2017 AIME II Problems/Problem 10
Problem
Rectangle has side lengths
and
. Point
is the midpoint of
, point
is the trisection point of
closer to
, and point
is the intersection of
and
. Point
lies on the quadrilateral
, and
bisects the area of
. Find the area of
.
Solution 1
Impose a coordinate system on the diagram where point
is the origin. Therefore
,
,
, and
. Because
is a midpoint and
is a trisection point,
and
. The equation for line
is
and the equation for line
is
, so their intersection, point
, is
. Using the shoelace formula on quadrilateral
, or drawing diagonal
and using
, we find that its area is
. Therefore the area of triangle
is
. Using
, we get
. Simplifying, we get
. This means that the x-coordinate of
. Since P lies on
, you can solve and get that the y-coordinate of
is
. Therefore the area of
is
.
Solution 2 (No Coordinates)
Since the problem tells us that segment bisects the area of quadrilateral
, let us compute the area of
by subtracting the areas of
and
from rectangle
.
To do this, drop altitude onto side
and draw a segment
parallel to
\overline{AD}
\overline{ND}
M
\overline{AD}
\overline{OE}
a
\triangle{MOQ}~\triangle{COD}
\triangle{DOC}
\triangle{AND}
BCON
P
\overline{MC}
\triangle{BPC}
\overline{PF}
b
\overline{BC}
\overline{PG}
\overline{DC}
\overline{PF}=\overline{GC}
\triangle{MDC}
\triangle{PGC}
\triangle{CDP}$ is given by
~blitzkrieg21 and jdong2006
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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