Difference between revisions of "User:Azjps/1951 AHSME Problems/Problem 3"

m (1951 AMC 12 Problems/Problem 3 moved to 1951 AHSME Problems/Problem 3: This problem isn't multiple-choice -- can someone check if it's correct?)
(ugh, this needs a lot of work...)
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==See Also==
 
==See Also==
* [[1951 AMC 12]]
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* [[1951 AHSME]]
* [[1952 AMC 12 Problems/Problem 2 | Next problem]]
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* [[1952 AHSME Problems/Problem 4 | Next problem]]
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 11:21, 10 January 2008

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 $|A_x| + |B_x| = 4\sqrt{3}$. We can now easily verify that this triangle indeed is equilateral.

See Also

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