# Challenge Math Challenge - October 2019

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#### etotheipi

Firstly for Io orbiting Jupiter, applying NII for circular motion:$$\frac{GM_{J}M_{I}}{r_{I}^{2}}=M_{I}r_{I}\omega ^{2}$$$$M_{J} = \frac{4\pi^{2}r_{I}^3}{GT^{2}}=1.9\times10^{27} kg$$Now for the Earth, using the definition of gravitational field strength:$$g=\frac{GM_{E}}{r_{E}^{2}}\implies M_{E} = \frac{gr_{E}^{2}}{G}=6.0 \times 10^{24} kg$$Finally for a general body of mass m orbiting the sun in circular motion of radius r:$$\frac{GM_{S}m}{r^{2}}=mr\omega^{2}=\frac{4\pi^{2}mr}{T^{2}}$$$$C = \frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM_{S}} \implies M_{S} = \frac{4\pi^{2}}{GC} = 2.0 \times 10^{30} kg$$

#### StoneTemplePython

Gold Member
The proof without this method can be given in a single line.
It took me 3 lines, courtesy of $\text{GM} \leq \text{AM}$, with a little clever padding

$\text{min: } 2\pi \big(rh + r^2\big) \text{ subject to: } r^2 h = c$ (with all numbers positive)

$\big(\frac{2}{c}\big)^\frac{1}{3} = \big(\frac{1}{r}\frac{1}{r}\frac{2}{h}\big)^\frac{1}{3} \leq \frac{1}{3}\big(\frac{1}{r} + \frac{1}{r} + \frac{2}{h}\big) = \frac{2}{3}\big(\frac{h +r}{rh}\big) \cdot \frac{r}{r}= \frac{2}{3}\cdot \frac{rh +r^2}{r^2 h} = \frac{2}{3}\cdot \frac{rh +r^2}{c}$

Rescale by $3 c \pi$ for standard form. Lower bound equality is met iff $\frac{1}{r} = \frac{2}{h}\longrightarrow h = 2r$

#### fresh_42

Mentor
2018 Award
It took me 3 lines, courtesy of $\text{GM} \leq \text{AM}$, with a little clever padding

$\text{min: } 2\pi \big(rh + r^2\big) \text{ subject to: } r^2 h = c$ (with all numbers positive)

$\big(\frac{2}{c}\big)^\frac{1}{3} = \big(\frac{1}{r}\frac{1}{r}\frac{2}{h}\big)^\frac{1}{3} \leq \frac{1}{3}\big(\frac{1}{r} + \frac{1}{r} + \frac{2}{h}\big) = \frac{2}{3}\big(\frac{h +r}{rh}\big) \cdot \frac{r}{r}= \frac{2}{3}\cdot \frac{rh +r^2}{r^2 h} = \frac{2}{3}\cdot \frac{rh +r^2}{c}$

Rescale by $3 c \pi$ for standard form. Lower bound equality is met iff $\frac{1}{r} = \frac{2}{h}\longrightarrow h = 2r$
Or a bit more compact:
$$\dfrac{A}{3\pi} = \dfrac{2r^2+rh+rh}{3}\stackrel{AM\geq GM}{\geq}\sqrt[3]{2r^2\cdot rh\cdot rh}= \sqrt[3]{\dfrac{2V^2}{\pi^2}} =: \text{ const. } > 0$$
with equality iff $2r^2=rh$.

#### etotheipi

Because it is a 0/0 indeterminate form, L'Hospital's rule gives\begin{align} \lim_{x\to0} \frac{\tan{(\sin{x})}}{\sin{(\tan{x})}} &= \lim_{x\to0} \frac{\sec^2{(\sin{x})}\cos{x}}{\cos{(\tan{x})}\sec^2{x}} \\ &= \lim_{x\to0} (\cos^3{x}) \cdot \lim_{x\to0} (\sec^2{(\sin{x})}) \cdot \lim_{x\to0} \frac{1}{\cos{(\tan{x})}} \\ &= \lim_{x\to0} (\cos^3{x}) \cdot \sec^2{(\lim_{x\to0} \sin{x})} \cdot \frac{1}{\cos{(\lim_{x\to0} \tan{x})}} \\ &= 1 \cdot \sec^2{0} \cdot \frac{1}{\cos{0}} = 1 \cdot 1 \cdot 1 = 1 \end{align}Sorry if I've made a mistake somewhere, I haven't done too much with limits as of yet.

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#### etotheipi

Is this correct for the change in frequency, with c = 340 and v = 20,
$$\Delta f = f_{0}(\frac{c}{c+v} - \frac{c}{c-v}) = -\frac{2cv}{c^{2}-v^{2}}f_{0} = -\frac{17}{144} f_{0}$$

#### fresh_42

Mentor
2018 Award
Is this correct for the change in frequency, with c = 340 and v = 20,
$$\Delta f = f_{0}(\frac{c}{c+v} - \frac{c}{c-v}) = -\frac{2cv}{c^{2}-v^{2}}f_{0} = -\frac{17}{144} f_{0}$$
No. And the question asks for the acoustic(!) interval, which is a ratio, not a difference. There is no way to deduce an initial condition to calculate the absolute frequencies.

