Problem: In â–³ A B C , A B = 10 , A C = 8 \triangle A B C, A B=10, A C=8â–³ A B C , A B = 1 0 , A C = 8 and B C = 6 B C=6B C = 6 . Circle P PP is the circle with smallest radius which passes through C CC and is tangent to A B A BA B . Let Q QQ and R RR be the points of intersection, distinct from C CC , of circle P PP with sides A C A CA C and B C B CB C , respectively. The length of segment Q R Q RQ R is
Answer Choices:
A. 4.75 4.754 . 7 5
B. 4.8 4.84 . 8
C. 5 55
D. 4 2 4 \sqrt{2}4 2 ​
E. 3 3 3 \sqrt{3}3 3 ​
Solution:
Let N NN be the center of circle P PP . If T TT is the point of tangency of tangent A B A BA B and C H C HC H is the altitude to side A B A BA B of △ A B C \triangle A B C△ A B C , then C N + N T ⩾ C T > C H C N+N T \geqslant C T>C HC N + N T ⩾ C T > C H . Since radius C N C NC N is minimal, T = H T=HT = H and N NN is the midpoint of C H C HC H ; ie. C H C HC H is a diameter of circle P PP . Since inscribed ∠Q C R \angle Q C R∠Q C R is 9 0 ∘ , Q R 90^{\circ}, Q R9 0 ∘ , Q R is also a diameter. Finally, since △ C B H ∼ △ A B C \triangle C B H \sim \triangle A B C△ C B H ∼ △ A B C ,
C H 6 = 8 10 Q R = C H = 4.8. \begin{aligned}
\dfrac{C H}{6} & =\dfrac{8}{10} \\
Q R=C H & =4.8 .
\end{aligned}
6 C H ​ Q R = C H ​ = 1 0 8 ​ = 4 . 8 . ​