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

Home > Just Math
Halved triangle (Posted on 2020-09-11) Difficulty: 4 of 5
Consider a line that simultaneously bisects the area and the perimeter of a given triangle.

Only two such lines exists for a triangle of side lengths 7,8,9.

Find the acute angle θ between the two lines.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Possible solution | Comment 1 of 2
In triangle ABC, let AB=7, AC=8, and BC=9.

Draw a circle of radius 6 on C, crossing AC at D and BC at E. DE is then the first line.

Draw two circles of radius (6-3sqrt(2)) and (6+3sqrt(2)) on B, crossing AB at F and BC at G respectively. FG is the second line.

The acute angle between these lines is of 85.6082 degrees.

Note: The controlling equation is (t-(P/4))^2-((P/2)^2-2*ab)/4, where P is the perimeter, and ab are the sides adjoining the angle on which the circles are to be drawn. The solutions to this quadratic are the radii of the circles. The remaining, hypothetical, solution would involve circles of radius (6-3/sqrt(2), 6+3/sqrt(2)) on A, but the larger circle does not intersect a side of ABC. Hence the only solutions are those given above.

Edited on September 12, 2020, 8:05 am
  Posted by broll on 2020-09-12 07:52:22

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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information