3. Point $O$ is the center of the circumcircle of triangle $A B C$. Points $Q$ and $R$ are chosen on sides $A B$ and $B C$ respectively. Line $Q R$ intersects the circumcircle of triangle $A B R$ again at point $P$ and intersects the circumcircle of triangle $B C Q$ again at point $S$. Lines $A P$ and $C S$ intersect at point $K$. Find the angle between lines $K O$ and $Q R$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-42.jpg?height=497&width=577&top_left_y=271&top_left_x=754)

Solution. Note that $\angle C S Q=\angle C B Q$ and $\angle A P R=\angle A B R$ as inscribed angles subtending the same arc. Therefore, $\angle K S P=\angle K P S$, which means triangle $P K S$ is isosceles. Since

$$
180^{\circ}-\angle A K C=2 \angle K S P=2 \angle A B C=\angle A O C
$$

quadrilateral $A O C K$ is cyclic. Since triangle $A O C$ is isosceles, we have

$$
\angle A K O=\angle A C O=\angle C A O=\angle C K O.
$$

Therefore, line $K O$ is the angle bisector of isosceles triangle $P K S$, and thus also its altitude. Hence, $K O \perp Q R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-42.jpg?height=52&width=1587&top_left_y=1276&top_left_x=243)
different decimal digits. This product is a six-digit number, its extreme digits are equal, and between them are two pairs of identical adjacent digits. What is written on the board?

Answer: $633556=847 \cdot 748$.

Solution. Since the problem does not change when I and K are swapped, we will assume $I \geqslant K \cdot 10000 > m \cdot 100000$.

From this, $d \geqslant m$, and equality is not possible, otherwise $I \cdot K \vdots 11$. Additionally, the inequality $p < (I+1)(K+1) \cdot 10000$ holds, from which we get

$$
m < d \leqslant \left[\frac{I \cdot K + I + K + 1}{10}\right] = m + \left[\frac{d + I + K + 1}{10}\right]
$$

From the condition, it follows that $p$ is divisible by 11. Then $\overline{KC I}$ or $\overline{I C K}$ is divisible by 11, which is equivalent to $I + K - C \vdots 11$. Therefore, either $C = I + K$ or $C = I + K - 11$. Let's consider these cases.

1) Suppose $C = I + K$. Since $d + I + K + 1 = d + 1 + C < 20$, from (*) we get $d = m + 1$, which means $I \cdot K = 11m + 1$. Additionally, $I + K \leqslant 9$, from which $I \cdot K \leqslant 20$. Therefore, $I \cdot K = 12$, and the pair (I, K) is $(2,6)$ or $(3,4)$. Since $682 \cdot 286 = 195052$ and $473 \cdot 374 = 176902$, these cases do not fit.
2) Suppose $C = I + K - 11$. Since $d + I + K + 1 \leqslant 27 < 30$, from (*) we get $d = m + 1$ or $d = m + 2$, from which $I \cdot K \bmod 11$ is 1 or 2. Additionally, $I + K \geqslant 11$, from which $I \cdot K \geqslant 18$. Therefore, $I \cdot K$ can take values $24, 35, 45, 56$, and the pair (I, K) can be $(3,8), (5,7), (5,9), (7,8)$. Since

$$
308 \cdot 803 = 247324, \quad 715 \cdot 517 = 369655, \quad 935 \cdot 539 = 503965 \quad 847 \cdot 748 = 633556
$$

only the last case fits.