1961 IMO Problems/Problem 2

Problem

Let $a$ , $b$ , and $c$ be the lengths of a triangle whose area is S. Prove that

$a^2 + b^2 + c^2 \ge 4S\sqrt{3}$

In what case does equality hold?

Solution

By Heron's formula, we have $S = \sqrt{s(s-a)(s-b)(s-c)}.$ This can be simplified to $S = \sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)\left(\frac{a-b+c}{2}\right)\left(\frac{a+b-c}{2}\right)}.$ Next, we can factor out all of the $2$ s and use a clever difference of squares: $S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$ $S = \frac{1}{4}\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}$ $S = \frac{1}{4}\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}.$ We can now use difference of squares again: $S = \frac{1}{4}\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}$ $4S\sqrt{3} = \sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}$ We must prove that the RHS of this equation is less than or equal to $a^2 + b^2 + c^2$ .

Let $a^2 = A$ , $b^2 = B$ , $c^2 = C$ . Then, our inequality can be reduced to $A + B + C \geq \sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.$ We now have to prove $(A + B + C)^2 \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.$ We can simplify: $A^2 + B^2 + C^2 + 2AB + 2BC + 2CA \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2$ $4A^2 + 4B^2 + 4C^2 \geq 4AB + 4BC + 4CA$ $A^2 + B^2 + C^2 \geq AB + BC + CA.$ Finally, we can apply AM-GM: $\frac{A^2 + B^2}{2} \geq AB$ $\frac{B^2 + C^2}{2} \geq BC$ $\frac{C^2 + A^2}{2} \geq CA$ Adding these all up, we have the desired inequality $A^2 + B^2 + C^2 \geq AB + BC + CA,$ and so the proof is complete. $\square$

To have $A + B + C = 4S\sqrt{3}$ , we must satisfy $\frac{A^2 + B^2}{2} = AB,$ $\frac{B^2 + C^2}{2} = BC,$ $\frac{C^2 + A^2}{2} = CA.$ This is only true when $A = B = C$ , and thus $a = b = c$ . Therefore, equality happens when the triangle is equilateral.

~mathboy100

Solution 2 (duality principle)

We firstly use the duality principle. $a=x+y~~b=x+z~~c=y+z$ The LHS becomes $(x+y)^2+(x+z)^2+(y+z)^2$ and the RHS becomes $4\sqrt{3}\sqrt{(x+y+z)xyz}$ If we use Heron's formula. By AM-GM $\frac{(x+y+z)^3}{27} \ge xyz$ Making this substitution $[ABC]$ becomes $\sqrt{(x+y+z)^4\frac{1}{27}}$ and once we take the square root of the area then our RHS becomes $\frac{4\sqrt{3}}{3\sqrt{3}}(x+y+z)^2=\frac{4}{3}(x+y+z)^2$ Multiplying the RHS and the LHS by 3 we get the LHS to be $3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).$ Our RHS becomes $4(x^2+y^2+z^2)+8(xy+yz+xz).$ Subtracting $4(x^2+y^2+z^2)+6(xy+yz+xz)$ we have the LHS equal to $(2(x^2+y^2+z^2))$ and the RHS being $2(xy+xz+yz)$ If LHS $\ge$ RHS then LHS-RHS $\ge 0$ LHS-RHS= $2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.$ $(x-y)^2+(x-z)^2+(y-z)^2 \ge 0$ by the trivial inequality so therefore, $a^2 + b^2 + c^2 \ge 4S\sqrt{3}$ and we're done.

~PEKKA

Solution 3 (Trigonometry)

We name the angle between vertices $a$ and $b$ as $\theta$ . We use two formulas:

1. $c^2=a^2+b^2-2\cos\theta$ , and

2. $T=\frac12ab\sin\theta$ .

By working from there, we can find a proof. Going backwards,

$\begin{rcases}a^2-2ab+b^2\leq0\\-2\leq2\sin x\geq2\end{rcases}\forall x\in\Bbb R:a^2-2ab\sin x+b^2\geq0$

$a^2-2ab\sin(\theta+30^\circ)+b^2\geq0$

$a^2-2ab(\sin\theta\cos30^\circ+\cos\theta\sin30^\circ)+b^2\geq0$

$a^2-ab\sin\theta\sqrt3-ab\cos\theta+b^2\geq0$

$a^2-ab\cos\theta+b^2\geq ab\sin\theta\sqrt3$

$a^2-ab\cos\theta+b^2\geq2\left(\frac12ab\sin\theta\right)\sqrt3$

$a^2-ab\cos\theta+b^2\geq2T\sqrt3$

$2a^2-2ab\cos\theta+2b^2\geq4T\sqrt3$

$a^2+b^2+(a^2-2ab\cos\theta+b^2)\geq4\sqrt3T$

$\boxed{a^2+b^2+c^2\geq4\sqrt3T}\ \text{QED}$

From $a^2-2ab\sin x+b^2\geq0$ ,

Equality occurs when $a=b$ and $\sin(\theta+30^\circ)=1\implies\theta=60^\circ$ , which is when $a=b=c$ so that they form an equilateral triangle.

~ztilB

Video Solution

https://www.youtube.com/watch?v=ZYOB-KSEF3k&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=4 - AMBRIGGS

1961 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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