Difference between revisions of "2011 AMC 12B Problems/Problem 14"
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Now using the law of cosines on <math>\triangle AVB</math> we have: | Now using the law of cosines on <math>\triangle AVB</math> we have: | ||
− | <math>AB^2 = 2AV^2-2AV\cos(\angle AVB) = 2AV^2(1-\cos(\angle AVB))</math> | + | <math>AB^2 = 2AV^2-2AV^2\cos(\angle AVB) = 2AV^2(1-\cos(\angle AVB))</math> |
<math>1 = \frac{5}{8} (1-\cos(\angle AVB))</math> | <math>1 = \frac{5}{8} (1-\cos(\angle AVB))</math> | ||
Line 34: | Line 34: | ||
==Solution 3== | ==Solution 3== | ||
− | After assuming that the parabola is x^2,find the points A and B, which are +/- 1,2,1/4.Now treat them as vectors,take the dot product,then find the magnitudes and multiply them.A well known definition of the dot product says that the quotient of the two is the cosine of the angle between them.This will give you D. | + | After assuming that the parabola is <math>x^2</math>,find the points A and B, which are +/- 1,2,1/4.Now treat them as vectors,take the dot product,then find the magnitudes and multiply them.A well known definition of the dot product says that the quotient of the two is the cosine of the angle between them.This will give you D. |
==Solution 4== | ==Solution 4== | ||
− | As we know, an expression for the equation of a parabola is <math>y-k=\frac{1}{4a}(x-h)^2</math>, where <math>(h,k)</math> is the vertex and <math>a</math> is the distance from the focus to the vertex, here <math>F</math> to <math>V</math>. The length of the latus rectum, or <math>\overline{AB}</math> here, is equal to <math>|4a|</math>. This means that <math>AF=BF=2a</math> since <math>F</math> is the midpoint of <math>AB</math>. Then we can use right Triangle <math>VAF</math> to figure out that <math>VA= | + | As we know, an expression for the equation of a parabola is <math>y-k=\frac{1}{4a}(x-h)^2</math>, where <math>(h,k)</math> is the vertex and <math>a</math> is the distance from the focus to the vertex, here <math>F</math> to <math>V</math>. The length of the latus rectum, or <math>\overline{AB}</math> here, is equal to <math>|4a|</math>. This means that <math>AF=BF=2a</math> since <math>F</math> is the midpoint of <math>AB</math>. Then we can use right Triangle <math>VAF</math> to figure out that <math>VA=\frac{a\sqrt{5}}{4}</math>. Now we can use the fact that <math>\cos{\angle AVB}=\cos{2 \angle AVF}</math> and use the double angle formula. This results in <math>\cos{2 \angle AVF}=2{\cos}^2{\angle AVF}-1</math>. We can find <math>\cos{\angle AVF}</math> from right triangle <math>AVF</math> using the Pythagorean Theorem, which is <math>\frac{a}{\sqrt{5}}</math>. Evaulating the expression, we find that <math>\cos{2 \angle AVF}=\frac{-3}{5}</math>. |
-Indefintense | -Indefintense | ||
− | + | == 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:15, 11 May 2021
Problem
A segment through the focus of a parabola with vertex
is perpendicular to
and intersects the parabola in points
and
. What is
?
Solution 1
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 .
Solution 2
WLOG we can assume that the parabola is . Therefore
and
. Also
and
.
and
by the pythagorean theorem.
Now using the law of cosines on we have:
Thus,
(solution by mihirb)
Solution 3
After assuming that the parabola is ,find the points A and B, which are +/- 1,2,1/4.Now treat them as vectors,take the dot product,then find the magnitudes and multiply them.A well known definition of the dot product says that the quotient of the two is the cosine of the angle between them.This will give you D.
Solution 4
As we know, an expression for the equation of a parabola is , where
is the vertex and
is the distance from the focus to the vertex, here
to
. The length of the latus rectum, or
here, is equal to
. This means that
since
is the midpoint of
. Then we can use right Triangle
to figure out that
. Now we can use the fact that
and use the double angle formula. This results in
. We can find
from right triangle
using the Pythagorean Theorem, which is
. Evaulating the expression, we find that
.
-Indefintense
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.