Difference between revisions of "2014 AMC 12B Problems/Problem 21"
m (→Solution 3) |
(→Solution 3) |
||
Line 32: | Line 32: | ||
==Solution 3== | ==Solution 3== | ||
− | Let <math>BE = x</math>, <math>EK = a</math>, and <math>EJ = b</math>. Then <math>x^2 = a^2 + b^2</math> and because <math>\triangle KEJ \cong GDH</math> and <math>\triangle KEJ \sim \triangle JAG</math>, <math>\frac{GA}{1} = 1 - a = \frac{b}{x}</math>. Furthermore, the area of the four triangles and the two rectangles sums to 1: | + | Let <math>BE = x</math>, <math>EK = a</math>, and <math>EJ = b</math>. Then <math>x^2 = a^2 + b^2</math> and because <math>\triangle KEJ \cong \triangle GDH</math> and <math>\triangle KEJ \sim \triangle JAG</math>, <math>\frac{GA}{1} = 1 - a = \frac{b}{x}</math>. Furthermore, the area of the four triangles and the two rectangles sums to 1: |
<cmath>1 = 2x + GA\cdot JA + ab</cmath> | <cmath>1 = 2x + GA\cdot JA + ab</cmath> |
Revision as of 20:29, 9 March 2014
Problem 21
In the figure, is a square of side length
. The rectangles
and
are congruent. What is
?
Solution 1
Draw the attitude from to
and call the foot
. Then
. Consider
. It is the hypotenuse of both right triangles
and
, and we know
and
, so we must have
.
Notice that all four triangles in this picture are similar and thus we have . This means
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
.
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.