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

m (1951 AHSME Problems/Problem 3 moved to User:Azjps/1951 AHSME Problems/Problem 3: well, this obviously isn't the right question ... move to userspace until we find where it does belong)
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== Solution ==
 
== Solution ==
The parabola <math>y=-1/2x^2</math> opens downward, and by symmetry we realize that the y-coordinates of <math>A,B</math> are the same. Thus the segments <math>\overline{AO}, \overline{BO}</math> will have slope <math>\pm \tan{60^{\circ}} = \pm \sqrt{3}</math>. [[Without loss of generality]] consider the equation of <math>AO</math> (we let <math>A</math> be in the third quadrant), which has equation <math>y = \sqrt{3}x</math>. This intersects the graph of <math>y = -\frac{1}{2}x^2</math> at <math>-\frac{1}{2}x^2 = \sqrt{3}x \Longrightarrow x(x + 2\sqrt{3}) = 0</math>; we drop zero so <math>A_x = -2\sqrt{3}</math>. The length of a side of the triangle is <math>|A_x| + |B_x| = 4\sqrt{3}</math>. We can now easily verify that this triangle indeed is equilateral.  
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The parabola <math>y=-1/2x^2</math> opens downward, and by symmetry we realize that the y-coordinates of <math>A,B</math> are the same. Thus the segments <math>\overline{AO}, \overline{BO}</math> will have slope <math>\pm \tan{60^{\circ}} = \pm \sqrt{3}</math>. [[Without loss of generality]] consider the equation of <math>AO</math> (we let <math>A</math> be in the third quadrant), which has equation <math>y = \sqrt{3}x</math>. This intersects the graph of <math>y = -\frac{1}{2}x^2</math> at <math>-\frac{1}{2}x^2 = \sqrt{3}x \Longrightarrow x(x + 2\sqrt{3}) = 0</math>; we drop zero so <math>A_x = -2\sqrt{3}</math>. The length of a side of the triangle is <math>|A_x| + |B_x| = 4\sqrt{3}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 12:10, 27 August 2020

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 $|A_x| + |B_x| = 4\sqrt{3}$.

See Also

1951 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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