Difference between revisions of "1994 AHSME Problems/Problem 18"
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME box|year=1994|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 02:08, 28 May 2021
Problem
Triangle 
is inscribed in a circle, and 
. If 
and 
are adjacent vertices of a regular polygon of 
sides inscribed in this circle, then 
![[asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy]](http://latex.artofproblemsolving.com/6/5/f/65fef671e3625f7b69731ed1d6f1df3b84df2f51.png)
Solution
We solve for 
as follows: ![\[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\]](http://latex.artofproblemsolving.com/7/1/4/714c24df1b316ce45a460bd26f0317d11d92a380.png)
That means that minor arc 
has measure 
. We can fit a maximum of 
of these arcs in the circle.
--Solution by TheMaskedMagician
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
