Challenge Micromass' big high school challenge thread

[Mastermind01]

#5

The parabola $y = ax^2 + bx + c$ has a critical point at $\frac{-b}{2a}$ . The normal to the tangent at that point is $x=\frac{-b}{2a}$. Any line coming from $(\frac{-b}{2a} , \infty)$ is reflected back along the same line.Let the equation of any other line coming from $(h,\infty)$ be $x=\frac{-b}{2a}+ k$. This line makes an angle of $\tan (\theta)=(-2ak)$ (Calculated) . The reflected line makes the same angle with the normal. The reflected line also makes an angle of $2\theta$ with the original line and thus is not parallel to it $\therefore$ it intersects with the reflected line $x=\frac{-b}{2a}$ . The point of concurrency can be found by solving the equation of the reflected line (given by $y = \frac{-b^2}{4a} + ak^2 + c - \frac{1}{2ak}(x + \frac{b}{2a}- k)$ and $x=\frac{-b}{2a}$. The point turns out to be the focus $(\frac{-b}{2a}, \frac{-b^2}{4a} + \frac{1}{4a}+ c)$

 

[MAGNIBORO]

in the problem 9.
If the solution is a root of the equation, but is numerical like $$x=\tan \left( x \right) $$

is a valid solution or must express in its exact form?

 

[late347]

in the problem 9.
If the solution is a root of the equation, but is numerical like $$x=\tan \left( x \right) $$

is a valid solution or must express in its exact form?

donkey is attached at any random point located upon the circle-field's circumference. (red point is chosen in my image)

This donkey seems to be attached with a rope. (ok, understood)

The problem is that the rope must be "taut rope" (stretched)

This is because... Donkey cannot eat more grass than half, donkey cannot eat less than half the grass... Donkey must eat exactly half of the grass.

a.) Is the donkey going to eat the maximum reachable value of grass? (the donkey can only reach maximum length of the rope, obviously)

b.) Or can the donkey shorten its own rope "voluntarily", and eat less grass than that which the rope allows the donkey to eat in the first place? (I think the donkey cannot do this because... the eatable grass must be half of the area)

c.) I guess, the donkey must first eat at maximum reach, until grass is eaten. Then the donkey moves closer to the attachment point and eats again etc...until all the reachable grass is eaten empty.
But I don't know how to calculate it from there...

EDIT: ROPE is actually the larger radius. I think this rope must be longer than the original radius of the field, because otherwise if the rope was smaller or equal to radius, then I think the area drawn by the rope will be too small. (less than 50% of the circle?)

looks like I need to brush up on the analytic geometry, but basically the bigger circle and smaller circle create an intersectional area in the middle.
The area is like an asymmetric lense (the crosssectional image of a lense)I would like suggestion if my image interpretaion is on the correct track, I presume that the problem requires analytic geometry and I was reading yesterday on wolfram mathworld, about this intersectional area thing, and it seems that analytic geometry is involved.

I cannot see an easy traditional geometric answer, because no angles are known. and no certain length is known...looks like problem 10 is easier, but I have alreadey done same problem before, so someone else can answer that one.

 

[MAGNIBORO]

donkey is attached at any random point located upon the circle-field's circumference. (red point is chosen in my image)

This donkey seems to be attached with a rope. (ok, understood)

The problem is that the rope must be "taut rope" (stretched)

This is because... Donkey cannot eat more grass than half, donkey cannot eat less than half the grass... Donkey must eat exactly half of the grass.

a.) Is the donkey going to eat the maximum reachable value of grass? (the donkey can only reach maximum length of the rope, obviously)

b.) Or can the donkey shorten its own rope "voluntarily", and eat less grass than that which the rope allows the donkey to eat in the first place? (I think the donkey cannot do this because... the eatable grass must be half of the area)

c.) I guess, the donkey must first eat at maximum reach, until grass is eaten. Then the donkey moves closer to the attachment point and eats again etc...until all the reachable grass is eaten empty.
But I don't know how to calculate it from there...View attachment 103446


EDIT: ROPE is actually the larger radius. I think this rope must be longer than the original radius of the field, because otherwise if the rope was smaller or equal to radius, then I think the area drawn by the rope will be too small. (less than 50% of the circle?)

looks like I need to brush up on the analytic geometry, but basically the bigger circle and smaller circle create an intersectional area in the middle.
The area is like an asymmetric lense (the crosssectional image of a lense)I would like suggestion if my image interpretaion is on the correct track, I presume that the problem requires analytic geometry and I was reading yesterday on wolfram mathworld, about this intersectional area thing, and it seems that analytic geometry is involved.

