Difference between revisions of "1961 IMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | Let | + | Let <math>a</math>, <math>b</math>, and <math>c</math> be the lengths of a triangle whose area is ''S''. Prove that |
<math>a^2 + b^2 + c^2 \ge 4S\sqrt{3}</math> | <math>a^2 + b^2 + c^2 \ge 4S\sqrt{3}</math> | ||
Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
+ | By Heron's formula, we have | ||
+ | <cmath>S = \sqrt{s(s-a)(s-b)(s-c)}.</cmath> | ||
+ | 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> | ||
+ | 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> | ||
+ | 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>4S\sqrt{3} = \sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}</cmath> | ||
+ | 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 can be reduced to |
+ | <cmath>A + B + C \geq \sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.</cmath> | ||
+ | We now have to prove | ||
+ | <cmath>(A + B + C)^2 \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.</cmath> | ||
+ | We can simplify: | ||
+ | <cmath>A^2 + B^2 + C^2 + 2AB + 2BC + 2CA \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2</cmath> | ||
+ | <cmath>4A^2 + 4B^2 + 4C^2 \geq 4AB + 4BC + 4CA</cmath> | ||
+ | <cmath>A^2 + B^2 + C^2 \geq AB + BC + CA.</cmath> | ||
+ | Finally, we can apply AM-GM: | ||
+ | <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> | ||
+ | Adding these all up, we have the desired inequality | ||
+ | <cmath>A^2 + B^2 + C^2 \geq AB + BC + CA, </cmath> | ||
+ | and so the proof is complete.<math>\square</math> | ||
+ | To have <math>A + B + C = 4S\sqrt{3}</math>, we must satisfy | ||
+ | <cmath>\frac{A^2 + B^2}{2} = AB,</cmath> | ||
+ | <cmath>\frac{B^2 + C^2}{2} = BC,</cmath> | ||
+ | <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. | ||
− | + | ~mathboy100 | |
− | |||
− | [ | + | ==Solution 2 (duality principle)== |
+ | We firstly use the duality principle. | ||
+ | <math>a=x+y~~b=x+z~~c=y+z</math> | ||
+ | The LHS becomes <math>(x+y)^2+(x+z)^2+(y+z)^2</math> | ||
+ | and the RHS becomes <math>4\sqrt{3}\sqrt{(x+y+z)xyz}</math> If we use Heron's formula. | ||
+ | By AM-GM <math>\frac{(x+y+z)^3}{27} \ge xyz</math> | ||
+ | Making this substitution <math>[ABC]</math> becomes | ||
+ | <math>\sqrt{(x+y+z)^4\frac{1}{27}}</math> | ||
+ | and once we take the square root of the area then our RHS becomes | ||
+ | <math>\frac{4\sqrt{3}}{3\sqrt{3}}(x+y+z)^2=\frac{4}{3}(x+y+z)^2</math> | ||
+ | Multiplying the RHS and the LHS by 3 we get the LHS to be | ||
+ | <math>3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).</math> | ||
+ | Our RHS becomes | ||
+ | <math>4(x^2+y^2+z^2)+8(xy+yz+xz).</math> | ||
+ | Subtracting <math>4(x^2+y^2+z^2)+6(xy+yz+xz)</math> we have the LHS equal to <math>(2(x^2+y^2+z^2))</math> and the RHS being <math>2(xy+xz+yz)</math> | ||
+ | If LHS <math>\ge</math> RHS then LHS-RHS<math>\ge 0</math> | ||
+ | LHS-RHS=<math>2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.</math> | ||
+ | <math>(x-y)^2+(x-z)^2+(y-z)^2 \ge 0</math> by the trivial inequality so therefore, <math>a^2 + b^2 + c^2 \ge 4S\sqrt{3}</math> and we're done. | ||
+ | |||
+ | ~PEKKA | ||
+ | |||
+ | ==Solution 3 (Trigonometry)== | ||
+ | Let <math>\theta</math> be the angle between vertices <math>a</math> and <math>b</math>. We use two formulae: | ||
+ | |||
+ | <cmath>\begin{align}c^2=a^2+b^2-2\cos\theta\\T=\frac12ab\sin\theta\end{align}</cmath> | ||
+ | |||
+ | Going backwards, | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | From <math>a^2-2ab\sin x+b^2\geq0</math>, | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ~ztilB | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=ZYOB-KSEF3k&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=4 | ||
+ | - AMBRIGGS | ||
+ | {{IMO box|year=1961|num-b=1|num-a=3}} |
Revision as of 13:12, 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 |