Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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Then by the rational root theorem, this has roots <math>\frac{1}{4}</math>, <math>2 - \sqrt{3}</math>, and <math>2 + \sqrt{3}</math>. The first and last roots are extraneous because they imply <math>a = 0</math> and <math>x > 1</math>, respectively, thus <math>x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</math>. | Then by the rational root theorem, this has roots <math>\frac{1}{4}</math>, <math>2 - \sqrt{3}</math>, and <math>2 + \sqrt{3}</math>. The first and last roots are extraneous because they imply <math>a = 0</math> and <math>x > 1</math>, respectively, thus <math>x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>\angle FKH</math> = <math>k</math> and <math>CF</math> = <math>a</math>. It is shown that all four triangles in the picture are similar. From the square side lengths: | ||
+ | |||
+ | |||
+ | [center]<math>a + sin(k)*1 + cos(k)a = 1</math> | ||
+ | |||
+ | <math>sin(k)a + cos(k) = 1</math>[/center] | ||
+ | Solving for <math>a</math> we get: | ||
+ | |||
+ | |||
+ | [center]<math>a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}</math> | ||
+ | |||
+ | <math>(1-sin(k))*sin(k) = (1 - cos(k))*(cos(k) + 1)</math> | ||
+ | |||
+ | <math>sin(k)-sin(k)^2 = cos(k) + 1 - cos(k)^2 - cos(k)</math> | ||
+ | |||
+ | <math>sin(k)-sin(k)^2 = sin(k)^2</math> | ||
+ | |||
+ | <math>1-sin(k) = sin(k)</math> | ||
+ | |||
+ | <math>sin(k) = \frac{1}{2}, cos(k) = \frac{\sqrt 3}{2}</math> | ||
+ | |||
+ | <math>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</math> | ||
+ | [/center] | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:56, 29 November 2016
Problem 21
In the figure, is a square of side length
. The rectangles
and
are congruent. What is
?
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
Solving for
in the second equation, we get
Plugging this into the first equation, we get
Plugging into the previous equation 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:
[center]
[/center]
Solving for
we get:
[center]
[/center]
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 |
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