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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.

Then 

   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:

So 

   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!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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