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

(→‎Solution 4)
(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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dot(D);
 
dot(D);
 
</asy>
 
</asy>
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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution 1 (Drawing an Accurate Diagram)==
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==Solution 1 (Similar Triangles and Area Ratios)==
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We can draw an accurate diagram by using centimeters and scaling everything down by a factor of <math>2</math>. The centroid is the intersection of the three medians in a triangle.
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<b>IN CONSTRUCTION. NO EDIT PLEASE</b>
  
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After connecting the <math>4</math> centroids, we see that the quadrilateral looks like a square with side length of <math>7</math>. However, we scaled everything down by a factor of <math>2</math>, so the length is <math>14</math>. The area of a square is <math>s^2</math>, so the area is: <cmath>\boxed{\textbf{(C) } 200}.</cmath>
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~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
  
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==Solution 2==
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==Solution 2 (Coordinate Geometry)==
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The centroid of a triangle is <math>\frac{2}{3}</math> of the way from a vertex to the midpoint of the opposing side. Thus, the length of any diagonal of this quadrilateral is <math>20</math>. 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 <math>\frac{20\cdot20}{2} = 200</math>, <math>\boxed{(C)}</math>.
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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 P 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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==Solution 3==
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==Solution 3 (Accurate Diagram)==
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The midpoints of the sides of the square form another square, with side length <math>15\sqrt{2}</math> and area <math>450</math>. Dilating the corners of this square through point <math>P</math> by a factor of <math>3:2</math> results in the desired quadrilateral (also a square). The area of this new square is <math>\frac{2^2}{3^2}</math> of the area of the original dilated square. Thus, the answer is <math>\frac{4}{9} * 450 = \boxed {C}</math>
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We can draw an accurate diagram by using centimeters and scaling everything down by a factor of <math>2.</math> The centroid is the intersection of the three medians in a triangle.  
  
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==Solution 4==
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After connecting the <math>4</math> centroids, we see that the quadrilateral looks like a square with side length of <math>7.</math> However, we scaled everything down by a factor of <math>2,</math> so the length is <math>14.</math> The area of a square is <math>s^2,</math> so the area is <math>\boxed{\textbf{(C) }200}.</math>
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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 P are <math>(x,y).</math> By using the centroid formula, we find that the coordinates of the centroids are <math>(\frac{x}{3},10+\frac{y}{3}),(10+\frac{x}{3},\frac{y}{3}),(20+\frac{x}{3},10+\frac{y}{3}),</math> and <math>(10+\frac{x}{3},20+\frac{y}{3}).</math> Shifting the coordinates down by <math>(\frac x3,\frac y3)</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{(C) 200.}</math>
 
  
 
== Video Solution (Meta-Solving Technique) ==
 
== Video Solution (Meta-Solving Technique) ==

Revision as of 13:40, 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)

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 (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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