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Let $ n\ge 3$ be an integer. Let $ f(x)$ and $ g(x)$ be polynomials with real coefficients such that the points $ (f(1),g(1)),(f(2),g(2)),\dots,(f(n),g(n))$ in $ \mathbb{R}^2$ are the vertices of a regular $ n$-gon in counterclockwise order. Prove that at least one of $ f(x)$ and $ g(x)$ has degree greater than or equal to $ n\minus{}1.$

Define $ P(x) \equal{} f(x) \plus{} ig(x).$ $ P$ is a polynomial with complex coefficients. The claim that at least one of $ f$ and $ g$ has degree at least $ n \minus{} 1$ is precisely the claim that the degree of $ P$ is at least $ n \minus{} 1.$

By hypothesis, $ P(k) \equal{} z_0 \plus{} re^{2\pi i\frac {(k \minus{} 1)}m \plus{} i\delta}$ for some fixed $ z_0$ (the center of the polygon), some $ r > 0$ (the radius of the circle that circumscribes the polygon) and some $ \delta$ (a measure of how far the polygon is rotated from standard position.) Subtracting a constant such as $ z_0$ does not change the degree of a nonconstant polynomial (and $ P$ is surely nonconstant). Multiplying or dividing by a nonzero constant such as $ r$ or $ e^{i\delta}$ does not change the degree of a polynomial. Hence WLOG we may assume that $ P(k) \equal{} e^{2\pi i(k \minus{} 1)/n} \equal{} \zeta^{k \minus{} 1}$ for $ \zeta \equal{} e^{2\pi i/n}.$

We now turn to finite differences. Define $ \Delta f(x) \equal{} f(x \plus{} 1) \minus{} f(x).$ By the binomial theorem, $ \Delta x^m$ equals $ mx^{m \minus{} 1}$ plus terms of degree lower than $ m \minus{} 1.$ It follows that if $ f$ is a polynomial of degree exactly $ m,$ for $ m\ge 1,$ then $ \Delta f$ is a polynomial of degree exactly $ m \minus{} 1.$ (And if $ f$ is constant, then $ \Delta f \equal{} 0.$) By iteration, we see that if $ f$ is a polynomial and if $ \Delta^mf(x)\ne 0$ at any point, then the degree of $ f$ must be at least $ m.$
\[ P(k) \equal{} \zeta^{k \minus{} 1}\text{ for }0\le k\le n \minus{} 1\]\[ \Delta P(k) \equal{} \zeta^{k \minus{} 1}(\zeta \minus{} 1)\text{ for }0\le k\le n \minus{} 2
\]\[ \Delta^2 P(k) \equal{} \zeta^{k \minus{} 1}(\zeta \minus{} 1)^2\text{ for }0\le k\le n \minus{} 3\]\[ \vdots\]\[ \Delta^{n \minus{} 1}P(0) \equal{} (\zeta \minus{} 1)^{n \minus{} 1}\]
Since $ \zeta \minus{} 1\ne 0,$ this shows that $ \Delta^{n \minus{} 1} P(0)\ne 0,$ which means that the degree of $ P$ is at least $ n \minus{} 1.$

Go me for writing a big spiel about finite differences in my blog, then proceeding to miss that solution at the end of this problem and turning to Lagrange Interpolation to show $ P$ had degree at least $ n\minus{}1$. At least it worked.

Palmer, I thought for sure you'd be elated once you saw this problem; I immediately thought of your post :)

The explicit formula for the $ (n\minus{}1)^{th}$ difference of any sequence with $ n$ terms is

$ \sum_{k\equal{}0}^{n\minus{}1} (\minus{}1)^{n\minus{}1\minus{}k} {n\minus{}1 \choose k} p(k)$

which gives Kent Merryfield's answer immediately by the binomial theorem.

Well, I was elated; polynomials and roots of unity tend to be something I'm good at. Doesn't mean I learned my own lessons. :D

You can do it without finite differences:

We can translate $ P(x)$ without changing its degree so that the polygon is centered at the origin.  Call the new polynomial $ Q(x)$.  Now if we let $ z\equal{}e^{\frac{2 \pi i}{n}}$, notice that for $ n\equal{}1,2,3,...,n\minus{}1$, $ Q(n\plus{}1)\equal{}zQ(n)$.

So now we define a new polynomial $ R(x) \equal{} Q(x\plus{}1) \minus{} zQ(x)$.  This has degree at most the degree of $ Q$, and it has $ n\minus{}1$ roots.  Therefore the degree of $ Q$ is at least $ n\minus{}1$, as desired.

Ahhh! I almost had this one! It seems that I made a computational error in my finite difference calculations. Oh well.

[quote="Sly Si"]You can do it without finite differences:

We can translate $ P(x)$ without changing its degree so that the polygon is centered at the origin.  Call the new polynomial $ Q(x)$.  Now if we let $ z \equal{} e^{\frac {2 \pi i}{n}}$, notice that for $ n \equal{} 1,2,3,...,n \minus{} 1$, $ Q(n \plus{} 1) \equal{} zQ(n)$.

So now we define a new polynomial $ R(x) \equal{} Q(x \plus{} 1) \minus{} zQ(x)$.  This has degree at most the degree of $ Q$, and it has $ n \minus{} 1$ roots.  Therefore the degree of $ Q$ is at least $ n \minus{} 1$, as desired.[/quote]

dang this is far more nice than the lagrange interpolation / finite difference route

pretty sweet

So, what I got to was proving that both of these could not be true:
\[ \sum_{k \equal{} 1}^{n \minus{} 1} ( \minus{} 1)^{k \minus{} n \plus{} 1} \left(\!\!\begin{array}{c}n \minus{} 1 \\
k \minus{} 1\end{array}\!\!\right) \cos{\left(\theta \plus{} \frac {2\pi (k \minus{} 1)}{n}\right)} \equal{} \cos{\left(\theta \minus{} \frac {2\pi}{n}\right)}
\]

\[ \sum_{k \equal{} 1}^{n \minus{} 1} ( \minus{} 1)^{k \minus{} n \plus{} 1} \left(\!\!\begin{array}{c}n \minus{} 1 \\
k \minus{} 1\end{array}\!\!\right) \sin{\left(\theta \plus{} \frac {2\pi (k \minus{} 1)}{n}\right)} \equal{} \sin{\left(\theta \minus{} \frac {2\pi}{n}\right)}
\]
Did anybody else somehow get to that?  Or, rather - does it look possible to prove?

Looking over these posts, I can't believe I didn't think of using complex numbers, but hey, whatcha gonna do...

The first equation plus $ i$ times the second equation is $ e^{ i \theta} (e^{\frac {2 \pi i}{n} } \minus{} 1)^{n \minus{} 1}$ on the LHS and $ e^{ i \theta} e^{ \minus{} \frac {2 \pi i}{n} }$ on the RHS.  The factor on the LHS has absolute value $ 1$ if and only if $ n \equal{} 3, 6$.

It looks like you Lagrange interpolated through the first $ n \minus{} 1$ points and you're trying to show that you can't get to the last point, correct?

Here's a thought.  The conclusion no longer holds if we drop the requirement that the points be in order.  What is the smallest degree required to produce $ n$ points on an $ n$-gon in [i]some[/i] order?

Yeah, that's exactly what I did.  Would what I did get any points, do you think?

Again, I just can't believe I didn't think to use complex numbers.   :(

Its leading coefficient is $ 1 \minus{} z$ times the leading coefficient of $ Q$.  I suppose you're right that that issue does need to be addressed, though; that is, $ R$ has degree [i]exactly[/i] the degree of $ Q$.

[quote="Nukular"][quote="Sly Si"]You can do it without finite differences:

We can translate $ P(x)$ without changing its degree so that the polygon is centered at the origin.  Call the new polynomial $ Q(x)$.  Now if we let $ z \equal{} e^{\frac {2 \pi i}{n}}$, notice that for $ n \equal{} 1,2,3,...,n \minus{} 1$, $ Q(n \plus{} 1) \equal{} zQ(n)$.

So now we define a new polynomial $ R(x) \equal{} Q(x \plus{} 1) \minus{} zQ(x)$.  This has degree at most the degree of $ Q$, and it has $ n \minus{} 1$ roots.  Therefore the degree of $ Q$ is at least $ n \minus{} 1$, as desired.[/quote]

You have to be careful to show that $ R$ is not identically zero, but nice.[/quote]

*kicking self*  How many points do you think I lost for not doing that?

you probably lost 1 point at the most. The solution is really nice and the graders are more lenient in the A(B)5 A(B)6 region.

[quote="Kent Merryfield"]
\[ P(k) \equal{} \zeta^{k \minus{} 1}\text{ for }0\le k\le n \minus{} 1
\]

\[ \Delta P(k) \equal{} \zeta^{k \minus{} 1}(\zeta \minus{} 1)\text{ for }0\le k\le n \minus{} 2
\]

\[ \Delta^2 P(k) \equal{} \zeta^{k \minus{} 1}(\zeta \minus{} 1)^2\text{ for }0\le k\le n \minus{} 3
\]

\[ \vdots
\]

\[ \Delta^{n \minus{} 1}P(0) \equal{} (\zeta \minus{} 1)^{n \minus{} 1}
\]
Since $ \zeta \minus{} 1\ne 0,$ this shows that $ \Delta^{n \minus{} 1} P(0)\ne 0,$ which means that the degree of $ P$ is at least $ n \minus{} 1.$[/quote]

Just a minor detail...did you mean to write, for example, $ 1\le k\le n$ instead of $ 0\le k\le n \minus{} 1$ and $ \Delta^{n \minus{} 1}P(1) \equal{} (\zeta \minus{} 1)^{n \minus{} 1}$ instead of $ \Delta^{n \minus{} 1}P(0) \equal{} (\zeta \minus{} 1)^{n \minus{} 1}$?

The notation in this problem works out nicer if you let $ P(x) \equal{} f(x \plus{} 1) \plus{} i g(x \plus{} 1)$ instead, although that's a cosmetic distinction.

I fixed that in the handout I gave my class afterwards - I guess I forgot to fix it here. Read what I meant to say, not what I actually said.

Hmmm...someone above mentioned a Lagrange Interpolation solution. The Lagrange Interpolant is (after transforming the polygon to the roots of unity)

\[ L(z) \equal{} \sum_{k\equal{}1}^n e^{2 \pi i (k\minus{}1)/n}
\frac{
(z\minus{}1)
(z\minus{}2)
\cdots
(z\minus{}(k\minus{}1))
(z\minus{}(k\plus{}1))
\cdots
(z\minus{}n)}
{(k\minus{}1)
(k \minus{} 2)
\cdots(k \minus{} (k\minus{}1) )
(k \minus{} (k\plus{}1))
\cdots (k \minus{} n)
}\]

so we would need to show that the coefficient of $ z^{n\minus{}1}$

\[ \sum_{k\equal{}1}^n e^{2 \pi i (k\minus{}1)/n}
\frac{1}
{(k\minus{}1)
(k \minus{} 2)
\cdots(k \minus{} (k\minus{}1) )
(k \minus{} (k\plus{}1))
\cdots (k \minus{} n)
}\]

is not zero. How would one do that?

That would have been me. :D Multiply that giant fraction by $ (n\minus{}1)!$ to make it a binomial coefficient times -1 to the something, then apply the binomial theorem.

As is usually the case when the $ x$ coordinates are well-spaced, the interpolation is much easier using Newton polynomials than using Lagrange interpolation.  See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url], especially the first practice problem, which is an alternate approach - you can interpolate through the first $ n\minus{}1$ points and show that you can't reach the last point.

Hmm does this work?

First we can translate the polygon so that its center is the origin. This is because setting $f_2(x) = f(x)-k$ doesn't affect its degree.

Let $a = 2\pi/n$ be the angle of rotation. Then we have the identities:

$f(x+1) = \cos(a)f(x) - \sin(a)g(x)$ and $g(x+1) = \sin(a)f(x)+cos(a)g(x)$ for $x=1,2...n-1$.

Define $d_1(x) = f(x+1) - \cos(a)f(x) + \sin(a)g(x)$ and $d_2(x) = g(x+1) - \sin(a)f(x) - cos(a)g(x)$. Since $d_1$ has at least $n-1$ roots, it is either of degree $n-1$ or greater or the zero polynomial. In the first case, since the degree of $f(x+1)$ is the same as the degree of $f(x)$, at least one of $f(x)$ or $g(x)$ have degree at least $n-1$. Hence we can assume $d_1(x) = 0$, similarly we can assume $d_2(x) = 0$.

Then $f(x+1) = \cos(a)f(x) - \sin(a)g(x)$ and $g(x+1) = \sin(a)f(x)+cos(a)g(x)$ holds for all $x$, squaring both equations and adding, we get

$(f(x+1))^2+(g(x+1))^2 = (f(x))^2+(g(x))^2$ for all $x$. Thus $(f(k))^2+(g(k))^2 = (f(1))^2+(g(1))^2$ for all integer $k$, which implies that $f(x)$ cannot tend to plus or minus infinity as $x$ tends to infinity. Yet this implies that $f(x)$ is a constant polynomial, which of course can't work for regular polygons of three or more vertices (the $x$ coordinate can't be constant). Contradiction.

