Contribution for Puzzlet #062
From: Denis Borris [borrisd@ca.inter.net]
AB = x+1, AC = x-1, BC = x, AD = x-2:
A
B (x-a) D (a) C
(not to scale! worse than yours)
Let DC = a; then BD = x-a;
from the 2 right triangles ABD and ACD:
(x+1)^2 - (x-a)^2 = (x-1)^2 - a^2
solve to get: a = (x-4) / 2
Using triangle ADC and substituting above:
(x-2)^2 + [(x-4)/2]^2 = (x-1)^2
solve to get: x^2 - 16x + 28 = 0
factor that sumbitch: (x - 14)(x - 2) = 0
So: x = 14 and x = 2
So answer is 14: 2 rights, 9-12-15 and 5-12-13,
forming a 13-"14"-15 triangle of height 12.
By the way, 2 is also valid solution, since you did not
specify that area 0
Well, I'll be darned! I just didn't see that one coming! Very neat analysis. I'm almost tempted to withdraw the Puzzlet as the idea is to set problems that need a computer program to solve them. However, I think a lot of readers will still enjoying solving it the hard way. Thanks for the excellent input - I'll add it to the contributions page.
Dave.
| Site design/maintenance: Dave Ellis | E-mail
me! |
Last Updated: February 22nd, 2004. |