Difference between revisions of "2013 AMC 12B Problems/Problem 20"
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− | 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 | + | 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. The three possible ways are <math>\sin x+\cos x = \tan x+\cot x</math>, <math>\sin x+\tan x = \cos x+\cot x</math>, and <math>\sin x+\cot x = \cos x+\tan x</math>. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1</math>, so the first two of those possible ways do not work because the LHS and the RHS have | + | 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. The three possible ways are <math>\sin x+\cos x = \tan x+\cot x</math>, <math>\sin x+\tan x = \cos x+\cot x</math>, and <math>\sin x+\cot x = \cos x+\tan x</math>. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1</math>, so the first two of those possible ways do not work because the LHS and the RHS have non-overlapping ranges. Thus, it must be the last possible way, <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 21:36, 22 February 2013
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. The three possible ways are , , and . On the interval , we have , so the first two of those possible ways do not work because the LHS and the RHS have non-overlapping ranges. Thus, it must be the last possible way, .
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 |