Micromass' big high school challenge thread

[micromass]

Alright, I will use the common notation then in the future. Thanks for the remark. For 3a) I have the following: https://3.bp.blogspot.com/-buAaztZp...ZB3aAtuWU8X17DgCLcB/s1600/20160717_201042.jpg . The limit is calculated in the same way as in 3c

Again: you probably want to say $x_i\geq x_s$ at the end and not the strict inequality. But this is ok!

 

[ProfuselyQuarky]

Don't know how to make tables here so I'm going to steal Math_QED's brilliant idea. Here's a starter attempt at #8.

https://65.media.tumblr.com/ae02036e5e43220c38cd1de6c9173ce6/tumblr_oah2a4DrfM1vag9ebo1_1280.jpg

After step 3, it was more of a trial and error sort of process. Perhaps there's a set of rules that can be followed, but I haven't figured it out yet. I look forward to (hopefully) seeing somebody's more rigorous solution since I'm sure that's what micro wants .

 

[micromass]

Don't know how to make tables here so I'm going to steal Math_QED's brilliant idea. Here's a starter attempt at #8.

https://65.media.tumblr.com/ae02036e5e43220c38cd1de6c9173ce6/tumblr_oah2a4DrfM1vag9ebo1_1280.jpg

After step 3, it was more of a trial and error sort of process. Perhaps there's a set of rules that can be followed, but I haven't figured it out yet. I look forward to (hopefully) seeing somebody's more rigorous solution since I'm sure that's what micro wants .

a goes with c both on friday and saturday

 

[ProfuselyQuarky]

a goes with c both on friday and saturday

Oops...The c in that row for Saturday was supposed to be uppercase :/

 

[member 587159]

Again: you probably want to say $x_i\geq x_s$ at the end and not the strict inequality. But this is ok!

Note too that I wrote: xi ∈ {x1, ..., xn} \ {xs}. So I think the strict inequality holds? Or am I missing something?

 

[micromass]

Oops...The c in that row for Saturday was supposed to be uppercase :/

Seems to be all correct! Well done.

https://en.wikipedia.org/wiki/Kirkman's_schoolgirl_problem

If you're willing to learn finite geometry and projective geometry, then I can tell you a more methodical solution.

 

[micromass]

Note too that I wrote: xi ∈ {x1, ..., xn} \ {xs}. So I think the strict inequality holds? Or am I missing something?

You're right.

 

[member 587159]

3b) https://1.bp.blogspot.com/-gJTcl_3i...2IdCVSqhuEX6t-gCLcB/s1600/20160717_213659.jpg

 

[member 587159]

It looks like problems 2, 6, 8, 9 and 10 are slightly easier problems at first glance.
I think those are typical textbook style exercises.

I think the other problems looked quite tough because I'm quite bad and inexperienced
at proofs.

I think that personally 1, 3, and 5 look really toughsimply because there is so many instructions. Like a wall of text. :D

Ive done number 10 type of problems earlier. They seem to be a somewhat common geometry textbook exercise.

Did you encounter those previously? Otherwise feel free to solve one ;)

 

[member 587159]

I think I don't fully understand question 1. The proportion of blue eyed people is p^2 and the proportion of heterozygotes is 2p(1-p). Does that mean there are other people who are not blue or brown-eyed because I think that otherwise p^2 + 2p(1 -p) should be equal to 1 no? Probably this is a dumb question, but I wanted to be sure before I start thinking. Thanks.

 

[micromass]

Sorry, there was a part missing. Check it now!

 

[ProfuselyQuarky]

If you're willing to learn finite geometry and projective geometry, then I can tell you a more methodical solution.

Yes, I'd like to see that solution. Thanks!

 

[Cosmophile]

i try the 2.
a)
if we try to delimit the sum 1/n we can see that

View attachment 103320

I'm not familiar with "delmitting the sum." Where can I find some information on this?

 

[MAGNIBORO]

I'm not familiar with "delmitting the sum." Where can I find some information on this?

