Squares and Triangles: a ProofDate: 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/ |
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