Given vectors $u,v, i$, find vector $w$ such that $i$ is incenter of triangle $[u,v,w]$

Question

Given $u=(3,1)$, $v=(3,6)$, $i=(2,3)$, find $w\in \mathbb{R}^2$ such that $i$ is the incenter of the triangle $[u,v,w]$.

To my understanding, in order to come up with the coordinate pair $(w_1, w_2)$, one needs to come up with six equations. I was only able to come up with four, so I'm asking to help me find the other two equations.

The distances from the incenter $i$ to the three sides of $[u,v,w]$ are equal, so I find the point $r$ on the line $[u,v]$ as follows:

$$(v-u)\cdot (i-r)=0\iff (0,5)\cdot (2-r_1, 3-r_2)=0$$ $$\implies 5(3-r_2)=0\implies r_2=3$$ $r_1 = 3$, since the difference between the first coordinates of $u$ and $v$ is zero. So we have that $r=(3,3)$, so that $\|i-r\|=1$.

So, now we can come up with the equations:

1. Let $s=(s_1,s_2)$ be the point on $[w,u]$ and $t=(t_1,t_2)$ be the point on $[v,w]$. Now, $$(w_1-3, w_2-1)\cdot(2-s_1,3-s_2)=0$$
2. Also, $(3-w_1)(2-t_1)+(6-w_2)(3-t_2)=0$ (from the other dot product).
3. $(2-s_1)^2+(3-s_2)^2=1$ (from the fact that the incenter is equidistant from all sides of $[u,v,w]$.
4. $(2-t_1)^2+(3-t_2)^2=1$

Now, to find $w=(w_1, w_2)$ precisely, I need two more equations. I would appreciate some hints. Moreover, I think this procedure is rather tedious, and I was wondering if there's a more elegant analytic (or maybe axiomatic) approach to this problem.

Answer 1

There's a simpler geometrical way to do the job. Consider angles $\alpha$ and $\beta$ in diagram below: from your data we know that $$ \tan\alpha={||i-r||\over||v-r||}={1\over3}, \quad \tan\beta={||i-r||\over||u-r||}={1\over2}, $$ whence: $$ \tan2\alpha={2\tan\alpha\over1-\tan^2\alpha}={3\over4}, \quad \tan2\beta={2\tan\beta\over1-\tan^2\beta}={4\over3}. $$ If $h$ is the projection of $w$ on line $[u,v]$, set: $x=||v-h||$ and $y=||u-h||=5-x$. We have then: $$ ||w-h||=x\tan2\alpha=y\tan2\beta, \quad\hbox{that is:}\quad {3\over4}x={4\over3}(5-x), \quad\hbox{whence:}\quad x={16\over5} $$ and $\displaystyle||w-h||={12\over5}$.

It is then straightforward to find the coordinates of $w$: $$ w=v+x{r-v\over||r-v||}+||w-h||{i-r\over||i-r||}= (3,6)+{16\over5}(0,-1)+{12\over5}(-1,0) =\left({3\over5},{14\over5}\right). $$

Answer 2

Drop a perpendicular from $i$ to the line $uv$ let us say the base of the perpendicular is $x$ . Draw a circle centered at $i$ with radius $ix$. Since the line $uv$ is already a tangent to the circle there is precisely one more tangent for each points $u$ and $v$. If the two tangents are not parallel then they intersect at a point $w$ is the required point. Only case the tangents will be parallel when the distances between the points $i,u,v$ are same. Then such a point does not exist.

You already found the point $x$ =(3,3). The radius of the circle centered at $i$ tangent to the line $uv$ at $x$ is 1.

Let the tangent to the circle from the point $u$ be incident on the circle $A=(a,b)$. Then the triangles $\triangle Aui$ is congruent to the triangle $\triangle iux$. So the distance $Au$ is 2. we get equations: $$(a-2)^2+(b-3)^2=1$$ $$(a-3)^2+(b-1)^2=4$$ solving these we get $(a,b)=(-\frac{7}{5}, \frac{11}{5})$. Similarly we get the point of incidence of the tangent to the circle from the point $v$ say $B=(c,d)$ by equations: $$(2-c)^2+(d-3)^2=1$$ $$(3-c)^2+(d-6)^2=9$$ we solve these to get $(c,d)=(\frac{6}{5}, \frac{18}{5})$. Then once has to intersect the lines $uA$, $3x+4y-13=0$ and $vB$, $-4x+3y-6=0$ to get the desired point $w=(\frac{3}{5}, \frac{14}{5})$. One can use Wolframalpha to check the calculations.

