Can You Solve These Summer Math Challenges?

[FelixLudi]

Showing that 4 blocks will be left is not sufficient, we have to show that 8 will be left.

8 will be left yes, that was a mistake on my part of the math (an area of 2 x 2 x 2).

 

[fresh_42]

Solution to #1.

a) Prove $(e+x)^{e-x}>(e-x)^{e+x}$ for $0<x<e.$

We define
$$f\, : \,]0,e[ \longrightarrow \mathbb{R}\, , \,f(x):=(e-x)\log(e+x)-(e+x)\log(e-x)$$
and observe $\lim_{x \searrow 0}f(x)=0\,.$ Then
\begin{align*}
f\,'(x)&=\dfrac{e-x}{e+x}+\dfrac{e+x}{e-x}-\left(\log(e+x)+\log(e-x)\right)\\[6pt]
&=2\cdot \underbrace{\dfrac{e^2+x^2}{e^2-x^2}}_{>1}-\underbrace{\log(e^2-x^2)}_{<2}\\[6pt]
&> 0
\end{align*}
and $f(x)>0\,$, resp. $(e-x)\log(e+x)>(e+x)\log(e-x)\,$ resp. $(e+x)^{e-x}>(e-x)^{e+x}$

b) Show that for $0 < b < a$ we have
$$\dfrac{1}{a} < \dfrac{2}{a+b} < \dfrac{\log (a) - \log (b)}{a-b} < \dfrac{1}{\sqrt{ab}} < \dfrac{1}{b}$$
For $x=\dfrac{a}{b}>1$ we have
\begin{align*}
\log(x^2)&=2 \log(x)\\
&=2 \int_1^x \frac{1}{t}\,dt\\
&< \int_1^x \left(1+\frac{1}{t^2}\right)\,dt\\
&= x-\frac{1}{x}\\
&\text{or}\\
\log(x)&<\sqrt{x}-\frac{1}{\sqrt{x}}
\end{align*}
With $\log(x)=\sum_{k=0}^\infty \frac{2}{2k+1}\left(\dfrac{x-1}{x+1}\right)^{2k+1}> 2\cdot\dfrac{x-1}{x+1}$ we get
$$
2\dfrac{\frac{a}{b}-1}{\frac{a}{b}+1}=2\dfrac{a-b}{a+b}< \log(\frac{a}{b})=\log(a)-\log(b)<\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}=\dfrac{a-b}{\sqrt{ab}}
$$
c) Let $f,g\, : \,[a,b]\longrightarrow \mathbb{R}$ be two monotone integrable functions, either both increasing or both decreasing. Show that $\int_a^bf(x)g(x)\,dx \ge \int_a^bf(x)\,dx \cdot \int_a^bg(x)\,dx$

We have $\left(f(x)-f(y)\right)\left(g(x)-g(y)\right)>0$ and therefore
\begin{align*}
\int_a^bf(x)g(x)\,dx +\int_a^bf(y)g(y)\,dy &\ge \\ \int_a^bf(x)\,dx \, \int_a^bg(y)\,dy &+ \int_a^bf(y)\,dy \, \int_a^bg(x)\,dx\\
&\text{or} \\
2 \int_a^bf(x)g(x)\,dx &\ge 2\int_a^bf(x)\,dx \, \int_a^bg(x)\,dx
\end{align*}

 

[QuantumQuest]

Solution for Problem $8$:

Spoiler

$f(x)$ and $g(x)$ have $M(x_0,f(x_0))$ as a (common) point of contact if

$f(x_0) = g(x_0)$ and $f'(x_0) = g'(x_0)$.

We have $f(x) = e^{-x}$ so $f'(x) = -e^{-x}$ and $g(x) = e^{-x}\cos x$ so $g'(x) = -e^{-x}\cos x - e^{-x}\sin x$.

We have to solve the system of equations

$$ \begin{cases}
f(x) = g(x)\\
f'(x) = g'(x)
\end{cases} \iff \begin{cases}
e^{-x} = e^{-x}\cos x\\
-e^{-x} = -e^{-x}\cos x - e^{-x}\sin x
\end{cases} \iff \begin{cases}
\cos x = 1\\
1 = \cos x + \sin x
\end{cases} \iff$$
$$\begin{cases}
\cos x = 1\\
\sin x = 0
\end{cases} \iff x = 2k\pi, k\in \mathbb{Z}$$

So, $f$ and $g$ are tangent to each other at points $(2k\pi, e^{-2k\pi})\space\space k\in \mathbb{Z}$.
Now, one function that is tangent to $f$ and $g$ is the tangent line

$y = f(x_0) + f'(x_0)\cdot (x - x_0)$.

For $x_0 = 2\pi$, $f(x_0) = e^{-2\pi}$, $f'(x_0) = -e^{-2\pi}$. So,

$y = e^{-2\pi} - e^{-2\pi} (x - 2\pi)$

 

[StoneTemplePython]

Problem 5:

The answer is no. I think people have collectively figured this out, though explicitly partitioning the volume enclosed in the box to be like a chessboard ties out loose ends.
- - - -
similar to the chess board problem given as a hint, but with a 3rd dimension, suppose we have black and white cubes alternating each of size 2x2x2 -- this makes up the airspace inside the box. Without loss of generality suppose it is 14 black cubes and 13 white cubes. (Note $3^3 = 27 = 14+13$.) All we've done so far is partition or 'tile' the volume of our 6x6x6 box in a way that looks like a chessboard.

Now whenever we place one of our 1x1x4 bricks into the box, half of it is in the space of a black cube and the other half in a white cube. The issue is that at best each 2x2x2 cube can be occupied by 4 bricks. This allows us to place 52 bricks in the box. However once we've done this we must have used up the 13 white cubes (i.e. $4(13) =52$). Adding in the 53rd brick requires 1 free black cube (available) and 1 free white cube (which doesn't exist), hence packing 53 bricks in the box is impossible.