Math Challenge - December 2020

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  • #51
lekh2003
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##\geq ## is sufficient, since we have all inequalities with the possibility of being equal. And ##\geq 0## is obvious.
Oh yeah, of course its ##\geq 0##, I'm a fool.
There is another little thing to consider:
How is this related?
 
  • #52
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Time for another half-assed proof :P I hope I interpreted the problem correctly.
To maximise the area of the tetrahedron would involve maximising the area of the face ABC, which can maximise the lengths of BC, BA, and AC. This would involve taking the limit of the tetrahedron as it approaches a flat tetrahedron where the area of ABC is equal to the area of ABD+BCD+ADC. Any other type of tetrahedron will slowly reduce the area since AD, BD, CD are constant and the more spread out they are, the larger ABC's area is.

Hence, this is simply a problem of finding the area of an equilateral triangle with a distance of 1 from the center to each of the vertices. Using simple trigonometry and law of cosines, you can find the side length of this equilateral triangle. I used a triangle with tau/3 and a side of length 1 on either side. The opposite is ##\sqrt 3##. That's the side length of the equilaterqal triangle. Area of an equilateral triangle is ##\frac{\sqrt 3}{4}s^2##. Hence, the triangle has area ##\frac{3\sqrt 3}{4}##.

Since the surface area of the tetrahedron must be double the area of this equilateral triangle (when flat, this being the limit), then the maximum limit of area is ##\frac{3\sqrt 3}{2}.
This is too hand wavy with too many unproven statements. Here is an image along which you (hopefully) formalize your argument:

tetrahedron.png
 
  • #53
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Oh yeah, of course its ##\geq 0##, I'm a fool.

How is this related?
Well, since you already got the credit and the solution is in any case a kind of brute force, I think I could simply post my solution:
##0\leq (xyz-1)^2=x^2y^2z^2-2xyz+1## and thus ##2\leq xyz+\dfrac{1}{xyz}.## Therefore
\begin{align*}
6&+2x+2y+2z+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+xy+yz+zx\\
&+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\\
&\leq xyz+\dfrac{1}{xyz}+4+2x+2y+2z+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+xy+yz+zx\\&+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\\
&\Longleftrightarrow \\
&\left(1+x+\dfrac{1}{y}\right)\cdot \left(1+y+\dfrac{1}{z}\right)+\left(1+y+\dfrac{1}{z}\right)\cdot\left(1+z+\dfrac{1}{x}\right)\\&+\left(1+z+\dfrac{1}{x}\right)\cdot\left(1+x+\dfrac{1}{y}\right)\\
&\leq \left(1+x+\dfrac{1}{y}\right)\cdot\left(1+y+\dfrac{1}{z}\right)\cdot\left(1+z+\dfrac{1}{x}\right)
\end{align*}
 
  • #54
189
92
I know it's been solved, but I just want to give it a shot myself for fun.

8d:

The denominator is almost a perfect square. Adding "zero" we can obtain:
$$I = \int_{-\infty}^{\infty} \frac{1}{x^2-2x+2}dx = \int_{-\infty}^{\infty} \frac{1}{x^2-2x+2 - 1 + 1}dx = \int_{-\infty}^{\infty} \frac{1}{(x-1)^2+1}dx$$
Using u-substitution, we obtain (the bounds don't change in this particular case).
$$I = \int_{-\infty}^{\infty} \frac{1}{u^2+1}dx$$
This is just a standard integral:
$$I = \left [\arctan(u) \right ]^\infty_{-\infty}$$
which leaves us with the two limits
$$I = \lim_{u \rightarrow \infty} \arctan(u) - \lim_{u \rightarrow -\infty} \arctan(u) = \frac{\pi}{2} - \left(- \frac{\pi}{2}\right) = \pi$$

For my own edification, how do I prove that ##\int \frac{1}{x^2+1} dx = \arctan(x) + C## and the value of the two limits? These are the things my professor would just tell us to "look up in a table" as our math course doesn't dig very deep.
 
  • #55
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For my own edification, how do I prove that ##\int \frac{1}{x^2+1} dx = \arctan(x) + C##...
You can differentiate ##x \longmapsto \tan(x)=\dfrac{\sin(x)}{\cos(x)}## and use it to differentiate ##x\longmapsto x=(\tan \circ \operatorname{arctan})(x).## This would be a proof by differentiation instead of a direct integration. For a direct integration I'd try to do it with a Weierstraß substitution:
https://en.wikipedia.org/wiki/Weierstrass_substitution
... and the value of the two limits?
This depends on what you take for granted. E.g. if you know the graph of the tangent function then you automatically have the graph of the inverse function by reflection at the diagonal ##y=x##. Another possibility is to use approximation formulas like
$$
\operatorname{arctan} x \approx \begin{cases}
\dfrac{x}{1+0.28x^2} &\text{ if } |x|\leq 1\\
\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x>1\\
-\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x<-1
\end{cases}
$$
 
  • #56
189
92
You can differentiate ##x \longmapsto \tan(x)=\dfrac{\sin(x)}{\cos(x)}## and use it to differentiate ##x\longmapsto x=(\tan \circ \operatorname{arctan})(x).## This would be a proof by differentiation instead of a direct integration. For a direct integration I'd try to do it with a Weierstraß substitution:
https://en.wikipedia.org/wiki/Weierstrass_substitution

This depends on what you take for granted. E.g. if you know the graph of the tangent function then you automatically have the graph of the inverse function by reflection at the diagonal ##y=x##. Another possibility is to use approximation formulas like
$$
\operatorname{arctan} x \approx \begin{cases}
\dfrac{x}{1+0.28x^2} &\text{ if } |x|\leq 1\\
\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x>1\\
-\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x<-1
\end{cases}
$$
Can arctan(x) be differentiated using the definition of the derivative? Then you could use F'(x) = f(x) as a "proof".
 
