Proof Concerning Central Line X5X6X_5X_6 of Triangle

Image
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...
Post a Comment

Proof that an integer is even if its square is even

 Hello everyone. Today we will prove that an integer is even if its square is even. We will use proof by contraposition as proof method.

Theorem: Let nn be an integer. If n2n ^ 2 is an even number, then nn is also an even number. 

Proof: 

The contrapositive of this statement is: Let nn be an integer. If nn is an odd number, then n2n ^ 2 is also an odd number. 

Let us now prove the contrapositive. Since nn is an odd number, we can write it in the form n=2k+1n = 2k + 1 where kk is an integer. By squaring this equation we get: n2=(2k+1)2n ^ 2 = (2k + 1) ^ 2 n2=4k2+4k+1n ^ 2 = 4k ^ 2 + 4k + 1 n2=2(2k2+2k)+1n ^ 2 = 2 \left(2k ^ 2 + 2k \right) +1 Since kk is an integer then 2k2+2k2k ^ 2 + 2k must be an integer due to the closedness of addition, multiplication and addition operations on the set of integers. Denote 2k2+2k2k ^ 2 + 2k by ll , then we have that n2=2l+1n ^ 2 = 2l + 1 , from which we conclude that n2n ^ 2 is an odd number. Since we have proved that the contrapositive of the original statement is correct, it  means that the original statement must be correct as well. 

■\blacksquare

Comments

Post a Comment

Popular posts from this blog

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 \[\frac{1}{5} \cdo...
Read more

Proof of the formula for the area of a circle

Hello everyone.  Today we will prove the formula for the area of ​​a circle.  We will use the method of direct proof as a proof method.  Theorem: Denote by PP the area of ​​a circle and by rr the radius of a circle.  Then the following equation holds: P=r2πP = r ^ 2 \pi   Proof:   The equation of a circle in Cartesian  coordinate system is x2+y2=r2x ^ 2 + y ^ 2 = r ^ 2 .  Hence we have that y=±r2−x2y = \pm \sqrt {r ^ 2-x ^ 2} .  Based on the geometric interpretation of a certain integral, it follows: P=∫−rr(r2−x2−(−r2−x2)) dxP = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx   So, P=∫−rr2r2−x2 dxP = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx P=∫−rr2r2(1−x2r2) dxP = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx P=∫−rr2r1−x2r2 dxP = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx   Let us now introduce the substitution x=rcos⁡θx = r \cos \theta .  Hence, we have t...
Read more

Proof of the formula for the sum of the first n Fibonacci numbers

Hello everyone.  Today we will prove the formula for the sum of the first nn Fibonacci numbers.  We will use the method of mathematical induction as a proof method.  Theorem: ∀n∈N0,∑j=0nFj=Fn+2−1\forall n \in \mathbb {N} _0, \quad \displaystyle \sum_ {j = 0} ^ nF_j = F_ {n + 2} -1   Proof:   1. Base case (n = 0)  ∑j=00Fj=F2−1\displaystyle \sum_ {j = 0} ^ 0F_j = F_ {2} -1 F0=F2−1F_ {0} = F_ {2} -1 0=1−10 = 1-1 0=00 = 0   2. Induction hypothesis (n = m)  Suppose that: ∑j=0mFj=Fm+2−1\displaystyle \sum_ {j = 0} ^ mF_j = F_ {m + 2} -1   3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that: ∑j=0m+1Fj=Fm+3−1\displaystyle \sum_ {j = 0} ^ {m + 1} F_j = F_ {m + 3} -1   So,  ∑j=0m+1Fj=∑j=0mFj+Fm+1=\displaystyle \sum_ { j = 0} ^ {m + 1} F_j = \displaystyle \sum_ {j = 0} ^ {m} F_j + F_ {m + 1} = Fm+2−1+Fm+1=F_ {m + 2} -1 + F_ {m +1} = Fm+1+Fm+2−1=F_ {m + 1} + F_ {m + 2} -1 = Fm+3−1F_ {m + 3} -1 ■\blacksquare
Read more