Problem:
In the figure below, semicircles with centers at A and B and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter JK. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at P is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at P?
Answer Choices:
A. 43
B. 76
C. 213
D. 852
E. 1211 Solution:
Let C be the center of the largest semicircle, and let r denote the radius of the circle centered at P. Note that PA=2+r, PC=3−r,PB=1+r,AC=1,BC=2, and AB=3. Let F be the foot of the perpendicular from P to AB, let h=PF, and let x=CF. The Pythagorean Theorem in △PAF,△PCF, and △PBF gives
h2=(2+r)2−(1+x)2=(3−r)2−x2=(1+r)2−(2−x)2
This reduces to two linear equations in r and x, whose solution is r=76,x=79.
OR
With the notation and observations above, apply Heron's Formula to △PCB and △PAC, noting that the former has twice the area of the latter and each has semiperimeter 3 . Thus
3⋅1⋅r⋅(r−2)=23⋅2⋅r⋅(r−1)
from which it follows that r=76.
OR
With the notation and observations above, let θ=∠PAB. The Law of Cosines applied to △PAC gives
(3−r)2=(2+r)2+12−2⋅(2+r)⋅1⋅cosθ
and simplifying yields (2+r)cosθ=5r−2. Applying the Law of Cosines to △PAB gives
(1+r)2=(2+r)2+32−2⋅(2+r)⋅3⋅cosθ
and simplifying yields 3(2+r)cosθ=r+6. Hence r+6=3(5r−2), so r=76.
