American mathematical monthly No.2(1999) Problem 10713 . Given a triangle with angles $A\ge B\ge C$ ,let $a,b$ and $c$ be the lengths of the corresponding opposite sides , let $r$ be the radius of the inscribed circle ,and let $R$ be the radius of the circumscribed circle . Show that $A$ is acute if and only if $R+r<\frac{b+c}{2}$.

My solution: According to Carnot's theorem (https://en.m.wikipedia.org/wiki/Carnot%27s_theorem, which can be proved by Ptolemy's theorem), we have: $R+r=R(cosA+cosB+cosC)$ Let $S=\frac{a+b-2R-2r}{2R}$, it suffices to find the sign og $S$ by the measure of angle $C$. By sine law, we have $a=2R.sinA, b=2R.sinB$, combine with the consequence of Carnot's theorem as above, we conclude that: $S=sinA+sinB-(cosA+cosB+cosC)$ Note that $cosC=-cos(A+B)=sinAsinB-cosAcosB$ hence: $S=(1-cosA)(1-cosB)-(1-sinA)(1-sinB)$ Let $g(x)=\frac{1-sinx}{1-cosx}$ and $y=90-A$ then: $S=(g(y)-g(B))(1-cosB)(1-cosy)$ Since $1-cosB>0, 1-cosC>0$ (since $b$ is not the largest length and $A>0$) then $S$ has the same sign as $g(y)-g(B)$. Note that in the interval $(0,90)$, $sinx$ is an increase function, $cosx$ is a decrease function hence $g(x)$ is a decrease function.So: $f>0\Leftrightarrow 90-A<B\Leftrightarrow C<90$. $f=0\Leftrightarrow 90-A=B\Leftrightarrow C=90$. $f<0\Leftrightarrow 90-A>B\Leftrightarrow C>90$. Q.E.D

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