I'm doing a presentation on Renaissance as my final for English class. My whole group (4 including myself) is doing Renaissance and my topic is "Importance of Renaissance."

I basically know what Renaissance is but I want to see some discussion going on about that topic so I can see whether I'm on the right track.

Here is my not-yet-edited document of the importance of Renaissance.

[hide="Here"]

Renaissance is a period from 14th century to 16th century where there occurred the huge development of painting, sculpture, architecture, music, literature, philosophy, science, and technology.  It was also the time of foundation change on economic, basic European structure, and the organization of states. This word "Renaissance" used to be known as Latin word "rinascita."

Italy during the Renaissance was made up of many diffferent countries that are not happy with each other.  Each had [i]Condottieri[/i], an army that fights by money, and this caused the relationship to get worse.

Anyway, this point, which was ending of Middle Ages and the beginning of the modern times, was the start of the Renaissance. In Italy, the Renaissancd is divided into three eras: Early, High, and Late. In Early Renaissance, the ancient things are revived.  When retracing these, the artists tried to create things that are consistent with nature and human personality than imitating the things in the past. In High Renaissance, the most concept was unity and it was destined to break up if one element is favored over another. In Late Renaissance, it was highly concentrates on individaul competition and the sack of Rome in 1527 ended Italy's Renaissance.

Although sack of Rome in 1527 [b]was[/b] important event on ending the Italian Renaissance, it wasn't only reason. Despite its development in literature/art and having new prospective of humanism and Greek with the invention of printing, which made learning of ancient things faster, the Renaissance period didn't last forever because of its neighboring countries. Italy was attacked by France in 1494 and as mentioned, sack of Rome in 1527, with Spanish domination of Italy at 16th century ended the Italy's Renaissance completely.

From France's first invasion at 1494, however, opened a door to Renaissance in England and other countries. Even thgouh there were students' movement from Germany, France, and England to Italy, the invasion of France made the Northern Europe to see highly skilled works of architect and art on Rome.  After those works were shown, the Northern Europe also began its learning of Greek and Latin with humanist through the Bible and Church Fathers.  [/hide]

I notice that when I write this.. I'm missing a lot of point (but I'm thinking of using PowerPoint so I'll do the talking --which I do better than writing from the note) and especially in the fact that I need more information on England Renaissance than Italian Renaissance. I think I have enough on Italy but I need quite more..

Anyone knows good book or site I can use for this? Thanks.. And post your comment on my writing or just about Renaissance. :)

There are a lot of important events that occurred in the Renaissance. These may include the advance in technology, art, economy, government, and many others. It is very difficult to make a good report about the Renaissance in general because so many things happened throughout its duration.

I would recommend that you give more information about the Renaissance in your report, since most of the report is setting the stage for cogent information to be given.

[quote]I would recommend that you give more information about the Renaissance in your report, since most of the report is setting the stage for cogent information to be given.[/quote]

Other three members are doing on important figures, clothing, and entertainment.

So what should I add? :)


Let S be a square pizza of side s. Let C be a circular pizza of radius r. The pizzas S and C have the same perimeter. Let AS be the area of S, and let AC be the area of C. What is AS/AC?

(A)π2/4 (B) 4/π2 (C) 1 (D) 4/π (E) π/4

[hide] From the given, $ 4s\equal{}2\pi r \implies s\equal{}\frac{\pi r}{2}$. Now, $ \frac{s^2}{\pi r^2}\equal{}\frac{(\frac{\pi r}{2})^2}{\pi r^2}\equal{}\boxed{\frac{\pi}{4}}$. Hence, the answer is E.[/hide]

All we have to do is find $ s$ in terms of $ r$. The perimeter of the square pizza is $ 4s$. The perimeter of the circle is its circumference, or $ 2\pi r$. Thus, $ 4s=2 \pi r$, so $ s=\frac{\pi r}{2}$. We now have
\begin{eqnarray*} \frac{AS}{AC}&=& \frac{s^2}{\pi r^2} \\ \\
&=& \frac{ \left( \frac{\pi r}{2} \right)^2}{\pi r^2} \\ \\
&=& \frac{\frac{\pi^2 r^2}{4}}{\pi r^2} \\ \\
&=& \frac{\pi^2 r^2}{4 \pi r^2} \\ \\
&=& \frac{\pi}{4} \end{eqnarray*}
Thus, the answer is $ \boxed{\mathbf{E}}$.

EDIT: Beaten.


AMC
AIME
ratio
function
geometric series

Q) Two distinct , real,infinite geometric series each have a sum of $ 1$ and have the same second term. The third term of one of the series is $ \frac{1}{8}$. If the second term of both the series can be written in the form $ \frac {\sqrt{m} \minus{}n}{p}$ , where m,n, and p are positive integers and m is not divisible by the square of any prime, find the value of $ 100m \plus{}10n \plus{}p$ .

looks like an AIME :D

[hide]suppose the second term is s and the ratios of the sequences are $ r_1,r_2$ respectively
so because the sum of each sequene is 1, we have:
$ s \cdot \frac{\left(r_1\right)^{\minus{}1}}{1\minus{}r_1}\equal{}1, s \cdot \frac{\left(r_2\right)^{\minus{}1}}{1\minus{}r_2}\equal{}1$
and since the third term of 1 sequence (WLOG the first), we have $ s \cdot r_1\equal{}\frac18$
now from the first 2 equations, we can see that
$ \frac{\left(r_1\right)^{\minus{}1}}{1\minus{}r_1}\equal{}\frac{\left(r_2\right)^{\minus{}1}}{1\minus{}r_2}$
$ \frac{1}{\left(1\minus{}r_1\right)r_1}\equal{}\frac{1}{\left(1\minus{}r_2\right)r_2}$
$ \minus{}r_1^2\plus{}r_1\equal{}\minus{}r_2^2\plus{}r_2$
$ r_1^2\minus{}r_1\equal{}r_2^2\minus{}r_2$
since the ratios must be different, we want 2 values x which give the same output for $ x^2\minus{}x$
and since this function is reflective about the line x=.5, $ r_1, r_2$ must average to .5, meaning $ r_1\plus{}r_2\equal{}1$
but we know
$ s \cdot \frac{\left(r_1\right)^{\minus{}1}}{1\minus{}r_1}\equal{}1$
$ s \cdot \frac{1}{\left(1\minus{}r_1\right)r_1}\equal{}1$
$ \frac{\minus{}r_1^2\plus{}r_1}{s}\equal{}1$
$ \minus{}r_1^2\plus{}r_1\equal{}s$
but we also know $ s \cdot r_1\equal{}\frac{1}{8}$
from these 2 equations, we have
$ r_1(\minus{}r_1^2\plus{}r_1)\equal{}\frac{1}{8}$
$ \minus{}r_1^3\plus{}r_1^2\equal{}\frac18$
$ \minus{}8r_1^3\plus{}8r_1^2\minus{}1\equal{}0$
$ 8r_1^3\minus{}8r_1^2\plus{}1\equal{}0$
$ (2r_1\minus{}1)(4r_1^2\minus{}2r_1\minus{}1)$[/hide]there are 3 real solutions so you will have to find them and test into the original equations to see which one works

[hide="Solution."]
=====
Let the two series have first terms $ x,y$ and common ratios $ p,q$, respectively. Since the second terms are equal, $ xp \equal{} yq$. Since the sum of the series are equal, $ \frac {x}{1 \minus{} p} \equal{} \frac {y}{1 \minus{} q} \implies x \minus{} xq \equal{} y \minus{} yp \implies xp \minus{} xpq \equal{} yp \minus{} yp^2 \implies yq \minus{} yq^2 \equal{} yp \minus{} yp^2 \implies p(1 \minus{} p) \equal{} q(1 \minus{} q)$. Since the two series are distinct, $ p\neq q$, so we must have $ p \equal{} 1 \minus{} q$.