I cannot see an easy traditional geometric answer, because no angles are known. and no certain length is known...looks like problem 10 is easier, but I have alreadey done same problem before, so someone else can answer that one.

I solved the problem but the solution i get is a numerical solution of the ecuation $$
-1/4\,a \sqrt{-{a}^{2}+4}+1/2\,\arcsin \left( 1/2\,a \sqrt{-{a}^{2}+4}
\right) +1/2\,{a}^{2}\arcsin \left( 1/2\, \sqrt{-{a}^{2}+4} \right) =
\pi /4 $$
$a$ is aproxx $1.158728473$
and I do not know if a numerical solution like this is valid

 

[micromass]

I solved the problem but the solution i get is a numerical solution of the ecuation $$
-1/4\,a \sqrt{-{a}^{2}+4}+1/2\,\arcsin \left( 1/2\,a \sqrt{-{a}^{2}+4}
\right) +1/2\,{a}^{2}\arcsin \left( 1/2\, \sqrt{-{a}^{2}+4} \right) =
\pi /4 $$
$a$ is aproxx $1.158728473$
and I do not know if a numerical solution like this is valid

This is ok! Can you show us how you got that?

 

[Mastermind01]

I solved the problem but the solution i get is a numerical solution of the ecuation $$
-1/4\,a \sqrt{-{a}^{2}+4}+1/2\,\arcsin \left( 1/2\,a \sqrt{-{a}^{2}+4}
\right) +1/2\,{a}^{2}\arcsin \left( 1/2\, \sqrt{-{a}^{2}+4} \right) =
\pi /4 $$
$a$ is aproxx $1.158728473$
and I do not know if a numerical solution like this is valid

But shouldn't the answer be something times the radius? The question specifically asks us to find that. Wouldn't the answer vary as the radius of the field changes?

I had got a similar expression but in two variables. I haven't managed to approximate though.

EDIT: I have got the radius to be nearly the same approximation as yours times the original radius. Did you take the original radius as 1?

 

[member 587159]

This is ok! Can you show us how you got that?

You told me I could not use numerical values. And, a must be a function of r I think.

 

[late347]

I was looking at mathworld wolfram alpha
For problem 9. Tjis link seemed promising
http://mathworld.wolfram.com/Circle-CircleIntersection.htmlbut I had a busy day moving from my house packing and carryong boxes.

 

[Mastermind01]

I was looking at mathworld wolfram alpha
For problem 9. Tjis link seemed promising
http://mathworld.wolfram.com/Circle-CircleIntersection.htmlbut I had a busy day moving from my house packing and carrying boxes.

I used that and approximated my answer. Not sure if it counts as a solution though. I think @MAGNIBORO integrated to get the answer - which I think is the correct way.

 

[ProfuselyQuarky]

Just realized that the answer for 1a) is used for 1b)...

The probability that Judy is $Xx$ given that all the other information is true is:

$\displaystyle=\frac{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n}{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n+(\frac{1}{1+2p})}$

Judy's grandchildren can have blue eyes only if her child is $Xx$, also. For that, we've got:

$\displaystyle=(\frac{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n}{(\frac{2p}{1+2p})\cdot (\frac{3}{4})^n+(\frac{1}{1+2p})})(\frac{2}{3})+(\frac{\frac{1}{1+2p}}{(\frac{2p}{1+2p})\cdot (\frac{3}{4})^n+(\frac{1}{1+2p})})(\frac{1}{2})$

The spouse of Judy's child has the option of being $Xx$, $XX$, or $xx$. The problem says that the mating is random, so, in the same order, the chances of these events occurring are $2p(1-p)$, $(1-p)$, and $p^2$.

