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 A point in a square (Posted on 2004-03-17)
Given a square ABCD, let P be such that AP=1, BP=2 and CP=3.

* What is the length of DP?
* What is the angle APB?

 Submitted by Federico Kereki Rating: 3.1250 (8 votes) Solution: (Hide) First: √6. Applying Pythagoras, it's easy to prove that AP²+CP²= BP²+DP². Second: Assuming P is within ABCD (if it's not, see below) rotate the square 90°, with center B, so that A moves to A', D moves to D', C moves to A, and P moves to P'. Angle PBP'=90° by definition. Since PB=P'B, triangle PBP' is isosceles, and thus angle P'PB=45°. From Pythagoras, PP'=√8. As AP'=CP=3, it follows that APP' is a rectangular triangle, and angle PP'A=90°. Summing both results, angle APB=135°.If P lies outside the square, the same reasoning would apply, but then APB would be 90°-45°=45°.

 Subject Author Date answer K Sengupta 2007-05-12 13:41:24 an answer logischer Verstand 2004-03-22 18:56:25 Two possibilities Brian Smith 2004-03-18 09:22:22 Solution (Excel-Aided) np_rt 2004-03-17 22:19:04 the other half Charlie 2004-03-17 22:03:40 re: Half a solution Barrett Hasseldine 2004-03-17 20:53:19 Half a solution e.g. 2004-03-17 14:54:52

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