Difference between revisions of "1961 IMO Problems/Problem 2"
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We can now use difference of squares again: | We can now use difference of squares again: | ||
<cmath>S = \frac{1}{4}\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}</cmath> | <cmath>S = \frac{1}{4}\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}</cmath> | ||
− | <cmath>4S\sqrt{3} = \sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2) | + | <cmath>4S\sqrt{3} = \sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}</cmath> |
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We must prove that the RHS of this equation is less than or equal to <math>a^2 + b^2 + c^2</math>. | We must prove that the RHS of this equation is less than or equal to <math>a^2 + b^2 + c^2</math>. | ||
− | Let <math>a^2 = A</math>, <math>b^2 = B</math>, <math>c^2 = C</math>. Then, our inequality | + | Let <math>a^2 = A</math>, <math>b^2 = B</math>, <math>c^2 = C</math>. Then, our inequality can be reduced to |
− | <cmath>A + B + C \geq | + | <cmath>A + B + C \geq \sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.</cmath> |
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We now have to prove | We now have to prove | ||
<cmath>(A + B + C)^2 \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.</cmath> | <cmath>(A + B + C)^2 \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.</cmath> | ||
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Adding these all up, we have the desired inequality | Adding these all up, we have the desired inequality | ||
<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>\ | + | and so the proof is complete.<math>\square</math> |
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 | ||
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~PEKKA | ~PEKKA | ||
− | {{ | + | |
+ | ==Solution 3 (Trigonometry)== | ||
+ | Let <math>\theta</math> be the angle between vertices <math>a</math> and <math>b</math>. We use two formulae: | ||
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+ | <cmath>\begin{align}c^2=a^2+b^2-2\cos\theta\\T=\frac12ab\sin\theta\end{align}</cmath> | ||
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+ | Going backwards, | ||
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+ | <math>\begin{rcases}(a-b)^2=a^2-2ab+b^2\leq0\\-2\leq2\sin x\geq2\end{rcases}\forall a,b,x\in\Bbb R^+:a^2-2ab\sin x+b^2\geq0</math> | ||
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+ | <cmath>\begin{align*}a^2-2ab\sin(\theta+30^\circ)+b^2&\geq0\\ | ||
+ | a^2-2ab(\sin\theta\cos30^\circ+\cos\theta\sin30^\circ)+b^2&\geq0\\a^2-ab\sin\theta\sqrt3-ab\cos\theta+b^2&\geq0\\a^2-ab\cos\theta+b^2&\geq ab\sin\theta\sqrt3\\a^2-ab\cos\theta+b^2&\geq2\left(\frac12ab\sin\theta\right)\sqrt3\\a^2-ab\cos\theta+b^2&\geq2T\sqrt3\\2a^2-2ab\cos\theta+2b^2&\geq4T\sqrt3\\a^2+b^2+(a^2-2ab\cos\theta+b^2)&\geq4\sqrt3\ T\\\boxed{a^2+b^2+c^2\geq4\sqrt3\ T}\ \blacksquare\end{align*}</cmath> | ||
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+ | From <math>a^2-2ab\sin x+b^2\geq0</math>, | ||
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+ | Equality occurs when <math>a=b</math> and <math>\sin(\theta+30^\circ)=1\implies\theta=60^\circ</math>, which is when <math>a=b=c</math> so that they form an equilateral triangle. | ||
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+ | ~ztilB | ||
==Video Solution== | ==Video Solution== | ||
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https://www.youtube.com/watch?v=ZYOB-KSEF3k&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=4 | https://www.youtube.com/watch?v=ZYOB-KSEF3k&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=4 | ||
- AMBRIGGS | - AMBRIGGS | ||
+ | {{IMO box|year=1961|num-b=1|num-a=3}} |
Revision as of 12:12, 13 February 2024
Contents
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 vertices and . We use two formulae:
Going backwards,
From ,
Equality occurs when and , which is when so that they form an equilateral triangle.
~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 |