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Problem of the Week #52 - March 25th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is an equilateral triangle.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
This week's problem was correctly answered by the following members:

Ackbach
anemone
BAdhi
Barioth
MarkFL
Sudharaka

You can find BAdhi's solution below:

If $a$,$b$,$b$ are the sides of the isosceles triangle, the height of the triangle $h$ would be,$$h=\sqrt{b^2-\left(\frac a 2\right)^2}$$
The perimeter $C$ and the area $A$ are given as,
$$C=a+2b\qquad (1)$$
$$A=\frac{a\times h}{2}=\frac{a\times \sqrt{b^2-\left(\frac a 2\right)^2}}{2}=\frac{a\sqrt{4b^2-a^2}}{4}$$
From (1),
$$\begin{align*}
A&=\frac{a\sqrt{4\left(\frac{C-a}{2}\right)^2-a^2}}{4}\\
&=\frac{a\sqrt{(C-a)^2-a^2}}{4}\\
&=\frac{a\sqrt{(C-a+a)(C-a-a)}}{4}\\
&=\frac{a\sqrt{C(C-2a)}}{4}\\
&=\frac{\sqrt{C(Ca^2-2a^3)}}{4}
\end{align*}$$
$$A^2=\frac{C(Ca^2-2a^3)}{16}$$
by differentiating with respect to a,
$$\begin{align*}
2A\frac{dA}{da}&=\frac{C(2aC-6a^2)}{16}\\
\frac{dA}{da}&=\frac{C(2aC-6a^2)}{32A}
\end{align*}$$
To find max/min $\frac{dA}{da}=0$
$$\frac{2aC-6a^2}{32A}=0$$
since $A>0$,
$$\begin{align*}
2aC-6a^2&=0\\
a(C-3a)&=0\\
a&=\frac{C}{3} \quad (\text{since $a\not=0$})
\end{align*}$$
by taking, second derivative,
$$2\left(\frac{dA}{da}\right)^2+2A\frac{d^2A}{da^2}=\frac{2C-12a}{16}$$
when $\frac{dA}{da}=0$, $a=\frac{C}{3}$
$$\begin{align*}
2(0)+2A\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}&=\frac{C-2C}{8}\\
\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}&= \frac{-C}{16A}\\
\end{align*}$$
with $\text{$C>0$ and $ A>0$}$
$$\left(\frac{d^2A}{da^2}\right)_{a=\frac{C}{3}}<0$$
Therefore the point $a=\frac{C}{3}$ is a maximum.
Therefore the area becomes maximum when $a=\frac{C}{3}$
from (1),
$$3a=a+2b\implies a=b$$
therefore the area of the isosceles triangle becomes maximum when when the triangle is an equilateral triangle with a given constant perimeter

I'm also going to post anemone's solution since it nicely utilizes Heron's formula.

By using the Heron's formula to find the area (A) of any isosceles triangle with sides a, a, and b with a given perimeter, says P, we have:

$\displaystyle A= \sqrt {\frac{p}{2}\cdot (\frac{p}{2}-a)^2\cdot(\frac{p}{2}-b)^2} $ (*)

Since $\displaystyle p=a+a+b$, we rewrite it to make b the subject and obtain $\displaystyle b=p-2a$. We then substitute it to the equation (*) and this gives:

$\displaystyle A= \sqrt {\frac{p}{2}\cdot (\frac{p}{2}-a)^2\cdot(\frac{p}{2}-(p-2a))^2} $

Further simplification yields

$\displaystyle A= \sqrt {\frac{p}{2}\cdot(2a^3-\frac{5pa^2}{2}+p^2a-\frac{p^3}{8} )} $

Now, if we let $\displaystyle y=2a^3-\frac{5pa^2}{2}+p^2a-\frac{p^3}{8} $, we see that if we maximize y, we will also maximize A at the same time.

Proceed to maximize y using differentiation method, we have:

$\displaystyle y'=6a^2-5pa+p^2=(3a-p)(2a-p) $

i.e. $\displaystyle y'=0$ iff $\displaystyle a=\frac{p}{2}$ or $\displaystyle a=\frac{p}{3}$.

Second derivative test $\left(\displaystyle y''=12a-5p=12(\frac{p}{3})-5p=-p \right)$ tells us y is maximum when $\displaystyle a=\frac{p}{3}$.

Therefore, A is also maximum when $\displaystyle a=\frac{p}{3}$, i.e. $\displaystyle p=3a$ or $\displaystyle b=a$ or if the triangle is an equilateral triangle.
 
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