1997 JBMO Problems/Problem 4


Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$.


Solution 1

Solving for $R$ yields $R = \tfrac{a\sqrt{bc}}{b+c}$. We can substitute $R$ into the area formula $A = \tfrac{abc}{4R}$ to get \begin{align*} A &= \frac{abc}{4 \cdot \tfrac{a\sqrt{bc}}{b+c} } \\ &= \frac{abc}{4a\sqrt{bc}} \cdot (b+c) \\ &= \frac{(b+c)\sqrt{bc}}{4}. \end{align*} We also know that $A = \tfrac{1}{2}bc \sin(\theta)$, where $\theta$ is the angle between sides $b$ and $c.$ Substituting this yields \begin{align*} \tfrac{1}{2}bc \sin(\theta) &= \frac{(b+c)\sqrt{bc}}{4} \\ 2\sqrt{bc} \cdot \sin(\theta) &= b+c \\ \sin(\theta) &= \frac{b+c}{2\sqrt{bc}} \end{align*} Since $\theta$ is inside a triangle, $0 < \sin{\theta} \le 1$. Substitution yields \[0 < \frac{b+c}{2\sqrt{bc}} \le 1.\] Note that $2\sqrt{bc}$, so multiplying both sides by that value would not change the inequality sign. This means \[0 < b+c \le 2\sqrt{bc}.\] However, by the AM-GM Inequality, $b+c \ge 2\sqrt{bc}$. Thus, the equality case must hold, so $b = c$ where $b, c > 0$. When plugging $b = c$, the inequality holds, so the value $b=c$ truly satisfies all conditions.

That means $\sin(\theta) = \frac{2b}{2\sqrt{b^2}} = 1,$ so $\theta = 90^\circ.$ That means the only truangle that satisfies all the conditions is a 45-45-90 triangle where $a$ is the longest side. In other words, $(a,b,c) \rightarrow \boxed{(n\sqrt{2},n,n)}$ for all positive $n.$

See Also

1997 JBMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All JBMO Problems and Solutions
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