(5) - Outcoming

Geometry Level 4

M \text{M} and N \text{N} are points respectively on line segment BC \text{BC} and CD \text{CD} of ABCD \square{\text{ABCD}} , side a a such that MC = 6 \text{MC} = 6 and CN = 8 \text{CN} = 8 . If MN \text{MN} is a tangent of ( A , a ) (\text{A}, a) at P \text{P} , the value of CP \text{CP} is m n \dfrac{m}{\sqrt n} with n n is square - free. What is the value of m n 2 \dfrac{m}{n^2} .


The answer is 0.48.

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1 solution

MN=√(64+36)=10 MP=a-6, NP=a-8. This implies 2a-14=10, or a=12. The position coordinates of P are (7.2, 2.4) Therefore CP=12/√5. Thus m=12 and n=5. Therefore the required quantity is 12/25 or 0.48

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