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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 30 In the coordinate plane, letABbe a segment of length 10 √ 2, and consider the circle with diameterAB. PointsPandQlie on this circle and satisfy ( − →PA+− →PB)·(− →PQ+− →PB) = 2

− →PQ

2 . If

− →PB

= 14 and

− →QB

0, then

− →PA· − →QB

= q p, wherepandqare coprime natural numbers. Computep+q. [4 points] 19 더보기

Let the circle have center 𝑂 O at the origin and radius 5 2 5 2 ​

. Place 𝐴

( − 5 2 , 0 ) A=(−5 2 ​

,0), 𝐵

( 5 2 , 0 ) B=(5 2 ​

,0). From ∣ 𝑃 𝐵 ∣

14 ∣PB∣=14 and 𝑃 P on the circle, we get 𝑃

( − 24 2 5 , ± 7 2 5 ) P=(− 5 24 2 ​

​

,± 5 7 2 ​

​

).

Using 𝑃 𝐴 ⃗ + 𝑃 𝐵 ⃗

− 2 𝑂 𝑃 ⃗ PA + PB =−2 OP and 𝑃 𝑄 ⃗ + 𝑃 𝐵 ⃗

𝑄 ⃗ + 𝐵 ⃗ − 2 𝑃 ⃗ PQ ​

PB

Q ​

B −2 P , the given equation reduces to

𝑃 ⃗ ⋅ 𝑄 ⃗

5 2   𝑥 𝑃

− 48. P ⋅ Q ​

=5 2 ​

x P ​

=−48.

Solving for 𝑄 Q on the circle yields (discarding 𝑄

𝐵 Q=B)

𝑄

( 527 2 125 , − 336 2 125 ) Q=( 125 527 2 ​

​

,− 125 336 2 ​

​

)

when 𝑃 P is in the upper half-plane. Then

𝑃 𝐴 ⃗ ⋅ 𝑄 𝐵 ⃗

( − 2 5 , − 7 2 5 ) ⋅ ( 98 2 125 , 336 2 125 )

− 196 25 . PA ⋅ QB ​

=(− 5 2 ​

​

,− 5 7 2 ​

​

)⋅( 125 98 2 ​

​

, 125 336 2 ​

​

)=− 25 196 ​

.

Thus 𝑃 𝐴 ⃗ ⋅ 𝑄 𝐵 ⃗

𝑞 𝑝 PA ⋅ QB ​

= p q ​

with 𝑞

196 q=196, 𝑝

25 p=25 (coprime).

𝑝 + 𝑞

25 + 196

p+q=25+196=221.

\boxed{221}

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