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

Home > Shapes
Equilateral Exercise (Posted on 2016-07-15) Difficulty: 3 of 5
PQR is an equilateral triangle with an area of 7 square units.

S and T are points on sides PQ and PR , respectively, such that PT = QS.

O denotes the intersection of QT and RS.

Given that the triangle QOR has an area of 2 square units:

(a) Find the possible values of SQ/PQ and, prove that no other values are possible.

(b) Find angle POQ.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1
Given that we only need ratios and not lengths, I decided to scale the PQR to side length 1 and area √3/4 so area QOR is 2/7 or this or √3/14.  Part (a) is now just asking for SQ which I'll call x below.

OR=z, OQ=y, angle QRO=θ. 
angle QOR=120, RQO=60-θ.
So by law of sines
sin120/1 = sin(60-θ)/z=sinθ/y
and y=2√3sinθ/3
The area of QOR is then
.5*1*2√3sinθ/3*sin(60-θ)

Equating the expressions for the area and simplifying gives
.5sinθ√(1-(sinθ)²)=√3(sinθ)²/6+√3/14=0
squaring gives the nice biquaratic
0=196(sinθ)^4-105(sinθ)^2+9
whose solutions are
[I] (sinθ)^2=3/7 or [II] (sinθ)^2=3/28

Case [I]
Using the formula for y above
y=√(1/7)
x=ysin(60)/sin(120-θ)=1/3
To find angle POQ we know SAS of triangle POQ so it isn't hard to find the angle=120.

Case [II]
Similar to case [I] we instead end up with
x=2/3 and the angle again comes out to 120.


  Posted by Jer on 2016-07-15 17:40:35
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 (3)
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