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A triangle is given by radius-vectors of its vertices: $r_1,r_2,r_3$. The task is to write the radius-vector of the center of the inscribed circle.

The context of the task suggests that we're taking about 3-dimensional vectors, but nothing says $r_1,r_2,r_3$ can't be co-planar.

I've tried using the fact that it's the intersection of bisectors, and thus

$$\dfrac{(\vec{x}-\vec{r_1})\cdot(\vec{r_3}-\vec{r_1})}{|\vec{r_1}-\vec{r_3}|}=\dfrac{(\vec{x}-\vec{r_1})\cdot(\vec{r_2}-\vec{r_1})}{|\vec{r_1}-\vec{r_2}|}$$

and so on, but that didn't lead to anything useful.

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  • $\begingroup$ HINT: Somehow use the Angle Bisector Theorem to obtain the relative position of the incentre. $\endgroup$
    – najayaz
    Commented Oct 1, 2015 at 9:19
  • $\begingroup$ Should a relation between $r_I, r_1, r_2, r_3 $ be obtained finally? $\endgroup$
    – Narasimham
    Commented Oct 1, 2015 at 17:14

1 Answer 1

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Corresponding to the vector notation, let the three vertices of the triangle be $R_1$, $R_2$, $R_3$.

When calculating the angle bisectors we need the unit vectors pointing from a vertex to the two other vertices. In the case of $R_1$ we have

$$\frac{\vec r_2-\vec r_1}{|\vec r_2-\vec r_1|}\text { and }\ \frac{\vec r_3-\vec r_1}{|\vec r_3-\vec r_1|}$$

so

$$\vec d_1=\frac{\vec r_2-\vec r_1}{|\vec r_2-\vec r_1|}+\frac{\vec r_3-\vec r_1}{|\vec r_3-\vec r_1|}$$

points to the direction of the angle bisector at $R_1$.

Similarly, for $R_2$ the bisecting direction vector is

$$\vec d_2=\frac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}+\frac{\vec r_3-\vec r_2}{|\vec r_3-\vec r_2|}.$$

The parametric equations of the angle bisector lines are then

$$\vec b_1(u)=\vec r_1 +u\vec d_1 \text { and }\ \vec b_2(s)=\vec r_2 +s\vec d_2 \tag 1$$

if $u=0$ and $s=0$ then we are at $r_1$ and $r_2$ respectively. If $u$ and $s$ are such that $\vec b_1(u)=\vec b_2(s)$, then we are at the incenter.

From $(1)$ we have then the following equation

$$\vec r_1 +u\vec d_1=\vec r_2 +s\vec d_2$$

or

$$\vec r_1-\vec r_2=s\vec d_2+u(-\vec d_1).\tag2$$

If the vertices of our triangle are not collinear (if we do have a triangle) then $\vec d_1$ and $-\vec d_2$ are linearly independent and form a basis. As it is shown in $(2)$ $u$ and $s$ are the unique coordinates of $\vec r_1-\vec r_2$ in the basis $\{\vec d_1,-\vec d_2\}$.

Note that the incenter is in the plane spanned by the said basis.

The following figure depicts the geometric meaning of the computations above:

enter image description here

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