Construct a triangle given base, inradius and exradius

Question

The following question can be found in a geometric text:

1. Construct triangle ABC, given BC, inradius r and exradius r1, which is opposite to vertex A.

I tried making the auxiliary figure for the above construction. I was pretty much able to find the perimeter in terms of the ratio of the radii and given one side (I can share the same if you’re interested). Is there any way to build more on this?

Answer 1

I will use hereafter classical notations for triangle $ABC$: $a=BC,b=CA,c=AB$ for the sides

$p=\tfrac12(a+b+c) \ \iff \ \ (p-b)+(p-c)=a\tag{1}$

for its semi-perimeter, $S$ for its area, $r$ for its inradius, $r_A$ for its exradius relative to vertex $A$.

The purpose is, knowing $a,r,r_A$ to be able to compute $b$ and $c$.

Here is a solution, inspired by the solution given by Cuman, but more simple and - important as well - giving constraint

$$a \ge 2 \sqrt{r_Ar}\tag{2}$$

that must be fulfilled between the 3 given quantities for a solution to exist.

(a constraint that @Oscar Lanzi has also found with his geometrical solution).

We have formulas :

$$r=\frac{S}{p}, \ \ r_A=\frac{S}{p-a}\tag{3}$$

which are established here.

Remark: from (3), it is easy to establish that :

$$p=\frac{ar_A}{r_A-r}\tag{3'}$$

Heron's formula can be expressed in the following way :

$$S^2=p(p-a)(p-b)(p-c)\tag{4}$$

Otherwise said :

$$\frac{S}{p}\frac{S}{p-a}=(p-b)(p-c)\tag{5}$$

Using (3) and setting $\beta=p-b$ and $\gamma=p-c$ :

$$rr_A= \beta \gamma\tag{6}$$

As (1) can be written under the form

$$\beta + \gamma = a \tag{7}$$

Knowing the product (6) and the sum (7) of $\beta$ and $\gamma$, we can say they are roots of the quadratic equation :

$$X^2-aX+rr_A=0$$

which has real roots iff its discriminant $\Delta = a^2-4rr_A \ge 0$, i.e., iff condition (2) is fulfilled.

Knowing roots $\beta=p-b$ and $\gamma=p-c$, one easily gets the values of $b$ and $c$ using relationship (3') for $p$.

Answer 2

If we let $s$ and $p$ be the area and perimeter of $\Delta ABC$ respectively, then we have the following formulas for exradius and inradius:

$$ \begin{cases} r = \frac{2s}{p} \\ r_1 = \frac{s}{\frac{p}{2}-BC} \end{cases} $$

Since $r, r_1, BC$ are given, we can solve for $s$ and $p$. After that, we can get $AB + AC = p - BC$. Now if we look at Heron's formula for triangle's area, we can get $AB \cdot AC$:

$$ s = \sqrt{\frac{p}{2}\left(\frac{p}{2} - AB\right)\left(\frac{p}{2} - AC\right)\left(\frac{p}{2} - BC\right)} \\ \Leftrightarrow \left(\frac{p}{2} - AB\right)\left(\frac{p}{2} - AC\right) = \frac{s^2}{\frac{p}{2} \left(\frac{p}{2} - BC\right)} \\ \Leftrightarrow \frac{p^2}{4} - \frac{p}{2} \left(AB + AC\right) + AB \cdot AC = \frac{s^2}{\frac{p}{2} \left(\frac{p}{2} - BC\right)} \\ \Leftrightarrow \frac{p^2}{4} - \frac{p}{2} \left(p - BC\right) + AB \cdot AC = \frac{s^2}{\frac{p}{2} \left(\frac{p}{2} - BC\right)} \\ \Leftrightarrow AB \cdot AC = \frac{s^2}{\frac{p}{2} \left(\frac{p}{2} - BC\right)} - \frac{p^2}{4} + \frac{p}{2} \left(p - BC\right) $$

Now we have the following system:

$$ \begin{cases} AB + AC = p - BC \\ AB \cdot AC = \frac{s^2}{\frac{p}{2} \left(\frac{p}{2} - BC\right)} - \frac{p^2}{4} + \frac{p}{2} \left(p - BC\right) \end{cases} $$

The solution to this system can be obtained through the quadratic formula.

Answer 3

Below is a sketch of the construction. (It's set up with $BC=5$, inradius $1$ and exradius $6$, which yields the $3-4-5$ right triangle.)

Start by identifying a pair of lines parallel to $\overline{BC}$. Line $L$ contains the incenter $I$ at a distance from $\overline{BC}$ equal to the given inradius. Opposite this is line $L'$, which contains the excenter $I'$ at a distance from $\overline{BC}$ equal to the given exradius.

Next observe that $I$ lies on the bisector of the interior angle of $\triangle ABC$ at $B$, while $I'$ lies on the bisector of the adjacent exterior angle. Since the bisected angles are supplementary, thus $\overline{BI}$ is perpendicular to $\overline{BI'}$. By similar reasoning $\overline{CI}$ is perpendicular to $\overline{CI'}$.

From these perpendicularities it follows that quadrilateral $BICI'$ is inscribed in a circle with $\overline{II'}$ as a diameter. We can now identify the center of this circle $O$ thusly:

• Since $I$ and $I'$ are opposite ends of a diameter and these points are known to lie on $L$ and $L'$ respectively, $O$ must lie on the midline $M$ between $L$ and $L$. $M$ is determined by passing it through the midpoints if any two transverse segments whose midpoints are distinct.

• Since the circumscribed circle also passes through both $B$ and $C$, its center must lie on the perpendicular bisector $N$ of $\overline{BC}$. (Apologies: I found the label $N$ was cropped off after I uploaded and imported the picture. Is there an easy way to edit or do I have to start over?)

Thus $O$ is determined as the intersection of $M$ and $N$, and a circle centered there passing through $B$ (thus also through $C$) must pass through $I$ where it intersects line $L$ (and through $I'$ where it intersects line $L'$). If the triangle exists (which requires the base $BC$ to be at least twice the geometric mean of the given inradius and extadius), in general there will be two points of intersection of this circle with $L$, either of which may be identified as $I$.

With $I$ determined, $\triangle IBC$ is defined. Since $I$ lies on the interior angle bisectors of $\triangle ABC$, that triangle is now obtained by doubling the angles $B$ and $C$ in $\triangle IBC$.