Now $ \frac {x}{1 \minus{} p} \equal{} 1 \implies x \equal{} 1 \minus{} p$. WLOG, let the first series have third term $ \frac {1}{8}$. Then $ xp^2 \equal{} (1 \minus{} p)p^2 \equal{} \frac {1}{8}$. Knowing that one solution to this is $ \frac {1}{2}$, we can use synthetic division to find the other two solutions, which will be $ p$ and $ q$. Taking either of these, we find $ p \equal{} \frac {1 \plus{} \sqrt {5}}{4} \implies x \equal{} 1 \minus{} p \equal{} \frac {3 \minus{} \sqrt {5}}{4} \implies xp \equal{} \frac {\sqrt {5} \minus{} 1}{8}$, so our answer is $ \boxed{518}$.
=====
[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$ be a finite abelian group, and $A$ and $B$ be two subsets of $G$ such that $ A\cap\{\minus{}B\}\equal{}\{0\}$. Prove:
\[ |A\plus{}B|\ge |A|\plus{}|B|\minus{}1\]

It's false. A counterexample: Let $ G\equal{}\mathbb{Z}\times (\mathbb{Z}/2\mathbb{Z})$, represented by ordered pairs $ (a,b)$. If $ A\equal{}B\equal{}\{(x,y): 1\le x\le 3\}$, $ A\plus{}B\equal{}\{(x,y): 2\le x\le 6\}$, so $ |A|\equal{}|B|\equal{}6$ and $ |A\plus{}B|\equal{}10$.
Or, more generally, if $ A$ and $ B$ are cosets of the same subgroup $ H$, $ |A|\equal{}|B|\equal{}|A\plus{}B|$. As long as $ H$ isn't the whole thing, it's always possible to choose those cosets so that their sum is a coset not containing zero.

For this result to have a chance, we need every nontrivial proper subgroup to be infinite. On $ \mathbb{Z}$ or $ \mathbb{Q}$ or $ \mathbb{R}$, the order structure gives a nice proof.

@Jmerry: The question states $ A\cap\{ \minus{} B\} \equal{} \{0\}$, not $ A\cap\{ \minus{} B\} \equal{} \emptyset$. See

L. Moser and P. Scherk, [i]Distinct elements in a set of sums[/i], Amer. Math. Monthly, 62 (1955), pp. 46-47.

  darij

How can I find the article?Can you help me?

I've attached it. Note that the solution is somewhat misstated at one point: When it says "Let $ A_1\equal{}\left\{a_1\right\}$ and $ B_1\equal{}\left\{b_1\right\}$", it actually wants to say "Let $ A_1$ be the set of all such $ a_1$, and $ B_1$ be the set of all such $ b_1$".

  darij

Thank you very much,darij.That's a really nice proof.


Find the formular of this sequence in terms of n:
  $u_{0}=0$
  $4u_{n+1}=u^{2}_{n}+4$
have a good day.

But can be give asimtotic formul. Let $v_{n}=2-u_{n}, v_{0}=2$, then $v_{n+1}=v_{n}-\frac{v_{n}^{2}}{4}$. Therefore $v_{n}\le \frac{4}{n+2}$ and exist c, suth that $v_{n}>\frac{4}{n+2}-\frac{c}{(n+2)^{2}}.$

I don't understand . Because I known little english . Sorry you if I write in correct . Thankyou. :blush:


For every $ 1\leq a,b,c,d\leq 2$, prove that

$ \frac{4}{3}\leq \frac{a}{b\plus{}cd}\plus{}\frac{b}{c\plus{}da}\plus{}\frac{c}{d\plus{}ab}\plus{}\frac{d}{a\plus{}bc}\leq 2$.

$ \sum_{cyc} {\frac {a}{b \plus{} cd}}\geq \sum_{cyc} {\frac {a}{b \plus{} 2c}} \equal{} \sum_{cyc} {\frac {a^2}{ab \plus{} 2ac}}\geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{ab \plus{} bc \plus{} cd \plus{} da \plus{} 2ac \plus{} 2bd}\geq\frac {4}{3}$

[quote="Marius Mainea"]$ \sum_{cyc} {\frac {a}{b \plus{} 2c}} \equal{} \sum_{cyc} {\frac {a^2}{ab \plus{} ac}}$[/quote]

I think you're wrong here...

[quote="discredit"]For every $ 1\leq a,b,c,d\leq 2$, prove that

$ \frac {4}{3}\leq \frac {a}{b \plus{} cd} \plus{} \frac {b}{c \plus{} da} \plus{} \frac {c}{d \plus{} ab} \plus{} \frac {d}{a \plus{} bc}\leq 2$.[/quote]
$ \sum_{cyc}\frac {a}{b \plus{} cd}\equal{}\sum_{cyc}\frac {a^2}{ab \plus{} acd}\geq\sum_{cyc}\frac {a^2}{ab \plus{} 2cd}\geq\frac{(a\plus{}b\plus{}c\plus{}d)^2}{3(ab\plus{}bc\plus{}cd\plus{}da)}\equal{}$
$ \equal{}\frac{(a\plus{}b\plus{}c\plus{}d)^2}{3(a\plus{}c)(b\plus{}d)}\geq\frac{4}{3}.$
The $ RS:$
Let $ f(a,b,c,d)\equal{}\sum_{cyc}\frac {a}{b \plus{} cd}.$ Then $ \frac{\partial^2 f}{\partial a^2}>0,$ $ \frac{\partial^2 f}{\partial b^2}>0,$ $ \frac{\partial^2 f}{\partial c^2}>0$ and $ \frac{\partial^2 f}{\partial d^2}>0.$
Thus, $ \max f\equal{}\max\{f(2,2,2,2),f(2,2,2,1),f(2,2,1,2),f(2,1,2,2),f(2,2,1,1),f(2,1,2,1),f(2,1,1,2),f(2,1,1,1),f(1,1,1,1)\}.$

I think the simplest way would be to use what arqady has said:

We know that a,b,c,d are independent variables (there are no constraints that relate one another)

If we consider LHS as a function in a, then this function is convex (it is a sum of a linear function and the reciprocal of a linear function - both of which are convex).

Therefore, it attains its maximal values at its extremal points - where a = 1,2. (Any other maximal point would be a contradiction to the convexity of this function)

If we do the same for b,c,d, we obtain that this function's maximum can be achieved by considering {a,b,c,d} = {1,2}.

And this is easily 2.

[quote="discredit"]For every $ 0 < a,b,c,d\leq 2$, prove that

$ \frac {4}{3}\leq \frac {a}{b + cd} + \frac {b}{c + da} + \frac {c}{d + ab} + \frac {d}{a + bc}.$[/quote]${ \sum_{cyc} {\frac {a}{b + cd}}\geq \sum_{cyc} {\frac {a}{b + 2d}} \geq \dfrac {\left(\stackrel{\sum}{_{cyc}} {a}\right)^2}{\stackrel{\sum}{_{cyc}}{a(b + 2d)}}} = \frac {(a + b + c + d)^2}{3(a + c)(b + d)}\geq\frac {4}{3}$.
[quote="Marius Mainea"]$ \sum_{cyc} {\frac {a}{b + 2c}} \geq\frac {4}{3}$[/quote]A counterexample : $ a = c = 1,b = d = 2$.
[quote="discredit"]For every $ 1\leq a,b,c,d\leq 2$, prove that

$ \frac {a}{b + cd} + \frac {b}{c + da} + \frac {c}{d + ab} + \frac {d}{a + bc}\leq 2.$[/quote]$ \sum_{cyc} {\frac {a}{b + cd}}\leq\sum_{cyc} {\frac {a(6a +b+ c + d)}{3\left(a^2 + b^2 + c^2 + d^2\right)+ab+bc+cd+da+ac+bd}} = 2.$


A cube has a volume of 343 cubic cm. What is the number of square
centimeters in the surface area of the cube?

