Formula for finding the coordinates of the Gergonne point of a triangle

Given any triangle with sides $a$ , $b$ and $c$ , and coordinates of vertices $(x_{1},y_{1})$ , $(x_{2},y_{2})$ and $(x_{3},y_{3})$ . What are the coordinates of its Gergonne point?

#Geometry

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Xuming Liang

@Nihar Mahajan

Swapnil Das - 5 years, 4 months ago

Its easy but bit tedious. What have you tried?

Nihar Mahajan - 5 years, 4 months ago

Tried infinitely many methods, but led to zero conclusions :P

Swapnil Das - 5 years, 4 months ago

Aah, I need help. @Ishan Singh

@Aditya Kumar

Swapnil Das - 5 years, 4 months ago

There is an expression in terms of s,a,b,c. First find $T_{a}$ by section formula. Then compute $\frac{AG_{e}}{G_{e}T_{A}}$ by menelaus. Use the section formula again to find the gergonne point. The formula comes out to be $\frac {\displaystyle \sum_{cyc} (s-b)(s-c)A}{\displaystyle \sum_{cyc} (s-b)(s-c)}$ . Here A is the complex number for the point A (this formula can easily be converted into cartesian form)

Pranav Kirsur - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$