function
trigonometry
set theory
advanced fields
advanced fields theorems

I have been interested in nonstandard analysis for some time, and I have some trouble with a few parts of it because my background in math is not sufficient. In particular, I am aware that the ideas of nonstandard analysis aren't specific to some particular universe enlargement, but work for any nonstandard model of the reals.

The problem is: I don't really know what is a "model." Is anyone here familiar enough with mathematical logic to explain to me a little about this concept?

(As usual I have read Wikipedia but it didn't help much since it's just a summary. My current understanding is pretty much that a model is just a realization of some axioms but I am doubtful about teh accuracy and utility of that...)

A model really is just an object in which certain axioms hold, usually chosen so that you understand it well.

For instance, if you take a set theory class, one thing you're likely to do is prove that the axiom of choice doesn't cause new contradictions in set theory with the other axioms.  The way one does this is to take the ZF system and to find a model (in my example, a "well-behaved" class of sets) in which the axiom of choice can be proven true (as a theorem, from the properties of the class and the other axioms).  Then that model provides a model for ZFC.  If ZF had no contradictions, this new system has no contradictions.  But if ZFC had contradictions, this model must.  So Choice can't add any new contradictions.

A "model of the reals" is a set (or class, I suppose) which obeys whatever set of axioms you're slapping on the real numbers.  You want to choose particular models where you have a grasp of what is happening so that you can prove things about them.

Thanks, that kinda helps. I guess I don't really know what I don't understand about models, which makes it hard for me to ask...  :blush:

I guess one thing I could ask is this: How do you distinguish two models of the same thing? Or alternatively, how can you tell if two models of the same axioms constructed in different ways are actually the same? Does this only work on a case-by-case basis?

[quote="Xevarion"]I guess one thing I could ask is this: How do you distinguish two models of the same thing? Or alternatively, how can you tell if two models of the same axioms constructed in different ways are actually the same? Does this only work on a case-by-case basis?[/quote]

So, given that I have no actual model-theoretic experience, I have 2 potential answers for you:

Potential answer 1: Yes, it only works on a case-by-case basis.  (In set theory, at least, models tend to be explicitly defined by transfinite recursion, and you can show inductively that two constructions are different.  I really know nothing about nonstandard analysis, so I have nothing helpful to tell you about it.)
Potential answer 2: What do you mean, "the same?"

I'll try to informally describe non-standard analysis.

Let $\hat{\mathbb R}$ be a set; and $\hat\cdot$ be a transformation that maps every $n$-ary function $f:{\mathbb R}\times\dots\times{\mathbb R}\to{\mathbb R}$ to a function $\hat f:{\hat{\mathbb R}}\times\dots\times{\hat{\mathbb R}}\to{\hat{\mathbb R}}$ (for every $n \geq 0$). In particular, $\hat\cdot$ maps all real numbers ($0$-ary functions) to $\hat{\mathbb R}$; it also maps the arithmetic operations (binary functions) $+$, $-$, $\cdot$, $/$ to "arithmetic operations" on $\hat{\mathbb R}$: $\hat+$, $\hat-$, $\hat\cdot$, $\hat /$.

Now for every first order sentence $\psi$ that can use arbitrary functions but has only quantifiers on real variables, let $\hat \psi$ be the sentence in which every function $f$ is replaced with $\hat f$, and $\mathbb R$ is replaced with $\hat{\mathbb R}$. For instance, for $\psi = (\sin(\pi) = 0)$, we have $\hat \psi = (\widehat\sin(\hat\pi) = \hat 0)$; for $\forall x\in{\mathbb R}\,\exists y \in{\mathbb R}\, \sin(x) = 1-y\cdot y$, we have $\forall x\in{\hat{\mathbb R}}\,\exists y \in \hat{\mathbb R}\ \widehat{\sin}(x) = \hat{1}\,\hat{-}\, y\, \hat{\cdot}\, y$.

We say that $\hat{\mathbb R}$ is elementary equivalent to $\mathbb R$ (in a proper language) if for every true sentence $\psi$ the sentence $\hat \psi$ is also true.

If $\hat{\mathbb R}$ is elementary equivalent to $\mathbb R$, but it is not an instance of real numbers, then we say that $\hat{\mathbb{R}}$ is a non standard model of real numbers.

Hope this helps. BTW, IMHO wikipedia articles on the subject are not very good.


On an 8 × 8 chessboard whose squares are colored black and white in an
arbitrary way we are allowed to simultaneously switch the colors of all squares in any
3 × 3 and 4 × 4 region. Can we transform any coloring of the board into one where all
the squares are black?

No

Observation #1 in applying transformations, order doesn't matter
Observation #2 if it is possible to go from position A to position B, it is possible to go from position B to position A, therefore we need only prove that from the position where all squares are black, we cannot move into any position.

Note that there are exactly 25 4 x 4 transformations, and 36 3 x 3 transformations. Also applying the same transformation twice is the same as applying it 0 times, so each of these 61 transformations is applied either 0, or 1 time. Therefore, from the position of all black squares there are at most 2^61 positions that can be obtained. But, there are 2^64 total positions, so not all can be obtained.

This solution works for an n x n board for all n such that (n-3)^2 + (n-2)^2 < n^2, which is not true when n =  9, so a good question is: For which n can we transform any coloring of an n x n board into one where all squares are black?


[quote="king_23"]A formula for calculating the durg dose for young children is given as a(t3-4t2-5t) over 22t2-110t. Some of the numbers are suppose to be exponents, i think you'll find them. [b]t[/b] is the age of the child in years and [b]a[/b] is the adult dose of the drug. If the adult dose for a certain drug is 200 mg, what dose does the formula indicate for a child of:

a) age 1 yr    b) age 5 yrs[/quote]

This seems like homework? =)


Two different circles are each tangent to both the x -axis and the
y -axis, as shown in the figure. One of the points of intersection of
the circles is (1, 5). What is the product of the lengths of their
radii?

Perhaps the picture would be nice. There are quite a few different cases that the circles could be placed, but I didn't check if it matters where the circles are.

Yeah, uh, no clue what I was thinking.

There's actually only one possible case, so the diagram's not needed. :P

[hide="Solution"]
Let the centers be $ (a,a)$ and $ (b,b)$, since that's the only way the circles can be tangent to the axes. These circles satisfy $ (x \minus{} a)^2 \plus{} (y \minus{} a)^2 \equal{} a^2$ and $ (x \minus{} b)^2 \plus{} (y \minus{} b)^2 \equal{} b^2$. Plugging in $ (x,y) \equal{} (1,5)$ and simplifying and stuff, we get $ a^2 \minus{} 12a \plus{} 26 \equal{} 0$ and $ b^2 \minus{} 12b \plus{} 26 \equal{} 0$. Clearly, since $ x^2 \minus{} 12x \plus{} 26 \equal{} 0$ has two roots, the roots must be $ a$ and $ b$, with a product, by Vieta's, of $ ab \equal{} \boxed{26}$.
[/hide]


linear algebra
matrix
algebra
polynomial

$B^{6}= 0$, so $B$ is nilpotent, hence $B^{3}= 0$. Now let $B = \left( \begin{matrix}a & b & c \\ x & y & z \\ u & v & w \end{matrix}\right)$ and solve the equation $\begin{pmatrix}0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}\cdot B = 0$. Looking at the square of the $B$ you obtain, you get a contradiction.

[quote="rustam"]Prove that there doesn't exist a  matrix B with real elements such that $B^{2}= \begin{pmatrix}0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}$. What if B is complex?[/quote]

More simple? By formal reasons any such $B=TB^{2}T^{-1}$ with shown jordan block and unknown $T$ (diagonal structures lead to diagonal ones) , follows (!) $B^{2}=TB^{4}T^{-1}$, but $B^{4}$ obviously 0. We have the contradiction right here.

[quote="cnbjy"]why use Jordan form?
direct computation says $B^{4}=E_{1,3}\neq 0$,
but from the char. polynomial, $B^{3}=0$, so $B^{4}=0$, contradiction.[/quote]

u have something against Jordan form? 'Direct computation', what u mean, in this case, where matrix doesn't exist??? How u will 'directly' computate IT???  I'm afraid, there are some doubts in this way whole of so many senses. It has some a problem inside, this consideration.. mmmm.. so a contradiction we have any such way.

[quote="perfect_radio"]People usually want to avoid Jordan form in such simple problems.[/quote]

People, whum u mean? Yourselves? I kindly ask u do not go to personalities, and the more - to meaning of unknown me 'people' in such simple cases (discussion). Or u will observe your 'tasks' in proud surrounding of 'people' i do not belong, naturally.

People want, mainly, to avoid a pressure of 'right way thinking', or i do not know people well.

Especially i like there $B$^3, this $BB*B$. Let everybody share this pleasure.  :lol:

PS
Matrix algebra, is manual covered by dust of centuries, different 'proves' observed here consist an elementar logical mistake by u cannot 'prove' a fact from itself actually (taken with no shadow of doubts reasons have some logical cycle inside being proved, in its turn, from the 'proved' fact), u cannot write here this or that statement in whole observation - so, these 'proving' - a constatation, taken from your notebook. You have, as less, to understand it.  Not serious and not interesting. Real proves were made much before. And other people, so i ask to respect their work, appeared much before, also, my modern and too hot conversators. And the more, works of Jordan u all use.


I think you mean [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=aczel&t=68963]aczel's inequality[/url]  :wink:


calculus
integration
trigonometry
calculus computations

Let $a$ be a real number.
Evaluate

\[\int _{-\pi+a}^{3\pi+a} |x-a-\pi|\sin \left(\frac{x}{2}\right)dx\]

This is not very hard I think :

$I=\int_{-\pi +a}^{3\pi +a}\left| x-a-\pi \right| \sin \left( \frac{x}{2}\right) dx$

$=\int_{-\pi +a}^{\pi +a}\left( -x+a+\pi \right) \sin \left( \frac{x}{2}\right) dx+\int_{\pi +a}^{3\pi +a}\left( x-a-\pi \right) \sin \left( \frac{x}{2}\right) dx$

$=8\left( \sin \frac{1}{2}a\right) a+8\left( \sin \frac{1}{2}a\right) \pi -\int_{-\pi +a}^{\pi +a}x\sin \left( \frac{x}{2}\right) dx+\int_{\pi +a}^{3\pi +a}x\sin \left( \frac{x}{2}\right) dx$

$=8\left( \sin \frac{1}{2}a\right) a+8\left( \sin \frac{1}{2}a\right) \pi -8\cos \frac{1}{2}a-4\left( \sin \frac{1}{2}a\right) a\allowbreak -8\cos \frac{1}{2}a-8\left( \sin \frac{1}{2}a\right) \pi -4\left( \sin \frac{1}{2}a\right) a$

$\allowbreak =\allowbreak -16\cos \frac{1}{2}a$

Setting $ x \minus{} a \minus{} \pi \equal{} t$ gives more shorter solution.
That is to say, parallel transformation.


If a rooster laid 11 eggs and the farmer took 5 of them and another rooster laid 14 eggs and 5 of them were rotten, how many eggs were left :huh:  :?:

20 eggs in nest and if u count ones farmer took then 25

[quote="bigboy82892"][hide="Contradiction!"]Roosters can't lay eggs. So $0$ are left.[/hide][/quote]

good point, I know a similiar problem

Yeah, unless its a chicken dressed up as a rooster  :blush:

[quote="mathnerd314"][quote="moogra"]good point, I know a similiar problem[/quote]

:rotfl:

Answer in 5 seconds: If a daddy bull can eat 10 bales of hay and a baby bull can eat 4 bales of hay, how much hay can a mommy bull eat?[/quote]

[hide]A bull is male...[/hide]


analytic geometry
graphing lines
slope
algebra
system of equations
calculus
calculus computations

A curve is given by $y = f(x) = x^3 - \frac{4}{3}x$. Give the equation for the tangent in the point $(a, f(a))$ and then determine for which points $a$ the tangent is also a normal to the curve.

We can give the tangent by $y = a^3 - \frac{4}{3}a + (3a^2 - \frac{4}{3})(x - a) = -2a^3 + x(3a^2 - \frac{4}{3})$.

Similarly, a general normal in the point $(b, f(b))$ is given by
$y = b^3 - \frac{4}{3}b - \frac{1}{3b^2 - 4/3}(x - b) = b^3 - \frac{4}{3}b + \frac{b}{3b^2 - 4/3} - \frac{x}{3b^2 - 4/3}$

The only solution that I can think of is to compare the intercepts and the gradients and solve the system of equations in a and b. However, the system is too difficult to solve. So, what to do?

(There's also another question that I can't figure out: "There is a line that tangents the curve $y = x + 2x^2 - x^4$ in two different points. Determine the equation of this line". However, if it requires similar reasoning as the problem above, it doesn't need to be solved, I suppose).

Thanks!

edit: Is this in the wrong forum? I thought maybe it was a bit too simple for the unsolved problems forums, but at the same time no one has replied :)

In the first question, the slope of the tangent is $3a^2-4/3$, so we want $f'(x) = -\frac{1}{3a^2-4/3} \Rightarrow x= \pm\frac{1}{3}\sqrt{\frac{25-36a^2}{4-9a^2}}$

Edit: Sorry, my first solution (for problem 2) was wrong. (see post #4)

Can you please explain more carefully the solution to the "second problem" (the one you solve first)? I've thought about it several times, and I don't get your solution. How do you arrive at that $\left(x_1,x_2\right) = \left(\pm1, \mp 1\right)$, and how do you know that these solutions will be consistent with the solutions to $y' = k$ (I realize that they are, but how did you know that)? Also, how do you get that the line is $y = x+1$ (which is the right answer), doesn't plugging in the values give you $y=x-1$. I am really misunderstanding something. I don't get this problem :(

Ok, here's a better solution:

The slope along the tangent line with points $x_1$ and $x_2$ is $\frac{\left(x_2+2x_2^2-x_2^4\right)-\left(x_1+2x_1^2-x_1^4\right)}{x_2-x_1}$ which we know is equal to $y'$, so now we have the equations:
\[ \frac{\left(x_2+2x_2^2-x_2^4\right)-\left(x_1+2x_1^2-x_1^4\right)}{x_2-x_1}=1+4x_1-4x_1^3  \]
and\[ \frac{\left(x_2+2x_2^2-x_2^4\right)-\left(x_1+2x_1^2-x_1^4\right)}{x_2-x_1} =1+4x_2-4x_2^3  \]

and after a bunch of algebra, theses equations give the solutions $(x_1,x_2) = (k,k)\text{ and } (\pm1, \mp1)$.