The cube has side length $ \sqrt[3]{343}\equal{}7$, so its surface area is $ 6(7^2)\equal{}294$.


hey there,

im looking to create a t-shirt for my g/f (silly i know) that has some mathematical equation or theory on it...but uses the number 10.....my last name is DIEZ

and in spanish (which i am) DIEZ is the number 10

so really im gunna replace the number 10 with DIEZ

or even if the letters D I E Z had some role in the equation or design..that would be cool too.

so if anyone could come up with something funny or clever...that would fricken cool!

perhaps you could have some quadratic formuls out into use where a=5 or something....

There is a commercial running on TV here (israel) of a guy buying a T shirt out of many t shirts, where it has the picture of his girl friend's face on it... i think its cool

maybe you can print her face and around it 10 ballons or 10 hearts

go for something simple - like 10 out of 10. It works well.


graph theory
combinatorics unsolved
combinatorics

[b](i)[/b] Determine the smallest number of edges which a graph of $ n$ nodes may have given that adding an arbitrary new edge would give rise to a 3-clique (3 nodes joined pairwise by edges).
[b](ii)[/b] Determine the smallest number of edges which a graph of $ n$ nodes may have given that adding an arbitrary new edge would give rise to a 4-clique (4 nodes joined pairwise by edges).

I've edited the statement - now it should be clearer (I had troubles with understanding the original problem statement, too).

  darij

[b](i)[/b] The answer is $ n\minus{}1$. The graph is clearly connected (otherwise adding an edge between two vertices in different components cannot create a triangle) so it has at least $ n\minus{}1$ edges, and the graph $ K_{1,n\minus{}1}$ which has $ n\minus{}1$ vertices all connected to a single root (and no other edges) clearly satisfies the requirement.

(An alternative description of the problem: determine the least number of edges in a graph with $ n$ vertices and diameter $ 2$).

The statement of the problem is still a bit vague, in that it's not clear if the graph is allowed to contain a clique to begin with. Presumably it may not, so my interpretation of the problem is as follows: Let $ k \geq 3$ be an integer. A graph is called $ k$-critical if it does not contain a $ k$-clique, but every graph obtained by adding an edge does. For a given number of vertices $ n$, determine the least possible number of edges in a $ k$-critical graph.

One simple type of $ k$-critical graphs is the family of complete $ (k \minus{} 1)$-partite graphs. These are defined by taking $ k \minus{} 1$ sets $ A_1, A_2, \ldots, A_{k \minus{} 1}$ of vertices, and connecting every vertex in $ A_i$ to every vertex in $ A_j$ for $ i \neq j$. Thus the sets $ A_i$ themselves remain independent. Such a graph cannot contain a $ k$-clique since any set  of $ k$ vertices necessarily contains two in the same $ A_i$, so they're not connected. On the other hand, the only edges we can add are those that connect two vertices in the same $ A_i$, and as soon as we add such an edge we get a $ k$-clique (usually, many of them) by selecting an arbitrary vertex in each of the other $ A_j$s.

If we focus on such graphs, the way to make them have the least number of edges is easily seen to be letting all the $ A_i$ for $ i > 1$ have size $ 1$ and putting all the other vertices in $ A_1$.

[b](ii)[/b] For $ k \equal{} 4$ this yields the following graph: There are two vertices $ a,b$ connected by an edge, and a set of $ n \minus{} 2$ vertices $ x_1, x_2, \ldots x_{n \minus{} 2}$ each of which is connected to both $ a$ and $ b$ but nothing else. This graph has $ 2n \minus{} 3$ edges, and we contend that this is the smallest possible number of edges in a $ 4$-critical graph. So, we need to prove that a graph $ G$ with $ n$ vertices and $ 2n \minus{} 4$ edges cannot be $ 4$-critical: if it doesn't contain a $ 4$-clique, then we may add some edge to it and obtain a graph which still doesn't contain a $ 4$-clique.

Let $ d_i$, $ i \equal{} 1,2,\ldots, n$ be the degrees of the vertices in $ G$. We obviously have $ \sum d_i \equal{} 2(2n \minus{} 4) \equal{} 4n \minus{} 8$ since adding the degrees counts each edge twice. We draw two conclusions from this simple result: First, $ G$ contains a vertex of degree at most $ 3$. Second, if there is a vertex of degree at least $ n \minus{} 2$, then $ G$ contains a vertex of degree at most $ 2$.

Obviously, $ G$ cannot contain a vertex of degree $ \leq 1$, since adding an edge adjacent with such a vertex would make it have degree $ \leq 2$ which is not enough for a $ 4$-clique. Suppose that $ G$ contains a vertex $ z$ of degree $ 2$, and let $ a$ and $ b$ be $ z$'s neighbors. Take any other vertex $ x$ in $ G$. Then $ zx$ is a missing edge, so adding it must yield a $ 4$-clique containing $ z$. Since $ z$'s neighbors are now $ a,b$ and $ x$, the $ 4$-clique must be spanned by no other than these $ 4$ vertices. Thus we have that $ a$ and $ b$ are connected, and $ x$ is connected to both of them. Since $ x$ was arbitrary, we have found $ 2n \minus{} 3$ edges in $ G$ (the three edges $ za, zb, ab$ and two more for any other vertex $ x$). Note that we just reconstructed the complete $ 3$-partite graph described above (it's easy to see that any two vertices $ x, y$ different from $ z,a,b$ cannot be connected to each other, since that would yield a $ 4$-clique in $ G$).

Finally, suppose that $ G$ has no vertices of degrees $ 1$ or $ 2$. It must then contain a vertex $ z$ of degree $ 3$. Let $ a,b,c$ be its neighbors. The edges $ ab, ac, bc$ cannot all be present, lest we get a $ 4$-clique, so suppose $ bc$ is not an edge of $ G$. As before, let $ x$ be any other vertex, so the edge $ zx$ is missing. Adding it results in a $ 4$-clique spanned either by $ \{z,x,a,b\}$ or $ \{z,x,a,c\}$: it cannot be spanned by $ \{z,x,b,c\}$ since $ bc$ is not an edge. In any case, $ x$ must be connected to $ a$. Thus $ a$ is connected to $ z$, either $ b$ or $ c$ (or both), and every other $ x$, so it has degree at least $ n \minus{} 2$. We saw that this forces $ G$ to contain a vertex of degree $ 2$, and we already dealt with this case, so we're done. QED

As a final note, we point out that the problem of determining the [i]largest[/i] number of edges in a $ k$-critical graph is a famous theorem in extremal graph theory, Turan's theorem. The answer is again obtained by looking at complete $ (k \minus{} 1)$-partite graphs, only instead of making the sets $ A_i$ as imbalanced as possible like we did, we need to make them as balanced as possible in order to obtain as many edges as possible. So both the maximum number of edges and the minimum number of edges of $ k$-critical graphs are achieved by $ (k \minus{} 1)$-partite graphs (we only proved the minimum for $ k \equal{} 3,4$. I'm not sure if this holds for higher $ k$). However, we note that not all $ k$-critical graphs are $ (k \minus{} 1)$-partite. For example, take the cycle of length $ 5$: it is a $ 3$-critical graph, but it's not bipartite.

As randomgraph suggested, this does generalize:

If a graph with $n$ vertices has the property that adding any arbitrary new edge would increase the number of $k$-cliques in the graph, then the graph must have $\geq n\left(k-2\right) - \binom{k-1}{2}$ edges. Moreover, if it has exactly $\geq n\left(k-2\right) - \binom{k-1}{2}$ edges and has this property, then it must consist of a $k-2$-clique and all possible edges connecting the vertices of this $k-2$-clique with the remaining $n-k+2$ vertices of the graph.

Source: Erdös, P., Hajnal, A. and Moon, J.W., [i]A problem in graph theory[/i], Amer. Math. Monthly 71 (1964), pp. 1107-1110. http://www.renyi.hu/~p_erdos/1964-02.pdf

Two remarks on randomgraph's proofs of the $k=4$ case:

[quote="randomgraph"]One simple type of $ k$-critical graphs is the family of complete $ (k \minus{} 1)$-partite graphs. These are defined by taking $ k \minus{} 1$ sets $ A_1, A_2, \ldots, A_{k \minus{} 1}$ of vertices,[/quote]

These sets must be nonempty for the construction to work.

