Maximum Area

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Maximum Area

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area of a triangle

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Source: BrMO 2012 R1 P5

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2280 posts

#1 h Jun 11, 2014, 6:04 AM • 2 Y

Y by Adventure10, Mango247

A triangle has sides of length atmost $2,3$ & $4$ respectively.
Determine with proof the max. possible area of the triangle.

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247 posts

#2 h Jun 11, 2014, 7:54 AM • 2 Y

Y by Adventure10 and 1 other user

Let $s$ be the semi-perimeter and $p$ the perimeter of triangle $ABC$ . Hence, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}\le \sqrt{s[(\frac{s-a+s-b+s-c}{3})^3]}= \sqrt{s*\frac{s^3}{27}}=\frac{s^2}{3\sqrt{3}}=\frac{p^2}{12\sqrt{3}}$ .
(The inequality is by $A.M.-G.M.$ ). So, if maximum $p=2+3+4=9$ , maximum area $A=\frac{81}{12\sqrt{3}}=\frac{9\sqrt{3}}{4}$ .

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2280 posts

#3 h Jun 11, 2014, 9:09 AM • 1 Y

Y by Adventure10

For such a solution, to consolidate your answer do always provide the equality condition.
The question is not as easy as it looks.

According to your solution the maximum will occur when $s-a=s-b=s-c$ but this never happens as $a \neq b \neq c$
Please retry the question.

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#4 h Jun 11, 2014, 9:18 AM • 1 Y

Y by Adventure10

Using Mitrinovic's inequality, we have :
$s^{2}\geq 3S\sqrt{3}\Rightarrow S\leq \frac{s^{2}\sqrt{3}}{9}=\frac{9\sqrt{3}}{4}$

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#5 h Jun 11, 2014, 9:46 AM • 2 Y

Y by Adventure10, Mango247

utkarshgupta wrote:

For such a solution, consolidate your answer by providing the equality condition.

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#6 h Jun 11, 2014, 6:07 PM • 2 Y

Y by Adventure10, Mango247

I have got a logic ......... hope I'm correct. Let $\Delta(a,b,c)$ denote the area of a triangle with sides $(a,b,c)$ . To get equality, we can have maximum $p=6$ , where $(a,b,c)=(2,2,2)$ . Now, increase $b$ and $c$ by $1$ . So, $\Delta(2,3,3)=2\sqrt{2}>\sqrt{3}=\Delta(2,2,2)$ . Note that for $2<b=c\le 3$ and $a=2$ the altitude steadily increases and hence the area too. This can be generalized to all $2<b,c\le 3$ and $a=2$ considering the fact that area of all triangles with the same base and between same parallels have the same area. So, till now $\Delta(2,3,3)$ is maximum. Now, increase $b$ or $c$ by 1. Say, $c$ is increased to $4$ . Note that when $c$ is increased from $3$ to $4$ , the altitude of the triangle increases upto
$c=\sqrt{13}$ when triangle $ABC$ is right-angled with sides $(2,3,\sqrt{13})$ . But after this triangle starts becoming obtuse; hence the altitude decreases. So, maximum area $A=\Delta(2,3,\sqrt{13})=\frac{1}{2} *2 *3=3$ .

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#7 h Jun 11, 2014, 6:26 PM • 3 Y

Y by ARCH999, Adventure10, Mango247

Finally something correct, although somewhat less than rigorous. A proof: $[ABC]=\tfrac{ab\sin C}{2}\le \tfrac{2\cdot 3\cdot 1}{2}=3.$ This gives $c=\sqrt{13}<4$ , so the maximal area is $3$ , achieved at the triangle with sides of length $2,3,\sqrt{13}$ .

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#8 h Jun 11, 2014, 7:56 PM • 1 Y

Y by Adventure10

Oh.......yes.......it was so simple.........

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