# Understand the problem

Let $$\Omega_1$$ be a circle with center O and let AB be a diameter of $$\Omega_1$$. Let P be a point on the segment OB different from O. Suppose another circle $$\Omega_2$$ with center P lies in the interior of $$\Omega_1$$. Tangents are drawn from A and B to the circle $$\Omega_2$$ intersecting $$\Omega_1$$ again at $$A_1$$ and $$B_1$$ respectively such that $$A_1$$ and $$B_1$$ are on the opposite sides of AB. Given that $$A_1B = 5, AB_1 = 15$$ and $$OP = 10$$, find the radius of $$\Omega_1$$.

##### Source of the problem

Pre Regional Math Olympiad India 2017, Problem 27

Geometry

Medium

##### Suggested Book

Challenges and Thrills of Pre- College Mathematics.

Do you really need a hint? Try it first!

$$\Delta APC \sim \Delta AA_1B$$

Why?

Notice that AC is perpendicular to $$AA_1$$ as the radius is perpendicular to the tangent.

Also $$\angle A$$ is common to both triangles. Hence the two triangles are similar (equiangular implies similar).

Similarly $$\Delta BPD \sim \Delta BAB_1$$.

Use the ratio of sides to find OA.

Suppose OA = R (radius of the big circle).

OC = r (radius of the small circle).

We already know OP = 10, $$A_1 B = 5, AB_1 = 15$$ Since $$\Delta AA_1B$$ and $$ACP$$ are similar we have $$\frac{AP}{AB} = \frac{PC}{A_1B}$$. This implies $$\frac{R+10}{2R} = \frac{r}{5}$$ (1)

Similarlly since $$\Delta BPD$$ and $$BAB_1$$ are similar we have $$\frac{BP}{BA} = \frac{PD}{AB_1}$$. This implies $$\frac{R-10}{2R} = \frac{r}{15}$$ (2)

Multiply the reciprocal of (2) with (1) to get R = 20.

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