Find all positive integers x, y, and z satisfying x 2 + y 2 + z 2 = 2xyz. (As usual, give solutions in comments to this post. On Friday I'll post some combination of my own solution and the most elegant approaches suggested by others.) Edited to add solution : If x 2 + y 2 + z 2 = 2xyz, then x 2 + y 2 + z 2 must be even. Thus there are two cases to consider: 1. One of x, y, z is even and two are odd. 2. All three of x, y, and z are even. Start with case 1. Assume (without loss of generality) that x is even, so x = 2a. Then we have 4a 2 + y 2 + z 2 = 4ayz. But then y 2 + z 2 must be a multiple of 4. But odd numbers are always 1 mod 4, so this is impossible. Thus case 1 can't be achieved. Now consider case 2. We have x=2a, y=2b, and z=2c. Thus 4a 2 + 4b 2 + 4c 2 = 16abc, or a 2 + b 2 + c 2 = 4abc. So let's now focus on this new equation. Again we need the sum to be even, so we need either one or three even numbers. But one even and two odd...
My quick sketch of a proof, which hopefully is correct:
ReplyDelete(1) Use SSS congruence to show that triangles ACF and CAD are congruent, which (along with BAC being isosceles) means that angles A and C of the hexagon are congruent. From symmetry, it follows that C and E are also congruent. So, angles A, C, and E of the hexagon are congruent.
(2) Use SAS congruence to show that BD=DF=FB, which means that triangle BDF is equilateral. Similarly, triangle ACE is also equilateral.
(3) Use symmetry to show that the incenter of BDF is equidistance from the 6 sides of hexagon ABCDEF.