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

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

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.

 

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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Real Surds - Problem 2 Pre RMO 2017

Problem

Suppose (a, b) are positive real numbers such that (a \sqrt{a}+b \sqrt{b}=183 . a \sqrt{b}+b \sqrt{a}=182). Find (\frac{9}{5}(a+b)).

Hint 1

This problem will use the following elementary algebraic identity:

$(x+y)^3=x^3+y^3+3 x^2 y+3 x y^2$

Can you identify what is x and what is y?

Hint 2

background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none"] Set $x=\sqrt{a}, y=\sqrt{b}$. Then the given information translates to

$$
x^3+y^3=183, x^2 y+x y^2=182
$$

This implies $(x+y)^3=(\sqrt{a}+\sqrt{b})^3=x^3+y^3+3\left(x^2 y+x y^2\right)=183+3 \times 182=729$ Finally taking cube root on both sides, we have $\sqrt{a}+\sqrt{b}=9$

Hint 3

Note that $\sqrt{a} b+a \sqrt{b}=182 \Rightarrow \sqrt{a} \sqrt{b}(\sqrt{a}+\sqrt{b})=182 \Rightarrow \sqrt{a b} \times 9=182$ So at this point we know $(\sqrt{a}+\sqrt{b})=9, \sqrt{a b}=\frac{182}{9}$. It should be easy to find the value of $\left.\frac{9}{( } 5\right)(a+b)$ from these relations.

Final Answer

$a+b=(\sqrt{a}+\sqrt{b})^2-2 \sqrt{a b}=9^2-2 \times \frac{182}{9}=\frac{365}{9}$

Hence $\frac{9}{5}(a+b)=\frac{9}{5} \times \frac{365}{9}=73$