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

Home > Probability
Pentagon and Acute Probability (Posted on 2016-08-02) Difficulty: 3 of 5
A point O is selected at random from the interior of the pentagon with vertices
P = (0, 2), Q = (4, 0), R = (2pi + 1, 0), S = (2pi + 1, 4), and T = (0, 4).

Determine the probability that ∠POQ is acute.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
The area of pentagon PQRST is that of rectangle Origin,RST minus that of triangle Origin,QP.

Angle POQ is acute if O lies outside the semicircle with diameter PQ. That semicircle falls entirely within the given pentagon, so we need only subtract the area of that semicircle from the area of the pentagon to get the area in which O must lie for a success (acute angle).

Pentagon PQRST has area 4*(2*pi+1) - 4 = 8*pi.

The radius of the semicircle of obtuseness is sqrt(5) The area of a full circle would be 5*pi, so the semicircle has area 5*pi/2.

The requested probability is (8*pi - 5*pi/2) / (8*pi) = 1 - 5/(2*8) = 1 - 5/16 = 11/16.

  Posted by Charlie on 2016-08-02 11:54:27
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
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