Answer 3

You’ve got the center of the incircle and one of the sides of the triangle, which is tangent to that circle. The third vertex of the triangle is the intersection of the other tangents to the incircle that pass through the two known vertices. This point can be found without explicitly constructing the incircle or solving any equations by using homogeneous coordinates. In the following, homogeneous vectors (points) are represented by column vectors are are indicated by boldface variables, while covectors (lines) are row vectors indicated with a script font. To save space, I’ll also use the convention that the cross product of column vectors is a row vector and vice-versa. Recall that covectors (covariant vectors) transform differently than do vectors: if $M$ is a matrix that represents a projective transformation, so that the image of a point $\mathbf p$ is $M\mathbf p$, then the image of the line $\mathscr l$ is $\mathscr lM^{-1}$.

The line through $\mathbf u$ and $\mathbf v$ is $\mathscr t_{\mathbf u\mathbf v}=\mathbf u\times\mathbf v$. Let $\mathscr b_{\mathbf u}=\mathbf u\times\mathbf i$ and $\mathscr b_{\mathbf v}=\mathbf v\times\mathbf i$ be the lines through the incenter and the two known vertices. Since the angle bisector of a pair of lines tangent to a circle passes through the circle’s center, we can find the other tangent lines by reflecting $\mathscr t_{\mathbf u\mathbf v}$ in $\mathscr b_{\mathbf u}$ and $\mathscr b_{\mathbf v}$, respectively. This can be accomplished in several ways. Here, I’ll use transformation matrices. Reflection in the line $\mathscr l=(a,b,c)$ is represented by the matrix $$R_{\mathscr l}=\pmatrix{b^2-a^2 & -2ab & -2ac \\-2ab & a^2-b^2 & -2bc \\ 0 & 0 & a^2+b^2}.$$ (You can find derivations of this matrix elsewhere.) A reflection is its own inverse, so $\mathscr t_{\mathbf u\mathbf w}=\mathscr t_{\mathbf u\mathbf v}R_{\mathscr b_{\mathbf u}}$ and $\mathscr t_{\mathbf v\mathbf w}=\mathscr t_{\mathbf u\mathbf v}R_{\mathscr b_{\mathbf v}}$. Finally, $\mathbf w=\mathscr t_{\mathbf u\mathbf w}\times\mathscr t_{\mathbf v\mathbf w}$.

For example, let $u=(-2,4)$, $v=(-1,-2)$ and $i=(0,2)$. The first three lines are $$\mathscr t_{\mathbf u\mathbf v}=(-2,4,1)^T\times(-1,-2,1)^T=(6,1,8) \\ \mathscr b_{\mathbf u}=(-2,4,1)^T\times(0,2,1)^T=(2,2,-4) \\ \mathscr b_{\mathbf v}=(-1,-2,1)^T\times(0,2,1)^T=(-4,1,-2).$$ Plugging the components of the two bisector lines into the above formula for the reflection matrices and applying them to $\mathscr t_{\mathbf u\mathbf v}$ results in $$(6,1,8)\pmatrix{0&-8&16\\-8&0&16\\0&0&8}=(-8,-48,176) \\ (6,1,8)\pmatrix{-15&8&-16\\8&15&4\\0&0&17}=(-82,63,44)$$ and $(-8,-48,176)\times(-82,63,44)=(-13200,-14080,-4440)^T$. Converting back to Cartesian coordinates by dividing through by the last component yields $w=\left(\frac{110}{37},\frac{352}{111}\right)$. This computation and its results are illustrated below:

Another way to approach this problem is to use the polar lines of the vertices. Once you’ve found the incircle, compute the polar lines of $u$ and $v$ with respect to this circle and find their intersections with it. This will give you the three points of tangency of the incircle with the sides of the triangle. From here, you can construct the two tangents and find their intersection via cross products as above, but there are a couple of more interesting ways to go. The polar points of the lines through pairs of tangent points to the incircle are the vertices of the triangle, so the polar of the line through the two tangent points of the unique tangent lines through $u$ and $v$ is $w$. As well, the polar line of a point on a circle is the tangent through that point, so you can get the tangent lines by computing the polars of the tangent points.

