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Is there any way to calculate $3$ sides of any triangles $(a,b,c)$, If we know Area $S$, Perimeter $P$, Circumradius $R$ and inradius $r$.

I took a deep look in to Wikipedia page, but no clue: https://en.wikipedia.org/wiki/Triangle

Thanks for reading my question.

Vinyl_cape_jawa
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Incpt.Mobis
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2 Answers2

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The three values, semiperimeter $\rho=\tfrac12(a+b+c)$, inradius $r$ and circumradius $R$ define a cubic equation,

\begin{align} x^3+a_2x^2+a_1x+a_0=0 , \end{align}

which roots are the values of the three sides $a,b,c$ of the triangle:

the coefficients are \begin{align} a_2&=-(a+b+c)=-2\rho ,\\ a_1&=ab+bc+ca=\rho^2+4rR+r^2 ,\\ a_0&=-abc=-4r\rho R . \end{align}

Edit:

Some light on how the expression for $a_1=ab+bc+ca$ in terms of $\rho, r$ and $R$ can be derived.

From Heron's formula for the area $S$ of triangle we have: \begin{align} &\rho^4-(a+b+c)\rho^3+(ab+bc+ca)\rho^2-abc \rho -S^2=0 ,\\ &\rho^4-2\rho\rho^3+(ab+bc+ca)\rho^2-abc \rho -S^2=0 ,\\ &-\rho^4+(ab+bc+ca)\rho^2-abc \rho -S^2=0 ,\\ a_1&= ab+bc+ca = \rho^2+abc \rho^{-1} +S^2\rho^{-2} . \end{align}

Now, using well-known identities $S=\rho\,r$ and $R=\frac{abc}{4S}=\frac{abc}{4\rho\,r}$, we obtain the final result:

\begin{align} a_1= ab+bc+ca &= \rho^2+abc \rho^{-1} +S^2\rho^{-2} \\ &= \rho^2+4\,\rho\,r\,R \rho^{-1} +(\rho\,r)^2\rho^{-2} \\ &= \rho^2+4r\,R +r^2 . \end{align}

In fact, more cubics can be constructed to solve the triangle, based on this known triple $\rho,r,R$ of linear measures. The cubic for $\cot\tfrac\alpha2,\cot\tfrac\beta2,\cot\tfrac\gamma2$ is the most beautiful and symmetric:

\begin{align} a_0&=a_2= -\frac{\rho}{r} ,\\ a_1&= \frac{4R}{r}+1 ,\\ y^3&-\frac{\rho}{r} y^2 + \left(\frac{4R}{r}+1\right)y -\frac{\rho}{r} =0. \end{align}

g.kov
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    @Raffaele: Thanks. Of course, $4RS=4R,r\rho$. It just somehow looks nicer to use the three input measures of the same dimension (linear). Also, $\rho,r,R$ can be equivalently replaced with $\rho,S,d=a,b,c$ with dimensions of length, area and volume. – g.kov Nov 29 '17 at 11:42
  • @g.kov Thank you, Nice trick! BTW, If you have time, would you mind to tell me how do you know the equation x^3 + a2x^2+a1x + a0 = 0 ? And the formula of a1, a2, a3 ? I'm really curious about that. – Incpt.Mobis Dec 01 '17 at 02:10
  • @IncptMobis: If we have two unknowns, say, $a,b$ and their sum $p$ and their product $q$ is known, then the roots of quadratic equations $x^2-px+q=0$ are the values of $a,b$. Similarly, for 3 unknowns $a,b,c$, if we know their sum $u$, their product $w$ and the sum of their pairwise products, $v=ab+bc+ca$, then roots of the cubic equation $x^3-ux^2+vx-w$ provide the answer. Hint: expand $(x-a)(x-b)(x-c)$. – g.kov Dec 01 '17 at 03:13
  • @IncptMobis: See added explaination for $a_1$. – g.kov Dec 01 '17 at 12:07
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Inradius $r=\dfrac{S}{P/2}$ so this information is useless

$R=\dfrac{a b c }{4S}$

Let $p=P/2$ by Heron formula

$p(p-a)(p-b)(p-c)=S^2$

which leads to

$\begin{cases} a+b+c=P \\ abc=4RS \\(b+c-a) (a + b - c) (a - b + c) (a + b + c)=16S^2\\\end{cases}$

Then solve the $12$th degree system and you are done!

Hope this helps

Raffaele
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