK$ is congruent to the size of the internal angle at vertex $P$ in the right triangle $CPV_{a}$, and is therefore equal to $90^{\circ}-\frac{1}{2} \gamma$. Analogously, angle $BQK$ also has the same size.

Triangles $AKP$ and $BKP$ have the same area, congruent sides $AK$ and $BK$, and thus also the heights to these sides, and they also coincide in the angle opposite them. From the construction of a triangle given a side, the height to this side, and the opposite internal angle, and from the symmetry of the constructed solutions, it follows that triangle $AKP$ is congruent either to triangle $KBQ$ or to triangle $BKP$. Since triangle $ABC$ is not isosceles (i.e., $\alpha \neq \beta$), triangle $AKP$ is congruent to triangle $KBQ$. The size of the internal angle at vertex $A$ of triangle $PAK$ is $180^{\circ}-\beta-\left(90^{\circ}-\frac{1}{2} \gamma\right)=90^{\circ}-\beta+\frac{1}{2} \gamma$, so from the stated congruence it follows that

$$
90^{\circ}-\beta+\frac{\gamma}{2}=\alpha \quad \text{or} \quad 90^{\circ}+\frac{\gamma}{2}=\alpha+\beta=180^{\circ}-\gamma
$$

From this, we get $\gamma=60^{\circ}$. Conversely, if $\gamma=60^{\circ}$, then $|\angle APK|=|\angle BQK|=60^{\circ}$ and triangles $AKP$ and $KBQ$ are congruent by the theorem $usu$, so they have the same area.

Conclusion: The angle $ACB$ has a size of $60^{\circ}$.

For a complete solution, award 6 points, of which 1 point for determining that $K$ is the midpoint of arc $AB$ (proof is not necessary if the student states that it is a known fact). Another 1 point for expressing the sizes of the internal angles in triangles $AKP$ and $BKP$ in terms of $\alpha, \beta, \gamma$. Award 2 points for the proof of the congruence of triangles $AKP$ and $BKP$, 1 point for calculating the angle $\gamma$, and 1 point for proving that the condition $\gamma=60^{\circ}$ implies the congruence (of areas) of triangles $AKP$ and $KBQ$.