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Difference between revisions of "1961 IMO Problems/Problem 2"
(→‎Solution 3 (Trigonometry))
 
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==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>.
  
−
Substitute <math>S=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}</math>
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Let <math>a^2 = A</math>, <math>b^2 = B</math>, <math>c^2 = C</math>. Then, our inequality can be reduced to
−
 
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<cmath>A + B + C \geq \sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.</cmath>
−
This shows that the inequality is equivalent to <math>a^2b^2+b^2c^2+c^2a^2\le a^4+b^4+c^4</math>.
+
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>
  
−
This can be proven because <math>a^2b^2\le\frac{a^4+b^4}{2}</math>. The equality holds when <math>a=b=c</math>, or when the triangle is equilateral.
+
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.
  
−
==Solution 2 (Heron Bash)==
+
~mathboy100
−
As in the first solution, 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>
 
  
−
==Solution 3 By PEKKA==
+
==Solution 2 (duality principle)==
 
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>
Line 39: Line 63:
 
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>
 
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.
 
<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.
−
{{IMO box|year=1961|num-b=1|num-a=3}}
+
 
 +
~PEKKA
 +
 
 +
==Solution 3 (Trigonometry)==
 +
Let <math>\theta</math> be the angle between edges <math>a</math> and <math>b</math> subtending <math>c</math>.
 +
 
 +
We notice that <math>0\le(a-b\cos x)^2=a^2-2ab\cos x+b^2\cos^2x\le a^2-2ab\cos x+b^2</math> holds for all <math>a,b,x\in\mathbb R</math>.
 +
 
 +
If we let <math>x=\theta+30^\circ</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>
 +
 
 +
For equality, it is necessary for <math>a^2-2ab\sin x+b^2=0\implies a=b</math> and <math>\sin x=\sin(\theta+30^\circ)=1\implies\theta=60^\circ</math>, which implies that <math>\triangle ABC</math> is equilateral.
 +
 
 +
~ztilB
  
 
==Video Solution==
 
==Video Solution==
Line 45: Line 83:
 
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}}

Latest revision as of 06:29, 30 July 2024

Problem

Let $a$, $b$, and $c$ be the lengths of a triangle whose area is S. Prove that

$a^2 + b^2 + c^2 \ge 4S\sqrt{3}$

In what case does equality hold?

Solution

By Heron's formula, we have \[S = \sqrt{s(s-a)(s-b)(s-c)}.\] This can be simplified to \[S = \sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)\left(\frac{a-b+c}{2}\right)\left(\frac{a+b-c}{2}\right)}.\] Next, we can factor out all of the $2$s and use a clever difference of squares: \[S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\] \[S = \frac{1}{4}\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}\] \[S = \frac{1}{4}\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}.\] We can now use difference of squares again: \[S = \frac{1}{4}\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}\] \[4S\sqrt{3} = \sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}\] We must prove that the RHS of this equation is less than or equal to $a^2 + b^2 + c^2$.

Let $a^2 = A$, $b^2 = B$, $c^2 = C$. Then, our inequality can be reduced to \[A + B + C \geq \sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.\] We now have to prove \[(A + B + C)^2 \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.\] We can simplify: \[A^2 + B^2 + C^2 + 2AB + 2BC + 2CA \geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2\] \[4A^2 + 4B^2 + 4C^2 \geq 4AB + 4BC + 4CA\] \[A^2 + B^2 + C^2 \geq AB + BC + CA.\] Finally, we can apply AM-GM: \[\frac{A^2 + B^2}{2} \geq AB\] \[\frac{B^2 + C^2}{2} \geq BC\] \[\frac{C^2 + A^2}{2} \geq CA\] Adding these all up, we have the desired inequality \[A^2 + B^2 + C^2 \geq AB + BC + CA,\] and so the proof is complete.$\square$

To have $A + B + C = 4S\sqrt{3}$, we must satisfy \[\frac{A^2 + B^2}{2} = AB,\] \[\frac{B^2 + C^2}{2} = BC,\] \[\frac{C^2 + A^2}{2} = CA.\] This is only true when $A = B = C$, and thus $a = b = c$. Therefore, equality happens when the triangle is equilateral.

~mathboy100

Solution 2 (duality principle)

We firstly use the duality principle. $a=x+y~~b=x+z~~c=y+z$ The LHS becomes $(x+y)^2+(x+z)^2+(y+z)^2$ and the RHS becomes $4\sqrt{3}\sqrt{(x+y+z)xyz}$ If we use Heron's formula. By AM-GM $\frac{(x+y+z)^3}{27} \ge xyz$ Making this substitution $[ABC]$ becomes $\sqrt{(x+y+z)^4\frac{1}{27}}$ and once we take the square root of the area then our RHS becomes $\frac{4\sqrt{3}}{3\sqrt{3}}(x+y+z)^2=\frac{4}{3}(x+y+z)^2$ Multiplying the RHS and the LHS by 3 we get the LHS to be $3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).$ Our RHS becomes $4(x^2+y^2+z^2)+8(xy+yz+xz).$ Subtracting $4(x^2+y^2+z^2)+6(xy+yz+xz)$ we have the LHS equal to $(2(x^2+y^2+z^2))$ and the RHS being $2(xy+xz+yz)$ If LHS $\ge$ RHS then LHS-RHS$\ge 0$ LHS-RHS=$2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.$ $(x-y)^2+(x-z)^2+(y-z)^2 \ge 0$ by the trivial inequality so therefore, $a^2 + b^2 + c^2 \ge 4S\sqrt{3}$ and we're done.

~PEKKA

Solution 3 (Trigonometry)

Let $\theta$ be the angle between edges $a$ and $b$ subtending $c$.

We notice that $0\le(a-b\cos x)^2=a^2-2ab\cos x+b^2\cos^2x\le a^2-2ab\cos x+b^2$ holds for all $a,b,x\in\mathbb R$.

If we let $x=\theta+30^\circ$, \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*}

For equality, it is necessary for $a^2-2ab\sin x+b^2=0\implies a=b$ and $\sin x=\sin(\theta+30^\circ)=1\implies\theta=60^\circ$, which implies that $\triangle ABC$ is equilateral.

~ztilB

Video Solution

https://www.youtube.com/watch?v=ZYOB-KSEF3k&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=4 - AMBRIGGS

1961 IMO (Problems) • Resources
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