[quote="randomgraph"] So, we need to prove that a graph $ G$ with $ n$ vertices and $ 2n \minus{} 4$ edges cannot be $ 4$-critical: [/quote]

Not $2n-4$ edges, but [i]at most[/i] $2n-4$ edges. (But it's easy to modify your solution so that it works.)


Prove that for any n>2 there are prime numbers p1, p2, ...,pn such that p1<p2<...<pn and p1+p2>pn.

I wonder if there is a more elementary way to see this than mine:

It is known (for instance it follows from the prime number theorem) that for each $ c>0$ there exists some positive integer $ N$ such that for all $ k>N$, there is a prime between $ k$ and $ (1\plus{}c)k$.

Taking $ c$ to be $ \sqrt[n]2\minus{}1$, we find that for all sufficiently large $ m$, there are $ n$ primes between $ m$ and $ 2m$. In this case, the sum of any two such primes must be bigger than the largest such prime.


مرحله ی دوم بیست و سومین المپیاد ریاضی ایران
تمام توابع f:R+→R+ را بیابید که برای هر  ،

(x+y)f(f(x)y)=x2f(f(x)+f(y))

منظور از R+ مجموعه ی اعداد حقیقی مثبت است (توجه کنید صفر عددی مثبت نیست!).

sabet kon $f(x)$ yek be yeke hala $x={{1+\sqrt 5 }\over 2}, y=1$ ...


rotation
integration
LaTeX
calculus
calculus computations

Hey guys, I have three questions that I need the answers checked for before I turn it in (seriously, I did them). They are all the same type of problem, take a graph and rotate it around an axis and find the volume of the solid created.

Sqrt(9-(x^2))+3 Around the x-axis (should be an arc with the point x=0, y=6) from 0 to 3sqrt(2)/2

(-x+7.24) rotated around x-axis from 0 to 7.24

The semi circle Sqrt(9-((x-3)^2) rotated around the y-axis to make a half-bagel like shape

Use whatever method you want, I used slices for the first two and cylinders for the last one. I really need to make sure my answer is correct. Thanks a lot!

For the first two questions, we can use
Volume $ V\equal{}\int_a^b \pi f(x)^2 dx$,
which gives the answers
$ \int_0^{\frac{3\sqrt{3}}{2}}\pi\left( \sqrt{9\minus{}x^2}\plus{}3\right)^2 dx\approx 254$
and $ \int_0^{7.24}\pi(x\minus{}7.24)^2 dx\approx 397$

I didn't get the third question. A half-bagel (or half-torus) isn't a closed surface, is it?

If you did them, why not write up your solutions (use [[LaTeX]] -- see that link or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this one[/url]) and ask others to check them for you?

well, the semi circle is cut off my the x-axis. Sorry I forgot to mention that. The also got 397 for the second one, but for the first one you integrated to 3Sqrt(3)/2 instead of 3Sqrt2)/2. For that one I got 218.993. And for the bagel I got 266.479 for a total of 883.22. Anyone please confirm?


I have five apples and ten oranges. If a fruit basket must contain at least one piece of fruit, how many kinds of fruit baskets can I make?  (The apples are identical and the oranges are identical.  A fruit basket consists of some number of pieces of fruit, and it doesn't matter how the fruit are arranged in the basket.)

You can put $ 0$, $ 1$, $ 2$, $ 3$, $ 4$, or $ 5$ apples, and $ 0$, $ 1$, $ 2$, $ 3$, $ 4$, $ 5$, $ 6$, $ 7$, $ 8$, $ 9$, or $ 10$ oranges, so there $ 6$ choices for the number of apples and $ 11$ choices for the number of oranges, giving $ 6\cdot11 \equal{} 66$ choices, but a basket with no fruits at all doesn't count, so we subtract the $ 1$ possibility ($ 0$ apples, $ 0$ oranges) for an empty basket: $ 66 \minus{} 1 \minus{} \boxed{65}$.


superior algebra
superior algebra unsolved

Let $ K$ be a comutative field with $ |K| \geq 3$ and $ x^2 \equal{} x \plus{} 1 , \forall x \in K$\{0,1} . Find $ |K|$ .

Rather trivial: That field has at most $ 4$ elements (the roots of that equation and $ 0,1$). And in fact, it's the field with $ 4$ elements.


factorial
function
algebra
domain
polynomial
number theory
Diophantine equation

Define $ F(x)$ as follows:

$ F^1 (x) \equal{} x!$
$ F^2 (x) \equal{} F(F(x)) \equal{} (x!)!$
$ F^n (x)$ is $ F(F(F(.....(x))))....$ for $ n$ times.

If $ F^{2008} (x) \minus{} F^{1} (x) \equal{} 0,$ find the sum of all roots of $ x$ that satisfy this equation.

Level: Easy Medium

If $ x\ge 2$, $ x! > x\implies F_{n \plus{} 1}(x) > F_n(x)$ . We check: $ x \equal{} 0,1$. These work...so $ 1$.

[quote="t0rajir0u"]The notation $ F(x)$ and the use of the word "roots" is misleading.  $ F(x)$ is understood to have domain the non-negative integers, so this is really a Diophantine equation and here we do not usually speak of "roots."[/quote]

What is the suitable word then? I'm not yet familiar with diophantine equation yet (I need to learn this when I get chance!).

http://mathworld.wolfram.com/Root.html

"The roots of an equation f(x)=0 are the values of x for which the equation is satisfied..."

So I think it is okay. But I see your point. Perhaps another phrase would be more appropriate.

Lol, it's incorrect because 2!=2

The correct answer is 0+1+2=3

The issue is not in the definition of "root" so much as the confusion over what values $ x$ is allowed to take.  Usually we say "find the sum of all non-negative integers $ x$ such that...."

[quote="The Zuton Force"]	Interesting problem. However, if one notices that the fixed points of F(x) are 1 and 2, and forgets to consider 0, that person would still get the right answer. If you use this problem anywhere else, you might want to change that :P[/quote]
You could do $ F^{2008}(x\minus{}1)\minus{}F^1(x\minus{}1)\equal{}0$ instead to fix that.

The way I solved it was to set $ F^{2008} (x) \equal{} F^{1} (x)$. I noticed that 1 is an immediate answer while 2, once finding $ F^2, F^3,...$, repeats so it must be an answer. I actually did not anticipate 0 to be an answer just because I presumed factorial to be for positive integers.

I'll remember to say "find all non-negative numbers" for the next time. Hmm, so the "root," in technical terminology, is the answer to the equation, not like a recursive function as in this one?

There's nothing technically wrong with using the word "root," but the notation $ F(x)$ suggests a polynomial and the word "root" along with the variable $ x$ suggests real answers (at least).  Usually for integer-valued variables we use, say, $ n$.


Find all positive integens $n$ for which $n^5-1$ is prime. Please show solution and without guess and check, because just by looking at it you can easily tell that 2 is a solution.

[hide]To be prime, a number must have factors 1 and itself.

$n^5-1$can be factored as $(n-1)(n^4 + n^3 + n^2 + n + 1)$ so either $n-1$ or $n^4 + n^3 + n^2 + n + 1$ must be equal to 1.

The latter obviously cannot = 1 in the natural numbers and $n-1$ is equal to 1 at $n=2$

Plug in $2$ to $n^4 + n^3 + n^2 + n + 1$ and you get $31$ which is prime.

So $n^5 -1$is prime when $\boxed{n=2}$[/hide]

Just in case someone didn't get shoefactory's quick factoring,its....

$x^n-y^n=(x-y)(x^{n-1} + x^{n-2}y+........+y^{n-1})$

For $x^n+y^n$,put $y=-y$

Sorry, my bad.Messed up the signs.

