# Regular pentagon and tangent lines

We consider a regular pentagon $A_1A_2A_3A_4A_5$ and $(C)$ is its inscribed circle. We then, taking as centres the points $Α_1$,$Α_2$,$Α_3$,$Α_4$,$Α_5$, draw the circles $(C_1)$,$(C_2)$,$(C_3)$,$(C_4)$,$(C_5)$ which are tangent to $(C)$ at the points $B_1,B_2,B_3,B_4,B_5$ respectively. Consider now a randomly chosen point $M$ of the arc $B_1B_5$. From $M$ we draw the tangents $MT_1,MT_2,MT_3,MT_4,MT_5$ of the circles $(C_1)$,$(C_2)$,$(C_3)$,$(C_4)$,$(C_5)$ repsectively. Show that $MT_1+MT_3+MT_5=MT_2+MT_4$

Here is a picture to help people to solve your problem. By the way, nice problem.

Note that you can take a second tangency point $T_i'$ but we know that the segments $MT_i'$ and $MT_i$ are congruent.

For convenience, I'll take the subscripts to run from $0$ to $4$.

Let the radius of the pentagon's incircle be $1$, and let the radius of the smaller circles be $r$. With the pentagon's center at the origin of the complex plane, we can align its vertices with the fifth roots of unity $$A_k = ( 1 + r ) \; e^{ik\alpha} \qquad \text{where} \qquad \alpha := \frac{2\pi}{5}$$

Write $M:=e^{i\theta}$ with $0 \geq \theta \geq -\alpha$, so that $M$ lies between $A_0$ and $A_4$.

Now, note that each $\triangle MA_kT_k$ has a right angle at $T_k$, so that \begin{align} |MT_k|^2 = |MA_k|^2 - r^2 &= \left|\;(1 + r )\; e^{ik\alpha}- e^{i\theta}\;\right|^2 - r^2 \\ &= \left( (1 + r )\; e^{ik\alpha} - e^{i\theta} \right)\left( (1 + r )\; e^{-ik\alpha} - e^{-i\theta} \right) - r^2 \\ &= 1 - ( 1 + r )\left( e^{i(k\alpha-\theta)} + e^{-i(k\alpha-\theta)} \right) + ( 1 + r )^2 - r^2 \\ &= \left( 1 + r \right) \left( 2 - \left( e^{i(k\alpha-\theta)} + e^{-i(k\alpha-\theta)} \right) \right)\\ &= -\left( 1 + r \right) \left( e^{i(k\alpha-\theta)/2} - e^{-i(k\alpha-\theta)/2} \right)^2 \\ &= 4 \left( 1 + r \right) \sin^2 \frac{k \alpha - \theta}{2} \end{align} (The simplicity of this formula makes me think that it must have a nice geometric derivation.)

Given our bounds on $\theta$, we have $0 \leq \frac{k \alpha}{2} \leq \frac{k \alpha -\theta}{2} \leq \frac{(k+1) \alpha}{2} \leq \pi$, which implies $$|MT_k| = 2 \sqrt{1+r} \; \sin\frac{k\alpha - \theta}{2} = -i\;\sqrt{1+r}\;\left( e^{i(k\alpha-\theta)/2} - e^{-i(k\alpha-\theta)/2} \right)$$ Consequently, the alternating sum of the $|MT_k|$s is a combination of geometric series: \begin{align} \frac{i}{\sqrt{1+r}}\; \sum_{k=0}^{4} (-1)^k|MT_k| &= e^{-i\theta/2}\; \sum_{k=0}^{4} \left(-e^{i\pi/5}\right)^k - e^{i\theta/2}\; \sum_{k=0}^{4} \left(-e^{-i\pi/5}\right)^k \\ &= e^{-i\theta/2}\;\frac{1-\left(-e^{i\pi/5}\right)^5}{1+e^{i\pi/5}} - e^{i\theta/2}\; \frac{1-\left(-e^{-i\pi/5}\right)^5}{1+e^{-i\pi/5}} \\ &= 0 - 0 = 0 \end{align} which gives the result.

Interestingly, the exact value of radius $r$ is irrelevant here. The result is valid for externally-tangent circles of any size; taking $r$ negative in the definition of $A_k$, we see that the result is also true for internally tangent circles (of radius less than $1$).

Come to think of it, the number of sides only becomes relevant in the computation $\left(-e^{i\pi/5}\right)^5 = 1$, which isn't so much a property of five-ness as of odd-ness. The result holds for any $n$-gon with odd $n > 1$.