Proof Concerning Central Line \(X_5X_6\) of Triangle

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

Hello everyone. Today we will prove Archimedes' theorem. We will use the method of contradiction as a proof method. 

Theorem: For every two real numbers \(a \) and \(b \) where \(a> 0 \) there exists a natural number \(n \) such that \(a \cdot n> b \). 

Proof: 

Considering that \(a> 0 \)  we can divide the inequality \(a \cdot n> b \)  by the number \(a \) so that we get a new inequality \(n> \frac{b}{a} \) . Let us denote the real number \(\frac{b}{a} \) by \(x \), then we can reformulate the theorem we need to prove as follows:

 For any real number \(x \) there exists a natural number \(n \) such that \(n> x \). 

Suppose the opposite, ie. that there exists a real number \(x \) such that \(x \ge n \) for any natural number \(n \). This would mean that the set of natural numbers \(\mathbb {N} \) is bounded from above. Let us denote this least upper bound by \(\operatorname{sup}(\mathbb{N}) = S \). Now consider the difference \(S-1 \). It is clear that \(S-1 \) is not the upper bound of the set \(\mathbb{N} \), because if it were then \(S \) that is greater than \(S-1 \) would not be the least upper bound. This means that there exists a natural number \(m \) such that \(m> S-1 \), that is, we have \(m + 1> S \). As the addition operation is closed on the set of natural numbers we have that \(m + 1 \in \mathbb{N} \), ie. the number \(m + 1 \) is a natural number greater than \(\operatorname{sup} (\mathbb{N}) \). This contradicts the initial assumption which means that for any real number \(x \) there exists a natural number \(n \) such that \(n> x \). 

\(\blacksquare\)

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