2018 AMC 12B Problems/Problem 13
Contents
Problem
Square has side length
. Point
lies inside the square so that
and
. The centroids of
,
,
, and
are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
Solution 1 (Drawing an Accurate Diagram)
We can draw an accurate diagram by using centimeters and scaling everything down by a factor of . The centroid is the intersection of the three medians in a triangle.
After connecting the centroids, we see that the quadrilateral looks like a square with side length of
. However, we scaled everything down by a factor of
, so the length is
. The area of a square is
, so the area is:
Solution 2
The centroid of a triangle is of the way from a vertex to the midpoint of the opposing side. Thus, the length of any diagonal of this quadrilateral is
. The diagonals are also parallel to sides of the square, so they are perpendicular to each other, and so the area of the quadrilateral is
,
.
Solution 3
The midpoints of the sides of the square form another square, with side length and area
. Dilating the corners of this square through point
by a factor of
results in the desired quadrilateral (also a square). The area of this new square is
of the area of the original dilated square. Thus, the answer is
Solution 4
We put the diagram on a coordinate plane. The coordinates of the square are and the coordinates of point P are
By using the centroid formula, we find that the coordinates of the centroids are
and
Shifting the coordinates down by
does not change its area, and we ultimately get that the area is equal to the area covered by
which has an area of
Solution 5 Lazy solution
Since the problem does not constrain , it indicates that the result does not depend on the location of
. So let's put it in the center. Then it becomes obvious the quadrilateral is a square side length equal 2/3 of the diagonal of each quarter of the original square, which is
, so area is
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 AMC 12 Problems and Solutions |
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