Difference between revisions of "2020 AMC 12B Problems/Problem 18"
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We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | ||
− | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math> | + | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. |
− | Notice that since the area of triangle <math>AEH</math> is 1 and <math>AE=AH</math> , <math>AE=AH=\sqrt{2}</math>, therefore <math>BE=HD=2-\sqrt{2}</math> | + | Notice that since the area of triangle <math>AEH</math> is 1 and <math>AE=AH</math> , <math>AE=AH=\sqrt{2}</math>, therefore <math>BE=HD=2-\sqrt{2}</math>. |
− | Let <math>CG=GF=m</math>, then <math>BF=DG=2-m</math> | + | Let <math>CG=GF=m</math>, then <math>BF=DG=2-m</math>. |
− | Also notice that <math>KB=BE=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math> | + | Also notice that <math>KB=BE=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>. |
Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation: | Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation: | ||
<math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math> | <math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math> | ||
− | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> | + | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>. |
Now notice that | Now notice that | ||
<math>FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> | <math>FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> | ||
<math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math> | <math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math> | ||
− | <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math> | + | <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>. |
− | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math> -HarryW | + | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW |
Revision as of 02:07, 9 February 2020
In square , points
and
lie on
and
, respectively, so that
Points
and
lie on
and
, respectively, and points
and
lie on
so that
and
. See the figure below. Triangle
, quadrilateral
, quadrilateral
, and pentagon
each has area
What is
?
Solution 1
Plot a point such that
and
are collinear and extend line
to point
such that
forms a square. Extend line
to meet line
and point
is the intersection of the two. The area of this square is equivalent to
. We see that the area of square
is
, meaning each side is of length 2. The area of the pentagon
is
. Length
, thus
. Triangle
is isosceles, and the area of this triangle is
. Adding these two areas, we get
. --OGBooger
Solution 2
Draw the auxiliary line . Denote by
the point it intersects with
, and by
the point it intersects with
. Last, denote by
the segment
, and by
the segment
. We will find two equations for
and
, and then solve for
.
Since the overall area of is
, and
. In addition, the area of
.
The two equations for and
are then:
Length of
:
Area of CMIF:
.
Substituting the first into the second, yields
Solving for gives
~DrB
Solution 3(HARD Calculation)
We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1.
Extend
and let the intersection with
be
. Connect
, and let the intersection of
and
be
.
Notice that since the area of triangle
is 1 and
,
, therefore
.
Let
, then
.
Also notice that
, thus
.
Now use the condition that the area of quadrilateral
is 1, we can set up the following equation:
We solve the equation and yield
.
Now notice that
.
Hence
. -HarryW
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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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