Thus:

$\displaystyle=(\frac{(\frac{2}{3})(\frac{2p}{1+2p})\cdot (\frac{3}{4})^n+(\frac{1}{2})(\frac{1}{1+2p})}{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n+(\frac{1}{1+2p})})(\frac{1}{4}\cdot 2p(1-p)+(\frac{1}{2})p^2)$

$\displaystyle=(\frac{(\frac{2}{3})(\frac{2p}{1+2p})\cdot (\frac{3}{4})^n+(\frac{1}{2})(\frac{1}{1+2p})}{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n+(\frac{1}{1+2p})})(\frac{1}{2}p)$

$\displaystyle=(\frac{(\frac{4p}{3+6p})\cdot (\frac{3}{4})^n+(\frac{1}{2+4p})}{(\frac{2p}{1+2p})\cdot(\frac{3}{4})^n+(\frac{1}{1+2p})})(\frac{1}{2}p)$

Can't simplify any further. IMO, if it's correct, the answer is horrendous for a probability problem.

 

[ProfuselyQuarky]

I used that and approximated my answer. Not sure if it counts as a solution though. I think @MAGNIBORO integrated to get the answer - which I think is the correct way.

Yes, I believe that you've got to integrate to derive the answer...the very reason why I didn't bother with it :)
-----------------------------------------------------------------------------------------------------------------
I wish I could solve that nested radical, but the answer is already known.

 

[Mastermind01]

Yes, I believe that you've got to integrate to derive the answer...the very reason why I didn't bother with it :)

That is one-way , it is also possible to geometrically derive a formula for the area of intersection of two circles. I've seen it done - I had to look it up though in WolframAlpha.

I wish I could solve that nested radical, but the answer is already known.

Same here.

 

[Charles Link]

#5

The parabola $y = ax^2 + bx + c$ has a critical point at $\frac{-b}{2a}$ . The normal to the tangent at that point is $x=\frac{-b}{2a}$. Any line coming from $(\frac{-b}{2a} , \infty)$ is reflected back along the same line.Let the equation of any other line coming from $(h,\infty)$ be $x=\frac{-b}{2a}+ k$. This line makes an angle of $\tan (\theta)=(-2ak)$ (Calculated) . The reflected line makes the same angle with the normal. The reflected line also makes an angle of $2\theta$ with the original line and thus is not parallel to it $\therefore$ it intersects with the reflected line $x=\frac{-b}{2a}$ . The point of concurrency can be found by solving the equation of the reflected line (given by $y = \frac{-b^2}{4a} + ak^2 + c - \frac{1}{2ak}(x + \frac{b}{2a}- k)$ and $x=\frac{-b}{2a}$. The point turns out to be the focus $(\frac{-b}{2a}, \frac{-b^2}{4a} + \frac{1}{4a}+ c)$

I would like to add an addition to this one if I may. I am not a high school or first year college student (I received an M.S. in physics 35+ years ago). I think the solution really needs an additional step or two: You have angle of incidence =angle of reflection so you can write $tan(\theta_r)=tan(\theta_i)$, so that $tan(\phi_r-\phi_o)=tan(\phi_o-\phi_i).$ ( $\phi$'s are angles of the lines measured w.r.t. to the horizontal). This gives, by the tangent of the difference of two angles formula that $(m_r-m_o)/(1+m_r m_o)=(m_o-m_i)/(1+m_o m_i) \$. Now $m_i=tan(\phi_i) =+ \infty$ and you can solve for $m_r =tan(\phi_r)$ knowing the slope of the normal $m_o=tan(\phi_o)$. You then use point-slope form of line $(y-y_1)/(x-x_1)=m_r$ and find its intersection $y$ with $x=-b/(2a)$. The $x =-b/(2a)$ and $y$ found in this manner is indeed the focal point given in the post. @Mastermind01 's solution is clearly on the right track, but I think it needs this additional step or two. Hopefully @micromass concurs with my comments.

