Valentin Vornicu wrote: Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$. Romania $\frac{b+c}{\sqrt{bc}}\geq 2$ while $a\leq 2R$ (chord-diameter) so $b=c$ and $A=90^0$

$a\geq 2R$ (chord-diameter) so $b=c$ and $A=90^0$

socrates wrote $a\leq2R$. M4RIO wrote $a\geq2R$. Who's wrong???

José wrote: socrates wrote $a\leq2R$. M4RIO wrote $a\geq2R$. Who's wrong??? Try this by yourself. Do some sketch.... I think i am not the liar..

José wrote: socrates wrote $a\leq2R$. M4RIO wrote $a\geq2R$. Who's wrong??? obviously $a \le 2R$ is correct because if you have a circle, the largest chord is the diameter

This thing follows directly from AM-GM: $R(b+c)>=2R\sqrt{bc}>=a\sqrt{bc}$. This means that $a=2R$ which means that $a$ is the diameter. Furthermore, the triangle is right. Also, equality holds in AM-GM when all the terms are equal so $b=c$. Therefore the triangle is isosceles right triangle which is of the for $(a,a,a\sqrt{2})$. Nice problem.

using sine rule yields , $sinB+sinC=2sinA\sqrt{sinBsinC}\le 2\sqrt{sinBsinC}$ on the other hand , $sinB+sinC\ge 2\sqrt{sinBsinC}$ so , $sinB+sinC=2\sqrt{sinBsinC} \implies B=C $ and $A=\pi/2$

We know $R^2=a^2bc/(b+c)^2$ We have $R^2=a^2/4sin^2A$ thus $sin^2A=(b+c)^2/4bc$≥$1$ also we have $1$≥$sin^2A$ thus $A=90$ and $b=c$

hello, with $R=\frac{abc}{4A}$ we get $\frac{(b+c)bc}{4\sqrt{bc}}=A$ $\frac{b+c}{2\sqrt{bc}}=\sin(\alpha)\le 1$ thus we obtain $b+c\le 2\sqrt{bc}$ and we get $b=c$ and $\alpha=90^{\circ}$ Sonnhard.

(a*√(b*c))/(b+c)=(a*b*c)/4S=(a*b*c)/(2*b*c*sinA)→sin A=(b+c)/(2*√(2&b*c))>=1 and sin A<=1→sin A=1→m(<A)=90 and b=c

Valentin Vornicu wrote: Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$. Romania $a+b \geq 2 \sqrt{bc}$ we get $a\sqrt{bc} \geq 2R\sqrt{bc}$ => $a \geq 2R$ but $a \leq 2R$ and we get $a = 2R$ and $b = c$.

we know that by law of sines R = a / 2sinA, so a * sqr(bc)/(b+c) = a/2sinA, dividing by a we have sinA = (b+c)/[2sqr(bc)] = 1/2(sqr(b/c)+sqr(c/b)) and MA => MG inequality we have 1/2(sqr(b/c)+sqr(c/b)) => 1 but sinA is less or equal to 1 so sinA=1 and A=90º. now we know R = a/2 and now substituting in the equation we have b=c so B = C = 45º FINAL ANSWER : A = 90º, B = C = 45º ( this problem is also 2013 Brazil Cono Sur Training List 4.3 )