Point in a Square
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Inside a square ABCD, a point P is selected such that AP = 1, BP = 2, CP = 3. Find  APB.
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The following solution is by Murray Klamkin.
Rotate the square through 90o around point B. Let P' be the image of P, A' that of A, and D' of D. Connect P and P'. In ΔBPP', BP = BP' and P'BP is right. The triangle is also isosceles so that BPP' = 45°. From the Pythagorean Theorem, PP'2 = 8.
In ΔAPP', PP'2 + AP2 = AP'2 (8 + 1 = 9). Therefore, by the converse of the Pythagorean Theorem, the triangle is right and P'PA = 90°.
Summing up, APB = BPP' + P'PA = 45° + 90° = 135°.
Reference
- R. Honsberger, More Mathematical Morsels, MAA, 1991, pp. 4-5.
Copyright © 1996-2008 Alexander Bogomolny
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