Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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Similarly, we deduce that <math>\angle G_2G_3G_4 = \angle G_3G_4G_1 = \angle G_4G_1G_2 = 90^\circ.</math> | Similarly, we deduce that <math>\angle G_2G_3G_4 = \angle G_3G_4G_1 = \angle G_4G_1G_2 = 90^\circ.</math> | ||
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− | <li>Note that <math>M_1M_2M_3M_4</math> is a square of side-length <math>15\sqrt2.</math> Since <math>G_1G_2=\frac23M_1M_2, G_2G_3=\frac23M_2M_3, G_3G_4=\frac23M_3M_4,</math> and <math>G_4G_1=\frac23M_4M_1,</math> we | + | <li>Note that <math>M_1M_2M_3M_4</math> is a square of side-length <math>15\sqrt2.</math> Since <math>G_1G_2=\frac23M_1M_2, G_2G_3=\frac23M_2M_3, G_3G_4=\frac23M_3M_4,</math> and <math>G_4G_1=\frac23M_4M_1,</math> we have <math>G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\sqrt2.</math></li><p> |
</ol> | </ol> | ||
Together, quadrilateral <math>G_1G_2G_3G_4</math> is a square of side-length <math>10\sqrt2,</math> so its area is <math>\left(10\sqrt2\right)^2=200.</math> | Together, quadrilateral <math>G_1G_2G_3G_4</math> is a square of side-length <math>10\sqrt2,</math> so its area is <math>\left(10\sqrt2\right)^2=200.</math> |
Revision as of 17:37, 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)
As shown below, let be the midpoints of respectively, and be the centroids of respectively. By SAS, we conclude that and By the properties of centroids, the side-length ratio for each pair of triangles is so the area ratio for each pair of triangles is
Let the brackets denote areas. Since it follows that
Remarks for Quadrilateral
- By angle addition, we have Similarly, we deduce that
- Note that is a square of side-length Since and we have
Together, quadrilateral is a square of side-length so its area is
~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.