Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \[ M_2 = \left(\f...
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Proof of the product formula for π23\dfrac{\pi}{2\sqrt{3}}

 Hello everyone. Today we will prove the product formula for  π23\dfrac{\pi}{2\sqrt{3}}. We will use the method of direct proof as a proof method. 

Theorem 1: π23=∑n=1∞χ(n)n\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}whereχ(n)={1,if n≡1(mod6)−1,if n≡−1(mod6)0,otherwise\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}

Theorem 2: We haveπ23=5⋅7⋅11⋅13⋅17⋅19⋅23⋅29⋯6⋅6⋅12⋅12⋅18⋅18⋅24⋅30⋯\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}expression whose numerators are the sequence of the odd prime numbers greater than 33 and whose denominators are even–even numbers one unit more or less than the corresponding numerators.


Proof:

By Theorem 1 we know that
π23=1−15+17−111+113−117+119−⋯\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots
we will have
15⋅π23=15−125+135−155+⋯ ,\frac{1}{5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5}-\frac{1}{25}+\frac{1}{35}-\frac{1}{55}+\cdots ,

and adding up both series,
65⋅π23=1+17−111+113−117+⋯ .\frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=1+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots .
We also have
17⋅65⋅π23=17+149−177+191−⋯ ,\frac{1}{7} \cdot \frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{7}+\frac{1}{49}-\frac{1}{77}+\frac{1}{91}-\cdots ,

that subtracted from the former leads to
67⋅65⋅π23=1−111+113−117+⋯ ,\frac{6}{7} \cdot \frac{6}{5} \cdot \frac{\pi}{2\sqrt{3}}=1-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots ,
series where there are no denominators divisible either by 5 or by 7.

In a similar way we remove all those divisible by 11,
111⋅6⋅67⋅5⋅π23=111−1121+1143−⋯ ;\frac{1}{11} \cdot \frac{6 \cdot 6}{7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{11}-\frac{1}{121}+\frac{1}{143}-\cdots ;
that added to the former gives
12⋅6⋅611⋅7⋅5⋅π23=1+113−117+119−⋯ .\frac{12 \cdot 6 \cdot 6}{11\cdot 7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=1+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots .

We notice that the denominators which are divisible by a prime number of the form 6n−16n - 1 are removed by addition, from which this new factor gets added 6n6n−1\dfrac{6n}{6n-1} , while the denominators divisible by a prime number of the form 6n+16n + 1 are removed by subtraction from which this new factor gets added, 6n6n+1\dfrac{6n}{6n+1} .


Thus the denominators of these new factors successively added will be prime numbers while the numerators will be even–even numbers, one unit more or less than the denominators. Consequently if all the terms of the series considered at the beginning are subtracted in this way, we will finally have
⋯30⋅24⋅18⋅18⋅12⋅12⋅6⋅6⋯29⋅23⋅19⋅17⋅13⋅11⋅7⋅5⋅π23=1.\frac{\cdots 30 \cdot 24 \cdot 18 \cdot 18 \cdot 12 \cdot 12 \cdot 6 \cdot 6}{\cdots 29 \cdot 23 \cdot 19 \cdot 17 \cdot 13 \cdot 11 \cdot 7 \cdot 5} \cdot \frac{\pi}{2\sqrt{3}}=1 .

From this we will get
π23=5⋅7⋅11⋅13⋅17⋅19⋅23⋅29⋯6⋅6⋅12⋅12⋅18⋅18⋅24⋅30⋯\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}

■\blacksquare



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