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{ | + | 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> |
==See Also== | ==See Also== |
Revision as of 20:53, 16 February 2018
Contents
[hide]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
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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