Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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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=\boxed{\textbf{(C) }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=\boxed{\textbf{(C) }200}.</math> | ||
− | <b>Remark</b> | + | <u><b>Remark</b></u> |
This solution shows that, if point <math>P</math> is within square <math>ABCD,</math> then the shape and the area of quadrilateral <math>G_1G_2G_3G_4</math> are independent of the location of <math>P.</math> Let the brackets denote areas. More generally, <math>G_1G_2G_3G_4</math> is always a square of area <cmath>[G_1G_2G_3G_4]=\left(\frac23\right)^2[M_1M_2M_3M_4]=\frac49[M_1M_2M_3M_4]=\frac29[ABCD].</cmath> On the other hand, the location of <math>G_1G_2G_3G_4</math> is dependent on the location of <math>P.</math> | This solution shows that, if point <math>P</math> is within square <math>ABCD,</math> then the shape and the area of quadrilateral <math>G_1G_2G_3G_4</math> are independent of the location of <math>P.</math> Let the brackets denote areas. More generally, <math>G_1G_2G_3G_4</math> is always a square of area <cmath>[G_1G_2G_3G_4]=\left(\frac23\right)^2[M_1M_2M_3M_4]=\frac49[M_1M_2M_3M_4]=\frac29[ABCD].</cmath> On the other hand, the location of <math>G_1G_2G_3G_4</math> is dependent on the location of <math>P.</math> |
Revision as of 16:05, 25 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)
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 ratio of similitude for each pair of triangles is
Note that quadrilateral is a square of side-length It follows that:
- Since and by the Converse of the Corresponding Angles Postulate, we have
- Since and by the ratio of similitude, we have
Together, quadrilateral is a square of side-length so its area is
Remark
This solution shows that, if point is within square then the shape and the area of quadrilateral are independent of the location of Let the brackets denote areas. More generally, is always a square of area On the other hand, the location of is dependent on the location of
~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 |
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