#### etotheipi

No. And the question asks for the acoustic(!) interval, which is a ratio, not a difference. There is no way to deduce an initial condition to calculate the absolute frequencies.
In that case would the factor instead be$$\frac{f_{2}}{f_{1}} = \frac{c-v}{c+v} = \frac{8}{9}$$which corresponds to a decrease by a major second as per the harmonic series?

#### fresh_42

Mentor
2018 Award
In that case would the factor instead be$$\frac{f_{2}}{f_{1}} = \frac{c-v}{c+v} = \frac{8}{9}$$which corresponds to a decrease by a major second as per the harmonic series?
Yes. It is $\nu'= \dfrac{\nu}{1-\dfrac{v}{s}}$ versus $\nu''= \dfrac{\nu}{1+\dfrac{v}{s}}$

#### YoungPhysicist

Number 15:
(1) & (2)
if (1) is wrong, then (2) must also be wrong, so (1) is correct.
(3) & (4)
since (1) told us that $a$ is rational, then $a$ can't be $\sqrt13$ or $-\sqrt13$, so (4) is correct.

Now, if (6) is correct, $a$ can be any even natural number divisible by $7$, which is wrong.

if (5) is right otherwise, $a$ can only be $7$ or $14$, but since (6) is wrong at the same time, $a$ can only be $7$, which is the answer.

#### archaic

I don't understand question 14. Are we asked to pick two points on the sphere and find the midpoint of the segment that connects them?

#### fresh_42

Mentor
2018 Award
I don't understand question 14. Are we asked to pick two points on the sphere and find the midpoint of the segment that connects them?
Yes and no. Yes, as this is part of the question, and no, since this procedure has to be done with any second point. We have a fixed single point $P$ and a set $\mathbb{S}^2$. Then we consider all lines $\overline{PQ}$ from $P$ to $Q$ to another point $Q$ on the sphere. This line has a center $M_Q$. The problem asks for $\{\,M_Q\in \overline{PQ}\,|\,Q\in \mathbb{S}^2\,\}\,.$

#### archaic

Yes and no. Yes, as this is part of the question, and no, since this procedure has to be done with any second point. We have a fixed single point $P$ and a set $\mathbb{S}^2$. Then we consider all lines $\overline{PQ}$ from $P$ to $Q$ to another point $Q$ on the sphere. This line has a center $M_Q$. The problem asks for $\{\,M_Q\in \overline{PQ}\,|\,Q\in \mathbb{S}^2\,\}\,.$
Like this?
Supposing that $P=(x,y,z)$, the set of all midpoints between $P$ and $Q$ by varying $Q$ would be $\{(\frac{x+x_Q}{2},\frac{y+y_Q}{2},\frac{z+z_Q}{2})|(x_Q,y_Q,z_Q)\in\mathbb{S}^2\}$.

#### fresh_42

Mentor
2018 Award
Like this?
Supposing that $P=(x,y,z)$, the set of all midpoints between $P$ and $Q$ by varying $Q$ would be $\{(\frac{x+x_Q}{2},\frac{y+y_Q}{2},\frac{z+z_Q}{2})|(x_Q,y_Q,z_Q)\in\mathbb{S}^2\}$.
Yes, but which equation describes those points? What is it what you have written down?

#### lekh2003

Gold Member
Question 14
Isn't this just a scaling of the original sphere? So this would be a sphere at point P whose slope at P is tangent to the original sphere and has radius half of the original sphere.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.

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#### fresh_42

Mentor
2018 Award
Question 14
Isn't this just a scaling of the original sphere? So this would be a sphere at point P whose slope at P is tangent to the original sphere and has radius half of the original sphere.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.
Well, it's not a scaling anymore if the center is shifted. Your description is right, as is the set which @archaic defined. What I wanted to see is a simple transformation from one description to the other one. As you both have basically solved it, let me give the solution I had in mind. (Please note the two directions! The centers are on the smaller ball, and all points of the smaller ball are indeed a center of some chord!)

The variable endpoint $X$ of the chord is on the sphere, so for its position vector we have $\vec{x}^2=r^2$. The position vector of the center of the chord $\overline{PX}$ is thus
$$\vec{c}=\dfrac{\vec{p}+\vec{x}}{2} \;\Longleftrightarrow \; \vec{x}=2\vec{c}-\vec{p}$$
hence $r^2=(2\vec{c}-\vec{p})^2$ or $\left( \vec{c}-\dfrac{\vec{p}}{2} \right)^2=\dfrac{r^2}{4}$.

So the set of points we were looking for are all on a sphere with center $\overline{OP}/2=\vec{p}/2$ and radius $r/2$. All points of this sphere are on the other hand a center of some chord of the original sphere with endpoint $P$, since we can go back. The point $P$ itself is the center of the chord $\overline{PP}$.

#### lekh2003

Gold Member
@fresh_42 ahh, thanks for the explanation, that makes sense.

#### Greg Bernhardt

4 challenge problems left, who can solve them?

New ones on Friday!

#### lekh2003

Gold Member
4 challenge problems left, who can solve them?

New ones on Friday!
I would try, but i don't fancy speed learning topology right now :P