Mathemarium

Hello everyone. Today we will prove theorem about area relation in right-angled triangle and extouch triangles. Theorem Statement Let $ABC$ be a right-angled triangle with the right angle at $C$. Consider its associated special triangles: The intouch triangle $A_0B_0C_0$ formed by the contact points of the incircle. The extouch triangles $A_1B_1C_1$, $A_2B_2C_2$, and $A_3B_3C_3$, corresponding to the excircles touching $BC$, $AC$, and $AB$ respectively. Denote their respective areas as $T_0, T_1, T_2,$ and $T_3$. Then, the areas satisfy the relation: $$T_3 = T_0 + T_1 + T_2.$$ Proof Basic Notation Let: $a, b, c$ be the sides of $\triangle ABC$, where $c$ is the hypotenuse. $s$ be the semiperimeter: $$s = \frac{a + b + c}{2}.$$ $r$ be the inradius (radius of the incircle). $r_A, r_B, r_C$ be the exradii opposite to $A, B, C$, respectively. The area ...

 Hello everyone. Today we will prove theorem regarding construction of the golden ratio. Theorem: Given isosceles triangle $ABC$ with angle of $30^{\circ}$ at vertex $C$ , inscribed rectangle $DEFG$ ,whose side $DE$ is twice side $EF$, with a side $FG$ along the base side $AB$ . If the side $DE$ is extended to intersect the circumcircle at $P$, then $E$ divides $DP$ in the golden ratio. Proof: We prove that $E$ divides $DP$ in the golden ratio, i.e., $$\frac{DP}{DE} = \frac{DE}{EP} = \phi,$$ where $\phi = \frac{1+\sqrt{5}}{2}$. Coordinate System We place the circumcircle of $\triangle ABC$ with center at the origin $O(0,0)$ and radius $R$.     $C$ is at $(0, R)$.     $A$ and $B$ lie symmetrically along the $x$-axis. Using trigonometry, the coordinates of $A$ and $B$ are: \[A = \left(-\frac{\sqrt{3}}{2}R, -\frac{1}{2}R \right), \quad B = \...

  Hello everyone.  Today we will prove the product formula for  $\dfrac{\pi}{2\sqrt{3}}$.  We will use the method of direct proof as a proof method.  Theorem 1:  $$\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}$$$$\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}$$ Theorem 2:  We have$$\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}$$expression whose numerators are the sequence of the odd prime numbers greater than $3$ and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that $$\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots$$ we will have \[\frac{1}{5} \cdo...

   Hello everyone.  Today we will prove that   $I_0(\sqrt{2})$ is an irrational number, where $I_0$ denotes a modified Bessel function of the first kind. We will use the method of contradiction as a proof method. Theorem 1:    $I_0(\sqrt{2})=\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n!)^22^n}$ Theorem 2:    $I_0(\sqrt{2})$   is an irrational number. Proof: Suppose that  $I_0(\sqrt{2})$  is a rational number. Then  $I_0(\sqrt{2})$  can be written in the form $\dfrac{p}{q}$ where $p$  and $q$ are coprime integers such that $q \ge 1$. By Theorem 1 we can write the following equality: $q!(q-1)!p2^q=\displaystyle\sum_{n=0}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}$ . Since left hand side of this equality is an integer the sum $\displaystyle\sum_{n=q+1}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}$ which is greater than zero $0$  also must be an integer. Clearly for $n \ge q+1$ we have , \(\...

Hello everyone. Today we will prove that the set of prime numbers is infinite.  We will use the method of contradiction as a proof method.  We'll also use the theorem of French mathematician Eduardo Lucas.  Theorem (Lucas): Every prime factor of Fermat number $F _ n = 2 ^ {2 ^ n} + 1$; ($n > 1$) is of the form $k2 ^{n + 2} + 1$.  Theorem: The set of prime numbers is infinite. Proof: Suppose opposite, that there are just finally many prime numbers and we denote the largest prime by $p$. Then $F_p$ must be a composite number because $F_p>p$. By Lucas theorem we know that there is a prime number $q$ of the form $k2 ^{p + 2} + 1$ that divides $F_p$. But $q>p$ , thus we arrived at  a contradiction. Hence, the set of prime numbers is infinite. $\blacksquare$

Hello everyone.  Today we will prove that any natural number greater than $1$ can be expressed as a product of a prime number and number one or as a product of few prime numbers.  We will use the method of mathematical induction as a proof method.  Theorem: Let $n$ be a natural number greater than $1$.  Then $n$ can be expressed as the product of one prime number and number one or as the product of few prime numbers. Proof: Note that if $n$ is a prime number, the statement is automatically proved because any number can be written as the product of that number and number one.  1. Base case (n = 2)  Since $2$ is a prime number the statement is automatically proved.  2. Induction hypothesis (n = m)  Suppose that $\forall k \in \mathbb {N}, \quad 2 \le k \le m$, $k$ can be expressed as the product of a prime number and number one or as a product of few prime numbers.  3. Inductive step (n = m + 1)  Using the assum...