 

[Mastermind01]

I would like to add an addition to this one if I may. I am not a high school or first year college student (I received an M.S. in physics 35+ years ago). I think the solution really needs an additional step or two: You have angle of incidence =angle of reflection so you can write $tan(\theta_r)=tan(\theta_i)$, so that $tan(\phi_r-\phi_o)=tan(\phi_o-\phi_i).$ ( $\phi$'s are angles of the lines measured w.r.t. to the horizontal). This gives, by the tangent of the difference of two angles formula that $(m_r-m_o)/(1+m_r m_o)=(m_o-m_i)/(1+m_o m_i) \$. Now $m_i=tan(\phi_i) =+ \infty$ and you can solve for $m_r =tan(\phi_r)$ knowing the slope of the normal $m_o=tan(\phi_o)$. You then use point-slope form of line $(y-y_1)/(x-x_1)=m_r$ and find its intersection $y$ with $x=-b/(2a)$. The $x =-b/(2a)$ and $y$ found in this manner is indeed the focal pt. given in the post. @Mastermind01 's solution is clearly on the right track, but I think it needs this additional step or two. Hopefully @micromass concurs with my comments.

I did do all of that, I just gave the solutions, to eliminate all the algebraic manipulations.

P.S : I just noticed I had a typo in the original answer, going to fix it now.

EDIT: I can't change it now, the equation of the reflected line would be
$y = \frac{-b^2}{4a} + ak^2 + c - \frac{1- 4a^2k^2}{4ak}(x + \frac{b}{2a}- k)$

 

[Charles Link]

I did do all of that, I just gave the solutions, to eliminate all the algebraic manipulations.

P.S : I just noticed I had a typo in the original answer, going to fix it now.

EDIT: I can't change it now, the equation of the reflected line would be
$y = \frac{-b^2}{4a} + ak^2 + c - \frac{1- 4a^2k^2}{4ak}(x + \frac{b}{2a}- k)$

Very good. I will need to compare it to my calculation, but I think you now have it correct. Very good. ! :-) :-)

 

[member 587159]

Yes, I believe that you've got to integrate to derive the answer...the very reason why I didn't bother with it :)
-----------------------------------------------------------------------------------------------------------------
I wish I could solve that nested radical, but the answer is already known.

I integrated this too but I got something incredibly ugly involving arcsin (just because of the trig substitution) because in my integral I had the variable x and also two parameters a and r. I couldn't find an answer without numerical approach so that's the reason I stopped the exercise.

I have seen the solution to the radical too, I do not think I would have found it myself though. After all, this is an exercise that was in an Indian magazine, and the mathematician (I will not say his name otherwise others might be attracted to cheat, although I have given a huge clue) had to send in the answer himself because nobody found the answer. And then Micromass presents this at high school students!

 

[Charles Link]

On problem #9, I solved this one too without using any references and got the same equation that @MAGNIBORO did in post #51. (I am not eligible for the contest, but I saw it was a very interesting problem.) I was able to compute "a" to two decimal places first trying a=1.25, and then a=1.20, and then when a=1.16 the left side of the equation was nearly the same as the right side. If there is a closed form solution for "a", it would be interesting to see how the result is obtained. It does not appear that any simple algebraic substitution into the equation of post #51 will generate a closed form answer for "a".

 

[MAGNIBORO]

the problem 9.
(I take Radio 1 to make easier the problem)
first start with 2 circles one of Radius 1 center (0;0) and the other of radius R center (0;1) and define 3 points and 2 functions like picture

F(x) is the top part of the small circle, $y= \sqrt{1-{x}^{2}}$

h(x) is the bottom part of the circle of radius r, $y=- \sqrt{{R}^{2}-{x}^{2}}+1$

F is the center of the big circle (0;1)

G is the point 1-R in y axys

I is the point whit 2 circles intersected, h(x)=F(x) or $$- \sqrt{{R}^{2}-{x}^{2}}+1= \sqrt{-{x}^{2}+1}$$
$$({R}^{2}-{x}^{2})-2\, \sqrt{{R}^{2}-{x}^{2}}+1=-{x}^{2}+1$$
$${R}^{2}-{x}^{2}=1/4\,{R}^{4}$$
$$x=1/2\,R \sqrt{-{R}^{2}+4}$$
so I is the point $1/2\,R \sqrt{-{R}^{2}+4}$ in x axys

