Difference between revisions of "2013 AMC 12B Problems/Problem 20"
m (Fixed broken Latex (answer choices)) |
(→Solution) |
||
(5 intermediate revisions by 2 users not shown) | |||
Line 2: | Line 2: | ||
For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? | For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? | ||
− | <math>\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}</math> | + | <math>\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}</math> |
==Solution== | ==Solution== | ||
− | Let <math>f,g,h,j</math> be <math>\ | + | Let <math>f,g,h,</math> and <math>j</math> be <math>\cos{x}, \cot{x}, \sin{x},</math> and <math>\tan{x}</math>, 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>. |
Latest revision as of 22:19, 11 August 2025
Problem
For , points
and
are the vertices of a trapezoid. What is
?
Solution
Let and
be
and
, 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.