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May 25th, 2015, 03:15 PM   #1
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inscribed circle of a triangle

Can anyone help me with this problem? I have no idea how to solve it.
Thanx in advance!
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May 26th, 2015, 09:09 AM   #2
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As triangle $\displaystyle ADF$ is equlateral, $\displaystyle \triangle{ADF}=\frac{x^2\sqrt3}{4}$. By the secant-tangent theorem $\displaystyle x^2=\overline{AG}\cdot\overline{AE}$ and $\displaystyle \frac{\triangle{ADF}}{\overline{AG}\cdot\overline{ AE}}=\frac{\sqrt3}{4}$.

If $\displaystyle \overline{BD}=4$ and $\displaystyle \overline{CF}=2,\overline{BC}=6$, as the distance from a point outside a circle to the points of tangency is equal.

Using the fact that the area of a triangle is equivalent to the radius of the incircle ($\displaystyle r$) times the semiperimeter ($\displaystyle s$) we may write

$\displaystyle \frac{\sqrt3}{2}(x+2)\left(\frac{x+4}{2}\right)=rs = \frac{x}{\sqrt3}(6+x)$ which can be simplified and rearranged to $\displaystyle x^2+6x-24=0$.

I used the law of cosines to determine the radius of the incircle in terms of $\displaystyle x$.
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