All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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.

See The Solution Submitted by Bractals    
Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Vectorial solution, plus new theorems | Comment 3 of 10 |
Let's call x=B-A; y=C-B, and z=A-C. It holds that x+y+z=0. Let's also write v' for vector v rotated by 90 degrees.

Then, we have I=B+x', J=A+x', K=C+y', L=B+y', M=A+z', and N=C+z'. Looking at the parallelograms, since L-B=P-I, we have P=L+I-B; similarly, Q=K+N-C, and R=J+M-A.

We want to prove that C-R is equal and ortogonal to P-C. First, C-R= C-J-M+A= C-A-x'-A-z'+A= C-A+y'= -z+y'. Then, P-C= L+I-B-C= B+y'+B+x'-B-C= B-C-z'= -y-z'.

Now, rotating -z produces -z', and rotating y'  produces -y, so... QED.

PS. By the way, out of the intermediate results we can find other interesting "theorems". For example, A-R=y', so AR equals BC, and is ortogonal to it, and similar conditions apply to the other points.

  Posted by Federico Kereki on 2005-10-19 19:59:27
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information