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Difference between revisions of "1997 JBMO Problems/Problem 4"

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== Problem ==
 
== Problem ==
  
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== Solution ==
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Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>.
  
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== See also ==
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== Solutions ==
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===Solution 1===
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Solving for <math>R</math> yields <math>R = \tfrac{a\sqrt{bc}}{b+c}</math>.  We can substitute <math>R</math> into the area formula <math>A = \tfrac{abc}{4R}</math> to get
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<cmath>\begin{align*}
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A &= \frac{abc}{4 \cdot \tfrac{a\sqrt{bc}}{b+c} } \\
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&= \frac{abc}{4a\sqrt{bc}} \cdot (b+c) \\
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&= \frac{(b+c)\sqrt{bc}}{4}.
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\end{align*}</cmath>
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We also know that <math>A = \tfrac{1}{2}bc \sin(\theta)</math>, where <math>\theta</math> is the angle between sides <math>b</math> and <math>c.</math>  Substituting this yields
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<cmath>\begin{align*}
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\tfrac{1}{2}bc \sin(\theta) &= \frac{(b+c)\sqrt{bc}}{4} \\
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2\sqrt{bc} \cdot \sin(\theta) &= b+c \\
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\sin(\theta) &= \frac{b+c}{2\sqrt{bc}}
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\end{align*}</cmath>
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Since <math>\theta</math> is inside a triangle, <math>0 < \sin{\theta} \le 1</math>.  Substitution yields
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<cmath>0 < \frac{b+c}{2\sqrt{bc}} \le 1.</cmath>
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Note that <math>2\sqrt{bc}</math>, so multiplying both sides by that value would not change the inequality sign.  This means
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<cmath>0 < b+c \le 2\sqrt{bc}.</cmath>
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However, by the [[AM-GM Inequality]], <math>b+c \ge 2\sqrt{bc}</math>.  Thus, the equality case must hold, so <math>b = c</math> where <math>b, c > 0</math>.  When plugging <math>b = c</math>, the inequality holds, so the value <math>b=c</math> truly satisfies all conditions.
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<br>
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That means <math>\sin(\theta) = \frac{2b}{2\sqrt{b^2}} = 1,</math> so <math>\theta = 90^\circ.</math>  That means the only truangle that satisfies all the conditions is a 45-45-90 triangle where <math>a</math> is the longest side.  In other words, <math>(a,b,c) \rightarrow \boxed{(n\sqrt{2},n,n)}</math> for all positive <math>n.</math>
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== See Also ==
  
 
{{JBMO box|year=1997|num-b=3|num-a=5}}
 
{{JBMO box|year=1997|num-b=3|num-a=5}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 14:48, 23 February 2021

Problem

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

Solutions

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