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Difference between revisions of "1961 IMO Problems/Problem 2"
(→‎Solution 2 (Heron Bash))
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This can be simplified to
 
This can be simplified to
 
<cmath>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)}.</cmath>
 
<cmath>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)}.</cmath>
 +
Next, we can factor out all of the <math>2</math>s and use a clever difference of squares
 +
<cmath>S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</cmath>
 +
<cmath>S = \frac{1}{4}\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}</cmath>
 +
<cmath>S = \frac{1}{4}\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}</cmath>
  
 
==Solution 3 By PEKKA==
 
==Solution 3 By PEKKA==

Revision as of 22:27, 2 December 2022

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

Substitute $S=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$

This shows that the inequality is equivalent to $a^2b^2+b^2c^2+c^2a^2\le a^4+b^4+c^4$.

This can be proven because $a^2b^2\le\frac{a^4+b^4}{2}$. The equality holds when $a=b=c$, or when the triangle is equilateral.

Solution 2 (Heron Bash)

As in the first solution, 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 \[S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\] \[S = \frac{1}{4}\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}\] \[S = \frac{1}{4}\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}\]

Solution 3 By PEKKA

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.

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

Video Solution

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

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