and that x+y+z=m/n, where m and n are positive integers, and n is not divisible by the square of any prime, find m+n.
Solution:
The radicals on the right side of the first equation are reminiscent of the Pythagorean Theorem. Each radical represents the length of a leg of a right triangle whose other leg has length 1/4, and whose hypotenuse has length y or z. Adjoin these two triangles along the leg of length 1/4 to create △XYZ with x=YZ,y=ZX, and z=XY, and with altitude to side YZ of length 1/4. Because of similar considerations in the other two equations, let the lengths of the altitudes to sides XZ and XY be 1/5 and 1/6, respectively. In the △XYZ thus created, the lengths x,y, and z of the sides satisfy the given equations, provided the altitudes of △XYZ are inside the triangle, that is, provided △XYZ is acute.
In general, a triangle the lengths of whose sides are a,b, and c is acute if and only if a2+b2>c2, where a≤b≤c. Denote the area of the triangle by K and the lengths of the altitudes to the sides of lengths a,b, and c by ha,hb, and hc, respectively. Then K=21aha=21bhb=21chc, so the condition a2+b2>c2 is equivalent to (1/ha)2+(1/hb)2>(1/hc)2, where 1/ha≤1/hb≤1/hc. Thus △XYZ is acute because 42+52>62.
Let K be the area of △XYZ. Then x=8K,y=10K, and z=12K, so x+y+z=30K. Apply Heron's Formula to obtain K2=15K⋅7K⋅5K⋅3K. Because K>0, conclude that K=1/(157). Then x+y+z=30K=2/7, so m+n=9.