1965 AHSME Problems/Problem 35
Problem
The length of a rectangle is inches and its width is less than inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is , then the width is:
Solution
Let the rectangle be with and , as in the diagram. We desire a line such that reflecting point across that line yields point . For this to happen, the line must be perpendicular to the diagonal , and it must go through the midpoint of (let it be point ). Let the intersection of this line with be point and with be point . From the problem, we know that . By HL congruence, , so , where is some number. Furthermore, . By AA similarity, , so . , , , and , so we can rewrite this proportion to solve for in terms of : \begin{align*} \frac{\sqrt{6}/2}{x}&=\frac{w}{5} \\ \frac{5\sqrt{6}}{2}&=xw \\ x&=\frac{5\sqrt{6}}{2w} \end{align*} By the Pythagorean Theorem on , we know that , and we can plug in our new expression for into this equation to solve for : \begin{align*} w^2+25&=4(\frac{5\sqrt{6}}{2w})^2 \\ w^2+25&=\frac{25*6}{w^2} \\ w^4+25w^2-150&=0 \\ (w^2-5)(w^2+30)&=0 \end{align*} Because , , and so , which corresponds to answer choice .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
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