Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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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 ratio of similitude for each pair of triangles is <math>\ | + | 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 ratio of similitude for each pair of triangles is <math>\frac{2}{3}.</math> |
Note that quadrilateral <math>M_1M_2M_3M_4</math> is a square of side-length <math>15\sqrt2.</math> It follows that: | Note that quadrilateral <math>M_1M_2M_3M_4</math> is a square of side-length <math>15\sqrt2.</math> It follows that: | ||
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~Pi31415926535897 ~MRENTHUSIASM | ~Pi31415926535897 ~MRENTHUSIASM | ||
− | ==Solution 4 ( | + | ==Solution 4 (Homothety)== |
− | Let <math> | + | Let <math>X,Y,Z,W</math> be the midpoints of sides <math>AB,BC,CD,DE</math>, respectively. |
+ | |||
+ | Notice that a homothety centered at P with ratio <math>\frac{2}{3}</math> will send <math>XYZW</math> to <math>G_{1}G_{2}G_{3}G_{4}</math>, so <math>G_{1}G_{2}G_{3}G_{4}</math> is a square with area <math>\left(\frac{2}{3}\right)^2 [XYZW]</math>, but <math>[XYZW]=\frac{1}{2}[ABCD]</math> so our desired area is <cmath>\frac{4}{9}\cdot\frac{1}{2}\cdot900=\boxed{\textbf{(C) }200}</cmath> | ||
+ | |||
+ | ~chrisdiamond10 | ||
== Video Solution (Meta-Solving Technique) == | == Video Solution (Meta-Solving Technique) == |
Latest revision as of 05:53, 31 August 2024
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 (Similar Triangles)
This solution refers to the diagram in Solution 1.
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 diagonal-length so Since and by the Converse of the Corresponding Angles Postulate, we have
Therefore, the area of quadrilateral is ~Funnybunny5246 ~MRENTHUSIASM
Solution 3 (Coordinate Geometry)
This solution refers to the diagram in Solution 1.
We place the diagram in the coordinate plane: Let and
Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that and
Note that and Therefore, the area of quadrilateral is
~Pi31415926535897 ~MRENTHUSIASM
Solution 4 (Homothety)
Let be the midpoints of sides , respectively.
Notice that a homothety centered at P with ratio will send to , so is a square with area , but so our desired area is
~chrisdiamond10
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.