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# 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!

Draw a diagram carefully.

Suppose the point of tangencies are at C and D. Join PC and PD.

Can you find two pairs of similar triangles?

$\Delta APC \sim \Delta AA_1B$

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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