Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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==Solution 4 (Extremely Risky)== | ==Solution 4 (Extremely Risky)== | ||
− | Let <math>G_1,G_2,G_3,G_4</math> be the centroids of <math>\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},</math> respectively. Looking at the answer choices, it seems as if < | + | Let <math>G_1,G_2,G_3,G_4</math> be the centroids of <math>\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},</math> respectively. Looking at the answer choices, it seems as if <cmath><cmath>G_1G_2G_3G_4</cmath></cmath> doesn't depend on where point <math>P</math> is placed. Therefore, we can ignore the condition set on <math>P</math> and instead place it at the center of square <math>ABCD</math>. This makes quadrilateral <math>G_1G_2G_3G_4</math> a square with a diagonal of length <math>20</math> since the centroids cut the medians into two segments in the ratio <math>2 : 1</math>, so its area is <math>\tfrac{20^2}{2} = 200.</math> |
== Video Solution (Meta-Solving Technique) == | == Video Solution (Meta-Solving Technique) == |
Revision as of 15:03, 1 May 2022
Contents
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 (Extremely Risky)
Let be the centroids of
respectively. Looking at the answer choices, it seems as if
</cmath> doesn't depend on where point
is placed. Therefore, we can ignore the condition set on
and instead place it at the center of square
. This makes quadrilateral
a square with a diagonal of length
since the centroids cut the medians into two segments in the ratio
, so its 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.