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

(Created page with "== Problem == <asy> draw(circle((0,0),10),black+linewidth(.75)); MP(")",(0,0),S); </asy> Let <math>S</math> be the set of points on the rays forming the sides of a <math>120^\ci...")
 
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== Problem ==
 
== Problem ==
 
<asy>
 
<asy>
−
draw(circle((0,0),10),black+linewidth(.75));
+
draw((0,0)--(1,sqrt(3)),black+linewidth(.75));
−
MP(")",(0,0),S);
+
draw((0,0)--(1,-sqrt(3)),black+linewidth(.75));
 +
draw((0,0)--(1,0),dashed+black+linewidth(.75));
 +
dot((1,0));
 +
MP("P",(1,0),E);
 
</asy>
 
</asy>
  
Line 17: Line 20:
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1993|num-b=1|num-a=2}}   
+
{{AHSME box|year=1993|num-b=24|num-a=26}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:09, 27 September 2014

Problem

[asy] draw((0,0)--(1,sqrt(3)),black+linewidth(.75)); draw((0,0)--(1,-sqrt(3)),black+linewidth(.75)); 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

$\fbox{E}$

See also

1993 AHSME (Problems • Answer Key • Resources)
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

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