Difference between revisions of "2017 AIME II Problems/Problem 10"
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Rectangle <math>ABCD</math> has side lengths <math>AB=84</math> and <math>AD=42</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>, point <math>N</math> is the trisection point of <math>\overline{AB}</math> closer to <math>A</math>, and point <math>O</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{DN}</math>. Point <math>P</math> lies on the quadrilateral <math>BCON</math>, and <math>\overline{BP}</math> bisects the area of <math>BCON</math>. Find the area of <math>\triangle CDP</math>. | Rectangle <math>ABCD</math> has side lengths <math>AB=84</math> and <math>AD=42</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>, point <math>N</math> is the trisection point of <math>\overline{AB}</math> closer to <math>A</math>, and point <math>O</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{DN}</math>. Point <math>P</math> lies on the quadrilateral <math>BCON</math>, and <math>\overline{BP}</math> bisects the area of <math>BCON</math>. Find the area of <math>\triangle CDP</math>. | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
pair A,B,C,D,M,n,O,P; | pair A,B,C,D,M,n,O,P; | ||
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Solution Altered By conantwiz2023 and IronicNinja~ | Solution Altered By conantwiz2023 and IronicNinja~ | ||
+ | |||
+ | ==Solution 2 (No Coordinates)== | ||
+ | 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 horizontal segment <math>\overline{MQ}</math> from side <math>\overline{AD}</math> to <math>\overline{ND}</math>. Since <math>M</math> is the midpoint of side <math>\overline{AD}</math>, <math>\overline{MQ}=14</math>. Denote <math>\overline{OE}</math> as <math>a</math>. Noting that <math>\triangle{MOQ}</math> and <math>\triangle{COD}</math> are similar, we can write the statement <math>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}\implies \frac{84}{a}=\frac{14}{21-a}\implies a=18</math>. Using this information, the area of <math>\triangle{DOC}</math> and <math>\triangle{AND}</math> are <math>\frac{18\cdot 84}{2}=756</math> and <math>\frac{28\cdot 42}{2}=588</math>, respectively. Thus, the area of quadrilaterial <math>BCON</math> is <math>84\cdot 42-588-756=2184</math>. 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 <math>\frac{2184}{2}=1092</math>. Given this, drop altitude <math>\overline{PF}</math> (let's call it <math>b</math>) onto <math>\overline{BC}</math>. Therefore, <math>\frac{42b}{2}=1092\implies b=52</math>. 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 <math>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}\implies \frac{\overline{PG}}{52}=\frac{21}{84}\implies \overline{PG}=13</math>. The area of <math>\triangle{CDP}</math> is given by <math>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</math> ~blitzkrieg21 | ||
=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 17:43, 18 February 2019
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 Altered By conantwiz2023 and IronicNinja~
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 horizontal segment
from side
to
. Since
is the midpoint of side
,
. Denote
as
. Noting that
and
are similar, we can write the statement
. Using this information, the area of
and
are
and
, respectively. Thus, the area of quadrilaterial
is
. Now, it is clear that point
lies on side
, so the area of
is
. Given this, drop altitude
(let's call it
) onto
. Therefore,
. From here, drop an altitude
onto
. Recognizing that
and that
and
are similar, we write
. The area of
is given by
~blitzkrieg21
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.