2003 AMC 10B Problems/Problem 20
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Problem
In rectangle and
. Points
and
are on
so that
and
. Lines
and
intersect at
. Find the area of
.
![[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=0; pair A=(0,0), B=(5,0), C=(5,3), D=(0,3); pair F=(1,3), G=(3,3); pair E=(5/3,5); draw(A--B--C--D--cycle); draw(A--E); draw(B--E); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$1$",midpoint(D--F),N); label("$2$",midpoint(G--C),N); label("$5$",midpoint(A--B),S); label("$3$",midpoint(A--D),W); [/asy]](http://latex.artofproblemsolving.com/8/b/4/8b4fceed0d7e03e19b1507f86778bc77a8b7f334.png)
Solution 1
because
The ratio of
to
is
since
and
from subtraction. If we let
be the height of
The height is so the area of
is
.
Solution 2
We can look at this diagram as if it were a coordinate plane with point being
. This means that the equation of the line
is
and the equation of the line
is
. From this we can set of the follow equation to find the
coordinate of point
:
We can plug this into one of our original equations to find that the coordinate is
, meaning the area of
is
Solution 3
At points and
, segment
is 5 units from segment
. At points
and
, the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.
Then calculate the area of trapezoid and triangle
separately and add them. The area of the trapezoid is
and the area of the triangle is
.
Solution 4
Since then
, where
and
are ponts on
and
respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of
, or something like this
we have to find the ratio of the areas when the sides have shrunk by length
Let be the area of the shape whose length is
Now comparing the ratios of
to
we get
By applying an infinite summation
Solution 5
Drop a perpendicular from to
and call the intersection point
and
are similar and so are
and
(You can prove this with
by observing that
is congruent to
and so on). This means that
so
which means that
Then,
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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