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Circle in Circle - PRMO 2017 | Problem 27

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 \( \Omega_1 \).

Start with hints

Do you really need a hint? Try it first!

Hint 1

Draw a diagram carefully.

PRMO 2017 Problem 27

Hint 2

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

Can you find two pairs of similar triangles?

PRMO 2017 Problem 27 Hint 2

Hint 3

\( \Delta APC \sim \Delta AA_1B \)


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.


Hint 4

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

OC  = r (radius of the small circle).

We already know OP = 10, \( A_1 B = 5, AB_1 = 15\)

PRMO 2017 Problem 27 Hint 2

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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