Pythagorean Condition in An Isosceles Right Triangle: What is this about?
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Copyright © 1996-2012 Alexander Bogomolny
The applet attempts to suggest the following problem:
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Consider the right isosceles triangle ABC and the points M, N on the hypothenuse BC in the order B, M, N, C such that |
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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,
ΔNCM' being right, we have
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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
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Copyright © 1996-2012 Alexander Bogomolny
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