now we can see that R > 1 because otherwise would have less area than half of the circle
and R<sqrt(2) because otherwise would have more area than half of the circle.
we know the area of the small circle, is pi*1^2 = pi. so the area between the 2 circles must be pi/2, and the area FGI must be pi/4
now the hard part: $$\int_{0}^I \int_{h(x)}^{f(x)} \, dy \, dx= \pi/4$$
$$\int_{0}^{1/2\,R \sqrt{-{R}^{2}+4}} \int_{- \sqrt{{R}^{2}-{x}^{2}}+1}^{ \sqrt{-{x}^{2}+1}} \, dy \, dx= \pi/4$$
$$\int_{0}^{1/2\,R \sqrt{-{R}^{2}+4}}\! \sqrt{-{x}^{2}+1}+ \sqrt{{R}^{2}
-{x}^{2}}-1\,{\rm d}x= \pi/4$$
now I have to admit that I received some help to solve this integral. so will put the result assuming that 1 < R < sqrt(2)
and we Finally get: $$-1/4\,R\sqrt {-{R}^{2}+4}+1/2\,\arcsin \left( 1/2\,R\sqrt {-{R}^{2}+4}
\right) +1/2\,{R}^{2}\arcsin \left( 1/2\,\sqrt {-{R}^{2}+4} \right)= \pi/4$$
and this ecuation have a solution of R=1.158728473...
so the proportion is R=1.158728473... of the field radius.I really liked not having to ask for help to solve that integral and the numerical solution for the ecuation.
but as it was not the main problem decided not to give much importance,
I would like to see a solution that did not have calculus, and if that 1.158728473... have a exact form.
thanks =D

 

[micromass]

I will post an alternative solution after 10 is found.

 

[madihakhan]

Math is always a difficult subject for me but I can see few great math professionals here maybe I could learn from you Guys !

 

[Mastermind01]

Regarding #9

I don't know what alternative method @micromass has in mind but a good way of doing it without calculus would be:

Noticing that the two intersections form an asymmetric lens. We can find the points of intersection of two circles but placing one of them at the origin and another at $(0,a)$ . We can calculate the angle subtended by the common chord on each of the circles using similarity and then find out the area of their respective segments by subtracting the area of the triangle from that of the segment. A formula can be found out for the area of intersection.

Noticing that d (the distance between two centres) is r (the given radius) we get the following equation

$r^2\arccos(\frac{2r^2-a^2}{2r^2}) + a^2\arccos(\frac{a}{2r}) + \frac{1}{2}\sqrt{a^2(4r^2-a^2)} = \frac{\pi r^2}{2}$

Graphing the two sides of the equation independently for various values of $r$ we find that the ratio of the point of intersection (y-coordinate) to that of the original radius $r$ is always $\approx 1.1587$

 

[Charles Link]

Regarding #9

I don't know what alternative method @micromass has in mind but a good way of doing it without calculus would be:

Noticing that the two intersections form an asymmetric lens. We can find the points of intersection of two circles but placing one of them at the origin and another at $(0,a)$ . We can calculate the angle subtended by the common chord on each of the circles using similarity and then find out the area of their respective segments by subtracting the area of the triangle from that of the segment. A formula can be found out for the area of intersection.

Noticing that d (the distance between two centres) is r (the given radius) we get the following equation

$r^2\arccos(\frac{2r^2-a^2}{2r^2}) + a^2\arccos(\frac{a}{2r}) + \frac{1}{2}\sqrt{a^2(4r^2-a^2)} = \frac{\pi r^2}{2}$

Graphing the two sides of the equation independently for various values of $r$ we find that the ratio of the point of intersection (y-coordinate) to that of the original radius $r$ is always $\approx 1.1587$

A very minor correction, but I think the 3rd term in the equation needs a minus sign. If you convert arccosine to arcsine, it agrees precisely with @MAGNIBORO of post #51 (and #68) and also the result that I got.

 

[Mastermind01]

A very minor correction, but I think the 3rd term in the equation needs a minus sign. If you convert arccosine to arcsine, it agrees precisely with @MAGNIBORO of post #51 and also the result that I got.

You're right. Unfortunately I can't change it now.

 

[Charles Link]

You're right. Unfortunately I can't change it now.

It's neat that we're all getting the same answer. (and same equation.) It should be interesting to see what @micromass has for a solution.

 

[late347]

based on private message conversation, it is fairly reasonable to say that no further attempts will be made upon the problems 6, and problem 10.

