Folded paper - June 17-21, 1996
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A 6-inch by 8-inch piece of paper is folded so that opposite vertices touch. How long is the fold?
This week we received Web pages from a number of Aussies including Erin McPartland, Steven Brown, and Clint Joseph and Ryan Cayford, who changed the problem to metric measurements and made diagrams. Below are other solutions from Edward Early, an "amateur mathematician" who just graduated from the Science Academy of Austin at LBJ in Austin, TX, and Ling Qi, LSMSA:Edward Early Science Academy of Austin That's one of the nicest POW's I've seen in quite a while. A similar problem (with 12 and 24 instead of 6 and 8) was on some contest practice material I did a few weeks ago. Anyway, on with the solution: What we are looking for is the length of the perpendicular bisector of a diagonal of the rectangle (the part which is in the rectangle, of course), call it 2x. Drawing in the diagonal (which has length sqrt(6^2+8^2)=10), the perpendicular bisector, and the lines from the endpoints of the bisector to the ends of the diagonal yields four congruent (by SAS) right triangles, which together form a rhombus with sides of length sqrt(x^2+5^2). The area of the rhombus is half the product of the diagonals, which is 10x, the area of the rectangle is 6*8=48, and the area of the two right triangles outside of the rhombus (together) is 6*(8-sqrt(x^2+5^2)), thus 48 = 10x + 6*(8-sqrt(x^2+5^2)) => 10x = 6*sqrt(x^2+5^2) => 100x^2 = 36x^2 + 30^2 => 8^2*x^2 = 30^2 => 8x = 30 => 2x = 15/2, which is what we were looking for. More generally, if we replace 6 and 8 by a and b (a<b) then the answer is (a/b)*sqrt(a^2+b^2). Note: This problem can also be done with coordinates. Placing the side of length 8 along the x-axis and 6 along the y-axis, the diagonal is y=(3/4)x, so the bisector has slope -4/3 and passes through (4,3), and with a bit of algebra to find intersections of lines we find that the answer is sqrt((25/4-7/4)^2+(0-6)^2) = 15/2, as before.
Ling Qi Grade: 11 School: LSMSA When we look at the folded piece, it is a 5-sided figure. We number the point where the 2 vertices touch A, and clockwise B,C,D,and E respectively. A /\ / \ E / \B \ / D \__/C There is an overlapping triangle, ACD. First, angles ABC and AED are right angles, thus the following applies to triangles ABC and AED, (only look at ABC): AB^2 + BC^2 = AC^2, and let AC = x. Then the relation between AC and BC can be established: BC = 8-x. Change the above equation into: 6^2+(8-x)^2=x^2 and solve: we get x = 25/4. Draw a perpendicular line, CD, passing point A and intersecting CD at point P (not on graph). Look at triangle ACP, with relation AP^2 + CP^2 = AC^2. We know that AP = 5 (from the theorem we are using all too often) and AC = 25/4; therefore CP^2 = 225/16 and CP, is half of the length we are looking for, = 15/4, and CD therefore equals 15/2. Done.
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