Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(Undo revision 249411 by Fries (talk))
(Tag: Undo)
 
(2 intermediate revisions by the same user not shown)
Line 20: Line 20:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
<math>\fbox{E}</math>
  
 
Take the "obvious" equilateral triangle <math>OAP</math>, where <math>O</math> is the vertex, <math>A</math> is on the upper ray, and <math>P</math> is our central point.  Slide <math>A</math> down on the top ray to point <math>A'</math>, and slide <math>O</math> down an equal distance on the bottom ray to point <math>O'</math>.
 
Take the "obvious" equilateral triangle <math>OAP</math>, where <math>O</math> is the vertex, <math>A</math> is on the upper ray, and <math>P</math> is our central point.  Slide <math>A</math> down on the top ray to point <math>A'</math>, and slide <math>O</math> down an equal distance on the bottom ray to point <math>O'</math>.

Latest revision as of 22:12, 29 June 2025

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

$\fbox{E}$

Take the "obvious" equilateral triangle $OAP$, where $O$ is the vertex, $A$ is on the upper ray, and $P$ is our central point. Slide $A$ down on the top ray to point $A'$, and slide $O$ down an equal distance on the bottom ray to point $O'$.

Now observe $\triangle AA'P$ and $\triangle OO'P$. We have $m\angle A = 60^\circ$ and $m \angle O = 60^\circ$, therefore $\angle A \cong \angle O$. By our construction of moving the points the same distance, we have $AA' = OO'$. Also, $AP = OP$ by the original equilateral triangle. Therefore, by SAS congruence, $\triangle AA'P \cong \triangle OO'P$.

Now, look at $\triangle A'PO'$. We have $PA' = PO'$ from the above congruence. We also have the included angle $\angle A'PO'$ is $60^\circ$. To prove that, start with the $60^\circ$ angle $APO$, subtract the angle $APA'$, and add the congruent angle $OPO'$.

Since $\triangle A'PO'$ is an isosceles triangle with vertex of $60^\circ$, it is equilateral.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_25&oldid=252869"