A lagrange interpolation solution has already been discussed, but here is a more complete one:

Like other solutions, we consider the polynomial $P(x) = f(x) + ig(x)$, with complex coefficients. Our goal is now to show that $P$ has degree at least $n-1$, as this will imply that either $f$ or $g$ have degree at least $n - 1$. Now, we know that $P$ takes on the value of a regular n-gon somewhere in the complex plane at the values 1,2...,n. There is a polynomial of minimal degree that does this, and clearly this polynomial must be a factor of $P$. Thus, it suffices to show that the minimal degree polynomial that takes on these values has degree at least $n - 1$. But we know how using lagrange interpolation to explicitly construct this polynomial.

Let $\omega = e^{\frac{2\pi i}{n}}$. Then we know that $P(k) = r\omega ^{k-1} + z_0$, where $r$ and $z_0$ are some complex numbers. for k = 1,2,...,n. Now lagrange interpolation tells us that

\[ P(x) = \sum_{k=1}^n P(k) \cdot \prod_{j = 1, j \neq k}^n \frac{x - j}{k - j}\]

It suffices to show that the coefficient of $x^{n-1}$ in this polynomial is non-zero, which will imply the result. It is not hard to see that coefficient of $x^{n-1}$ in this polynomial is

\[ \sum_{k = 1}^n P(k) \cdot \prod_{j = 1, j \neq k}^n \frac{1}{k-j}\]

Now, we note that we have that
\[ \prod_{j = 1, j \neq k}^n \frac{1}{k-j} = \frac{(-1)^{n-k}}{(k-1)!(n - k)!}\]

If the coefficient of $x^{n-1}$ is not zero, then certainly it will not be zero after we multiply if by $(n-1)!$, so after we multiply the coefficient of $x^{n-1}$ by $(n-1)!$, we get

\[ \sum_{k = 1}^n P(k) \cdot (-1)^{n-k} \cdot \binom{n-1}{k-1}\]

Subtracting a constant from $P(x)$ and multiplying it by some non-zero complex number clearly does not change the degree of the polynomial, so we may just assume that $P(k) = \omega^{k-1}$ for k = 1,2,...,n. Now, all we have to show is that
\[\sum_{k=1}^n \omega^{k-1} \cdot (-1)^{n-k} \cdot \binom{n-1}{k-1}\] is nonzero. But by the binomial theorem, this is just $(\omega - 1)^{n-1}$, which is clearly non-zero as $\omega \neq 1$, so we are done.

Let $P(x)=f(x)+ig(x)$. Then via a suitable [hide=linear transformation] Suppose the center of the polygon in the complex plane is $k\in \mathbb{C}$ and the vertices are $k+t, k+t\omega, \cdots, k+t\omega^{n-1}$, then we take the transformation $Q(x) = \frac{P(x)-k}{t}$ [/hide] we can assume $P(x)=\omega^x$ for $x=0,1,\cdots,n-1$ where $\omega=e^{\frac{2\pi i}{n}}$

I will show if $P(x)=k^x$ for some $k\ne 1$ and for all $x=0,\cdots,n-1$ then $\deg P\ge n-1$.

Induct on $n$; consider $Q(x)=\frac{P(x+1)-P(x)}{k-1}$, which has degree at least $n-2$ by inductive hypothesis.

Yet another way. Repeating a lot of what previous users have said, using some transformations, we see that we must prove that given a polynomial such that for $x=0,1,\ldots,n-1$ we have $P(x)=\omega^x$ where $\omega=e^{2\pi i/n}$ must have degree at least $n-1$. It suffices to find a degree $n-1$ polynomial $Q(x)$ that satisfies $Q(x)=\omega^x$ for $x=0,1,\ldots,n-1$. This is because if we found such a polynomial, then $Q(x)-P(x)$ would have $n$ roots, so it must either be the zero polynomial or a polynomial of degree at least $n$. We can see that if $P(x)$ had degree less than $n-1$ then neither case would be true.

Now to construct our $Q$ consider $Q(x)=\sum_{k=0}^{n-1} \binom{x}{k}(\omega-1)^k$. Note that $\binom{x}{k}=\frac{1}{k!}x(x-1)(x-2)\cdots(x-k+1)$ is a polynomial in $x$ of degree $k$ so this sum is indeed a polynomial of degree $n-1$. Moreover, by binomial theorem, for any $x=0,1,\ldots,n-1$ we have $Q(x)=(1+\omega-1)^x=\omega^x$.

Let $\zeta=e^{2\pi i/n}$, and assume FTSOC that $\max(\deg f, \deg g)< n-1$. We may take the linear transformation mapping this polygon to the one centered at the origin with the $n$th roots of unity, so that we result in $(\widetilde f(0),\widetilde g(0))$, $(\widetilde f(1),\widetilde g(1))$, etc. Now let $g(X)=\widetilde f(X)+i\widetilde g(X)$. Then $g(0)=1$, $g(1)=\zeta$, $g(2)=\zeta^2$, etc, so \[\zeta g(X)-g(X+1)=0\qquad\text{for }X=0, 1, 2,\dots,n-2.\] Therefore $X(X-1)(X-2)\cdots(X-(n-2))$ divides $\zeta g(X)-g(X+1)$, so $g$ must have degree at least $n-1$.
However, $\max(\deg f(x),\deg(g(x))$ implies that $\deg g<n-1$, contradiction.

Without loss of generality let the polygon be centered at the origin and on the unit circle. We can do this since we can just shift by the centroid of the polygon and scale down, neither of which change the degree of the polynomials. Letting $a = \cos \frac{2\pi}{n}$ and $b = \sin \frac{2\pi}{n}$ we obtain the following:

\begin{align*}
a\cdot f(x)-b\cdot g(x) &= f(x+1), \\
b \cdot f(x)+a \cdot g(x) &= g(x+1),
\end{align*}

for $x = 0, 1, \ldots, n-2$. We get this by multiplying the complex number $f(k) + i \cdot g(k)$ by $e^{2 \pi i /n}$ and equating real and imaginary parts with the complex number $f(k+1) + i \cdot g(k+1)$, again for $k = 0, 1, \ldots, n-2$.

Now suppose $\max (\deg f, \deg g) < n-1$. Rewriting both equations by moving everything to left side, we see that the degree of the left is at most $n-2$. However, both polynomials have $n-1$ solutions, which is a contradiction. Thus, $\max (\deg f, \deg g) \geq n-1$.

This is such an amazing problem.
Let $(s, t)$ be a center of regular $n$-gon with vertices $(f(0), g(0)), ... , (f(n - 1), g(n - 1))$ and let $r$ be a its circumcircle radius. Replacing $f(X), g(X)$ to $\frac{1}{r}(f(X) - s), \frac{1}{r}(g(X) - t)$, we can assume that circumcircle of regular $n$-gon with vertices $(f(0), g(0)), ... , (f(n - 1), g(n - 1))$ is unit circle. Let $a$ be a real number such that $f(0) = \cos{a}, g(0) = \sin{a}$. Then for $0 \leq k \leq n - 1$, we have $f(k) = cos(\frac{2\pi k}{n} + a), g(k) = sin(\frac{2\pi k}{n} + a)$. Assume deg$f$, deg$g$ $\leq n - 1$. Then $f$ is the interpolation polynomial of $n$ points $(0, cos(a)), (1, cos(\frac{2\pi}{n} + a)), ... , (n - 1, cos(\frac{2\pi(n - 1)}{n} + a))$, so $f(X) = \sum_{k=0}^{n-1}cos(\frac{2\pi k}{n} + a)\prod_{i=0,i\neq k}^{n-1}\frac{X-i}{k-i}$. Therefore the coefficient of $X^{n-1}$ of $f$ is $\sum_{k=0}^{n-1}cos(\frac{2\pi k}{n} + a)\prod_{i=0, i\neq k}^{n-1}\frac{1}{k-i} = \sum_{k=0}^{n-1}cos(\frac{2\pi k}{n} + a)\frac{(-1)^{n - 1 - k}}{k!(n - 1 - k)!} = \frac{1}{(n-1)!}\sum_{k=0}^{n-1}cos(\frac{2\pi k}{n} + a)(-1)^{n-1-k}\binom{n-1}{k}$. Similarly the coefficient of $X^{n-1}$ of $g$ is
$\frac{1}{(n-1)!}\sum_{k=0}^{n-1}sin(\frac{2\pi k}{n} + a)(-1)^{n-1-k}\binom{n-1}{k}$. Let $a$ and $b$ be coefficient of $X^{n-1}$ of $f, g$, respectively. Then $a + bi = \frac{1}{(n-1)!}\sum_{k=0}^{n-1}(cos(\frac{2\pi k}{n} + a) + isin(\frac{2\pi k}{n}+a))(-1)^{n-1-k}\binom{n-1}{k} = \frac{e^{ia}}{(n-1)!}\sum_{k=0}^{n-1}e^{\frac{2\pi k}{n}}(-1)^{n-1-k}\binom{n-1}{k} = \frac{e^{ia}}{(n-1)!}(e^{\frac{2\pi}{n}} - 1)^{n-1} \neq 0$. Therefore one of the $a,b$ is not equal to 0 means $\max(\deg $f$, \deg $g$) \geq n - 1$, as desired.

We can throw this on the complex plane by defining $m(x) = f(x) + g(x) \cdot i$. Since $m(x)$ forms a regular $n$-gon the complex plane for $x = 0, 1, \ldots, n-1$, we have
\[m(x+1) = \operatorname{cis } \left(\frac{2\pi}{n}\right) \cdot m(x)\]

for $x = 0, 1, \ldots, n-2$. As a result, we know
\[n(x) = m(x+1) - m(x) \cdot \operatorname{cis } \left(\frac{2\pi}{n}\right)\]

has roots $0, 1, \ldots, n-2$. Hence
\[n-1 \leq \deg(n) \leq \deg(m) \leq \max(\deg (f), \deg (g)).~\blacksquare\]


integration
trigonometry
calculus
calculus computations

$\int \frac{a}{\cos^{2}x}\ dx=a\int \sec^{2}x\ dx=a\tan+C$


AMC
AIME
AMC 12
trigonometry
trig identities

Um... well, Algebra II is sufficient, I'd guess. But if you want to do well, you need to get actual contest-problem-solving-oriented (whew) books.

The highest level of math needed is Algebra II. That doesn't say anything about difficulty...

Hmm...I think the AMC website says precalculus but that might only be for USAMO...it might be for the AIME as well, someone should check.

Well, knowing some basic trig identities is helpful, and that's not usually taught until the end of AII or middle of PreCalc.

We never did any trig identities until the middle of Pre-Calc at my school, and there was a problem on the AMC-12 last year that was only possible with the angle addition identity for cosine.  And there's definitely problems that are a lot easier to do with stuff you don't learn in your standard classes through Algebra 2...

You definitely need 'precalculus' (what does this really mean anyways...) knowledge, as there are often questions about sums, trig, on the AIME

[quote="Amirsan"]I have heard of AIME many times on other forums but never took time to look into it.  Now I have and its very interesting.  Personally I am just interested in challenges.

I just want to know, what do you guys think is adequate math background to be able to do well on AMC12 and maybe beyond?  As in, what level math, in high school, do the ones who do well take?

Do you think it is too late for a Junior in high school taking Algebra 2 to even consider competing?[/quote]

If you're interested in challenges then I really don't think it should matter what math you are taking. While having math through precalc is useful, most of the problems are based on solving problems, not how much math you know. It is more based on applications of techniques, etc.

Additionally, problem solving does not end in high school. It will follow you throughout your entire life, and so the fact that you started perhaps one or two years later than many other people is going to have very little impact 5-10 years down the road, when your problem solving skills will finally be applied to "real" problems.

A knowledge of common counting techniques such as permutations and combinations is essential.


I'd like to applaud the very busy and hardworking folks at AMC Lincoln for putting together the annual summary book and making it available in PDF this year.

http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/2007-1012Summary.pdf

This is a very handy resource for all those who may need their scores in later years, since students can just use Adobe's utility to search for their last name and school in the book.   So you do not need to bother the hard-working AMC Lincoln staff when you are applying to college and need your scores.

Making the document available in PDF (vs. the old system of publishing books and mailing them out to all contest organizers) also saves lots of trees and all the added fuel consumed by all that weight going out in postal service planes and trucks, cutting down on greenhouse gases, air pollution, etc.

Has anyone noticed that people who got 5s on the AIME (like me) also made it onto the AIME - Scores of 6+ section?  :P

Maybe others should use technology and save more trees in the process!

I think there are Ebooks that are available online for people who want to read some books, but I think you have to sign up at the library. Well, at least that is how it is here.

Maybe school textbooks should all be online, and everyone should have a labtop? Then, during class, everyone can view their labtop. OR, a teacher can use the computer as a projection and use it to teach in class. That would be awesome! Haha that was such a random thought.

Well, I applaud how it has been available online for anyone to view, at least anyone who has Adobe handy!

oh haha i got the highest aime score in oregon! not much of an accomplishment but i did beat lars at aime... :P


What is the ratio of the numerical value of the area, in square units, of an equilateral triangle of side length 4 units to the numerical value of its perimeter, in units? Express your answer as a common fraction in simplest radical form.

