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Isosceles Right Triangles (Posted on 2005-10-19) Difficulty: 3 of 5
Let BAC be an arbitrary triangle with external squares ABIJ, BCKL, and CAMN. IBLP, KCNQ, and MAJR are parallelograms. Prove that PAQ, QBR, and RCP are isosceles right triangles.

  Submitted by Bractals    
Rating: 3.2500 (4 votes)
Solution: (Hide)
Let all the specified polygons be labeled clockwise.
Consider the capital letters as complex numbers and i = sqrt(-1).
We will make use of the fact that the product of a complex number
and plus i (minus i) is the complex number rotated CCW (CW) by a
right angle.

  Q - A = (Q - K) + (K - C) + (C - A)
        = (N - C) + (K - C) + (C - A)
        = (A - C)(-i) + (B - C)(i) + (C - A)
        = (C - A) + (B - A)(i)
        = [(B - A) - (C - A)(i)](i)
        = [(C - B)(-i) + (A - B)(i) + (B - A)](i)
        = [(L - B) + (I - B) + (B - A)](i)
        = [(P - I) + (I - B) + (B - A)](i)
        = (P - A)(i)

Therefore, PAQ is an isosceles right triangle. Similar arguments
can be made for QBR and RCP.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
hmmAlex Korn2005-10-30 21:16:58
re(4): Parting shotMcWorter2005-10-28 19:56:29
re(3): Parting shotCharlie2005-10-27 14:04:30
re(2): Parting shotMcWorter2005-10-27 13:10:52
re: Parting shotCharlie2005-10-26 14:55:59
SolutionParting shotMcWorter2005-10-25 19:55:10
SolutionA complex wayOld Original Oskar!2005-10-19 22:01:31
SolutionVectorial solution, plus new theoremsFederico Kereki2005-10-19 19:59:27
re: Analytic solution -- typosCharlie2005-10-19 14:02:19
Analytic solutionJer2005-10-19 13:36:12
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