Edit: I forgot two pairs of parentheses in my old solution, so that is why you got $y=x-1$, also I don't think that solution was right anyway (sorry about that).


function
analytic geometry
symmetry
induction
arithmetic sequence
arithmetic series

Here's a different solution that DP didn't mention in the math jam: (in outline form)

Think of the points as geometric points in the Euclidean/Cartesian plane. We express their label as a linear function of their coordinates, which then causes consecutive points to have labels that form arithmetic sequences. We then prove a lemma that the average of the linear function evaluated at each of a set of points is equal to the linear function of the average of those points' coordinates. By symmetry, the average of the coordinates of the points must be the centroid of the triangle, meaning the centroid's label function is the average of all the label function. We furthermore know that the centroid's coordinates are the average of those of the vertices, meaning that the label function of the centroid is $\frac{a+b+c}{3}$, which is the average of all the labels. We then find the number of labels to find their sum.

here's mine (yea i know it's not pretty):

We want to generalize part $a)$ with vertex labels $a,b,c$ with $n^{2}$ small triangles.

Let the labels $a,b,c$ be where the labels $4,9,1$ were in part $a)$ respectively, and let the common difference of each successive term on the line segment with vertices $a$ and $b$ be $d_{1}$, and let the common difference of each successive term on the line segment with vertices $a,c$ be $d_{2}$. So this means that $a+nd_{1}=b$ and $a+nd_{2}=c$.

The easiest way to solve this problem is to look at each of the "slanted" line segments, and add up all the labels on that line segment. For example, the first point is $a$ of course, and the first line segment then has endpoints $a+d_{1}$ and $a+d_{2}$ and these are the two total points, the second line segment has endpoints $a+2d_{1}$ and $a+2d_{2}$ and has three total points, and so on.

We add these slanted lines as such. First we find the sum of all the $a$ values, and then focus on $d_{1}$ and $d_{2}$. Because the big triangle that is composed of $n^{2}$ triangles contains $n+1$ labels on its sidelengths, the number of times $a$ shows up is $1+2+3+4+\cdots+(n+1)=\frac{(n+1)(n+2)}{2}$. So the sum for all the $a$s is $\frac{(n+1)(n+2)}{2}a$.

Now we want to add up all the $d_{1}$ and $d_{2}$ values. Now for each slanted lines, (note that we don't include any of the $a$s now because we already found the sum for all those), we have that the first slanted segment has the vertex labels (and these are the only labels for this segment) $d_{1}+d_{2}$.

Continuing in this manner, and noting that all the values of the labels on these slanted segments are in an arithmetic progression, we can add up the values through the arithmetic series formula (and letting $d_{1}+d_{2}=x)$, thus getting:

${\frac{1}{2}(2x+6x+12x+\cdots (n+1)(nx)}=x(1+3+6+10+\cdots+\frac{n(n+1)}{2})$

Now using the fairly well known formula $1+3+6+10+\cdots+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}$, we simplify the above sum to: $x(\frac{n(n+1)(n+2)}{6})$.

Now using the fact that $a+nd_{1}=b$ and $a+nd_{2}=c$, substituting in for $x=d_{1}+d_{2}$, we have that the sum evaluates to: $(\frac{b+c-2a}{n})(\frac{n(n+1)(n+2)}{6})=(b+c-2a)(\frac{(n+1)(n+2)}{6})$

Now after combining this with the sum of all the $a$s, we have that the formula is $S=\frac{(n+1)(n+2)}{2}a+(b+c-2a)(\frac{(n+1)(n+2)}{6})$, where $S$ represents the sum of all the labels.

Now we can simplify this formula by combining the terms together and simplifying to get $S=\frac{(n+1)(n+2)(a+b+c)}{6}$ and plugging in our values, we get $S=\frac{(6)(7)(14)}{6}=98$, and indeed satisfies our conditions.

Here is my painstakingly convoluted solution:

PROBLEM 3
PART A
Since the vertices on the left edge form an arithmetic sequence that starts with 1 and ends with 4, we can easily figure out every number on those vertices. Common difference $d = \frac{4-1}{5}= \frac35$, thus the numbers (from top to bottom) are $1+\frac35$, $1+\frac65$, $4-\frac65$, and $4-\frac35$. Likewise, the numbers on the right edge are $1+\frac85$, $1+\frac{16}{5}$, $9-\frac{16}{5}$, and $9-\frac85$. The numbers on the bottom are 5, 6, 7, and 8. Now the unknown vertices are the 6 in the interior of the triangle. However, we only wish to compute the sum, and we can achieve this by noting the fact that vertices on every horizontal line form an arithmetic sequence.

For example, on the second horizontal line from the top, the first vertex is $1+\frac65$, while the last is $1+\frac{16}{5}$. We can find the sum by using the formula for the sum of arithmetic sequence. $ \Sigma = \frac{[(1+\frac56)+(1+\frac{16}{5})] \cdot 3}{2}$. Using the same method, we compute the sum for every line:

Summing the results yeilds $\boxed{98}$.

PART B
LEMMA: $ \sum_{i=1}^{n}\frac{i(i+1)}{2}= \frac{(n^{2}+2n)(n+1)}{6}$\\
{\bf Proof: } \[ \sum_{i=1}^{n}\frac{i(i+1)}{2}\]  \[= \sum_{i=1}^{n}\frac{i^{2}+i}{2}\]  \[= \frac12\left( \sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}i \right)\]  \[= \frac12\left( \frac{n(n+1)(n+2)}{6}+\frac{n(n+1)}{2}\right)\]  \[= \frac{n(n+1)(n+2)+3n(n+1)}{12}\]  \[=  \frac{(2n^{2}+4n)(n+1)}{12}\]  \[=\frac{(n^{2}+2n)(n+1)}{6}\]  \[--\text{End Of Lemma}--\] To generalize part (a), we still employ the method of summing individual vertices on every horizontal line (including the tip at the top). Without loss of generality, suppose $a$ is at the top of the triangle, $b$ is at the lower left, and $c$ is at the lower right. Denote $D_{ab}$ as the common difference between two consecutive vertices on the left edge. Similarly, denote $D_{ac}$ as the common difference on the right edge. \[D_{ab}= \frac{b-a}{n}\]  \[D_{ac}= \frac{c-a}{n}\] Now, the numbers on the vertices along the left edge would be (from top to bottom): $a$, $a+D_{ab}$, $a+2D_{ab}$, $a+3D_{ab}$,  . . .  $a+(n-2)D_{ab}$, $a+(n-1)D_{ab}$ ($b$). The numbers on vertices along the right edge would be (from top to bottom): $a$, $a+D_{ac}$, $a+2D_{ac}$, $a+3D_{ac}$,  . . .  $a+(n-2)D_{ac}$, $a+(n-1)D_{ac}$ ($c$).
Let $\Sigma_{i}$ denote the sum of vertices on the $i$th horizontal line, while $\Sigma_{0}$ being the tip at the top. Also, $\Sigma_{n}$ is the last horizontal line. Thus, \[\Sigma_{0}= a\]  \[\Sigma_{1}= \frac{[(a+D_{ab})+(a+D_{ac})]\cdot 2}{2}\]  \[\Sigma_{2}= \frac{[(a+2D_{ab})+(a+2D_{ac})]\cdot 3}{2}\]  \[\Sigma_{3}= \frac{[(a+3D_{ab})+(a+3D_{ac})]\cdot 4}{2}\]  \[\vdots\]  \[\Sigma_{n-1}= \frac{[(a+(n-1)D_{ab})+(a+(n-1)D_{ac})]\cdot n}{2}\]  \[\Sigma_{n}= \frac{[(a+nD_{a}b)+(a+nD_{a}c)](n+1)}{2}= \frac{(b+c)(n+1)}{2}\] To simplify them, we break each sum into two parts: one part containing $a$, and the other part containing $D_{a}b+D_{a}c$: \[\Sigma_{0}= a\]  \[\Sigma_{1}= 2a+(D_{ab}+D_{ac})\cdot 2 /2\]  \[\Sigma_{2}= 3a+(D_{ab}+D_{ac})\cdot 6 /2\]  \[\Sigma_{3}= 4a+(D_{ab}+D_{ac})\cdot 12 /2\]  \[\vdots\]  \[\Sigma_{n}= (n+1)a+(D_{a}b+D_{a}c)\cdot n(n+1) /2\] Now we sum all the sums: \[ \sum_{i = 0}^{n}\Sigma_{i}= [1+2+\dots+(n+1)]a+(D_{a}b+D_{a}c)\sum_{i = 1}^{n}\frac{i(i+1)}{2}\] Using the lemma, \[=\frac{[1+(n+1)](n+1)}{2}a+\frac{(n^{2}+2n)(n+1)}{6}(D_{ab}+D_{ac})\]  \[= \frac{(n+1)(n+2)a}{2}+\frac{(n^{2}+2n)(n+1)}{6}(\frac{b-a}{n}+\frac{c-a}{n})\]  \[ = \frac{3(n+1)(n+2)a}{6}+\frac{n(n+1)(n+2)}{6}\cdot \frac{b+c-2a}{n}\]  \[ = \frac{3n(n+1)(n+2)a}{6n}+\frac{n(n+1)(n+2)(b+c-2a)}{6n}\]  \[ = \frac{n(n+1)(n+2)[3a+(b+c-2a)]}{6n}\]  \[ = \boxed{\frac{(n+1)(n+2)(a+b+c)}{6}}\]

My proof was basically the same as pianoforte's, except I went from n-1 to n in the inductive step.

[quote="asquared"]All of these proofs are good, and mine isn't too bad either (it's fairly similar to the first one in the math jam), but I was kicking myself when I saw the solution nallenusn posted. It's so easy, it practically trivialized the problem. I can't believe I didn't see it.[/quote]
Hmm, was it something like this? :D
[hide]It is not difficult to see that each label is a weighted average of the three corner labels.  Then (to abbreviate slightly) by symmetry the mean of the labels is $\frac{a+b+c}{3}$, which gives us the sum as $\frac{(n+1)(n+2)(a+b+c)}{6}$.[/hide]
I was kind of thinking, "Oh, I can bruteforce that with sums when I sit down to write the solution" for a few days, when all of a sudden, I though, "Aha!  Symmetry!"  (I made it a little more explicit when I wrote the solution, though.)

[quote="calc rulz"]Here's a different solution that DP didn't mention in the math jam: (in outline form)
Think of the points as geometric points in the Euclidean/Cartesian plane...[/quote]

That's exactly what I did!  It's much simpler than messy algebra.

Calc, I bet that will get you a commended solution, since it's much simpler than the one Mr. Patrick used in the math jam. (Assuming you wrote it out nicely.)

[quote="draperp"]I think that it's technically incorrect to write $n$ instead of $\sqrt{n^{2}}$ or $|n|$.  $n$ could be a negative integer and the situation would still be valid (although many of the solutions that people have stated wouldn't be).[/quote]
Well, by convention, $n$ usually represents a non-negative integer, and the point is pretty trivial anyhow, so...

I saw the "weighted average" solution, but couldn't think of a way to get it to "fee" rigorous to me so I did it the long-handed summation way.

Maybe that's why I got a gold last year and only one commended -- I always pick the longhand solution =)

JB

[quote="jackb1115"]I always pick the longhand solution =) JB[/quote]
Yeah, my proofs are too long.  I usually make them overly rigorous.

[quote="pianoforte"]I found the formula $\frac{(n+1)(n+2)(a+b+c)}{6}$ (1) from some insight and looking at simpler cases.

My proof by induction:
-For n=1, the only vertices are a,b, and c. The formula gives $\frac{(2)(3)(a+b+c)}{6}=a+b+c$, which is clearly the answer.

-For our inductive step, consider the triangle with $n^{2}$ triangles, or $n+1$ rows of vertices, and let the top vertex be $a$, the bottom left vertex be $b$, and the bottom right be $c$, and assume that the sum of all the vertices is given by $\frac{(n+1)(n+2)(a+b+c)}{6}$. (2)

Now consider adding a $n+2$ row of vertices below the triangle to create a triangle with $(n+1)^{2}$ triangles.
The common difference for the terms on the left side of the triangle is $\frac{b-a}{n}$, so the first term in the $n+2$th row is $b+\frac{b-a}{n}=\frac{nb+b-a}{n}$.
The common difference for the terms on the right side is $\frac{c-a}{n}$, so the last term in the $n+2$th row is $c+\frac{c-a}{n}=\frac{nc+c-a}{n}$.
The sum of the $n+2$ terms in the $n+2$th row is given by the sum of an arithmetic series and is $\frac{(n+2)(nb+b+nc+c-2a)}{2n}$. (3)

We now show that the sum of (2)+(3) must equal (3) for when n=n+1, or
$(1)+\frac{(n+2)(nb+b+nc+c-2a)}{2n}=\frac{(n+2)(n+3)(a+\frac{nb+b-a}{n}+\frac{nc+c-a}{n})}{6}$
$\implies (1)+\frac{(n+2)3(nb+b+nc+c-2a)}{6n}=\frac{(n+2)(n+3)((nb+b+nc+c-2a)+na)}{6n}$
rearranging, $\implies (1)=\frac{(n+2)n((nb+b+nc+c-2a)+(n+3)a)}{6n}$
cancelling n's, $\implies (1)=\frac{(n+2)(nb+b+nc+c+na+a)}{6}$
and factoring, $\implies (1)=\frac{(n+1)(n+2)(a+b+c)}{6}$, which is true.[/quote]
You and SamE managed to find that formula with just insight!!! amazing...


