1993 AHSME Problems/Problem 25


[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

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



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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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