  • #57
15,540
13,636
Can arctan(x) be differentiated using the definition of the derivative? Then you could use F'(x) = f(x) as a "proof".
It might work if you write it as a power series, but I guess it is quite hard if you only have
$$
\lim_{h \to 0} \dfrac{\operatorname{arctan}(x+h)-\operatorname{arctan}(x)}{h}
$$
The problem with any proof is always: 'Where are we allowed to start?', means what do we know already. If you can show ##F'(x)=f(x)## then you have automatically proven ##\int F'(x)\,dx = f(x)+C##. And even this uses the fundamental theorem of calculus.
 
  • #58
189
92
It might work if you write it as a power series, but I guess it is quite hard if you only have
$$
\lim_{h \to 0} \dfrac{\operatorname{arctan}(x+h)-\operatorname{arctan}(x)}{h}
$$
The problem with any proof is always: 'Where are we allowed to start?', means what do we know already. If you can show ##F'(x)=f(x)## then you have automatically proven ##\int F'(x)\,dx = f(x)+C##. And even this uses the fundamental theorem of calculus.
I think any starting point is fine as long as the starting point is 1) well defined 2) doesn't result in circular reasoning. Using the fact that ##(\arctan(x))' = 1/x^2 + 1 ## isn't circular, but in practice it is a bad way thinking, as guess-and-check quickly becomes unfeasible.
 
  • #59
15,540
13,636
I think any starting point is fine as long as the starting point is 1) well defined 2) doesn't result in circular reasoning. Using the fact that ##(\arctan(x))' = 1/x^2 + 1 ## isn't circular, but in practice it is a bad way thinking, as guess-and-check quickly becomes unfeasible.
My first suggestion should work. ##(\tan x)'=\left(\dfrac{\sin x}{\cos x}\right)'=1+\tan^2 x.## So
\begin{align*}
(id)'(x)=1&=(\tan \circ \operatorname{arctan})'(x)=(\tan y)'\cdot \operatorname{arctan }'(x)\\
&\Longrightarrow \\
\operatorname{arctan}'(x)&=\dfrac{1}{1+\tan^2(y)}=\dfrac{1}{1+\tan^2(\operatorname{arctan}(x))}=\dfrac{1}{1+x^2}
\end{align*}
but this uses ##(\sin)'=\cos\, , \,(\cos)'=-\sin## and ##1=\sin^2+\cos^2## in order to differentiate the tangent function.

I think that the direct way for ##\int\frac{dx}{1+x^2}## with the Weierstraß substitution works as well: ##t:=\tan\left(\dfrac{x}{2}\right)## which yields ##dx=\dfrac{2dt}{1+t^2}.## At least it immediately connects the rational polynomial with the tangent function.
 
  • #60
189
92
My first suggestion should work. ##(\tan x)'=\left(\dfrac{\sin x}{\cos x}\right)'=1+\tan^2 x.## So
\begin{align*}
(id)'(x)=1&=(\tan \circ \operatorname{arctan})'(x)=(\tan y)'\cdot \operatorname{arctan }'(x)\\
&\Longrightarrow \\
\operatorname{arctan}'(x)&=\dfrac{1}{1+\tan^2(y)}=\dfrac{1}{1+\tan^2(\operatorname{arctan}(x))}=\dfrac{1}{1+x^2}
\end{align*}
but this uses ##(\sin)'=\cos\, , \,(\cos)'=-\sin## and ##1=\sin^2+\cos^2## in order to differentiate the tangent function.

I think that the direct way for ##\int\frac{dx}{1+x^2}## with the Weierstraß substitution works as well: ##t:=\tan\left(\dfrac{x}{2}\right)## which yields ##dx=\dfrac{2dt}{1+t^2}.## At least it immediately connects the rational polynomial with the tangent function.
I'm not super familiar with Weierstrass substitution, but it does seem to yield a useful result.
 
  • #61
124
48
Answer: 5
Explanation: The modulo w.r.t. 7 of any natural number belongs to one of the 4 sets from ##A=\{\{0\}, \{1, 6\}, \{2, 5\}, \{3, 4\}\}##, i.e. sets of the form ##\{r, (7-r) \mod 7\}## where ##r## belongs to the set of all possible remainders w.r.t. 7, i.e. {0, 1, 2, ..., 6}. Consider 2 distinct natural numbers ##a, b##. It is easy to see that if ##(a \mod 7)## and ##(b\mod 7)## belong to 2 different sets in ##A##, then neither their sum nor difference is divisible by 7. On the other hand, if they belong to the same set from within ##A##, then 2 possibilities arise:
  1. ##(a \mod 7) = (b \mod 7)##, i.e. the remainders w.r.t. 7 of both ##a## and ##b## correspond the same element in the set. In this case, ##(a - b) \equiv 0 \mod 7##, i.e. the difference is divisible by 7.
  2. ##(r_{a} \equiv (a \mod 7)) \neq (r_{b} \equiv (b \mod 7))##. Then, by definition of the sets, ##r_{a} = (7 - r_{b}) \mod 7##. Therefore, ##(r_{a} + r_{b}) = 7 \equiv 0 \mod 7 \Rightarrow (a+b) \equiv 0 \mod{7}##, i.e. the sum is divisible by 7.
Thus, for neither sum nor difference of ##a, b## to be divisible by 7, their modulo w.r.t. 7 must belong to 2 different sets from ##A##. By pigeonhole principle, there can be at most 4 distinct natural numbers in a set such that the modulo w.r.t. 7 of no two of those belong to the same set from ##A##. Thus, if we have a set of 5 (or more) natural numbers, at least two of those numbers will have either their sum or their difference divisible by 7.
 

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