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Let A B C be a triangle with A B=5, A C=4, B C=6 . The internal angle bisector of C intersects the side A B at D . Points M and N are taken on sides B C and A C , respectively, such that D M | A C and D N | B C . If (M N)^2= p q where p and q are relatively prime positive integers then what is the sum of the digits of |p-q| ?

Correct answer

02

Step-by-step solution

A D B D = A C B C = 2 3 = C M B M = A N C N So, C M= 2 5 6= 12 5 and, C M= 3 5 4= 12 5 aligned & C= B C^2+A C^2-A B^2 2 B C A C = C M^2+C M^2-M N^2 2 C N C M & 6^2+4^2-5^2 2 4 6 = ( 12 5 )^2+ ( 12 5 )^2-M N^2 2 ( 12 5 ) ( 12 5 ) & M N^2= 126 25 = p q aligned Sum of digits of |p-q|=2

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