$u=(3,1)$, $v=(3,6)$, $i=(2,3)$, need find $w\in \mathbb{R}^2$ such that $i$ is the incenter of the triangle $[u,v,w]$.
I have looked at some other people's approaches to this problem, but I think those approaches are too complicated, and are more like for some advanced intuitive geometer. With the fact that I'm just taking a course in geometry, my approach is probably not so elegant, but it should work. The problem is that I'm still not getting the correct answer.
Let $\alpha, \beta, \gamma$ be the three angles of $[u,v,w]$. Then bisect the three angles to assume $\alpha/2, \beta/2, \gamma/2$.
We can find that $a=(3,3)$ is the point closest to $i$ on the edge $[u,v]$.
We can now set up the following equations:
$$\alpha/2 = \arccos\left(\frac{(v-u)\cdot (v-a)}{\|v-u\|\|v-a\|}\right)=\arccos\left(\frac3{\sqrt{10}}\right)\implies \alpha = 2\arccos\left(\frac3{\sqrt{10}}\right)$$
$$\gamma/2 = \arccos\left(\frac{(v-w)\cdot (w-u)}{\|v-w\|\|w-u\|}\right)=\arccos\left(-\frac{9}{10}\right)\implies \gamma = 2\arccos\left(-\frac9{10}\right)$$ $$\beta = \pi-\alpha-\beta$$
$$ \cos\alpha = \frac{(v-u)\cdot (w-u)}{\|v-u\|\|w-u\|} $$
Can someone please suggest what is wrong with my approach? Is there an additional equation missing? Also, I'm getting some "crazy" expression for $\beta$, which even WolframAlpha finds difficult to deal with. Is this geometric problem really that complicated?
I would use the double able formula for tangent to get the slopes of $uw$ and $vw$, where $w$ is the third vertex. Here I do $AC$, as $BC$ would be similar.
The counterclockwise angle from $uv$ to $ui$ is given by $\tan \theta=1/2$ which you can easily see from the slopes of the two rays ($uv$ is straight up, $ui$ has slope $-2$. Then the counterclockwise angle from $uv$ to $uw$ is $2\theta$ and from the double angle formula for tangent we get $\tan 2\theta=4/3$. With $uv$ directed straight up the slope of $uw$ is $-3/4$.
The reader can similarly compute the slope of $vw$. The answer will be a $3-4-5$ right triangle.