# Toy boat on a string

Little Johnny Red is standing on the edge of a lake, and the place where he stands is $1$ yard above the water level. Johnny's toy boat is floating in the lake, and the boy is pulling it ashore with a string (that is more than $2$ yards long). When Johnny pulls in $1$ yard of string, will the boat advance towards the shore by (i) more than one yard, (ii) exactly one yard, or (iii) less than one yard?

(Note: This puzzle is tagged as [math]. The solution does neither require lateral thinking nor physics.)

• yard vs meter? or this is kinda conversion question? :) – Oray Mar 17 '16 at 12:08
• Where is the string tied? Is it exactly at the water level, or is it tied to the mast at some height above the water? Also, is he holding the string with his feet (i.e. at the ground), or at some height above where he is standing? – dpwilson Mar 17 '16 at 12:17
• @Gamow could you clarify if it's possible for the rope to be less than 2 yard to begin with? In those cases the boat would reach the shore and would be lifted in the air I guess – Ivo Beckers Mar 17 '16 at 12:51
• @Ivo Beckers: The string should be at least 2y at the beginning, so that the situation can take place. – Gamow Mar 17 '16 at 12:59
• A shame about the two-yards stipulation. I liked the complexity of it. – goodguy5 Mar 17 '16 at 13:30

Proof by self-explanatory picture:

The answer is on the pic below:

• Ha -- I had a very similar idea and we posted almost the exact same time =). – Tyler Seacrest Mar 17 '16 at 14:49

It has to be (mathematically)

(i) This is our boat and geometry of it.

First of all we know that

$x-y=1$

since the Little Johhny pulled 1 yard the boat and he is 1 meter above from the sea level.

$b^2+1=y^2$ and $1+a^2+2ab+b^2=x^2$

and these are the simple the Pythagorean theorem of an triangle, so by using both equations;

$x^2-y^2=(x-y)(x+y)=1+a^2+2ab+b^2-(b^2+1)=a^2+2ab$

so;

$x+y=a(a+2b)$

The question is asking about $a$ whether if greater, less or equal to $1$;

1. $y$ has to be greater than $1$ since $b$ should be greater than $0$.
2. $x$ has to be greater than $2$ since $x-y=1$.

Therefore,

$a(a+2b)>3$

the largest value of $a$ is possible only when $b\rightarrow0$, which makes $x$ and $y$ value as $2$ and $1$ sequently, that makes $a$ value as;

$a^2=3$ or $a=\sqrt3$

the smallest value of $a$ is possible only while $b\rightarrow \infty$ that makes $a$ value;

$a\geqslant1$

so the range of a becomes;

$\sqrt3 > a\geqslant1$

• I'm not sure what you've proved; if $b=150$, then your final equation only shows that $a\geq0.01$ or so. – Zandar Mar 17 '16 at 12:55
• @Zandar there are some typos I am on it :) – Oray Mar 17 '16 at 12:56
• @Oray is there reference on x-y = 1? – Alex Mar 17 '16 at 14:28
• @Alex x-y=1 is the total length of the string which John pulled. – Oray Mar 17 '16 at 14:29
• @Oray ok thanks I was thinking you meant it's part of the geometry – Alex Mar 17 '16 at 14:31

More than 1 yard.

Imagine if instead of pulling the boat in by 1 yard of rope, the rope broke 1 yard from the end. The 1 yard falls to the water (pivoting at the original boat location), and the remainder's tip falls to the water. As the shortest distance was the original rope, the two tips have to move away from each other. Hence, more than 1 yard. Ignoring catenary curvature, and momentum of the boat, I would say

More than a yard.

Because

Imagine Johnny's hand is directly above a point A on the shore. Boat to A and A to Johnny's hand and Johnny's hand to the boat forms a right-angled triangle with the string as hypotenuse. Imagine that Johnny is 0 yards tall and the string is 2 yards in length. The boat is $sqrt(2^2 - 1^2) = sqrt(3)$ from the shore. When Johnny reels in a yard of string, the boat is right at the bank. It has traveled $sqrt(3)$ yards, which is more than one.

More generally,

If the distance from Johnny's hand to A is x and the length of the string is y then the square of the distance of the boat from the shore is $sqrt(y^2 - x^2)$. Because the bank is raised, we know both that x is constant and > 0. Therefore the distance from the shore must decrease at a rate strictly greater than the decrease in the length of the string.

Just to note,

We don't actually know the height of Johnny's hand above water level. The point is only that since the shore is a yard above water level and Johnny is standing, then the distance above water level is somewhere between 1 yard and however tall it is reasonable for Johnny to be. The point being that under any reasonable definition of the word "standing" and reasonable geometry of Johnny (e.g. his arms don't hang below his feet) we know that the length of the string is strictly greater than the distance of the boat to shore. So, one more try at clarity.

Call the length of the string y and the distance from shore x. The height of Johnny's hand above water is a constant $n >= 1$. We have:
$y^2 = x^2 + n^2$
When y decreases by 1, the left hand side of the equation decreases by 2y - 1
the right hand side must decrease by more than 2x -1 to maintain equality since y is strictly greater than x.

