I Micromass' big October challenge

[QuantumQuest]

ADVANCED CHALLENGES: Problem 9

We will utilize letters a, b, c, d, e, f, g, h, i, j for digits. We'll begin with divisibility (i.e b) part) and take into account a) part along the way.

According to part b) constraint, the whole number denoted $abcdefghij$, must be divisible by 10, so $j = 0$. Now, we'll proceed to the first five digits. The reason for this, is that we can quickly conclude that $e$ is $0$ or $5$. Now, taking into account part a) constraint (no digit repetitions) and because $0$ is already taken, we conclude that $e = 5$.

At this point, it would be wise to divide remaining candidate digits into odd and even. So, numbers ending in $b$, $d$, $f$ and $h$ must be divisible by 2, 4, 6, 8 respectively, so the previous letters are even "spots". This also means that in accordance to part a), $a$, $c$, $g$, $i$ are odd "spots" (with the same previous meaning). Proceeding to the first four digits $abcd$, because 4 divides them, it also divides $cd$. But because $c$ is odd, $d$ must be 2 or 6$^{*}$. In a similar fashion, $abcdefgh$ is divisible by 8, so $gh$ is divisible by 4. Because $g$ is odd, this leaves us with $h$ to be 2 or 6. Taking into account all the above, we conclude that $b$ and $f$ can be 4 or 8.

Now, going to $abcdef$, this must be divisible by 6. But from the odd parts of the number, $abc$ must be divisible by 3, so $def$ must also be divisible by 3. We already know from the above that $d$ is 2 or 6, $e = 5$ and $f$ is 4 or 8. So, from the four numbers that are formed ($def$ must be divisible by 3), only 258 and 654 survive. Also, according to the above analysis, whatever values $d$ and $f$ take, the values of $h$ and $b$ are fixed accordingly.

Proceeding to $abcdefgh$, this must be divisible by 8, so from divisibility rules, $fgh$ must be divisible by 8. But as we stated earlier, $f$ is even, so it has to be that $gh$ is divisible by 8. Now, $g$ is odd (excluding of course the value 5) and $h$ can be 2 or 6 as we stated earlier, so from all the forming numbers for $gh$ only 4 numbers (16,32,72,96) survive.

From this point on, it would be a good idea to create some sketches for each situation.
So far:

$_4_25816_0$ or $_4_25896_0$ and $_8_65432_0$ or $_8_65472_0$ are the four candidate patterns.

Starting from the first two, we examine the first three digits. Now, we know that $a$ and $c$ can take all the odd values from 1 to 10 except the value 5. Also, $b = 4$ and $abc$ divisible by 3. Taking also into account the constraint a), we have twelve candidate numbers for $abc$, but only 147 and 741 survive the divisibility by 3. So:

$14725816-0$, $14725896-0$, Now, we turn our attention to the divisibility by 7 for $abcdefg$ and we see that it fails for both. So, we can safely exclude the first two candidate patterns.

Now, we examine the last two patterns. In the same fashion as before we examine the first three digits pattern: $abc$. Now, $a$ and $c$, can take the same values as in the previous case of the first two patterns, but $b = 8$ here. Also, taking into account divisibility by 3, constraint a) and the fact that $g$ can be only 3 or 7 (according to the form of the examined patterns), from the twelve possible values, only six survive: 183, 189, 381, 789, 981, 987.

Now, we examine each one of the formed patterns:

$18365432-0$ fails due to repetition of 3

$18965432-0$ fails to the divisibility of $1896543$ by 7

$38165432-0$ fails due to repetition of 3

$78965432-0$ fails to the divisibility of $7896543$ by 7

$98165432-0$ fails to the divisibility of $9816543$ by 7

$98765432-0$ fails to the divisibility of $9876543$ by 7

$18365472-0$, $18965472-0$, $78965472-0$, $98165472-0$, $98765472-0$ fail to the divisibility of the first seven digits by 7.

