Problem of the Week

Problem 5 solution

Congratulations to Ivan at St Bede’s Inter-church School, Vedant Joshi, and Rayyan Ansari at Goldington Academy for solving this problem!

(i) Since $A = \frac{1}{2}ab\sin\theta$ ,

$\begin{aligned} 16A^2 &= 4a^2b^2(\sin\theta)^2 \\ &= 4a^2b^2(1-(\cos\theta)^2) \text{ since $(\sin \theta)^2 + (\cos \theta)^2 = 1$} \\ &= 4a^2b^2 - 4a^2b^2(\cos\theta)^2 \\ &= 4a^2b^2 - (2ab\cos\theta)^2 \text{.} \end{aligned}$

By the cosine rule, $c^2 = a^2 + b^2 - 2ab\cos \theta$ and hence $2ab\cos \theta = a^2 + b^2 - c^2$ . Substituting this into the above expression for $16A^2$ gives $16A^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2$ .

(ii)

$\begin{aligned} 16A^2 &= 4a^2b^2 - (a^2 + b^2 - c^2)^2 \\ &= (2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2) \\ &= ((a+b)^2-c^2)(c^2-(a-b)^2) \\ &= (a+b+c)(a+b-c)(c+a-b)(c-a+b) \\ &= 2s(2s-2c)(2s-2b)(2s-2a) \text{ since $2s = a+b+c$} \\ &= 16s(s-a)(s-b)(s-c) \text{} \end{aligned}$

So $A^2 = s(s-a)(s-b)(s-c)$ and hence $A = \sqrt{s(s-a)(s-b)(s-c)}$ .

(iii) If the area $A$ is maximized for a given perimeter $2s$ , then $\frac{A^2}{s} = (s-a)(s-b)(s-c)$ must also be maximized since $s$ is fixed. Since $2s = a + b + c$ , $s-c = a+b-s$ and hence $\frac{A^2}{s} = (s-a)(s-b)(a+b-s)$ . Now $a$ and $b$ can vary without changing the perimeter $2s$ , so $\frac{A^2}{s}$ must be maximized as $a$ varies.

Thus $\frac{A^2}{s(s-b)} = (s-a)(a+b-s)$ must be maximized as $a$ varies with $b$ and $s$ fixed ( $s-b$ can’t be negative as no side length can be greater than half the perimeter). This is a quadratic in $a$ with a maximum turning point, which must be half-way between the roots $s$ and $s-b$ . So for $a$ to be maximized, we must have $a = \frac{s + s-b}{2} = \frac{2s-b}{2}$ and hence $2a = 2s-b = (a+b+c) - b = a+c$ , which implies that $a=c$ .

Then by symmetry (it doesn’t matter which sides’ lengths are called $a$ and $c$ ), we must also have $a=b$ and hence $a=b=c$ , so the triangle must be equilateral.

The most common approach to part (iii) was to use the AM-GM inequality to show that an equilateral triangle has an area greater than or equal to any other triangle of the same perimeter, but nobody taking this approach proved that the inequality is strict. This could be done by using the fact that $\text{AM} = \text{GM}$ if and only if all numbers in the list are equal.

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