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Difference between revisions of "2018 AMC 12B Problems/Problem 13"

(Added in Sol 2--we don't need to prove that G1G2G3G4 is a square.)
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==Solution 3 (Coordinate Geometry)==
 
==Solution 3 (Coordinate Geometry)==
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We put the diagram on a coordinate plane. The coordinates of the square are <math>(0,0),(30,0),(30,30),(0,30)</math> and the coordinates of point <math>P</math> are <math>(x,y).</math> By using the centroid formula, we find that the coordinates of the centroids are <math>\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),</math> and <math>\left(10+\frac{x}{3},20+\frac{y}{3}\right).</math> Shifting the coordinates down by <math>\left(\frac x3,\frac y3\right)</math>does not change its area, and we ultimately get that the area is equal to the area covered by <math>(0,10),(10,0),(20,10),(10,20)</math> which has an area of <math>\boxed{\textbf{(C) }200}.</math>
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This solution refers to the diagram in Solution 1.
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We place the diagram in the coordinate plane: Let <math>A=(0,30),B=(0,0),C=(30,0),D=(30,30),</math> and <math>P=(3x,3y).</math>
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Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that <math>G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),</math> and <math>G_4=(x+10,y+20).</math>
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Note that <math>G_1G_3=G_2G_4=20</math> and <math>\overline{G_1G_3}\perp\overline{G_2G_4}.</math> Therefore, the area of quadrilateral <math>G_1G_2G_3G_4</math> is <cmath>\frac12\cdot G_1G_3\cdot G_2G_4=\boxed{\textbf{(C) }200}.</cmath>
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~Pi31415926535897 ~MRENTHUSIASM
  
 
== Video Solution (Meta-Solving Technique) ==
 
== Video Solution (Meta-Solving Technique) ==

Revision as of 23:01, 4 October 2021

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)

As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ 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] By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,$ and $\triangle G_4G_1P\sim\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac23.$

Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\sqrt2.$ It follows that:

  1. Since $\overline{G_1G_2}\parallel\overline{M_1M_2},\overline{G_2G_3}\parallel\overline{M_2M_3},\overline{G_3G_4}\parallel\overline{M_3M_4},$ and $\overline{G_4G_1}\parallel\overline{M_4M_1}$ by the Converse of the Corresponding Angles Postulate, we have $\angle G_1G_2G_3=\angle G_2G_3G_4=\angle G_3G_4G_1=\angle G_4G_1G_2=90^\circ.$

  2. Since $G_1G_2=\frac23M_1M_2, G_2G_3=\frac23M_2M_3, G_3G_4=\frac23M_3M_4,$ and $G_4G_1=\frac23M_4M_1$ by the ratio of similitude, we have $G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\sqrt2.$

Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\sqrt2,$ so its area is $\left(10\sqrt2\right)^2=\boxed{\textbf{(C) }200}.$

Remark

This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1G_2G_3G_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1G_2G_3G_4$ is always a square of area \[[G_1G_2G_3G_4]=\left(\frac23\right)^2[M_1M_2M_3M_4]=\frac49[M_1M_2M_3M_4]=\frac29[ABCD].\] On the other hand, the location of $G_1G_2G_3G_4$ is dependent on the location of $P.$

~RandomPieKevin ~Kyriegon ~MRENTHUSIASM

Solution 2 (Similar Triangles)

This solution refers to the diagram in Solution 1.

By SAS, we conclude that $\triangle G_1G_3P\sim\triangle M_1M_3P$ and $\triangle G_2G_4P\sim\triangle M_2M_4P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac23.$

Note that quadrilateral $M_1M_2M_3M_4$ is a square of diagonal-length $30,$ so $\overline{M_1M_3}\perp\overline{M_2M_4}.$ Since $\overline{G_1G_3}\parallel\overline{M_1M_3}$ and $\overline{G_2G_4}\parallel\overline{M_2M_4}$ by the Converse of the Corresponding Angles Postulate, we have $\overline{G_1G_3}\perp\overline{G_2G_4}.$

Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \[\frac12\cdot G_1G_3\cdot G_2G_4 = \frac12\cdot\left(\frac23\cdot M_1M_3\right)\cdot\left(\frac23\cdot M_2M_4\right)=\boxed{\textbf{(C) }200}.\] ~Funnybunny5246 ~MRENTHUSIASM

Solution 3 (Coordinate Geometry)

This solution refers to the diagram in Solution 1.

We place the diagram in the coordinate plane: Let $A=(0,30),B=(0,0),C=(30,0),D=(30,30),$ and $P=(3x,3y).$

Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that $G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),$ and $G_4=(x+10,y+20).$

Note that $G_1G_3=G_2G_4=20$ and $\overline{G_1G_3}\perp\overline{G_2G_4}.$ Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \[\frac12\cdot G_1G_3\cdot G_2G_4=\boxed{\textbf{(C) }200}.\]

~Pi31415926535897 ~MRENTHUSIASM

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

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