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Difference between revisions of "2005 Canadian MO Problems/Problem 4"

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Let <math>ABC</math> be a triangle with circumradius <math>R</math>, perimeter <math>P</math> and area <math>K</math>. Determine the maximum value of <math>KP/R^3</math>.
 
Let <math>ABC</math> be a triangle with circumradius <math>R</math>, perimeter <math>P</math> and area <math>K</math>. Determine the maximum value of <math>KP/R^3</math>.
  
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==Solution==
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==Solution Outline==
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Let the sides of triangle <math>ABC</math> be <math>a</math>, <math>b</math>, and <math>c</math>. Thus <math>\dfrac{abc}{4K}=R</math>, and <math>a+b+c=P</math>. We plug these in:
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Use the formula <math>K=\dfrac{abc}{4R}</math> to get <math>KP/R^3=\dfrac{abc(a+b+c)}{4R^4}</math>. Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish.  (Jensen's is used for <math>\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}</math>.
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<math>\dfrac{K(a+b+c)}{\dfrac{a^3b^3c^3}{64K^3}}=\dfrac{64K^4(a+b+c)}{a^3b^3c^3}</math>.
 
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Now Heron's formula states that <math>K=\sqrt{(\dfrac{a+b+c}{2})(\dfrac{-a+b+c}{2})(\dfrac{a-b+c}{2})(\dfrac{a+b-c}{2})}</math>. Thus,
 
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<cmath>\dfrac{KP}{R^3}=\dfrac{(a+b+c)^3(-a+b+c)^2(a-b+c)^2(a+b-c)^2}{4a^3b^3c^3}</cmath>
 
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{{incomplete|solution}}
 
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==See also==
 
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{{CanadaMO box|year=2005|num-b=3|num-a=5}}
 
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[[Category:Olympiad Geometry Problems]]
 

Latest revision as of 08:13, 1 November 2022

Problem

Let $ABC$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $KP/R^3$.

Solution Outline

Use the formula $K=\dfrac{abc}{4R}$ to get $KP/R^3=\dfrac{abc(a+b+c)}{4R^4}$. Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish. (Jensen's is used for $\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}$.

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