The area of an equilateral triangle with side length four units is: $ 2\sqrt{3}*4/2\equal{}4\sqrt{3}$
The perimeter of an equilateral triangle with side length four units is: $ 3*4\equal{}12$

$ \frac{4\sqrt3}{12}\equal{}\frac{\sqrt3}3$

I think the formula for an equilateral triangle's area is this:

$ A \equal{} \frac{s^2 \sqrt{3}}{4}$

I was playing and I put [code]\frac{\sqrt3}{3}[/code]

and it said wrong, the answer is a / instead of frac.  I think my answer should have been accepted also.

just to be safe put $ \frac{\sqrt{3}}{3}$

[quote="mathemagician1729"]just to be safe put $ \frac {\sqrt {3}}{3}$[/quote]

If you read my Code, that is what I put...

[quote="FTW How To Play"]Fractions should always be reduced. A fraction should be entered as x/y.
[code]1/2[/code][/quote]

So there.  No bug.


Let $f(a,b,c)=2\sum\frac{a}{b+c}-3$.Prove that $\forall a,b,c>0$ we have
$f(a,b,c)\ge 3f(a+b,b+c,c+a)$
And $3$ is the best constant. ;)
[b][u]Other forms[/u]:[/b]$\forall a,b,c>0$ we have
a)$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+6(\frac{a}{2a+b+c}+\frac{b}{2b+c+a}+\frac{c}{2c+a+b})\ge 6$
b)$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{9} {a+b+c}\ge 6.(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b})$
 ;)

S.O.S is surely work here. The problem is that
\[ \sum (x-y)^2(\frac{1}{xy}-\frac{3}{x+z)(y+z)}) \ge 0 \]
in which $x,y,z$ are three lengths of a triangle.
Notice that
\[ \frac{1}{xy}-\frac{3}{x+z)(y+z)}+\frac{1}{xz}-\frac{3}{y+x)(y+z)} \ge 0 \]
Notice that $x\le y+Z$, hence we only need to check if $x=y+z$, or
\[ \frac{1}{y}+\frac{1}{z} \ge \frac{3}{2y+z} +\frac{3}{2z+y} \]
Which is obviously true.

It's actually the problem 4 in Viet Nam IMO TST 2006.

[quote="nthd"]
No  ;)[/quote]
Thank you! :)

[quote="nthd"][quote="hungkhtn"]

It's actually the problem 4 in Viet Nam IMO TST 2006.[/quote]
No,I created it 2months ago. ;)
I think we can compare $f(a,b,c)$ and $kf((a+b)^2,(b+c)^2,(c+a)^2)$ ;)[/quote]

Hi, nhtd. I only mean that your problem and VN TST 2006 is exactly the same in solution. First I don't realize that, but after solving your problem, I've just known,

Prove that $\forall a,b,c>0$ we have :
$f(a,b,c)\ge \frac{5}{9}.f((a+b)^2,(b+c)^2,(c+a)^2)$
And $\frac{5}{9}$ is the best constant .
 ;)  ;)
(I use S.O.S but with another form) ;)

The first pro isn't VN TST 2006,but the following pro is VN TST 2006:
Prove that$\forall a,b,c>0$ we have
$f(a,b,c)\le 6f(\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a})$
and $6$ is the best constant. :)


We should have a blog report thread.

Reporting anne1's blog. [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1697]link[/url]

Reporting dentohpue's blog. [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1694]link[/url]

There's a general reporting thread in the moderators' forum. I've added these there as well as five more, so an administrator should notice soon.
The blogs aren't watched as much, but they get far less spam than the forum- at least until the last few days.

How do we report a blog?
And what are some reasons to report a blog?

These blogs were reported for being commercial spam. Since EggyLv.999's post, about ten such blogs have been deleted, and their makers banned. A new policy has also been implemented that should prevent such spam in the future.


limit
trigonometry
function
analytic geometry
geometry
vector
calculus

I'm new to limits...can someone help me with these problems...(use of l'hospital or Taylor's expansion strictly not allowed)
Plz also explain how you worked through them

1. Let $ f(x)\equal{}\lim_{m\rightarrow \infty} [\lim_{n\rightarrow \infty} (\cos^{2m} (n! \pi x))]$, where $ x\in R$.
Prove: f(x)=1 if x is rational or 0 if x is irrational.

2. Find $ \lim_{n\rightarrow \infty} n^2[\sqrt {(1\minus{}\cos\frac 1n) \sqrt {(1\minus{}\cos\frac 1n) \sqrt {(1\minus{}\cos\frac 1n)....}}}]$

3.Evaluate $ \lim_{x\rightarrow \frac {\pi}{2}} [\sin(<x>\plus{}1)]$ where [.] is the greatest integer function and <.> is the fractional part of x.

4. $ \lim_{n\rightarrow \infty} \prod_{r\equal{}3}^n (\frac {r^3\minus{}8}{r^3\plus{}8})$

1.if $ x \in \mathbb{Q}$ then clearl at some $ n$ $ \cos (n!\pi x) \equal{}$ here on we have $ \cos^{2m}(n!\pi x)$ also + or -$ 1$ so the limit converes to 1.
if $ x$ is irrational then for no $ n$ $ \cos (n! \pi x) \equal{} 1$ hence we have $ | \cos (n! \pi x)| < 1$ thus the limit converges to $ 0$ since $ \lim _{x\to \infty} a^{x} \equal{} 0$ iff $ |a| < 1$

2. assuming the thing in the radical converges to areal number...
let $ A \equal{} n^{2}[\sqrt {(1 \minus{} \cos\frac 1n) \sqrt {(1 \minus{} \cos\frac 1n) \sqrt {(1 \minus{} \cos\frac 1n)....}}}]$ this gives
$ \frac {A^{2}}{n^{4}} \equal{} (1 \minus{} \cos (\frac {1}{n}))\frac {A}{n^{2}}$ thus giving $ A \equal{} n^{2}(1 \minus{} \cos( \frac {1}{n})) \equal{} \frac {2n^{2}}{\sin^{2} \frac {1}{2n}}$
hence $ \lim _{n\to \infty} A \equal{} \lim _{n\to \infty} \frac {(\frac {\sin(\frac {1}{2n})}{\frac {1}{2n}})^{2}}{2} \equal{} \frac {1}{2}$

[quote="hell_ever"]I am sorry but i don't quite follow...[/quote]
is there anything specific u couldn't follow

3.is the thing inside $ \sin(...)$ in radian or real number

[quote="hell_ever"]Why does $ \cos^{2m}(n!\pi x)$ converge to +1 or -1??[/quote]
it converges only to $ \plus{}!$ ...but since $ x \in \mathbb{Q}$ for some $ n_{0}$ we will ahve $ n_{0}! x$ is a n integer in tha cas e$ \cos(\pi n_{0}!x)$ will be $ \plus{}1$ or $ \minus{}1$....hence on for all $ n \geq n_{0}$ we will have $ \cos(\pi n! x)\equal{}$+1 or -1 hence $ \cos^{2m}(n!\pi x) \equal{}1$ for all $ m \in {\mathbb{N}}$hence the limit converges to 1

4. for this observe that $ (r\plus{}4)\minus{}2\equal{}r\plus{}2$ and that $ r^{2}\plus{}r\plus{}4\equal{}(r\plus{}1)^{2}\minus{}(r\plus{}1)\plus{}4$
henc the question can be written as $ \lim_{n\to \infty}\prod_{r\equal{}3}^{n} \frac{(r\minus{}2)(r^{2}\plus{}r\plus{}4)}{(r\plus{}2)(r^{2}\minus{}r\plus{}4)}$ hence we pair off
hence pair $ (r,r\plus{}4)$ to cancel out the linear factors and similar;y do it for other ...
now look at what cannot be paired $ 1.2.3.4.\equal{}4!$ also looking below we can't pair off $ 3^{2}\minus{}3\plus{}4\equal{}10$ so the answer is $ \frac{12}{5}$

Its rather surprising why people sometimes specify things such as 'L'Hospital's rule not allowed'. Now its actually Taylor's Expansion not being allowed! These are quite mathematically correct tools and there is no reason why one should not use them when it is appropriate. It is something like preventing one from using Trigonometry or Coordinate Geometry to solve a Geometry problem. Of course, a pure Geometry solution is always elegant and welcome. But, then its not like Trigonometry or Coordinate Geometry are non-mathematical in any sense of the word.
Sathej

$ \lim_{x\rightarrow \frac {\pi}{2}} [\sin( < x > \plus{} 1)]$

Hmmm...Consider x tending to Pi/2. Pi/2 = 1.5...... <Pi/2> = .57... <x> +1 is again Pi/2.

As x --> Pi/2 from LHL or RHL Sinx tends to 1 but is slightly less than 1. So Limit is Zero.

:-)

Sathej, You have a valid point but this was the trend followed even in JEE when they had subjective. L'Hospital rule is a superb shortcut method to find the soln and many people blindly use it to find solns. But it has a very solid proof (to hell_ever: Try prooving it, good excercise). Anyway, I guess they want to encourage people applying other creative methods than just using Lhospital.

Well, 'this was the trend even in JEE'. Surely, JEE is not the be-all and end-all as far as Mathematics is concerned. Definitely, there have been several instances where JEE Math questions have been on the flip side. Further, there is no question of encouraging creativity. Then JEE must be asking questions on vector calculus and multi-variable calculus to see how well a student can tackle hitherto unseen areas. And as for the proof of L'Hospital's rule, it is found in almost all books on Differential Calculus. If it is the intention of the examiners that such rules have to be used only after proving them, then its a tricky situation as to where to draw the line. Some people might say you have to just derive the coordinates of focus and directrix every time you do a problem in Coordinate Geometry. A similar problem is encountered in Olympiads but they have a fairly acceptable way of dealing with it.
 And Taylor's Series being forbidden is really ridiculous to say the least.
Sathej

Well, you have a valid point, but MO does not have calculus, so for school students, the only top contest which requires calculus is JEE. :) And since we are preparing for JEE, we have to reorient our minds for that purpose only :-( which may be the wrong thing to do but sadly under such cut-throat competition this is only thing we can do. After all you need to enter a top college and JEE(Current paper) is definitely not the best test of one's talent.

[quote="Sathej"]Of course, yes, re-orienting towards JEE is necessary :) All that I said was its a ridiculous proposition 'not to allow L'Hospitals' Rule or Taylor's Series or any mathematically correct technique' in any exam or wahtever. Dismissing something as shortcut is not the right way to look at it. Using the quadratic formula is also a shortcut then :) You may have to derive it each time by completing the squares....
Sathej[/quote]

THen l'hospital \ Taylor's it is!!!  :10:

o would like to disagree a bit...i suppose this problem was given to hell_ever in a tution or he himself solving it....
well till the time he faimiliarizes himselves with basic proofs using L hospital can be a bit degrading....it's like directly integrations without knowing limit sum rule.ofcours ethis doesn't mean one has to use  taht each time...but it tells u the definition of riemann integra...so before jumping into something which even if u understand the proof be sure u 'understnand' it by heart and soul and why the appoach was so and so.....


superior algebra
superior algebra unsolved

Let $ a\in \mathbb Z^*$, and $ n\in \mathbb N$.
It seems that $ X^{2^n}(X\minus{}a)^{2^n}\plus{}1$ is irreducible over $ \mathbb{Q}[X]$. Is it true? If yes, how to prove it? (I know how to prove that  $ X^{2^n}\plus{}1$ is irreducible over $ \mathbb{Q}[X]$ but my method does not generalize here).

[quote="Portland"]Let $ a\in \mathbb Z^*$, and $ n\in \mathbb N$.
It seems that $ X^{2^n}(X \minus{} a)^{2^n} \plus{} 1$ is irreducible over $ \mathbb{Q}[X]$. Is it true? If yes, how to prove it? (I know how to prove that  $ X^{2^n} \plus{} 1$ is irreducible over $ \mathbb{Q}[X]$ but my method does not generalize here).[/quote]It is false: take n = 0 and a = 2.

[quote="LaChouetteRieuse"]It is false: take n = 0 and a = 2.[/quote]
Correct, but I considered ${ n\in \{1,2,3,\ldots}$.
As you can read here http://en.wikipedia.org/wiki/Natural_numbers  there are two conventions for the set of natural numbers: it is either the set of positive integers $ \{1, 2, 3, \ldots \}$ according to the traditional definition or the set of non-negative integers $ \{0, 1, 2, \ldots\}$ according to a definition first appearing in the nineteenth century.

Any other interesting observation you'd like to share LaChouetteRieuse?

[quote="Portland"]Any other interesting observation you'd like to share LaChouetteRieuse?[/quote]No, I'm sorry.
It's just I've alays used the convention $ \mathbb{N}$ for $ \{0,1,\dots\}$ and did not think that you had eliminated n = 0.
Suddenly, that's much harder...

$ f(X)$ is irreducible iff $ f(X\plus{}c)$ is irreducible for any $ c \in \mathbb{Q}$.

so how about

$ f(X) \equal{} X^{2^n}(X\minus{}a)^{2^n}\plus{}1$

then

$ f(X\plus{}\frac{a}{2}) \equal{} (X\plus{}\frac{a}{2})^{2^n}(X\minus{}\frac{a}{2})^{2^n}\plus{}1 \equal{} (X^2\minus{}\frac{a^2}{4})^{2^n}\plus{}1$

can you generalize from your previous result now?