A racer travels 5000 meters in $ 1\frac{2}{3}$ minutes. What is the
racer's average speed in kilometers per hour?

5000 meters is 5 kilometers,  $ 1 \frac23$ minutes is the same as $ \frac {1}{36}$ hours. Divide the distance by the time to get the rate, which is $ 180 \frac {\text{km}}{\text{h}}$.

(How do you make the km and h not italic?)

[size=75][color=red]I fixed your $ \text{\LaTeX}$. You can use the \text{}. -Isabella2296[/color][/size]

Always keep in mind $ velocity\equal{}\frac{distance}{time}$

In this case, keeping track of units, we're told he goes 5000 meters in 5/3 minutes, so
$ \frac{5000}{5/3}\equal{}5000\cdot\frac{3}{5}\equal{}3000\frac{meters}{minute}$

As for conversion to km/h, I found the idea of "conversion factors" especially useful when I was in highschool.

60 minutes=1 hour, right?  So $ \frac{60 minutes}{1 hour}\equal{}1$.
1000 meters=1 km, So $ \frac{1 km}{1000 meters}\equal{}1$.  Multiplying by 1 doesn't change anything, so basically we can insert these 1's and cancel unit labels!

$ 3000\frac{meters}{minute}\equal{}3000\frac{meters}{minute}\cdot 1\cdot 1$
$ 3000\frac{meters}{minute}\cdot \frac{60 minutes}{1 hour}\cdot\frac{1 km}{1000 meters}\equal{}\frac{3000\cdot 60 km}{1000 hour}$
$ 180\frac{km}{hour}$


For  a,b,c is not less than 0 and a+b+c=1   Prove that
1/(a^2+b^2)+1/(b^2+c^2)+1/(c^2+a^2) is not less than 10[/quote]

[quote="mychrom"]First use Bergstrom. Than, observe that iff 0<=a,b,c<=1, a*a+b*b+c*c<=2(ab+bc+ac).You can have this form iff you see that
(1-a)(1-b)<=1. Starting from here the ineq is very simple :lol:[/quote]
I can't understand you method. Notice that (1-a)(1-b)<=1 has made the 10 impossible. Can you make it clearer?
 :)

[quote="arqady"][quote="tanpham90"]
We need to prove that :
$\frac{1}{x^{2}+y^{2}}+\frac{1}{x^{2}}+\frac{1}{y^{2}}\geq\frac{1}{(x+y)^{2}}$
( Very simple With AM-GM  :D )[/quote]
Nice proof, tanpham90.
But, $\frac{1}{x^{2}+y^{2}}+\frac{1}{x^{2}}+\frac{1}{y^{2}}\geq\frac{10}{(x+y)^{2}}.$ :wink:[/quote]
Thank you !  :blush:  :oops:  :D

[quote="tanpham90"]Very Ugly !  :D
Suppose that $c=min (a,b,c)$
We have :

$a^{2}+b^{2}\leq\ (a+\frac{c}{2})^{2}+(b+\frac{c}{2})^{2}$

$a^{2}+c^{2}\leq\ (a+\frac{c}{2})^{2}$

$b^{2}+c^{2}\leq\ (b+\frac{c}{2})^{2}$

With $x=a+\frac{c}{2}$
       $y=b+\frac{c}{2}$
We need to prove that :
$\frac{1}{x^{2}+y^{2}}+\frac{1}{x^{2}}+\frac{1}{y^{2}}\geq\frac{1}{(x+y)^{2}}$
( Very simple With AM-GM  :D )[/quote]
tanpham90 thank you for you nice solution


algebra
polynomial
linear algebra
matrix
geometry
algorithm
vector

consider the $ n \times n$ Vandermonde determinant of $n$ elements $a_1 ,..., a_n$ in a field $K$ :
\[ det \begin{pmatrix} 1 & a_1 & {a_1}^2 & .... & {a_1}^{n \minus{} 1} \\
1 & a_2 & {a_2}^2 & .... & {a_2}^{n \minus{} 1} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & a_n & {a_n}^2 & .... & {a_n}^{n \minus{} 1} \end{pmatrix}
\]
prove that this determinant is nonzero when the $ a_i$'s are distinct.
hint: if the determinant is zero then the columns are linearly dependent. show that the coefficients of the linear relation determine  a polynomial of degree $ \le n-1$ which has $n$ roots.

this question is from the "geometry, algebra and algorithms" by  cox_little_o'shea
I only want to find the polynomial.

Maybe this is cheating, but the determinant of the Vandermonde matrix is given by $ \det(V)\equal{}\prod_{1\leq i\leq j\leq n}(a_{j}\minus{}a_{i})$.  Obviously, two of the $ a_{i}$'s have to be distinct or that product is 0.

If you want to actually prove this, it's just an exercise in matrix operations and induction.

Another proof - if $ a_i \equal{} a_j$ for some $ i \neq j$, there are 2 equal rows and thus the determinant is 0.
When all the $ a_i's$ are distinct, assume for contradiction's sake that the determinant is zero. Thus, there are constant $ c_0, \cdots c_{n \minus{} 1}$, not all zero ($ \vec{c} \equal{} (c_0,\cdots,c_{n \minus{} 1})$ is the vector that satisfies $ V\vec{c} \equal{} \vec{0}$), such that:
$ \sum_{i \equal{} 0}^{n \minus{} 1} c_ia_k^i \equal{} 0$ for $ k \equal{} 0,\cdots,{n \minus{} 1}$. So we have a polynomial with degree $ n \minus{} 1$ with $ n$ roots, a contradiction.

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]here[/url].  I like to say that the Vandermonde determinant is invertible because polynomial interpolation works, not the other way around.  This viewpoint is fruitful not least of all because it generalizes:  you can replace the exponents $ i$ with a more general sequence of exponents, which corresponds to solving a more general interpolation problem, and arrive at the definition of the [url=http://en.wikipedia.org/wiki/Schur_polynomial]Schur polynomials[/url].


[quote="kunny"]Solve the following triangular equations.

(1) $ \sin (x + \frac {\pi}{3}) = \frac {1}{2} \ (0\leq x < 2\pi)$

(2) $ \cos 2x + 5\cos x - 2 = 0$

(3) $ \cos 3x - \sin x = 0\ (0\leq x < 2\pi)$

(4) $ \sqrt {3}\sin x - 2 = 2\ (0\leq x < 2\pi)$

(5) $ \sin x + \frac {1}{2}\sin 2x + \cos x = 1\ ( - \pi \leq x\leq \pi)$

(6) $ 2\sin 2x - 2(\sin x + \cos x) + 1 = 0$

(7) $ \sin 2x + \sin 3x + \sin 4x = 0$[/quote]
[hide="hint for the 7"]
Use the $ sin3x=3sinx-4sin^3x$ and then express everything in terms of $ cosine$[/hide]

[hide="6"]
Rewrite it as $ (2sinx \minus{} 1)(2cosx \minus{} 1) \equal{} 0$[/hide]

[hide="2"]
$ 2(cosx\plus{}3)(cosx\minus{}\frac{1}{2})\equal{}0$[/hide]

[quote="kunny"]Solve the following triangular equations.

(1) $ \sin (x + \frac {\pi}{3}) = \frac {1}{2} \ (0\leq x < 2\pi)$

(2) $ \cos 2x + 5\cos x - 2 = 0$

(3) $ \cos 3x - \sin x = 0\ (0\leq x < 2\pi)$

(4) $ \sqrt {3}\sin x - \cos x = 2\ (0\leq x < 2\pi)$

(5) $ \sin x + \frac {1}{2}\sin 2x + \cos x = 1\ ( - \pi \leq x\leq \pi)$

(6) $ 2\sin 2x - 2(\sin x + \cos x) + 1 = 0$

(7) $ \sin 2x + \sin 3x + \sin 4x = 0$[/quote]

I might wright up solutions later...
[hide="Answer for 1"]$ \frac{11\pi}{6}$ and $ /frac{5/pi}{6}$
[hide="Partial Answer for 2"]=$ 2y^2+5y-3=0$ (where y=cos x, from here on its just grinding)[/hide]

[/hide]

[hide="4"]
$ \frac{\sqrt{3}}{2}\sin x\minus{}\frac{1}{2}\cos x \equal{} 1$
$ \cos 30^\circ\sin x\minus{}\sin 30^\circ\cos x\equal{}1$
so now $ \sin(x\minus{}30^\circ)\equal{}1$
[/hide]

hello, to 3) applying the addition formulas we get
$ 4\cos(x)^3-3\cos(x)-\sin(x)=0$
converting everthing in $ \tan(1/2x)$ we get
$ 4\,{\frac { \left( 1- \left( \tan \left( 1/2\,x \right)  \right) ^{2}
 \right) ^{3}}{ \left( 1+ \left( \tan \left( 1/2\,x \right)  \right) ^
{2} \right) ^{3}}}-3\,{\frac {1- \left( \tan \left( 1/2\,x \right)
 \right) ^{2}}{1+ \left( \tan \left( 1/2\,x \right)  \right) ^{2}}}-2
\,{\frac {\tan \left( 1/2\,x \right) }{1+ \left( \tan \left( 1/2\,x
 \right)  \right) ^{2}}}=0$
factoring
$ {-{\frac { \left(  \left( \tan \left( 1/2\,x \right)  \right) ^{2}-2\,
\tan \left( 1/2\,x \right) -1 \right)  \left(  \left( \tan \left( 1/2
\,x \right)  \right) ^{4}+4\, \left( \tan \left( 1/2\,x \right)
 \right) ^{3}-6\, \left( \tan \left( 1/2\,x \right)  \right) ^{2}-4\,
\tan \left( 1/2\,x \right) +1 \right) }{ \left( 1+ \left( \tan \left(
1/2\,x \right)  \right) ^{2} \right) ^{3}}}=0}$
substituting $ t=\tan(1/2x)$ we get
$ - \left( {t}^{2}-2\,t-1 \right)  \left( {t}^{4}+4\,{t}^{3}-6\,{t}^{2}-
4\,t+1 \right)=0$.
Sonnhard.

hello to 7) write your equation in the form $ \sin(x)(2\cos(x)\minus{}1)(1\plus{}2\cos(x))^2\equal{}0$.
Sonnhard.

[hide="For 7"]
Directly the following can be used as the angles are in A.P
$ \frac{\sin(\alpha\plus{}\frac{(n\minus{}1)\beta}{2})\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}$

[/hide]

[quote="Poincare"]I don't think there is any other way to solve 3 unless you bash it which i'm not willing to do.[/quote]

No, you take the $ \sin x$ over to the other side and let $ \sin x\equal{}\cos \frac{\pi}{2}\minus{}x$.

[hide="3)"]$ \cos 3x\equal{}\cos \frac{\pi}{2}\minus{}x$

$ 3x\equal{}\pm(\frac{\pi}{2}\minus{}x)\plus{}2\pi k$

If we choose the plus: $ 4x\equal{}\frac{\pi\plus{}4\pi k}{2}$

$ x\equal{}\frac{\pi(1\plus{}4k)}{8}$

If we choose the minus: $ 2x\equal{}\minus{}\frac{\pi}{2}\plus{}2\pi k$

$ x\equal{}\frac{\pi(4k\minus{}1)}{4}$.

The values between $ 0$ and $ 2\pi$ are $ \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}, \frac{3\pi}{4}, \frac{7\pi}{4}$.  We can verify that these all work.[/hide]


inequalities
induction
inequalities proposed

$ a_1 \geq a_2 \geq \ldots \geq a_n > 0$     Prove the inequality
$ a_1a_2\ldots a_{n\minus{}1} \plus{} (2a_2\minus{}a_1)(2a_3\minus{}a_2) \ldots (2a_n\minus{}a_{n\minus{}1}) \geq 2a_2a_3 \ldots a_n$

[quote="ricardo4"]$ a_1 \geq a_2 \geq \ldots \geq a_n > 0$     Prove the inequality
$ a_1a_2\ldots a_{n \minus{} 1} \plus{} (2a_2 \minus{} a_1)(2a_3 \minus{} a_2) \ldots (2a_n \minus{} a_{n \minus{} 1}) \geq 2a_2a_3 \ldots a_n$[/quote]
Let
$ X\equal{}a_1a_2...a_{n\minus{}1}$
$ Y\equal{}(2a_2\minus{}a_1)(2a_3\minus{}a_2)...(2a_{n}\minus{}a_{n\minus{}1})$
$ Z\equal{}2a_2a_3...a_n$
from
$ a_i\geq 2a_{i\plus{}1}\minus{}a_i$
so $ X\geq Y$
i will prove your inequality by induction
for $ n\equal{}2$ we got that $ LHS\equal{}RHS$
for $ a_1\geq a_2\geq a_3...\geq a_{n\plus{}1}\geq 0$
if $ X\plus{}Y\geq Z$
it remains to prove that
$ a_nX\plus{}(2a_{n\plus{}1}\minus{}a_n)Y\geq a_{n\plus{}1}Z$
since  $ a_{n\plus{}1}X\plus{}a_{n\plus{}1}Y \geq a_{n\plus{}1}Z$
it remain to prove that $ (a_n\minus{}a_{n\plus{}1})X\geq (a_n\minus{}a_{n\plus{}1})Y$ which is true from $ X\geq Y$


Prove that 111111111 is not a product of three consecutive integers in ANY base.

[quote="t0rajir0u"]I'm not really sure how this is classroom math.  It seems to me a somewhat difficult problem and should probably be moved to Intermediate Topics.[/quote]

Got this from MOEM

It is my first time to come across the concept of consecutive numbers in other bases.

