Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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− | Note that <math>HJ</math> is a diagonal of <math>JKHG</math>, so it must be equal in length to <math>FB</math>. Therefore, quadrilateral <math> | + | Note that <math>HJ</math> is a diagonal of <math>JKHG</math>, so it must be equal in length to <math>FB</math>. Therefore, quadrilateral <math>FHJB</math> has <math>FH\parallel BJ</math>, and <math>FB=HJ</math>, so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of <math>FB</math> and <math>HJ</math>, we see that it must be a parallelogram. Therefore, <math>FH=BJ</math>. But by the symmetry in rectangle <math>FEAD</math>, we see that <math>FH=JA</math>. Therefore, <math>BJ=FH=JA</math>. We also know that <math>BJ+JA=1</math>, hence <math>BJ=JA=\frac12</math>. |
As <math>JG=1</math> and <math>JA=\frac12</math>, and as <math>\triangle GJA</math> is right, we know that <math>\triangle GJA</math> must be a 30-60-90 triangle. Therefore, <math>GA=\sqrt{3}/2</math> and <math>DG=1-\sqrt{3}/2</math>. But by similarity, <math>\triangle DHG</math> is also a 30-60-90 triangle, hence <math>DH=\sqrt{3}-3/2</math>. But <math>\triangle DHG\cong\triangle EJK</math>, hence <math>EJ=\sqrt{3}-3/2</math>. As <math>BJ=1/2</math>, this implies that <math>BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}</math>. Thus the answer is <math>\boxed{\textbf{(C)}}</math>. | As <math>JG=1</math> and <math>JA=\frac12</math>, and as <math>\triangle GJA</math> is right, we know that <math>\triangle GJA</math> must be a 30-60-90 triangle. Therefore, <math>GA=\sqrt{3}/2</math> and <math>DG=1-\sqrt{3}/2</math>. But by similarity, <math>\triangle DHG</math> is also a 30-60-90 triangle, hence <math>DH=\sqrt{3}-3/2</math>. But <math>\triangle DHG\cong\triangle EJK</math>, hence <math>EJ=\sqrt{3}-3/2</math>. As <math>BJ=1/2</math>, this implies that <math>BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}</math>. Thus the answer is <math>\boxed{\textbf{(C)}}</math>. |
Revision as of 17:33, 7 January 2017
Contents
[hide]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:
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.