Pythagorean Condition in An Isosceles Right Triangle: What is this about?
A Mathematical Droodle

Explanation

Copyright © 1996-2008 Alexander Bogomolny

The applet attempts to suggest the following problem:

Consider the right isosceles triangle ABC and the points M, N on the hypothenuse BC in the order B, M, N, C such that BM² + NC² = MN². Prove that MAN = 45°.

The configuration appears in a problem introduced in the National Olympiad 2001, in Romania, by Mircea Fianu.

Let's rotate ΔABC 90° counterclockwise around A. Since, by construction, BAC = 90° and AB = AC, ΔABC will map onto ΔACC', where CC' is equal and perpendicular to BC. M will map on M' on CC' such that MAM' = 90°.

ΔNCM' being right, we have

Therefore, NM' = NM and, by SSS, ΔAMN = ΔAM'N. So that angles MAN and NAM' are equal and both are equal 45°.

(This problem admits a nice generalization.)

References

1. W. G. Boskoff and B. D. Suceava, A Projectivity Characterized by the Pythagorean Relation, Forum Geometricorum, Volume 6 (2006) 187–190.

Copyright © 1996-2008 Alexander Bogomolny