Difference between revisions of "1963 AHSME Problems/Problem 10"
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| − | == Problem | + | == Problem == |
Point <math>P</math> is taken interior to a square with side-length <math>a</math> and such that is it equally distant from two | Point <math>P</math> is taken interior to a square with side-length <math>a</math> and such that is it equally distant from two | ||
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\textbf{(C)}\ \frac{3a}{8}\qquad | \textbf{(C)}\ \frac{3a}{8}\qquad | ||
\textbf{(D)}\ \frac{a\sqrt{2}}{2}\qquad | \textbf{(D)}\ \frac{a\sqrt{2}}{2}\qquad | ||
| − | \textbf{(E)}\ \frac{a}{2} </math> | + | \textbf{(E)}\ \frac{a}{2} </math> |
==Solution== | ==Solution== | ||
Latest revision as of 07:22, 5 June 2018
Problem
Point 
is taken interior to a square with side-length 
and such that is it equally distant from two
consecutive vertices and from the side opposite these vertices. If 
represents the common distance, then equals:
Solution
![[asy] draw((0,0)--(16,0)--(16,16)--(0,16)--(0,0)); draw((0,0)--(8,6)); draw((16,0)--(8,6)); draw((8,16)--(8,6)); draw((8,0)--(8,6),dotted); label("$d$",(4,3),NW); label("$d$",(12,3),NE); label("$d$",(8,11),W); label("$\frac{a}{2}$",(4,0),S); label("$\frac{a}{2}$",(12,0),S); label("$a-d$",(8,2),W); [/asy]](http://latex.artofproblemsolving.com/5/0/9/509c25bf5568eacab3b98fb7cc86e0c1f5011a97.png)
Draw a diagram and label it as shown. Because of SSS Congruency, the two bottom triangles are right triangles. By the Pythagorean Theorem,
![\[d^2 = (a-d)^2 + (\frac{a}{2})^2\]](http://latex.artofproblemsolving.com/d/1/9/d195736d5ff3cb2ffec1396a29af354e90f67faf.png)
![\[d^2 = a^2 - 2ad + d^2 + \frac{a^2}{4}\]](http://latex.artofproblemsolving.com/7/1/5/715fff45402cae2f7d6e57189e12b7b40e0a632e.png)
![\[2ad = \frac{5a^2}{4}\]](http://latex.artofproblemsolving.com/0/a/6/0a686135c8e5b3bd48abebc173c0a3d8e2c0cf45.png)
![\[d = \frac{5a}{8}\]](http://latex.artofproblemsolving.com/3/2/f/32f9537c5a0abfd67fb80a4c25d234ddc749a819.png)
Thus, the answer is 
.
See Also
| 1963 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
