Prove that triangle is right if $s=2R+r$, where $R$, $r$, $s$ are its circumradius, inradius, and semiperimeter

Question

Let $R$ and $r$ be circumradius and inradius respectively of a triangle with semiperimeter $s$. Prove that it is right if $$s=2R+r$$

It is not hard to find $$s = {a\over \sin \alpha} + (s-a)\tan{\alpha \over 2}$$ Labeling $x=\tan{\alpha\over 2}$ we have $$(2s-a)x^2-2sx +a=0 $$ which has solution $x=1$ and $x= \displaystyle{a\over b+c}$. If $x=1$ we are done else $${a\over b+c} = \tan{\alpha\over 2} = {r\over s-a}\hspace{2cm} (1)$$ We can do the same for other two angles and if some is right then we are done, else we have also $${b\over c+a} = \tan{\beta\over 2} = {r\over s-b}\hspace{2cm} (2)$$ and $${c\over a+b} = \tan{\gamma\over 2} = {r\over s-c}\hspace{2cm} (3)$$

Multiplying all equations (1),(2) and (3) we have $$s(s-a)(s-b)(s-c)abc = r^3s(a+b)(b+c)(c+a)$$ $$S^2abc = r^3s(a+b)(b+c)(c+a)$$ $$sabc = r(a+b)(b+c)(c+a)$$ $$sbc = (a+b)(s-a)(c+a)$$ $$(a+b+c)bc = (a(a+b+c)+bc)(b+c-a)$$ $$ abc+(b+c)bc = a(b^2+c^2+2bc-a^2)+ bc(b+c)-abc$$ $$ 0 = a(b^2+c^2-a^2)$$ and we are done.

Is there more nice synthetic solution without trigonometry?

Answer 1

Let $A$ be the area of the triangle. Note that $$ \begin{aligned} \cos {\alpha} = {} & \frac{b^2 + c^2 - a^2}{2bc} \\ = {} & \frac{(b + c)^2 - a^2}{2bc} - 1 \\ = {} & \frac{(b + c - a)(b + c + a)}{2bc} - 1 \\ = {} & \frac{(2s - 2a)(2s)}{2bc} - 1 \\ = {} & \frac{2s(s - a)}{bc} - 1 \\ = {} & \frac{2sr(s - a)}{rbc} - 1 \\ = {} & \frac{2A(s - a)}{rbc} - 1 \\ = {} & \frac{2Aa(s - a)}{rabc} - 1 \\ = {} & \frac{2Aa(s - a)}{4ARr} - 1 \\ = {} & \frac{a(s - a)}{2Rr} - 1 \\ = {} & \frac{2R (s - 2R \sin {\alpha}) \sin {\alpha} }{2Rr} - 1 \\ = {} & \frac{(s - 2R \sin {\alpha}) \sin {\alpha} }{r} - 1. \end{aligned} $$ Hence $$ r(1 + \cos {\alpha}) = (s - 2R \sin {\alpha}) \sin {\alpha}, $$ which is $$ r \cdot 2\cos {\alpha \over 2} \cos {\alpha \over 2} = (s - 2R \sin {\alpha}) \cdot 2\cos {\alpha \over 2} \sin {\alpha \over 2}, $$ which is $$ r \cdot 2\cos {\alpha \over 2} \sin {\alpha \over 2} = (s - 2R \sin {\alpha}) \cdot 2\sin {\alpha \over 2} \sin {\alpha \over 2}, $$ which is $$ r \sin {\alpha} = (s - 2R \sin {\alpha}) (1 - \cos {\alpha}), $$ which is $$ \sin {\alpha} = \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}}. $$ Hence $$ r(1 + \cos {\alpha}) = \left( s - 2R \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}} \right) \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}}, $$ which is $$ 4R^2 \cos^3 {\alpha} - 4R (R + r) \cos^2 {\alpha} + (s^2 + r^2 - 4R^2) \cos {\alpha} - (s^2 - (2R + r)^2) = 0. $$

---

Hence $\cos {\alpha}$, $\cos {\beta}$, $\cos {\gamma}$ are the roots of the eqaution $$ 4R^2 x^3 - 4R (R + r) x^2 + (s^2 + r^2 - 4R^2)x - (s^2 - (2R + r)^2) = 0. $$ Hence $$ \cos {\alpha} \cos {\beta} \cos {\gamma} = \frac{s^2 - (2R + r)^2}{4R^2} = 0. $$ Hence one of $\cos {\alpha}$, $\cos {\beta}$, $\cos {\gamma}$ must be zero.

