Since we're approaching the Christmas season, I'm calculating how many feet of lights I need for a few decorations.
Let's say I have X feet of lights, is it possible to calculate the height/width of an isosceles triangle so that it's outlined completely in lights? I can always divide the isosceles triangle in two right triangles, and then cover each one separately.
So in terms of angles, the outer rectangle should either be 70, 90, & 20 degrees (if divided into two rectangles) or 70, 90, and 40.
Three questions: If I have X feet of string, how high & wide should the isosceles triangle be so that it's covered completely by the string?
If I have X feet of string, how high & wide should the right triangle be so that it's covered completely by the string?
Instead of using one piece of string for the isosceles triangle (which has 6 sides total), would I use less string if I divide it in two right triangles? I would need 2 strings, but maybe the sum of these two strings is less than the one string I would use if I covered the isosceles triangle.
Anyways I can buy the string in 30 or 60 feet. And if I buy the lights, I don't want to be short or over-buy. And since they're a specific length, I want the triangle to use the as much of the lights (ie. not buy a 60' string so that the whole triangle only uses 50').
Thanks.
Let's denote lengths of the sides of isosceles triangle with $a$ and $b$ ($a$ being basis).
Note that first picture cannot be realized "in one go" without overlapping. So what you want to know is which picture "overlaps less", so you get more coverage with the same length.
Consider the first picture:
If you cover this with string starting in $F$ in order:
$F$-$A$-$C$-$F$-$B$-$E$-$F$-$D$
you will cover the whole triangle in one go, with overlapping red part $BC$, which is of the length $\frac a 4$.
If you consider the second picture, you have two pieces which you can cover in one go, but obviously the height is doubled, that is, overlapping is of the length $h$.
So, as long $\frac a 4<h$, the first picture is more optimal. And that will certainly be the case if $\alpha \approx 26.5651^\circ$ or greater.
Thus, we dismiss the second picture, and calculate for the first one.
Height of the triangle is given by Pythagoras theorem (hypotenuse being of length $b$ and one leg $\frac{a}2$):
$h = \sqrt{b^2 - \frac{a^2}4}$
Medians of the right triangles are again given by Pythagoras theorem (legs are $h$ and $\frac a 4$):
$m = \sqrt{h^2 + \frac{a^2}{16}} = \sqrt{b^2 - \frac{a^2}{4} + \frac{a^2}{16}} = \sqrt{b^2 - \frac{3a^2}{16}}$
Total length is thus:
$X = a + 2b + h + 2m +\frac a 4= \frac{5a}4 + 2b + \sqrt{b^2 - \frac{a^2}4} + 2\sqrt{b^2 - \frac{3a^2}{16}}$
Let $\alpha$ be the angle between $a$ and $b$. Then we have $\cos\alpha=\frac{\frac a2}{b}$, i.e. $a=2b\cos\alpha$.
We can simplify $X$ easily using this substitution to get:
$X = b (2 + \frac 5 2 \cos\alpha + \sqrt{4 - 3 \cos^2\alpha} + \sin\alpha)$
So, set $$c = 4 + 5 \cos\alpha + 2\sqrt{4 - 3 \cos^2\alpha} + 2\sin\alpha$$ to get:
So, for $\alpha = 70^\circ$ we have:
$a\approx 0.119902 X,\ h\approx 0.164714 X$
and for $X=30$:
$a\approx 3.6,\ h\approx 4.94$
In this case, overlapping would be $\approx 0.9$, and useful part $30-0.9 = 29.1$.
(the question was edited, by I will leave this part here anyway)
Let us now do some calculations for fixed ratio of width and height, i.e. $k = \frac a h$. Since we want tree to be more tall than wide, we want $k<1$. I will skip over the particular technicalities, as they are bit gruesome, and give you the result immediately:
where $$c=4 + 5 k + 4 \sqrt{4 + k^2} + 2 \sqrt{16 + k^2}$$
So, for example, if $k=\frac 1 2$ (the height being two times the width) and $X=30$ we get:
$a \approx 2.63,\, h \approx 5.26$
You can play with it until you get something you like.
Also, overlapping this time would be $\approx 0.66$.
Closing words:
Useful part of string will be $X-\frac a 4$, so lesser the basis, more useful the string. Take note, though, that making $a$ too little will give awkward looking tree, so you might want to play with ratios or angles with the formulas I've given you until you get what seems the best for you.