Difference between revisions of "1994 AHSME Problems/Problem 18"
(Created page with "==Problem== Triangle <math>ABC</math> is inscribed in a circle, and <math>\angle B = \angle C = 4\angle A</math>. If <math>B</math> and <math>C</math> are adjacent vertices of a ...") |
Logsobolev (talk | contribs) m (→Solution) |
||
(One intermediate revision by one other user not shown) | |||
Line 13: | Line 13: | ||
<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] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1994|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:08, 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
Solution
We solve for as follows:
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.