#30 of Sept 2017 Grade 10 CSAT (Korean SAT) Mock test

Geometry Level 5

There are two circles on the coordinate plane:

{ C 1 : x 2 + ( y − 4 ) 2 = 4 ; C 2 : ( x − 6 ) 2 + ( y − 4 + 6 3 ) 2 = 16. \cases{C_1~:~x^2+(y-4)^2=4; \\ \\ C_2~:~(x-6)^2+(y-4+6\sqrt{3})^2=16.}

P ( x 1 , y 1 ) P(x_1,~y_1) and Q ( x 2 , y 2 ) Q(x_2,~y_2) are points that move around on C 1 C_1 and C 2 , C_2, respectively. R ( 2 , y 3 ) R(2,~y_3) is a point that internally divides P Q ‾ . \overline{PQ}. The three points satisfy:

{ 0 ≤ x 1 ≤ 1 ; 2 P R ‾ = R Q ‾ ; y 1 ≤ 4 , y 2 ≥ 4 − 6 3 . \cases{0\le x_1 \le 1; \\ \\ 2\overline{PR}=\overline{RQ}; \\ \\ y_1 \le 4,~~y_2 \ge 4-6\sqrt{3}.}

The area of the locus of P Q ‾ \overline{PQ} is a − b π a-b\pi for some rational numbers a a and b . b.

Find the value of a + 9 b . a+9b.

This problem is a part of <Grade 10 CSAT Mock test> series .


The answer is 25.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Sep 22, 2017

It is clear that R ( 2 x 1 + x 2 3 , 2 y 1 + y 2 3 ) R\left(\dfrac{2x_1+x_2}{3},~\dfrac{2y_1+y_2}{3}\right) since R R divides P Q ‾ \overline{PQ} with a ratio of 1 : 2. 1:2.

Therefore, 2 x 1 + x 2 = 6 , 2x_1+x_2=6, leading to x 2 = 6 − 2 x 1 . x_2=6-2x_1.

Then, since P ( x 1 , y 1 ) P(x_1,~y_1) is on C 1 C_1 and Q ( 6 − 2 x 1 , y 2 ) Q(6-2x_1,~y_2) is on C 2 , C_2,

x 1 2 + ( y 1 − 4 ) 2 = 4 , ( − 2 x 1 ) 2 + ( y 2 − 4 + 6 3 ) 2 = 16. {x_1}^2+(y_1-4)^2=4,~(-2x_1)^2+(y_2-4+6\sqrt{3})^2=16.

Divide the second equation by 4 4 and then you'll see that

( y 1 − 4 ) 2 = ( y 2 2 − 2 + 3 3 ) 2 . (y_1-4)^2=\left(\dfrac{y_2}{2}-2+3\sqrt{3}\right)^2. Since y 1 ≤ 4 , y 2 ≥ 4 − 6 3 , y_1 \le 4,~~y_2 \ge 4-6\sqrt{3},

4 − y 1 = y 2 2 − 2 + 3 3 , 4-y_1=\dfrac{y_2}{2}-2+3\sqrt{3}, leading to 2 y 1 + y 2 3 = 4 − 2 3 . \dfrac{2y_1+y_2}{3}=4-2\sqrt{3}.

Then we know that R ( 2 , 4 − 2 3 ) . R(2,~4-2\sqrt{3}).

The picture on the left is the illustration of the locus of P Q ‾ . \overline{PQ}. R R is shown as C C in the picture.

A bit of geometry proves that ∠ A O 1 B = ∠ C O 2 D = 3 0 ∘ . \angle AO_1B=\angle CO_2D=30^{\circ}.

S 1 = △ O 1 A C − sector O 1 A B = 2 − π 3 . S_1=\triangle O_1AC - \text{sector }O_1AB=2 - \dfrac{\pi}{3}.

Since S 1 S_1 and S 2 S_2 are similar at a ratio of area of 1 : 4 1:4 , S 2 = 8 − 4 π 3 . S_2=8-\dfrac{4\pi}{3}.

Therefore, S 1 + S 2 = 10 − 5 π 3 . S_1+S_2=10-\dfrac{5\pi}{3}.

∴ a + 9 b = 10 + 9 ⋅ 5 3 = 25 . \therefore~a+9b=10+9\cdot\dfrac{5}{3}=\boxed{25}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Your solution:

R R divides P Q ‾ \overline{PQ} with a ratio of 1 : 2 1:2

indicates that

2 P R ‾ = R Q ‾ 2 \overline{PR}=\overline{RQ}

not

2 P R ‾ = P Q ‾ 2 \overline{PR}=\overline{PQ}

Could you please edit the problem accordingly.

Maria Kozlowska - 3 years, 7 months ago

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I can swear I wrote 2 P R ‾ = R Q ‾ 2\overline{PR}=\overline{RQ} ... oh well, sorry, and thank you for pointing out! ^^

Boi (보이) - 3 years, 7 months ago

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