I meant to find an "upper and lower limit " to the sum, like $x<{e}^{x}$ or ${x}^{2}<{x}^{2}+3$
Here I explain better:
https://www.physicsforums.com/threa...-school-challenge-thread.879086/#post-5522162

 

[MAGNIBORO]

7 b .
if we know that there are infinitely many Pythagorean Triples of the form $$
(2\,h)^{2}= \left( {h}^{2}+1 \right) ^{2}- \left( {h}^{2}-1 \right) ^{
2}$$
for all h.

so:
$x_{0}$ = ${\frac {{h}^{2}-1}{{h}^{2}+1}}$
to get $y_{0}$ just put $x_{0}$ in ecuation $\sqrt{1-{x}^{2}}$.$y_{0}=\sqrt{1-{\frac { \left( {h}^{2}-1 \right) ^{2}}{ \left( {h}^{2}+1\right) ^{2}}}}= {\frac { \sqrt{ \left( {h}^{2}+1 \right) ^{2}- \left( {h}^{2}-1 \right) ^{2}}}{{h}^{2}+1}}$here replace $\sqrt{ \left( {h}^{2}+1 \right) ^{2}- \left( {h}^{2}-1 \right) ^{2}}$ with $2\,h$
and we get

$y_{0}=2\,{\frac {h}{{h}^{2}+1}}$. and this is a rational number for all integer h.

the point on the circle with coordinates $x_{0}$ and $y_{0}$ is a rational point

 

[late347]

In real-life it's possible that people have green eyes (but I always guessed that green is maybe just a variant of blue eyes genetically, I don't know for sure...)

Also it is quite rare but still it is possible that a person has one eye different colour than the other eye. In real life that is...

Mathematically I suopse we assume either blue or brown?

Problem 1 certainly looked tough, though...
I know that I myself am heterozygote(?)Because my sisters have blue eyes and I have brown eyes.

My mother has blue eyes and my father has brown eyes. But I think that my father's mother must have had blue eyes. Because I remember that my fathwrs father actually had brown eyes. Therefore it seems that my father is heterozugote. And I myself seemingly am heterozygote. Yet my two sisters both of them received blue eyes (lucky!)

I vaguely remember that when both parents are brown eyed. It follows that all children will be brown eyed.

However if one parent is blue eyed and the other parent is brown eyed heterozygote. It follows that there ought to be "higher proportion" of brown eyed and smaller number of blue eyedd kids.

 

[ProfuselyQuarky]

Nevermind. Post deleted.

 

[member 587159]

Just to be clear. In 1) there are

In real-life it's possible that people have green eyes (but I always guessed that green is maybe just a variant of blue eyes genetically, I don't know for sure...)

Also it is quite rare but still it is possible that a person has one eye different colour than the other eye. In real life that is...

Mathematically I suopse we assume either blue or brown?

Problem 1 certainly looked tough, though...
I know that I myself am heterozygote(?)Because my sisters have blue eyes and I have brown eyes.

My mother has blue eyes and my father has brown eyes. But I think that my father's mother must have had blue eyes. Because I remember that my fathwrs father actually had brown eyes. Therefore it seems that my father is heterozugote. And I myself seemingly am heterozygote. Yet my two sisters both of them received blue eyes (lucky!)

I vaguely remember that when both parents are brown eyed. It follows that all children will be brown eyed.

However if one parent is blue eyed and the other parent is brown eyed heterozygote. It follows that there ought to be "higher proportion" of brown eyed and smaller number of blue eyedd kids.

In reality, things are more complex. Two blue-eyed people can have a brown eyed child.

 

[ProfuselyQuarky]

For 1a) we're looking for $Pr($the kid is heterozygote $|$ the parents have brown eyes and their child has brown eyes$)$.

$=\displaystyle\frac {0\cdot(1-p)^4+(1/2)\cdot4p(1-p)^3+1/2\cdot4p^2(1-p)^2}{1\cdot(1-p)^4+1\cdot4p(1-p)^3+(3/4)\cdot4p^2(1-p)^2}$

$=\displaystyle\frac {1/2\cdot4p(1-p)^3+(1/2)\cdot4p^2(1-p)^2}{(1-p)^4+1\cdot4p(1-p)^3+(3/4)\cdot4p^2(1-p)^2}$

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

$=\displaystyle\frac{2p}{1+2p}$

 

[micromass]

That's correct!

 

[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.