Triangle construction given semiperimeter and radii of inscribed and circumscribed circles.

Question

Given $$ \rho,r,R $$ of a scalene triangle

how to construct it geometrically, say when side $c$ is parallel to $x-$ axis... with no Ruler/Compass restriction?

Answer 1

You need to solve a cubic equation whose roots are the sides of the triangle. To do so generally you need first to establish that the triangle exists, then use an angle trisector along with your straightedge and compasses.

I shall first derive the cubic equation, from which the conditions for the triangle to exist -- three real roots for the sides -- can be derived. Then a case will be identified where the given quantities are whole numbers yet the cubic equation is irreducible, which prevents construction with an unmarked straightedge and compasses alone. That an angle trisector is a sufficient additional tool follows from the trigonometric formula for solution of cubic equations with three real roots.

Deriving the Cubic Equation

Define $r$ as the inradius, $R$ as the circumradius, $s$ as the semiperimeter, and $x,y,z$ as the unknown sides. From standard formulas for the intadius and circumradius we have

$r=\sqrt{(s-x)(s-y)(s-z)/s}$ Eq. 1

$R=xyz/4\sqrt{(s-x)(s-y)(s-z)}$ Eq. 2

Multiplying Eq. 1 by $\sqrt{s}$ leads to

$\sqrt{(s-x)(s-y)(s-z)}=r\sqrt{s}$ Eq. 3

and upon substituting Eq. 3 into Eq. 2:

$xyz=4Rrs$ Eq. 4

Now square Eq. 3 and expand the left side. The coefficient of $s^2$ is $-(x+y+z)$ which equals $-2s$ from the definition of the semiperimeter. The constant term $-xyz$ is known from Eq. 4. We then get

$s^3-(2s)s^2+(xy+xz+yz)s-xyz=r^2s$

$xy+xz+yz=r^2+s^2+4Rr$ Eq. 5

Using Vieta's formulas and Eqs. 4 and 5 we derive the cubic equation whose roots are the sides:

$P(w)=w^3-d_2w^2+d_1w-d_0=0$ Eq. 6

$d_2=x+y+z=2s$

$d_1=xy+xz+yz=r^2+s^2+4Rr$

$d_0=xyz=4Rrs$

The reader can verify that if we insert $r=1,R=2.5, s=6$ corresponding to a 3-4-5 right triangle, the cubic equation will in fact have roots $\{x,y,z\}=\{3,4,5\}$.

Condition for Existence

Descartes' Rule of Signs assures that all real roots of Eq. 6 are positive, so the triangle exists if all three roots are in fact real (including multiple counting of double or triple roots, which corresponds to isosceles or equilateral triangles). These existence conditions are written here as a requirement that the cubic polynomial $P(w)$ have a critical points in the real domain and that the polynomial not be one-signed between these points:

$d_2^2\ge 3d_1$ for critical points to exist

$P((d_2+\sqrt{d_2^2-3d_1})/3)P((d_2-\sqrt{d_2^2-3d_1})/3)\le 0$ to prevent critical values from having the same sign.

If these conditions are met, then the triangle exists and, as noted above, the formulas for roots of cubic equations with three real roots guarantees that it may be constructed using an unmarked straightedge, compasses and an angle trisector. But we will in general need the trisector.

And Here is That Proof

It is well known that if a cubic equation with integer coefficients is to be solved with unmarked straightedge and compasses alone, it must be reducible having a rational root. If $r, R, s$ are whole numbers then Eq. 6 will have integer coefficients. Yet for the following case the cubic equation meets the conditions for existence of all three roots, the coefficients are integers and yet by trial there are no rational roots that would enable unaided straightedge/compasses construction:

$r=1, R=6, s=12$

$w^2-24w^2+169w-288=0$

Rational roots must be positive whole numbers dividing $288$. Actual roots lie strictly between $2$ and $3$, between $10$ and $11$ and between $11$ and $12$.