Sum of the area of infinite similar equilateral triangles

Question

How would I solve for the side depicted in the picture?

Answer 1

Call $S_n$ the side of the $n^{th}$ equilateral triangle from the right side. The area is

$$A = \sum_{n=1}^{\infty} \dfrac{\sqrt{3}}{4}S_n^2 = 277$$

If you can get $S_n$ in terms of $S_1$ for all $n$, and evaluate the series, then you can find $S_1$ as desired.

If you look at the upper triangle in between the two right-most equilateral triangles, you can see (using standard techniques) that the angles are $88^o, 60^o, 32^o$. Using the law of sines, you find that

$$\dfrac{S_2}{\sin(32^o)} = \dfrac{S_1}{\sin(88^o)} \iff S_2 = \dfrac{\sin(32^o)}{\sin(88^o)}S_1$$

and similarly that

$$S_n=\left(\dfrac{\sin(32^o)}{\sin(88^o)}\right)^{n-1}S_1$$

From here, after making the substitution in the above series, you will get a geometric series in which $S_1^2$ may be factored out.

Answer 2

Assume that
- angle of incline is $\alpha$
- width of first* triangle is $2a_1$
- width of second* triangle is $2a_2$
etc.
*counting from the right, and using this convention throughout

Consider the apex of the first two triangles.
Their respective heights are $a_1\sqrt3$ and $a_2\sqrt3$.
Their horizontal separation is $a_1+a_2$.
Their vertical separation (difference in height) is $(a_1-a_2)\sqrt3$.

Hence we have $$\tan\alpha=\frac{(a_1-a_2)\sqrt3}{a_1+a_2}\\ r=\frac{a_1}{a_2}=\frac{\sqrt3-\tan\alpha}{\sqrt3+\tan\alpha}=\frac{a_i}{a_{i+1}}\\ \Rightarrow a_i=a_1r^{i-1}$$ i.e. heights of triangles are in GP.

Also, we are given that the total the area of all triangles is 277, i.e. $$\begin{align} 277&=\sum_{i=1}^\infty A_i =\sum_{i=1}^\infty\sqrt3a_i^2\\ &=a_1^2\sqrt3\sum_{i=1}^\infty r^{2(i-1)}\\ &=\frac{a_1^2\sqrt3}{1-r^2}\\ a_1^2&=\frac{277}{\sqrt3}(1-r^2) \end{align}$$

Putting $\alpha=28^\circ$ gives $r=0.684079, a_1=9.22422$.
Hence side of first triangle is $2a_1=\color{red}{18.4485}$.

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NB - it would have been a "neat" problem if, e.g. $\alpha$ had been set to $30^\circ$ (or $\frac\pi6$ radians), and the total area to, say, $4\sqrt 3$, as that would result in $r=\frac 12$ and $a_1=3$.