Prove that for all triangles with angles α,β,γ\alpha,\beta,\gammaα,β,γ that cosαcosβcosγ≤18\cos\alpha\cos\beta\cos\gamma\le \dfrac{1}{8}cosαcosβcosγ≤81
I just want to see as many solutions as possible.
Note by Daniel Liu 6 years, 12 months ago
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Show that lncosx \ln \cos x lncosx is a concave function and apply Jensens.
This is, in essence, combining the 2 steps of your solution into 1.
"As many solutions as possible"? Okay, here's another approach. Let xxx be the incremental variable, and we find the series expansion of Cos(α+x)Cos(β+x)Cos(γ−2x)Cos(\alpha +x)Cos(\beta +x)Cos(\gamma -2x)Cos(α+x)Cos(β+x)Cos(γ−2x), which works out to, after simplification
Cos(α)Cos(β)Cos(γ)+(12Sin(2α)+12Sin(2β)−Sin(2γ))x+Ox2+...Cos(\alpha )Cos(\beta )Cos(\gamma )+(\frac { 1 }{ 2 } Sin(2\alpha )+\frac { 1 }{ 2 } Sin(2\beta )-Sin(2\gamma ))x+O{ x }^{ 2 }+...Cos(α)Cos(β)Cos(γ)+(21Sin(2α)+21Sin(2β)−Sin(2γ))x+Ox2+...
where OOO indicates higher order coefficients. For the expression to be an extrema, the 2nd2nd2nd term must vanish, that is
12Sin(2α)+12Sin(2β)−Sin(2γ))=0\frac { 1 }{ 2 } Sin(2\alpha )+\frac { 1 }{ 2 } Sin(2\beta )-Sin(2\gamma ))=021Sin(2α)+21Sin(2β)−Sin(2γ))=0
But for symmetry reasons, this must vanish also
12Sin(2α)−Sin(2β)+12Sin(2γ))=0\frac { 1 }{ 2 } Sin(2\alpha )-Sin(2\beta )+\frac { 1 }{ 2 } Sin(2\gamma ))=021Sin(2α)−Sin(2β)+21Sin(2γ))=0, or
Sin(2α)−2Sin(2β)+Sin(2γ))=0Sin(2\alpha )-2Sin(2\beta )+Sin(2\gamma ))=0Sin(2α)−2Sin(2β)+Sin(2γ))=0
Adding this to the first, we quickly find that
Sin(2α)=Sin(2β)Sin(2\alpha )=Sin(2\beta )Sin(2α)=Sin(2β)
and that eventually α=β=γ=π3\alpha =\beta =\gamma =\dfrac { \pi }{ 3 } α=β=γ=3π, and the rest more or less follows. At this point, we've proven that this is the only extrema possible (with some caveats), but we haven't proven that it's a maxima. This is just a different approach, which can be used in other inequality problems like this. This is all part of "finding where the extrema are hiding?" problem, if you can't quite plot the landscape. If you can't see them, then feel them.
ADDED: In the spirit of "coming up with more proofs", here's a short proof. The following two expressions are equivalent, provided α+β+γ=π\alpha +\beta +\gamma =\pi α+β+γ=π
1−8Cos(α)Cos(β)Cos(γ)1-8Cos(\alpha )Cos(\beta )Cos(\gamma )1−8Cos(α)Cos(β)Cos(γ)
(Sin(α−β))2+(Cos(α−β)−2Cos(γ))2{ (Sin(\alpha -\beta )) }^{ 2 }+{ (Cos(\alpha -\beta )-2Cos(\gamma )) }^{ 2 }(Sin(α−β))2+(Cos(α−β)−2Cos(γ))2
But notice that the 2nd expression, being a sum of squares, has to be positive or 000. Hence, the inequality follows.
use AP > GP..
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Solution:
We see that cos(α+β+γ3)≥cosα+cosβ+cosγ3≥cosαcosβcosγ3\cos \left(\dfrac{\alpha+\beta+\gamma}{3}\right)\ge \dfrac{\cos\alpha+\cos\beta+\cos\gamma}{3}\ge \sqrt[3]{\cos\alpha\cos\beta\cos\gamma}cos(3α+β+γ)≥3cosα+cosβ+cosγ≥3cosαcosβcosγ
where the first inequality is Jensen's and the second is AM-GM.
But α+β+γ=π\alpha+\beta+\gamma=\piα+β+γ=π socosαcosβcosγ3≤cosπ3=12\sqrt[3]{\cos\alpha\cos\beta\cos\gamma}\le \cos \dfrac{\pi}{3}=\dfrac{1}{2}3cosαcosβcosγ≤cos3π=21
Thus cosαcosβcosγ≤18\cos\alpha\cos\beta\cos\gamma \le\dfrac{1}{8}cosαcosβcosγ≤81 as desired.
