Q.Prove that $r_1r_2+r_2r_3+r_3r_1=s^2$.

$\dfrac1r-\dfrac1{r_1}=\dfrac{a}{\Delta}$ (and cyclically), so the product is $\dfrac{abc}{\Delta^3}$; then $abc=4R\Delta$ and $\Delta=rs$ give $\dfrac{4R}{r^2 s^2}$.

Recall the standard results $r=\dfrac{\Delta}{s}$, $r_1=\dfrac{\Delta}{s-a}$, $r_2=\dfrac{\Delta}{s-b}$, $r_3=\dfrac{\Delta}{s-c}$. Then

$\frac1r=\frac{s}{\Delta},\qquad \frac1{r_1}=\frac{s-a}{\Delta},$

so

$\frac1r-\frac1{r_1}=\frac{s-(s-a)}{\Delta}=\frac{a}{\Delta}.$

Similarly $\dfrac1r-\dfrac1{r_2}=\dfrac{b}{\Delta}$ and $\dfrac1r-\dfrac1{r_3}=\dfrac{c}{\Delta}.$ Multiplying,

$\left(\frac1r-\frac1{r_1}\right)\left(\frac1r-\frac1{r_2}\right)\left(\frac1r-\frac1{r_3}\right)=\frac{a}{\Delta}\cdot\frac{b}{\Delta}\cdot\frac{c}{\Delta}=\frac{abc}{\Delta^3}.$ …

Use the identity $\Delta^2 = r\,r_1r_2r_3$ to find the triangle's area, then $s=\dfrac{\Delta}{r}$ for the semi-perimeter, and finally $s-a=\dfrac{\Delta}{r_1}$ (and similarly for $b,c$) to recover each side.

Given: $r_1=2,\ r_2=3,\ r_3=6,\ r=1$ (in-radius and ex-radii of $\triangle ABC$). Show $a=3,\ b=4,\ c=5$.

Step 1. Use the standard identity $\Delta^2 = r\cdot r_1\cdot r_2\cdot r_3$ (where $\Delta$ is the triangle's area):

$\Delta^2 = 1\times2\times3\times6 = 36 \Rightarrow \Delta = 6$

Step 2. Use $r=\dfrac{\Delta}{s}$ to find the semi-perimeter $s$:

$s = \dfrac{\Delta}{r} = \dfrac{6}{1} = 6$

Step 3. Use $r_1=\dfrac{\Delta}{s-a}$, so $s-a=\dfrac{\Delta}{r_1}$:

$s-a = \dfrac{6}{2}=3 \Rightarrow a = s-3 = 6-3 = 3$

…

Heron gives $\Delta=84$ with $s=21$; then each radius follows from its standard formula.

Semi-perimeter: $s=\dfrac{13+14+15}{2}=21$.

Area (Heron): $\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}=84$.

Circumradius: $R=\dfrac{abc}{4\Delta}=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{2730}{336}=\dfrac{65}{8}$.

Inradius: $r=\dfrac{\Delta}{s}=\dfrac{84}{21}=4$.

Exradii: …

Use the standard excircle identities $r_1r_2+r_2r_3+r_3r_1=s^2$ and $rr_1r_2r_3=\Delta^2$ to find $s$ and $\Delta$, then recover $a,b,c$ from $s-a=\Delta/r_1$, $s-b=\Delta/r_2$, $s-c=\Delta/r_3$.

Given $r_1=2,\ r_2=3,\ r_3=6,\ r=1$.

Step 1 — find $s$ using the identity $r_1r_2+r_2r_3+r_3r_1=s^2$:

$s^2 = (2)(3)+(3)(6)+(6)(2) = 6+18+12 = 36 \ \Rightarrow\ s=6$

Step 2 — find $\Delta$ (area) using the identity $r\,r_1\,r_2\,r_3=\Delta^2$ (this follows from $r=\Delta/s$, $r_1=\Delta/(s-a)$, etc. and $\Delta^2=s(s-a)(s-b)(s-c)$):

$\Delta^2 = r\,r_1\,r_2\,r_3 = (1)(2)(3)(6) = 36 \ \Rightarrow\ \Delta=6$

Check: $r=\Delta/s = 6/6=1$ ✓, consistent with the given $r=1$.

Step 3 — recover the sides. Since $r_1=\dfrac{\Delta}{s-a}$, $r_2=\dfrac{\Delta}{s-b}$, $r_3=\dfrac{\Delta}{s-c}$: …

Use the standard identities $r_1-r=4R\sin^2(A/2)$; substituting the given ratio forces $\sin^2(A/2)=1/2$, i.e. $A=90^\circ$.

Given $r:R:r_1=2:5:12$, write $r=2k,\ R=5k,\ r_1=12k$ for some $k>0$.

Step 1 — recall the standard results:

$r=4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}, \qquad r_1=4R\sin\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$

Step 2 — subtract:

$r_1-r=4R\sin\dfrac{A}{2}\left[\cos\dfrac{B}{2}\cos\dfrac{C}{2}-\sin\dfrac{B}{2}\sin\dfrac{C}{2}\right]=4R\sin\dfrac{A}{2}\cos\dfrac{B+C}{2}$

Step 3. Since $B+C=\pi-A$, $\dfrac{B+C}{2}=\dfrac{\pi}{2}-\dfrac{A}{2}$, so $\cos\dfrac{B+C}{2}=\sin\dfrac{A}{2}$. Hence:

$r_1-r=4R\sin^2\dfrac{A}{2}$

Step 4 — substitute the given ratio. $r_1-r=12k-2k=10k$ and $R=5k$: …