Question

In triangle DEF, DE = DF = 100, and EF = 56. Circle Q has radius 16 and is tangent to segment DF and segment EF. Circle S is externally tangent to Circle Q and tangent to segments DE and EF. No point of circle S lies outside of triangle DEF. The radius of circle S can be expressed in the form a-b√c where a, b, and c are positive integers and c is the product of distinct primes. Find bc - a. a) 35 b) 166 c) 195 d) 254 e) None of the above

Answer 1

In triangle DEF, DE = DF = 100, and EF = 56. Circle Q has radius 16 and is tangent to segment DF and segment EF. Circle S is externally tangent to Circle Q and tangent to segments DE and EF. The radius of circle S can be found using similar triangles and the Pythagorean Theorem.

Step-by-step explanation:

To find the radius of circle S, we can use the concept of tangents and the properties of similar triangles. Let's label the center of circle Q as O and the center of circle S as P. Since circle Q is tangent to segment DF and EF, angle DFE is a right angle. Thus, triangle DEF is isosceles with DE = DF = 100.

Since circle S is tangent to DE and EF, OP will be perpendicular to EF and will bisect EF. Let M be the midpoint of EF. Triangles OMP and EMO are similar. Since OM = 16, we can find EM by using the Pythagorean Theorem on triangle EMD:

EM^2 + DM^2 = DE^2
EM^2 + (EM+56)^2 = 100^2
2EM^2 + 112EM + 56^2 - 100^2 = 0
EM = -14 or EM = 50

Since EM cannot be negative, EM = 50.

Now, since OMP and EMD are similar triangles, we have:
(EM/OP) = (DM/OM)
(50/OP) = 56/16
50OP = 56*16
OP = 56/16 * 1/50 = 7/25

Therefore, the radius of circle S, which is OP, is 7/25. Thus, a=7, b=25, and c=1, so bc - a = 25*1 - 7 = 18.

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