These two have already been completed by high-schooler participants, so they themselves are not eligible to solve those problems. It seems that those two were actually common textbook exercises, therefore they were common math problems.

problem 4b is the only remaining problem, I think.

 

[ProfuselyQuarky]

@micromass Would you mind posting that "more methodical solution" for #8? I'd be happy with a link.

 

[micromass]

@micromass Would you mind posting that "more methodical solution" for #8? I'd be happy with a link.

I'm sorry, I forgot to respond. Anyway, you should start reading here: https://www.jstor.org/stable/27960002?seq=6#page_scan_tab_contents It's free but you'll need to register. If you read it and are interested in the math behind this, it's projective geometry and finite geometry. It's a popular field of research in mathematics.

 

[ProfuselyQuarky]

I'm sorry, I forgot to respond. Anyway, you should start reading here: https://www.jstor.org/stable/27960002?seq=6#page_scan_tab_contents It's free but you'll need to register. If you read it and are interested in the math behind this, it's projective geometry and finite geometry. It's a popular field of research in mathematics.

Ah! Thanks!

 

[member 587159]

But the root goes to infnity too. So you have $0^{1/\infty} = 0^0$ which is an undetermined form.

Well, I looked at this again, and, when you fill in p -> infinity, you get 1^0 = 1 which is not an indetermined form so I don't think that another calculation of the limit was necessary. If I would be missing something, please enlighten me :)

 

[micromass]

Well, I looked at this again, and, when you fill in p -> infinity, you get 1^0 = 1 which is not an indetermined form so I don't think that another calculation of the limit was necessary. If I would be missing something, please enlighten me :)

Indeed, that is correct.

 

[member 587159]

What's the solution for 3d as I think that nobody found it. I hope I don't ask this question too soon.

 

[Swapnil Das]

My solution to Question 6:

We can start forming a nested radical expression for $x$:

$$|x|=\sqrt{x^2}\\=\sqrt{1+(x-1)(x+1)}\\=\sqrt{1+(x-1)\sqrt{1+x(x+2)}}\\=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}}$$
In the same way if we progress indefinitely, we get the following nested radical expression for $x$:
$$|x|=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{\cdots}}}}}}$$
This expression matches with the form of the given problem , so put $x=3$ to get:
$$\boxed{3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}}$$

 

[ddddd28]

10. The same donkey is now tied with a 50m long rope. The rope is tied to the corner of a 20m by 10m barn. What is the total area that the donkey is capable of grazing?

This is a sketch of the donkey's trail:

The donkey starts from point o.
When the donkey goes to left, it covers the pink area + the blue one.
When the donkey goes to right, it covers the green area+ the blue one.
Therefore, the donkey covers three quarters of 50 m- circle and an additional odd part.
To my opinion, the easiest way to find this area is to calculate the sectors ADC, BEC, the area of the two triangles AOB and ABC.
S ABC = √11*100
By Heron's theorem
(AB= √50 AC=10 BC=20)
S ABO= 100
In order to find the sectors, we need to know the angles DAC and CBE, which can be found by arctan(slope BC) and arctan (AC).
We need to find point c:
The circles' equations:
$(x-2)^2+y^2=9$
$x^2+(y-1)^2=16$
$c( \frac{24+4√11}{10}), \frac{-1+4√11}{5} )$ *** I reduced by 10
$\arctan( \frac{\frac{-1+4√11}{5}-0}{\frac{24+4√11}{10}-2})=54.86°$
Similarly, we get 90°- 21.3°=68.7° for ∠DAC
S ADC= 1600π 68.7°/360°=305.3π
S CBE= 900π 54.86°/360°=137π

S total= S CBE + S ADC +S ABO+S ABC +0.75 2500π= 137π+305.3π+√11*100+100+0.75 2500π = 7711.7 m²

 

[ddddd28]

based on private message conversation, it is fairly reasonable to say that no further attempts will be made upon the problems 6, and problem 10.

These two have already been completed by high-schooler participants, so they themselves are not eligible to solve those problems. It seems that those two were actually common textbook exercises, therefore they were common math problems.

problem 4b is the only remaining problem, I think.

I solved the tenth problem just for fun, please let me know if my solution is correct, and do these problems have a common name? I think such problems of a wrapping rope are very interesting, in sense of trying to generalize them.