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?
![[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 1 (Similar Triangles and Area Ratios)
As shown below, let 
be the midpoints of 
respectively, and 
be the centroids of 
respectively.
![[asy] unitsize(210); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); pair M1 = midpoint(A--B); pair M2 = midpoint(B--C); pair M3 = midpoint(C--D); pair M4 = midpoint(D--A); pair G1 = centroid(A,B,P); pair G2 = centroid(B,C,P); pair G3 = centroid(C,D,P); pair G4 = centroid(D,A,P); filldraw(M1--M2--P--cycle,red); filldraw(M2--M3--P--cycle,yellow); filldraw(M3--M4--P--cycle,green); filldraw(M4--M1--P--cycle,lightblue); draw(A--B--C--D--cycle); draw(M1--M2--M3--M4--cycle); draw(G1--G2--G3--G4--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); label("$M_1$", M1, W); label("$M_2$", M2, S); label("$M_3$", M3, E); label("$M_4$", M4, N); label("$G_1$", G1, 1.5S); label("$G_2$", G2, 1.5E); label("$G_3$", G3, 1.5NE); label("$G_4$", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); [/asy]](http://latex.artofproblemsolving.com/3/d/d/3dd915ae976caf94a9afce0b62467cdffdaeb655.png)
By SAS, we conclude that 
and 
By the properties of centroids, the side-length ratio for each pair of triangles is 
so the area ratio for each pair of triangles is 
Let the brackets denote areas. Since ![$[ABCD]=30^2=900,$](http://latex.artofproblemsolving.com/b/f/0/bf03e8d70257033282c41fa68fdf8bc7c63a753c.png)
it follows that
![\begin{align*} [G_1G_2G_3G_4]&=[G_1G_2P]+[G_2G_3P]+[G_3G_4P]+[G_4G_1P] \\ &=\frac49[M_1M_2P]+\frac49[M_2M_3P]+\frac49[M_3M_4P]+\frac49[M_4M_1P] \\ &=\frac49([M_1M_2P]+[M_2M_3P]+[M_3M_4P]+[M_4M_1P]) \\ &=\frac49[M_1M_2M_3M_4] \\ &=\frac29[ABCD] \\ &=\boxed{\textbf{(C) }200}. \end{align*}](http://latex.artofproblemsolving.com/7/7/2/772d8c72550b6b4e35ded8a8db81e8bcc7c925ee.png)
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
Solution 2 (Coordinate Geometry)
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 3 (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
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1439
~ pi_is_3.14
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. 
