2018 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2018 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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Concepts:centroid coordinate geometry area

Difficulty rating: 1810

13.

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?

$100\sqrt{2}$

$100\sqrt{3}$

$200$

$200\sqrt{2}$

$200\sqrt{3}$

Solution:

Place $A=(0,30),$ $B=(0,0),$ $C=(30,0),$ $D=(30,30),$ and $P=(3x,3y).$ Averaging the vertices, the four centroids are $(x,\,y+10),\ (x+10,\,y),\ (x+20,\,y+10),\ (x+10,\,y+20).$

These form a square whose diagonals, one horizontal and one vertical, each have length $20.$ Its area is $\tfrac12\cdot20\cdot20=200,$ independent of where $P$ lies.

Thus, the correct answer is C.

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