Joy of Mathematics

Sum of segments

March 26, 2023 Tolaso

Given the figure below

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evaluate $\mathrm{AB}^2 + \Gamma \Delta^2$ .

Solution

Solution to be migrated by this link.

Generalization of Pythagoras’ theorem

November 24, 2022 Tolaso

In the following figure the triagle $\mathrm{AB} \Gamma$ is right angled. Using the sides of the triangle we draw equilateral triangles as shown in the picture.

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Prove that

$\mathrm{E}_1 = \mathrm{E}_2 + \mathrm{E}_3$

Solution

Since $\mathrm{AB} \Gamma$ is right angled , we know that Pythagoras’ theorem applies; hence:

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2$

On the other hand the area of an equilateral triangle of side $\alpha$ is given by the formula $\displaystyle \mathrm{E} = \frac{\alpha^2 \sqrt{3}}{4}$ . Hence,

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2 \Leftrightarrow \frac{\sqrt{3}\mathrm{AB}^2}{4} + \frac{\sqrt{3} \mathrm{A} \Gamma^2}{4} = \frac{\sqrt{3} \mathrm{B}\Gamma^2}{4} \Leftrightarrow \mathrm{E}_3 + \mathrm{E}_2 = \mathrm{E}_1$

and the result follows.

Exterior angle bisector

September 10, 2022September 10, 2022 Tolaso

In the following figure $\mathrm{AE}$ is the exterior angle bisector of the angle $\mathrm{A}$ of the triangle $\mathrm{AB} \Gamma$ .

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Prove that $\displaystyle \overrightarrow{\mathrm{AE}} = \frac{1}{\beta-\gamma} \left ( \beta \overrightarrow{\mathrm{AB}} -\gamma \overrightarrow{\mathrm{A} \Gamma} \right )$ .

Solution

It follows from the exterior angle bisector theorem that

(1) $\begin{equation*} \overrightarrow{\Gamma \mathrm{E}} = \frac{\beta}{\gamma} \cdot \overrightarrow{\mathrm{BE}} \end{equation*}$

Hence,

$\begin{align*} \overrightarrow{\mathrm{BE}} - \overrightarrow{\Gamma \mathrm{E}} = \overrightarrow{\mathrm{B} \Gamma} = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} &\overset{(1)}{\Leftrightarrow} \frac{\gamma - \beta}{\gamma} \cdot \overrightarrow{\mathrm{BE}} = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} \\ &\Leftrightarrow \frac{\gamma - \beta}{\gamma} \cdot \left ( \overrightarrow{\mathrm{AE}} - \overrightarrow{\mathrm{AB}}\right ) = \overrightarrow{\mathrm{A} \Gamma} - \overrightarrow{\mathrm{AB}} \\ &\Leftrightarrow \frac{\gamma - \beta}{\gamma} \cdot \overrightarrow{\mathrm{AE}} = \left ( \frac{\gamma - \beta}{\gamma} -1 \right ) \cdot \overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{A} \Gamma} \\ &\Leftrightarrow \overrightarrow{\mathrm{AE}} = \frac{1}{\beta - \gamma} \left ( \beta \cdot \overrightarrow{\mathrm{AB}} -\gamma \cdot \overrightarrow{\mathrm{A} \Gamma} \right ) \end{align*}$

Relation on a triangle

August 21, 2022August 21, 2022 Tolaso

Given the following figure

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prove that $a^3 + c^3 = 3ac^2$ .

Circumscribed and inscribed circle

August 18, 2022 Tolaso

Let $ABC$ be an acute triangle. Let us denote as $(O, R)$ its circumscribed circle and as $(I, r)$ its inscribed circle. If the segment $AI$ intersects the circle $(O, R)$ at $D$ prove that

$AI \cdot ID = 2Rr$

Solution

We simply that

$AI\cdot ID=R^2 - OI^2 =R^2 -\left( R^2 - 2Rr \right)=2Rr$