The only surviving pattern is $38165472-0$. We can now simply complete it with the missing digit 9, so the number is $3816547290$.
$^{*}$

This is trivial but for completeness, the candidate numbers for $cd$ (given the constraints) are 12, 14, 16, 18, 32, 34, 36, 38, 72, 74, 76, 78, 92, 94, 96, 98. From these, only 12, 32, 72, 92, 16, 36, 76, 96 survive the divisibility by 4, so $d$ can be 2 or 6. In the next, for brevity, I will omit showing similar cases.

 

[QuantumQuest]

I'll also give some missing explanation - sorry but my head spun around a bit during posting ;)

Starting from the first two, we examine the first three digits. Now, we know that $a$ and $c$ can take all the odd values from 1 to 10 except the value 5. Also, $b=4$ and $abc$ divisible by 3. Taking also into account the constraint a), we have twelve candidate numbers for $abc$, but only 147 and 741 survive the divisibility by 3. So:

$14725816−0$ $14725896−0$

741 survives the divisibility by 3, but if we insert it into the above pattern in position $abc$, both $74125816-0$ and $74125896-0$ fail to the divisibility by 7. So, we're left with the above two patterns beginning with 147 in $abc$.

 

[EnumaElish]

For the last question, "Show that for 0<λ<1 and α,β≥0 holds ...", isn't there a short proof using a well-known single-variabled function and its properties?

 

[micromass]

For the last question, "Show that for 0<λ<1 and α,β≥0 holds ...", isn't there a short proof using a well-known single-variabled function and its properties?

It's a question for high schoolers.

 

[ibkev]

Nice solution @Biker! I was working on finding the area of the red spots too but like the others took an algebraic approach. I got to the point where all I had left to solve was:

$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx = \int \sqrt {x(10 - x)} \ dx $$

This integral is not as easy as it looks... LOL!

 

[Pepper Mint]

That looks nice actually.
It is $$ \int \sqrt {25 - (x - 5)^2}\ dx$$
not
$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx$$

So if you let $$u=x-5$$ then $$du=dx$$
You set it again by letting $$u=5sint$$ and go on with with it.

 

[ibkev]

It is $$ \int \sqrt {25 - (x - 5)^2}\ dx$$
not
$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx$$

Tru dat - whoops! On the bright side, the $x^2-5$ term is a Latex typo but it seems I can't fix it anymore.

So if you let $u=x-5$ then $du=dx$. You set it again by letting $u=5sint$ and go on with with it.

Thanks for the tip - I'm in the process of brushing up on my integration techniques.

 

[Chestermiller]

I almost have Problem 4 solved, but I need some help evaluating an integral. This is basically the last step:

$$\int{\frac{\sec^2{\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}d\theta}$$

 

[Charles Link]

I almost have Problem 4 solved, but I need some help evaluating an integral. This is basically the last step:

$$\int{\frac{\sec^2{\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}d\theta}$$

I looked at problem #4 just now, and I wrote a vector equation for $d \vec{R}/dt$ which can be split into two components: $dx/dt$ and $dy/dt$ both of which had the same, but rather complex denominator. I then got a little clever and wrote $dy/dx=(dy/dt)/(dx/dt)$ which gave something rather simple: $dx/x=dy/(y-v_o t )$. I thought I should be able to integrate this, (natural logs), but it isn't giving sensible results. editing... And I just figured out what the problem is with this method=The t needs to be replaced by some function of x and y in order for this to be valid...t is not a constant.

 

[Chestermiller]

I looked at problem #4 just now, and I wrote a vector equation for $d \vec{R}/dt$ which can be split into two components: $dx/dt$ and $dy/dt$ both of which had the same, but rather complex denominator. I then got a little clever and wrote $dy/dx=(dy/dt)/(dx/dt)$ which gave something rather simple: $dx/x=dy/(y-v_o t )$. I thought I should be able to integrate this, (natural logs), but it isn't giving sensible results. editing... And I just figured out what the problem is with this method=The t needs to be replaced by some function of x and y in order for this to be valid...t is not a constant.

Right. If I have some time today, I'll write out what I have done so far to get to the point that I'm at.

