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Monday, August 5, 2013

Fractional inequality

Let $x, y > 0$, $x \neq y$, and let $n > 1$ be an integer.

If $x^n - y^n = x^{n+1} - y^{n+1}$, show that $$1 < x+y < \frac{2n}{n+1}$$

Solution Let $f(x) = x^n ( 1 - x)$ and we have $f(x) = f(y)$. WLOG, we may assume that $x If $y > 1$ then $f(y)$ is negative, so that $f(x)$ is negative, which means $x > 1$ as well. But $f(x)$ is monotonically decreasing for $x > 1$ (because $x^n$ is monotonically increasing and $1-x$ is monotonically decreasing). Contradiction.

If $y=1$ then $f(x) = f(y) = 0$ impossible for $x>0$. Contradiction.

If $0 < x < y < 1$, we draw the graph of $f(x)$. It starts off at (0,0), going up to a maximum between 0 and 1, and goes back down to (1,0). This graph can be deduced easily be realizing that $f(x) = x(1-x) . x^{n-1}$. Because $x(1-x)$ only has one turning point and $x^{n-1}$ is monotonically increasing (or constant), then $f(x)$ also only has one turning point.

We can find that turning point by using AM-GM: $$f(x) = x.x\dots x.(1-x)$$ $$ = \frac{1}{n} x.x\dots x.(n-nx)$$ $$ \leq \frac{1}{n} (\frac{n}{n})^n = \frac{1}{n}$$ Equality happens when $x = n-nx$ or $x = \frac{n}{n+1}$. Note that for $n>1, \frac{n}{n+1} = 1 - \frac{1}{n+1}> 1/2$

Monotonically increasing sequences with sums

Let $a_1, \dots, a_{1007}$ and $b_1, \dots, b_{1007}$ be monotonically increasing sequences such that:

For all $i$, $0 \leq a_i \leq 2013$ and $0 \leq b_i \leq 2013$.

For $i \neq j$, $a_i + a_j \neq 2013$ and $b_i + b_j \neq 2013$

$a_0 = b_0$

There exists a real number S such that for all $i$, $a_i + b_{1008-i} = S$.

What are the possible values of $S$?

Solution

Note that for $a_i$ and $b_i$, we need to choose 1007 numbers from 2014 numbers (0,1,...,2013) and no two of those chosen ones may add up to 2013. Thus, from each of the following pairs of numbers $(0,2013), (1,2012), \dots, (1006,1007)$ we must choose exactly one number. Thus, at most one of 0 or 2013 may be chosen. For now, let's assume that $a_1 = b_1 = 0$, which means 2013 is not chosen in both $a_i$ and $b_i$. We'll deal with the other cases later.

Let $k$ be the first number that is not chosen in $a_i$. That means $0, \dots, k-1$ are all chosen, so we have $a_1 = 0, a_2 = 1, \dots, a_k = k-1$.

This also means that $2013, \dots, 2013-(k-1)$ are not chosen but $2013-k$ is chosen. In other words, $2013-k$ is the largest chosen number in $a_i$, so $a_{1007} = 2013-k$. Because $b_1 = 0$ then $S = 2013-k$. That means $b_{1007}=2013-k - a_0 = 2013-k$

Because $2013-k$ is the largest chosen number in $b_i$, then $2013-k+1,\dots,2013$ are not chosen in $b_i$, therefore $0,\dots,k-1$ are chosen in $b_i$ and $k$ is not chosen in $b_i$.

Because $0,\dots,k-1$ are chosen in $b_i$, and $S = 2013-k$, that means $2013-k, 2013-k-1, \dots, 2013-(2k-1)$ are chosen in $a_i$, which means $k,k+1,\dots,2k-1$ are not chosen in $a_i$. Likewise, because $0,\dots,k-1$ are chosen in $a_i$, and $S = 2013-k$, that means $2013-k, 2013-k-1, \dots, 2013-(2k-1)$ are chosen in $b_i$, which means $k,k+1,\dots,2k-1$ are not chosen in $b_i$.

