Difference between revisions of "2005 AMC 12B Problems/Problem 24"
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== Problem == | == Problem == | ||
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− | + | All three vertices of an equilateral triangle are on the parabola <math>y = x^2</math>, and one of its sides has a slope of <math>2</math>. The <math>x</math>-coordinates of the three vertices have a sum of <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is the value of <math>m + n</math>? | |
− | + | <math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math> | |
− | + | == Solution == | |
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− | + | <center><asy> | |
+ | import graph; | ||
+ | real f(real x) {return x^2;} | ||
+ | unitsize(1 cm); | ||
+ | pair A, B, C; | ||
+ | real a, b, c; | ||
+ | a = (-5*sqrt(3) + 11)/11; | ||
+ | b = (5*sqrt(3) + 11)/11; | ||
+ | c = -19/11; | ||
+ | A = (a,f(a)); | ||
+ | B = (b,f(b)); | ||
+ | C = (c,f(c)); | ||
+ | draw(graph(f,-2,2)); | ||
+ | draw((-2,0)--(2,0),Arrows); | ||
+ | draw((0,-0.5)--(0,4),Arrows); | ||
+ | draw(A--B--C--cycle); | ||
+ | label("$x$", (2,0), NE); | ||
+ | label("$y$", (0,4), NE); | ||
+ | dot("$A(a,a^2)$", A, S); | ||
+ | dot("$B(b,b^2)$", B, E); | ||
+ | dot("$C(c,c^2)$", C, W); | ||
+ | </asy></center> | ||
+ | Using the slope formula and differences of squares, we find: | ||
− | + | <math>a+b</math> = the slope of <math>AB</math>, | |
− | + | <math>b+c</math> = the slope of <math>BC</math>, | |
+ | <math>a+c</math> = the slope of <math>AC</math>. | ||
+ | So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by <math>2</math>. Without loss of generality, let <math>AB</math> be the side that has the smallest angle with the positive <math>x</math>-axis. Let <math>J</math> be an arbitrary point with the coordinates <math>(1, 0)</math>. Translate the triangle so <math>A</math> is at the origin. Then <math>\tan(BOJ) = 2</math>. Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>. | ||
+ | Using <math>\tan(BOJ) = 2</math>, and the tangent addition formula, this simplifies to <math>\dfrac{3}{11}</math>, so the answer is <math>3 + 11 = \boxed{\mathrm{(A)}\ 14}</math> | ||
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}} | |
+ | {{MAA Notice}} |
Latest revision as of 14:12, 15 July 2018
Problem
All three vertices of an equilateral triangle are on the parabola , and one of its sides has a slope of . The -coordinates of the three vertices have a sum of , where and are relatively prime positive integers. What is the value of ?
Solution
Using the slope formula and differences of squares, we find:
= the slope of ,
= the slope of ,
= the slope of .
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by . Without loss of generality, let be the side that has the smallest angle with the positive -axis. Let be an arbitrary point with the coordinates . Translate the triangle so is at the origin. Then . Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is .
Using , and the tangent addition formula, this simplifies to , so the answer is
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.