5. Given is $\triangle A B C$, and a circle $\gamma$ is noted which contains point $A$, touches the circumcircle of $\triangle A B C$, and touches the line $B C$ at point $D$, where $B$ is between $C$ and $D$. If $\measuredangle B A C=\pi-\arcsin \frac{35}{37}, B C=70$ and $B D=10$, determine the radius of the circle $\gamma$.

## Fourth Grade - B Category
5. Let $O$ be the center and $r$ the radius of the circumcircle of $\triangle ABC$. Since $\arcsin \frac{35}{37} < \frac{\pi}{2}$, the angle at vertex $A$ is obtuse, so $O$ and $A$ are on different sides of line $BC$. Let $O_{1}$ be the center and $r_{1}$ the radius of the circle $\gamma$. Since $\gamma$ contains point $A$ and is tangent to the circumcircle of $\triangle ABC$, the point of tangency is precisely $A$, so we have $O O_{1} = r + r_{1}$. Applying the Law of Sines to $\triangle ABC$, we get $2r = \frac{BC}{\sin \angle BAC} = \frac{70}{\frac{35}{37}} = 74$, i.e., $r = 37$. Let $M$ be the foot of the perpendicular from $O$ to line $BC$ (which is also the midpoint of side $BC$), and let $K$ be the foot of the perpendicular from $O$ to line $O_{1}D$. Applying the Pythagorean Theorem to $\triangle OMB$, we get $OM^2 = 37^2 - 35^2 = 144$, i.e., $OM = 12$. We have $OK \parallel BC$, and $O$ and $K$ are on opposite sides of line $BC$ relative to point $O_{1}$. Since $OMDK$ is a rectangle, we have $DK = MO = 12$, so we calculate $O_{1}K = O_{1}D + DK = r_{1} + 12$, and $OK = MD = MB + BD = 35 + 10 = 45$. Applying the Pythagorean Theorem to $\triangle O_{1}KO$, we get $O_{1}O^2 = OK^2 + O_{1}K^2$, i.e., $(37 + r_{1})^2 = 45^2 + (12 + r_{1})^2$, which simplifies to $1369 + 74r_{1} = 2025 + 144 + 24r_{1}$, from which we get $r_{1} = \frac{800}{50} = 16$.

## Fourth Grade - B Category