Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| + | The centroid of a triangle is <math>\frac{2}{3}</math> of the way from a vertex to the midpoint of the opposing side. Thus, the length of any diagonal of this quadrilateral is <math>20</math>. 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 <math>\frac{20\cdot20}{2} = 200</math>, <math>\boxed{(E)}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 18:11, 16 February 2018
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?
![[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]](http://latex.artofproblemsolving.com/4/a/7/4a7552d7af26cb2be88658376dd5934a70ef16c6.png)
Solution
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 
, 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
