6. Around an acute isosceles triangle $A M T$ with base $M T$, a circle $\Omega$ is described. The distance from the midpoint of the arc $A T$, not containing point $M$, to the side $A T$ is 3, and the distance from the midpoint of the arc $M T$, not containing point $A$, to the side $M T$ is 1.6. Find the radius of the circle $\Omega$ and the area of triangle $A M T$.
Answer: $R=5, S=\frac{168 \sqrt{21}}{25}$.

Solution. Let $Q$ and $L$ be the midpoints of arcs $M T$ and $A T$; $B$ and $H$ be the midpoints of segments $A T$ and $M T$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O L \perp A T, O Q \perp M T ; O B=O L-L B=R-3, O H=O Q-Q H=R-\frac{8}{5}$.

From the Pythagorean theorem for triangles $O B T$ and $O H T$, we get that $B T^{2}=R^{2}-(R-3)^{2}=6 R-9$, $H T^{2}=R^{2}-\left(R-\frac{8}{5}\right)^{2}=\frac{16 R}{5}-\frac{64}{25}$. We find the sides of triangle $A T H: A H=A O+O H=2 R-\frac{8}{5}$, $A T=2 B T=2 \sqrt{6 R-9}, D H=\sqrt{\frac{16 R}{5}-\frac{64}{25}}$.

By the Pythagorean theorem for triangle $A T H$, we get $\left(2 R-\frac{8}{5}\right)^{2}+\left(\frac{16 R}{5}-\frac{64}{25}\right)=24 R-36$, from which $5 R^{2}-34 R+45=0, R=5$ or $R=\frac{9}{5}$. Only $R=5$ fits (in the second case, $B O=R-3<0$).

Thus, $A H=\frac{42}{5}, D H=\frac{4 \sqrt{21}}{5}, S_{A M T}=A H \cdot T H=\frac{168 \sqrt{21}}{25}$.

“Phystech-2016”, mathematics, instructions for checking, tickets 17-20

The problem is considered fully solved (and the maximum possible number of points is awarded) only if all necessary transformations are provided in the solution text and all logical steps are fully explained, with the answers reduced to their simplest form.

The presence of a correct answer does not guarantee a positive score for the problem.
1.(4) It is written that $b_{2}^{2}=b_{1} b_{3}$
1 point, - no selection is made ......................................................................................................................
2.(6) The set of points is constructed ..... 4 points,
- its area is found ..... 2 points.
3.(5) Incorrect trigonometric formulas are applied
0 points for all subsequent actions.
4.(6) All possible variants for the last digit of the number are indicated ..... 1 point,
- for the formulation of divisibility criteria by 6 (by 15) ..... points are not added,
- if the answer is obtained by enumeration and is incorrect ..... no more than 1 point for the problem,
- if the answer is written in the form $6^{k} .8$ and so on, points are not deducted.
5.(6) The first equation is factored ..... 2 points,
- for each of the two resulting cases ..... 2 points.
6.(8) Only the radius or only the area is found ..... 6 points.