Nice!
How could we generalize to $ X^{2^n}(X\minus{}a)^{2^n}(X\minus{}b)^{2^n}$, with $ a,b\in \mathbb{Z}$, $ ab\neq0$ and $ a\neq b$?


Hi, I am trying to include images in my TeX document, but I found out that the images must be in the postscript format. Could someone please inform me of what exactly a postscript image is (just another image file format, like bmp, jpg, gif, etc...?) and how I could possibly convert a bmp/jpg/gif image to the postscript format? thx

Fierytycoon

The format you want is extension .eps. You can get a program called Image Alchemy, and you can get a fully functional trial for some amount of time (I forget how long) on the web. Try Googling for it, since I don't remember the URL.

I believe this is the site that you want: [url]http://www.handmadesw.com/Products/Image_Alchemy_PS.htm[/url]

I personally like PostScript a lot.  That is, the .ps file format that is somewhat popular among math things.  It has better print quality than PDFs, anyway.  I don't think I've used .eps much, but you may find it useful.

Thanks for the information. But, two things. Firstly, in some TeX examples, I saw, the pictures were of type .ps and not .epsi (for example, '\includegraphics[height=2.0in,width=5.0in]{pp1.ps}'). Also, do you know where I could get a free utility rather than a trial version of a selling product?

Fierytycoon

Although I think .ps can be a type of picture, in my experience I've seen it look a lot like a PDF, except you need a special app like Ghostscript + GSView to view it.  And I wish I could tell you where to find a free app, but it's a secret!   :twisted:  Not.  I don't know where to find one, but good luck.

Yeah, I use gsview, but I need to find a piece of freeware that allows me to convert to .epsi format. I mean, what do you use now after your trial period ran out?

Fierytycoon

[quote="ComplexZeta"]From my experiences, gsview is the best way to view postscript files. It can be found for free download at [url]http://www.cs.wisc.edu/~ghost/gsview/[/url].[/quote]

Yes, I think I mentioned that in the Resources section, under "large file of downloadable problems."  Either that, or it was under "Suggestions", when I was doing the Website List...

Either purchase it or uninstall it and then install it again, unless it puts something in the registry. But if you like a program, and you think you would use it, it's good to support the company and purchase the program. Is it even that expensive?

Er, let's not get into the "cheating the AMC" type topic again.  But yes, if you really like something, why not pay the $5 or $10 to use it!  Some hard-working computer programmer (read: former math nerd, like you and me) would sure like it if you would...

[quote]why not pay the $5 or $10 to use it! [/quote]

yes, yes, of course..as in the case of AMC/AIME exams. But you see, the product in question is priced at $395...far more than I could afford for a program that I would only use to convert to .epsi format.

Fierytycoon

Oh.  Okay.  I see what you mean, then.

I wasn't sure what the price was, but $395 is [i]just a little[/i] excessive, don't ye think?

Hmm...well, perhaps some of the fewer members could help me out here. I'm still looking for a free program to convert to images to .epsi format.

You might want to check this out:

http://www.imagemagick.org/

A guide to including graphics in TeX:

http://tex.loria.fr/graph-pack/epslatex.pdf

I found the above at http://www.tug.org/ .

Hmm, the software looks quite promising. I will check it out sometime. Thanks for the tip.

..on second look, it seems quite troublesome to actually build the binaries on windows, since the binaries aren't available for download. Did you have to build it before?

Actually, I have jsut discovered htat there is a very common program called jpeg2ps.exe that you is available for download everywhere on the web (just google it), and it serves my purposes quite nicely (very small .epsi files).

what are you using all of this software for?

I didn't download it; it did provide a web-based tool to where you can upload your image to manipulate though.

Me, I just used photoshop.

EPS files are needed for TeX (a math typeset language), that's why we're converting JPG files to EPS, which we need a software to do.

no what i meant was what are you trying to do?  Write a paper, or a proof, or ?  Im just curious as to the need of all of this during high school.  I can understand its use in college, but in high school...

Oh. I participate in MOCP (Math Olympiads Correspondence Program). I send solutions to Prof. Edward Barbeau at Univ. of Toronto. I have messy handwriting, so I type my solutions up in TeX, print them and send them to him.

I'm typing up my notes from Mathcamp 2003 as well...

i see.  I am going to use it to type up USAMTS problems ( i  have awful handwriting too), but i was just curious to see why other people use it, since virtually everyone here does.

Yeah, USAMTS problems and for the math journal at my school.


Construct the line $ d$ be the perp bisector of $ E_aH_b$; $ O'$ be the symmetric point of $ O$ wrt $ d$. Draw the line $ d'$ such that $ d'$ though $ O'$ and parallel to $ d$. The intersection of $ d'$ and the circle $ (E_a;E_aH_b)$ is just the orthocenter $ H$ of triangle $ ABC$. Let $ A$ be the symmetric to $ H$ wrt $ E_a$. $ HH_b$ meets the circle $ (O;OA)$ at $ B$. Construct $ C$ easily.


So like posts 2, 3, 5, 7, etc. win.

But posts 1, 4, 6, 8 etc. lose.

You lose, I win. Yey.

h_s_potter2002: 0
1=2: 1

I win too!  Yay!

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1

A point: People don't want to lose, do they? So if there has been a prime number of osts, and the next post will make the number  a composite, the no one will post will they?

P.S. h_s_potter2002: 0
1=2: 1
lingomaniac88: 1

Maybe, but it should just be based on # of prime# posts.

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1

[quote="Treething"]i suck

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0[/quote]

But I dont!

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1

Post #9
h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 0

Yeah, otherwise there's no point to the game. This is actually a fun game.

By the way, mind putting the post # you're at in your post? That should be a rule, it would make things a lot easier. Like this would be post #9, whoever writes post #11 gets a point.

We could organize it like hurt and heal, where a few people sign up, but anyone can randomly post. The rules would be like:
- List all scores and post # in your post
- Only people who sign up have score
- No Double Posting
etc

or hey you could do it like this, 1 point if you're not on a prime and 5 points if you hit a prime

#10

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 0

Nah, it's more pointless this way.

I'm good at this game!

#11

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 0
randomdragoon: 1

I'm still at 0
#12

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 0
randomdragoon: 1
Protestant: 0

#13 :D

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0

I lose.

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0
vishalarul: 0

i lost too

#17
h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0
vishalarul: 0
junggi:1

I actually don't like the idea that I keep seeing myself lose in every post.

#19
h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0
vishalarul: 0
junggi:2

and junggi takes the lead!

20

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0
vishalarul: 0
junggi:2
perfect628:0

eh, w/e

I suck too...

[b]21[/b]

h_s_potter2002: 0
1=2: 1
lingomaniac88: 1
baseballfreak2: 1
Treething: 0
Jackdillon:1
Calc Rulz: 1
randomdragoon: 1
Protestant: 0
vishalarul: 0
junggi:2
perfect628:0


In Allen Hatcher's Algebraic Topology (http://www.math.cornell.edu/~hatcher/AT/ATpage.html) he places two vertical bars to the left of theorems. On his web page he explains how to do this in $\text{\TeX}$, but does anyone know how to do this in $\text{\LaTeX}$?

The $\text{\TeX}$ code he uses is below, but not knowing $\text{\TeX}$ I have no idea how it works.

\long\def\proclaim #1. #2\par{\bigbreak\vbox{\hbox {\vrule\kern1.5pt\vrule
\everypar{\leftskip 5pt\rightskip 4pt}\vbox{\noindent {\bf #1.\enspace}{\sl#2}}}}
\par\ifdim\lastskip<\medskipamount \removelastskip\penalty55\medskip\fi}

Basically, $\text{\LaTeX}$ adds a bunch of higher-level macros to $\text{\TeX}$. Anything written in plain $\text{\TeX}$ should work with $\text{\LaTeX}$. (Although not necessarily with the simplified $\text{\LaTeX}$ of this forum.)

Thank you for the information. I would still like to know how to do this with $\LaTeX$ commands. So far I have come up with

\documentclass[10pt,letterpaper]{article}
\begin{document}
\noindent \rule{.04em}{[color=red]2em[/color]}\hspace{1.5pt}\rule{.04em}{2em}\hspace{2pt}
\begin{minipage}[b]{[color=blue].98[/color]\textwidth}
This is a two line test. This is a two line test. This is a two line test. This is a two line test. This is a two line test. This is a two line test.
\end{minipage}
\end{document}

Is there a way to auto set the red number based on the text in the minipage? And is there a way to more acurately set the blue number? I just went by eye to get .98.

In the end I would like to create a doublebar environment that would place the bars automaticly on the side of whatever is put in the environment, e.g. theorem enviroment, paragraph, etc.

Have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=changebar]changebar package[/url] (included with MikTeX) which may do what you want. The [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=changebars]TeX FAQ[/url]suggests other packages.

This is what I ended up with

\newlength{\dvbar}
\setlength{\dvbar}{\textwidth}
\addtolength{\dvbar}{-6.5pt}
\newenvironment{doublebar}
{\noindent\vrule\hspace{1.5pt}\vrule\hspace{2pt}\begin{minipage}{\dvbar}}
{\end{minipage}}

I am still working on getting tha space on the top and bottom of the environment to behave like other AMS theorem environments, i.e., leaving the right amount of space above and bellow the environment. Any ideas on how to improve the code or get the spacing working are welcome.

You need to be clearer about the spacing you want eg by giving a complete example with indications of where the vertical spacing is too little or too much.

Meanwhile \vspace is very flexible allowing you to use as much space (including negative values) as you wish.

The final solution is

\newlength{\dvbar}
\setlength{\dvbar}{\textwidth}
\addtolength{\dvbar}{-6.5pt}
\newenvironment{doublebar}{\par\addvspace{10pt}\noindent\vrule\hspace{1.5pt}\vrule\hspace{2pt}  \begin{minipage}[t]{\dvbar}}{\end{minipage}\par\addvspace{10pt}}

After looking through the forum I found the command \addvspace which adds space but will not double add space.


topology
advanced fields
advanced fields unsolved

I want to prove that if G is a k-colour-critical graph then G is finite.

Any idea about this?
Thank you in advance.

Ah, that Erdos-deBruijn theorem again! You need to show that if every finite subgraph can be properly colored using $ k\minus{}1$ colors, then the whole (infinite) graph can also be properly colored using $ k\minus{}1$ colors. Let $ V$ be the set of vertices. Consider the set $ X\equal{}\{1,\dots,k\minus{}1\}^V$ with the product topology (the topology on $ \{1,\dots,k\minus{}1\}$ is discrete). Now, for every finite set $ U\subset V$, consider the set $ Y_U$ of all colorings of $ V$ that use $ k\minus{}1$ colors and are proper on $ U$. These sets are compact and enjoy the finite intersection property. Hence, there is a point common to all these sets. This point gives the desired proper coloring of $ V$.

The key step was Tihonov's theorem about compactness of the product. It is highly non-trivial, of course, but its proof can be found in any decent topology textbook so I do not think it would make much sense to post it here.

A little side remark: it is always a good habit to explain the terminology you are using when posting a problem, especially when it takes just a couple of lines or a single link. We know how to use google and mathworld, of course, but it is still much better to have the explanation of unfamiliar terms either right in the post or one mouse click away ;).


After the Olympiad today (October 28), schools have until October 29 to mail their answer sheets for processing. Therefore, please refrain from discussing the contest on AOPS until October 30.

Thanks.

Rick Yanco
Worcester Academy


A volleyball coach has three setters and eight other players on her team.
Without assigning any of the positions, in how many ways can she choose
six starters if exactly one of the three setters is selected as a
starter?

[hide]
Is it just $3 \cdot \frac{8!}{5!3!} = 3 \cdot \frac{8 \cdot 7 \cdot 6}{6} = 8 \cdot 7 \cdot 3 = 168$?
[/hide]

I still doubt you would have gotten it faster than Neal.

I don't quite remember how fast he got it, since I was in a post-written round drunken stupor. You're probably right.

How long did it take Neal?  It took me about 4 seconds to get 3*C(8,5)... so it took me about a total of 8 or 9 seconds.

Well, Neal took a long time because mark answered in really quickly and missed it.

[quote="09husbri"]So could I! That is, if I had placed twelfth written and beaten number five and then gotten into the semis and beaten somebody in the semis - somebody who is already a national champion, mind you - and then lasted long enough in the finals to get to the fifth question... ;)  :lol:[/quote]
Yeah! If 48 people died strategically, I could have won first place at Written Competition!

I was amazed at how fast some of the questions were answered, and disappointed as I saw obvious questions being missed.

I know, it is much more different to be up onstage than watching it from a table at the floor level.

What a shame Mark missed it. What answer did he give? Maybe he did
3 * 8C6?

[quote]So could I! That is, if I had placed twelfth written and beaten number five and then gotten into the semis and beaten somebody in the semis - somebody who is already a national champion, mind you - and then lasted long enough in the finals to get to the fifth question...  [/quote]

yea...well if 227 ppl died and i was left i could have won... :?

-rorri :?

If I got 1 more Sprint right I could be in the CDR.

