Problem 1185

Answer: b) $\frac{39}{49}$.

Solution. It is not difficult to show that $ABCD$ is an isosceles trapezoid or a rectangle, so a circle (let's call it $\Omega$) can be circumscribed around $ABCD$. The diagonals of quadrilateral $CODT$ are bisected by their point of intersection, so it is a parallelogram, and $\angle CTD = \angle COD = 180^\circ - \angle AOD = 120^\circ$. Since $\angle CAD = 60^\circ$, in quadrilateral $CADT$ the sum of opposite angles is $180^\circ$, and a circle can also be circumscribed around it. Therefore, all 5 points $A, B, C, T$, and $D$ lie on the circle $\Omega$.

Angles $\angle ATB$ and $\angle ACB$ are inscribed in $\Omega$ and subtend the same arc, so they are equal, and $\angle ATB = 60^\circ$. Next, we note that

$$\begin{gathered} \angle DBT = \angle DCT \text{ (inscribed, subtend the same arc), } \\ \angle DCT = \angle BDC \text{ (since } BD \parallel CT), \\ \angle BDC = \angle BAC \text{ (the trapezoid is isosceles). } \end{gathered}$$

From this, it follows that $\angle ABT = \angle ABD + \angle DBT = \angle ABD + \angle BAC = 180^\circ - \angle AOB = 60^\circ$. Thus, it is proven that in triangle $ABT$ two angles are $60^\circ$, so it is equilateral.

b) By the cosine rule in triangle $ABO$, we find that $AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos 120^\circ = 2^2 + 5^2 + 2 \cdot 5 = 39$. Then the area $S_1$ of triangle $ABT$ is $AB^2 \cdot \frac{\sqrt{3}}{4} = \frac{39 \sqrt{3}}{4}$. The area of the trapezoid $S_2$ is found as half the product of the diagonals, multiplied by the sine of the angle between them: $S_2 = \frac{1}{2} \cdot 7 \cdot 7 \cdot \frac{\sqrt{3}}{2} = \frac{49 \sqrt{3}}{4}$. Therefore, $S_1 : S_2 = 39 : 49$.

## VARIANT 12. PART 1