 

[person_random_normal]

6) Thanks to Math_QED Consider the integrals I and J.

I=π2∫0sinxcosxx+1dxI=∫0π2sin⁡xcos⁡xx+1dxI = \int\limits_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{x+1}dx
J=π∫0cosx(x+2)2dxJ=∫0πcos⁡x(x+2)2dxJ = \int\limits_0^{\pi}\frac{\cos x}{(x+2)^{2}}dx

What is I in function of J?

https://mail.google.com/mail/u/0/?u...68608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662

https://mail.google.com/mail/u/0/?ui=2&ik=8afd5aa52a&view=fimg&th=15794b3f1927dbe0&attid=0.1&disp=inline&realattid=1547350341991792640-local0&safe=1&attbid=ANGjdJ9AL_gRE5sXmUYaevAhKSEVWeCDGMrM2iIlWdWoI_HQXuUtcdRj5gRNTsNVhx1fD7uUlFXVMOekFVnhgzPXCNVO8-rWTr1U5G2r1pmTcTPjii6g4adAZZBM6Ic&ats=1475668608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662
Is it right ??

 

[mfb]

You cannot link to your private mails. Even if you can see the image (I don't know), no one else can. You can use the "Upload" button to upload images here.

 

[Charles Link]

Continuing problem #4 with my post 39 above, I think I now have most of the solution: Writing it as $x(dy/dx)=y-v_o t$ and letting $dy/dx=tan(\phi)$ and taking $d/ds$ on both sides gives $(dx/ds)tan(\phi)+x \, d \, tan(\phi)/ds=dy/ds-v_o/v_1$. Now $dx/ds=cos(\phi)$ and $dy/ds=sin(\phi)$. This gives
$x sec^2(\phi)(d \phi/ds)=-v_o/v_1$ Now $ds=dx/cos(\phi)$ so that $d \phi/cos(\phi)=-(v_0/v_1) dx/x$. The solution of this is $x=x_o e^{(-v_o/v_1)ln|sec(\phi)+tan(\phi)|}$. It will need a little more work to finish up, but I think I'm almost there. editing... $\\$ $x=x_o |sec(\phi)+tan(\phi)|^{-v_o/v_1}$. ( $v_o$ is speed of good ship; $v_1$ is speed of pirate ship.) $\\$ Using $x (dy/dx)=y-v_o t$ we have $\\$ $y=v_o t +x \, tan(\phi)$. $\\$ If $v_o=v_1$ we have $\\$ $y=v_o t -x_o (|tan(\phi)/(sec(\phi)+tan(\phi))|)$ $\\$ where $\phi$ goes from $\pi$ to $\pi/2 +\delta$ in this problem. We see in this case that $y$ is always less than $v_o t$ so it doesn't catch up to the other ship. $\\$ In our equation for x above, we can put in for $tan(\phi)=(y-v_o t)/x$ and we can solve for $sec(\phi)$ as well to get an equation with just x and y. Result is $x=x_o|-\sqrt{((y-v_o t)/x)^2+1)}+(y-v_o t)/x|^{-v_o/v_1}$. I still have a "t" in the equations, so I don't quite have just x and y in the equation.

 

[Chestermiller]

PROBLEM 4

I did something very similar to Charles Link.

Parametric equations for coordinates of pirate ship as a function of time:
$$\frac{dx}{dt}=\frac{(x_0-x)}{\sqrt{(x_0-x)^2+(vt-y)^2}}V$$
$$\frac{dy}{dt}=\frac{(vt-y)}{\sqrt{(x_0-x)^2+(vt-y)^2}}V$$

Change of variable:
$$vt-y=r\sin{\theta}$$
$$x_0-x=r\cos{\theta}$$
In these equations, r(t) is the distance between the pirate ship and the merchant ship at time t, and $\theta (t)$ is the direction of the pirate ship velocity relative to the x axis.

