Problem:
Square ABCD has side length 30 . Point P lies inside the square so that AP=12 and BP=26. The centroids of β³ABP,β³BCP, β³CDP, and β³DAP are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
Answer Choices:
A. 1002β
B. 1003β
C. 200
D. 2002β
E. 2003β Solution:
Let E and F be the midpoints of sides BC and CD, respectively. Let G and H be the centroids of β³BCP and β³CDP, respectively. Then G is on PE, a median of β³BCP, a distance 32β of the way from P to E. Similarly, H is on PF a distance 32β of the way from P to F. Thus GH is parallel to EF and 32β the length of EF. Because BC=30, it follows that EC=15,EF=152β, and GH=102β. The midpoints of AB,BC,CD, and DA form a square, so the centroids of β³ABP,β³BCP,β³CDP, and β³DAP also form a square, and that square has side length 102β. The requested area is (102β)2=200β.
OR
Place the figure in the coordinate plane with A=(0,30),B=(0,0), C=(30,0),D=(30,30), and P=(3x,3y). Then the coordinates of the centroids of the four triangles are found by averaging the coordinates of the vertices: (x,y+10),(x+10,y),(x+20,y+10), and (x+10,y+20). It can be seen that the quadrilateral formed by the centroids is a square with center (x+10,y+10) and vertices aligned vertically and horizontally. Its area is half the product of the lengths of its diagonals, 21ββ 20β 20=200β.
Note: As the solutions demonstrate, the inner quadrilateral is always a square, and its size is independent of the location of point P. The location of the square within square ABCD does depend on the location of P.