Difference between revisions of "1994 AHSME Problems/Problem 18"
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<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 </math> | <math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 </math> | ||
==Solution== | ==Solution== | ||
| + | We solve for <math>\angle A</math> as follows: <cmath>4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.</cmath> That means that minor arc <math>\widehat{BC}</math> has measure <math>40^\circ</math>. We can fit a maximum of <math>\frac{360}{40}=\boxed{\textbf{(C) }9}</math> of these arcs in the circle. | ||
| + | |||
| + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
Revision as of 20:28, 20 July 2014
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