I totally agree with 168. I was at Nationals and I got the answer when Neal buzzed in. So sad I didn't make Nationals. Woe is me. Woe is me. Woe is me and 215 other people at Nationals who didn't make Countdown. My friend from my national team got Neal's autograph and plans to sell it on Ebay.

I got the answer way before Mark buzzed in, but I had no pressure on me what so ever.

I don't really remember what happened, I gather from these posts that Mark buzzed in like 4 seconds after the problem was showed. Did you mean Neal?

No, I definetly meant Mark, the first one to buzz in.

Yes, but again, I was under no stress at all.


inequalities
trigonometry
algebra unsolved
algebra

Find all real solns to the system of inequalties:
sinx+cosy>=sqrt2
siny+cosz>=sqrt2
sinz+cosx>=sqrt2

Bomb

Add the inequalities and notice that equality must hold since $\sin x + \cos x \leq \sqrt{2}$ for all real numbers $x$. Do you see how to finish it?


Do a search for mass point geometry on the forums.  There have been a few posts on it.


function
calculus
integration
complex analysis
topology
complex analysis unsolved

I'm trying to solve the following problem - show that if we define
$ f(z) \equal{} \bar{z}$ (z conjugate) on the unitc circle $ S^{1}$, then there is no analytic
function $ g$ defined on the whole unit disc $ D^{2}$ such that $ f \equal{} g$ on $ S^{1}$.

Can it be shown that if such a g existed, it would have to be one-to-one?

If it's true, the rest follows from Schwarz's lemma, because then g would be
a conformal disc self-map.

So you want to show basically, that $ f(z)\equal{}\bar{z}$ is not analytic, right?

Easy.  Use Cauchy-Riemann.  Write $ f(z)\equal{}u(x,y)\plus{}iv(x,y)\equal{}x\minus{}iy$.  Then
$ \frac{\partial u}{\partial x}\equal{}1\neq \minus{}i\equal{}\frac{\partial v}{\partial y}$
$ \frac{\partial u}{\partial y}\equal{}0\equal{}\minus{}\frac{\partial v}{\partial x}$

But the first equation failed so $ f(z)\equal{}\bar{z}$ is not analytic.  Analytic basically means "no dependence on $ \bar{z}$, no naturally $ f$ can't be analytic.

That was not the question: the OP wants to show that there is no analytic function on the unit disk that agrees with $ \bar z$ [b]on the boundary[/b].

Depends a little bit on what it means by being equal on the boundary. There are analytic functions whose boundary values, in some weak sense, are equal to the boundary values of an anti-analytic function.

But suppose that the sense in which the boundary values are equal to $ \bar{z}$ is strong enough to say that $ g$ is the cauchy integral of its boundary values (e.g. if $ g$ extends continuously to the boundary). The cauchy integral of $ \bar{z}$ is zero, so it's a contradiction.

The most natural sense here: $ f$ is defined on some open set which contains the closed disk, and agrees with $ \overline{z}$ on the boundary.

The natural choice for an analytic function which agrees with $ \overline{z}$ on the circle is $ \frac1z$ - but that's not analytic on the whole disk. The Cauchy integral detects this need for a singularity somewhere in the middle.

The way I look at Cauchy integrals is that they take the boundary values' fourier series and removes the negative coefficients. In the present case $ \bar{z}$ has its -1:th coefficient as 1 and all others zero. So the cauchy integral is 0.

OK, we were using different language. I was thinking of Cauchy's theorem that $ \int_C f(z)\,dz\equal{}0$ if $ f$ is analytic on the region enclosed by $ C$.

If $ C$ is the unit circle, $ \frac1{2\pi i}\int_C z^{\minus{}n\minus{}1}f(z)\,dz$ gives the $ z^n$ Fourier coefficient. The $ \minus{}1$ coefficient is $ \frac1{2\pi i}\int_C f(z)\,dz$, which is 1 in this case, and zero for all functions analytic on the disk. Since $ \int_C z^kf(z)\,dz\equal{}0$ whenever $ k\ge 0$ for functions analytic on the disk, these functions have zero Fourier coefficients at all negative indices.

That's OK, but my question was: how can we show that any holomorphic function with such a boundary condition must necessarily be one-to-one?

Well, in a sense we did show that because we showed no such function existed, but why do you really want to know how this can be shown directly somehow? Questions about univalent functions are not easy.

If you really want to use Schwarz-lemma type stuff, you can show that $ g$ will be a finite Blaschke product (a product of conformal disc self-maps). Just factor out the zeros with conformal self-maps and use a Schwarz lemma typ argument.

The most natural approach is not this however


The initial speed of a missile is 330m/s. At what angle should it be fired so that it hits a target 10km away? You should find that there are two possible angles. Graph the missile paths for both angles. For which angle is the target hit sooner?

$ x\equal{}v_0t\cos \theta\ , y\equal{}v_0t\sin \theta \minus{}\frac{1}{2}gt^2$

$ y\equal{}0\Longleftrightarrow t\equal{}0,\ \frac{2v_0\sin \theta}{g}$

$ \therefore x\equal{}\frac{v_0^2}{g}\sin 2\theta$


geometry
circumcircle
perpendicular bisector

Points O and H are the circumcenter and orthocenter of acute triangle ABC, respectively. The perpendicular bisector of segment AH meets sides AB and AC at D and E, respectively. Prove that $\angle$DOA=$\angle$EOA.

Does anybody have a solution to this question?  :maybe:

We solved this in Blue with Zuming...  :cursing:

We did.....and for some reason I can't call it up either :(

That's why I posted it...I didn't understand it at all


What is the largest proper divisor of 225?

Largest proper divisor = 225/smallest divisor besides 1 = 225/3 = 75

Note that a proper factor is a factor of a number that is not the number itself.


[quote="polarpan523"]im stupid

javascript:emoticon(':(')
Sad

can you help me???

javascript:emoticon(':(')
Sad

1) How many perfect squares are less than 10,000? :(  :([/quote]

[hide="Hint"]
Consider that $100^2=10000$
[/hide]

[quote="math92"][hide]well, since 100^2=10000 the answer is 100.[/hide][/quote]

[hide]
wouldn't it be 99, it said less than 10000. :?
[/hide]

think about how many squares are less than 10000. since $ 100 X 100 = 10000 $, then 99 squares are less than 10000.

And don't be so hard on yourself.  :(  :D  :(  :D

[quote="shinwoo"]extension :D

how many perfect squares are in 1000?[/quote]

Do you mean how many perfect squares are less than 1000?

[hide]Because $100^2=10000$, the perfect squares that are less than $10,000$ are $1^2, 2^2, 3^2, \ldots , 99^2$. There are $99$ of these.[/hide]


There are an equal number of pennies, nickels, dimes, and quarters in a bag.  What is the probability that the combined value of four coins randomly selected with replacement will be 41 cents?  Express your answer as a common fraction.

[hide]The only way that 41 cents can be made is with one penny, one nickel, one quarter, and one dime. The probability of picking those in that order is (1/4)^4 but it can be picked in 4! = 24 different ways. So the answer should be 1/256 * 24 = 3/32.[/hide]


Two runners jog around an oval path at constant rates, but in opposite directions. One runner completes a lap every 10 minutes,
and the other does a lap every 12 minutes. How many minutes
elapse between consecutive meetings of the runners? Express your
answer as a mixed number.

The ratio of A's time to B's time is 5:6, so their rates is 6:5. Thus, their first meet is 6/11 of the track (in A's perspective) and the second meet is 12/11 or 1/11 of the track. The distance is 6/11, which is 360/11=32 8/11 minutes.

One runs 1/10 per minute. Other runs 1/12.

1/10+1/12=11/60. THus the answer is the reciprocal or 60/11=5 5/11

Wait how is the reciprocal related to this rate


Graph y = 4 - x^2 and y = |4 - x^2| on the same xy-plane.

Then explain the relationship between the graphs of y = f(x) and y = |f(x)|.

Both graphs intercept at the parabola im too lazy to figure out  :D

parabola maximum: $(0,4)$ and has roots $\pm 2$

its the same two functions, just one is absolute valued, so basicly all the negative values for y just become positive.


Let $ ABC$ be a triangle whose circumcircle is $ (O)$ and such that $ AC<AB$. The tangent lines to $ (O)$ at $ B,C$ intersect at $ T$. The lines through $ A$ and perpendicular to $ AT$ meet $ BC$ at $ S$. Choose the points $ B_1$ and $ C_1$ on ST such that $ TB_1\equal{}TC_1\equal{}TB$ and that $ C_1$ lies between $ S$ and $ T$. Prove that $ ABC$ and $ A_1B_1C_1$ are similar

http://www.mathlinks.ro/viewtopic.php?t=284159
this topic should be locked


I am in desperate need of a book on combinatorics book that I could learn how to do the harder AIME combinatorics problems, especially some thing that would teach me how to use recursion to do these types of problems. What book would you suggest? Or maybe there's some online notes on this topic?

You could try the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=2]Art of Problem Solving Volume 2[/url], although I've heard it may not be as hard as the harder AIME problems.  You could also just practice on [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]old competitions[/url].

The Art and Craft of Problem Solving, by Paul Zeitz, has a good chapter on combinatorics that should help you with the tougher AIME counting problems.

[quote] I have AOPS 2 and its combinatorics section isn't very good and practicing on old competitions doesn't make sense if i don't have enough theory to do the problems. Is there like some good MAA publication, like lets say Geometry Revisited, but about combinatorics?
[/quote]

i suggest Mathematics of Choice by Ivan Morton Niven; this is part of maa and i in fact have this book. i agree in aops 2 that the combinatorics chapter is a bit identities and theorems while than problem solving.


geometry
inradius
trigonometry
inequalities
inequalities proposed

Let $R$ be the cicumradius and $r$ be the inradius in triangle $\triangle ABC$ prove that  $\frac 12 (\frac {R+r}{R})^2-1\leq \prod \cos A\leq \frac {r^2}{2R^2}\leq \frac 18$

Ehsan2004, one of the inequalities follows from this one:

 $(b^2+c^2-c^2)(c^2+a^2-b^2)(b^2+c^2-a^2)\leq (b+c-a)^2\cdot(c+a-b)^2\cdot(a+b-c)^2 $ ;)


Construct the [[Brocard point]] of a given triangle.

Is this a lame way to expand a stub? Sure, I'll help.

Um, no. The link was because no sane person (i.e. people who don't go around looking up random geometry things) would know what a brocard point was.

Really.

Well, I guess so. Not sure though, where did I see this? Geometry? Maybe. Oh yeah, because the Brocard circle is the seven point one. Aha!  :blush: Conway mentioned this in a lecture(I think.)

So you only want one Brocard point? Triangle ABC(shamefully not labeled by me) could give you the next one with the reverse orientation(ACB[though I'm not sure about a third point, nor willing to take the time.])  :rotfl:

So basically it's the intersection of these epicycles of ABC. And basically the center of those are like the perpendicular bisectors and then the next lines perpendicular through a vertex.  And then the circles radius is to that vertex(or the other of that line for that matter.)

Temperal, I don't know much about the images here, so if you would be so kind to look at my gallery and explain....


Find all $n$ is positive integer such that :  $P(x)=x^n + 64$ is reducible (over $\mathbb{Z}[x]$ ).
(Don't use complex)


7x+5(x-3)=69

I know this is basic but can you explain the steps.
I need to know how to simplify and solve this algebraric expression using demonstrative associative and communitive properties.
Can I call someone to help me with this? I am having a mental block!! help!!

hello, your next step must be
$ 7x\plus{}5x\minus{}15\equal{}69$
Sonnhard.

The correct names for the three properties are:

Commutative property
Associative property
Distributive property

This way, you will be able to look them up on Google or Wikipedia to see how to use them.

Here's how to distribute.

Ok, say you have $ 5 (x \plus{} 2)$.

Multiply the first term in the parentheses by $ 5$, and do the same with the second.

Thus, you should get $ 5x \plus{} 10$.

Try These:

$ 7 (x \minus{} 4)$

$ \minus{}4 (x \minus{} 8)$

$ 2x (x \plus{} 2)$

Given the order of operation rules, there is nothing we can do with 7x+5(x-3),
unless we start applying those properties

The distributive property tells you that 5(x-3)= 5x – 15
Knowing that,
7x + 5(x - 3) = 69  translates into
7x + (5x – 15) = 69
Then, the associative property tells you that 7x + (5x – 15) = (7x + 5x) – 15
So, an equivalent expression is:
(7x + 5x) – 15 = 69
The distributive property says that 7x + 5x = (7 + 5)x, which simplifies to 12x, so the equation simplifies to:
12x – 15 = 69

You did not need the commutative property here. The equation is now simplified. If you go further, I would call that solving the equation.

