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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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