User:Azjps/1951 AHSME Problems/Problem 3

Problem

Points $A$ and $B$ are selected on the graph of $y=-1/2x^2$ so that triangle $ABO$ is equilateral. Find the length of one side of triangle $ABO$ (point $O$ is at the origin).

Solution

The parabola $y=-1/2x^2$ opens downward, and by symmetry we realize that the y-coordinates of $A,B$ are the same. Thus the segments $\overline{AO}, \overline{BO}$ will have slope $\pm \tan{60^{\circ}} = \pm \sqrt{3}$ . Without loss of generality consider the equation of $AO$ (we let $A$ be in the third quadrant), which has equation $y = \sqrt{3}x$ . This intersects the graph of $y = -\frac{1}{2}x^2$ at $-\frac{1}{2}x^2 = \sqrt{3}x \Longrightarrow x(x + 2\sqrt{3}) = 0$ ; we drop zero so $A_x = -2\sqrt{3}$ . The length of a side of the triangle is $|A_x| + |B_x| = 4\sqrt{3}$ .

1951 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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User:Azjps/1951 AHSME Problems/Problem 3

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