And yes,when $n$ is odd.Thanks for pointing that out you two. :)

You can only factor like that when n is odd i believe.

[quote="José"][quote="edriv"][quote="shadysaysurspammed"]Just in case someone didn't get shoefactory's quick factoring,its....

$x^n+y^n=(x+y)(x^{n-1} + x^{n-2}y+........y^{n-1})$

For $x^n-y^n$,put $y=-y$[/quote]

I should be $x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-\ldots-xy^{n-2}+y^{n-1})$ (only when n is odd)[/quote]

There are some special cases when you CAN factorice n, when n is even: $n=q*2^k$, with q odd. Then, $x^{q*2^k}+y^{q*2^k}=(x^{2^k})^q+(y^{2^k})^q$[/quote]

Well, ultimately you are tansforming it to odd powers and then using the identity

[quote="shadysaysurspammed"]Well, ultimately you are tansforming it to odd powers and then using the identity[/quote]

Yes, but it was to show that you can for same even n's.


The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits, any of which could be zero.  It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$  What is the number of possible values for $r$?

[hide="Solution"]
Note that any fraction with 1 as a numerator can be written as a fraction with 2 as a numerator, so we will only look at the set of fractions that can be written with 2 as the numerator and an integer denominator.

The largest fraction satisfying your requirements $< \frac{2}{7} = .28571428...$ is $\frac{2}{8} = .25$, and the smallest fraction satisfying your requirements $> \frac{2}{7}$ is $\frac{2}{6} = .33333...$

In order that $\frac{2}{7}$ be the best approximation, r must be closer to $\frac{2}{7}$ than to either $\frac{2}{8}$ or $\frac{2}{6}$, so r can range from $\displaystyle\frac{ \frac{2}{7} + \frac{2}{8} }{2} = \frac{15}{56} = .267857...$ to $\displaystyle\frac{ \frac{2}{7} + \frac{2}{6} }{2} = \frac{13}{42} = .309523...$.

This tells us $.2679 \le r \le .3095$ (because if $r = .2678$ it would be closer to $\frac{2}{8}$, and if $r = .3096$ it would be closer to $\frac{2}{6}$), so there are $3095 - 2679 + 1 = \boxed{417}$ possibilities.
[/hide]

[hide=Solution] The number r needs to be at least the average of $\frac{2}{7}$ and $\frac{1}{4}$, which is  about .26785. The number r needs to be less than the average of $\frac{2}{7}$ and $\frac{1}{3}$ which is about .30952. Hence we have the inequalities $2679 \leq 1000r \leq 3095$, the answer is $3095-2679+1=\boxed{417}$.[/hide]

p10: If $a = 2, b = 6, c = 7$ then we have $1$ possibility for $d$. Otherwise if $c = 8,9$ there are $10$ ways, for a total of $21$. If $b = 7, 8, 9$ then there are $300$ ways total, for a total of $321$ with $a = 2$. If $a = 3, b=0$ then $c$ going from $0$ to $8$ gives $90$ ways. If $c = 9$, $d$ can be from $0$ to $5$ inclusive, for $6$ more ways, totaling $96$. Adding gives $321+96 = \boxed{417}$.


If $ x,y,z$ be nonnegative numbers, then

$ 24565\sum_{sym}x^4yz\plus{}25600\sum_{sym}y^3z^3\plus{}726398x^2y^2z^2\geq 126945\sum_{sym}x^3yz(y\plus{}z),$

with equality if $ x \equal{}5,y \equal{}1, z \equal{} 2.$

[quote="Ji Chen"]If $ x,y,z$ be nonnegative numbers, then

$ 24565\sum_{sym}x^4yz \plus{} 25600\sum_{sym}y^3z^3 \plus{} 726398x^2y^2z^2\geq 126945\sum_{sym}x^3yz(y \plus{} z),$

with equality if $ x \equal{} 5,y \equal{} 1, z \equal{} 2.$[/quote]

Assume $ a \plus{} b \plus{} c \equal{} 1$ and rewrite the inequality
\[ Axyz\left( {\sum {x^3 } } \right) \plus{} B\left( {\sum {x^3 y^3 } } \right) \plus{} Cx^2 y^2 z^2 \minus{} Dxyz\left( {\sum {xy\left( {x \plus{} y} \right)} } \right) \ge 0
\]
with
\[ A \equal{} 24565,B \equal{} 25600,C \equal{} 726398,D \equal{} 126945
\]
. Setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ then the inequality becomes
\[ Bq^3 \plus{} \left( {A \minus{} \left( {3A \plus{} 3B \plus{} D} \right)q} \right)r \plus{} \left( {3A \plus{} 3B \plus{} C \plus{} 3D} \right)r^2 \ge 0
\]
We see that if
\[ q \le \frac {A}{{3A \plus{} 3B \plus{} D}} \equal{} 0,0885
\]
then the inequality is proved. So we just have to prove with
\[ q \ge 0,0885
\]
. We have
\[ \Delta \equal{} A^2 \plus{} \left( {3A \plus{} 3B \plus{} D} \right)^2 q^2 \minus{} 2A\left( {3A \plus{} 3B \plus{} D} \right)q \minus{} 4B\left( {3A \plus{} 3B \plus{} C \plus{} 3D} \right)q^3
\]

\[ \Rightarrow \Delta \le 0
\]
with
\[ q \ge 0,0664
\]
So the inequality is proved. Equality occur with $ x \equal{} 5/8,y \equal{} 1/8$ and $ z \equal{} 2/8$.


At school, I'm a little weak in English subject, Could anyone tell me anyway to learn English easier.

[quote="Valentin Vornicu"][quote="paganinio"]learn Japanese, German
it does nothing to improve your English, but it improves your overall language skills[/quote]This does not help him at all.
What you need to do is to watch as many movies as possible without tranlations (or with at most subtitles). That's how I learned English (I learned almost nothing in school :D).[/quote]
I tried, but I had only headache. :(

[quote="Myth"]I tried, but I had only headache. :([/quote]Depends on what movies you watch, or like to watch. I loved the Dusk till Dawn series :P

Is it the grammar thats getting you?  Because if its that just watch movies or anything to help you get a grip on the language then write.  Anything keep a diary or write short stories.  If you can write then you can do all the english stuff.

Listen to English-language radio broadcasting as much as you possibly can. That is even better than English-language television for learning English. That's what I did to learn a lot of Chinese: listen to the radio.

Zhi ni hao yun.

Maybe it is easy for you to say that, but I have great difficulties :(

[quote="eryaman"]I would mostly read, and listen to English.  I am Chinese but was born in America, so I can speak (not well) but not read or write in Chinese.  The reason I can speak even though my last formal Chinese lessons were like 5-6 years ago is through listening to my parents, and sometimes speaking.
So listening really helps, if they speak too fast for you, you should try to find a slower person or whatever.
I don't know if this method is possible, but you can go to an English-speaking country for a few weeks.  My cousins, who lived in Taiwan but now live in the United States, had bad English before they came but after a few weeks they spoke so-so English.  It's still not as good as mine, but they can generally communicate.  And they've started saying mostly English things like darn, shute, cool, and even AOL speak, like "u" and "lol".  Living in a country forces you to learn it quickly because you can't survive without communicating iwht others.[/quote]

I think you are right, listening is really important.
     "go to an English-speaking country for a few weeks",hmm, I think that's impossible for me, my family isn't rich enough to go abroad. Maybe I can listen to tapes and the radio.
     But I shoud point ou that the English Education in China is so bad. The teachers require the student cecite as much grammar as they can, most questions in the exam paper is strange and hard, as I know, even the ones from English speaking countries could not choose the right answer, the English exam in China isn't examining the students' English ability, It't examining the luck and the skills of how to make more marks, so that the student can make high maks in the college attending exam, so he could go to a good college, but in fact, maybe his English ability is not good . In my daily time, I have to spend most time reciting English grammar and words.
 (:blush: In fact, I truly want to make high maks)

[quote="qxy"][quote="Valentin Vornicu"][quote="paganinio"]learn Japanese, German
it does nothing to improve your English, but it improves your overall language skills[/quote]This does not help him at all.
What you need to do is to watch as many movies as possible without tranlations (or with at most subtitles). That's how I learned English (I learned almost nothing in school :D).[/quote]
To watch movie, I think it is a good way, but the speed of the actors speak is always too fast, I can't catch it, what can I do then?[/quote]In real life people speak as fast as the actors do. So you have to adapt ;)

I can recommend the series Friends. Excelent comedy, and in the same time, very clear jokes and talking. You learn a lot of English and also have a lot of fun watching them (as opposed to, say, Married with Children - aka The Bundy Family, where many of the jokes rely on slang and local culture (Oprah, etc.) :) ).

i say probably kids shows like cartoons or power rangers and stuff are the best.. i learned english from TV, and make sure u watch commercials too, its all good.

