Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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Let <math>BE = x</math>, <math>KE = y</math> | Let <math>BE = x</math>, <math>KE = y</math> | ||
− | <math>DG = y</math>, <math>AG = 1-y</math>, <math>KJ = x</math>, <math>AJ = \frac{y}{x}</math>, <math>EJ = 1 - x - \frac{y}{x}</math>, <math>JG = 1</math> | + | <math>DG = y</math>, <math>AG = 1-y</math>, <math>KJ = x</math>, <math>AJ = JG \cdot \frac{KE}{JK} = \frac{y}{x}</math>, <math>EJ = 1 - x - \frac{y}{x}</math>, <math>JG = 1</math> |
− | <math>\frac{AG}{JG} = \frac{EJ}{ | + | <math>\frac{AG}{JG} = \frac{EJ}{JK}</math>, <math>1-y = \frac{1 - x - \frac{y}{x}}{x}</math>, <math>x - xy = 1 - x - \frac{y}{x}</math>, <math>2x^2 - x^2y + y - x = 0</math> <math>(1)</math> |
− | <math> | + | <math>AJ^2 + AG^2 = JG^2</math>, <math>\left( \frac{y}{x} \right) ^2 + (1-y)^2 = 1</math>, <math>\frac{y^2}{x^2} + 1 - 2y + y^2 = 1</math>, <math>x^2 y -2x^2 + y = 0</math> <math>(2)</math> |
− | <math> | + | <math>(1) + (2)</math>, <math>2y - x = 0</math>, <math>y = \frac{x}{2}</math> |
− | + | Substitute <math>y</math> into <math>(2)</math> we get <math>x^2 \cdot \frac{x}{2} - 2 x^2 + \frac{x}{2} = 0</math>, <math>x^2 - 4x + 1 = 0</math> | |
<math>x = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}</math>, as <math>x < 1</math>, <math>x = \boxed{\textbf{(C)}2 - \sqrt{3}}</math> | <math>x = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}</math>, as <math>x < 1</math>, <math>x = \boxed{\textbf{(C)}2 - \sqrt{3}}</math> |
Revision as of 01:02, 1 January 2023
Contents
[hide]Problem
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 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
.
Solution 6
Let ,
,
,
,
,
,
,
,
,
,
,
,
,
,
Substitute into
we get
,
, as
,
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.