Difference between revisions of "2005 AMC 12B Problems/Problem 24"
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== Problem == | == Problem == | ||
| − | All three vertices of an equilateral triangle are on the parabola y = x^2, and one of its sides has a slope of 2. The x-coordinates of the three vertices have a sum of m/n, where m and n are relatively prime positive integers. What is the value of m + n? | + | 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>? |
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| + | <math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math> | ||
== Solution == | == Solution == | ||
| − | Let the points be (a,a^2), (b,b^2) and (c,c^2). Using elementary calculus concepts and the fact that they lie on y = x^2 | + | Let the points be <math>(a,a^2)</math>, <math>(b,b^2)</math> and <math>(c,c^2)</math>. Using elementary calculus concepts and the fact that they lie on <math>y = x^2</math>, |
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| − | + | <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 simply the sum of the slopes of the three sides of the triangle divided by <math>2</math>. WLOG, 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>. Let us translate the triangle so <math>A</math> is at the origin. Then <math>tan(BOJ) = 2</math>. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply <math>\dfrac{tan(BOJ) + tan(BOJ+60) + tan(BOJ-60)}{2}</math>. | ||
| + | Using <math>tan(BOJ) = 2</math>, and basic trigonometric identities, 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}} | |
Revision as of 12:52, 4 July 2011
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
Let the points be 
, 
and 
. Using elementary calculus concepts and the fact that they lie on ,
= the slope of 
,
= the slope of 
,
= the slope of 
.
So the value that we need to find is simply the sum of the slopes of the three sides of the triangle divided by . WLOG, let be the side that has the smallest angle with the positive -axis. Let 
be an arbitrary point with the coordinates 
. Let us translate the triangle so 
is at the origin. Then 
. Using the fact that the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the value we now need to find is simply 
.
Using , and basic trigonometric identities, 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 | |
