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Angle $BAC$ = $60$ degrees. $D$, $E$ and $F$ are points at which the circle $O$ is tangent to the sides $AB$, $BC$ and $CA$. Let $G$ be the point of intersection of the line segment $AE$ and the circle $O$.

Let $AD=x$

$\frac{\text{Area of triangle ADF}}{AG\cdot AE} = \alpha$

Let $BD=4$ and $CF=2$, then $BC= \beta$. Given that $x=AD$ satisfies the equation $x^2+\gamma x-\tau=0$ where $\gamma$ and $\tau$ are constants, find the value of $\gamma$, $\tau$ and $x$.

How would I find the values of $\alpha$, $\beta$, $\gamma$, $\tau$ and $x$? Im really stuck.

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  1. $AD=AF,\ BD=BE,\ CE=CF$ because they are the tangent segments to the same circle.
  2. $AG\cdot AE=x^2$ because they are both the powers of $A$ w.r.t. the circle. (Or, if you prefer, the triangles $AGD$ and $ADE$ are similar.)
  3. Area of triangle $ADF\ =\ \displaystyle\frac{AD\cdot AF\cdot \sin(DAF\angle)}2$.
  4. Let $y:=BD$ and $z:=CF$. Then $AB=x+y$ and $AC=x+z$, and use the theorem of cosine (for example) to calculate the possible values of $x$, knowing that $BAC\angle=60^\circ$.

I bet, you will get $2$ solutions for $x$, their sum will be good for $-\gamma$ and their product for $-r$, because the normalized quadratic polynomial with roots $a$ and $b$ is $(x-a)(x-b)=x^2-(a+b)x+ab$.

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Ah, you don't even need to solve the equation that you get by the law of cosines. Unless I miscalculated, $\gamma=6$ and $r=48$. – Berci Feb 14 '13 at 17:39
Thanks, solved it! :) – NL49 Feb 23 '13 at 13:43

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