Segments
P
A
,
P
B
, and
P
C
are all chords of the same circle,
Γ
, such that
P
A
=
5
,
P
B
=
8
, and
P
C
=
3
. In addition,
∠
A
P
B
≅
∠
B
P
C
.
The area of
Γ
can be written as
b
a
π
, where
a
and
b
are positive, coprime integers. Find
a
+
b
.
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The question asks to find
R
2
where
R
is the radius.
image
In the figure, ∠ O C P = ∠ O P C = θ and ∠ O P A = ∠ O A P = α .
Now,
2 R c o s ( θ ) = 3
2 R c o s ( α ) = 5
2 R c o s ( 2 θ − α ) = 8
So,
2 c o s 2 ( 2 θ − α ) − 1 = c o s ( θ − α ) = 2 ( R 4 ) 2 − 1
⟹ c o s ( θ ) c o s ( α ) + s i n ( θ ) s i n ( α ) = R 2 3 2 − 1
⟹ 2 R 3 2 R 5 + 2 R 4 R 2 − 9 2 R 4 R 2 − 2 5 = R 2 3 2 − 1
⟹ ( 4 R 2 − 9 ) ( 4 R 2 − 2 5 ) = 1 1 3 − 4 R 2 ⟹ R 2 = 7 6 8 1 2 5 4 4 = 3 4 9
⟹ π R 2 = π 3 4 9
Hence, 4 9 + 3 = 5 2
WOW!
Since angle APB = angle BPC, line AB = line BC. Let angle APB = x. Using cosine rule on triangle PAB and PBC, we get AB^2 = 73 - 48cosx = 89 - 80cosx. Solving it, we get x = 60 degrees and AB = 7. Using the extended sine rule on triangle PAB, we get 7/sin(60) = 2R where R is the circumradius of triangle PAB. Finally, we get a/b = R^2 = 49/3. a+b = 52.
Wow.
A neat solution
Correction: 7/cos(60) = 2R is incorrect. It should be 7/sin(60 degrees) = 2R
Thanks for the correction! :) My bad...
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∠ A P B = ∠ B P C ⇒ A B = B C ⇒ A B = B C
Let, A B = B C = x .
Then, by Ptolemy's Theorem,
P A ⋅ C B + P C ⋅ A B = P B ⋅ A C
⇒ 5 x + 3 x = 8 A C
⇒ 8 x = 8 A C
⇒ x = A C
∴ A B = B C = A C
So, △ A B C is equilateral.
∴ ∠ A P B = ∠ A C B = 6 0 ∘
Now, applying Cosine-Rule to △ A P B , we get,
A B 2 = P B 2 + P A 2 − 2 ⋅ P B ⋅ P A ⋅ cos ∠ A P B
= 8 2 + 5 2 − 2 . 8 . 5 cos 6 0 ∘
= 4 9
∴ A B = 7 .
Now, A B = 2 R ⋅ sin 6 0 ∘ ⇒ R = 3 7
∴ Area of Γ = π R 2 = π ⋅ ( 3 7 ) 2 = 3 4 9 π
∴ a + b = 4 9 + 3 = 5 2