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

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==Problem==
 
==Problem==
 
 
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>.
  
==Solution==
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==Solution Outline==
 
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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>.
It would be convenient if this were an equilateral triangle, so we try an equilateral triangle first:
 
 
 
<math>\dfrac{KP}{R^3}=\dfrac{27}{4}</math>
 
 
 
Now we just need to prove that that is the maximum.
 
 
 
{{incomplete|solution}}
 
 
 
==See also==
 
*[[2005 Canadian MO]]
 
 
 
{{CanadaMO box|year=2005|num-b=3|num-a=5}}
 
 
 
[[Category:Olympiad Geometry Problems]]
 

Latest revision as of 07:13, 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 $\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}$.