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

Home > Shapes > Geometry
Perpendicular Area (Posted on 2005-12-08) Difficulty: 4 of 5
The altitude from one of the vertices of an acute-angled triangle ABC meets the opposite side at D. From D, perpendiculars DE and DF are drawn to the other two sides. Prove that the length EF is the same whichever vertex is chosen.

For extra credit: if a=|BC|, b=|CA|, c=|AB|, and d=|EF|, show that Area(ABC)=½√(abcd).

2002 British Mathematical Olympiad, Round 2, Problem 1.

See The Solution Submitted by Bractals    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution No extra credit Comment 2 of 2 |
Start the construction at the vertex A with angles Alpha1 and Alpha2 (Alpha1+Alpha2=Alpha) created by the altitude AD. Let Beta and Gamma be the other two acute angles. The law of cosines aplied to the inscribed triangle gives
|EF|^2 = |DE|^2+|DF|^2-2|DE||DF|Cos(Beta+Gamma).
Substituting Beta+Game=Pi-Alpha1-Alpha2, |AD|Sin(Alpha1)=|DE| and |AD|Sin(Alpha2)=|DF| and simplifying (Ha Ha) gives
|EF|^2 = (|AD|Sin(Alpha))^2
|EF|^2 = (|AD||BC|)^2 (Sin(Alpha)/|BA|)^2
|EF|^2 = (2*Area)^2 (Sin(Alpha)/|BA|)^2

Note that the right hand side is constant; by the law of sines, regardless of the beginning vertex the second factor is constant.
  Posted by owl on 2005-12-09 10:40:05
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 (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information