Difference between revisions of "1961 IMO Problems/Problem 2"
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This can be proven because <math>a^2b^2\le\frac{a^4+b^4}{2}</math>. The equality holds when <math>a=b=c</math>, or when the triangle is equilateral. | This can be proven because <math>a^2b^2\le\frac{a^4+b^4}{2}</math>. The equality holds when <math>a=b=c</math>, or when the triangle is equilateral. | ||
− | ==Solution 2 By PEKKA== | + | ==Solution 2 (Heron Bash)== |
+ | As in the first solution, we have | ||
+ | <cmath>S = \sqrt{s(s-a)(s-b)(s-c)}.</cmath> | ||
+ | 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> | ||
+ | |||
+ | ==Solution 3 By PEKKA== | ||
We firstly use the duality principle. | We firstly use the duality principle. | ||
<math>a=x+y~~b=x+z~~c=y+z</math> | <math>a=x+y~~b=x+z~~c=y+z</math> |
Revision as of 23:23, 2 December 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 (Heron Bash)
As in the first solution, we have
This can be simplified to
Solution 3 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