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

Home > Shapes > Geometry
T-N-T (Posted on 2017-11-25) Difficulty: 3 of 5

Let ABC be an arbitrary triangle with points D, E, and F
lying on rays AB, BC, and CA respectively such that

   AD/AB = BE/BC = CF/CA = x.

Prove that Area(ΔDEF)/Area(ΔABC) = 3x2 - 3x + 1.

See The Solution Submitted by Bractals    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2
The three corner triangles have the same area because if you take any two of them their bases and heights are in inverse proportion.

Let AC=b be the base of ΔABC and the height=h, area = bh/2 
The base and height of ΔAFD are then (1-x)b and xh, area = x(1-x)bh/2

Area(ΔDEF) = bh/2 - 3x(1-x)bh/2

The ratio sought is then 1 - 3x(1-x) = 3x2 - 3x + 1

  Posted by Jer on 2017-11-26 11:16:42
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information