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

Home > Shapes > Geometry
Take The Midpoints, Get Equilateral? (Posted on 2007-08-01) Difficulty: 2 of 5
The circumcircle of the hexagon PQRSTU has a radius of length 1. It is known that PQ = RS = TU = 1.

Let J, K and L correspond respectively to the mid points of the sides QR, ST and UP.

Is it always the case that the triangle JKL is equilateral?

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 3 of 4 |

Let cis(theta) denote cos(theta) + i sin(theta).
WOLOG let the vertices be
    P = cis(alpha+30)
    Q = cis(alpha+90)
    R = cis(-[beta+90])
    S = cis(-[beta+30])
    T = cis(-30)
    U = cis(30)
The midpoints then become
    J = (Q + R)/2
    K = (S + T)/2
    L = (U + P)/2
It is easy, but tedious to show that
    (K - J) cis(60) = L - J
Therefore, JKL is an equilateral triangle.
 

  Posted by Bractals on 2007-08-02 03:48:07
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 (2)
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