We have a triangle made from three arbitrary points:
- $A = (α, δ)$,
- $B = (β, ε)$, and
- $C = (γ, θ)$.
The side of the triangle opposite point $A$ has a length of $a$, the side opposite $B$ has length $b$, and the side opposite $C$ has length $c$, where
- $a = \sqrt{(β − γ)^2 + (ε − θ)^2}$,
- $b = \sqrt{(γ − α)^2 + (θ − δ)^2}$, and
- $c = \sqrt{(α − β)^2 + (δ − ε)^2}$.
The incircle of a triangle is the largest circle that can be inscribed within that triangle, and which is tangent to all three sides of the triangle. The center of the incircle is called the incenter.
The radius squared of the incircle is $R^2 = \frac{(−a + b + c) × (a − b + c) × (a + b − c)}{4 × (a + b + c)}$. The incenter has coordinates $I = (X_I, Y_I)$, where
- $X_I = \frac{a × α + b × β + c × γ}{a + b + c}$ and
- $Y_I = \frac{a × δ + b × ε + c × θ}{a + b + c}$.
Thus the equation for the incircle is $(x − X_I)^2 + (y − Y_I)^2 = R^2$.
The incenter is also the point of intersection between the three interior angle bisectors of the triangle. The equations for the interior angle bisectors are:
- $\frac{(δ − ε) × x + (β − α) × y + α × ε − β × δ}{c} = \frac{(ε − θ) × x + (γ − β) × y + β × θ − γ × ε}{a}$
- $\frac{(ε − θ) × x + (γ − β) × y + β × θ − γ × ε}{a} = \frac{(θ − δ) × x + (α − γ) × y + γ × δ − α × θ}{b}$
- $\frac{(θ − δ) × x + (α − γ) × y + γ × δ − α × θ}{b} = \frac{(δ − ε) × x + (β − α) × y + α × ε − β × δ}{c}$
We can find the intersections between a circle and a line by rearranging their respective equations to get $y$ on its own in each equation, setting those equations equal to one another, solving for $x$, and then plugging the resulting $x$-value into one of the equations to solve get the $y$-value.
However, in this case the equation for the circle expands into a huge complicated wall of symbols, and as a result the expressions I get for the $x$−values are exceedingly long. For example:
$$ x = \frac{ ± \left(\frac{α − γ}{b} − \frac{β − α}{c}\right) × (a − b − c) \sqrt{\frac{(a + b − c) × (a − b + c)}{4 b c}} + \left(\frac{α − γ}{b} − \frac{β − α}{c}\right)^2 × \frac{a × α + b × β + c × γ}{a + b + c} + \left(\frac{δ − ε}{c} + \frac{δ − θ}{b}\right) × \frac{a δ + b ε + c θ}{a + b + c} \left(\frac{α − γ}{b} − \frac{β − α}{c}\right) − \left(\frac{δ − ε}{c} + \frac{δ − θ}{b}\right) × \left(\frac{α ε − β δ}{c} + \frac{α × θ − γ × δ}{b}\right) }{ \left(\frac{δ − ε}{c} + \frac{δ − θ}{b}\right)^2 + \left(\frac{α − γ}{b} − \frac{β − α}{c}\right)^2 } $$
This is not at all pleasing to look at. I've been trying to simplify it, but what you see above is as simple as I've been able to make it. I've been trying to condense as many of the Greek-letter point-coordinate variables into the side-length variables $a$, $b$, and $c$ as possible, which helped somewhat, but it's still not pretty.
Can this be simplified further? Is there an easier way to derive the $x$ and $y$ coordinates of the intersections given the three vertices of the triangle?
As we know the distance $R$ between the intersection points $P_A$, $Q_A$ of angle bisector $AI$ with the incircle and its center $I$, we have: $$ \left.{P_A\atop Q_A}\right\}=I\pm{R\over AI}(A-I) $$ and analogous formulas for the other angle bisectors $BI$, $CI$.