Difference between revisions of "2011 AMC 12B Problems/Problem 14"
m (→Problem 15) |
(took me a while to realize that you didn't need to memorize a bunch of equations to solve this) |
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== Solution == | == Solution == | ||
− | {{ | + | Name the directrix of the parabola <math>l</math>. Define <math>d(X,k)</math> to be the distance between a point <math>X</math> and a line <math>k</math>. |
+ | |||
+ | Now we remember the geometric definition of a parabola: given any line <math>l</math> (called the directrix) and any point <math>F</math> (called the focus), the ''parabola'' corresponding to the given directrix and focus is the locus of the points that are equidistant from <math>F</math> and <math>l</math>. Therefore <math>FV=d(V,l)</math>. Let this distance be <math>d</math>. Now note that <math>d(F,l)=2d</math>, so <math>d(A,l)=d(B,l)=2d</math>. Therefore <math>AF=BF=2d</math>. We now use the [[Pythagorean Theorem]] on triangle <math>AFV</math>; <math>AV=\sqrt{AF^2+FV^2}=d\sqrt{5}</math>. Similarly, <math>BV=d\sqrt{5}</math>. We now use the [[Law of Cosines]]: | ||
+ | |||
+ | <cmath>AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}</cmath> | ||
+ | |||
+ | <cmath>\Rightarrow \cos{\angle AVB}=-\frac{3}{5}</cmath> | ||
+ | |||
+ | This shows that the answer is <math>\boxed{\textbf{(D)}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2011|ab=B|num-b=13|num-a=15}} |
Revision as of 12:09, 1 September 2011
Problem
A segment through the focus of a parabola with vertex
is perpendicular to
and intersects the parabola in points
and
. What is
?
Solution
Name the directrix of the parabola . Define
to be the distance between a point
and a line
.
Now we remember the geometric definition of a parabola: given any line (called the directrix) and any point
(called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from
and
. Therefore
. Let this distance be
. Now note that
, so
. Therefore
. We now use the Pythagorean Theorem on triangle
;
. Similarly,
. We now use the Law of Cosines:
This shows that the answer is .
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |