Given the circle Γ with P Q is the diameter of the circle Γ . The point R lies outside the circle Γ so that the P R and Q R lines intersect circles Γ successively at points X and Y . Given P Q = 3 0 units and P X : P R = Q Y : Q R = 1 : 3 . If the area of the △ P Q R can be expressed in the form a b unit area where a , b ∈ N , find the value of a + b .
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First, we show that P R = Q R . By the problem condition, we have R X = 3 2 P R and R Y = 3 2 R Q . Applying Power of a Point to R yields
R X ( R P ) 3 2 R P 2 R P = R Y ( R Q ) = 3 2 R Q 2 = R Q .
Thus, △ P Q R is isosceles. Let P X = Q Y = a , so that X R = Y R = 2 a . Draw X Q , and note that ∠ P X Q = 9 0 ∘ , since it subtends a semicircle arc. By applying Pythagorean Theorem on right triangle R X Q , we can find X Q in terms of a :
R X 2 + X Q 2 ( 2 x ) 2 + X Q 2 X Q = Q R 2 = ( 3 x ) 2 = x 5 .
Doing the same for △ P X Q gives us a value for a 2 :
P X 2 + X Q 2 a 2 + ( a 5 ) 2 6 a 2 a 2 = P Q 2 = 3 0 2 = 9 0 0 = 1 5 0 .
Therefore, the area of △ P Q R is
2 ( P R ) ( X Q ) = 2 3 a 2 5 = 2 3 ( 1 5 0 ) 5 = 2 2 5 5 ,
and m + n = 2 2 5 + 5 = 2 3 0 .
notice by symmetry that Δ P Q R is iscoceles. let O be the center of the circle, then O R = ( 3 x ) 2 − 1 5 2 = 3 x 2 − 5 2 is the altitude. applying power of a point to R, 2 x ( 3 x ) = ( 3 x 2 − 5 2 − 1 5 ) ( 3 x 2 − 5 2 + 1 5 ) → 6 x 2 = 9 x 2 − 2 2 5 − 2 2 5 → x = 1 5 0 we get the area [ Δ P Q R ] = 1 5 × 3 x 2 − 5 2 = 2 2 5 5
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C h o r d X Y i s ∣ ∣ P Q s i n c e R X / R P = R Y / R Q = 2 / 3 . ∴ X Y = 2 0 . ⟹ ⊥ d i s t a n c e b e t w e e n X Y a n d P Q = 1 5 2 − 1 0 2 = 5 5 . S o a l t i t u d e f r o m R t o P Q = 3 ∗ 5 5 . A r e a Δ R P Q = 2 1 ∗ 3 0 ∗ 1 5 5 = 2 2 5 5 = a b . a + b = 2 3 0 .