The figure above shows an infinite series of triangles ,in which $r_1 >r_2>r_3>..............$

Question

I am stuck on the following problem that says:

The figure above shows an infinite series of triangles ,in which $r_1 >r_2>r_3>..............$ What is the total length of the solid line segments in the figure? The options are:

1. $\frac{r_1}{r_2}+\frac{r_2}{r_3}+.......$

2. $\frac{r_1^2}{r_1-r_2}$

3. $\frac{r_2^2}{r_1+r_2}$

4. $\frac{r_1-r_2}{r_1^2}$

I have no clue how to tackle it. Can someone explain in details ? Thanks in advance for your time.

Answer 1

Let $T$ be the clockwise rotation of angle $\theta$ followed by a dilation of ratio $k=r_2/r_1$. The second triangle $\triangle_{2}$ is the image under $T$ of the first triangle $\triangle_{1}$. More generally, the $n+1^\mathrm{st}$ triangle $\triangle_{n+1}$is the image of $\triangle_{n}$ under $T$. So the $n^\mathrm{th}$ segmant is the image under $T^{n-1}$ of the first one. Consequently, $ r_n=k^{n-1}r_1 $. It follows that $$\sum_{n=1}^\infty r_n=r_1\sum_{n=1}^\infty k^{n-1}=\frac{r_1}{1-k}=\frac{r_1}{1-r_2/r_1} =\frac{r_1^2}{r_1-r_2}.$$