Difference between revisions of "1961 IMO Problems/Problem 2"
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In what case does equality hold? | In what case does equality hold? | ||
− | ==Solution | + | ==Solution 1 (Heron Bash + Nice Alg)== |
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As in the first solution, we have | As in the first solution, we have | ||
<cmath>S = \sqrt{s(s-a)(s-b)(s-c)}.</cmath> | <cmath>S = \sqrt{s(s-a)(s-b)(s-c)}.</cmath> | ||
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<cmath>A^2 + B^2 + C^2 \geq AB + BC + CA, </cmath> | <cmath>A^2 + B^2 + C^2 \geq AB + BC + CA, </cmath> | ||
and so the proof is complete.<math>\blacksquare</math> | and so the proof is complete.<math>\blacksquare</math> | ||
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+ | To have <math>A + B + C = 4S\sqrt{3}</math>, we must satisfy | ||
+ | <cmath>\frac{A^2 + B^2}{2} \geq AB,</cmath> | ||
+ | <cmath>\frac{B^2 + C^2}{2} \geq BC,</cmath> | ||
+ | <cmath>\frac{C^2 + A^2}{2} \geq 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. | ||
~mathboy100 | ~mathboy100 | ||
− | ==Solution | + | ==Solution 2 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 23:56, 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 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 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.
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