Difference between revisions of "2021 USAJMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | I will use the word "center" to refer to the centroid of any equilateral triangle. | ||
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+ | Consider the center of each triangle we place and the surrounding area that is "denied" placement for any centers. With a little bit of testing and sliding the equilateral triangle along each side of the center triangle, it is easy to see that this region forms a hexagon of side length <math>1</math> where any other center being in this hexagon would mean that the two equilateral triangles at those centers would overlap. | ||
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+ | Let us define a region for each center as its "personal space", where its personal space intersecting with any other center's personal space results in the two equilateral triangles intersecting. Given the above fact, it is easy to see that this personal space region encapsulates a hexagon of sidelength <math>\frac{1}{2}</math> centered at each center by simply scaling each side down by a factor of <math>\frac{1}{2}</math>. The area of each personal space would simply be <math>6(\frac{\sqrt{3}}{4}(\frac{1}{2})^2)=\frac{3\sqrt{3}}{8}</math>. Since each personal space cannot intersect, the maximum amount of equilateral triangles would be bounded by <math>\frac{\text{Area of Big Triangle}}{\text{Area of individual personal space}}=\frac{L^2\frac{\sqrt{3}}{4}}{\frac{3\sqrt{3}}{8}}=\frac{2}{3}L^2</math>, which completes the problem. (Unrigorous solution that I hope someone else can improve on ~ hyxue) | ||
==See Also== | ==See Also== |
Revision as of 03:22, 20 March 2022
Problem
An equilateral triangle of side length
is given. Suppose that
equilateral triangles with side length 1 and with non-overlapping interiors are drawn inside
, such that each unit equilateral triangle has sides parallel to
, but with opposite orientation. (An example with
is drawn below.)
Prove that
Solution
I will use the word "center" to refer to the centroid of any equilateral triangle.
Consider the center of each triangle we place and the surrounding area that is "denied" placement for any centers. With a little bit of testing and sliding the equilateral triangle along each side of the center triangle, it is easy to see that this region forms a hexagon of side length where any other center being in this hexagon would mean that the two equilateral triangles at those centers would overlap.
Let us define a region for each center as its "personal space", where its personal space intersecting with any other center's personal space results in the two equilateral triangles intersecting. Given the above fact, it is easy to see that this personal space region encapsulates a hexagon of sidelength centered at each center by simply scaling each side down by a factor of
. The area of each personal space would simply be
. Since each personal space cannot intersect, the maximum amount of equilateral triangles would be bounded by
, which completes the problem. (Unrigorous solution that I hope someone else can improve on ~ hyxue)
See Also
2021 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.