Difference between revisions of "2013 AMC 12B Problems/Problem 20"
(→Solution) |
(→Solution) |
||
Line 8: | Line 8: | ||
Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,g^2),(h,h^2),(j,j^2)</math>, and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>. | Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,g^2),(h,h^2),(j,j^2)</math>, and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>. | ||
− | Now, we must find how to match up <math>\sin, \cos, \tan, \cot</math> to <math>f,g,h,j</math> so that the above equation has a solution. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>\cot x <-1<\cos x<0<\sin x</math>, and <math>\cot x <-1<\tan x<0<\sin x</math>so the sum of the largest and the smallest is equal to the sum of the other two, namely, <math>\sin x+\cot x = \cos x+\tan x</math>. | + | Now, we must find how to match up <math>\sin, \cos, \tan, \cot</math> to <math>f,g,h,j</math> so that the above equation has a solution. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>\cot x <-1<\cos x<0<\sin x</math>, and <math>\cot x <-1<\tan x<0<\sin x</math> so the sum of the largest and the smallest is equal to the sum of the other two, namely, <math>\sin x+\cot x = \cos x+\tan x</math>. |
Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>: | Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>: |
Revision as of 11:23, 29 January 2014
Problem
For , points and are the vertices of a trapezoid. What is ?
$\textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D}}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}$ (Error compiling LaTeX. ! Undefined control sequence.)
Solution
Let be (not respectively). Then we have four points , and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that , or .
Now, we must find how to match up to so that the above equation has a solution. On the interval , we have , and so the sum of the largest and the smallest is equal to the sum of the other two, namely, .
Now, we perform some algebraic manipulation to find :
Solve the quadratic to find , so that .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.