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; | ||
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); | A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); | ||
− | fill(C--D--P--cycle, | + | fill(C--D--P--cycle,blue); |
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(C--M); | draw(C--M); | ||
Line 31: | Line 31: | ||
label("56",(28,42)--(84,42),N); | label("56",(28,42)--(84,42),N); | ||
</asy> | </asy> | ||
− | <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>N</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>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>. |
+ | |||
+ | ==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 segment <math>\overline{MQ}</math> parallel to <math>\overline{AN}</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>, <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}</math> 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}} |
Latest 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
from side
to
. Since
is the midpoint of side
,
Denote
as
. Noting that
, we can write the statement
Using this information, the area of
and
are
and
respectively. Thus, the area of quadrilateral
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 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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