#20 of June 2017 Grade 10 CSAT (Korean SAT) mock test

Like the picture above, let $\text{Q, R}$ be midpoints of $\overline{\text{AP}}, \overline{\text{BP}}$ respectively.

Also, $\overline{\text{AP}}$ , $\overline{\text{BP}}$ and $\overline{\text{AB}}$ contact with circle $O$ at $\text{S, T, H}$ respectively.

Midpoint of $\overline{\text{AB}}$ is $\text{C}$ .

Radii of $O$ , $O_1$ and $O_2$ are $r$ , $r_1$ and $r_2$ respectively.

Let $\overline{\text{PQ}}=a$ and $\overline{\text{PR}}=b$ .

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Since $\overline{\text{CQ}}=\overline{\text{PR}}$ and $\overline{\text{CR}}=\overline{\text{PQ}}$ ,

$r_1=\dfrac{5-b}{2}$ , $r_2=\dfrac{5-a}{2}$ .

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Meanwhile, since $\overline{\text{AS}}=\overline{\text{AH}}$ and $\overline{\text{BT}}=\overline{\text{BH}}$ ,

$(2a-r)+(2b-r)=5\times2;$

$r=a+b-5$ .

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The sum of the areas of $O$ , $O_1$ and $O_2$ is equal to:

$\pi\left\{\left(\frac{5-b}{2}\right)^2+\left(\frac{5-a}{2}\right)^2+(a+b-5)^2\right\}\text{ }\cdots\text{ }[A]$

Let $a+b=t\quad (5<t\leq5\sqrt{2})$ , and the expression $[A]$ is equal to:

$\begin{aligned} & \pi\left(\frac{a^2+b^2-10(a+b)+50}{4}+(t-5)^2\right) \\ & =\pi\left(\frac{75-10t}{4}+t^2-10t+25\right) \quad (\because a^2+b^2=\overline{\text{CP}}^2=25)\\ & =\pi\left(t^2-\frac{25}{2}t+\frac{175}{4}\right) \\ & =\pi\left(t-\frac{25}{4}\right)^2+\frac{75}{16}\pi \end{aligned}$

Therefore, $S=\dfrac{75}{16}\pi$ .

$\dfrac{144S}{\pi}=144\times\dfrac{75}{16}=\boxed{675}$ .