• if the boat is 1y away from the shore, how can it travel more than 1y ? Sorry if this is a strange question, but I don't understand – Marius Mar 17 '16 at 12:31
• @Marius It reaches the shore before you pull the rope 1y. I think starting with a hypotenuse of 2y might be a better example, though. – Zandar Mar 17 '16 at 12:33
• @question_asker. I was quoting from the answer "...and the boat is 1 yard from the shore" – Marius Mar 17 '16 at 12:35
• @Marius ah, gotcha – question_asker Mar 17 '16 at 12:36

The boat is pulled

more than one yard.

Suppose the boat begins $10$ yards from shore. The string forms a hypotenuse of a right triangle with legs $10$ yards and $1$ yard, resulting in a length of $\sqrt{10^2+1^2}=\sqrt{101}$. When the string is pulled one yard, the hypotenuse is reduced by $1$ to $\sqrt{101}-1$. We now have a new right triangle with legs $x$ and $1$, yielding $\sqrt{101}-1=\sqrt{x^2+1}$. Rearranging for yields $x^2=101-2\sqrt{101}$, so $x\approx{}8.994$. Therefore,

the boat has been pulled ever so slightly more than one yard.

More than 1yard.

Why?

Let's take a simple case. The rope is 6y long. this means the boat is $\sqrt{6^2 - 1^2} = \sqrt{35}$ that is aproximately $5.916$ yards from the shore.
Pull 1y of rope, the rest of the rope is 5y. The boat is $\sqrt{5^2 - 1^2} = \sqrt{24}$ that is aproximately $4.899$ yards from the shore.
The difference is more than 1y.

• That's not the distance travelled by the boat - the triangles before and after pulling are not similar. – Zandar Mar 17 '16 at 12:31
• @Zandar. Hmmm...true. I'll fix it in a few minutes. Thanks for the heads up – Marius Mar 17 '16 at 12:33
• @Zandar. Fixed it. – Marius Mar 17 '16 at 13:05

(iii) Less than one yard. The water surface, the string and the quay form a right triangle, with the string being the hypothenuse. Imagine an one-yard-long segment of the hypothenuse casting a "shadow" onto the cathetus, the shadow would be shorter than the original pieve (by the ratio of hypothenuse/cathetus).

This is valid as long as

the string is still longer than sqrt(2) yards; from there on, it is more than one yard (see @Hugh Meyers' answer).

• Actually, the rope should be over 2 yards. If it's less than that, Johnny will start lifting the boat when the rope length goes under 1y. – Marius Mar 17 '16 at 13:46
• He'll lift it at 1y, but for (iii) to be true, a length greater than sqrt(2) does suffice. With a rope length of 2, there would be sqrt(3) yards of water between the boat and the quay. – Verzweifler Mar 17 '16 at 13:50

Define $D(r)$ as the distance from the boat to the shore when the rope is of length $r$. From the Pythagorean theorem, $D(r) = \sqrt{r^2-1}$. Now, if the derivative of this function is always greater than $1$, then $D(r)$ will change faster than $r$; that is, decreasing $r$ by $1$ will always decrease $D(r)$ by more than $1$. This would mean that pulling the rope $1$ yard would always bring the boat more than $1$ yard closer to the shore.

Now, we see that

\begin{align} D'(r) & =\frac{d}{dr}(r^2-1)^\frac{1}{2} \\ & =\frac{1}{2}(r^2-1)^{-\frac{1}{2}} \frac{d}{dr}(r^2-1) \\ & = \frac{2r}{2\sqrt{r^2-1}} \\ & = \frac{r}{\sqrt{r^2-1}}\\ \end{align}

Since $r>\sqrt{r^2-1}$, this derivative is always greater than $1$. Thus when Johnny pulls in $1$ yard of string, the boat will always move more than $1$ yard towards shore.

More than a yard

Because

Imagine the scenario the other way around. The boat is $n$ yards from the shore, so the rope is $\sqrt{1+n^2}$ yards long. Johnny wants to pull the boat in $1$yd, so the rope will be $\sqrt{1+(n-1)^2}=\sqrt{1+n^2-2n+1}$ yards long.

Maths bit

$\sqrt{1+n^2}-\sqrt{1+n^2-2n+1}=x\to 1+n^2-2\sqrt{(1+n^2)(1+n^2-2n+1)}+1+n^2-2n+1=x^2$, which implies $x\lt1$ by the AM-GM inequality.

I would say

More than one yard - even if the yard of string only accounted for less than one yard of distance traveled, surely momentum would carry the boat further forward.

(This answer was posted before multiple changes were made to the question)

• I was thinking this as well. Unfortunately there seem to be a lot of assumptions that can be made that yield different answers. – dpwilson Mar 17 '16 at 12:14
• @dpwilson Oh, I'm positive this wasn't the intended answer, but until it's asserted that this is a lake unlike any that exists in the real world, I'm leaving this here as a totally reasonable and realistic answer, heh – question_asker Mar 17 '16 at 12:18
• If momentum is kept at bay, I trivially would say that for one yard of rope the distance travelled by boat in water will be less than one yard. I'm not good with mathematics but that seems like it to me. – Prashant Mar 17 '16 at 12:18
• After a few hit and trials, I take my words back. It will always be more than a yard. I tested with boat being two yards and four yards in water respectively. – Prashant Mar 17 '16 at 12:31 