2018 AMC 12B Problems/Problem 13

Revision as of 14:40, 24 September 2021 by MRENTHUSIASM (talk | contribs) (The original Solutions 2 and 3 are very similar and should be combined. I will combine them into Sol 1, as it is the most educational. Credits are retained to the original authors)

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 (Similar Triangles and Area Ratios)

IN CONSTRUCTION. NO EDIT PLEASE

~RandomPieKevin ~Kyriegon ~MRENTHUSIASM

Solution 2 (Coordinate Geometry)

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 $\left(\frac{x}{3},10+\frac{y}{3}\right),\left(10+\frac{x}{3},\frac{y}{3}\right),\left(20+\frac{x}{3},10+\frac{y}{3}\right),$ and $\left(10+\frac{x}{3},20+\frac{y}{3}\right).$ Shifting the coordinates down by $\left(\frac x3,\frac y3\right)$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 $\boxed{\textbf{(C) }200}.$

Solution 3 (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 $\boxed{\textbf{(C) }200}.$

Video Solution (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=1439

~ pi_is_3.14

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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