Difference between revisions of "1961 IMO Problems/Problem 2"
Mathboy100 (talk | contribs) |
Mathboy100 (talk | contribs) (→Solution 2 (Heron Bash)) |
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This can be simplified to | 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> | <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> | ||
+ | Next, we can factor out all of the <math>2</math>s and use a clever difference of squares | ||
+ | <cmath>S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}</cmath> | ||
+ | <cmath>S = \frac{1}{4}\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}</cmath> | ||
+ | <cmath>S = \frac{1}{4}\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}</cmath> | ||
==Solution 3 By PEKKA== | ==Solution 3 By PEKKA== |
Revision as of 23:27, 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
Next, we can factor out all of the
s and use a clever difference of squares
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