Yes, that is a beautiful puzzle. It actually works if you replace 'open' with 'closed'. For a triangle to be acute, we require x^2, y^2 and z^2 to be the sidelengths of a triangle. So, let X' be a subset of size 12 of the interval [1, 144], obtained by squaring each element of X. Order X' and call the elements x_1 < x_2 < ... < x_12. Assume there are no triangles with sidelengths in X'. This gives us the following set of weak inequalities: x_1 + x_2 <= x_3 x_2 + x_3 <= x_4 x_3 + x_4 <= x_5 ... x_10 + x_11 <= x_12 By substituting the first inequality into the second, and the first two into the third, and so on, we obtain this: 89 x_1 + 55 x_2 <= x_12 But, we still have another strict inequality! x_1 < x_2 --> 144 x_1 < x_12 As x_1 is at least 1, x_12 is strictly greater than 144. Contradiction. Sincerely, Adam P. Goucher http://cp4space.wordpress.com
----- Original Message ----- From: Dan Asimov Sent: 12/26/12 05:10 AM To: math-fun Subject: [math-fun] Puzzle I heard
Not a killer, but cute:
Suppose X is a subset of size 12 of the open interval (1,12).
Show there must exist distinct x, y, z in X such that there is an acute-angled triangle of sides x, y, z.
--Dan
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