If we substitute these variable changes into the differential equations, we obtain:
$$\frac{dr}{dt}=v\sin{\theta}-V\tag{1}$$
$$r\frac{d\theta}{dt}=v\cos{\theta}\tag{2}$$

If we divide Eqn. 1 by Eqn. 2, we get:
$$\frac{d\ln{r}}{d\theta}=\tan{\theta}-\frac{V}{v}\sec{\theta}$$
The solution to this equation, subject to the initial condition $r=x_0$ @ $\theta = 0$, is:
$$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
Substitution of this into Eqn. 2 gives:
$$\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{V/v}}=\frac{vdt}{x_0}$$

That's where I'm stuck for now. I don't know how to evaluate the integral on the left-hand side.

 

[Chestermiller]

The solution to part (c) of Problem 4 is easy. For, V/v = 1, the distance of the pirate ship from the merchant ship in the end is equal to the distance r evaluated to $\theta = \pi/2$ in the equation: $$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
The value is $$r=\frac{x_0}{2}$$

 

[Charles Link]

The solution to part (c) of Problem 4 is easy. For, V/v = 1, the distance of the pirate ship from the merchant ship in the end is equal to the distance r evaluated to $\theta = \pi/2$ in the equation: $$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
The value is $$r=\frac{x_0}{2}$$

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

 

[Chestermiller]

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

If I had an expression for the integral, I could get t, which would then give us everything we need. But so far, I've only been able to solve the integral for V/v = 3. In that case, the distance the pirate ship had to travel to catch the merchant ship is (9/8)x₀. The tangential distance the pirate ship travels in any of the cases is just Vt.

 

[Charles Link]

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[ (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}]$. This gives $\\$ $y=(1/2)x_o[(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}]+x_o v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration. I've got a wrong sign or two somewhere, but I get that at x=0, $y=x_o v_0 v_1/(v_1^2-v_o^2)$. This gives distance traveled by the pirate ship is D=x_o v_1^2/(v_1^2-v_o^2) ##.

 

[Twigg]

#7:

Per the given diagram, $X = -b\sin\theta$ defined on the interval $[-\pi,\pi]$.. Normalization of the probability density function yields $\rho = \frac{1}{2*\pi}$, as $\int_{-\pi}^{\pi} \frac{1}{2\pi} d\theta = \frac{\pi - (-\pi)}{2\pi} = 1$. $$\langle X \rangle = \int_{-\pi}^{\pi} \rho X d\theta = \frac{-b}{2\pi}\int_{-\pi}^{\pi}\sin\theta d\theta = 0$$ This last step (=0) can be justified since sine is an odd function, or since the integral of sine is cosine which is periodic with a period of $2\pi$. Either argument is sufficient. $$Var(X) = \langle X^{2} \rangle - \langle X \rangle ^{2} = \langle X^{2} \rangle = \int_{-\pi}^{\pi} \rho X^2 d\theta = \frac{b^2}{2\pi} \int_{-\pi}^{\pi} \sin^{2} \theta d\theta = \frac{b^{2}}{2\pi} (\frac{\theta}{2} - \frac{\sin(2\theta)}{4})|_{-\pi}^{\pi})$$

The integral of sine squared I took from an integral table found here: http://integral-table.com/. The $\sin(2\theta)$ term makes no difference because it is a periodic function with a period of $\pi$, and the interval from $-\pi$ to $\pi$ is a distance of $2\pi$. By its periodicity, $\sin(2\theta)$ takes the same value at both points. That leaves, $$Var(X) = \frac{b^{2}}{2\pi} (\frac{\theta}{2})|_{-\pi}^{\pi} = \frac{b^{2}}{2\pi} (\frac{\pi - (-\pi)}{2}) = \frac{b^{2}}{2} $$

 

[Erland]

Some preliminary work on Advanced Problem 3.