At this point, we have established the following fact:

  • The first $k$ numbers in $a_i$ and $b_i$ are chosen, and the next $k$ numbers are not chosen.
  • The last $k$ numbers in $a_i$ and $b_i$ are not chosen, and the previous $k$ numbers are chosen.
We would like then to claim that this pattern would have to continue itself all the way to the "middle" of the sequences, which finally means that $2k$ divides 2014. Formally, we claim that for each $m \geq 0$, we have the following:
  • $2km, \dots, (2m+1)k-1$ are chosen in $a_i$ and $b_i$.
  • $(2m+1)k, \dots, (2m+2)k-1$ are not chosen in $a_i$ and $b_i$.
  • $2013-((2m+1)k-1), \dots, 2013-2km$ are not chosen in $a_i$ and $b_i$.
  • $2013-((2m+2)k-1), \dots, 2013-(2m+1)k$ are chosen in $a_i$ and $b_i$.
The case of $m=0$ has been proven above. Now suppose it holds for $m$.

Because $(2m+1)k, \dots, (2m+2)k-1$ are not chosen in $a_i$ and $S = 2013-k$ then $2013-k-((2m+2)k-1), \dots, 2013-k-(2m+1)k$ are not chosen in $b_i$. In other words, $2013-((2m+3)k-1), \dots, 2013-(2m+2)k$ are not chosen in $b_i$. Similarly, they're not chosen in $a_i$ as well.

Now because those are not chosen in $b_i$, then $(2m+2)k,\dots,(2m+3k)-1$ are chosen in $b_i$. Similarly, chosen in $a_i$ as well.

Because $S=2013-k$ and $(2m+2)k,\dots,(2m+3k)-1$ are chosen in $b_i$ then $2013-((2m+4)k-1), \dots, 2013-(2m+3)k$ are chosen in $a_i$, and similarly, chosen in $b_i$ as well.

Because the above are chosen, then $(2m+3)k, \dots, (2m+4)k-1$ are not chosen in $a_i, b_i$, and this completes the induction step.

So now that our claim is proven, we note that this pattern must continue for all values of $m$ as long as no index becomes negative. There is no restriction of the value of $m$ in our proof of the claim. That means that towards the end of the sequence, our pattern must "line up" with what we know about $a_{1007}$ and $b_{1007}$, namely, that $2013-k$ is the largest chosen number in the sequence. This means that the sequence of numbers from 0 to 2013 is divided into an even number of groups, each having $k$ elements. Each group would alternate between all being chosen and all being not chosen. It's easy to check that this configuration matches the conditions above.

From there, we know that $2k | 2014$ or $k | 1007$. Because $1007 = 19 \times 53$, then we have four possible values of $k$.

If $k=1$, then both sequences look like this: $0,2,4,\dots, 2012$. Then $S=2012$.

If $k=19$, then both sequences look like this: $0,1,2,\dots,17,18,38,39,\dots,55,56,\dots, 988$. Then $S=1994$.

If $k=53$, then both sequences look like this: $0,1,2,\dots,51,52,106,107,\dots,158,159,\dots, 988$. Then $S=1960$.

If $k=1007$, then both sequences look like this: $0,1,2,\dots,1006$. Then $S=1006.

Now, we have to address the case where 0 is not chosen in $a_i$ and $b_i$. We define new sequences $c_i = 2013-a_{1008-i}$ and $d_i = 2013-b_{1008-i}$. It's easy to see that $c_i,d_i$ satisfy the conditions of the problem, and since $0$ is not chosen in $a_i,b_i$ then 2013 must be chosen in them, which means $0$ is chosen in both $c_i,d_i$. This reduces $c_i,d_i$ to the case above, except $S$ is replaced with $4026-S$.

Because the possible values for $S$ in the first case is $1007,1960, 1994, 2012$ then the possible values for $S$ in this case is $2014,2032,2134,3020$.

