Difference between revisions of "1961 IMO Problems/Problem 2"
Mathboy100 (talk | contribs) (→Solution 1 (Heron Bash + Nice Alg)) |
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To have <math>A + B + C = 4S\sqrt{3}</math>, we must satisfy | To have <math>A + B + C = 4S\sqrt{3}</math>, we must satisfy | ||
− | <cmath>\frac{A^2 + B^2}{2} | + | <cmath>\frac{A^2 + B^2}{2} = AB,</cmath> |
− | <cmath>\frac{B^2 + C^2}{2} | + | <cmath>\frac{B^2 + C^2}{2} = BC,</cmath> |
− | <cmath>\frac{C^2 + A^2}{2} | + | <cmath>\frac{C^2 + A^2}{2} = CA.</cmath> |
This is only true when <math>A = B = C</math>, and thus <math>a = b = c</math>. Therefore, equality happens when the triangle is equilateral. | This is only true when <math>A = B = C</math>, and thus <math>a = b = c</math>. Therefore, equality happens when the triangle is equilateral. | ||
Revision as of 00:00, 3 December 2022
Problem
Let ,
, and
be the lengths of a triangle whose area is S. Prove that
In what case does equality hold?
Solution 1 (Heron Bash + Nice Alg)
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 this to the equation, we obtain
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 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