Author Topic: University Problem Solving Thread (Read 5720 times) Tweet Share

Ahmad · « **on:** February 28, 2008, 09:39:22 pm »

The purpose of this thread is to generate a lively discussion of mathematics and for FSNers to learn from each other by solving challenging problems!

Anyone can post a problem, but they have to be somewhat problem-solving type problems i.e. not the usual homework problem. I'm not sure about restricting what sort of problems you can pose, for example, posing something that uses sophisticated maths, but for now anything goes.

As the problems get posted I'll edit this post and add them in.

1. Can you tile an 18 x 18 board (i.e. one divided into 324 squares) using only T-shaped pieces each made of 4 squares?

2. Prove that for reals $x + y + z = 1, \; xy + yz + zx \le \frac{1}{3}$ .

3. A bug crawls along the edges of a cube; at each vertex it has probability $\frac{1}{3}$ of going to any adjacent vertex. When it reaches the vertex opposite its starting one, enlightenment is achieved. What is the average number of edges it must crawl on to do this?

4. Any right triangle contains an isosceles triangle whose area is at least $\alpha$ times the area of the original triangle. Find the maximum value for $\alpha$ .

5. Find $\int_{0}^\pi \frac{x}{1 + \cos^2 x} \mbox{d}x$

6. A certain Kingdom of Neobeoland introduces the following family-planning scheme: A couple will keep bearing children until a daughter is born, after which they will stop having children. Assume that they have an equal chance of birthing either gender. What will be the ratio of sons to daughters in this kingdom?

7. Find the smallest prime number which is a factor of 99! - 1.

Ahmad · « **Reply #1 on:** February 29, 2008, 08:43:09 pm »

Problems added, source: Paradox.

Neobeo · « **Reply #2 on:** March 01, 2008, 09:26:00 pm »

BUMP'd

I solved question 1 to be "yes". Now to draw a to-scale diagram.

dcc · « **Reply #3 on:** March 01, 2008, 10:31:08 pm »

Question 6:

First look at the first few cases to get an idea for this:

$\mbox{Let n = number of males}$

$\mbox{Pr(n = 0) = 0.5 (i.e., 0 males, 1 female, since the chance of having a female straight off is 0.5)}$

$\mbox{Pr(n = 1) = 0.25, Pr(n = 2) = 0.125, PR(n = k) = 2^{-k-1}}$

So we want to find $E(n)$ , that is, the mean number of males that we will have in Neobeoland:

$E(n) = 0 \times 0.5 + 1 \times 0.25 + 2 \times 0.125 ....$

which we can write as:

$E(n) = \sum^\infty_{k=1} 2^{-k-1} \times k$

now rewriting this sum differently (THANKS AHMAD

:

$E(n) = \dfrac{1}{2} \times \sum^\infty_{k=1} k(\dfrac{1}{2})^k$

now the summation on the right is a variation of infinite geometric sequence, so we can rewrite $E(n)$ as:

$E(n) = \dfrac{1}{2} \times \dfrac{\dfrac{1}{2}}{(1 - \dfrac{1}{2})^2}$

$E(n) = 1$

therefore if the expected number of males is 1, and there will ALWAYS be 1 female, so the ratio of males to females should be 1!

Neobeo · « **Reply #4 on:** March 01, 2008, 10:59:30 pm »

Objection! Let's take a look at the problem from another angle. Instead of letting n be number of males, we let it be the proportion of females out of both genders. Note that I used dcc's template since I was too lazy to come up with my own.

