Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 12B Problems/Problem 13"
(→‎See Also)
Line 43: Line 43:
 
{{AMC12 box|year=2018|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2018|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 16:30, 18 June 2018

Problem

Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ 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]


$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

Solution 1 (Drawing an Accurate Diagram)

We can draw an accurate diagram by using centimeters and scaling everything down by a factor of $2$. The centroid is the intersection of the three medians in a triangle.

After connecting the $4$ centroids, we see that the quadrilateral looks like a square with side length of $7$. However, we scaled everything down by a factor of $2$, so the length is $14$. The area of a square is $s^2$, so the area is: \[\boxed{\textbf{(C) } 200}.\]

Solution 2

The centroid of a triangle is $\frac{2}{3}$ of the way from a vertex to the midpoint of the opposing side. Thus, the length of any diagonal of this quadrilateral is $20$. 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 $\frac{20\cdot20}{2} = 200$, $\boxed{(C)}$.

Solution 3

The midpoints of the sides of the square form another square, with side length $15\sqrt{2}$ and area $450$. Dilating the corners of this square through point $P$ by a factor of $3:2$ results in the desired quadrilateral (also a square). The area of this new square is $\frac{2^2}{3^2}$ of the area of the original dilated square. Thus, the answer is $\frac{4}{9} * 450 = \boxed {C}$

Solution 4

We put the diagram on a coordinate plane. The coordinates of the square are $(0,0),(30,0),(30,30),(0,30)$ and the coordinates of point P are $(x,y).$ By using the centroid formula, we find that the coordinates of the centroids are $(\frac{x}{3},10+\frac{y}{3}),(10+\frac{x}{3},\frac{y}{3}),(20+\frac{x}{3},10+\frac{y}{3}),$ and $(10+\frac{x}{3},20+\frac{y}{3}).$ Shifting the coordinates down by $(\frac x3,\frac y3)$does not change its area, and we ultimately get that the area is equal to the area covered by $(0,10),(10,0),(20,10),(10,20)$ which has an area of $\boxed{(C) 200.}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_13&oldid=95363"