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

(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)
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==Solution 1 (Similar Triangles and Area Ratios)==
 
==Solution 1 (Similar Triangles and Area Ratios)==
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<b>IN CONSTRUCTION. NO EDIT PLEASE</b>
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As shown below, let <math>M_1,M_2,M_3,M_4</math> be the midpoints of <math>\overline{AB},\overline{BC},\overline{CD},\overline{DA},</math> respectively, and <math>G_1,G_2,G_3,G_4</math> be the centroids of <math>\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},</math> respectively.
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<asy>
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unitsize(210);
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pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
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pair M1 = midpoint(A--B);
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pair M2 = midpoint(B--C);
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pair M3 = midpoint(C--D);
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pair M4 = midpoint(D--A);
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pair G1 = centroid(A,B,P);
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pair G2 = centroid(B,C,P);
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pair G3 = centroid(C,D,P);
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pair G4 = centroid(D,A,P);
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filldraw(M1--M2--P--cycle,red);
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filldraw(M2--M3--P--cycle,yellow);
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filldraw(M3--M4--P--cycle,green);
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filldraw(M4--M1--P--cycle,lightblue);
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draw(A--B--C--D--cycle);
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draw(M1--M2--M3--M4--cycle);
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draw(G1--G2--G3--G4--cycle);
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dot(P);
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defaultpen(fontsize(10pt));
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draw(A--P--B);
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draw(C--P--D);
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label("$A$", A, W);
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label("$B$", B, W);
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label("$C$", C, E);
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label("$D$", D, E);
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label("$P$", P, N);
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label("$M_1$", M1, W);
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label("$M_2$", M2, S);
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label("$M_3$", M3, E);
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label("$M_4$", M4, N);
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label("$G_1$", G1, 1.5S);
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label("$G_2$", G2, 1.5E);
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label("$G_3$", G3, 1.5NE);
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label("$G_4$", G4, 1.5E);
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dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(M1);
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dot(M2);
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dot(M3);
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dot(M4);
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dot(G1);
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dot(G2);
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dot(G3);
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dot(G4);
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</asy>
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By SAS, we conclude that <math>\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,</math> and <math>\triangle G_4G_1P\sim\triangle M_4M_1P.</math> By the properties of centroids, the side-length ratio for each pair of triangles is <math>\frac23,</math> so the area ratio for each pair of triangles is <math>\left(\frac23\right)^2=\frac49.</math>
  
 +
Let the brackets denote areas. Since <math>[ABCD]=30^2=900,</math> it follows that
 +
<cmath>\begin{align*}
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[G_1G_2G_3G_4]&=[G_1G_2P]+[G_2G_3P]+[G_3G_4P]+[G_4G_1P] \\
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&=\frac49[M_1M_2P]+\frac49[M_2M_3P]+\frac49[M_3M_4P]+\frac49[M_4M_1P] \\
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&=\frac49([M_1M_2P]+[M_2M_3P]+[M_3M_4P]+[M_4M_1P]) \\
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&=\frac49[M_1M_2M_3M_4] \\
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&=\frac29[ABCD]
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&=\boxed{\textbf{(C) }200}.
 +
\end{align*}</cmath>
 
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
 
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
  

Revision as of 15:20, 24 September 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 and Area Ratios)

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 side-length ratio for each pair of triangles is $\frac23,$ so the area ratio for each pair of triangles is $\left(\frac23\right)^2=\frac49.$

Let the brackets denote areas. Since $[ABCD]=30^2=900,$ it follows that \begin{align*} [G_1G_2G_3G_4]&=[G_1G_2P]+[G_2G_3P]+[G_3G_4P]+[G_4G_1P] \\ &=\frac49[M_1M_2P]+\frac49[M_2M_3P]+\frac49[M_3M_4P]+\frac49[M_4M_1P] \\ &=\frac49([M_1M_2P]+[M_2M_3P]+[M_3M_4P]+[M_4M_1P]) \\ &=\frac49[M_1M_2M_3M_4] \\ &=\frac29[ABCD] &=\boxed{\textbf{(C) }200}. \end{align*} ~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 (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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