Base and height of equilateral triangle that are closest to integers

Question

I know that formula for the height of an equilateral triangle is height = (1/2) * √3 * base, so the height and base can never both be integers. But how could I find values that are as close as possible to integers? I happen to know that a base of 15 results in a height of 12.9903811, which is really close to 13. Is a better base that results in an even closer height?

Answer 1

Use the continued fraction expansion for $\frac {\sqrt{3}}2$. The first $10$ convergents are $$\left \{ 0, 1, \frac 6 {7}, \frac {13}{ 15},\frac {84}{97}, \frac {181}{ 209}, \frac {1170}{1351}, \frac {2521}{ 2911}, \frac {16296}{ 18817}, \frac {35113}{ 40545}\right \} $$

Note that your $13,15$ pair occurs early on.

Using the final one, we note that $$\frac {35113}{ 40545}=0.866025403872240\dots$$ while $$\frac {\sqrt 3}2=0.86602540378443864676\dots$$ and $$40545\times \frac {\sqrt 3}2=35112.99999644\dots$$

Answer 2

Yes, you can get arbitrarily close. A good way is to look at the convergents of the continued fraction for $\frac {\sqrt 3}2$ In this case it is $[0;1,\overline{6,2}]$ or $$\frac 1{1+\frac 1{6+\frac 1{2+\frac 1{6+\frac 1{2+\ldots}}}}}$$ $\frac {13}{15}$ comes from taking just the first $6$ and first $2$, so $$\frac 1{1+\frac 1{6+\frac 12}}=\frac 1{1+\frac 2{13}}=\frac{13}{15}$$ They are particularly close if you stop before a large term, so you could next look at $$\frac 1{2+\frac 1{6+\frac 12}}=\frac 1{2+\frac 2{13}}=\frac{13}{28}\\ \frac 1{1+\frac 1{6+\frac {13}{28}}}=\frac 1{1+\frac {28}{181}}=\frac {181}{209}\\ \left(\frac{181}{209}\right)^2\approx 0.7500057$$