1
$\begingroup$

There is an equilateral triangle with sides $a$ and another triangle with sides $p,q,r$, both having the same perimeter $S$.

How can we mathematically show which of them has a larger area?

$\endgroup$
2
$\begingroup$

Have you heard of Heron's formula? Especially on the form $$ A = \sqrt{\frac S2\left(\frac S2-a\right)\left(\frac S2-b\right)\left(\frac S2-c\right)},\quad a, b, c\text{ are the sides of the triangle} $$we come very close to a full solution. What we need to get all the way is the AM-GM inequality, which states that for any three positive numbers $k,l,m$, we have $$ \frac{k+l+m}{3}\geq \sqrt[3]{klm} $$


Since you asked, here is a full work-out:

For the equilateral triangle, we have $k = l = m = (S/2 - a)$, and the AM-GM inequality is actually an equality: $$ \frac{S/2 - a + S/2 - a + S/2 - a}{3} = \sqrt[3]{(S/2 - a)^3}\\ \frac{3S/2 - 3a}{3} = \sqrt[3]{(S/2 - a)^3}\\ \frac{3S/2 - S}{3} = \sqrt[3]{(S/2 - a)^3}\\ \frac S6 = \sqrt[3]{(S/2-a)^3}\\ \left(\frac S6\right)^3 = \left(\frac{S}2 - a\right)^3 $$ (Although, that could've been worked out without using AM-GM.)

Now, for a non-equilateral triangle, we set $k = (S/2 - p), l = (S/2 - q)$ and $m = (S/2 - r)$, and the AM-GM inequality is a strict inequality: $$ \frac{S/2 - p + S/2 - q + S/2 - r}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\ \frac{3S/2 - (p+q+r)}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\ \frac{3S/2 - S}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\ \frac S6 > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\ \left(\frac S6\right)^3 > \left(\frac S2 - p\right)\left(\frac S2 - q\right)\left(\frac S2 - r\right) $$ Now insert these two different products into Heron's formula, and you can see that the equilateral triangle gives the bigger area.

$\endgroup$
  • $\begingroup$ yeah i tried Heron's formula...How can I show which is larger? $(S-a)^3$ or $((S-p)(S-q)(S-r))$ ? $\endgroup$ – Hemand Nair Nov 6 '15 at 11:58
  • $\begingroup$ @HemandNair See my update. Try to guess which three numbers $k,l,m$ should be. $\endgroup$ – Arthur Nov 6 '15 at 12:02
  • $\begingroup$ can you show it for me? $\endgroup$ – Hemand Nair Nov 6 '15 at 12:05
  • $\begingroup$ I do know what AM_GM inequality is.Please help to find a solution to my question using it $\endgroup$ – Hemand Nair Nov 6 '15 at 12:14
0
$\begingroup$
We have M=√(A(A-a)(A-b)(A-c),A=(1/2)(a+b+c)=(1/2).circumference. Thus M=√(A(A-a)(A-b)(A-{2A-a-b})=√(A(A-a)(A-b)(-A+a+b)). Let N=((M²)/A)=(A-a)(A-b)(-A+a+b). We find ∂N/a=0=∂N/b. These translates to (A-b)(-1)(-A+a+b)+(A-b)(A-a)(1)=0,
(A-a)(-1)(-A+a+b)+(A-a)(A-b)(1)=0. Thus A-a=-A+a+b,A-b=-A+a+b. Thus 2A=2a+b=2b+a,a=b=c. Thus when the sides are equal we get an extremum. In fact a maximum.
$\endgroup$
0
$\begingroup$

(1) For a fixed base and the sum of the other two sides fixed the maximum height (and therefore maximum area) is when the triangle is isosceles. Prove this geometrically by the fact that the locus of a point which has the sum of distances from two fixed points constant is an ellipse: so the maximum height is when the two distances are equal.

(2) Then it follows that if the triangle is not equilateral we can find a larger one by making one of the equal sides the base and equalizing the other two with their sum constant.

(3) Hence (by contradiction) the equilateral triangle is the largest.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.