Difference between revisions of "1993 AHSME Problems/Problem 25"

m (Problem)
(Solution)
Line 20: Line 20:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
+
<math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:21, 29 July 2019

Problem

[asy] draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,0),dashed+black+linewidth(.75)); dot((1,0)); MP("P",(1,0),E); [/asy]

Let $S$ be the set of points on the rays forming the sides of a $120^{\circ}$ angle, and let $P$ be a fixed point inside the angle on the angle bisector. Consider all distinct equilateral triangles $PQR$ with $Q$ and $R$ in $S$. (Points $Q$ and $R$ may be on the same ray, and switching the names of $Q$ and $R$ does not create a distinct triangle.) There are

$\text{(A) exactly 2 such triangles} \quad\\ \text{(B) exactly 3 such triangles} \quad\\ \text{(C) exactly 7 such triangles} \quad\\ \text{(D) exactly 15 such triangles} \quad\\ \text{(E) more than 15 such triangles}$

Solution

$\fbox{B}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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. AMC logo.png