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)
(Deleted a wrong solution)
Line 19: Line 19:
 
\text{(E) more than 15 such triangles} </math>
 
\text{(E) more than 15 such triangles} </math>
  
−
== Solution ==
+
The answer is E, there are an infinite number
−
<math>\fbox{B}</math>
+
Please add to this answer, any explanation or anything. The previous answer was wrong ~inaccessibles
−
 
 
−
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>.
 
−
 
 
−
Now observe <math>\triangle AA'P</math> and <math>\triangle OO'P</math>.  We have <math>m\angle A = 60^\circ</math> and <math>m \angle O = 60^\circ</math>, therefore <math>\angle A \cong \angle O</math>.  By our construction of moving the points the same distance, we have <math>AA' = OO'</math>.  Also, <math>AP = OP</math> by the original equilateral triangle.  Therefore, by SAS congruence, <math>\triangle AA'P \cong \triangle OO'P</math>. 
 
−
 
 
−
Now, look at <math>\triangle A'PO'</math>.  We have <math>PA' = PO'</math> from the above congruence.  We also have the included angle <math>\angle A'PO'</math> is <math>60^\circ</math>.  To prove that, start with the <math>60^\circ</math> angle <math>APO</math>, subtract the angle <math>APA'</math>, and add the congruent angle <math>OPO'</math>.
 
−
 
 
−
Since <math>\triangle A'PO'</math> is an isosceles triangle with vertex of <math>60^\circ</math>, it is equilateral.
 
  
 
== See also ==
 
== See also ==

Revision as of 22:10, 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

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

The answer is E, there are an infinite number Please add to this answer, any explanation or anything. The previous answer was wrong ~inaccessibles

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=252867"