Difference between revisions of "1958 AHSME Problems/Problem 21"
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| + | Draw <math>OE</math>. Since triangles <math>AOB</math>, <math>COE</math>, and <math>DOE</math> are congruent, you can fit triangle <math>AOE</math> twice in <math>CED</math>. Thus, our answer is <math>2:1</math>, or <math>(E)</math>. | ||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
Revision as of 00:05, 1 January 2024
Problem
In the accompanying figure 
and 
are equal chords of a circle with center 
. Arc 
is a quarter-circle. Then the ratio of the area of triangle 
to the area of triangle 
is:
![[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(0,0)--(10,0)--(0,10)--cycle,dot); MP("O",(0,0),N);MP("C",(-10,0),W);MP("D",(10,0),E);;MP("E",(0,10),N); draw((-sqrt(70),-sqrt(30))--(sqrt(30),-sqrt(70))--(0,0)--cycle,dot); MP("A",(-sqrt(70),-sqrt(30)),SW);MP("B",(sqrt(30),-sqrt(70)),SE); [/asy]](http://latex.artofproblemsolving.com/b/6/a/b6a4d70cdcbe1bba1fade52a583f1bf849202b3d.png)
Solution
Draw 
. Since triangles , 
, and 
are congruent, you can fit triangle 
twice in . Thus, our answer is 
, or 
.
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
