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

Revision as of 10:16, 5 May 2008

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

Let the sides of triangle $ABC$ be $a$ , $b$ , and $c$ . Thus $\dfrac{abc}{4K}=R$ , and $a+b+c=P$ . We plug these in:

$\dfrac{K(a+b+c)}{\dfrac{a^3b^3c^3}{64K^3}}=\dfrac{64K^4(a+b+c)}{a^3b^3c^3}$ .

Now Heron's formula states that $K=\sqrt{(\dfrac{a+b+c}{2})(\dfrac{-a+b+c}{2})(\dfrac{a-b+c}{2})(\dfrac{a+b-c}{2})}$ . Thus,

$\dfrac{KP}{R^3}=\dfrac{(a+b+c)^3(-a+b+c)^2(a-b+c)^2(a+b-c)^2}{4a^3b^3c^3}$

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 ==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>.
 ==Solution==
+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:
-It would be convenient if this were an equilateral triangle, so we try an equilateral triangle first:
+<math>\dfrac{K(a+b+c)}{\dfrac{a^3b^3c^3}{64K^3}}=\dfrac{64K^4(a+b+c)}{a^3b^3c^3}</math>.
-<math>\dfrac{KP}{R^3}=\dfrac{27}{4}</math>
+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,
-Now we just need to prove that that is the maximum.
+<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>
 {{incomplete|solution}}
 ==See also==
-*[[2005 Canadian MO]]
 {{CanadaMO box|year=2005|num-b=3|num-a=5}}
 [[Category:Olympiad Geometry Problems]]

2005 Canadian MO (Problems)
Preceded by Problem 3	1 • 2 • 3 • 4 • 5	Followed by Problem 5