Question 6:

First look at the first few cases to get an idea for this:

$\mbox{Let n = proportion of females}$

$\mbox{Pr(n = 1) = 0.5 (i.e., 0 males, 1 female, so 1 of 1 children are female)}$

$\mbox{Pr(n = 1/2) = 0.25, (i.e., 1 male, 1 female, so 1 of 2 children are female)}$

$\mbox{Pr(n = 1/3) = 0.125, Pr(n = 1/k) = 2^{-k}}$

So we want to find $E(x)$ , that is, the overall proportion of females that we will have in Neobeoland:

$E(n) = 1 \times 0.5 + \frac{1}{2} \times 0.25 + \frac{1}{3} \times 0.125 ....$

which we can write as:

$E(n) = \sum^\infty_{k=1} \frac{1}{k 2^k}$

now solving this sum directly (THANKS MATHEMATICA

:

$E(n) = \log2$

now the proportion of males of the whole population will just be:

$E_{male}(n) = 1-\log2$

Thus simplifying the ratio of sons to daughters to be:
$\mbox{Ratio } = \frac{1-\log2}{\log2}$

dcc · « **Reply #5 on:** March 16, 2008, 10:43:37 am »

Another Problem:

Find the smallest prime number which is a factor of:
$99! - 1$

Ahmad · « **Reply #6 on:** March 16, 2008, 11:04:14 am »

Solution
$99! - 1$ leaves a remainder when divided by all primes less than 99.

However, 99! - 1 has the same smallest prime divisor as 100(99! - 1), but by Wilson's Theorem,

$100! + 1 \equiv 0 \pmod {101}$

$\implies 100! - 100 \equiv 0 \pmod {101}$

$\implies 101 | (99! - 1)$

Therefore 101 is the smallest prime divisor of 99! - 1

dcc · « **Reply #7 on:** March 28, 2008, 03:12:58 pm »

show that:
$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

Ahmad · « **Reply #8 on:** March 28, 2008, 09:39:34 pm »

Quote from: dcc on March 28, 2008, 03:12:58 pm

show that:
$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

Use the above result to show that,

$\frac{\sin x}{x} = \prod_{i=1}^\infty \cos \frac{x}{2^i}$

Hence, show that,

$\pi = \frac{2}{\sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}} )} \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}} )})} \cdots}$

Mao · « **Reply #9 on:** March 28, 2008, 09:54:13 pm »

Quote from: dcc on March 28, 2008, 03:12:58 pm

show that:
$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

this sucks:

let k be a natural number

by the double angle formular for sin(x):
$\sin(2x)=2\sin(x)\cos(x)$

$\sin\left(\frac{x}{2^{k-1}}\right) = 2\sin\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^k}\right)$

$\sin\left( \frac{x}{2^{k-2}}\right) = 2\sin\left(\frac{x}{2^{k-1}}\right) \cos\left(\frac{x}{2^{k-1}}\right)$

$\sin\left( \frac{x}{2^{k-2}}\right) = 2\left(2\sin\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^k}\right)\right) \cos\left(\frac{x}{2^{k-1}}\right)$

$\sin\left( \frac{x}{2^{k-2}}\right) = 2^2\sin\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^{k-1}}\right)$

$\sin\left( \frac{x}{2^{k-3}}\right) = 2^3\sin\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^k}\right) \cos\left(\frac{x}{2^{k-1}}\right) \cos\left(\frac{x}{2^{k-2}}\right)$

$\sin\left( \frac{x}{2^{k-c}}\right) = 2^c\sin\left(\frac{x}{2^k}\right) \prod_{i=0}^{c-1}\left(\cos\left(\frac{x}{2^{k-i}}\right) \right)$

if we let k=c=n

$sin\left( \frac{x}{2^{0}}\right) = 2^nsin\left(\frac{x}{2^n}\right) \prod_{i=0}^{n-1}\left(cos\left(\frac{x}{2^{n-i}}\right) \right)$

$sin\left( x\right) = 2^nsin\left(\frac{x}{2^n}\right) \prod_{i=1}^{n}\left(cos\left(\frac{x}{2^{i}}\right) \right)$

since the two are algebraically equal:

$\lim_{n \to \infty}\left(sin\left( x\right)\right)= \lim_{n \to \infty}\left(2^nsin\left(\frac{x}{2^n}\right) \prod_{i=1}^{n}\left(cos\left(\frac{x}{2^{i}}\right) \right)\right)$

$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

QED

*and VICTORY*

but i dont like that last step applying the limits to both sides........

