1961 IMO Problems/Problem 2
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
Next, we can factor out all of the
s and use a clever difference of squares
We can now use difference of squares again:
We know that
This is because the area of the triangle stays the same if we switch around the values of
,
, and
.
Thus,
We must prove that the RHS of this equation is less than or equal to
.
Let ,
,
. Then, our inequality is reduced to
We will now simplify the RHS.
For any real numbers ,
, and
,
and thus
Applying it to the equation, we obtain
We now have to prove
We can now simplify:
Finally, we can apply AM-GM:
Adding these all up, we have the desired inequality
and so the proof is complete.$\qed$ (Error compiling LaTeX. Unknown error_msg)
~mathboy100
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