5. The cosine of the angle between the lateral sides $A D$ and $B C$ of trapezoid $A B C D$ is 0.8. A circle is inscribed in the trapezoid, and the side $A D$ is divided by the point of tangency into segments of lengths 1 and 4. Determine the length of the lateral side $B C$ of the trapezoid.
Answer: 4 or $\frac{100}{7}$.

Solution: Let $S$ be the intersection point of lines $A D$ and $B C$; $K, L, M$ be the points of tangency of the inscribed circle with sides $A B, A D$, and $C D$ respectively, and $O$ be its center. Then $O K \perp A B, O M \perp C D$, as radii, and since $A B \| C D$, points $K, O, M$ lie on the same line, meaning $KM$ is a diameter. The problem's conditions correspond to two possible cases for the position of point $L$ on side $A D$.

1) In this case, $A L=1, D L=4$ (Fig.2). Then $A K=A L=1, D M=D L=4$. Drop $A N \perp C D$. Considering that $A N \| K M$, we get $N M=1, D N=3$. In the right triangle $A D N$, the hypotenuse is $A D=5$, and the leg $D N=3$, so $A N=4$. Therefore, $\cos \angle D A N = \frac{4}{5}$, and given the problem's conditions, we conclude that $\angle D A N = \angle D S C$, and line $S C \parallel A N$. Thus, $\angle C$ of the trapezoid is a right angle, and the length $B C = A N = 4$.
2) In this case, $A L=4, D L=1$ (Fig.3). As before, the length of the perpendicular $A N=4, D N=3$, $\cos \angle A D N=\frac{3}{5}, \sin \angle A D N=\frac{4}{5}$. The acute angle $A D N$ is an external angle for the trapezoid and triangle $D S C$, so $\angle A D N = \angle C + \angle S$. Given that $\cos \angle S=\frac{4}{5}$, we find $\sin \angle S=\frac{3}{5}$, $\sin \angle C=\sin (\angle A D N - \angle S)=\frac{7}{25}$. Since $A N = B C \cdot \sin \angle C$, then $B C = \frac{100}{7}$.

## MINISTRY OF SCIENCE AND HIGHER EDUCATION OF THE RF COUNCIL OF RECTORS OF UNIVERSITIES OF TOMSK REGION OPEN REGIONAL INTERUNIVERSITY OLYMPIAD 2018-2019 MATHEMATICS (9th GRADE) FINAL STAGE 2nd VARIANT (ANSWERS)