Micromass' big high school challenge thread

[micromass]

Here is a thread of challenges made especially for high school students and first year university students. All the following problems can be solved with algebra, trigonometry, analytic geometry, precalculus and single-variable calculus. That does not mean that the question are all easy.

For non-high schoolers, there is my summer challenges: https://www.physicsforums.com/threads/micromass-big-summer-challenge.879072/

RULES:

1. In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2. It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3. If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4. You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

1. Suppose that in each individual of a large population there is a pair of genes, each of which can either $x$ or $X$, that controls eye color: those with $xx$ have blue eyes, while heterzygotes (those with $Xx$ or $xX$) and those with $XX$ have brown eyes. The proportion of blue eyed individuals is $p^2$ and of heterzygotes is $2p(1-p)$ where $0<p<1$. Each parent transmits one of its own genes to the child; if the parent is a heterozygote, the probability that it transmiths the gene of type $X$ is $1/2$.
a) SOLVED BY ProfuselyQuarky Assuming random mating, find the expected proportion of heterzygotes in brown-eyed children of brown-eyed parents. Otherwise stated: consider the population of all brown-eyed people whose two parents are brown-eyed. Find the proportion of heterzygotes in that population.
b) SOLVED BY ProfuselyQuarky Suppose Judy, a brown-eyed child of brown-eyed parents marries a heterozygote, and they have $n$ children, all brown eyed. Find the probability that Judy is a heterzygote and the probability that her first grandchild has blue eyes.

2. a) SOLVED BY MAGNIBORO Prove that
1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{1000000} &lt; 20
b) SOLVED BY MAGNIBORO Find an explicit $n$ such that
1+ \frac{1}{2}+ \frac{1}{3} + ... + \frac{1}{n} &gt; 20

3. Let $p\neq 0$ be a real number. Let $x_1,...,x_n$ be positive real numbers, we define the $p$-mean as
M_p(x_1,...,x_n) = \sqrt[p]{\frac{1}{n}\sum_{i=1}^n x_i^p}
Note that $M_1(x_1,...,x_n)$ is the usual mean. Prove:
a) SOLVED BY Math_QED $\lim_{p\rightarrow -\infty} M_p(x_1,...,x_n) = \min\{x_1,...,x_n\}$. We define this as $M_{-\infty}(x_1,...,x_n)$.
b) SOLVED BY Math_QED $\lim_{p\rightarrow 0} M_p(x_1,...,x_n) = \sqrt[n]{x_1\cdot ... \cdot x_n}$, the geometric mean. We define this as $M_0(x_1,...,x_n)$.
c) SOLVED BY Math_QED $\lim_{p\rightarrow +\infty} M_p(x_1,...,x_n) = \max\{x_1,...,x_n\}$. We define this as $M_{+\infty}(x_1,...,x_n)$.
d) Prove that for all $p,q\in \mathbb{R}\cup \{- \infty,+\infty\}$ has that $p\leq q$ implies $M_p(x_1,...,x_n)\leq M_q(x_1,...,x_n)$.

4. Take rational numbers $\frac{a}{c}<\frac{b}{d}$ with $a,b,c,d\in \mathbb{N}$.
a) SOLVED BY Mastermind01 Prove
\frac{a}{c} &lt; \frac{a+b}{c+d} &lt; \frac{b}{d}
b) Prove that if $bc - ad = 1$, then $\frac{a+b}{c+d}$ is the simplest fraction in $\left(\frac{a}{c},\frac{b}{d}\right)$ in the sense of having the smallest denominator.

5. SOLVED BY Mastermind01 A light ray falling on a straight line is reflected in the following picture:

To find the reflected line, we use that $\theta_1 = \theta_2$.
If a light ray falls on a curve, then it is reflected on the tangent line of the curve:

Given the parabola $y = ax^2 + bx + c$. A light ray comes from $(h,+\infty)$ and follows the vertical line $x = h$. All such light rays are reflected on the parabola and give rise to reflected light rays. Show that the paths of all the reflected light rays are all concurrent (i.e. they all have a point in common). Find the common point.

6. Find
\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + ...}}}}

7. Let $C$ be the unit circle as defined by $x^2 + y^2 = 1$. Let $L_\alpha$ be the line through $(-1,0)$ with slope $\alpha$. A point $(x,y)\in C$ is called rational if both $x$ and $y$ are rational.
a) SOLVED BY Mastermind01 Show that $L_\alpha$ intersects $C$ in a rational point (except for $(-1,0)$) if and only if $\alpha$ is rational.
b) SOLVED BY MAGNIBORO Find all rational points on the circle. In particular, show there are infinitely such points.
c) SOLVED BY MAGNIBORO Find all positive integers $a$, $b$ and $c$ such that they can be combined to form a right-angled triangle.

8. SOLVED BY ProfuselyQuarky In a certain school there are $15$ girls. It is desired to make a seven-day schedule such that each day the girls can walk in the garden in five groups of three in such a way that each girl will be in the same group with each other girl just once in the week. How should the groups be formed each day?