Marble distribution

In a class, the teacher gives each student $n$ marbles, and each marble was either red, blue, or green. Each student gets a different combination of marbles (i.e. no two students get the same number of red, blue, and green marbles altogether). Also, no two students have a total of exactly $n$ marbles of any color between them. What is the maximum number of students in the class?

Tuesday, May 7, 2013

Integer solutions

Let $f(t) = t(t-1)$. Find all positive integers $x,y$ such that $$f(x) + f(y) = f(x+y) / 2$$ First Solution Expanding and canceling, we have: $$x^2+y^2-2xy = x+y$$ It's easy to see that there's no solution for $x=y$. The above equation is now equivalent to: $$16x^2+16y^2+1-8x-8y+32xy = 64xy+8x+8y+1$$ $$(4x+4y-1)^2 = (8x+1)(8y+1)$$ Because both quantities are positive, we have: $$4x+4y-1 = \sqrt{8x+1} \sqrt{8y+1}$$ Once again rearranging, $$(8x+1) - 2\sqrt{8x+1} \sqrt{8y+1} + (8y+1) = 4$$ $$(\sqrt{8x+1} - \sqrt{8y+1})^2 = 4$$ If $x=y$, then there is no solution WLOG, we may assume that $x > y$, so that $$\frac{\sqrt{8x+1}}{2} - \frac{\sqrt{8y+1}}{2} = 1$$ Note that $g(t) = \frac{1+\sqrt{8t+1}}{2}$ is the reverse triangular number function. Thus, any two consecutive triangular numbers (for example 3 and 6, 6 and 10, and so on) are a solution.

Now we show that there's no other solution. The equation is equivalent to $g(x) - g(y) = 1$. In order for $g(x)$, to be an integer, we must have $8x + 1 = k^2$ for some $k$. But $k$ is odd, so $8x+1 = 4m^2 + 4m + 1$ which means $x = m(m+1)/2$ is a triangular number

Now suppose $g(x)$ is not an integer, then it will be irrational, and so will $g(y)$. They both have form of $a+\sqrt{b}$ with $a,b$ rational, but with different $b$s, because $x \neq y$. The difference can't be rational, a contradiction.

Thus, the only solutions are two consecutive triangular numbers. Second Solution Let $d = \gcd(x,y)$, and $x = dm, y = dn$ with $m,n$ coprime. Upon substitution and rearranging, we have: $$d(m-n)^2 = m+n$$ Clearly $m=n$ is not a solution. Now wlog let $m < n$ and $n = m + k$ $$dk^2 = 2m+k$$ Because $k$ divides $dk^2$ and $2m$ then $k$ divides $2m$. Because $m$ and $n$ are coprime, then $k$ and $m$ are coprime. Thus we have $k=1$ or $k=2$.

For $k=1$, $d = 2m+1, n=m+1$, so $x = (2m+1)m,y=(2m+1)(m+1)$.

For $k=2$, $d = (m+1)/2, n=m+2$, and obviously we can only use odd values of $m$. Suppose $m = 2l+1$, then $d = l+1, n=2l+3$, so $x=(l+1)(2l+1), y =(l+1)(2l+3)$. Interestingly, this form is the same as the case for $k=1$, except we substitute $m = l + 1/2$

Combining the two cases, we can conclude that the solution must have form of $m(2m+1),(m+1)(2m+1)$ for $m$ multiples of 1/2, or equivalently, $k(k+1)/2, (k+1)(k+2)/2$ for $k$ any integers. In other words, they must be two consecutive triangular numbers.

Wednesday, March 6, 2013

Charged pendulum

From brilliant.org: Two point charges, each of charge q and mass m, are hung from 1 m long cords and constrained to move in the xy-plane. The charge and mass are chosen such that the when the charges are in their equilibrium position the angle $/theta$ the cords make with the vertical is 0.1 radians. The masses are placed in their equilibrium positions and then each is pulled out by some small angle 0.01 radians and released. What is the period of the resulting oscillation in seconds?