Difference between revisions of "2020 AMC 12B Problems/Problem 18"
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Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | ||
+ | |||
+ | ==Solution 3(a little bit complex)== | ||
+ | Extend <math>FI</math>, and denote the intersection with <math>AB</math> as <math>K</math> | ||
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:16, 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(a little bit complex)
Extend , and denote the intersection with
as
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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