Answer 2

There is an identity concerning the product of cosines of angles $A, B, C$ with $s,r, R$:

In a triangle $ABC,$ we have $$\cos A \cos B \cos C= \frac{s^2-(2R+r)^2}{4R^2}.$$

It is being shown here:

Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$

and additional reference:

How to prove that $\frac{r}{R}+1=\cos A+\cos B+\cos C$?

Hence, in any triangle $ABC$, $$\triangle ABC \textrm{ is }\begin{cases}\textrm{obtuse $\quad \quad $ if $s<2R+r$} \\ \textrm{right $\quad \quad \ \ \ $ if $s=2R+r$} \\ \textrm{acute $\quad \quad \ \ $ if $s>2R+r.$} \end{cases}$$

Answer 3

We have $$a+b+c=2s\tag1$$ $$abc=4rRs\tag2$$

Let us represent $ab+bc+ca$ by $r,R$ and $s$.

Representing the area of the triangle in two ways, we have $$\sqrt{s(s-a)(s-b)(s-c)}=rs$$ (LHS is Heron's formula)

Squaring the both sides and dividing by $s$, we get $$(s - a)(s - b)(s - c) = r^2s$$ i.e. $$s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc = r^2s$$

Substituting $(1)(2)$, we obtain $$ ab + bc + ca = s^2 + r^2 + 4rR\tag3$$

So, if $s=2R+r$, we have $$a+b+c=4R+2r\tag4$$ $$ab+bc+ca=4R^2+2r^2+8rR\tag5$$ $$abc=4rR(2R+r)\tag6$$

From $(4)(5)(6)$, we see that $a,b,c$ are the roots of $$x^3-(4R+2r)x^2+(4R^2+2r^2+8Rr)x-4rR(2R+r)=0$$ which can be written as $$(x-2R)\bigg(x^2-(2r+2R)x+4rR+2r^2\bigg)=0$$

Since one of $a,b,c$ is equal to $2R$, the triangle is right.$\ \ \blacksquare$

---

Added :

I'm going to add another proof which might be "less algebraic".

It is sufficient to prove that if $\triangle{ABC}$ is not right, then $s\not=2R+r$.

Proof :

We may suppose that $\angle{ACB}\lt 90^\circ$.

For fixed $R,AB$ and $\angle{ACB}$, let us consider $s-2R-r$.

$s-2R-r$ is of the form $f(x)=px+q$ where $p>0,q$ are constants and $x=CA+CB$ because we have $$\begin{align}s-2R-r&=\frac{AB+x}{2}-2R-\bigg(\frac{AB+x}{2}-AB\bigg)\tan\frac C2 \\\\&=\underbrace{\frac{1-\tan\frac C2}{2}}_{\text{positive constant}}x+\underbrace{\frac{AB}{2}\bigg(1+\tan\frac C2\bigg)-2R}_{\text{constant}}:=f(x)\end{align}$$

Let $P,Q$ be points on the circumcircle such that $\triangle{PAB},\triangle{QAB}$ are right where $f(PA+PB)=f(QA+QB)=0$.

Now, let us consider an ellipse $E$ whose foci are $A,B$ passing through $P,Q$.

We can see that if $\triangle{ABC}$ is not right, then $x\not=PA+PB$.

Since $f(x)$ is of the form $px+q$, we can say that if $x\not=PA+PB$, then $f(x)\not=f(PA+PB)=0$, i.e. $s-2R-r\not=0$.$\ \ \blacksquare$

(In the proof, I used $r=(s-c)\tan\frac C2$. If I find a proof without using trigonometry, I'll post it.)

Answer 4

Reflect $A$ across circumcenter $S$ into $A'$. And let $D$ be a midpoint of arc $AB$ which does not contain $C$. Then $A'D$ and $CD$ are angle bisectors of $\angle AA'B$ and $\angle ACB$.

Lemma: Circle with center at $D$ and radius $DA=DB$ cut angle bisector at incenter of a triangle. (This should be well know fact and easy to prove.)

Similarity of colored triangles give us $${r'\over r} = {s'-c\over s-c}$$ and since triangle $AA'B$ is right we have $s' = 2R+r'$. Eliminating $r$ we get $$ (2R-c)(s-s')=0$$ so if $c=2R$ we are done else $s=s'$. But then $r'=r$ and so $II'$ is parallel to $AB$ and thus cyclic quadrilater $ABI'I$ must be isosceles. But then $ABC$ and $ABA'$ are congruent and we are done.