Hi dear,
 You can bye some simple and clear english learning videos,and selected words,phrasal verb,idioms .Phrasal verb and idioms are most important in speaking english.

This is how I learned English (it is my 2nd language btw):
always take a dictionary with you; underline the words whose meaning that you are not aware of and check its meaning using the dictionary; always go back and review those words

yeah I know it's tiresome but that's how I learned it; I usually have about 500 words (I think) underlined when finishing a novel.
No pain, no gain.

English is my first language, but I use a dictionary when reading through novels to learn vocabulary too.  Be sure to use one with a pronunciation guide (and key?).  Do [b]NOT [/b]learn vocab from lists without knowing context.

A test if you truly know slightly more difficult words:  Spell it.  Say it.  Use it in a sentence.  Then get someone to check it over to make sure you're using it properly.

I learned English by listening to English songs and try to follow it with lyrics.... then find of what it means

That might not work since I still can't hear the lyrics to most songs.   :(

Very good idea if you limit yourself to medium speed (not slow) songs that don't have heavily accented or slurred words.

I learned english as a 2nd language, but pretty young (3-4ish), and I still have trouble catching most song lyrics, probably one of the reasons I don't like to listen to music  :D

I find this way very helpful:

I prefer the British accent. When I started to learn English, I read and recite all the passages from New Concept English book 2. When I say "read and recite", I actually mean "read to rectie". I read each passage (after the tape) for 20 times a day and it took me two to three days to finish a passage. During this process, I really made a huge progress in my pronunciation. Besides, reading and reciting a lot of passages helped me a lot with my writing, listening and reading skill.

This may take some time, but I think that it really helps. Now the New Concept English is out of date, I sometimes go to the [url=http://www.bbc.co.uk/worldservice/learningenglish/]BBC Learning English Home Page[/url]for some reading practice. It has a very large collection of short news passages read by BBC broadcasters with vocabulary. If you are really busy, just read for 5 times each day and after a few months you will feel the difference.

But learning English also requires extensive readings and listengs. Movies, novels, online-news, English textbooks.......there are so many choices. If you find the actors in the movies speak too fast, try some documentaries, they are easier to understand.

Good Luck

I learned my (alas, rather mediocre :() English by reading and writing a lot first. Then came the ability to speak. Then I learned to understand what other people were saying. The ability to understand radio broadcasts and TV shows came last and I still cannot understand the words in the songs (especially when a woman sings in a high pitched voice accompanied by loud music). In general, just read (both silently and aloud), listen, and write as much as you can (writing is very important but you'd better have somebody to check what you wrote for grammar mistakes) and use every opportunity to speak English to somebody who knows it better than you (speaking to a person who is at the same level doesn't really help...). Also, when reading, learn each new word you read (flash cards is a great way to do it). Don't get discouraged by the fact that it'll take you a few hours to read the first 10 pages this way. The rule of thumb is that after page 20 or so most authors will exhaust their vocabulary and the number of new words per page will start decaying exponentially. As to pronunciation, shobber gave you the best advice one can give already.

[quote]I still cannot understand the words in the songs (especially when a woman sings in a high pitched voice accompanied by loud music).[/quote]

There is a solid physical reason for being unable to understand words sung against a background of loud music, and by female singers singing at a high pitch.

The information about which vowel or consonant is being pronounced is provided by concentrations of sound energy at particular frequencies, called formants (labeled f0, f1, f2, etc. in order of increasing frequency.)  f0 carries pitch, and f1 and f2 (and to a much lesser extent f3) are the resonant frequencies of various parts of the vocal tract and indicate which segment is being pronounced.

So, if there is a lot of noise (= loud music) in the same frequencies as f1 and f2, then it greatly reduces your ability to pick out the formants and tell which segment is being pronounced.

Why is it so hard to understand female singers singing at a high pitch?  If f0 is at a higher frequency than f1 or f2, then you can't distinguish which segment is being produced.  At all.

(BTW, to overcome the problem of being heard over a loud orchestra, trained opera singers pitch a large amount of sound energy around 3000 Hz, where there is a dropoff in the amount of sound energy produced by the orchestra.  See [url]http://www.ncvs.org/ncvs/tutorials/voiceprod/tutorial/singer.html[/url])

[quote="fedja"]I learned my (alas, rather mediocre :() English by reading and writing a lot first[/quote]
your english is not mediocre...I do really think so...
[quote="fedja"]
I still cannot understand the words in the songs (especially when a woman sings in a high pitched voice accompanied by loud music).[/quote]
I find it rather hard to understand the lyrics of some rapidly sung songs,especially rap music...don't you have a problems with that?
[quote="fedja"]
use every opportunity to speak English to somebody who knows it better than you (speaking to a person who is at the same level doesn't really help...)
[/quote]
I'm truly convinced that nothing can help you more to improve your english than speaking and having conversations with english native speakers.In my opinion speaking skills is the weakest side of all persons who do not live in english-speaking country,unlike all other skills,which can be improved on your own.
So the best suggestion I could give concerning speaking skills,is to attend some english courses,or to be prepared by some english native speaker,of course,if he has the needed specialization.
Grammar is not so complicated as it is,for instance, in Russian Language,nevertheless speaking is fairly easy...(in case if you do not need to study russian deeply).
[quote="fedja"] Also, when reading, learn each new word you read (flash cards is a great way to do it). Don't get discouraged by the fact that it'll take you a few hours to read the first 10 pages this way. The rule of thumb is that after page 20 or so most authors will exhaust their vocabulary and the number of new words per page will start decaying exponentially. As to pronunciation, shobber gave you the best advice one can give already.[/quote]
actually the best way to improve all skills is to travel in english-speaking country and to spend over there at least half a year or maybe even less...it depends on your memory and on your initial skills.
P.S:
hmm...did I write the post above correctly? :)

[quote="Erken"]did I write the post above correctly? :)[/quote]  It's completely understandable, but it's also reasonably evident that you aren't a native speaker.  For example, the sentence [quote="Erken"]In my opinion speaking skills is the weakest side of all persons who do not live in english-speaking country,unlike all other skills,which can be improved on your own.[/quote] has features that aren't characteristic of sentences written by native speakers -- especially the phrase "in english-speaking country," with no article.

well I meant to download and print lyrics... and physically follow it when listenning songs...(I agreed that it is impossible to distinguish some lyrics just by hearing) because that in that way you dont bore easily (assuming you are listening songs you like), and you excersice your listenning ability.

hmm... I suggest either watching movies in YOUR native language, but with english subtitles, or watching englsh movies with native language subtitles. (that's how I learned japanese!)

If someone has already posted this, I'm sorry.

ONLY if you already know at least 95% of the words.  Otherwise your word order will be entirely screwed up, as well as vocabulary.  This kind of thing doesn't work for me since I get involved in the movie, and tend to ignore the language.

Wondering:  Shouldn't this be in the Linguistics forum?

Also, watching movies is a MUST.  ALthough it may teach you inaccurate english culture.


