A Perimeter Puzzle (Posted on 2006-10-31)

	Submitted by K Sengupta
Rating: 3.2500 (4 votes)
Solution:	(Hide)
For positive integers m, p and q with p> q; let m(p^2 – q^2) and 2mpq be the lengths of two sides of the triangle(in cm), excluding the hypotenuse which is m(p^2+q^2) cm . Then, m(p^2+q^2) = 65(given) Let r be the inradius of the triangle. Now, Area of the triangle (A) = m^2 * p*q*(p^2 – q^2) and semi-perimeter of the triangle(s) = m/2* (p^2 – q^2 + 2pq + p^2 + q^2) = mp(p+q)----------(#) So, r = A/s = mq(p+q) Or, mq(p-q) = 12, so that: p = 12/mq + q; with the restriction m(p^2+q^2) = 65. m , p and q are positive integers, so m must divide 65 and accordingly, m = 1, 5, 13, 65. Clearly, (m,p,q) = (1,7,4);(1,8,1);(5,3,2); (13,2,1) corresponds to all possible positive integer solution of m(p^2+q^2) = 65. Of the four triplets, only (m,p,q) = (1,7,4) satisfies p = 12/mq + q since, 12/4 +4 =7 Hence, from (#), the required perimeter of the triangle = 2mp(m+q) = 2*1*7*11 = 154 cm. --------------------------------------------------------------------------------- A much shorter methodology is submitted by Bractals in this location .

	Subject	Author	Date
	Solution	Bractals	2006-10-31 10:21:15
	Solution	JLo	2006-10-31 07:42:28