Difference between revisions of "1961 IMO Problems/Problem 2"
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LHS-RHS=<math>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.</math> | LHS-RHS=<math>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.</math> | ||
<math>(x-y)^2+(x-z)^2+(y-z)^2 \ge 0</math> by the trivial inequality so therefore, <math>a^2 + b^2 + c^2 \ge 4S\sqrt{3}</math> and we're done. | <math>(x-y)^2+(x-z)^2+(y-z)^2 \ge 0</math> by the trivial inequality so therefore, <math>a^2 + b^2 + c^2 \ge 4S\sqrt{3}</math> and we're done. | ||
+ | {{IMO box|year=1961|num-b=1|num-a=3}} | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:04, 1 August 2022
Problem
Let ,
, and
be the lengths of a triangle whose area is S. Prove that
In what case does equality hold?
Solution
Substitute , where
This shows that the inequality is equivalent to .
This can be proven because . The equality holds when
, or when the triangle is equilateral.
Solution 2 By PEKKA
We firstly use the duality principle.
The LHS becomes
and the RHS becomes
If we use Heron's formula.
By AM-GM
Making this substitution
becomes
and once we take the square root of the area then our RHS becomes
Multiplying the RHS and the LHS by 3 we get the LHS to be
Our RHS becomes
Subtracting
we have the LHS equal to
and the RHS being
If LHS
RHS then LHS-RHS
LHS-RHS=
by the trivial inequality so therefore,
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