Difference between revisions of "2017 AIME II Problems/Problem 10"
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Since the problem tells us that segment <math>\overline{BP}</math> bisects the area of quadrilateral <math>BCON</math>, let us compute the area of <math>BCON</math> by subtracting the areas of <math>\triangle{AND}</math> and <math>\triangle{DOC}</math> from rectangle <math>ABCD</math>. | Since the problem tells us that segment <math>\overline{BP}</math> bisects the area of quadrilateral <math>BCON</math>, let us compute the area of <math>BCON</math> by subtracting the areas of <math>\triangle{AND}</math> and <math>\triangle{DOC}</math> from rectangle <math>ABCD</math>. | ||
− | To do this, drop altitude <math>\overline{OE}</math> onto side <math>\overline{DC}</math> and draw a | + | To do this, drop altitude <math>\overline{OE}</math> onto side <math>\overline{DC}</math> and draw a segment <math>\overline{MQ}</math> parallel to <math>\overline{AN} from side </math>\overline{AD}<math> to </math>\overline{ND}<math>. Since </math>M<math> is the midpoint of side </math>\overline{AD}<math>, <cmath>\overline{MQ}=14</cmath> Denote </math>\overline{OE}<math> as </math>a<math>. Noting that </math>\triangle{MOQ}~\triangle{COD}<math>, we can write the statement <cmath>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}</cmath> <cmath>\implies \frac{84}{a}=\frac{14}{21-a}</cmath> <cmath>\implies a=18</cmath> Using this information, the area of </math>\triangle{DOC}<math> and </math>\triangle{AND}<math> are <cmath>\frac{18\cdot 84}{2}=756</cmath> and <cmath>\frac{28\cdot 42}{2}=588</cmath> respectively. Thus, the area of quadrilateral </math>BCON<math> is <cmath>84\cdot 42-588-756=2184</cmath> Now, it is clear that point </math>P<math> lies on side </math>\overline{MC}<math>, so the area of </math>\triangle{BPC}<math> is <cmath>\frac{2184}{2}=1092</cmath> Given this, drop altitude </math>\overline{PF}<math> (let's call it </math>b<math>) onto </math>\overline{BC}<math>. Therefore, <cmath>\frac{42b}{2}=1092\implies b=52</cmath> From here, drop an altitude </math>\overline{PG}<math> onto </math>\overline{DC}<math>. Recognizing that </math>\overline{PF}=\overline{GC}<math> and that </math>\triangle{MDC}<math> and </math>\triangle{PGC}<math> are similar, we write <cmath>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}</cmath> <cmath>\implies \frac{\overline{PG}}{52}=\frac{21}{84}</cmath> <cmath>\implies \overline{PG}=13</cmath> The area of </math>\triangle{CDP}$ is given by <cmath>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</cmath> ~blitzkrieg21 and jdong2006 |
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=9|num-a=11}} | {{AIME box|year=2017|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:21, 27 January 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.