Problem 716

2. Let circle $k$ touch sides $AD, CD$, and $BC$ at points $J, F$, and $K$, respectively. Let $C_{0}$ and $D_{0}$ be the feet of the perpendiculars from $C$ and $D$ to $AB$, respectively.

Let $r$ be the radius of circle $k$. Denote $x = DJ$. From the equality of tangent segments from a point to a circle, we have $AJ = AE = 15$, $BK = BE = 10$, $DF = DJ = x$, and $CK = CF = 8 - x$. We also have $D_{0}E = DF = x$ and $C_{0}E = CF = 8 - x$. From

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Ok 2018 2B 2

By applying the Pythagorean theorem to $\triangle ADD_{0}$ and $\triangle BCC_{0}$, we have $AD_{0}^{2} + DD_{0}^{2} = AD^{2}$ and $BC_{0}^{2} + CC_{0}^{2} = BC^{2}$, i.e.,

$$(15 - x)^{2} + (2r)^{2} = (15 + x)^{2}$$

and

$$(10 - (8 - x))^{2} + (2r)^{2} = (10 + (8 - x))^{2}$$

The first equation simplifies to $225 - 30x + x^{2} + 4r^{2} = 225 + 30x + x^{2}$, i.e., $4r^{2} = 60x$, and thus $x = \frac{r^{2}}{15}$. The second equation simplifies to $4 + 4x + x^{2} + 4r^{2} = 324 - 36x + x^{2}$, i.e., $40x + 4r^{2} = 320$, and thus $x = \frac{320 - 4r^{2}}{40} = \frac{80 - r^{2}}{10}$. Therefore, $\frac{r^{2}}{15} = \frac{80 - r^{2}}{10}$, which simplifies to $10r^{2} = 1200 - 15r^{2}$, and from this we calculate $r = \sqrt{\frac{1200}{25}} = \sqrt{48} = 4\sqrt{3}$.