Consider initial sequence $1,1$. Insert between any two neighbor terms their sum:
\[1,1 \to 1,2,1.\]
And again
\[1,2,1 \to 1,3,2,3,1 \to 1,4,3,5,2,5,3,4,1.\]
And so on...

Prove that every natural number $n>1$ will be written exactly $\varphi(n)$ times, where $\varphi(n)$ is an Euler's totient function.

We show by induction (after $n$) that for each $n$ and each $a<n$ with $(a,n)=1$ there will be exactly one turn when we'll have $a,n$ in our sequence (consecutive terms). When this happens, we also have $n,a$.

Assume we've shown it for $k\le n$, and we want it for $n+1$. Let $a<n+1$ with $(a,n+1)=1$. There is precisely one turn when we have $a,n+1-a$ and $n+1-a,a$, and when this happens, we get $a,n+1,n+1-a$ and $n+1-a,n+1,a$ respectively after performing the operation.

In other words, each number $n$ is obtained exactly once as $a+n-a$ for each $a<n,\ (a,n)=1$ (and it's easy to see that it can't be obtained as $n=a+n-a$ with $(a,n)\ne 1$, because at each step, each two consecutive terms in our sequence are coprime).

Something like that :).

The problem is Vietnam Mathematics Olympiad 2013, P4: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2901397&sid=b22a422d191cb2d7304543ec7a5dffde#p2901397 (just different by used words)

Those are just the denominator of the Farey sequence of irreducible fractions between $\frac{0}{1}$ and $\frac{1}{1}$, and there are exactly $ \varphi(n) $ irreducible fractions of denominator $n$.


geometry parallelogram perimeter trigonometry

the diagonals of a parallelogram have a measure of 4 and 6 metres. They cut, forming an angle of 60º. Find the perimetre of the parallelogram.

The peasant's solution:
[hide]Let $AC$ be the shorter diagonal, $BD$ be the longer, and $E$ the point where they meet. Let $\angle BEC = \angle AED = 60^{\circ}$. Draw $F$ on $BD$ so that $\angle EFC = 90^{\circ}$; $EF = CEcos60^{\circ} = 1$ and $CF = CEsin60^{\circ} = \sqrt{3}$.
Clearly, then, $DF = 4$ and $FB = 2$.

$CD = \sqrt{(CF)^2+(DF)^2} = \sqrt{3 + 16} = \sqrt{19}$
$CB = \sqrt{(CF)^2+(BF)^2} = \sqrt{3 + 4} = \sqrt{7}$
Therefore, the perimeter is $2(CD+CB) = \boxed{2\sqrt{19} + 2\sqrt{7}}$.[/hide]

And I just thought of the Cosine formula...
[hide]$CD = \sqrt{3^2 + 2^2 + 2(3)(2)cos60^{\circ}} = \sqrt{13 + 12(0.5)} = \sqrt{19}$
$CB = \sqrt{3^2 + 2^2 + 2(3)(2)cos120^{\circ}} = \sqrt{13 + 12(-0.5)} = \sqrt{7}$
Therefore, the perimeter is $2(CD+CB) = \boxed{2\sqrt{19} + 2\sqrt{7}}$.[/hide]

[hide]Remembering that a parallelogram's diagonals bisect each other, four triangles are made with $2$ and $3$ as their sides, 2 of which with a $60\circ$ angle, the others with a $120\circ$ angle. So the smaller and larger sides of the parallelogram, respectively, are:

$a^2=2^2+3^2-2(2)(3)(cos60)$, $a=\sqrt{7}$
$b^2=2^2+3^2-2(2)(3)(cos120)$, $b=\sqrt{19}$.

The perimeter is then $2a+2b=\boxed{2\sqrt{7}+2\sqrt{19}}$.[/hide]


abstract algebra
superior algebra
superior algebra unsolved

suppose $ R$ is a commutative ring with unity .
the R-module  $ M$ is finite generated .and  the homomorphism $ f: M \longrightarrow M$ is onto.
prove $ f$ is one-one.

[hide="Hint"]Consider $ M$ as $ R[X]$-module, where $ Xm \equal{} f(m)$ for $ m\in M,$ and apply Nakayama Lemma.[/hide]


[hide][quote="arkmastermind"][hide="hint/answer"]Notice:
\[ a^4+b^4+c^4 = \frac{(a+b+c)^4 - 6(a+b+c)^2(a^2+b^2+c^2) + 8(a+b+c)(a^3+b^3+c^3) + 3(a^2+b^2+c^2)^2}{6}  \]

Substituting, we obtain: $\boxed{69}$

...unless my arithmetic's off.[/hide][/quote]

How do you know that hint ?[/hide]

eh, i was being vague. use newton's sums to try and create a third degree polynomial that satisfies the above conditions for its roots...then extend newtons sums to find $a^4+b^4+c^4$.

yeah, i didnt pull that expression out of nowhere lol  :)


logarithms
calculus
derivative
integration
geometric series
algebra
binomial theorem

I'm doing a short presentation for my math elective class on infinite series/geometric series.  (for those of you who went to the Julia Robinson Math Festival, I'm using the Little Boxes problem as my main problem)

Any simple questions/tips under this topic would be appreciated.  Not too hard, because the class ranges from kids in Pre-Algebra to Geometry.

After introducing simple geometric series, you may present this:
$1+2+4+8+... =-1$

This is one of my favourites, because 1111... is the infinite precision binary representation of -1 !

Also, if you plan to talk about convergence, introduce this fallacy:

$\ln 2 = 1-\frac12+\frac13-\frac14+... = 1+\frac12+\frac13+\frac14-2\left(\frac12+\frac14+\frac16+\frac18+...\right) = \left(1+\frac12+\frac13+\frac14+...\right)-\left(1+\frac12+\frac13+\frac14+...\right) = 0$

to illustrate why non absolutely convergence series can be dangerous and why you should never change the order.

It's important to establish that for a geometric series
a + ar +....+ ar^n = a(1-r^(n+1))/(1-r)
the sum as n goes to infinity will be a/(1-r) only as long as l r l < 1
as the r^(n+1) term then vanishes.
It is interesting that this formula for ANY r can also be valid under certain special circumstances.
Simple trigonometric series are of great interest.
Also a final example to illustrate the power of the method of differences,
consider,
f(n) = n^r .(n+1)^r - n^r .(n-1)^r -----(1)

f(1) + f(2) +...+f(n) = n^r .(n+1)^r

and the binomial theorem gives,
f(n) /2 = rC1.n^(2r-1) + rC3.n^(2r-3)+....

putting (n-1),(n-2),...,2,1 in succession for n and adding we have,
n^r.(n+1)^r /2 = rC1 S(2r-1) + rC3 S(2r-3) +...

where S(2r -k) is a sum of odd powers of integers.
substitute a suitable r and get S1,S2,etc.
Now find a similar relation for sums of even powers of integers, all you have to do is change a negative sign in (1)

It's quite algorithmic but it has a great aesthetic appeal to it...

I realised that for even integers you need to consider a derivative too.

You should also talk about the Taylor Series: $f(x)=\sum_{n=0}^\infty{\frac{f^{(n)}(c)}{n!}(x-c)^{n}}$

which is really nice and can give you things like

$e^{x}=\sum_{n=0}^\infty\frac{x^{n}}{n!}$

and

$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}\cdot x^{n}}n$

and many others

[quote="star99"]Not too hard, because the class ranges from kids in [b]Pre-Algebra to Geometry.[/b][/quote]

[quote="nat mc"]You should also talk about the [b]Taylor Series[/b][/quote]

 :huh:

Personally, I would talk about really nice intuitive ways of thinking about geometric series.  Here is a great example from my Differential Equations teacher:

You want to distribute some item of size $1$ to a group of $n$ people.  Rather than giving each $\frac{1}{n}$, you give each of them $\frac{1}{n+1}$ so that there is $\frac{1}{n+1}$ left over.