Mao · « **Reply #10 on:** March 28, 2008, 10:19:47 pm »

Quote from: Ahmad on March 28, 2008, 09:39:34 pm

Quote from: dcc on March 28, 2008, 03:12:58 pm
show that:
$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

Use the above result to show that,

$\frac{\sin x}{x} = \prod_{i=1}^\infty \cos \frac{x}{2^i}$

$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

$\frac{\sin x}{\lim_{n\to \infty}x} = \frac{\lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)}{\lim_{n\to \infty}x}$

$\frac{\sin x}{x} = \frac{\lim_{n \to \infty} (2^n \sin \dfrac{x}{2^n})}{x} \right) \prod_{i=1}^{\infty} \left(\cos \dfrac{x}{2^i} \right)$

looking at that $\lim_{n \to \infty} (2^n \sin \dfrac{x}{2^n})$

$\implies \lim_{n \to \infty} \left(\frac{\sin \dfrac{x}{2^n}}{2^{-n}}\right)$

which yields an 0/0 indeterminant form, so we use l'Hospital's rule

$\implies \lim_{n \to \infty} \left(\frac{\dfrac{d}{dn}\sin \dfrac{x}{2^n}}{\dfrac{d}{dn}2^{-n}}\right)$

$\implies \lim_{n \to \infty} \left(\frac{\dfrac{d}{dn}\left(\dfrac{x}{2^n}\right) \cdot \cos \dfrac{x}{2^n}}{-ln(2)\cdot 2^{-n}}\right)$

$\implies \lim_{n \to \infty} \left(\frac{-ln(2)\cdot 2^{-n}\cdot x\cdot \cos \dfrac{x}{2^n}}{-ln(2)\cdot 2^{-n}}\right)$

$\implies \lim_{n \to \infty} \left(x\cdot \cos \dfrac{x}{2^n}\right)$

and for dcc:

$\implies \lim_{n \to \infty}(x) \cdot \lim_{n \to \infty} \left(\cos \dfrac{x}{2^n}\right)$

$\implies x \cdot \lim_{n \to \infty} \left(\cos \dfrac{x}{2^n}\right)$

if we let $u=\dfrac{x}{2^n}$

then u approaches 0 as n approaches infinity:

$\implies x \cdot \lim_{u \to 0} \left(\cos u \right)$

$\implies x \cdot 1$

$\implies x$

sooo....

$\therefore \frac{\sin x}{x} = \frac{x}{x} \right) \prod_{i=1}^{\infty} \left(\cos \dfrac{x}{2^i} \right)$

$\implies \frac{\sin x}{x} = \prod_{i=1}^\infty \cos \frac{x}{2^i}$

QED

*ANOTHER VICTORY*

dcc · « **Reply #11 on:** March 28, 2008, 10:29:07 pm »

Quote from: Ahmad on March 28, 2008, 09:39:34 pm

Quote from: dcc on March 28, 2008, 03:12:58 pm
show that:
$\sin x = \lim_{n \to \infty} \left( (2^n \sin \dfrac{x}{2^n}) \prod_{i=1}^{n} \left(\cos \dfrac{x}{2^i} \right) \right)$

Consider:

$\lim_ {n \to \infty} 2^n \sin \dfrac{x}{2^n}$

$\mbox{Let }u = \dfrac{1}{2^n}\mbox{, clearly, as n $\to \infty$, u $\to$ 0}$
so we can rewrite the limit as:

$\lim_ {u \to 0} \dfrac{x}{u} \sin u = \lim_ {u \to 0} x \dfrac{\sin u}{u} = x$

So:
so from this, we get:

$\dfrac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \dfrac{x}{2^i}$

Now, consider $x = \dfrac{\pi}{2}$

$\dfrac{1}{\dfrac{\pi}{2}} = \prod_{i=1}^{\infty} \cos \dfrac{\dfrac{\pi}{2}}{2^i} = \dfrac{2}{\pi}$

$\pi = \dfrac{2}{\prod_{i=1}^{\infty} \cos \dfrac{\dfrac{\pi}{2}}{2^i}}$

now expanding this out:

$\pi = \dfrac{2}{\sqrt{\dfrac{1}{2}} \cos \dfrac{\pi}{8} \cos \dfrac{\pi}{16} .....}$

now we know from our handy double angle formulae:

