The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Squares and Triangles: a Proof

Date: 8/23/96 at 4:59:13
From: Anonymous
Subject: Squares and triangles

Two points P and Q lie within or on the boundary of a square of side 
1 metre, one corner of which is at O. Prove that the length of at 
least one of the lines OP, PQ, and QO must be less than or equal to
   {sqrt(6 ) -  sqrt(2)} metres.

Date: 8/23/96 at 8:20:3
From: Doctor Anthony
Subject: Re: Squares and triangles

By symmetry we have the worst situation when all three sides of the 
triangle are the same length, for if any side is less than the others 
you would increase it to try to get beyond the limit that might apply.  
We therefore consider an equilateral triangle with one vertex at O, 
and points P and Q on the opposite pair of sides of the square from O.  
By simple trigonometry, we have 

   OP * cos(15) = 1,  so 
   OP = 1/cos(15)

and we must find cos(15) in surd form

Using the double angle formula 

   cos(2A) = 2cos^2(A) - 1
   cos(30) = 2cos^2(15) - 1
   sqrt(3)/2 = 2cos^2(15) - 1
   2cos^2(15) = 1 + sqrt(3)/2  and 
   cos^2(15) = 1/2 + sqrt(3)/4
             = {2+sqrt(3)}/4.


   OP = 1/cos(15) = 2/sqrt{2+sqrt(3)}.
At this stage we must simplify sqrt{2+sqrt(3)}.
We put this equal to sqrt(a) + sqrt(b).

Squaring both sides:

   2 + sqrt(3) = a + b + 2sqrt(ab).  

Now equate rational and irrational parts:


   a + b = 2 and 2sqrt(ab) = sqrt(3) or 4ab = 3.

Solving these equations for a and b, we get a = 3/2 and b = 1/2.
So now 

   OP = 2/{sqrt(3/2) + sqrt(1/2)}.

Multiply top and bottom by sqrt(3/2)-sqrt(1/2) and we get

   OP = 2{sqrt(3/2)-sqrt(1/2)}/{(3/2)-(1/2)}
      = sqrt(6) - sqrt(2).    QED

-Doctor Anthony,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.