Only beginners are required to show and explain every little step like that. Soon you will go without explanation to
7x + 5x – 15 = 69 (without brackets), and then you'll say that you get to
12x – 15 = 69 by "collecting like terms."
[hide="Unsolicited Algebra lesson"]Algebra is a foreign language. Luckily, the number of symbols whose meaning needs to be memorized is small. There is also a syntax, which includes rules about the order of operations. The order of operations rules say that you first calculate what’s inside bracketing symbols and that some operations have priority over other, including doing multiplications/divisions before addition/subtraction. When we read 5 X 10 + 5 X 2 we understand that it means we will multiply first to get 50 and 10 and then we will add 50+10=60.
The set of real numbers, with addition and multiplication as they have been defined, becomes more than just the sum of its parts, just like a room plus some equipment, utensils and a cook becomes a kitchen. The set of real numbers acquires a structure, with the properties (commutative, associative, and distributive) that you want to apply (and others).
Note that there is a [b]commutative property[/b] for addition and one for multiplication:
7 + 5 = 5 + 7
7 X 5 = 5 X 7
There are also two [b]associative properties[/b], that tell you that when adding a bunch of numbers, you can pair them up any way you want and you will get the same result,
for example    (7 + 2) + 8 = 7 + (2 + 8),
and that the same thing can be said about multiplying. (Once teachers are sure that you understand that, they will let you write just "7 + 2 + 8" and save some ink).
The [b]distributive property[/b] applies to a combination of addition and multiplication. For example
5(10 + 2) = 5 X 10 + 5 X 2   or
5(40 - 3) = 5 X 40 - 5 X 3[/hide]

[b]Temporarily locked until an issue is resolved.[/b]


Prove the inequality

$\frac{b^2+c^2-a^2}{a\left(b+c\right)}+\frac{c^2+a^2-b^2}{b\left(c+a\right)}+\frac{a^2+b^2-c^2}{c\left(a+b\right)}\geq\frac32$

for any three positive reals $a$, $b$, $c$.

[i]Comment.[/i] This was an attempt of creating a contrast to the (rather hard) inequality at the QEDMO before. However, it turned out to be more difficult than I expected (a wrong solution was presented during the competition).

  Darij

I will post a solution .I will use  sums .

The ineq is equivalent to show that

$\sum\frac{b^2+c^2}{a(b+c)}\geq\frac{3}{2}+\sum\frac{a}{b+c}$.

I will use the fact that  $\sum\frac{a}{b+c}\leq\frac{1}{4}(\sum\frac{a+b}{c})$

So it suffices to prove that

$\sum\frac{b^2+c^2}{a(b+c)}\geq\frac{3}{2}+\frac{1}{4}(\sum\frac{a+b}{c})$

which is equivalent to $\sum\frac{3b^2+3c^2-2bc}{a(b+c)}\geq 6$

We have $\sum\frac{3b^2+3c^2-2bc}{a(b+c)}\geq 2(\sum\frac{b^2+c^2}{a(b+c)})$

and by Cauchy in Engel form we have $2(\sum\frac{b^2+c^2}{a(b+c)})\geq 2\frac{(a+b+c)^2}{ab+bc+ca}\geq 6$

which is the desired

Although, Silouan solved this first, here is an alternative solution a little more directly to this nice inequality of Darij. :)

  [color=blue][b] Problem[/b]

   Let $a, b, c>0$ three positive real numbers. Prove the following inequality:

                  \[ \frac{b^{2}+c^{2}-a^{2}}{a(b+c)}+\frac{c^{2}+a^{2}-b^{2}}{b(c+a)}+\frac{a^{2}+b^{2}-c^{2}}{c(a+b)}\geq\frac{3}{2}.  \]

   [/color]

  [i] Solution. [/i]

      Anyway, the inequality is equivalent with this one:

                           \[ \frac{b^{2}+c^{2}}{a(b+c)}+ \frac{c^{2}+a^{2}}{b(c+a)}+\frac{c^{2}+a^{2}}{b(c+a)}-(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq\frac{3}{2}.  \]

        But, from $b^{2}+c^{2}\geq\frac{1}{2}(b+c)^{2}$, it remains to prove that

                              \[ \frac{1}{2}(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c})-(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq\frac{3}{2}.  \]

       But, now using Levinson inequality, \[ \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})  \], it remains to prove that \[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}  \], which is Nesbitt.

Nice solution, Cezar ! I like this sequence :
$.....(b+c)^2\le2( b^2+c^2).....Levinson.....Nesbit.....q.e.d.$

$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$

What is Levinson inequality??? (I don't know because I am 14 year old)

now, I proved. Maybe use inequality:  $\frac{1}{x} + \frac{1}{y} \geq \frac{4}{x+y}$
 :blush:
sorry everybody

Vasc's last inequality is true...

Clearing the denominators and eliminating thousands of terms our inequality turns into
$\sum_{sym}a^3b^2c+a^4b^2\geq \sum_{sym}a^4bc+a^2b^2c^2$, which is obviously true by Muirhead

does anyone have a nice solution to this inequality proposed by Vasc?  :?

[quote="Vasc"]I think the harder inequality holds:

$\frac{b^2+c^2-4a^2}{a\left(b+c\right)}+\frac{c^2+a^2-4b^2}{b\left(c+a\right)}+\frac{a^2+b^2-4c^2}{c\left(a+b\right)}+3\geq 0$.[/quote]Is $k=4$ best possible for
$\frac{b^2+c^2-ka^2}{a\left(b+c\right)}+\frac{c^2+a^2-kb^2}{b\left(c+a\right)} +\frac{a^2+b^2-kc^2} {c\left(a+b\right)}+(k-2)\frac32\geq 0$ to hold :?:

is this inequality called levinson's?
 $\sum \frac{S-x_i}{x_i}\geq (n-1)^2\sum \frac{x_i}{S-x_i}$

PS: the proof is very easy though...

The inequality

$\frac{b^2+c^2-4a^2}{a\left(b+c\right)}+\frac{c^2+a^2-4b^2}{b\left(c+a\right)}+\frac{a^2+b^2-4c^2}{c\left(a+b\right)}+3\geq 0$

is equivalent to

$\sum\frac{(b-c)^2(ab-bc+ca)^2}{b+c} \geq 0$. ;)


what happens after you finish all the problems on alcumus?

[quote="myyellowducky82"]That has not been possible, so far, but there are new problems coming out soon, so you'll just have to solve those. :wink:[/quote]

Not true. Many, including AIME15, isabella2296, and pinkmuskrat, have finished all of the problems.

when you finish all the problems, you are feeded the ones you got wrong
once you got all of those right, you have to wait for new problems to come out

[quote="vallon22"]when you finish all the problems, you are feeded the ones you got wrong
[/quote]

r15s11z55y89w21: That's what I meant. :)


[quote="MIA6"]1. Tangent and cotangent are cofunctions. They are reciprocal.  {What does 'reciprocal' mean here?}
2. tanA=sinA/cosA, cotA=cosA/sinA. As we can see, sinA=cosA. So can i say: tan A=cosA/sinA, cot A=sinA/cosA?
3. If 0< angle AOB<90, it is in the [i]first [/i]quadrant.
If 90<angle AOB<180, it is in the [i]second[/i] quadrant.
If 180<angle AOB<270, it is in the [i]third[/i] quadrant.
If 270 < angle AOB <360, it is in the [i]fourth[/i] quadrant.
Angle AOB in these ranges is a positive degree.  How about if angle AOB is a negative degree?  angle AOB=-40. So it must be in the  fourth quadrant.  I think there must be some ranges for the negative angles. It just doesn't list on the book.
4. If two angles are coterminal, the difference in their measures is 360 or a multiple of 360. For the angle, 300-(-60)=360. Can their difference be negative? As in, (-60)-300=-360.
5. IF we need to find angles coterminal with an angle of 910, is this formula right: 910+/-360k where k is a whole number. My point is an angle always has a lot coterminal angles. Is that right?

Thanks for your replies.[/quote]

1. Basically reciprocal things multiply together to get 1. $cot = \frac{1}{tan}$, and $\frac{1}{tan}* tan = 1$.

2. sinA does [b]not[/b] equal cosA.

3. If it's not between 0 and 360, just subtract or add 360 until it is.

4. Yes they can be.

5. Yes. But adding or subtracting 360 degress gets the same angle, it's just a different way of expressing it.

Hi, treething. Thanks for your answers, but i got a question: Can you prove why cot=1/tan, so that cot*tan=1?

I am pretty sure that $\cot=\frac1{\tan}$ by definition.

Multiply it out for $\cot\times\frac1{\tan}$.

$\frac1{\tan}\times\tan=1$.


The AoPS books which obviously are available at this site... and also Art and Craft of Problem Solving by Paul Zeitz is a great book. You should also see the books mentioned in the resources section. :)

chess64, I notice that link under your message. That is a great list of topics in which my school does not teach close to half of what is on there. You suggested going to the resource page for a list of problem solving books. They all seem interesting to me. I was wondering if I spend the time to read and go through every book on there except the ones under the list "For the Pros" because those are will be too difficult for me, do you think I will learn all those topics in the three pages?

[quote="chess64"]I think you probably will be able to. :) Don't rush through the books to finish all of them though; take your time in making sure you understand everything in them. ;)[/quote]

aside that the link doesnt' work anymore...

That's weird! Maybe they took it off...because of me...:?

Oh well, I'll just use google's cache :D

[quote="airelemental135"]Are there any books that can help me help me become a better math problem solver? Or one that can teach me how to write proofs?[/quote]

Proof from the books and Problem Solving Strategies are great books.

I want to learn how to write proofs too. Do you think I will be able to write them just by reading examples?

[quote="chess64"]resources section. [/quote]
What's this?

[quote="fourierseries"]To write proofs, the best way is to PRACTICE. Yes, reading proofs will help, but practice is most important. [/quote]
And practice works best when someone grades your solutions. In my experience, the USAMTS and other contests of similar format are the best way to get this feedback.

[quote="fourierseries"]For problem solving, the BEST is Arthur Engel's Problem Solving Strategies.

To write proofs, the best way is to PRACTICE.  Yes, reading proofs will help, but practice is most important.

Good luck![/quote]

I agree Arthur Engel's Problem Solving Strategies is the best, but it is quite difficult for beginners.

Also, you can have a look at Proof Techinques to see how to write some proof.


Let $a$ be a positive integer such that $5^{1994}-1\mid a$. Prove that the expression of $a$ in base $5$ contains at least $1994$ nonzero digits.

Problem of the day!!!
Firstly,we [b] Claim:[/b] $5^n-1$ has $n$ nonzero digits all equal to 4 in base 5.
[b]Proof:[/b] This is clear by induction.
Since the base case $n=1$ work,we use the inductive hypothesis for $n=k+1$
Now $5^{k+1}=5(5^k-1)+4=(44444.........4)_5$ in base 5 so our claim is correct.
So let $a=k(5^{1994}-1)$,and if $k>4$ it would be obvious,so if $k=1,2,3,4$ then:
$(5^{1994}-1)_{10}=(444.........4444)_5$ where there are 1994 4's all of which are non zero,so this satisfies.
$2(5^{1994}-1)=(14444.......4443)_5$ again satisfying the condition.
$3(5^{1994}-1)=(24444........42)_5$ again satisfying the condition.
$4(5^{1994}-1)=(344444........41)_5$ again satisfying the condition.
Hence proved.


How to find the $ Aut(S_n, S_{n\minus{}1})$. I know this is equal to $ \frac{N_{Aut(S_n)}(S_{n\minus{}1})}{C_{Aut(S_n)}(S_{n\minus{}1})}$ for all n.

Since $ Aut(S_n)\equal{}S_n$ unless $ n \not\equal{} 6$, $ Aut(S_n, S_{n\minus{}1})\equal{}\frac{N_{S_n}(S_{n\minus{}1})}{C_{S_n}(S_{n\minus{}1})}$ when $ n\not \equal{} 6$.

When $ n\equal{}6,  Out(S_6)\equal{}\frac{Aut(S_6)}{Inn(S_6)}$.

What is happening when $ n\equal{}6$ for $ Aut(S_n, S_{n\minus{}1})$?

$ n\geq 4\Longrightarrow C_{S_n}(S_{n\minus{}1})\equal{}\{id\},N_{S_n}(S_{n\minus{}1})\equal{}S_{n\minus{}1}$

Yea, that is correct. I want to know whether, when n is not equal to 6.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=16615]This thread[/url] may be of interest.


assume that $f(n)$ is the largest positive integer $m$ for which there exists some way of splitting the integers $1,\dots,m$ into n sum-free sets.(a set $S$ is sum free if $a,b\in S$ then $a+b$ is not in $S$).this is classical that $f(n)\leq [\frac{n!}{e}]-1.
$now first prove that $f(n+1)\geq 3f(n)+1$ and then prove $f(n)>89^{\frac{n}{4}-c.log(n)}$ for some absolute constant $c$ and all sufficiently large n.

The first part is fairly straightforward.  Suppose $f(n) = m$, and that $\{1,2,\cdots,m\}$ can be written as $\cup_{i=1}^na_i$.