Let $\frak L$ be the set of all generalized limits. We know from previous challenges that there are several distinct generalized limits. Let $L_0$ and $L_1$ be two distinct generalized limits. Then, there is a sequence $(y_n)_n\in X$ such that $L_0((y_n)_n)\neq L_1((y_n)_n)$. For each $t\in[0,1]$, define $L_t$ on $X$ by $L_t((x_n)_n)=(1-t)L_0((x_n)_n)+tL_1((x_n)_n)$ for all $(x_n)_n\in X$. It is easy to prove that $L_t$ is a generalized limit. This holds for all $t\in [0,1]$ ($L_0$ and $L_1$ will be the same as before). We see that the $L_t(y_n)_n$:s will be distinct for distinct $t\in[0,1]$, so the family $\{L_t\,|\,t\in[0,1]\}\subseteq\frak L$ has the same cardinality as $[0,1]$, i.e. $\frak c$ (continuum).
Thus, $\text {card}\,\frak L\ge \frak c$.

On the other hand, $X$ can be viewed as a subset of the set of all functions from $\mathbb N$ to $\mathbb R$, and the latter set has the cardinality $\frak c^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\frak c$. Since $X$ contains all real constant sequences, $\text {card} \,X=\frak c$.
Next, $\frak L$ is a subset of the set of all functions from $X$ to $\mathbb R$, and the latter set has the cardinality $\frak c^\frak c=(2^{\aleph_0})^\frak c =2^{\aleph_0\cdot\frak c}=2^\frak c$.
Thus $\text{card}\,\frak L\le 2^\frak c$.

So, ${\frak c}\le \text{card}\,\frak L\le 2^\frak c$.

Now, if we could prove in ZFC that ${\frak c}< \text{card}\,\frak L< 2^\frak c$, this would contradict the Generalized Continuum Hypothesis (GCH). But this would imply that ZFC is inconsistent (GCH is not disprovable in ZFC if ZFC is consistent). Since ZFC is not known to be inconsistent, and there should be a proof in ZFC that ${\text {card}}\,\frak L$ equals a specific cardinal (otherwise, our problem would be currently unsolved, which it obviously is not), this means that either $\text {card}\,\frak L=\frak c$ or $\text{card}\,\frak L=2^\frak c$ is provable in ZFC.

But which of them? I don't know, for the moment. I am inclined towards the latter one, since I don't think that the restrictons given in the definition of generalized limit are sufficient to shrink the set of all functions from $X$ to $\mathbb R$ down from cardinality $2^\frak c$ to $\frak c$.
But don't know this... I'm working on it...

 

[Chestermiller]

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx= (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}$. This gives $\\$ $y=(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}-2 v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration.

What does this predict for y at x = x0 when v1/v0 = 3?

 

[micromass]

$X = -b\sin\theta$

Are you sure of that?

Also, note that I made a correction of the problem.

 

[Twigg]

Are you sure of that?

Also, note that I made a correction of the problem.

Whoops! Thanks for the pointer.

 

[micromass]

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx= (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}$. This gives $\\$ $y=(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}-2 v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration.

Shouldn't $(0,0)$ be on the curve?

 

[Charles Link]

Shouldn't $(0,0)$ be on the curve?

I made a couple of algebraic mistakes. I will go back and edit my original if I still can. I had to go somewhere for about two hours, so I hurried the result and didn't get the chance to finish it.

 

[micromass]

I made a couple of algebraic mistakes. I will go back and edit my original if I still can. I had to go somewhere for about two hours, so I hurried the result and didn't get the chance to finish it.

No please don't edit your post after you have received replies. It makes it much harder to read. Just make a new post with the correct version.

 

[Charles Link]

No please don't edit your post after you have received replies. It makes it much harder to read. Just make a new post with the correct version.

I made too many mistakes in post #48. Here I will start with my equation from post #43 $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_1-v_o)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

 

[micromass]

I made too many mistakes in post #48. Here I will start with my equation from post #43 $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_o-v_1)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

Why would $(x_0,0)$ be on the curve?

 

[Charles Link]

Why would $(x_0,0)$ be on the curve?

I have the good ship sailing up the y-axis (starting at (0,0) and the pirate ship starting at $(x_o,0)$.

 

[micromass]

I have the good ship sailing up the y-axis (starting at (0,0) and the pirate ship starting at $(x_o,0)$.

That is not what I wrote in the OP.