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A gardener throws $18$ seeds into an equilateral triangle-shaped plot of land with sides of length $100$ centimeters. Then is it true that two seeds are within a distance of $25$ cm?

I found the area of equilateral triangle $${\sqrt{3}\over4}(\text{side})^2=2500 \sqrt{3}\ \text{cm}^2$$

Since ${\text{Area}\over 18}$ is greater than $25$, does it follow that the answer is yes? If not, how to proceed?

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  • $\begingroup$ Your comparing two quantities having different units. So, no. $\endgroup$
    – Ron Gordon
    Nov 11, 2014 at 15:35
  • $\begingroup$ Then how to do @RonGordon $\endgroup$
    – Learnmore
    Nov 11, 2014 at 15:36
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    $\begingroup$ that regular hexagons are made up of regular triangles is convenient here! $\endgroup$ May 25, 2015 at 22:33

4 Answers 4

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Here is a picture to accompany Fmonkey2001's answer. Each dot represents a seed. I have 15 of them. From each dot we draw a circle with radius 25.

enter image description here

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    $\begingroup$ How'd you make that picture? What program?! $\endgroup$ Nov 11, 2014 at 15:59
  • $\begingroup$ Geogebra. Don't take the picture as a proof though! $\endgroup$
    – Pburg
    Nov 11, 2014 at 15:59
  • $\begingroup$ I don't think it'd take much work though to make this a proof by picture! $\endgroup$ Nov 11, 2014 at 16:03
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Even with $17$ seeds instead of $18$ one can guarantee that there are two seeds within $25$ cm of each other:

Proof. Divide the large triangle into $16$ closed equilateral triangles $T_i$ ($1\leq i\leq16$) of side length $25$ cm in the obvious way. Number the seeds from $1$ to $17$ and define a function $f:\>[17]\to[16]$ by putting $$f(s):=\min\{i\>|\>s\in T_i\}\ .$$ By the pigeonhole principle there will be two seeds $s_1$, $s_2$ with $f(s_1)=f(s_2)$. These two seeds are at a distance $\leq25$ cm of each other.

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The answer is false. I believe my logic is correct.

1) Consider an equilateral triangle with side length of $100\ \text{cm}$

2) Since we're looking for two seeds that are with a distance of $25\ \text{cm}$ we can break the big triangle up into little equilateral triangles with side length of $25\ \text{cm}$.

3) Once we do this we can see that the above statement is false. Since you can fit $16$ triangles of side length $25\ \text{cm}$ inside the triangle of side length $100\ \text{cm}$

If you draw out the picture it'll give you a really nice visual of this!

You could also crunch numbers if you're more of a number person.

$${\text{Triangle with 100 cm side length}\over\text{Triangle with 25 cm side lenth}}={2500\sqrt3\over{625\sqrt3\over4}}={10000\over625}=16$$

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    $\begingroup$ But how many vertices do those triangles have? ;) $\endgroup$
    – N. S.
    Nov 11, 2014 at 15:55
  • $\begingroup$ You make a good point! Let me count! $\endgroup$ Nov 11, 2014 at 15:56
  • $\begingroup$ Am I wrong that no matter the shape of the garden, the approach to a problem like this is a triangular tiling of the plane? Followed by a recentering of the origin... $\endgroup$ May 25, 2015 at 22:44
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enter image description here

You can cover an equilateral triangle of side $4 d$ with $15$ regular hexagons of diameter $d$. So your problem also works for $16$.

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