Then let $b_i=\{x|x\in a_i\}\cup\{x|x-2m-1\in a_i\}$ for each $1\le i\le n$, and $b_{n+1}=\{m+1,m+2,\cdots,2m+1\}$.  It is easy to verify that each of the $b_i$ are sum free, and that $\cup_{i=1}^{n+1}b_i=\{1,2,\cdots,3m+1\}$, and hence $f(n+1)\ge3f(n)+1$ as required.


calculus
integration
logarithms
calculus computations

[quote="AndrewTom"]$ \int_{1}^{e} \frac {1}{t(2 \plus{} \ln t)^{2}}\ dt.$[/quote]

the given integral equals $ \int_{1}^{e}\frac{d\ln t}{(2\plus{}\ln t)^2}$,let $ u\equal{}\ln t$,
the given inegral can be rewritten as $ \int_{0}^{1}\frac{du}{(u\plus{}2)^2}\equal{}\frac{1}{6}$


[quote="lkryptonicl"]In right triangle $ABC$, $\angle{C}$ is right, $c=3$, and $a-b=\sqrt{17}$. Compute the area of the triangle.[/quote]
[hide]Umm...
c is a leg? a is the hypotenuse and b is a leg. So then $a^2-b^2=9$ and $a-b=\sqrt{17}$ so then multiply the second equation by $a+b$ and you get $9=\sqrt{17}(a+b)$, $a+b=\frac{9\sqrt{17}}{17}$ and then using $a-b=\sqrt{17}$ solve for a and the area is $\frac{3b}{2}=\frac{\frac{12\sqrt{17}}{17}}{2}=\boxed{\frac{6\sqrt{17}}{17}}$[/hide]

c is the hypotenuse buddy since angle C is 90... and a and b are the legs

a or b has to be negative :huh:

$a^2+b^2=9$
$a-b=\sqrt{17}$
$a^2-2ab+b^2=17$
$9-2ab=17$
$ab=-4$
so... a or b is negative.

[quote="lkryptonicl"]In right triangle $ABC$, $\angle{C}$ is right, $c=3$, and $a-b=\sqrt{17}$. Compute the area of the triangle.[/quote]
Uhhh...well if c is opposite $\angle{C}$ and you have equation $a=b+\sqrt{17}$...and $b$ is part of a triangle so it can't be negative...$a>3$ but $c$ is the hypotenuse...so i assumed $c$ wasn't opposite $\angle{C}$

Why cant you say that since $a^2+b^2=9$ then $a^2+2ab+b^2=17$ so you have $2ab=8$ and $ab=4$ making the area $\frac{ab}{2}=2$?  :starwars: O snap guys I messed up the problem. Its supposed to say $a+b=\sqrt{17}$.

:P

[quote="lkryptonicl"]In right triangle $ABC$, $\angle{C}$ is right, $c=3$, and $a+b=\sqrt{17}$. Compute the area of the triangle.[/quote]
[hide]
Set up the equations and you get a or b= $\sqrt{\frac{9+\sqrt{17}}{2}}$ and then $\sqrt{\frac{9-\sqrt{17}}{2}}$ so the area is those two multiplied by each other divided by two or $\boxed{2}$[/hide]

[hide]square both sides to get $17=a^2+b^2+2ab$.  Since this is a right triangle $a^2+b^2=c^2=9$ and the area is $\frac{ab}{2}$.  Thus since $2ab=8$, $\frac{ab}{2}=2$.  Notice that we don't even care about the values of a and b.[/hide]

[hide="My comment"]Yep, you don't need the values of a and b.  Just clever manipulation.

So that is why the answer is 2.
$2$[/hide]


limit
calculus
integration
geometry
3D geometry
inequalities
triangle inequality

Show that the number $r(n)$ of representations of $n$ as a sum of two squares has $\pi$ as arithmetic mean, that is \[\lim_{n \to \infty}\frac{1}{n}\sum^{n}_{m=1}r(m) = \pi.\]

We prove the following generalisation:
\[ \lim_{n\to\infty}\frac{1}{n}\sum^{n}_{m\equal{}1}r_{k}(m)\equal{}V(B_{1}^{k})\equal{}\frac{\pi^{\frac{n}{2}}}{\Gamma\left(1\plus{}\frac{n}{2}\right)}.\]
Here $ r_{k}(m)$ denotes the number of representations of $ m$ as sum of $ k$ squares (reordering and different signs counted as different representations), $ B_{1}^{k}$ is the unit ball (of radius $ 1$) in $ k$ dimensions, $ V(B_{1}^{k})$ is it's volume and $ \Gamma$ is the Gamma-Function.
In other words: the arithmetic mean of the number of representations is the volume of a unit ball.

Lemma 1: $ \sum_{m \equal{} 0}^{n}r_{k}(m)$ is the number $ L(n)$ of (integral) lattice points in $ B_{\sqrt n}^{k}$ ($ k$-dimensional ball around $ 0$ with radius $ \sqrt n$).
Proof:
This is just simple counting: a representation $ m \equal{} a_{1}^{2}\plus{}a_{2}^{2}\plus{}...\plus{}a_{k}^{2}$ of $ m\leq n$ gives a lattice point $ (a_{1}, a_{2}, ..., a_{k})$ inside the ball, and conversely every such point gives a representation the same way.

Lemma 2: We have $ V(B_{\sqrt{n}\minus{}c_{1}}^{k})\leq L(n)\leq V(B_{\sqrt{n}\plus{}c_{2}}^{k})$ for some positive real constants $ c_{1},c_{2}$ not depending on $ n$.
Proof:
We take $ c_{1}\equal{}\frac{\sqrt k}{2}\equal{} c_{2}$:
Around every lattice point $ a \equal{} (a_{1}, a_{2}, ..., a_{k})$ we can draw a [closed] cube $ C(a)$ of volume $ 1$ having the $ 2^{k}$ vertices $ (a_{1}\pm\frac{1}{2}, a_{2}\pm\frac{1}{2}, ..., a_{k}\pm\frac{1}{2})$ (all combinations of $ \pm$ used). Thus a point $ x \equal{} (x_{1}, x_{2}, ..., x_{k})$ is in $ C(a)$ iff $ |a_{i}\minus{}x_{i}|\leq\frac{1}{2}$. Especially, if $ a,b$ are two different lattice points, $ C(a)$ and $ C(b)$ do not share any interior points, thus $ V\left( C(a)\cup C(b)\right) \equal{} V\left( C(a)\right)\plus{}V\left( C(b)\right)$.
Denote by $ L_{n}$ the set of lattice points in $ B_{\sqrt n}^{k}$ (thus $ |L_{n}| \equal{} L(n)$). Define $ C_{n}\equal{}\bigcup_{a\in L_{n}}C(a)$. We get that $ V(C_{n}) \equal{} V\left(\bigcup_{a\in L_{n}}C(a)\right) \equal{}\sum_{a\in L_{n}}V\left( C(a)\right) \equal{}\sum_{a\in L_{n}}1 \equal{} |L_{n}| \equal{} L(n)$.
By this, it suffices to show that
\[ B_{\sqrt{n}\minus{}c_{1}}^{k}\subseteq C_{n}\subseteq B_{\sqrt{n}\plus{}c_{2}}^{k}.\]
Left $ \subset$:
Let $ [x]$ denote an integer closest to a real number $ x$ (taky any if there is more than one), especially $ \left| x\minus{}[x]\right|\leq\frac{1}{2}$.
If $ x \equal{} (x_{1}, x_{2}, ..., x_{k})\in B_{\sqrt{n}\minus{}c_{1}}^{k}\equal{} B_{\sqrt{n}\minus{}\frac{\sqrt k}{2}}^{k}$, we take $ a \equal{} (a_{1}, a_{2}, ..., a_{k}) : \equal{} ([x_{1}], [x_{2}] , ..., [x_{k}])$. It is a lattice point, and by $ |a_{i}\minus{}x_{i}| \equal{}\left| [x_{i}]\minus{}x_{i}\right|\leq\frac{1}{2}$ we get $ x\in C(a)$. The triangle inequality gives $ \parallel a\minus{}x\parallel\leq\frac{1}{2}\parallel (1,1,...,1)\parallel \equal{}\frac{\sqrt k}{2}$, giving $ \parallel a\parallel\leq\parallel x\parallel\plus{}\frac{\sqrt k}{2}\leq\sqrt n\minus{}\frac{\sqrt k}{2}\plus{}\frac{\sqrt k}{2}\equal{}\sqrt n$, thus that $ a\in L_{n}$ and that $ C(a)\subset C_{n}$, giving $ x\in C_{n}$.
Right $ \subset$:
If $ x\in C_{n}$, we get that $ x \equal{} a\plus{}c$ for some $ a\in L_{n}$ and some $ c \equal{} (c_{1}, c_{2}, ..., c_{k})$ with $ |c_{i}|\leq\frac{1}{2}$. Using triangle inequality again gives $ \parallel x\parallel\leq\parallel a\parallel\plus{}\parallel c\parallel\leq\sqrt n\plus{}\frac{\sqrt k}{2}$.

Using Lemma 2, we get
\[ \frac{(\sqrt{n}\minus{}c_{1})^{2}}{n}V(B_{1}^{k})\equal{}\frac{1}{n}V(B_{\sqrt{n}\minus{}c_{1}}^{k})\leq\frac{1}{n}L(n)\leq\frac{1}{n}V(B_{\sqrt{n}\plus{}c_{2}}^{k})\equal{}\frac{(\sqrt{n}\plus{}c_{2})^{2}}{n}V(B_{1}^{k}).\]
Taking limits, the result follows.

The second equality in the first statement is the standard formula for $ V(B_{1}^{k})$ which we won't prove now; for the original statement, only the area of a unit circle is needed, which is very well known to be $ \pi$.


calculus
integration
calculus computations

hello, i think this integral can only be solved by a numerical way.
Sonhard.


geometry
geometric transformation
rotation
complex numbers
geometry solved

Let ABCDEF by a hexagon inscribed in a circle, which has O as its
center so that: m(AOB)=m(COD)=m(EOF)=alpha and we denote the halves of the side BC, DE, AF with M,N,P. Prove that if the triangle MNP is
equilateral and ACE is not then m(alpha)=60.

cheers!

Is it like this?

MNP IS an equilateral triangle and AEC IS NOT one?

by "halves" you must mean middle points, do you Lagrangia? :D

oh yeah.. middle points! ... :D

grobber indeed:
MNP IS an equilateral triangle and ACE IS NOT one

Here's what I came up with:

Let's look at it in complex numbers: z=cos(alfa)-isin(alfa)
A(a), B(za), C(c), D(zc), E(e), F(ze), M((za+c)/2) etc. By considering the fact that M is obtained from N by a rotation of angle 60 with the center P, we get: m=p+(n-p)w, where w=cos60+isin60. By replacing m, n, p we get:
z=(c-(1-w)a-ew)/(-a+(1-w)e+cw). I think we use the fact that AEC is not an equilateral triangle so that the denominator is different from 0. We can easily see that 1-w=-w^2. By replacing 1-w in the expression of z from above, we get z=1/w=-w^2, so we get that b is obtained from a by a rotation of (-60) degrees, which is exactly what we want.


[img]http://gogeometry.com/problem/p274_triangle_80_80_20.gif[/img]
I prove it in a strange way  :rotfl: but I want to see your's then I will put mine.

EDIT:  :huh: I put in High School Basic how it's come here  :| the question is pretty easy. Anyway maybe I'm wrong  :P .

I'm not sure if this problem is so easy. It's very similar to this [url=http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf]notoriously hard geometry problem[/url].

No I belive it's easy. Anyway maybe my solution was wrong and I just came to the answer  :maybe:

Post your solution here (hidden of course).

[hide="I hate this problem"]
Draw the line parallel to $ AC$ through $ D$ and label the point of intersection of that line and $ AB$ as $ F$. Then, label the point of intersection Then, label the intersection of $ DA$ and $ FC$ as $ H$. Now, $ \triangle DFH$ and $ \triangle CAH$ are equilateral because $ \angle HAC \equal{} 60^\circ$. Then, draw the segment $ HG$, where $ G$ lies on $ AB$ and $ \angle FHG \equal{} 100^\circ$. Since $ \angle FHA \equal{} 120^\circ$, both $ \triangle FHG$ and $ \triangle GHA$ are isosceles. Therefore, $ AG\equal{}GH\equal{}HF\equal{}FD\equal{}HD$. $ \angle HCA$ is $ 60^\circ$, which means $ \angle FCB \equal{} 20^\circ$ and $ EC$ is the angle bisector. Using the angle bisector theorem, $ \frac{EF}{FC} \equal{} \frac{EB}{BC}$. $ \triangle BEC$ is similar to $ \triangle AGD$, so $ \frac{GA}{AD} \equal{} \frac{EB}{BC}$. If we substitute, we get $ \frac{GA}{AD} \equal{} \frac{EF}{FC}$. And then, we notice that $ DA \equal{} FC$ because of the isosceles trapezoid. THUS, $ EF \equal{} GA$. THEREFORE, $ EF \equal{} FD$. CONSEQUENTLY, $ \triangle EFD$ is isosceles. IT FOLLOWS THAT, (after taking note that $ \angle BFD \equal{} 80^\circ$ due to the parallel line) $ \angle FED \equal{} 50^\circ$ and we are done.
I hate this problem. :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:
[/hide]

[hide] Now I'm hisitaing about my solution  :maybe:  :( . It's easy to know that $ \angle BEC \equal{} 150^\circ$ and also it's easy to know that $ \angle BDA\equal{}140^\circ$.
Then I say that I want to calculate the angles of $ \angle BED$ so I say that $ 20\plus{}150\minus{}x\plus{}140\minus{}40\minus{}x\minus{}50\equal{}180$ and so $ x\equal{}20^\circ$ [/hide]

The $ 140 \minus{} 40 \minus{} x \minus{} 50$ is not correct. If $ \angle EDA \equal{} y$, then $ y \equal{} 180 \minus{} (x\plus{}50)$, and you want $ 140 \minus{} y \equal{} 140 \minus{} (180 \minus{} (x\plus{}50)) \equal{} 10 \plus{} x$. Then the $ x$ terms cancel and the equation is true trivially; this isn't surprising since all you're basically doing is using that the angles of $ BEFD$ ($ F$ the intersection of $ AD$ and $ CE$) add up to 360 when you already calculated every angle of that quadrilateral.

