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Difference between revisions of "1993 AHSME Problems/Problem 25"
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== Problem ==
 
== Problem ==
 
<asy>
 
<asy>
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draw((0,0)--(1,sqrt(3)),black+linewidth(.75));
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draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow);
−
draw((0,0)--(1,-sqrt(3)),black+linewidth(.75));
+
draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow);
 
draw((0,0)--(1,0),dashed+black+linewidth(.75));
 
draw((0,0)--(1,0),dashed+black+linewidth(.75));
 
dot((1,0));
 
dot((1,0));
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== Solution ==
 
== Solution ==
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
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 +
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>. 
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 +
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>.
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Since <math>\triangle A'PO'</math> is an isosceles triangle with vertex of <math>60^\circ</math>, it is equilateral.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:13, 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

$\fbox{E}$

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

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