Mathemarium

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Proof of the Area Relation in Right-Angled Triangle and Extouch Triangles

February 06, 2025

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 ...

Proof of the Theorem Regarding Golden Ratio

February 05, 2025

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 = \...

Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

July 09, 2023

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...

Proof that \(I_0(\sqrt{2})\) is an irrational number

June 19, 2023

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 , \(\...

Proof that the set of prime numbers is infinite

February 06, 2023

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. \(\blacksquare\)

Proof that a natural number can be expressed as a product of prime numbers

December 17, 2021

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...