1961 IMO Problems/Problem 2
Contents
[hide]Problem
Let ,
, and
be the lengths of a triangle whose area is S. Prove that
In what case does equality hold?
Solution
By Heron's formula, 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 must prove that the RHS of this equation is less than or equal to
.
Let ,
,
. Then, our inequality can be reduced to
We now have to prove
We can simplify:
Finally, we can apply AM-GM:
Adding these all up, we have the desired inequality
and so the proof is complete.
To have , we must satisfy
This is only true when
, and thus
. Therefore, equality happens when the triangle is equilateral.
~mathboy100
Solution 2 (duality principle)
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.
~PEKKA
Solution 3 (Trigonometry)
Let be the angle between edges
and
subtending
.
We notice that holds for all
.
If we let ,
For equality, it is necessary for and
, which implies that
is equilateral.
~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 |
All IMO Problems and Solutions |