9. SOLVED BY MAGNIBORO A donkey is attached by a rope to a point on the perimeter of a circular field. How long should the rope be in terms of the radius of the field so that the donkey can reach exactly half the field and eat half the grass.

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?

 

[MAGNIBORO]

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

so

b)
if we try to delimit the sum 1/n we can see that

so

 

[Mastermind01]

4. Take rational numbers $\frac{a}{b}<\frac{b}{d}$ with $a,b,c,d\in \mathbb{N}$.

There is no c in the given condition?

Let me try #7a

The equation of L_\alpha is y = \alpha x + \alpha . To satisfy the circle x^2 + (\alpha x + \alpha)^2 = 1 Solving gives us x = \frac{1-\alpha^2}{1+\alpha^2} , y = \frac{2\alpha}{1+\alpha^2}

While x may be rational even if \alpha be not an irational number y can be rational only if \alpha is rational because of the 2\alpha term.

Am I correct?

 

[micromass]

View attachment 103320

Why is this?

If we try to delimit the sum 1/n we can see that
View attachment 103322

Why?

 

[micromass]

There is no c in the given condition?

Let me try #7a

The equation of L_\alpha is y = \alpha x + \alpha . To satisfy the circle x^2 + (\alpha x + \alpha)^2 = 1 Solving gives us x = \frac{1-\alpha^2}{1+\alpha^2} , y = \frac{2\alpha}{1+\alpha^2}

While x may be rational even if \alpha be not an irational number y can be rational only if \alpha is rational because of the 2\alpha term.

Am I correct?

I agree that it follows that if $\alpha$ is rational, then so are $x$ and $y$. But if both $x$ and $y$ are rational, how exactly do you find that $\alpha$ is rational?

 

[Mastermind01]

I agree that it follows that if $\alpha$ is rational, then so are $x$ and $y$. But if both $x$ and $y$ are rational, how exactly do you find that $\alpha$ is rational?

Well combining the equations we get x/y = \frac{1-\alpha^2}{2\alpha} Solving for \alpha gives a rational value.

 

[micromass]

Well combining the equations we get x/y = \frac{1-\alpha^2}{2\alpha} Solving for \alpha gives a rational value.

Sorry, but it's not quite obvious to me how any solution to that equation must be rational.

 

[MAGNIBORO]

Why is this?
Why?

if we compare the 2 sums

a) 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9...

b) 1+1/2+1/2+1/4+1/4+1/4+1/4+1/8+1/8...

We note that a<b , and We stopped at a 2^k-1 term in b we get k

example: 1+1/2+1/2+1/4+1/4+1/4+1/4 = 3 (we stop in 7 term and we get 3)

so 1+1/2+1/3+...+1/(2^k-1) < k or https://www.physicsforums.com/attachments/103320/and the other
if we compare the 2 sums

a) 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9...

b) 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16...

We note that a>b , and We stopped at a 2^k term in b we get k/2 + 1

example: 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8 = 3/2 +1 (we stop in 8 term and we get 3/2 + 1)

so so 1+1/2+1/3+...+1/(2^k) > k/2 + 1 or https://www.physicsforums.com/attachments/103322/

 

[Mastermind01]

Sorry, but it's not quite obvious to me how any solution to that equation must be rational.

2\alpha x = y - y\alpha^2 Using the quadratic formula \alpha = \frac{2x +- sqrt(4x^2 + 4y^2)}{2y} which is \alpha = \frac{2x +- 2}{2y} [/tex] since x^2 + y^2 = 1 . This is rational if x and y are rational.

 

[micromass]

Alright thank you. I'll credit you with the solution as soon as the other parts are solved.

 

[Mastermind01]

Sorry for my bad Latex in the previous post.

#4a

Since a,b,c,d\in \mathbb{N}, bc &gt; ad \implies bc + bd &gt; ad + bd \implies \frac{b}{d} &gt; \frac{a+b}{c+d}
Similarly bc &gt; ad \implies bc + ac &gt; ad + ac \implies \frac{a}{c} &lt; \frac{a+b}{c+d} .

Combining the two gives us

\frac{a}{c} &lt; \frac{a+b}{c+d} &lt; \frac{b}{d}

 

[Mastermind01]

#4b

Trying this one:

bc = ad + 1 \implies b = d \frac{a+b}{c+d} + \frac{1}{c+d} and c \frac{a+b}{c+d} = a + \frac{1}{c+d} Combining the two we get \frac{b}{d} - \frac{a}{c} = 1 \implies bc - ad = cd \implies cd = 1 \implies c = d = 1 (\because<br /> a,b,c,d\in \mathbb{N}) \therefore c + d = 2 which is the smallest possible sum of two natural numbers. Hence the denominator in the fraction is the smallest.

