Difference between revisions of "2018 AMC 12B Problems/Problem 13"
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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 (Similar Triangles | + | ==Solution 1 (Similar Triangles)== |
As shown below, let <math>M_1,M_2,M_3,M_4</math> be the midpoints of <math>\overline{AB},\overline{BC},\overline{CD},\overline{DA},</math> respectively, and <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. | As shown below, let <math>M_1,M_2,M_3,M_4</math> be the midpoints of <math>\overline{AB},\overline{BC},\overline{CD},\overline{DA},</math> respectively, and <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. | ||
<asy> | <asy> | ||
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dot(G4); | dot(G4); | ||
</asy> | </asy> | ||
− | 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 | + | 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>\frac23.</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: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
− | <li> | + | <li>Since <math>\overline{G_1G_2}\parallel\overline{M_1M_2},\overline{G_2G_3}\parallel\overline{M_2M_3},\overline{G_3G_4}\parallel\overline{M_3M_4},</math> and <math>\overline{G_4G_1}\parallel\overline{M_4M_1}</math> by the Converse of the Corresponding Angles Postulate, we have <math>\angle G_1G_2G_3=\angle G_2G_3G_4=\angle G_3G_4G_1=\angle G_4G_1G_2=90^\circ.</math></li><p> |
− | < | + | <li>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> by the ratio of similitude, we have <math>G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\sqrt2.</math></li><p> |
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− | </li><p> | ||
− | <li>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 | ||
</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=\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> |
Revision as of 01:34, 25 September 2021
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 (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.