$\cos \dfrac{\pi}{4} = 2\cos^2 \dfrac{\pi}{8} - 1$
rearranging:

$\cos \dfrac{\pi}{8} = \sqrt{\dfrac{1}{2}(1 + \sqrt{\dfrac{1}{2}})}}}$

similiarly, the same thing can be done with the rest of the cos terms:

$\therefore \pi = \frac{2}{\sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}} )} \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}(1 + \sqrt{\frac{1}{2}} )})} \cdots}$

Mao · « **Reply #12 on:** March 28, 2008, 11:15:26 pm »

Quote from: dcc on March 28, 2008, 10:29:07 pm

similiarly, the same thing can be done with the rest of the cos terms:

i'll anti-cbf for u:

with our handy cosine double angle formula:

$\cos(x)=2\cos^2\left(\frac{x}{2}\right)-1$

let k be a natural number

$\cos \frac{\pi}{2^{k-1}}=2\cos^2\frac{\pi}{2^k}-1$

$\sqrt{\frac{\cos \frac{\pi}{2^{k-1}}+1}{2}}=\cos \frac{\pi}{2^k}$

$\cos \frac{\pi}{2^k}=\sqrt{\frac{1}{2}\left(1+ \cos \frac{\pi}{2^{k-1}}\right)}$

using the above identity, we can expand $\cos \frac{\pi}{2^{k-1}}$ to get:

$\implies \cos \frac{\pi}{2^{k-1}}=\sqrt{\frac{1}{2}\left(1+ \cos \frac{\pi}{2^{k-2}}\right)}$

substituting that back in and we'll get

$\cos \frac{\pi}{2^k}=\sqrt{\frac{1}{2}\left(1+ \sqrt{\frac{1}{2}\left(1+\cos\frac{\pi}{2^{k-2}}\right)}\right)}$

doing this repeatedly will yield:

$\cos \frac{\pi}{2^k}=\sqrt{\frac{1}{2}\left(1+ \sqrt{\frac{1}{2}\left(1+\cdots \sqrt{\frac{1}{2}\left(1+ \cos \frac{\pi}{2^{2}}\right)}\right)}\right)}$ , k-2 radicals

$\cos \frac{\pi}{2^k}=\sqrt{\frac{1}{2}\left(1+ \sqrt{\frac{1}{2}\left(1+\cdots \sqrt{\frac{1}{2}\left(1+ \sqrt{\frac{1}{2}}\right)}\right)}\right)}$ , k-1 radicals

now the last step makes sense

Ahmad · « **Reply #13 on:** March 28, 2008, 11:50:09 pm »

Find $\sin (10) \sin(30) \sin (50) \sin (70)$ .

dcc · « **Reply #14 on:** March 29, 2008, 02:09:36 am »

man Ahmad i did some wonderful things trying to figure out this problem. I managed to solve 2 cubics which gave 6 complex roots then matched them up and did all sorts of wonderful things with complex numbers and all i managed to show was that sin(10) = cos(80). ... quite definitively though, as my proof was quite comprehensive.

anyway i figured it out with algebra, no geometry required!

Consider firstly (as we know sin 30 already)
$\sin 10 \sin 50 \sin 70 = \cos20 \cos 40 \cos 80$ (look, they are doubles!)

we know also (by manipulating the double angle formulae)
$\cos 20 = \dfrac{\sin 40}{2 \sin 20}$

so our original product:

$\implies \dfrac{\sin 40}{2\sin 20} \cos 40 \cos 80$

applying the double angle formula again:

$\implies \dfrac{\sin 80}{4\sin 20} \cos 80$

and again:

$\implies \dfrac{\sin 160}{8\sin 20}$

But we know sin(180 - x) = sin(x), so:

$\implies \dfrac{\sin 20}{8\sin 20} = \dfrac{1}{8}$

Now, from this, we can ascertain:

$\sin 10 \sin 30 \sin 50 \sin 70 = \dfrac{1}{8} \times \dfrac{1}{2} = \boxed{\dfrac{1}{16}}$

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