Difference between revisions of "1988 AIME Problems/Problem 14"
(→Solution 3) |
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The matrix for a reflection about the polar line <math>\theta = \alpha, \alpha+\pi</math> is: | The matrix for a reflection about the polar line <math>\theta = \alpha, \alpha+\pi</math> is: | ||
<cmath>\[ \left( \begin{array}{ccc} | <cmath>\[ \left( \begin{array}{ccc} | ||
− | \cos(2\alpha) & sin(2\alpha) \\ | + | \cos(2\alpha) & \sin(2\alpha) \\ |
\sin(2\alpha) & -\cos(2\alpha) | \sin(2\alpha) & -\cos(2\alpha) | ||
\end{array} \right)\]</cmath> | \end{array} \right)\]</cmath> | ||
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Let <math>\alpha = \arctan 2</math>. Then <math>2\alpha = \arctan\left(-\frac{4}{3}\right)</math>, so <math>\cos(2\alpha) = -\frac{3}{5}</math> and <math>\sin(2\alpha) = \frac{4}{5}</math>. | Let <math>\alpha = \arctan 2</math>. Then <math>2\alpha = \arctan\left(-\frac{4}{3}\right)</math>, so <math>\cos(2\alpha) = -\frac{3}{5}</math> and <math>\sin(2\alpha) = \frac{4}{5}</math>. | ||
− | Therefore, if <math>(x, y)</math> is mapped to <math>(x | + | Therefore, if <math>(x, y)</math> is mapped to <math>(x', y')</math> under the reflection, then <math>x' = -\frac{3}{5}x+\frac{4}{5}y</math> and <math>y' = \frac{4}{5}x+\frac{5}{5}y</math>. |
− | The new coordinates <math>(x | + | The new coordinates <math>(x', y')</math> must satisfy <math>x'y' = 1</math>. Therefore, |
− | <cmath>\left(-\frac{3}{5}x+\frac{4}{5}y | + | <cmath>\left(-\frac{3}{5}x+\frac{4}{5}y\right)\left(\frac{4}{5}x+\frac{5}{5}y\right) = 1</cmath> |
<cmath>-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0</cmath> | <cmath>-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0</cmath> | ||
<cmath>12x^2 - 7xy - 12y^2 + 25 = 0</cmath> | <cmath>12x^2 - 7xy - 12y^2 + 25 = 0</cmath> |
Revision as of 02:36, 26 November 2015
Problem
Let be the graph of , and denote by the reflection of in the line . Let the equation of be written in the form
Find the product .
Solution 1
Given a point on , we look to find a formula for on . Both points lie on a line that is perpendicular to , so the slope of is . Thus . Also, the midpoint of , , lies on the line . Therefore .
Solving these two equations, we find and . Substituting these points into the equation of , we get , which when expanded becomes .
Thus, .
Solution 2
The asymptotes of are given by and . Now if we represent the line by the complex number , then we find the direction of the reflection of the asymptote by multiplying this by , getting . Therefore, the asymptotes of are given by and .
Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: . At this point, the right hand side of the equation will be determined by plugging the point , which is unchanged by the reflection, into the expression. But this is not necessary. We see that , , so .
Solution 3
The matrix for a reflection about the polar line is: This is not hard to derive using a basic knowledge of linear transformations. You can refer here for more information: https://en.wikipedia.org/wiki/Orthogonal_matrix
Let . Then , so and . Therefore, if is mapped to under the reflection, then and .
The new coordinates must satisfy . Therefore, Thus, and , so . The answer is .
See also
1988 AIME (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.