Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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==Solution 3== | ==Solution 3== | ||
The midpoints of the sides of the square form another square, with side length <math>15\sqrt{2}</math> and area <math>450</math>. Dilating the corners of this square through point <math>P</math> by a factor of <math>3:2</math> results in the desired quadrilateral (also a square). The area of this new square is <math>\frac{2^2}{3^2}</math> of the area of the original dilated square. Thus, the answer is <math>\frac{4}{9} * 450 = \boxed {C}</math> | The midpoints of the sides of the square form another square, with side length <math>15\sqrt{2}</math> and area <math>450</math>. Dilating the corners of this square through point <math>P</math> by a factor of <math>3:2</math> results in the desired quadrilateral (also a square). The area of this new square is <math>\frac{2^2}{3^2}</math> of the area of the original dilated square. Thus, the answer is <math>\frac{4}{9} * 450 = \boxed {C}</math> | ||
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+ | ==Solution 4== | ||
+ | We put the diagram on a coordinate plane. The coordinates of the square are <math>(0,0),(30,0),(30,30),(0,30)</math> and the coordinates of point P are <math>(x,y).</math> By using the centroid formula, we find that the coordinates of the centroids are <math>(\frac{x}{3},10+\frac{y}{3}),(10+\frac{x}{3},\frac{y}{3}),(20+\frac{x}{3},10+\frac{y}{3}),</math> and <math>(10+\frac{x}{3},20+\frac{y}{3}).</math> Shifting the coordinates down by <math>(\frac x3,\frac y3)</math>does not change its area, and we ultimately get that the area is equal to the area covered by <math>(0,10),(10,0),(20,10),(10,20)</math> which has an area of <math>\boxed{(C) 200.}</math> | ||
==See Also== | ==See Also== |
Revision as of 01:06, 19 February 2018
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
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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