Difference between revisions of "2017 AIME II Problems/Problem 10"
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label("56",(28,42)--(84,42),N); | label("56",(28,42)--(84,42),N); | ||
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− | <math>\boxed{546}</math> | + | Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>M</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2}</math> and the distance from <math>P</math> to line <math>PC</math> is <math>52</math> and its <math>x</math>-coordinate is <math>32</math>. Because <math>P</math> lies on the equation <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, <math>P</math>'s <math>y</math>-coordinate is <math>13</math>, which is also the height of <math>CDP</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2}\times13\times84=\boxed{546}</math>. |
=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 18:03, 23 March 2017
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
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
, its area is
. Therefore the area of triangle
is
and the distance from
to line
is
and its
-coordinate is
. Because
lies on the equation
,
's
-coordinate is
, which is also the height of
. Therefore the area of
is
.
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.