Here is my understanding about this new concept( please let me know if I have mis-interpreted the sentence):

0,1,10,11,100,101,111,.... are consecutive integers in base 2.

0,1,11,12,20,21,22,100,101,102,112,122,.... are consecutive integers in base 3.

How to describe the behavior of consecutive numbers in other bases??? :idea:

[quote="timwu"]Does "any base" include fractional/irrational base?[/quote]

Fractional / irrational bases are, in general, not well-defined.  In particular, it's not clear which digits to allow.

[quote="luimichael"]How to describe the behavior of consecutive numbers in other bases??? :idea:[/quote]

The behavior of consecutive numbers is base-independent.  The product of three consecutive numbers has the algebraic form $ (n\minus{}1)n(n\plus{}1)$ regardless of how numbers are represented.

[quote="timwu"]Does "any base" include fractional/irrational base? If not, here is my solution:

In base n (n is a integer),
$ 111111111$
$ \equal{} 1 \plus{} n \plus{} n^2 \plus{} ... \plus{} n^8$
$ \equal{} \frac {n^9 \minus{} 1}{n \minus{} 1}$
$ \equal{} \frac {(n^3 \minus{} 1)(n^6 \plus{} n^3 \plus{} 1)}{n \minus{} 1}$
$ \equal{} \frac {(n \minus{} 1)(n^2 \plus{} n \plus{} 1)(n^6 \plus{} n^3 \plus{} 1)}{n \minus{} 1}$
$ \equal{} [n(n \plus{} 1) \plus{} 1][n^3(n^3 \plus{} 1) \plus{} 1]$

Since $ n(n \plus{} 1)$ and $ n^3(n^3 \plus{} 1)$ are always even,
$ 111111111\equiv 1 (mod2)$
However, every product of three consecutive integers are also even. Hence, $ 111111111$ is not a product of three consecutive integers.

EDIT: In fact, for integer base, it's easy to argue that it's odd at the first line :huh: but for non-integer ones...I gave up when I saw cubic and radical equations...[/quote]
Nice solution, but I am not sure if the factorization was neccesary.  We can prove that 111111111 in any base is odd as follows:
[hide] If the base we choose is even, then it is obviously odd (since all of the terms except the final 1 are even).
If the base we choose is odd, then all of the terms are odd.  So, it is the sum of 9 odd numbers, which is odd.
So, in both cases, 111111111 is odd.  [/hide]

Actually it's unnecessary as you just say.The purpose of the factorization is trying to get a proof for non-integer bases; however, I failed and was totally confused then :( ...

Before you start trying to prove stuff in non-integer bases, make sure you have a well-defined definition for them.

Also, the statement is false anyway with non-integer bases. I can construct a non-integer base such that $ 11111111_{k} \equal{} 24$ by solving an 8th degree polynomial. And 24=2x3x4.


Congrats to all four of our Illinois Team Mathletes who competed in the 2008 MATHCOUNTS National Competition.
The team members were:

Joshua Dement (15th)
Adam Kalinich (48th)
John Park
Ravi Jagadeesan (53rd)

You guys did an extra-fabulous job as a team. You placed 13th and that has been really good for Illinois, as they had a long time for victory to come to them. Again, I want to tell you guys that you did a very nice job. Keep up the good work for next year!
O_o Josh Dement almost made Countdown Round! Congratulations for that as well!
And Ravi made the top 57 as a sixth grader. That was amazing too. :)

Congratulations on not getting beaten by Alaska, as we did last year!
REPRESENT!  :roll:

Idk, but he got lower than 57th so his name wasn't on the published list :(

oh... ok

by the way did u make top 25% @ illionios mc state

Yes, i took 6th. I belive you can recognize me...from Northbrook...I go to a stupid-at-math school :o

[quote="RussianRocket"]So were u the 3rd or 2nd best non-8th-grader? :maybe:[/quote]
3rd non-eighth grader and 2nd 7th grader.

Who was the next best non-eighth-grader after u?

After me was Patrick Liscio (7th place, Lincoln Junior High School, 7th grade)

Oh looks like all the non-8th graders had really close scores. Its going to be an exciting year!  :D

[quote="RussianRocket"]The 4th,5th,6th, and 7th people are all non 8 graders![/quote]
yup =)

and maybe u may b one of da strong competitors for next year/?

oh...would nats really be hard for me? i got a state score of 30 and got 6th place in Illinois...

Another thing I did bad on was reading mistakes, should have won state if I hadnt made so many and at least 5 more points at nats... This is all making me want to do good next year! :D

It all depends on your strengths and weaknesses. I got a 29 and 3rd place at state, and 17 at nats... but you and I may not necessarily be that close as our state scores may show... if you know a lot about c and p you probably make like a 30 or maybe higher... and did you feel like you did great or awful at state? (I felt like I did awful and told my mom not to bother to feel like there was a possibility I made cd... I didnt even see the last page of sprint.)

Same. The last page of Sprint is still haunting me... o.O lol jk. I think.

I relooked them over and I can understand them all! YIPPEE!!!!!! :D  :D  :D

Ya i looked at the explanations on the MC website.
They make sense!  :D

CD is not official in Illinois (i think...) and I do not know the scores sorry

Nope it's not official. I believe the first place individual got 36, and the fourth place individual got 32.

[quote="suma_milli"]Does CD count in Illinois, as in towards making it to nats or not?  This year will be my first year in mathcounts.  Does anybody know the scores of the top 4 in Illinois?[/quote]
u from IL? what school?

Good luck for next year, everyone! I am unofficially coaching a mathcounts team this year.

hey good luck everyone and .........................................
i am the one who has come first in one of the mc team1
 :D


quadratics
analytic geometry
conics
parabola
algebra

Let $ (h,k)$ be the coordinates of the axes of the parabola of a quadratic equation.

 Now, the quadratc equation is of the form: $ y\equal{}a(x\minus{}h)^2 \plus{}k$

which gives: $ y \equal{} ax^2 \minus{}2ahx \plus{} ah^2 \plus{}k$

Roots of the quadratic equation are: $ \frac{\minus{}b \pm \sqrt{ b^2 \minus{} 4ac } } {2a}$

$ \equal{} \frac{\minus{}b}{2a} \pm \sqrt { \frac{b^2\minus{} 4ac}{4a^2}}$

$ \equal{} h \pm \sqrt{[\frac{\minus{}b}{2a}]^2 \minus{} \frac{c}{a}}$

$ \equal{} h \pm \sqrt{ h^2 \minus{} \frac{ ah^2 \plus{} k}{a}}$

$ \equal{} h \pm \sqrt{ h^2 \minus{} h^2 \plus{}\frac{k}{a}}$

$ \equal{} h \pm \sqrt{\frac{k}{a}}$

 This means that the possibility of roots depends on $ k$ and $ a$.
Supposing $ a$ is positive, then the parabola will open upwards. So, if $ k$ increases parabola will go upwards  and if increased to sufficiently large constant then it will cross the $ x\minus{}axis$ and no roots will be possible.
But we know that a positive value under root would give real roots, which is a contradiction.

[quote="analytic"]...
$ \equal{} h \pm \sqrt { h^2 \minus{} \frac { ah^2 \plus{} k}{a}}$

$ \equal{} h \pm \sqrt { h^2 \minus{} h^2 \plus{} \frac {k}{a}}$
[/quote]

You forgot to distribute your minus sign through the second term under your square root. It should be $ \equal{} h \pm \sqrt { h^2 \minus{} h^2 \minus{} \frac {k}{a}}$

Ah! bad mistake.  :o
Anyways, thanks for pointing out . :)


calculus
integration
calculus computations

Hi all, im new here sorry if this is in the wrong section but i desparately need some help, so here goes...

A particle moves along the x-axis with velocity at time t, given by v=5t^2 + 2t. Find the distance it moves in the 2nd second.

...Now, integrating v to get s=5/3t^3 +t^2 + C.  Now how would i find out what C is equal to? I have no values to substitute for t and s, and have no idea what to do next.

 Any help would be greatly appreciated!

Problems of this nature belong in "Calculus Computations and Tutorials."

You don't need to find $C.$ Look at the question you've been asked: "How far does the particle move in the 2nd second?"

That's $s(2)-s(1).$ You can include any possible value for $C$ in your formula for $s,$ and after the subtraction, it makes no difference. Try it!

2nd second means between 1sec and 2sec.

v=5t^2 + 2t

dx/dt = 5t^2 + 2t

 x=integral (5t^2 + 2t) dt from t=1 to t=2

i.e x = 44/3 m[/quote]

Help me in my differential equation problem.
Anjani Kumar Singh


On a recent flight to San Francisco, the pilot announced that we were 139 miles from the city, flying at an altitude of 35,000 feet.  The pilot claimed that he could see the Golden Gate bridge and beyond.  How far could the pilot see?

to start, there are many flaws with this problem.  like that he only claims to see the brigde and such. but...

[hide]we have a right triangle with height 35000 ft, but we want that in miles.  so $\frac{35000}{5280}\approx6.63$.  the length is $139$.  so $a^{2}+b^{2}=c^{2}$.  so $6.63^{2}+139^{2}=c^{2}$.

$c^{2}\approx19365$
$c\approx139.16$

the problem states he can see to and beyond the bridge, so if his view length is $x$, he can see

$x\ge{139.16}$[/hide]

actually, I'm fairly sure this is the type of problem that involves the radius of the earth and other really messy numbers, so I'll leave this to someone with more time for busywork than me

think of this as a mathcounts level problem.  just ue pythag theroem


algorithm
number theory unsolved
number theory

Assume R is a positive integer and is known. Find m such that

$R \equiv r (mod \, m) \, and \, 2r+1 \equiv 0 (mod \, m)$

Probable unsolvable.

What is problem?
Obviosly m is odd and $r=\frac{m-1}{2}$.

Suppose, that such a $m$ exists.
Then you have $km=2r+1 \Longleftrightarrow r=\frac{km-1}{2}$ for a $k\in\mathbb{N}$, $m$ is odd and $R+R=2R\equiv R+r\equiv r+r=2r(mod m)$ what is equivalent to $2R\equiv km-1\equiv-1(mod m)$, i.e. $m$ can be every odd divisor of $2R+1$ and  then, you can also find $r$ and it holds even for every multiple of $r$.

[quote="kathi;)"]i.e. $m$ can be every odd divisor of $2R+1$ and  then, you can also find $r$ and it holds even for every multiple of $r$.[/quote]

But that means factoring $2R+1$ and what if that is difficult? Can m be found directly from $R$?

I have found a close approximation.

Let $m_{0}= [R/2]$

Let $x = [R/2]$

Now apply the following algorithm,
________________________________

$x = [x/2]$

$m_{1}= m_{0}+x$

$x = [x/2]$

$m_{2}= m_{1}-x$

$x = [x/2]$

$m_{3}= m_{2}+x$

$x = [x/2]$

$m_{4}= m_{3}-x...$

continue in this way until

$[x] = 0$

$m_{n}$, the last value of m, is very close to the required value.

but that's an algorithm which doens't even give always the right result.

No, but it is close. I'm trying to extrapolate something more general from it... It is concerned with integer factorization and I'm trying to find a value for R which is close to a multiple of $p$ or $q$ if $pq = n$, the number to be factored.

If $pq = n$, $(n-1)/2 \equiv (p-1)/2 \, (mod\, p)$ and $(n-1)/2 \equiv (q-1)/2 \, (mod\, q)$ so I'm beginning with $R = (n-1)/2$

[quote="mgb"] But that means factoring $2R+1$ and what if that is difficult? Can m be found directly from $R$?[/quote]

As any solution m is a factor of $2R+1$, finding a quick algorithm for this problem is exactly equivalent to find a quick algoritm for factoring odd numbers ...

--
Patrick

Yes, this is exactly what I'm trying to do!

then have fun with it because it's an open problem (just think of the RSA-encryption)...


geometry
circumcircle
incenter
geometric transformation
reflection
rotation
analytic geometry

Arbitrary line through the vertex $ A$ of a $ \triangle ABC$ cuts its circumcircle $ (O)$ again at $ K.$ Let $ X$ be an arbitrary point on the line $ AK,$ outside of the triangle incircle $ (I).$ Tangents to $ (I)$ from $ X$ cut $ BC$ at $ Y, Z.$ Show that the circumcircle of the $ \triangle KYZ$ cuts $ (O)$ again at the tangency point $ A^*$ of the mixtilinear incircle of the $ \triangle ABC$ in the angle $ \angle A.$

I am supposing the point A* is the point of tangency with (O) of a circle tangent to the sides of angle A and internally tangent to (O), right?

That is correct, see the attached figure.

This is beautiful, Vladimir! It generalizes our previous problem [url]http://www.mathlinks.ro/viewtopic.php?p=1022749#1022749[/url].

My "idea" (hard to call it an [i]idea[/i]) is to consider $ Y_{K}$, $ Z_{K}$ the intersection points of the tangents from $ K$ to $ (I)$ with the sideline $ BC$ and to prove that the circumcircles of triangles $ KY_{K}Z_{K}$, $ KYZ$ and $ ABC$ are concurrent. In this case, your problem follows from the one mentioned above. However, I couldn't find any synthetic proof for my assertion  :blush:.

What is your proof?

Regards,
Cosmin

I forgot the proof  :oops:, honestly, and I had to figure it out again :D . I do not have time today, give me a little more time (tomorrow :?: ) and I promise you will laugh out loud.

Yes, it is generalization of your problem. I got it from this triangle construction: [url]http://www.mathlinks.ro/viewtopic.php?t=207250[/url], even though I did not need it there, your problem was sufficient.

