Prove $r^2+r_1^2+r_2^2+r_3^2=16R^2-a^2-b^2-c^2$ for a triangle with sides $a$, $b$, $c$, circumradius $R$, inradius $r$, exradii $r_1$, $r_2$, $r_3$

Question

From SL Loney Trigonometry:

Prove that $$r^2+r_1^2+r_2^2+r_3^2=16R^2-a^2-b^2-c^2$$ for a triangle with sides $a$, $b$, $c$, circumradius $R$, inradius $r$, and exradii $r_1$, $r_2$, $r_3$.

This one is a really dirty question.

I have tried a lot of stuff like $r=4R\times \Pi(\sin\frac{A}{2}$) and $r_1=4R\times\sin\frac{A}{2}\cos\frac{B}2\cos\frac{C}2$ etc..

There are other ways like $-r+r_1+r_2+r_3=4R$, but all these are way too cumbersome or clumsy.

Anything more intuitive?

Answer 1

we Need the following formulas: $$r_1=\frac{\Delta}{s-a}$$ etc $$\Delta=rs$$, $$\Delta=\frac{abc}{4R}$$, $$s=\frac{a+b+c}{2}$$ so the right-hand side of your equation is given by $$\frac{8abc)^2}{s(s-a)(s-b)(s-c)}-a^2-b^2-c^2=-{\frac {{a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}+10\,{a}^ {2}{b}^{2}{c}^{2}-{a}^{2}{c}^{4}+{b}^{6}-{b}^{4}{c}^{2}-{b}^{2}{c}^{4} +{c}^{6}}{ \left( a+b+c \right) \left( a-b-c \right) \left( a-b+c \right) \left( a+b-c \right) }}={\frac {{x}^{4}{y}^{2}+{x}^{4}{z}^{2}+2\,{x}^{3}{y}^{3}+2\,{x}^{3}{y}^ {2}z+2\,{x}^{3}y{z}^{2}+2\,{x}^{3}{z}^{3}+{x}^{2}{y}^{4}+2\,{x}^{2}{y} ^{3}z+4\,{x}^{2}{y}^{2}{z}^{2}+2\,{x}^{2}y{z}^{3}+{x}^{2}{z}^{4}+2\,x{ y}^{3}{z}^{2}+2\,x{y}^{2}{z}^{3}+{y}^{4}{z}^{2}+2\,{y}^{3}{z}^{3}+{y}^ {2}{z}^{4}}{ \left( y+z+x \right) xyz}} $$ with $$a=y+z,b=x+z,c=x+y$$ the left-hand side is given by $$s(s-a)(s-b)(s-c)\left(\frac{1}{s^2}+\frac{1}{(s-a)^2}+\frac{1}{(s-b)^2}+\frac{1}{(s-c)^2}\right)$$ the result is the same as posted above

Answer 2

We need to prove that $$4S^2\left(\frac{1}{(a+b+c)^2}+\sum_{cyc}\frac{1}{(a+b-c)^2}\right)^2+a^2+b^2+c^2=\frac{a^2b^2c^2}{S^2}$$ or $$\frac{1}{4}\sum_{cyc}(2a^2b^2-a^4)\left(\tfrac{1}{(a+b+c)^2}+\tfrac{\sum\limits_{cyc}(a+b-c)^2(a+c-b)^2}{\prod\limits_{cyc}(a+b-c)^2}\right)=\tfrac{16a^2b^2c^2}{\sum\limits_{cyc}(2a^2b^2-a^4)}-\sum_{cyc}a^2$$ or $$\frac{1}{4}\left(\sum_{cyc}(2a^2b^2-a^4)\right)^2\left(\tfrac{1}{(a+b+c)^2}+\tfrac{\sum\limits_{cyc}(a+b-c)^2(a+c-b)^2}{\prod\limits_{cyc}(a+b-c)^2}\right)=16a^2b^2c^2-\sum_{cyc}a^2\sum_{cyc}(2a^2b^2-a^4)$$ or $$\prod_{cyc}(a+b-c)^2+(a+b+c)^2\sum_{cyc}(a^2-b^2-c^2+2bc)^2=$$ $$=4\sum_{cyc}\left(a^6-a^4b^2-a^4c^2+\frac{10}{3}a^2b^2c^2\right).$$

But $$\prod_{cyc}(a+b-c)^2=\left(\sum_{cyc}\left(a^3-a^2b-a^2c+\frac{2}{3}abc\right)\right)^2=$$ $$=\sum_{cyc}\left(a^6-2a^5b-2a^5c-a^4b^2-a^4c^2+4a^3b^3+6a^4bc-4a^4b^2c-4a^3c^2b+\frac{10}{3}a^2b^2c^2\right)$$ and $$(a+b+c)^2\sum_{cyc}(a^2-b^2-c^2+2bc)^2=$$ $$=\sum_{cyc}(a^2+2ab)\sum_{cyc}(3a^4-4a^3b-4a^3c+2a^2b^2+4a^2bc)=$$ $$=\sum_{cyc}(3a^6+2a^5b+2a^5c-3a^4b^2-3a^4c^2-6a^4bc-4a^3b^3+4a^3b^2c+4a^3c^2b+10a^2b^2c^2).$$ Done!

Also, we have the following way.

Since $r_1+r_2+r_3-r=4R$, we need to prove that $$r_1^2+r_2^2+r_3^2+r^2+a^2+b^2+c^2=(r_1+r_2+r_3-r)^2,$$ which is a bit of easier because

$r_1r_2=p(p-c)$, $r_1r_3=p(p-b)$ and $r_2r_3=p(p-a)$, where $p=\frac{a+b+c}{2}$. Thus, it's enough to prove that $$\sum_{cyc}(2ab-a^2)=4r(4R+r),$$ which is for you.

Answer 3

$$r^2+r_1^2=2R^2\left\{(1-\cos A)(1-\cos B)(1-\cos C)+(1-\cos A)(1+\cos B)(1+\cos C)\right\}=2R^2(1-\cos A)(2+2\cos B\cos C)$$

$$r_2^2+r_3^2=2R^2\left\{(1+\cos A)(1-\cos B)(1+\cos C)+(1+\cos A)(1+\cos B)(1-\cos C)\right\}=2R^2(1+\cos A)(2-2\cos B\cos C)$$

$$\implies r^2+r_1^2+r_2^2+r_3^2=2R^2[4-4\cos A\cos B\cos C]$$

Now use $\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$ (Proof)

and $a=2R\sin A$