1965 AHSME Problems/Problem 35: Difference between revisions
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== Solution == | == Solution == | ||
<asy> | |||
import geometry; | |||
point M; | |||
segment l; | |||
draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5))); | |||
dot((0,sqrt(5))); | |||
label("A", (0,sqrt(5)), NW); | |||
dot((0,0)); | |||
label("B", (0,0), SW); | |||
dot((5,0)); | |||
label("C", (5,0), SE); | |||
dot((5,sqrt(5))); | |||
label("D", (5, sqrt(5)), NE); | |||
M=(2.5,sqrt(5)/2); | |||
l=line((0,sqrt(5)),(5,0)); | |||
draw(l); | |||
draw(perpendicular(M,l)); | |||
dot(M); | |||
label("M",M,W); | |||
</asy> | |||
<math>\fbox{D}</math> | <math>\fbox{D}</math> | ||
Revision as of 12:23, 19 July 2024
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
![[asy] import geometry; point M; segment l; draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5))); dot((0,sqrt(5))); label("A", (0,sqrt(5)), NW); dot((0,0)); label("B", (0,0), SW); dot((5,0)); label("C", (5,0), SE); dot((5,sqrt(5))); label("D", (5, sqrt(5)), NE); M=(2.5,sqrt(5)/2); l=line((0,sqrt(5)),(5,0)); draw(l); draw(perpendicular(M,l)); dot(M); label("M",M,W); [/asy]](http://latex.artofproblemsolving.com/9/5/5/9551b8d8cc5ed0cf1c31c21a8e877857112ba607.png)
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
