Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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Phoenixfire (talk | contribs) (→Solution 2) |
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<cmath>FK = \sqrt{1-y^2}</cmath> | <cmath>FK = \sqrt{1-y^2}</cmath> | ||
<cmath>EJ = x\sqrt{1-y^2}</cmath> | <cmath>EJ = x\sqrt{1-y^2}</cmath> | ||
− | Because we know that <math>BE+EJ+AJ = EK + FK = 1</math>, we can set up a systems of equations | + | Because we know that <math>BE+EJ+AJ = EK + FK = 1</math>, we can set up a systems of equations and solving for <math>x</math>, we get |
− | <cmath>x + x\sqrt{1-y^2} + y = 1</cmath> | + | <cmath>x + x\sqrt{1-y^2} + y = 1 \implies x= \frac{1-y}{1+\sqrt{1-y^2}}</cmath> |
− | <cmath>xy + \sqrt{1-y^2} = 1 | + | <cmath>xy + \sqrt{1-y^2} = 1 \implies x= \frac{1-\sqrt{1-y^2}}{y}</cmath> |
− | + | Now solving for <math>y</math>, we get | |
− | + | <cmath>\frac{1-y}{1+\sqrt{1-y^2}}=\frac{1-\sqrt{1-y^2}}{y} \implies y(1-y)=(1-\sqrt{1-y^2})(1+\sqrt{1-y^2}) \implies y-y^2=y^2 \implies y=\frac{1}{2}</cmath> | |
− | + | Plugging into the second equations with <math>x</math>, we get | |
− | <cmath>\frac | ||
− | Plugging into the | ||
<cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | <cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | ||
Revision as of 08:28, 27 January 2021
Contents
Problem
In the figure, is a square of side length
. The rectangles
and
are congruent. What is
?
Solutions
Solution 1
Draw the altitude from to
and call the foot
. Then
. Consider
. It is the hypotenuse of both right triangles
and
, and we know
, so we must have
by Hypotenuse-Leg congruence. From this congruence we have
.
Notice that all four triangles in this picture are similar. Also, we have . So set
and
. Now
. This means
, so
is the midpoint of
. So
, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have
and subsequently
. This means
, which gives
, so the answer is
.
Solution 2
Let . Let
. Because
and
,
are all similar. Using proportions and the pythagorean theorem, we find
Because we know that
, we can set up a systems of equations and solving for
, we get
Now solving for
, we get
Plugging into the second equations with
, we get
Solution 3
Let ,
, and
. Then
and because
and
,
. Furthermore, the area of the four triangles and the two rectangles sums to 1:
By the Pythagorean theorem:
Then by the rational root theorem, this has roots ,
, and
. The first and last roots are extraneous because they imply
and
, respectively, thus
.
Solution 4
Let =
and
=
. It is shown that all four triangles in the picture are similar. From the square side lengths:
Solving for
we get:
Solution 5
Note that is a diagonal of
, so it must be equal in length to
. Therefore, quadrilateral
has
, and
, so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of
and
, we see that it must be a parallelogram. Therefore,
. But by the symmetry in rectangle
, we see that
. Therefore,
. We also know that
, hence
.
As and
, and as
is right, we know that
must be a 30-60-90 triangle. Therefore,
and
. But by similarity,
is also a 30-60-90 triangle, hence
. But
, hence
. As
, this implies that
. Thus the answer is
.
See Also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.