Sum of segments inside a right triangle.

Question

I am interested for a problem involving the sum of segments inside a right triangle. Consider a right triangle of hypotenuse $\overline{BC}$ and catheti $\overline{AB}$ and $\overline{AC}$. From the right angle, trace the segment $\overline{AP_1}$ forming an angle $\alpha$ with the cathetus $\overline{AB}$. $P_1$ is a point on the hypotenuse. Trace, then, the segment $\overline{P_1P_2}$ perpendicular to $\overline{AB}$ ($P_2$ is a point on this cathetus).

By continuing the process, trace the segment $\overline{P_2P_3}$, forming an angle $\alpha$ with the cathetus $\overline{AB}$, with $P_3$ on the hypotenuse, and trace $\overline{P_3P_4}$ perpendicular to $\overline{AB}$, with $P_4$ on the cathetus. And so on.

Does exist a closed form on $\overline{BC}, \overline{AB}, \overline{AC}$and $\alpha$ for the sum $$ \sum_{i=0}^{\infty}\overline{P_iP_{i+1}}?$$ Set $P_0 = A$.

Answer 1

Extend the first line $AP_1$ through the hypotenuse till it reaches a point $D$ such that $\angle{ABD}$ is a right-angle. If you add every odd-numbered segment ($P_0P_1,\; P_2P_3,\;\ldots$) it is the same as length of $AD$.

Similarly, if you add every even-numbered segment ($P_1P_2,\; P_3P_4,\;\ldots$) it is the same as length of $BD$.

Therefore, $$\sum_{i=0}^{\infty}\overline{P_iP_{i+1}} = \overline{AD} + \overline{BD} = \overline{AB}(\sec{\alpha} + \tan{\alpha}).$$

Answer 2

The triangles, $P_0P_1P_2$, $P_2P_3P_4$,...,$P_{2n}P_{2n+1}P_{2n+2}$,... are similar. Furthermore, the ratio of each triangle and the next one are always the same (can you find it with trigonometry?).

The lengths of the segments $\overline{P_0P_1}$, $\overline{P_2P_3}$,...,$\overline{P_{2n}P_{2n+1}}$,... form a geometric infinite sequence. Can you find its sum? Same for the segments $\overline{P_{2n+1}P_{2n+2}}$.

Answer 3

By similar triangles, you know that $\frac{P_5P_6}{P_3P_4}=\frac{P_4P_5}{P_2P_3}=\frac{P_3P_4}{P_1P_2}$ and so on... So all the ratios $\frac{P_{k+2}P_{k+3}}{P_{k+0}P_{k+1}}$ are equal to $\frac{P_1P_2}{AB}:=r$, which is known. Let $t=P_0P_1+P_1P_2$.

Hence the total length is $\sum_{k=0}^\infty tr^k=t\sum_{k=0}^\infty r^k=\frac{t}{1-r}$.