Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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Now we apply Barycentric Coordinates w.r.t. <math>\triangle ACD</math>. We let <math>A=(1,0,0),D=(0,1,0),C=(0,0,1)</math>. Then <math>E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)</math>. | Now we apply Barycentric Coordinates w.r.t. <math>\triangle ACD</math>. We let <math>A=(1,0,0),D=(0,1,0),C=(0,0,1)</math>. Then <math>E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)</math>. | ||
− | In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.</cmath> Let <math>[FEC]=x</math> so that <math>[ACD]=45+x</math>. Then we have <math>\frac{x}{x+45}=\frac 16 \Rightarrow x=9</math> so the answer is <math>2(45+9)=108</math>. | + | In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.</cmath> Let <math>[FEC]=x</math> so that <math>[ACD]=45+x</math>. Then we have <math>\frac{x}{x+45}=\frac 16 \Rightarrow x=9</math> so the answer is <math>2(45+9)=108</math>. |
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==Video Solution== | ==Video Solution== |
Revision as of 00:32, 27 December 2020
Contents
Problem 22
Point is the midpoint of side
in square
and
meets diagonal
at
The area of quadrilateral
is
What is the area of
Solution 1
Let the area of be
. Thus, the area of triangle
is
and the area of the square is
.
By AA similarity, with a 1:2 ratio, so the area of triangle
is
. Now consider trapezoid
. Its area is
, which is three-fourths the area of the square. We set up an equation in
:
Solving, we get
. The area of square
is
.
Solution 2
We can use analytic geometry for this problem.
Let us start by giving the coordinate
,
the coordinate
, and so forth.
and
can be represented by the equations
and
, respectively. Solving for their intersection gives point
coordinates
.
Now, ’s area is simply
or
. This means that pentagon
’s area is
of the entire square, and it follows that quadrilateral
’s area is
of the square.
The area of the square is then .
Solution 3
Note that triangle has half the area of the square and triangle
has
th. Thus the area of the quadrilateral is
th the area of the square. The area of the square is then
.
Solution 4
Extend and
to meet at
. Drop an altitude from
to
and call it
. Also, call
. As stated before, we have
, so the ratio of their heights is in a
ratio, making the altitude from
to
. Note that this means that the side of the square is
. In addition,
by AA Similarity in a
ratio. This means that the side length of the square is
, making
.
Now, note that . We have
and
Subtracting makes
We are given that
so
Therefore,
so our answer is
- moony_eyed
Solution 5
Solution with Cartesian and Barycentric Coordinates:
We start with the following
Claim: Given a square , let
be the midpoint of
and let
. Then
.
Proof. We use Cartesian coordinates. Let be the origin,
. We have that
and
are governed by the equations
and
, respectively. Solving,
. The result follows.
Now we apply Barycentric Coordinates w.r.t. . We let
. Then
.
In the barycentric coordinate system, the area formula is where
is a random triangle and
is the reference triangle. Using this, we find that
Let
so that
. Then we have
so the answer is
.
Video Solution
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
Set s to be the bottom left triangle.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.