Difference between revisions of "2020 AMC 12B Problems/Problem 18"
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==Video Solution 2 by the Beauty of Math== | ==Video Solution 2 by the Beauty of Math== | ||
− | + | https://youtu.be/VZYe3Hu88OA?t=189 | |
==See Also== | ==See Also== |
Revision as of 20:57, 11 February 2021
Contents
[hide]Problem
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
Since the total area is , the side length of square
is
. We see that since triangle
is a right isosceles triangle with area 1, we can determine sides
and
both to be
. Now, consider extending
and
until they intersect. Let the point of intersection be
. We note that
is also a right isosceles triangle with side
and find it's area to be
. Now, we notice that
is also a right isosceles triangle and find it's area to be
. This is also equal to
or
. Since we are looking for
, we want two times this. That gives
.~TLiu
Solution 2 (Lucky Measuring)
Since this is a geometry problem involving sides, and we know that is
, we can use our ruler and find the ratio between
and
. Measuring(on the booklet), we get that
is about
inches and
is about
inches. Thus, we can then multiply the length of
by the ratio of
of which we then get
We take the square of that and get
and the closest answer to that is
. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)
This cannot work if the problem says not to scale - awu2014
Note that this will only work if the diagram is to scale, and at the start of the test, they mention that all diagrams are not necessarily to scale (whether or not the problem states that). Therefore, if you are to use this strategy on a problem, you are betting on the fact that this diagram IS to scale, so only use it as a last resort.
Solution 3
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 4
Plot a point such that
and
are parallel 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 5 (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
-edit: annabelle0913
Solution 6
Easily, we can find that: quadrilateral
and
are congruent with each other, so we can move
to the shaded area (
and
,
and
overlapping) to form a square
(
=
,
=
,
=
=
so
). Then we can solve
=
=
,
=
,
=
.
=
of
of
of
=
--Ryan Zhang @BRS
Video Solution 1
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
Video Solution 2 by the Beauty of Math
https://youtu.be/VZYe3Hu88OA?t=189
See Also
2020 AMC 10B (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 10 Problems and Solutions |
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.