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Difference between revisions of "2021 USAJMO Problems/Problem 3"

(Created page with "==Problem== An equilateral triangle <math>\Delta</math> of side length <math>L>0</math> is given. Suppose that <math>n</math> equilateral triangles with side length 1 and with...")
 
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==Solution==
 
==Solution==
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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 the denied region 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==
 
{{USAJMO newbox|year=2021|num-b=2|num-a=4}}
 
{{USAJMO newbox|year=2021|num-b=2|num-a=4}}
  
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[[Category:Olympiad Number Theory Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:23, 20 March 2022

Problem

An equilateral triangle $\Delta$ of side length $L>0$ is given. Suppose that $n$ equilateral triangles with side length 1 and with non-overlapping interiors are drawn inside $\Delta$, such that each unit equilateral triangle has sides parallel to $\Delta$, but with opposite orientation. (An example with $n=2$ is drawn below.) [asy] draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle,linewidth(0.5)); filldraw((0.45,0.55)--(0.65,0.55)--(0.55,0.55-sqrt(3)/2*0.2)--cycle,gray,linewidth(0.5)); filldraw((0.54,0.3)--(0.34,0.3)--(0.44,0.3-sqrt(3)/2*0.2)--cycle,gray,linewidth(0.5)); [/asy] Prove that\[n \leq \frac{2}{3} L^{2}.\]

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 $1$ 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 $\frac{1}{2}$ centered at each center by simply scaling the denied region by a factor of $\frac{1}{2}$. The area of each personal space would simply be $6(\frac{\sqrt{3}}{4}(\frac{1}{2})^2)=\frac{3\sqrt{3}}{8}$. Since each personal space cannot intersect, the maximum amount of equilateral triangles would be bounded by $\frac{\text{Area of Big Triangle}}{\text{Area of individual personal space}}=\frac{L^2\frac{\sqrt{3}}{4}}{\frac{3\sqrt{3}}{8}}=\frac{2}{3}L^2$, 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

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