This solution works as long as the triangle is acute. However, when the triangle is obtuse, cosαcosβcosγ<0\cos\alpha\cos\beta\cos\gamma < 0cosαcosβcosγ<0 so the inequality still holds.
yeah.. this is exactly what i was talking about..
take the measure of the angles as 60 degrees . :P
Would this work?:
Let f(x)=cosxcos(K−x) f(x)=\cos{x}\cos{(K-x)} f(x)=cosxcos(K−x) where 0<K<180 0<K<180 0<K<180 is constant. We can rewrite f as f(x)=1/2cos(x+K−x)+1/2cos(x−(K−x))=1/2cosK+1/2cos(2x−K) f(x)=1/2\cos{(x+K-x)}+1/2\cos{(x-(K-x))}=1/2\cos{K}+1/2\cos{(2x-K)} f(x)=1/2cos(x+K−x)+1/2cos(x−(K−x))=1/2cosK+1/2cos(2x−K) which is maximized when x=K/2 x=K/2 x=K/2 (x is less than K for our intentions.)
Thus we should have at least some two of α,β,γ \alpha, \beta, \gamma α,β,γ equal say α=γ \alpha=\gamma α=γ to maximize the LHS of the inequality. This also means β=180−2α \beta=180-2\alpha β=180−2α
Now we wish to maximize cos2αcos(180−2α)=−cos2αcos2α \cos^2{\alpha}\cos{(180-2\alpha)}=-\cos^2{\alpha}\cos{2\alpha} cos2αcos(180−2α)=−cos2αcos2α.
We rewrite as −1+cos2α2⋅cos2α -\frac{1+\cos{2\alpha}}{2} \cdot \cos{2\alpha} −21+cos2α⋅cos2α. As a quadratic in cos2α\cos{2\alpha} cos2α the maximum is 1/8 when cos2α=−1/2 \cos{2\alpha}=-1/2 cos2α=−1/2. This is equivalent to an equilateral triangle.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Comments
Show that lncosx is a concave function and apply Jensens.
This is, in essence, combining the 2 steps of your solution into 1.
"As many solutions as possible"? Okay, here's another approach. Let x be the incremental variable, and we find the series expansion of Cos(α+x)Cos(β+x)Cos(γ−2x), which works out to, after simplification
Cos(α)Cos(β)Cos(γ)+(21Sin(2α)+21Sin(2β)−Sin(2γ))x+Ox2+...
where O indicates higher order coefficients. For the expression to be an extrema, the 2nd term must vanish, that is
21Sin(2α)+21Sin(2β)−Sin(2γ))=0
But for symmetry reasons, this must vanish also
21Sin(2α)−Sin(2β)+21Sin(2γ))=0, or
Sin(2α)−2Sin(2β)+Sin(2γ))=0
Adding this to the first, we quickly find that
Sin(2α)=Sin(2β)
and that eventually α=β=γ=3π, and the rest more or less follows. At this point, we've proven that this is the only extrema possible (with some caveats), but we haven't proven that it's a maxima. This is just a different approach, which can be used in other inequality problems like this. This is all part of "finding where the extrema are hiding?" problem, if you can't quite plot the landscape. If you can't see them, then feel them.
ADDED: In the spirit of "coming up with more proofs", here's a short proof. The following two expressions are equivalent, provided α+β+γ=π
1−8Cos(α)Cos(β)Cos(γ)
(Sin(α−β))2+(Cos(α−β)−2Cos(γ))2
But notice that the 2nd expression, being a sum of squares, has to be positive or 0. Hence, the inequality follows.
use AP > GP..
Log in to reply
Thanks!
Solution:
We see that cos(3α+β+γ)≥3cosα+cosβ+cosγ≥3cosαcosβcosγ
where the first inequality is Jensen's and the second is AM-GM.
But α+β+γ=π so3cosαcosβcosγ≤cos3π=21
Thus cosαcosβcosγ≤81 as desired.
This solution works as long as the triangle is acute. However, when the triangle is obtuse, cosαcosβcosγ<0 so the inequality still holds.
Log in to reply
yeah.. this is exactly what i was talking about..
take the measure of the angles as 60 degrees . :P
Would this work?:
Let f(x)=cosxcos(K−x) where 0<K<180 is constant. We can rewrite f as f(x)=1/2cos(x+K−x)+1/2cos(x−(K−x))=1/2cosK+1/2cos(2x−K) which is maximized when x=K/2 (x is less than K for our intentions.)
Thus we should have at least some two of α,β,γ equal say α=γ to maximize the LHS of the inequality. This also means β=180−2α
Now we wish to maximize cos2αcos(180−2α)=−cos2αcos2α.
We rewrite as −21+cos2α⋅cos2α. As a quadratic in cos2α the maximum is 1/8 when cos2α=−1/2. This is equivalent to an equilateral triangle.