Alright, let $ (I)$ be an arbitrary fixed circle with 2 fixed parallel tangent lines $ a, a'$ and let $ k$ be an arbitrary fixed line, intersecting the circle or not. Let $ X \in k$ be arbitrary point of this line and let the tangents to $ (I)$ from $ X$ cut $ a$ at $ Y, Z,$ so that $ (I)$ is incircle of the $ \triangle XYZ.$ (This is not completely general, $ X \in k$ has to be on the opposite side of $ a'$ than $ (I),$ but proofs for the other cases - $ X$ between $ a, a'$ outside of $ (I),$ or $ X$ on the opposite side of $ a$ than $ (I)$ - are the same, except some  incircle / excircles get switched.) Let $ S, S'$ be the intersections of $ k$ with $ a, a'$ and let a tangent to $ (I)$ from $ S'$ other than $ a'$ cut $ a$ at $ T.$ The tangent $ S'T$ is fixed. Let $ (I_x)$ be the X-excircle of the $ \triangle XYZ.$ Since $ S'T$ is a fixed tangent to $ (I),$ parallel $ s$ to $ S'T$ through $ S$ is tangent to $ (I_x).$ Thus the X-excircle is tangent to 2 fixed lines $ a, s,$ which means that it belongs to a file of circles with the common similarity center $ S$ and 2 common tangents $ a, s.$ (A file of circles is concept dual to a pencil of circles.) Anyway, it is centered on the bisector $ i$ of the angle $ \widehat{as}$ formed by the lines $ a, s.$ Let $ M$ be the midpoint of $ II_x.$ This is the circumcenter of the cyclic quadrilateral $ IYI_xZ$ and its circumcircle $ (M)$ is centered on a parallel $ m \parallel i$ at half the distance from $ I,$ than the angle bisector $ i.$ But one point of the circle $ (M)$ is fixed, the incenter $ I,$ therefore, the reflection $ J \in (M)$ of $ I$ in the line $ m$ is also fixed. Thus the circle $ (M)$ belongs to a pencil of circles defined by the 2 fixed points $ I, J.$ Let $ IJ$ cut $ a \equiv YZ$ at $ T'.$ Then $ \overline {T'Y} \cdot \overline {T'Z} \equal{} \overline {T'I} \cdot \overline {T'J} \equal{} \text{const}.$ When $ X$ coincides with $ S',$ $ Y$ coincides with $ T$ and $ Z$ is at infinity as $ Z^* \equiv a' \cap a.$ Since $ \overline {T'T} \cdot \overline {T'Z^*} \equal{} \text{const}$ and $ \overline {T'Z^*} \equal{} \infty,$ if follows that $ \overline {T'T} \equal{} 0$ and $ T' \equiv T$ are identical. Let $ K \in k$ be an arbitrary fixed point (as opposed to $ X \in k$ arbitrary movable) and let the fixed circumcircle of the $ \triangle KIJ$ cut the fixed line $ KT$ again at $ L.$ Then $ \overline {TY} \cdot \overline {TZ} \equal{} \overline {TI} \cdot \overline {TJ} \equal{} \overline {TK} \cdot \overline {TL}$ and it follows that the quadrilateral $ KYLZ$ is cyclic and moreover, its circumcircle passes through 2 fixed points $ K, L,$ i.e., belongs to a pencil defined by these 2 points. Almost done.

Let $ (I)$ be the incircle and $ (O)$ the circumcircle of a $ \triangle ABC.$ Let arbitrary fixed line $ k$ through $ A$ cut the circumcircle again at $ K$ (fixed point). Let $ X \in k$ be arbitrary and let tangents to $ (I)$ from $ X$ cut the sideline $ a \equiv BC$ at $ Y, Z.$ Circumcircles of all triangles $ \triangle KYZ$ form a pencil with 2 fixed points $ K, L.$ Let $ A^*$ be the tangency point with $ (O)$ of the A-mixtilinear incircle of the $ \triangle ABC$ in the angle $ \angle A.$ When $ X$ coincides with $ A,$ then $ Y, Z$ coincide with $ B, C,$ the circumcircle $ \odot (KYZ) \equiv \odot (KBC)$ coincides with the circumcircle $ (O)$ and obviously, $ A^* \in (O).$ When $ X$ coincides with $ K,$ then by the problem [url]http://www.mathlinks.ro/viewtopic.php?t=186117[/url], circumcircle of the $ \triangle KYZ$ passes through $ A^*.$ Therefore, $ K, L \equiv A^*$ are the 2 fixed points of the pencil of all circumcircles $ \odot(KYZ),$ where $ Y, Z \in BC$ are feet of the incircle tangents from $ X \in k.$

Beautiful problem and, as usual, impressive solution of Yetti. Although after all the posted it is hard say some really interesting things, to follow the idea of Cosmin I was thinking in try to prove that the power of A* with respect to the circumcircles of KY_KZ_K and KYZ is 0 (the third one is obvious), but I have not yet achieved this.
Congratulations for a such nice problem!

Dear Yetti and Mathlinkers,
a new way for a proof of this result can be found on on my website :
http://perso.orange.fr/jl.ayme   vol. 4   A new mixtilinear incircle adventure III, p. 24
Sincerely
Jean-Louis

Let $BC$、$XY$、$XZ$ touch $\odot I$ at $D$、$E$、$F$ resp，$XA$ intersec $BC$ at $J$ . Inversion with $\odot I$：
Delimit $XK \rightarrow \odot (IW),K \rightarrow K',Y \rightarrow Y', Z \rightarrow Z',J \rightarrow J'$ .Then $EF$ pass thought $W$，$W$ is a fixed point，
$Y$、$Z$ are the midpoint of $DE$、$DF$ resp，$K'$ is on $\odot (IW)$，$J'$ is the intersection of $\odot (ID)$ with $\odot (IW)$ . Now，consider
$\odot (K'Y'Z'),\odot (ID),\odot (DJ'K')$，their 3 radical axis $Y'Z',DJ',A'K'$($A'$ is on$\odot (K'Y'Z')$ and $\odot (DJ'K')$) are concurrent at $T$ .
Obviously，$T$ is the midpoint of $DW$，so $T$ is a fixed point，and $TK' \cdot TA'=TY' \cdot TZ'=\frac{WE \cdot WF}{4}=\text{const}$ .
It means that $\odot (K'Y'Z')$ pass thought a fixed point $A'$ where is on $\odot (DJ'K')$，i.e. $\odot (KYZ)$ pass thought a fixed point $A^*$ where is on $\odot (DJK)$ . and $A^*$ also is on $\odot (ABCK)$ . Let $A^*D$ intersect $\odot (ABC)$ at $A''$，then use [b]Reim's Theorem[/b]，we get $AA'' \parallel BC$，acording to this character，we conclude $A^*$ is the A-mixtilinear point . Done .

Wow，very short proof，Nice ！

I really appreciated this beautiful problem. :) I gave this problem to this year's Korean IMO team members, and they found following amazing solutions. Hope that everyone can enjoy this!

The first solution is by Dong Ryul Kim (KOR2), who got 577777 in IMO 2012.

Let $M$ be a mixtilinear point of triangle $ABC$, $T$ be a tangent point of the incircle and $BC$, and $A'$ be a point of reflection of $A$ by the perpendicular bisector of $BC$. Let $S=PA\cap BC$.
It is well-known that $M, T, A'$ is collinear. Since $AA'\parallel BC$, we can show that $TMDS$ is concyclic, so we have $TU\cdot US=MU\cdot UD$, where $U$ is the intersection of $MD$ and $BC$. Furthermore, $MU\cdot UD=BU\cdot UC$. Now the problem reduces to show $EU\cdot UF=TU\cdot US(=BU\cdot UC)$...(*).

Here comes a fresh idea. Rotate the line $BC$ slightly so that it still tangents to the incircle $(I)$. Let this line be $l$, and its intersection with $AB,AC, PE,PF$ be $B',C',E',F'$. Let $T'$ be the intersection of $BC$ and $l$. Then apply [i]Brianchon's theorem[/i] on a hexagon $AB'E'PFC$. It follows that $AP, B'F, CE'$ are concurrent. Let this point be $S'$ (denoted by $Q$ in the above picture). By Menelaus' theorem on triangle $\triangle B'T'F$, we have $\frac{B'E'}{E'T'}\cdot\frac{T'C}{CF}\cdot\frac{FS'}{S'B'}=-1$.
Now take the limit as $l$ approaches to $BC$. Maybe this part needs further verifications, but it seems almost certain that each point of the form $x'$ converges to $x$ and the length of each $x'y'$ converges to that of $xy$. As $S'$ lies on $PA$ and $B'F,CE'$ converges to the line $BC$, $S'$ converges to $S$. Therefore we have the identity $\frac{BE}{ET}\cdot\frac{TC}{CF}\cdot\frac{FS}{SB}=-1$.

This is enough to show (*). Regard the line $BC$ as the real line and take $U$ as its origin. Denote the coordinate of the point $X$ by $x$. Then we have $st=bc$ and need to show $ef=bc=st$. The above identity yields $\frac{b-c}{e-t}\cdot\frac{t-c}{c-f}\cdot\frac{f-s}{s-b}=-1\iff (b+c)(ef-st)+(s+t)(bc-ef)+(e+f)(st-bc)=0.$ From this we can easily conclude the proof.

I will post another solution later because of lack of time.

Let $D$ be the tangency point of the incircle $(I)$ with $BC.$ $AX$ cuts $BC$ at $P.$ $\odot(DPK)$ passes through the tangency point $M$ of the A-mixtilinear incircle with $(O),$ see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=527094]Tangent Circles+Centers Concyclic[/url] (the proof for the mixtilinear incircle is analogous).

By dual of Desargues theorem for the degenerate quadrilateral $ABDC$ circumscribed in the conic $(I),$ the pencil $XB \mapsto XC,$ $XY \mapsto XZ,$ $XD \mapsto XA$ is involutive regardless of $X$ $\Longrightarrow$ $B \mapsto C,$ $Y \mapsto Z,$ $D \mapsto P$ is an involution on the line $BC.$ The center $R$ of this involution is then the intersection of $BC$ with the radical axis $MK$ of $\odot(KBC)$ and $\odot(KMDP).$ Thus $\overline{RY} \cdot \overline{RZ}=\overline{RD} \cdot \overline{RP}=\overline{RM} \cdot \overline{RK}$ $\Longrightarrow$ $M \in \odot(KYZ).$


Hello,

find all $ (a,b)$ with $ a,b \in \mathbb{N}^{*}$ such that : $ \displaystyle \frac{a}{b} \plus{} \frac{{21b}}{{25a}} \in\mathbb{N}$

Have a look here [url]http://www.les-mathematiques.net/phorum/read.php?5,509122,509141#msg-509141[/url]


Let $ ABC$ be a triangle and incircle $ (I)$ touch $ BC$at $ D$, let $ E$ be projection of $ I$ on $ AD$ prove that if $ EB \equal{} EC$ then $ ABC$ is isoceles.

[quote] [color=darkblue][b][u]Lemma[/u].[/b] The incircle $ w \equal{} C(I,r)$ of $ \triangle ABC$ touches its sides in $ D\in BC,\ E\in CA,\ F\in AB$ .

Denote $ L\in BC\cap EF\ ,\ P\in LI\cap AD$ . Prove that $ LI\perp AD$ and $ \widehat{DPB}\equiv\widehat{DPC}$ (see [/color][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46146][color=red][b]here[/b][/color][/url][color=darkblue]).

[b][u]Proof[/u].[/b] $ LI\perp AD\Longleftrightarrow LA^2\minus{}LD^2\equal{}IA^2\minus{}ID^2\Longleftrightarrow$ $ LA^2\minus{}\left(LI^2\minus{}r^2\right)\equal{}IA^2\minus{}r^2\Longleftrightarrow$

$ LA^2\minus{}LI^2\equal{}IA^2\minus{}2r^2\stackrel {(LE\perp AI)}{\ \Longleftrightarrow\ }$ $ EA^2\minus{}r^2\equal{}IA^2\minus{}2r^2\Longleftrightarrow EA^2\plus{}r^2\equal{}IA^2$ , what is true.

[b][u]Remark[/u].[/b] $ LA^2\equal{}LD^2\plus{}AF^2$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] From upper lemma results $ PB \equal{} PC \Longleftrightarrow PD\perp BC\Longleftrightarrow AD\perp BC\Longleftrightarrow  AB \equal{} AC$ .[/color]


In $\triangle ABC$, $AB = AC$. On $BC$, choose $D$ such that $BD:DC = 2:1$. On $AD$, choose $P$ such that $\angle BAC = \angle BPD$. Find $\frac{\angle DPC}{\angle BAC}$

[quote="riddler"]is the answer $\frac{1}{2}$?[/quote]
Yes, you are right  :)

Yes, I got $\frac{1}{2}$ also. The way I did it was that I picked angles, calling angles $ABC, ACB, BAC$ to be $70,70, 40$ respectively. Thus $BPD=40$ and $DPC=20$ because the angles should be in the same ratio as the sides, thus $\frac{20}{40}=\frac{1}{2}$.

[quote="Iversonfan2005"][quote="riddler"]YES! I got it right even though my method is not all good.
Shobhit, Is it neccessary that the angles are in the same ratio as the sides???? ALWAYS????[/quote]

I believe so....however, I need to check on that... :? P.S. My first pre-olympiad problem :D  :first:  :10:  :clap2:  :clap:[/quote]
Same here, this is my $1$st pre-olympiad problem too.

No, I don't think the angles should always be in the same ratio as the sides!

[quote="shobber"]$\frac{BD}{CD}=\frac{BP \sin{\angle{BPD}}}{CP \sin{\angle{CPD}}}$[/quote]

How would that imply the solution?
(Actually I hadn't even thought much about the problem. The only reason I knew the answer is $\frac{1}{2}$ is because the original problem was to prove that $\angle DPC = \frac{\angle BAC}{2}$ but I changed it  :D  :D )

I was going to show that the above proofs are wrong.


How to find a measure space $(X,\Gamma,\mu)$ such that $\mu$ takes only values $0,1/4,3/4,1$. Is it possible that $\mu$ takes only values $0,1/4,1$?

A two-element set whose elements have measure $1/4$ and $3/4$.

No.


calculus
integration
logarithms
real analysis
real analysis unsolved

Hi,
Prove that the following integral formulae holds:
$\int_{0}^{1}\frac{\ln x}{1+x+x^{2}}dx=\frac{4\pi^{2}}{27}-\sum_{k=0}^{\infty}\frac{2}{(3k+1)^{2}}$.

What is interesting about this equality is that it can be proved by using a simple trick.
Didi


Εστω $ a,b,c$ ακεραιοι αριθμοι με $ (b,c)\equal{}1$
Να αποδειχθει οτι $ (a,bc)\equal{}(a,b)(a,c)$
Μια ασκηση πιο πολυ για τους junior φιλους μας......

[quote="Dimitris X"]Εστω $ a,b,c$ ακεραιοι αριθμοι με $ (b,c) \equal{} 1$
Να αποδειχθει οτι $ (a,bc) \equal{} (a,b)(a,c)$
Μια ασκηση πιο πολυ για τους junior φιλους μας......[/quote]

Ενα hint γι αυτους που θελουν να το προσπαθησουν:

[hide]δειξτε οτι οι παραγοντες του δεξιου μελους ειναι σχετικα πρωτοι, μετα δειξτε οτι καθε δυναμη πρωτου που διαιρει το ενα μελος διαιρει και το αλλο.[/hide]


[quote="joejia"]In a seven-story building on which floor would you be if the walk to the seventh floor would be twice as far as the walk to the first floor.[/quote]
[hide]
3rd story: Walk to 7th floor: 4 Walk to 1st floor: 2[/hide]

[hide]
By inspection, we see that it is the third floor.

2=d to bottom.
4= d to top

[/hide]

[hide]
3rd floor because you have to walk 4 stories which is twice as much as walking to 1st floor from the 3rd florr.[/hide]

[hide]
Let $x$ be the floor you are on. Thus
$7-x=2(x-1)$
$7-x=2x-2$
$9=3x$, $x=3$[/hide]


The problem is as follows --

Given three variables a1 , a2 , a3.
and following three equations ---

  p*q*r           =  a1
  p+q+r          =  a2
  p*q + q*r + p*q     =  a3

Compute p , q , r Efficiently.

How can we solve this ?

Thanks

If you meant "solve the system in $ \mathbb C$". Then $ p;q;r$  are solutions of the equation $ x^3 \minus{} a_2x^2 \plus{} a_3 \minus{} a_1 \equal{} 0$
(Viete law).
We know how to solve it here http://en.wikipedia.org/wiki/Cubic_equation


geometry
circumcircle
parallelogram
geometry unsolved

Given an Arbitrary non-isosceles and non-right triangle Δ$ ABC$，$ H$ is the orthocenter，$ E,F$ are the root of $ AC,AB$；
$ M$ is the midpoint of $ BC$ ，Let $ N$ be the intersection of $ MH$ and ⊙$ ABC$，and $ K,L$ are the midpoint $ AH,AN$。
$ AH,EF$ intersect at $ G$，and $ MG,KL$ intersect at $ P$.

Please Prove：$ AP\parallel{}BC$，and $ AP\equal{}1/4*AH*|tan$∠$ ABC\minus{}tan$∠$ ACB|$ .

[hide]Because of the line segment $ NHM$ ( see the shema 1.GIF, of the previus lym's post ), passes through the antidiametric (=diametric opposite) point of $ A,$ I think it is enough to prove that $ \frac{AH}{HD} \equal{} 2\cdot \frac{AG}{GD}$ and if I am not mistaken, I will post here later more details.[/hide]
Kostas Vittas.

We denote as $ P,$ the point of intersection of the line through the vertex $ A$ and parallel to $ BC,$ from the line segment $ KL.$

We denote also the points $ P'\equiv AP\cap MZ$ and $ Q\equiv AP\cap MH,$ where $ Z\equiv AD\cap EF$ ( instead of $ G$ ) and we will prove that $ P'\equiv P.$

Let $ A'$ be, the antidiametric (=diametric opposite) point of $ A$ and it is easy to show that the points $ H,\ M,\ A'$ are collinear, because of $ AH\parallel OM$ and $ \frac{A'O}{A'A} \equal{} \frac{1}{2} \equal{} \frac{OM}{AH},$ as well.

So, we conclude that $ \angle ANA' \equal{} 90^{o}$ and then, the line segment $ KL,$ connecting the midpoints of $ AN,\ AH,$ passes through the circumcenter $ O$ of $ \bigtriangleup ABC.$

$ \bullet$ From $ AP'\parallel DM,$ by [b][size=100]Thales theorem[/size][/b], we have that $ \frac {AP'}{DM} \equal{} \frac {AZ}{ZD}$ $ ,(1)$

Similarly, from $ AQ\parallel DM,$ we have that $ \frac {AQ}{DM} \equal{} \frac {AH}{HD}$ $ \Longrightarrow$ $ \frac {AP}{DM} \equal{} \frac {1}{2}\cdot \frac {AH}{HD}$ $ ,(2)$  ( because $ AQ \equal{} 2\cdot AP$ )

From $ (1),$ $ (2),$ in order to be $ \frac {AP'}{DM} \equal{} \frac {AP}{DM},$  it is enough to prove that $ \frac {AH}{HD} \equal{} 2\cdot \frac {AZ}{ZD}$ $ ,(3)$ which is true based on the below [b][size=100]Lemma[/size][/b].

Hence, we conclude that $ P'\equiv P$ and the proof of the proposed problem is completed.

$ \bullet$  [b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given with $ AD,\ BE,\ CF$ its altitudes  and orthocenter $ H$ and let be the point $ Z\equiv AD\cap EF.$ Prove that $ \frac {AH}{HD} \equal{} 2\cdot \frac {AZ}{ZD}$[/color][/size][/b].

Kostas Vittas.

PS. I will post here later the proof of the above Lemma I have in mind.

[quote="vittasko"] $ \bullet$  [b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given with $ AD,\ BE,\ CF$ its altitudes  and orthocenter $ H$ and let be the point $ Z\equiv AD\cap EF.$ Prove that $ \frac {AH}{HD} \equal{} 2\cdot \frac {AZ}{ZD}$[/color][/size][/b].[/quote]
[b][size=100]PROOF OF THE LEMMA.[/size][/b] - Let be the point $ S\equiv AA'\cap ZM$ and applying the [b][size=100]Menelaos theorem[/size][/b], in the triangle $ \bigtriangleup AHA'$ with transversal $ ZMS,$ we have that $ \frac {ZA}{ZH}\cdot \frac {MH}{MA'}\cdot \frac {SA'}{SA} \equal{} 1$ $ \Longrightarrow$ $ \frac {ZA}{ZH} \equal{} \frac {SA}{SA'}$ $ ,(1)$

We denote the points $ D'\equiv (O)\cap AD,\ X\equiv D'A'\cap OM,$ where $ (O)$ is the circumcircle of $ \bigtriangleup ABC$ and let be the point $ S'\equiv AA'\cap DX.$

Applying again the [b][size=100]Menelaos theorem[/size][/b], in the right triangle $ \bigtriangleup AD'A'$ with transversal $ DXS',$ we have that $ \frac {DA}{DD'}\cdot \frac {XD'}{XA'}\cdot \frac {S'A'}{S'A} \equal{} 1$ $ \Longrightarrow$ $ \frac {DA}{DD'} \equal{} \frac {S'A}{S'A'}$  $ ,(2)$

$ \bullet$ Because of now, the points $ A,\ Z,\ H,\ D,$ are in harmonic conjugation, we have that $ \frac {AZ}{AD} \equal{} \frac {HZ}{HD}$

$ \Longrightarrow$ $ \frac {ZA}{ZH} \equal{} \frac {DA}{HD} \equal{} \frac {DA}{DD'}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {SA}{SA'} \equal{} \frac {S'A}{S'A'}$ $ \Longrightarrow$ $ S'\equiv S$ $ ,(4)$

So, from the concurrency of $ AA',\ ZM,\ DX$ and $ OX\parallel AD,$ based on the [b][size=100]Thales theorem[/size][/b],

we have that $ \frac {AZ}{ZD} \equal{} \frac {OM}{MX} \equal{} \frac {OM}{DD'} \equal{} \frac {1}{2}\cdot \frac {AH}{HD}$ $ \Longrightarrow$ $ \frac {AH}{HD} \equal{} 2\cdot \frac {AZ}{ZD}$ $ ,(5)$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

Kostas Vittas.

It can be sovled with five point have a circle or Inversion.

[quote][color=darkred][b][u]Lemma[/u].[/b] Let $ d$ be a orientated line and $ \{A,B,C,D\}\subset d$. Then $ \frac {\overline {BA}}{\overline {BC}} \equal{} \minus{} \frac {\overline {DA}}{\overline {DC}}\ \Longleftrightarrow\ \frac {\overline {CA}}{\overline {CD}} \equal{} 2\cdot \frac {\overline{BA}}{\overline {BD}}$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Choose a origin $ O$ on the orientated line $ d$ . Then for any $ \{X,Y\}\subset d$ we define $ \overline {OX} \equal{} x\in \mathbb R$ and $ \overline {XY} \equal{} y \minus{} x$ .

Therefore, $ \left\|\begin{array}{ccccc} \frac {\overline {BA}}{\overline {BC}} \equal{} \minus{} \frac {\overline {DA}}{\overline {DC}} & \Longleftrightarrow & (a \minus{} b)(c \minus{} d) \plus{} (c \minus{} b)(a \minus{} d) \equal{} 0 & \Longleftrightarrow & (a \plus{} c)(b \plus{} d) \equal{} 2(ac \plus{} bd) \\
 \\
\frac {\overline {CA}}{\overline {CD}} \equal{} 2\cdot \frac {\overline{BA}}{\overline {BD}} & \Longleftrightarrow & (a \minus{} c)(d \minus{} b) \equal{} 2(d \minus{} c)(a \minus{} b) & \Longleftrightarrow & (a \plus{} c)(b \plus{} d) \equal{} 2(ac \plus{} bd)\end{array}\right\|$ .

[b][u]Particular case[/u].[/b] Let $ ABC$ be a triangle with the orthocenter $ H$ . Denote $ \left\|\begin{array}{c}
D\in AH\cap BC\\\
E\in BH\cap CA\\\
F\in CH\cap AB\end{array}\right\|$ and $ S\in EF\cap AH$ . Is well-known that

[u]the points[/u] $ S$ , $ D$ [u]are harmonical conjugate w.r.t. the points[/u] $ A$ , $ H$ i.e. $ \frac {\overline{SA}}{\overline {SH}}\equal{}\minus{}\frac {\overline {DA}}{\overline {DH}}$ . Using the upper lemma obtain $ \frac {\overline {HA}}{\overline {HD}}\equal{}2\cdot\frac {\overline {SA}}{\overline {SD}}$ .[/color]

Hello.
Here is my solution for the firt part. :)

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-101.jpg[/img]

We denote the point : $ T \equal{} AO \cap (O)$ .

We have :$ BHCT$ is a parallelogram ,so $ H,M,T$ are collinear $ \equal{} > \angle {ANM} \equal{} 90^0$ $ \equal{} >$ the lines $ AN,EF,BC$

concur at $ S$.    ( Because $ N$ is the Miquel point of  $ BFEC$ )

Since $ ME \perp KE, MF \perp KF$ ,then $ M$ is the pole of $ EF$ WRT $ (K)$    (1)

But $ (AGFE) \equal{} \minus{} 1$  (2)

Since (1) ,(2) and the La Hire's theorem ,then $ PM$ is the polar of $ S$ WRT $ (K)$  (3)

Since (3) ,La Hire's theorem and $ KP \perp AN$  ,then $ P$ is the pole of $ AN$ WRT $ (K)$.

Therefore, $ KA \perp AP \equal{} > AP//BC$.
.

let AP and MN meet each other at T, and AP & BC meet at Q . then we have:

AK = KH & AL = LN => AP=PT. And for AFHE we have (AHGD)=-1 and if we put the points on AT with the way of M we  have (ATPQ)=-1 and we found that, AP=PT, so AP||BC! All done! :)
Nice! H can be any point inside of ABC! :lol:


Let N be a three-digit number that is divisible by 3. One of N's digits is chosen at random and removed. Compute the porbablilty that the remaining number is divisible by 3 (Note if we have like 300 we can form either 30, 30 or 00=0. Or if we have 207 we can have 20, 07=7 or 27)

I dont quite understand why this solution is wrong.

[hide]Say we have a number ABC, then if A+B+C is 0 mod 3, then in order to be 0 mod 3, the removed digit must be 0 mod 3.

So its just 1/3(3/9+4/10+4/10)= 17/45.

However, the real answer is 19/50.[/hide]

Don't double-post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185130


quadratics
number theory
number theory unsolved

Find all the sets of positive integers n with the property: The set (n, n+1, n+2, n+3, n+4, n+5) can be splitted into two other sets, so the product of the numbers of the first new set is equal to the product of the numbers of the second new set.

below, assume that $ \{a_1, ..., b_3\} \equal{} \{1,...,6\}$. we're looking for $ n$ such that

$ (n \plus{} a_1)(n \plus{} a_2)(n \plus{} a_3) \equal{} (n \plus{} b_1)(n \plus{} b_2)(n \plus{} b_3)$, that is,

$ (\sum a_i \minus{} \sum b_i)n^2 \plus{} (\sum a_ia_j \minus{} \sum b_ib_j)n \plus{} (a_1a_2a_3 \minus{} b_1b_2b_3) \equal{} 0$ (*)

we have 10 essentially different choices for the $ a_1,...,b_3$, so we have 10 quadratics to check for integer solutions... use a computer  :wink:

edit: i see the part sizes don't have to be 3 each... to find other partitions, use the same method.

[hide="hint"]
There's a generalization to a set of the form
\[ \{ n,n\plus{}1,n\plus{}2,\cdots ,n\plus{}p\minus{}2\}\]
where $ p$ is a prime congruent to 3 modulo 4. Written like this...
[/hide]

Scorpius is telling you to consider the set $ \bmod 7$.  ;)

Edit:  I might as well give the full solution.  The set $ \{ n, n\plus{}1, ... n\plus{}p\minus{}2 \}$ either contains a single value divisible by $ p$ (in which case a solution is clearly impossible) or it contains the residues $ 1, 2, ... p \minus{} 1$.  By Wilson's Theorem, $ (p\minus{}1)! \equiv \minus{}1 \bmod p$, and for $ p \equiv 3 \bmod 4$ this is not a quadratic residue.


Find all functions $ f : \mathbb{R} \to \mathbb{R}$ such that
\[ \lim_{x \to 0 } \frac{f(x)}{x} =  1\]
and
\[ f(x\plus{}y) \equal{} f(x) \plus{} f(y) \plus{} 2x^2 \plus{} 3xy \plus{} 2y^2, \quad  \forall x , y  \in  \mathbb{R}.\]

If we fill in $ x\equal{}0$ in the second equation we get:
\[ f(y) \equal{} f(0) \plus{} f(y) \plus{} 2y^2,\]
so $ y^2 \equal{} \minus{}\frac{1}{2}f(0)$ for all $ y \in \mathbb{R}$, contradiction.
So no such functions exist.

[quote="x164"]If we fill in $ x \equal{} 0$ in the second equation we get:
\[ f(y) \equal{} f(0) \plus{} f(y) \plus{} 2y^2,\]
so $ y^2 \equal{} \minus{} \frac {1}{2}f(0)$ for all $ y \in \mathbb{R}$, contradiction.
So no such functions exist.[/quote]
indeed :)
Very quick !

Well , just in case there was a typo , i may assume that the function defined in R+ and have value on R+
putting x=y gives
$ f(2x) \equal{} 2f(x) \plus{} 7x^2$
$ \frac {f(2x)}{2x} \equal{} \frac {f(x)}{x} \plus{} \frac {7}{2}x$

So
$ \frac {f(x)}{x} \equal{} \frac {f(\frac {x}{2})}{\frac {x}{2}} \plus{} \frac {7}{2}\frac {x}{2} \equal{} ... \equal{} \frac {f(\frac {x}{2^n})}{\frac {x}{2^n}} \plus{} \frac {7x}{2}(\frac {1}{2} \plus{} \frac {1}{2^2} \plus{} ... \plus{} \frac {1}{2^n} ) \rightarrow 1 \plus{} \frac {7}{2}x$
So $ f(x) \equal{} x \plus{} \frac {7}{2}x^2$

We notice that this solution doesnt work so there is no solution


Find all integer solutions $ (x, y)$ to the inequalities:

$ x^{2} \minus{}y <1$ and $ x^{2} \plus{} y< 4$.

$ 2x^2<5\rightarrow x\in${$ -2,-1,0,1,2$},so from here we can know what are the values of $ y$.


[quote="varun"]I learn C++ for leisure. What can I do using the knowledge I gain by learning c++? Is it helpful in some field. I want to do Mechanical Engineering. Will comp knowledge help me!

Also what other languages can I learn for fun.

Please suggest me some good books for C++ and Java.

Varun[/quote]

If you're still in high school you can compete in the Computing Olympiad: http://www.usaco.org. (Training site at http://train.usaco.org has a large number of problems.)

Also knowledge of computer science is good in general (once you've learned it there's so many little problems you can do just by writing up a quick algorithm rather than having to get someone else to do it for you). Although if you want to do Mechanical Engineering, I believe Computational Physics is a fairly substantial subject and has a lot to do with Mechanical Engineering.

[quote="JSteinhardt"]
If you're still in high school you can compete in the Computing Olympiad: http://www.usaco.org. (Training site at http://train.usaco.org has a large number of problems.)[/quote]

Can't you participate, even if you're not in high school?  Because you can participate as an oberver, which isn't really official, but you can still do problems and stuff.


greatest common divisor
algebra open
algebra

\[a,b,c \in{\rm N}\]  \[ab+ac = bc\] prove that:

 (a;b)=(a;c)=(b;c)

What is that last part?  Is (a;b) supposed to represent $\gcd\left(a,b\right)$?

It cannot be a greatest common divisor symbol.

Look at this example:

If you assume it to be that symbol,

let a = 1, b = 2, c = 2 for an instance

1(2) + 1(2) = 2(2)

(1;2) = (1;2) = (2;2) becomes. . .
...1... = ...1.. = ...2... which is false.


Given $ a, b, c \geq\ 0$. Prove that:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{8\sqrt{2}}{5}\sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \geq\ \frac{18}{5}$
 :)

I had 2 proofs by Am-Gm for the result above. Besides, by Am-Gm, we also can prove the general problem:
Given $ a, b, c \geq\ 0$ and $ k\geq\ \frac{1}{2}$. Prove that:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{} 2\sqrt{2}k.\sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \geq\ 2(k\plus{}1)$
 :)


Question:
On my computer i built a simple input system. It is simply a grid composed of squares. Each square can be either black or white. I also created a simple virtual neuron which behaves (i hope) moderately close to a neuron. This virtual neuron has a collection of incoming virtual dendrites and an axon. Each neuron has its own "threshold value" and associated with each dendrite is a "weight value". This way each neuron receives potentials from other neurons which arrive at a neuron through the dendrites, the weight of the dendrite specifying the impact of that potential on the neuron's state value (representing the difference in concentration of some ion, Na+, K+, i don't know, inside the soma, and outside it). Hence, and finally, when a neuron's state value reaches its threshold value the neuron fires an action potential.

What i did then is connect each square of the input grid to a neuron. If the color of the square changes from white to black (accomplished by my clicking on the square), then the neuron attached to it fires an action potential immediatly).

I am interested in putting together some sort of network of neurons and maybe "teach" the network to recognize certain inputs and produce some outputs.
Of course i don't want to manually build the network to do this because it can obviously be the equivalent of a very hard math problem (setting the connections, the weights, and the thresholds appropriately).

Instead i was thinking of having each neuron randomly (or maybe in a simple deterministic way) connect itself to another existing neuron. Of course the connection may not make much sense, it may not give the network the desired behavior, so what i need is a protocol by which to decide which connections to strengthen and which to deteriorate. What protocol does the human brain use?

I was thinking of using simple pleasure and pain to achieve this. Suppose that i have two glands in the brain that secrete some sort of substance (a substance for "pain" and a substance for "pleasure"). These glands would be attached to the network of neurons. If a gland receives a certain amount of potential it produces its substance.
This way, say that a certain input causes a potential to propagate to a pain gland, this pain gland produces "pain" substance which interacts with dendrites which have just "fired" and reduces their respective weight as to prevent this from happening again. Pleasure glands would have the opposite effect.

Suppose that in my network i have something monitoring the levels of "pain" and "pleasure" substance. I would expect that the network would modify itself overtime and reduce the "pain" gland activity.
Hence, the connections that lead to an output A and "pain" gland activity are deteriorated over time and the connections that lead to an output B and "pleasure" gland activity are reinforced, causing the whole system to learn. By setting pain and pleasure receptors throughout the system i can teach the network to do certain things on a given input. it works on theory right?

(I have limited computational power in my computer which happens to be very fast, but with a thread per neuron i don't think i'll be able to have alot of neurons "running" at the same time.)

Hi! I wonder if you're still interested in this... I don't know too much about AI and how brain works but at least, I know some, and I find your project quite interesting. I myself once wanted to simulate some kind of neuron structure that would be capable of learning, however, I never got to doing this. Anyway, to your post:
How do you wand to connect the pain and pleasure glands? Directly to the neurons / output of your brain? I think it would make much more sense to make the brain produce some output, then compare it to the desired output and based on how close to this the brain got activate the pain or the pleasure gland (or none of them). I'd say that's indeed (somewhat) the way human brain works. (Or at least I always thought so. Comments anybody?)
And a question - why do you want to use one thread per neuron? Wouldn't it be more efficient / easier to code if you had the whole brain processing in one thread? Or at least, the communication between the neurons? Because I have a feeling that you're going to need quite a lot of them. (Neurons.)

I don't think having the brain "compare output" is realistic and is maybe a little ambiguous perhaps. I want the neurons controlling the pain/pleasure glands to be thrown in the brain. I don't particularly care how they're thrown in there, but i want them to be "around" to be excited by brain activity. I like this approach because i don't want to have to build the "brain" neuron by neuron, i can just randomly create the neurons and their connections and "see what happens".
The reason why it's very important to have a thread (or more) per neuron is that the human brain is asynchronous, the neuron's are not firing in harmony and the timing between the firing of action potentials is very relevant.
This is a project i haven't had much time to work on, but perhaps next semester i can do this in independent study.

Hm, that sounds quite reasonable. Although I don't know what's going to prevent all the connections going to pain gland from dying. Anyway, if you'll ever get to doing that project, let us (or me :) ) know - and good luck with it.


Find all integral solutions to the equation (x:^2:-y:^2:):^2: = 1 + 16y.

[hide]From the given equation it is evident that the left side is non-negative so y must be greater than or equal to 0. Solving for x we get x^2 = y^2  :pm:  :sqrt: (16y+1).

Case 1: x^2 = y^2  +  :sqrt: (16y+1)

We have y^2 < y^2  +  :sqrt: (16y+1) < (y+1)^2 -- the left inequality holds for all y and the right inequality holds for y greater than or equal to 4. So, for y^2  +  :sqrt: (16y+1) to be square, we must have y = 0, 1, 2, or 3. Checking these, we get the solutions ( :pm: 1,0) and ( :pm: 4,3).

Case 2: x^2 = y^2  -  :sqrt: (16y+1).

Similarly, we get y^2 > y^2  -  :sqrt: (16y+1) > (y-1)^2 when y is at least 6, so we must check for solutions with y between 0 and 5. We get ( :pm: 4,5).[/hide]

its an attempt to revive a dead thread, so you bump it to the top by responding to it in hopes that somebody will respond.


How do we prove that a nth catalan number

Cat(n)=C(2n,n)/(n+1) is an integer?
 n=1,2...
where C(n,k)=n!/(k!)(n-k)!

Note: [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan number[/url] has various combinatorial interpretations and it is clear that it is an integer by counting. I am looking for a proof based on divisibility properties and not an combinatorial argument.

Rewrite $ Cat(n) =\frac{(2n)!}{(n+1)!n!}$ . now assume that $ p^{k}\parallel n!$ so we have $ k = [\frac{n}{p}]+[\frac{n}{p^{2}}]+\cdots+[\frac{n}{p^{e}}]$ and assume that $ p^{m}\parallel (n+1)!$ so we have ${ m = [\frac{n+1}{p}}]+\frac{n+1}{p^{2}}]+\cdots+\frac{n+1}{p^{f}}]$
and assume that $ p^{l}\parallel 2n$ , similiary we get $ l = [\frac{2n}{p}]+[\frac{2n}{p^{2}}]+\cdots+[\frac{2n}{p^{g}}]$
now we want to prove that $ [\frac{n}{p^{i}}]+[\frac{n+1}{p^{i}}]\leq [\frac{2n}{p^{i}}]$
write $ n = p^{i}k+r$ where $ 0\leq r < p^{i}$ then $ [\frac{n}{p^{i}}] = k$
and we have $ n+1 = p^{i}k+(r+1)$ and $ 2n = 2kp^{i}+2r$ . now we have two case
1)$ r = p^{i}-1$ then $ [\frac{n+1}{p^{i}}] = k+1$ but we have $ 2r = 2p^{i}-2\geq p^{i}$ ($ p\geq2 , i\geq 1$) and we get : $ [\frac{2n}{p^{i}}] = 2k+1$ and in this case we have $ k+(k+1)\leq (2k+1)$
2)$ r < p^{i}-1$ then $ [\frac{n+1}{p^{i}}] = k$ and $ 2n = 2kp^{i}+2r$ which gives $ [\frac{2n}{p^{i}}]\geq 2k$
and $ k+k\leq 2k$ and the problem is solved :)


algebra
polynomial
function
search
combinatorics unsolved
combinatorics

Hello:

I like counting and probability. Here is a problem I think is a little tough. Can someone give a hand as to how to go about it?.

[i]"How many ways we can choose 8 letters from[/i] [b]aaaaa bbbb ccc dd e?".[/b]

Since we are choosing more letters than there are different groups, this is a little trickier. Or am I wrong?.

Surely it can't be as easy as choosing 8 from 15?. I know there are $ \frac {15!}{5!4!3!2!1!}$ different ways to arrange these, but I don't think that is really pertinent to the problem.

Does anyone have an idea how to go about a problem like this one?.  Thanks much.

Maybe we can consider the coefficient of $ x^8$ in the following polynomial
\[ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5)(1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4)(1\plus{}x\plus{}x^2\plus{}x^3)(1\plus{}x\plus{}x^2).\]

Thanks. I think I got it figured out but it requires the 8-permutation of the multiset {5a,4b,3c,2d,1e}.  There are quite a few if one starts trying to count them up. But I like your GF suggestion.

Thanks very much. Generating functions are very handy.

i dont really have much experience with generating functions. could someone direct me to a page with generating functions?

Put "generating function*" into the search function here; within the first 10 posts or so you will find a variety of advice.  For example, you will find a post of mine in which I suggest that someone else puts the word "generatingfunctionology" into his or her favorite search engine.  (Of course, putting the words "generating function" directly into a search engine also brings up a variety of useful resources.)

As a matter of fact, I downloaded generatingfunctionology a long time ago and forgot about it. Thank you. I have on my computer somewhere.


geometry
rectangle
perimeter
combinatorics unsolved
combinatorics

You are given a $n \times n$ square board in which some of the squares are coloured red and some white. If a square has two red neighbours then it can be coloured red. If initially there were $n-1$ red squares on the board, is it possible to arrive at the configuration with all squares red?

Are you sure $n-1$ is enough ?

If you place the red squares on the diagonal for instance, none has neighbours (I assume squares must have an edge in common to be called neighbours) : no move is possible

No, if you place squares on the diagonal, say leaving out the corner $(n,n)$ square, then you can colour $n-1 \times n-1$ square red. The answer I think is that you cannot colour all squares red.

We prove a stronger result: to be able to completely color an $a \times b$ rectangle with the above rules, we need an initial configuration with at least $\frac{a+b}{2}$ squares. Obviously this implies the problem.

The base cases are easy, so we assume our assertion true for all $c \times d$ rectangles with $c<a, d<b$. We consider the initial red squares as $1 \times 1$ rectangles. We say that two sub-rectangles painted all red $A, B$ are adjacent if there's a square in the board having a neighbor from $A$ and a neighbor from $B$. It's easy to see that, if this happens, we can get an all-red subrectangle containing both $A$ and $B$.

Also, whenever we can make a move, this can be considered as part of the above process, since a square we're going to paint red can't have two red neighbors from the same rectangle; then it must have a neighbor from some $A$ and one from some $B$, and we can just assume that after making our intended move we complete the rectangle that contains both $A$ and $B$ (obviously we can assume that we make any possible move when it is available).

But now it's easy to check that by joining an $x \times y$ rectangle and a $z \times w$ rectangle, we can get a rectangle with sum of sides no greater than $x+y+z+w$ (this is easier to look at with a diagram). Then, if the $x \times y$ rectangle and the $z \times w$ rectangle had at least $\frac{x+y}{2}$ and $\frac{z+w}{2}$ initial red squares respectively (by our hypothesis), our new rectangle contains at least $\frac{x+y+z+w}{2}$ initial red squares, which is at least half the sum of the sides of it. And since the coloring of our $a \times b$ rectangle can be considered as a sequence of such moves, we're done.

An alternative method:

[hide]
The perimeter of the set of red squares is nonincreasing
[/hide]


calculus
integration
limit
calculus computations

Find the limit $ \lim_{m\to\infty}\int_{\minus{}\infty}^{\infty} \frac{dx}{1\plus{}x^2\plus{}x^4\plus{}\cdots \plus{}x^{2m}}\ (m\equal{}1,\ 2,\ \cdots)$.

$ L\equal{}\int_{\minus{}\infty}^{\infty} \frac{dx}{1\plus{}x^2\plus{}x^4\plus{}\cdots \plus{}x^{2m}}\  \equal{}2\int_{0}^{\infty}\frac{x^2\minus{}1}{x^{2m\plus{}2}\minus{}1}\;dx$

$ \lim_{m\to\infty}\frac{x^2\minus{}1}{x^{2m\plus{}2}\minus{}1}\equal{}1\minus{}x^2$ if $ 0\leq x<1$ and $ 0$ if $ x\geq 1$.

$ L\equal{}2\int_{0}^{1}(1\minus{}x^2)\;dx\equal{}2\left(1\minus{}\frac{1}{3}\right)\equal{}\frac{4}{3}$.


I'm trying to teach myself mechanics and I have this problem but I can't solve it. I have the answer, but not the solution.

An inventive child named Chris wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley, Chris pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Chris's true weight is 320 N, and the chair weighs 160 N. (a) Show that the acceleration of the system is upward and find its magnitude. (b) Find the force Chris exerts on the chair.

And then there's a picture with Chris sitting in a chair, then a rope comes off the top and over a pulley, then there's a spring scale, and then it comes down further to where Chris is holding it.

So I have that there's a force of 480 N pulling the rope on Chris's end, and some other force from the end from Chris pulling on it, and I need to find that force, but I can't figure out how the spring scale works, and if I could get that, then I should be good...

Keep in mind that the tension is transmitted unaffected throughout the rope, and that the rope can't be exerting a force of 480 N on the Chris-chair system, since then it would be in equilibrium.

The reading on the spring scale simply gives the tension $ T$.  Call $ F_\perp$ the force exerted on Chris by the chair (or on the chair by Chris); then the forces acting on Chris are $ T \equal{} 250$ N (upward), $ F_\perp$ (upward), and $ W_\textrm{Chris} \equal{} 320$ N (downward).  Likewise, the forces on the chair are $ T$ (upward), $ F_\perp$ (downward), and $ W_\textrm{Chair}$ (downward).  The acceleration $ a$ of Chris and the chair are the same.  This gives us two equations that allow us to solve for $ a$ and $ F_\perp$:
\[ T \plus{} F_\perp \minus{} W_\textrm{Chris} \equal{} \frac{W_\textrm{Chris}}{g}a
\]
\[ T \minus{} W_\textrm{Chair} \minus{} F_\perp \equal{} \frac{W_\textrm{Chair}}{g}a
\]


Solve in $\mathbb{R}$ the following equation: $\left(2\cos^{2}{\frac{\pi}{24}}\right)^{x}+\left(4\cos{\frac{5\pi}{12}}\right)^{2}\leq \frac{5\sqrt{3}-3}{2\sqrt{2}}$.

Remember $2 \cos^2 y - 1 = \cos 2y$

This lets us write pi/24 and 10pi/24 in terms of the cosines of pi/12 and 10pi/12, which we know (30, 300 deg).


$ \{a_n\}$ is a sequence for which

$ a_1\equal{}1$ , $ a_{n\plus{}1}\equal{}\sqrt{1\plus{}(a_1\plus{}a_2\plus{}\cdots \plus{}a_n)^2}$  $ (n\equal{}1,2,\cdots )$.

Determine the range of positive real $ c$ such that $ \frac{a_n}{c^n}$ converges ($ n\to \infty$),and then find its limit value.

[hide="hint"]
Now,$ a_1\equal{}1,a_2\equal{}\sqrt{2},a_3\equal{}\sqrt{4\plus{}2\sqrt{2}}$,

you ought to notice we can make use of trig.

[/hide]


A jar contains 10 red, 7 blue and 5 yellow marbles. Blue
marbles are then added in order to change the probability of
randomly selecting a blue marble from the jar to ``greater
than $ \frac{1}{2}.$''  What is the least number of blue marbles that must
be added?


Ok, I was just confirming that by posting it.

[quote="Palytoxin"][quote="4everwise"]Solve $ 7^{t-3}=5t$[/quote]
this is graphing calculator question, you can't possibly solve it by hand...[/quote]

Good, cuz I was trying but couldn't get anything out of it.... :oops:

just for your information: roots correct to at least 7 digits: $5.837531058\times 10^{-4},4.612747414$

[quote="YooSam"]How do you know which question needs a graphing calculator, which doesn't?[/quote]
well it really depends on how 'insightful' you are when you look at the question. my reason is that

1. i was not able to get trivial solutions, by trivial i mean easy solutions that you can get right away when you try to solve it

2. look at the 7 and the 5. these are two different primes and yet 7 is the base of an exponent equation. since bases are different primes, more than likely the solutions are not integers

3. i tried using logarithms briefly and got something like of the form $\frac{t}{\log t}$. i know this type of equation cannot be solved by hand, can you? no one does

these are my beliefs that drove me to think this question is a graphing calculator question. yes you can prove my intuition can sometimes be wrong, but when they are right many times, it's bound to be believable

and again, it's a matter of opinion, so your question cannot be definitely answered, only you can answer it


[hide]First, notice that
$ \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\ge\frac{a\plus{}b\plus{}c}{3}\equal{}\frac{1}{3}$

$ \Rightarrow a^3\plus{}b^3\plus{}c^3\ge\frac{1}{9}$

Also, we have

$ \sqrt{\frac{\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}{3}}\ge\frac{3}{a\plus{}b\plus{}c}\equal{}3$

$ \Rightarrow\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\ge27$

Now, since the sequences $ \frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ and $ b^3\plus{}c^3,a^3\plus{}c^3,a^3\plus{}b^3$ are similarly sorted, by Chebychev we have

$ \frac{a^3\plus{}b^3}{c^2}\plus{}\frac{b^3\plus{}c^3}{a^2}\plus{}\frac{c^3\plus{}a^3}{b^2}\ge\frac{1}{3}\left(2a^3\plus{}2b^3\plus{}2c^3\right)\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)$

$ \Rightarrow\frac{a^3\plus{}b^3}{c^2}\plus{}\frac{b^3\plus{}c^3}{a^2}\plus{}\frac{c^3\plus{}a^3}{b^2}\ge\frac{1}{3}\left(\frac{2}{9}\right)\left(27\right)$

$ \Rightarrow\frac{a^3\plus{}b^3}{c^2}\plus{}\frac{b^3\plus{}c^3}{a^2}\plus{}\frac{c^3\plus{}a^3}{b^2}\ge2$[/hide]

$ \iff$    $ \frac{a^3}{c^2}\plus{}\frac{a^3}{b^2}\plus{}\frac{b^3}{c^2}\plus{}\frac{b^3}{a^2}\plus{}\frac{c^3}{a^2}\plus{}\frac{c^3}{b^2}\geq 2$

by AM-GM :

$ LHS \geq 2(\frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab})$

so w need to prove that :

$ \frac{a^3}{bc}\plus{}\frac{b^3}{ca}\plus{}\frac{c^3}{ab}\geq 1$

$ \iff$   $ a^4\plus{}b^4\plus{}c^4 \geq abc$

but by holder ;  $ a^4\plus{}b^4\plus{}c^4 \geq \frac{1}{27} \geq abc$

because  $ abc \leq \frac{1}{27}(a\plus{}b\plus{}c)^3\equal{}\frac{1}{27}$

:)

[hide="Easy Dumbassing"]We are going to use Muirhead and Schur (well not actually Schur but I had to finish the quote :lol:).  Multiply by $ a^{2}b^{2}c^{2}$ to get

$ \sum_{\text{cyc}}a^{2}b^{2}(a^{3}\plus{}b^{3})\ge 2a^{2}b^{2}c^{2}$

Now expand the LHS and multiply the RHS by $ a\plus{}b\plus{}c$ (which is just 1) to get

$ \sum_{\text{sym}}a^{5}b^{2}\ge\sum_{\text{sym}}a^{3}b^{2}c^{2}$

which is true by Muirhead (or weighted AM-GM if you really wanted).[/hide]


I am having a huge amount of trouble with problems regarding proportional line segments such as these:

1. In Triangle $ABC$, $D, E$ split $BC$ into 3 congruent parts. $M$ is the midpoint of $AC$. $BM$ intersects $AD$ at $G$, $AE$ at $H$. Then what is $BH: GH: HM$?

2.In Triangle $ABC$, $FE$ is parallel to $BC$, $AF=10$, $BF=12$. $LM$ is parallel to $FE$. Point $D$ is on $BC$, $BD=14$, $DC=6$. Connect $LK, FD, ED, MN$. $LK$ is parallel to $FD$. $ED$ is parallel to $MN$. Then what is the value of $KD: DN$?

I know that the general approach to these questions is to construct parallel lines or list the existing proportional line segments and then manipulate them or transform them. But I have no idea as to how(especially for the first question. I've connected $ACD$'s midsegment $EM$ and found that $BG=GM$, but then I collapsed horribly). Is there some kind of trick to this kind of problems or do I just search through the proportional line segments until I can find something useful?

And feel free to do those questions above :).

K I got number one. Can someone help with number 2 please?

Can you attach a picture for the second one? I tried making one in Visio, but messed up. :(

See attached file triangle1.

Here is another similar problem (see attached file trapezoid1):

In trapezoid $ABCD$, $AB$ is parallel to $CD$, and $AB=7$, $CD=4$. Extend $AD,BC$ intersecting at $E$. Through $E$, create a line that's parallel to $AB$, intersecting the extensions of $AC, BD$ at $N,M$. Then what is the length of $MN$?

by similar triangles...

[hide="answer"]
MN = $\boxed{\frac{56}{3}}$ if I'm not mistaken.

(look at segments ED and DA, then look at triangles ABD and EMD.)
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Thank you!

I think my problem is the inability to pick out proportional line segments or use the correct pair of similar triangles.

Do you know the solution to the other problem (the one with lots of parallel lines)?


trigonometry
real analysis
real analysis unsolved

find sum
$ \sum_{k\equal{}0}^{\infty} \frac{\sin (3^kx)}{3^k}$.

I am aware of the convergence only. I use Comparison test to prove it.

$ \left| \frac {sin(3^{k}x)}{3^{k}} \right| \le \frac {1}{3^{k}}$

Since $ \sum_{k \equal{} 0}^{\infty} \frac {1}{3^{k}}$ is convergent I use Comparison test. Since we have all absolutely convergent series is convergent we have,

$ \implies \sum_{k \equal{} 0}^{\infty} \frac {sin(3^{k}x)}{3^{k}}\le \frac {2}{3}$ :oops:

Dear aktyw19, have you already seen

http://www.mathlinks.ro/viewtopic.php?t=29562    ?


trigonometry
calculus
integration
function
calculus computations

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}}= \arcsin{\frac{x}{a}}+C$

$\int \frac{dx}{x\sqrt{x^{2}-a^{2}}}= \frac{1}{a}\sec^{-1}{\frac{|x|}{a}}+C$

$\int \frac{dx}{a^{2}+x^{2}}= \frac{1}{a}\arctan{\frac{x}{a}}+C$

But should $a$ be negative or positive?

I'm guessing they should be positive since the square root function's range $>0$ correct?

Except for the case of the $\frac{1}{a}\arctan\frac{x}{a}+C$, because the 2 negatives would cancel each other.

Is this right?

I'd put it this way: an antiderivative of a positive function should be increasing. Hence $a>0$ in the first and second formulas.

Well, $a$ cannot be zero, but it [b]can[/b] be negative.

But then, you need to write, $|a|$ rather than $a$.