There's a useful article at cut-the-knot [url=http://www.cut-the-knot.org/triangle/80-80-20/index.shtml]here[/url].

Well, I have actually an easier(?) solution:
As in the link, take $F$ on $|AB|$ so that $AF=AC$, and, as we know from the classic variant, $\widehat{ADF}=30^\circ$.
But $BE=AC$ as well (for this, take $P$ inside the triangle so that $\triangle PAC$ is equilateral and, easy to see, $\triangle BEC\equiv\triangle APB$, a.s.a. case, i.e. $BC=AB,\widehat{EBC}=20^\circ=\widehat{PAB}, \widehat{BCE}=10^\circ=\widehat{ABP}$).
Now, easily to see, $\triangle ABD$ is isosceles and points $E,F$ are symmetrical about the midpoint of the base $AB$, hence $\triangle DEF$ is isosceles with apex $D$, being $\angle EDF=80^\circ$, and the conclusion follows.

Best regards,
sunken rock


$ \tan\alpha$ is the gradient of a line. Why is it necessary to take the positve direction between the x-axis and the line?

If a line is in the 2nd quadrant, do you take the acute or obtuse angle as $ tan\theta.$

If a line is in 3rd quadrant, do you take the acute angle or obtuse angle as postive?

Definately not for Classroom Math. Can someone move this?  :)

This should probably be moved to High School Basics.

Always take the angle starting from the $ x\minus{}axis$ to the line;since tangent function's period is $ \pi$, so if the line pass 1st and 3rd quadrant, take the acute one, and if the line pass 2nd and 4th quadrant, take the obtuse one.

An interesting application of this is the directions of forces, velocities, and accelerations of objects in 2 dimensions.

For instance someone that accelerated due to gravity, after jumping from a plant would have an angle below a horizontal of

$ \displaystyle \theta \equal{} \arctan\left(\frac{V_y}{V_x}\right)$

after a certain time.

Just allow the tangent to care of that stuff for you, it will spit out the correct angle, probably with a $ \pm 180^{\circ}$ range. At least, thats what my TI-84 does.


quadratics algebra polynomial function algorithm

Let a, b, c be distinct numbers such that the equations $ x^2\plus{}ax\plus{}1 \equal{} 0$ and
$ x^2\plus{}bx\plus{}c \equal{} 0$ have a common real root, and the equations $ x^2\plus{}x\plus{}a \equal{} 0$
and $ x^2 \plus{}cx\plus{}b\equal{}0$ also have a common real root. Compute the sum $ a\plus{}b\plus{}c$.

[hide]question:

Doesn't the existence of a real root of the first quadtratic contradict the existence that of the 3rd ? :huh: [/hide]

[hide="First, to answer your question"]
No, there's no contradiction. The first equation has a real root iff $ a^2 \minus{} 4 \geq 0$, and the second iff $ 1 \minus{} 4a \geq 0$. Both are satisfied for $ a \leq \minus{} 2$.
[/hide]

[hide="Second, how to cheat"]
Some problems are easy to answer if you assume that they actually have a well-defined answer :) If this sum can be calculated at all given the data, it must equal $ \minus{} 3$, since that's what you get with the simplest choice $ a \equal{} \minus{} 2$. This choice is natural since it's the edge of the allowed region for $ a$, and it makes the first quadratic have a repeated root which makes it trivial to analyze what's going on.

Of course, this does nothing to prove that $ a \plus{} b \plus{} c$ is actually single-valued given the information provided.
[/hide]

[hide="Third, a complete solution"]
This is kind of brute-force and not very clever, but there's some stuff in here that worthwhile knowing. The answer is $ \minus{} 3$.

I'm going to change notation a bit to make things more symmetric and more general. Let

$ f(x) \equal{} x^2 \plus{} ax \plus{} b$

$ g(x) \equal{} x^2 \plus{} cx \plus{} d$

be two monic quadratic polynomials. How can we determine if they have a common root? Suppose $ f(x) \equal{} 0$ has roots $ \alpha_1, \alpha_2$ and $ g(x) \equal{} 0$ has roots $ \beta_1, \beta_2$. The expression

$ R(f,g) \equal{} (\alpha_1 \minus{} \beta_1)(\alpha_1 \minus{} \beta_2)(\alpha_2 \minus{} \beta_1)(\alpha_2 \minus{} \beta_2)$

is zero precisely when $ f$ and $ g$ have a common root. This is called the resultant of the polynomials, and it is a general fact (not just for degree 2) that it can be expressed as a polynomial in the coefficients; this is a consequence of the fact the it's a symmetric function of the roots.

There are various shortcuts and algorithms for computing resultants, but we don't want to assume that they're known. Since here the polynomials are quadratic we can take a direct approach. Let $ \Delta_1 \equal{} a^2 \minus{} 4b$ and $ \Delta_2 \equal{} c^2 \minus{} 4d$. Then $ \alpha_{1,2} \equal{} \frac {1}{2}( \minus{} a \pm \sqrt {\Delta_1})$ and $ \beta_{1,2} \equal{} \frac {1}{2}( \minus{} c \pm \sqrt {\Delta_2})$. Substituting those into the definition of $ R(f,g)$ we get

$ R(f,g) \equal{} \frac {1}{16}( \minus{} a \plus{} c \plus{} \sqrt {\Delta_1} \plus{} \sqrt {\Delta_2}) ( \minus{} a \plus{} c \minus{} \sqrt {\Delta_1} \minus{} \sqrt {\Delta_2})( \minus{} a \plus{} c \plus{} \sqrt {\Delta_1} \minus{} \sqrt {\Delta_2})( \minus{} a \plus{} c \minus{} \sqrt {\Delta_1} \plus{} \sqrt {\Delta_2})$

now we can use $ (A \minus{} B)(A \plus{} B) \equal{} A^2 \minus{} B^2$ twice to obtain

$ 16R(f,g) \equal{} \left((c \minus{} a)^2 \minus{} (\sqrt {\Delta_1} \plus{} \sqrt {\Delta_2})^2\right) \left((c \minus{} a)^2 \minus{} (\sqrt {\Delta_1} \minus{} \sqrt {\Delta_2})^2\right)$

now expand the squares of the $ \Delta$s:

$ 16R(f,g) \equal{} \left((c \minus{} a)^2 \minus{} \Delta_1 \minus{} \Delta_2 \minus{} 2\sqrt {\Delta_1\Delta_2}\right) \left((c \minus{} a)^2 \minus{} \Delta_1 \minus{} \Delta_2 \plus{} 2\sqrt {\Delta_1\Delta_2}\right)$

and once again we have $ (A \minus{} B)(A \plus{} B)$:

$ 16R(f,g) \equal{} \left((c \minus{} a)^2 \minus{} (\Delta_1 \plus{} \Delta_2)\right)^2 \minus{} 4\Delta_1\Delta_2$

Now $ \Delta_1 \plus{} \Delta_2 \equal{} a^2 \plus{} c^2 \minus{} 4(b \plus{} d)$ and $ \Delta_1\Delta_2 \equal{} a^2c^2 \minus{} 4bc^2 \minus{} 4a^2d \plus{} 16bd$. Substituting these there's a little bit of cancellation going on and we get

$ R(f,g) \equal{} b^2 \plus{} d^2 \minus{} 2bd \minus{} abc \minus{} adc \plus{} bc^2 \plus{} a^2d$, namely

$ \boxed{R(f,g) \equal{} (b \minus{} d)^2 \plus{} (a \minus{} c)(ad \minus{} bc)}$

which is kind of a nice expression for the resultant - I couldn't make it much simpler.

So far this was completely general. Now let's look at the problem. We have two polynomials with a common root (that is, resultant = 0), and then another pair which is identical with the first except that the roles of the coefficients of $ x^1$ and $ x^0$ have been reversed. We also know that $ b \equal{} 1$ but let's ignore this for the moment and work in greater generality. The conditions on the common roots are

1. $ (b \minus{} d)^2 \plus{} (a \minus{} c)(ad \minus{} bc) \equal{} 0$

2. $ (a \minus{} c)^2 \minus{} (b \minus{} d)(ad \minus{} bc) \equal{} 0$

where we obtained the second equation simply by taking the first and replacing $ a \leftrightarrow b$ and $ c \leftrightarrow d$. As a reminder, we changed notation so we know that $ b \equal{} 1$ and that $ a,c,d$ are distinct. We're trying to determine $ a \plus{} c \plus{} d$.

To make things simpler we once again assign names to some expressions: Let $ s \equal{} a \minus{} c$, $ t \equal{} b \minus{} d$ and $ \delta \equal{} ad \minus{} bc$. Our equations read

$ t^2 \plus{} s\delta \equal{} 0$

$ s^2 \minus{} t\delta \equal{} 0$

In particular, $ t^2 \plus{} s\delta \equal{} s^2 \minus{} t\delta$ so $ \delta(s \plus{} t) \equal{} (s \minus{} t)(s \plus{} t)$. There are therefore two options: $ s \plus{} t \equal{} 0$ or $ \delta \equal{} s \minus{} t$. Let's do the second one first.

Case A: $ \delta \equal{} s \minus{} t$ so $ ad \minus{} bc \equal{} a \minus{} c \minus{} b \plus{} d$. Since $ b \equal{} 1$ this is $ ad \minus{} c \equal{} a \minus{} c \minus{} 1 \plus{} d$ or $ ad \equal{} a \minus{} 1 \plus{} d$, so $ a(d \minus{} 1) \equal{} d \minus{} 1$. Either $ a \equal{} 1$ or $ d \equal{} 1$. The former is impossible since this would make the first quadratic $ x^2 \plus{} x \plus{} 1$ which has no real roots. If $ d \equal{} 1$ then equation 1 becomes $ (a \minus{} c)^2 \equal{} 0$ so $ a \equal{} c$ which is ruled out by the requirement of $ a,c,d$ being distinct. So case A is impossible.

Case B: $ s \plus{} t \equal{} 0$. This transforms the equation $ t^2 \plus{} s\delta \equal{} 0$ into $ s^2 \plus{} s\delta \equal{} 0$ so either $ s \equal{} 0$ or $ s \equal{} \minus{} \delta$. Now $ s \equal{} 0$ is impossible since $ a \neq c$. We're left with $ s \equal{} \minus{} \delta$, that is $ a \minus{} c \equal{} bc \minus{} ad$. Using now $ b \equal{} 1$ this is $ a \minus{} c \equal{} c \minus{} ad$. But we also know that $ s \plus{} t \equal{} 0$ so $ a \minus{} c \plus{} b \minus{} d \equal{} 0$, so $ c \equal{} a \plus{} 1 \minus{} d$. Putting these together we get $ a(d \plus{} 1) \equal{} 2(a \plus{} 1 \minus{} d)$ which is just $ a \plus{} 2 \equal{} (a \plus{} 2)d$. So either $ d \equal{} 1$ which we already ruled out or $ a \equal{} \minus{} 2$.

We just proved that our "cheat" solution from above is in fact the only solution: $ a$ must be $ \minus{} 2$, so the quadratic $ x^2 \plus{} ax \plus{} 1$ is just $ (x \minus{} 1)^2$, it has only 1 for a root, so 1 must be a root of $ x^2 \plus{} cx \plus{} d$ as well which makes $ 1 \plus{} c \plus{} d \equal{} 0$ so $ a \plus{} c \plus{} d \equal{} \minus{} 2 \plus{} c \plus{} d \equal{} \minus{} 3$, and this is our answer.

[/hide]


Suppose that $ f$ is a continuous real-valued function defined on the closed interval $ [0,1]$.  Which of the following must be true?

I.  There is a constant $ C > 0$ such that $ |f(x) \minus{} f(y)| \le C$ for all $ x$ and $ y$ in $ [0,1]$.

II.  There is a constant $ D > 0$ such that $ |f(x) \minus{} f(y)| \le 1$ for all $ x$ and $ y$ in $ [0,1]$ that satisfy $ |x \minus{} y| \le D$.

III.  There is a constant $ E > 0$ such that $ |f(x) \minus{} f(y)| \le E|x \minus{} y|$ for all $ x$ and $ y$ in $ [0,1]$.

Give explicit reasoning.

Well if $ f$ is a continuous function defined on a closed interval, then it must reach a maximum and it must reach a minimum in that interval.

If $ f$ is continuous in $ [0,1]$, then for each $ \varepsilon>0$, $ \exists\delta>0$ such that whenever $ \left|x\minus{}y\right|<\delta$, then $ \left|f(x)\minus{}f(y)\right|<\varepsilon$.

See where these two things get you.

I can only prove 1 by using arguments of boundedness. Cant prove 2 and 3.Help