Difference between revisions of "2018 AMC 12B Problems/Problem 13"
Pi is 3.14 (talk | contribs) (→Solution 4) |
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) |
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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> | ||
− | ==Solution 1 ( | + | ==Solution 1 (Similar Triangles and Area Ratios)== |
− | + | <b>IN CONSTRUCTION. NO EDIT PLEASE</b> | |
− | + | ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | |
− | ==Solution 2== | + | ==Solution 2 (Coordinate Geometry)== |
− | The centroid of | + | 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> |
− | ==Solution 3== | + | ==Solution 3 (Accurate Diagram)== |
− | + | 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. | |
− | + | 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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== Video Solution (Meta-Solving Technique) == | == Video Solution (Meta-Solving Technique) == |
Revision as of 13:40, 24 September 2021
Contents
[hide]Problem
Square has side length . Point lies inside the square so that and . The centroids of , , , and are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
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 and the coordinates of point P are By using the centroid formula, we find that the coordinates of the centroids are and Shifting the coordinates down by does not change its area, and we ultimately get that the area is equal to the area covered by which has an area of
Solution 3 (Accurate Diagram)
We can draw an accurate diagram by using centimeters and scaling everything down by a factor of The centroid is the intersection of the three medians in a triangle.
After connecting the centroids, we see that the quadrilateral looks like a square with side length of However, we scaled everything down by a factor of so the length is The area of a square is so the area is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.