9.5. Does there exist thirteen consecutive natural numbers such that their sum is a 2021st power of a natural number?
Answer. They exist.

Solution. Let's denote 13 consecutive numbers as $N-6, N-5$, ..., $N+5, N+6$. Then their sum is $13N$. If $N=13^{2020}$, then the sum will be $13N = 13 \cdot 13^{2020} = 13^{2021}$.

Comment. A correct answer without justification - 0 points.

9.6-1. On a given circle $\omega$, points $A, B$, and $C$ are chosen such that angle $ABC$ is $70^{\circ}$. Let $D$ be the point of intersection of the angle bisector of $ABC$ with the circle $\omega$, and let point $K$ be symmetric to point $A$ with respect to point $D$. The line $KC$ intersects the circle again at point $P (P \neq C$, point $C$ lies on the segment $KP)$. Find the angle $ADP$.

Answer. $90^{\circ}$.

Solution. We will prove that $PD$ is perpendicular to $AK. AD = DK$, so we need to prove that $\alpha = \angle APD = \beta = \angle KPD$. But $\alpha = \angle APD = \angle ABD$ (inscribed, subtend arc $AD$), $\beta = \angle CPD = \angle CBD$ (inscribed, subtend arc $CD$). Finally, $\angle ABD = \angle CBD$, since $BD$ is the bisector of $\angle ABC$. The statement is proved.

![](https://cdn.mathpix.com/cropped/2024_05_06_41077738e9467cb84c4cg-6.jpg?height=622&width=870&top_left_y=297&top_left_x=730)

Comment. A correct answer without justification - 0 points.

9.6-2. On a given circle $\omega$, points $A, B$, and $C$ are chosen such that angle $ABC$ is $65^{\circ}$. Let $D$ be the point of intersection of the angle bisector of $ABC$ with the circle $\omega$, and let point $K$ be symmetric to point $A$ with respect to point $D$. The line $KC$ intersects the circle again at point $P (P \neq C$, point $C$ lies on the segment $KP)$. Find the angle $ADP$.

Answer. $90^{\circ}$.

Solution. We will prove that $PD$ is perpendicular to $AK. AD = DK$, so we need to prove that $\alpha = \angle APD = \beta = \angle KPD$. But $\alpha = \angle APD = \angle ABD$ (inscribed, subtend arc $AD$), $\beta = \angle CPD = \angle CBD$ (inscribed, subtend arc $CD$). Finally, $\angle ABD = \angle CBD$, since $BD$ is the bisector of $\angle ABC$. The statement is proved.

![](https://cdn.mathpix.com/cropped/2024_05_06_41077738e9467cb84c4cg-6.jpg?height=619&width=874&top_left_y=2049&top_left_x=728)

Comment. A correct answer without justification - 0 points.