Difference between revisions of "2005 AMC 12B Problems/Problem 24"
Armalite46 (talk | contribs) (→Solution) |
Armalite46 (talk | contribs) (→Solution) |
||
Line 5: | Line 5: | ||
<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math> | <math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math> | ||
== Solution == | == Solution == | ||
− | |||
− | |||
<center><asy> | <center><asy> | ||
Line 20: | Line 18: | ||
B = (b,f(b)); | B = (b,f(b)); | ||
C = (c,f(c)); | C = (c,f(c)); | ||
+ | J = (1,0); | ||
draw(graph(f,-2,2)); | draw(graph(f,-2,2)); | ||
draw((-2,0)--(2,0),Arrows); | draw((-2,0)--(2,0),Arrows); | ||
Line 29: | Line 28: | ||
dot("$B(b,b^2)$", B, E); | dot("$B(b,b^2)$", B, E); | ||
dot("$C(c,c^2)$", C, W); | dot("$C(c,c^2)$", C, W); | ||
+ | dot("$J(1,0)$", 1,0); | ||
</asy></center> | </asy></center> | ||
Using the slope formula and differences of squares, we find: | Using the slope formula and differences of squares, we find: |
Revision as of 19:20, 30 August 2013
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
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)); J = (1,0); 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); dot("$J(1,0)$", 1,0); (Error making remote request. Unknown error_msg)
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.