Using the same example as above, we have $\mathscr t_{\mathbf u\mathbf v}=(6,1,8)$. The square of the distance of $i$ from this line is $$r^2={(\mathscr t_{\mathbf u\mathbf v}\mathbf i)^2\over \mathscr t_{\mathbf u\mathbf v}\,\operatorname{diag}(1,1,0)\,\mathscr t_{\mathbf u\mathbf v}^T}=\frac{100}{37}.$$ The matrix of the incircle is thus $$Q=\pmatrix{1&0&-i_x \\ 0&1&-i_y \\ -i_x&-i_y&i_x^2+i_y^2-r^2} = \pmatrix{1&0&0\\0&1&-2\\0&-2&\frac{48}{37}}.$$ The polar of $u$ is $\mathbf u^TQ=\left(-2,2,-\frac{248}{37}\right)$ and the polar of $v$ is $\mathbf v^TQ=\left(-1,-4,\frac{196}{37}\right)$. Solving for the intersections of these lines with the circle gives the respective pairs of points $\left(-\frac{60}{37},\frac{64}{37}\right)$, $\left(\frac{10}{37},\frac{134}{37}\right)$ and $\left(\frac{820}{629},\frac{628}{629}\right)$, $\left(-\frac{60}{37},\frac{64}{37}\right)$. The polars of the unique points of each pair are $$\mathscr t_{\mathbf u\mathbf w}=\left(\frac{10}{37},\frac{134}{37},1\right)Q=\left(\frac{10}{37},\frac{60}{37},-\frac{220}{37}\right) \\ \mathscr t_{\mathbf v\mathbf w}=\left(\frac{820}{629},\frac{628}{629},1\right)Q=\left(\frac{820}{629},-\frac{630}{629},-\frac{440}{629}\right)$$ and their cross product (after taking out obvious common factors from each) is $\mathbf w=\left(-1650,-1760,-555\right)^T$, which when converted to Cartesian coordinates gibes with the previous computation. Following the other method, the line through the unique points of each pair, found by taking the cross product of their homogeneous coordinates, is $\mathscr p_{\mathbf w}=\left(\frac{1650}{629},\frac{650}{629},-\frac{2800}{629}\right)$ and finally $\mathbf w=Q^{-1}\mathscr p_{\mathbf w}^T=\left(\frac{1650}{629},\frac{1760}{629},\frac{15}{17}\right)$.

Yet another way to find the points of tangency is to compute the distances from $u$ and $v$ to the tangent point on the known triangle side. This will give you the radii of circles centered at $u$ and $v$ that intersect the incircle at the points where the tangents from $u$ and $v$, respectively, touch the incircle. This will lead to the same equations as the polar route, above. Of course, you can also find the tangents to the incircle from the known vertices in the more usual way—solve quadratic equations for the slopes of the lines, and then, as above, compute the intersection of the two unique tangents.

Answer 4

In the figure, the dotted lines are the angle bisectors of $\triangle UVW$. They will also serve as the lines of symmetries for us to locate U’ and V’, the points of reflection of U (about VI) and of V (about UI) respectively.

$U’ = … = (0, 2)$ by the formula suggested by John Wayland Bales at How to find the reflection of a function about a straight line of the form $y = mx + c$?

Similarly, $V’ = (–1, 4)$.

Use two-point form to get the equations of UV’ and VU’.

Solve the above equations to get the co-ordinates W.

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Remark:- Instead of applying the given formula, we can find V’ (and also U') in a step by step process as:-

(1) Get the equation of UI.

(2) Get the equation of the green line which is perpendicular to UI and passing through V.

(3) Find K by solving the above two equations.

(4) Get V’ where K is the midpoint of VV’

Answer 5

Another answer specific to this numerical values is:

Note the angle $\angle xvi= \arctan \frac{1}{3}$ and $\angle xui= \arctan \frac{1}{2}$ so the sum $\angle xvi + \angle xui = \arctan \frac{1}{3}+\arctan \frac{1}{2}=\frac{\pi}{4}$. So the angle $\angle uwv = \pi - 2(\angle xvi + \angle xui )= \frac{\pi}{2}$. Now $\angle vwi=\angle iwu = \frac{\pi}{4}$. So the lenth of the segment $Bw$=1, where $B$ is the point the tangent $wv$ touches the circle. Similarly the length $wA$=1, where $A$ is the point the tangent $uw$ touches the cirlce.

From these informations we can calculate the coordinates of $w=(\alpha,\beta)$ by solving the equations we obtain by the lengths: $$(\beta-6)^2-(\alpha-3)^2=16$$ $$(\beta-1)^2+(\alpha-3)^2=9$$.

I am getting the point $(\frac{3}{5},\frac{14}{5})$.