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?) |
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==Problem== | ==Problem== | ||
− | + | [[Point]]s <math>A</math> and <math>B</math> are selected on the graph of <math>y=-1/2x^2</math> so that triangle <math>ABO</math> is [[equilateral triangle|equilateral]]. Find the length of one side of triangle <math>ABO</math> (point <math>O</math> is at the origin). | |
== 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> | + | 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== | ||
− | + | {{AHSME box|year=1951|num-b=2|num-a=4}} | |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Latest revision as of 13:10, 27 August 2020
Problem
Points and are selected on the graph of so that triangle is equilateral. Find the length of one side of triangle (point is at the origin).
Solution
The parabola opens downward, and by symmetry we realize that the y-coordinates of are the same. Thus the segments will have slope . Without loss of generality consider the equation of (we let be in the third quadrant), which has equation . This intersects the graph of at ; we drop zero so . The length of a side of the triangle is .
See Also
1951 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |