Art of Problem Solving

2005 Canadian MO Problems/Problem 4: Difference between revisions

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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==
==Solution Outline==
 
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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==See also==
 
*[[2005 Canadian MO]]
 
 
[[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 $\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}$.