2019 AMC 12B Problems/Problem 25: Difference between revisions
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==Solution== | ==Solution== | ||
Let <math>X,Y,Z</math> be the centroids of <math>ABC,BCD,CDA</math> | |||
<math>XY=AD/3,ZY=AB/3</math> | |||
<math>XYZ is equilateral therefore ABD is also equilateral</math> | |||
Rotate <math>ACD</math> to <math>AEB</math> | |||
Then <math>AEC</math> is also equilateral | |||
Which has a larger or equal area than <math>ABCD</math> | |||
<math>CE<=BC+EB =BC+CD =2+6=8</math> | |||
Area of <math>AEC </math> is at most <math>16\sqrt3</math> | |||
Therefore the maximum area for <math>ABCD</math> is<math>16\sqrt3</math> | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:47, 14 February 2019
Problem
Let 
be a convex quadrilateral with 
and 
Suppose that the centroids of 
and 
form the vertices of an equilateral triangle. What is the maximum possible value of ?
Solution
Let 
be the centroids of 
Rotate 
to 
Then 
is also equilateral
Which has a larger or equal area than
Area of is at most 
Therefore the maximum area for is
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Problem |
| 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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