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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?

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  • $\begingroup$ Do you mean that you have used the fact that the excircle is the image of the incircle by the dilation with center $A$ and ratio $r_1/r$ (as presented in this video ? $\endgroup$
    – Jean Marie
    May 22, 2023 at 20:56
  • $\begingroup$ You should consider the solution by Oscar Lanzi : it is a more direct way than the accepted solution. I have another solution too based on the comment just above. $\endgroup$
    – Jean Marie
    May 25, 2023 at 5:41

3 Answers 3

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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.)

enter image description here

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 $\triangle IBC$.

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  • $\begingroup$ What were those comments? $\endgroup$ May 25, 2023 at 23:00
  • $\begingroup$ I was saying in particular that I had at last taken time to understand your solution and that I find it excellent, more "natural" than the selected one. $\endgroup$
    – Jean Marie
    May 26, 2023 at 4:36
  • $\begingroup$ But there is a point that must be addressed : there are cases where no solution exist, i.e., if length $BC$ is too small with respect to radii $r$ and $r_1$ (see the comment I have made to the solution given by cuman). $\endgroup$
    – Jean Marie
    May 26, 2023 at 4:57
  • $\begingroup$ See my answer which is a large simplification of Cuman's solution including in particular the discussion about the "famous" condition of existence of a solution. $\endgroup$
    – Jean Marie
    May 28, 2023 at 19:03
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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 $p$.

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  • $\begingroup$ I like it. Now how about this: we know the common tangents of the relevant incircle and excircle are collinear with the unknown vertex. Using this we can work out $h=2rr_A/(r_A-r)$, and the unknown vertex is also collinear with $I$ and $I'$. Using these results, if $I$ exists then the unknown vertex is defined as the intersection of line $OI$ with a constructible line parallel to the base. $\endgroup$ May 28, 2023 at 20:13
  • $\begingroup$ Yes : it looks an alternative tractable way in the spirit of one circle being the homothetic image of the other. $\endgroup$
    – Jean Marie
    May 28, 2023 at 20:19
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    $\begingroup$ @Oscar Lanzi : Another possible connection is to consider this 2D issue as a cross section of a 3D classical figure : Dandelin spheres where side BC, which has length 2a (a = major semi-axis of the ellipse) is equal to the length of the common piece of generatrix of the cone. $\endgroup$
    – Jean Marie
    May 28, 2023 at 21:00
  • $\begingroup$ @Oscar Lanzi [Ctd] with for example interesting elements in this answer by Blue $\endgroup$
    – Jean Marie
    May 28, 2023 at 21:38
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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.

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    $\begingroup$ One has to distinguish the cases where the final quadratic has real/imaginary roots : condition "discriminant $\ge 0$" must surely be interesting to discuss. $\endgroup$
    – Jean Marie
    May 25, 2023 at 20:28
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    $\begingroup$ To get a root the base must be at least twice the geometric mean of the given inradius and exradius. If it isn't, the intended circumscribed circle in my construction misses $L$ and $L'$, so we can't get the inscribed quadrilateral and thus don't have $\triangle IBC$ from which to build up $\triangle ABC$. $\endgroup$ May 25, 2023 at 20:45
  • $\begingroup$ See my solution where I find back the constraint $a \ge \sqrt{rr_A}$ you have found in a geometrical way. $\endgroup$
    – Jean Marie
    May 28, 2023 at 5:04

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