Difference between revisions of "2020 AMC 12B Problems/Problem 18"
Runyangwang (talk | contribs) |
Runyangwang (talk | contribs) (→Solution 3(HARD Calculation)(Not Completed yet)) |
||
Line 55: | Line 55: | ||
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(HARD Calculation | + | ==Solution 3(HARD Calculation)== |
− | |||
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> | ||
+ | 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: | ||
+ | <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> | ||
+ | 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>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{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 | ||
+ | |||
+ | |||
{{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 01:54, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.