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I am looking for a proof of property as following:

Let $M_1M_2...M_n$ be a regular polygon. Let $N_1N_2...N_n$ be its tangential polygon. Let $P$ be arbitraty point inside $M_1M_2....M_n$. Let $2n$ ray through $P$, such that the angle of neighbor rays is $\pi/n$. The rays meet in-sidelines of $M_1M_2....M_n$ at $A_1, A_2,....,A_n$. The rays meet in-sidelines of $N_1N_2...N_n$ at $B_1, B_2, B_3,....,B_n$. Then show that:

$$\sum_{1}^{n}{ PB_i} = \sec{\frac{\pi}{n}}\sum_{1}^{n}{ PA_i} $$

enter image description here

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    $\begingroup$ If the angles between these rays are as stated, shouldn't there be $2n$ such rays? $\endgroup$ – Batominovski Jul 30 '16 at 19:10
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    $\begingroup$ I dont agree to close this question, which is not classical. The fact that the OP has spend time on her figure is a proof that she has worked on the question. $\endgroup$ – Jean Marie Jul 30 '16 at 19:34
  • $\begingroup$ Sometimes closed votes are made as a motivation for the OP to clarify unclear points in the question. $\endgroup$ – Batominovski Jul 30 '16 at 20:05
  • $\begingroup$ Dear @Batominovski Yes, you are right. I edited $\endgroup$ – Oai Thanh Đào Jul 31 '16 at 2:36
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This solution is only based on a regular 6-gon. I guess the reasoning used can be extended to a regular n-gon.

Wrt the 6-gon, let R and r be the circum-scribing and in-scribing radii respectively. Then, $R = \dfrac {2}{\sqrt 3}r$.

enter image description here

For a given interior point P, we can always rotate the 6-gon such that the ray $PB_1$ is perpendicular to a side of the 6-gon. Then, $PB_7$ is also perpendicular to the corresponding opposite side. It should clear that $B_1B_7 = B_3B_9 = B_5B_{11} = 2r$. Then, $B_1B_7 + B_3B_9 + B_5B_{11} = 6r$.

Next we want to find the sum of the 6 blue line segments. Let the offset distance of $B_6B_{12}$ from ST be $d_1$. The other offset distances $d_2$ and $d_3$ are similarly defined.

From $\triangle XB_6B_{12} \sim \triangle XST$, we have $\dfrac {B_6B_{12}}{2R} = \dfrac {\sqrt 3R – d_1}{\sqrt3R}$. That is, $B_6B_{12} = 2R – (\dfrac {2}{\sqrt 3})d_1$. The other two blue lines segment can be similarly found. Therefore, $B_2B_8 + B_4B_{10} + B_6B_{12} = 6R - (\dfrac {2}{\sqrt 3})(d_1 + d_2 + d_3)$.

Next, we are going to find the total lengths of the A's. First, note that $A_2A_8 = A_4A_{10} = A_6A_{12} = \dfrac {\sqrt 3}{2}2r$

From $\triangle ZA_1A_7 \sim \triangle ZMN$, we have $\dfrac {A_1A_7}{2r} = \dfrac {\sqrt 3r – d_4}{\sqrt3r}$. That is, $A_1A_7 = 2r – (\dfrac {2}{\sqrt 3})d_4$. The other two segments, $A_3A_9$ and $A_5A_{11}$ can be found in the same way. Therefore, $A_1A_7 + A_3A_9 + A_5A_{11} = 6r - \dfrac {2}{\sqrt 3})(d_4 + d_5 + d_6)$; where $d_5$ and $d_6$ are similarly defined offset distances.

I think $d_1 + d_2 + d_3 = d_4 + d_5 + d_6$.

Summing all these up and noting that (1) $R = \dfrac {2}{\sqrt 3}r$ and $\dfrac {2}{\sqrt 3} = \sec 30^0 = \sec (\dfrac {\pi}{n})$; for n = 6, we should get the required equality.

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