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

(WIP Solution)
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== Problem ==
 
== Problem ==
  
''(Romania)'' 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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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>.
  
 
== Solution ==
 
== Solution ==
  
== See also ==
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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}ab \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}ab \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>.
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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}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 14:58, 4 August 2018

Problem

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

Solution

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 $A = \tfrac{1}{2}ab \sin(\theta)$, where $\theta$ is the angle between sides $b$ and $c.$ Substituting this yields \begin{align*} \tfrac{1}{2}ab \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, $0 < \sin{\theta} \le 1$.


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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