 

[micromass]

#4b

Trying this one:

bc = ad + 1 \implies b = d \frac{a+b}{c+d} + \frac{1}{c+d} and c \frac{a+b}{c+d} = a + \frac{1}{c+d} Combining the two we get \frac{b}{d} - \frac{a}{c} = 1 \implies bc - ad = cd \implies cd = 1 \implies c = d = 1 (\because<br /> a,b,c,d\in \mathbb{N}) \therefore c + d = 2 which is the smallest possible sum of two natural numbers. Hence the denominator in the fraction is the smallest.

This can't possibly be correct. Take $a=4, b=3, c = 3, d = 2$. Then $bc - ad = 9 - 8 = 1$. But $c+d = 5\neq 2$.

 

[Mastermind01]

I found my mistake.

 

[member 587159]

For 9, are numerical methods allowed?

 

[micromass]

No.

 

[member 587159]

3c) https://3.bp.blogspot.com/-JbJ9MKNf...yD4nWkmKJ08yd0QCLcB/s1600/20160717_135441.jpg

If this is correct, I will try 3a) later.

 

[micromass]

You might want to go in detail as in how the last big limit is $1$.

 

[member 587159]

You might want to go in detail as in how the last big limit is $1$.

Each invidual term under the root except for the +1 term goes to zero because (xi/xs)^p with xs>xi goes to zero as p goes to infinity.

 

[micromass]

Each invidual term under the root except for the +1 term goes to zero because (xi/xs)^p with xs>xi goes to zero as p goes to infinity.

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

 

[ProfuselyQuarky]

8. In a certain school there are $15$ girls. It is desired to make a seven-day schedule such that each day the girls can walk in the garden in five groups of three in such a way that each girl will be in the same group with each other girl just once in the week. How should the groups be formed each day?

So if there are three girls $1$, $2$, $3$ in one group once a week, does that mean that girl $1$ can't be in the same group as $2$ or $3$ (with another third girl) or both $2$ and $3$?

Like, can there be $1$, $2$, $3$ and then on another day within the same week $1$, $2$, $4$?

 

[member 587159]

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

Here is the limit you were asking for.

https://1.bp.blogspot.com/-Ak0-OTqN...0vsGjHJHL4RAK2wCLcB/s1600/20160717_154618.jpg

 

[micromass]

Here is the limit you were asking for.

https://1.bp.blogspot.com/-Ak0-OTqN...0vsGjHJHL4RAK2wCLcB/s1600/20160717_154618.jpg

You can't have $p$ outside the limit if that is your limit variable.

 

[micromass]

So if there are three girls $1$, $2$, $3$ in one group once a week, does that mean that girl $1$ can't be in the same group as $2$ or $3$ (with another third girl) or both $2$ and $3$?

Like, can there be $1$, $2$, $3$ and then on another day within the same week $1$, $2$, $4$?

$1,2,3$ cannot happen in the same week as $1,2,4$ since then $1$ and $2$ walked twice with each other. The idea is that $1$ walks with every girl once.

 

[late347]

For 9, are numerical methods allowed?

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.

 

[late347]

for problem 6, this same type of problem was recently talked about in the pre-calculus subforum for math homework.

 

[member 587159]

You can't have $p$ outside the limit if that is your limit variable.

Even if I would solve the limit and show it's equal to 1, there is still another problem. What if there are 2 elements a = b = max{x1,x2, ..., xn}. I suppose that can't happen because {x1, x2, ..., xn} is a set.

 

[member 587159]

You can't have $p$ outside the limit if that is your limit variable.

I should have seen that!

Here is my final bid: https://2.bp.blogspot.com/-m8F8FBOn...XMvKfDMWopLWKmwCLcB/s1600/20160717_194309.jpg

 

[micromass]

I should have seen that!

Here is my final bid: https://2.bp.blogspot.com/-m8F8FBOn...XMvKfDMWopLWKmwCLcB/s1600/20160717_194309.jpg

It's right, except the thing you noticed earlier: that there might be multiple $x_i$ such that $\frac{x_i}{x_s} = 1$. In that case, you'll end up with something like $\lim_{p\rightarrow +\infty} \frac{1}{p} \log(n)$. But this doesn't change the answer of course.

Also, since you're going to study math at university, I just want to warn you that $\text{ln}$ to denote basis $e$ is a very high school thing to do. Real mathematicians use $\log$ to denote basis $e$. Nobody cares about base $10$ anyway.

 

[member 587159]

It's right, except the thing you noticed earlier: that there might be multiple $x_i$ such that $\frac{x_i}{x_s} = 1$. In that case, you'll end up with something like $\lim_{p\rightarrow +\infty} \frac{1}{p} \log(n)$. But this doesn't change the answer of course.

Also, since you're going to study math at university, I just want to warn you that $\text{ln}$ to denote basis $e$ is a very high school thing to do. Real mathematicians use $\log$ to denote basis $e$. Nobody cares about base $10$ anyway.

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