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A point in a square (Posted on 2004-03-17) Difficulty: 3 of 5
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°.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
answerK Sengupta2007-05-12 13:41:24
Solutionan answerlogischer Verstand2004-03-22 18:56:25
Two possibilitiesBrian Smith2004-03-18 09:22:22
Solution (Excel-Aided)np_rt2004-03-17 22:19:04
the other halfCharlie2004-03-17 22:03:40
re: Half a solutionBarrett Hasseldine2004-03-17 20:53:19
SolutionHalf a solutione.g.2004-03-17 14:54:52
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