Geometric maxima minima — a little question for IITJEE Mains

Problem:

Of all triangles with a given perimeter, find the one with maximum area.

Solution:

Consider an arbitrary triangle with side lengths a, b, c and perimeter $2s=a+b+c$ . By Heron’s formula, its area F is given by

$F = \sqrt{s(s-a)(s-b)(s-c)}$

Now, the arithmetic mean geometric mean inequality gives

$\sqrt[3]{s(s-a)(s-b)(s-c)} \leq \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3}$

Therefore, $F \leq \sqrt{s(\frac{s}{3})^{3}}=s^{2}\frac{\sqrt{3}}{9}$

where inequality holds if and only if $s-a=s-b=s-c$ , that is, when $a=b=c$ .

Thus, the area of any triangle with perimeter 2s does not exceed $\frac{s^{2}\sqrt{3}}{9}$ and is equal to $\frac{s^{2}\sqrt{3}}{9}$ only for an equilateral triangle. QED.

More later,

Nalin Pithwa

PS: Ref: Geometric Problems on Maxima and Minima by Titu Andreescu et al.

This entry was written by Nalin Pithwa, posted on May 13, 2016 at 12:44 am, filed under geometry, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY, RMO. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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