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 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 \(P \) the area of ​​a circle and by \(r \) the radius of a circle. Then the following equation holds: \(P = r ^ 2 \pi \) 

Proof: 

The equation of a circle in Cartesian  coordinate system is \(x ^ 2 + y ^ 2 = r ^ 2 \). Hence we have that \(y = \pm \sqrt {r ^ 2-x ^ 2} \). Based on the geometric interpretation of a certain integral, it follows: \[P = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx \] So, \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx \] \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx \] \[P = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx \] Let us now introduce the substitution \(x = r \cos \theta \). Hence, we have that \(dx = -r \sin \theta d \theta \). So we can write the following equation: \[P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\frac{r^2\cos^2\theta}{r^2}}\sin\theta \, d\theta\] After simplification we get:  \[P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\cos^2\theta}\sin\theta \, d\theta\]\[P=-\int\limits_{\pi}^0 2r^2\sin^2\theta \, d\theta\]\[P=-r^2\int\limits_{\pi}^0 2\sin^2\theta \, d\theta\]\[P=-r^2\int\limits_{\pi}^0 (1-\cos2\theta) \, d\theta\]\[P=-r^2\left(\int\limits_{\pi}^0 d\theta-\int\limits_{\pi}^0 \cos2\theta \, d\theta\right)\]Since \(\int\limits_{\pi}^0 \cos2\theta \, d\theta=0\) we have: \[P=-r^2\int\limits_{\pi}^0 d\theta\]\[P=-r^2(0-\pi)\]\[P=r^2\pi\]

\(\blacksquare\)

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