Difference between revisions of "1990 AHSME Problems/Problem 14"
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== Solution == | == Solution == | ||
<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
| + | We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{360}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{360}{7}</math> degrees, the answer is <math>\dfrac{2\pi}{7}</math>. | ||
== See also == | == See also == | ||
Revision as of 20:24, 20 November 2014
Problem
![[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]](http://latex.artofproblemsolving.com/b/7/6/b76c92e86b0567832a550d1a19f11df2bb73962c.png)
An acute isosceles triangle, 
, is inscribed in a circle. Through 
and 
, tangents to the circle are drawn, meeting at point 
. If $\angle{ABC=\angle{ACB}=2\angle{D}$ (Error compiling LaTeX. Unknown error_msg) and 
is the radian measure of 
, then 
Solution
We can make two equations (assume angle D is y): 
and 
. We find that 
. Now we have to convert this to radians. 360 degrees is 
radians, so since we have 
degrees, the answer is 
.
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
