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Let A B C D be a unit square. Suppose M and N are points on B C and C D respectively such that the perimeter of triangle M C N is 2 . Let O be the circumcentre of triangle M A N and P be the circumcentre of triangle M O N . If ( O P O A )^2= m n for some relatively prime positive integers m and n , find the value of m+n .

Correct answer

03

Step-by-step solution

O A= circumradius of A M N= M N 2 O P= circumradius of O M N= M N 2 2 So ( O P O A )^2= ( 1 2 )^2 Perimeter of M C N=2=(1-x)+(1-y)+M N M N=x+y Now rotate A B M about A so that A B overlaps with A D (by 90^ ) Clearly A M N= A M^ N So 2 =90^ =45^ Hence ( O P O A )^2= 1 2 = m n m+n=3

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