You then divide the leftovers in the same way; give each person $\frac{1}{(n+1)^{2}}$ so that there is $\frac{1}{(n+1)^{2}}$ left over.

And so forth.

In the end, each person will have

$\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+... = \frac{1}{n}$

Because there will be nothing left over!  This is even easier in table format; in each row you list the amount each person has and then the amount left over, and each new row is a new step.  Since the sum of the entries in each row must be $1$, the conclusion follows.

This is even easier to understand intuitively if $n$ is replaced by a concrete number.  My teacher used $n = 3$, but there are certain appeals to the cases $n = 1, 2$.

With regard to Taylor series, I don't think it would be appropriate to present something which you can't prove to the audience yet.
However if you manage to prove Cauchy's integral formula to them by all means go ahead and prove Taylor's and Laurent's theorems. All you need to do is, incidentally, use a geometric expansion! :rotfl:

The binomial theorem though (a special case of Taylor's) might be worth mentioning. Lots of fun with the combinations and all.
Plus a proof can be given with simple induction. Range of validity is again important.
If the method of differences doesn't appeal to you, I'd go with the intuitive, 'sans words' proofs for series like,
$\sum_{r=1}^{n}r^{2}$
that appear on the AoPS sidebar.

t0rajir0u (hope I spelled that right) gave me an idea, the n=1 case presents a simple solution to Zeno's paradox, the one that speaks of Achilles and the Tortoise if memory serves.

Recurrence relations are good fun, solving simple types of course. For example see the nCr series thread, but that particular example might be too complicated!
and putting x=1 in the binomial expansion gives,
$\sum_{r=0}^{n}nCr = 2^{n}$


algebra
polynomial
function
algebra unsolved

Find all polynomials $ P(x)\in \mathbb{R}[x]$ such that for all $ n\in \mathbb{N}$

$ f(P(\frac{1}{2}))\plus{}f(P(1\plus{}\frac{1}{2}))\plus{}...\plus{}f(P(n\plus{}\frac{1}{2}))\equal{}P(1)\plus{}P(2)\plus{}...\plus{}P(n)$,

where $ f$ denotes the floor function.

Let $ X(n)\equal{}[P(\frac 12)]\plus{}...\plus{}[P(n\plus{}\frac 12 )], Y(n)\equal{}P(1)\plus{}...\plus{}P(n)$.
From $ X(n)\equal{}Y(n),X(n\minus{}1)\equal{}Y(n\minus{}1)$ we get $ [P(n\plus{}\frac 12)]\equal{}P(n) \ \forall n>1$ and $ [P(\frac 12)]\plus{}[P(\frac 32)]\equal{}P(1).$
Therefore $ P(x)\equal{}ax\plus{}b$ and a,b integer (because $ P(n)$ integer for all n>1.
It give $ [P(n\plus{}\frac 12)]\equal{}an\plus{}b\plus{}[\frac{a}{2}]\equal{}an\plus{}b\to a\equal{}0 \ or \ a\equal{}1$. From $ [\frac a2 \plus{}b]\plus{}[\frac{3a}{2}\plus{}b]\equal{}a\plus{}b$ we get
1. $ a\equal{}0, b\equal{}0$ or
2. $ a\equal{}1, b\equal{}\minus{}1$.

[quote="Rust"]
Therefore $ P(x) \equal{} ax \plus{} b$ and a,b integer (because $ P(n)$ integer for all n>1.
[/quote]

Why is this?


geometry
analytic geometry
graphing lines
slope
trigonometry
ratio
angle bisector

A, B, and C are the points (1,2), (0,0), and (-1, 3) respectively. What is the slope of the bisector of angle ABC?

[hide="General approach"]Find $D$ on $AC$ such that $AD: DC=AB: BC$. Then $BD$ is the bisector.[/hide]

[hide="Particular approach"]Note that $\triangle ABC$ is an isosceles right, hence ${1\over 2}\angle ABC=22.5^\circ$. Use Tangent of a Sum formula to find the bisector's slope knowing the slope of $BA$.[/hide]

[hide="Result"]$7+5\sqrt{2}$[/hide]

How did you get the result of 8 + 5(2^(1/2)).
When I solved it, I used the half angle formula for tangent and got the result of 2-2^(1/2) / 2^(1/2).

Thank you.

You got $\frac{2-\sqrt{2}}{\sqrt{2}}$, right? First, you can simplify it to $\sqrt{2}-1$ (after you extract the common factor $\sqrt{2}$ from the numerator), and second, that's just $\tan 22.5^\circ$, but you need the tangent of $\angle DBx$, which is a sum of said angle and $\angle ABx$. That's where the tangent of a sum enters the picture.

[hide="General Solution"] $AC$ must be split in a ratio of $\sqrt2$ to $1$. We will try to find the x-coordinate of the point $D$, intersection of the angle bisector and $AC$. Since the horizontal length of $AC$ is $2$, $\frac{2}{\sqrt2+1}=\frac{x}{\sqrt2}\Longrightarrow x=4-2\sqrt2$, where $x=CD$. So, the x-coordinate of $D$ is $-1+4-2\sqrt2=\boxed{3-2\sqrt2}$.

The equation of line $AC$ is $y=-\frac{1}{2}x+\frac{5}{2}$, so the y-coordinate of $D$ is $\boxed{1+\sqrt2}$.

$\frac{1+\sqrt2}{3-2\sqrt2}=\boxed{\boxed{8+5\sqrt2}}$[/hide]

There is a mistake in your arithmetic lotrgreengrapes7926.

[1 + 2^(1/2)] / [3 - 2(2^(1/2)] = 7 + 5(2^1/2)), an answer that does not agree with Farenhajt's. :|

And why do you have to subtract one to get the x-coordinate of D?[/code]

Actually, both lotrgreengrapes7926 and myself have made a calculation error  :maybe: . Your result is correct, maestroGSW, and I have edited my post. My apologies.

[quote="maestroGSW"]There is a mistake in your arithmetic [b]lotrgreengrapes7926[/b] : $\frac{1+\sqrt 2}{3-2\sqrt 2}= 7+5\sqrt 2$, an answer that does not agree with [b]Farenhajt[/b]'s.[/quote] \[\underline{\mathrm{Two\ presents.}}\] [color=darkblue]$1.\blacktriangleright$ [b]Lemma ( [u]prove and/or memorize ![/u] ).[/b] Let $d_{1},d_{2}$ be two lines with the equations $D_{1}(x,y)\equiv a_{1}x+b_{1}y+c_{1}=0$ and $D_{2}(x,y)\equiv a_{2}x+b_{2}y+c_{2}=0$ respectively, so that $a_{1}a_{2}+b_{1}b_{2}\ne 0$, i.e. $d_{1}\not\perp d_{2}$. Then [b]the equation of the bisector of the [u]acute angle[/u] between the lines $d_{1}$ and $d_{2}$[/b] is \[\boxed{\ \frac{D_{1}(x,y)}{\sqrt{a^{2}_{1}+b^{2}_{1}}}+\mathrm{sign\ }(a_{1}a_{2}+b_{1}b_{2})\cdot\frac{D_{2}(x,y)}{\sqrt{a^{2}_{2}+b^{2}_{2}}}=0\ }\ .\] [b]Remark.[/b] In the our case, the equation of the line $BA$ is $2x-y=0$ and the equation of the line $BC$ is $3x+y=0$. Prove easily that the angle $\widehat{ABC}$ is acute. Thus, the equation of the bisector $b$ of the acute angle between the lines $BA$ and $BC$ is $\frac{2x-y}{\sqrt 5}+\frac{3x+y}{\sqrt 10}=0$, i.e. $(\sqrt 2-1)y=(2\sqrt 2+3)x\ ,$ i.e. the slope of the line $b$ is $\boxed{\ s(b)=7+5\sqrt 2\ }\ .$

$2.\blacktriangleright$ [b]Exercise ( [u]prove and/or memorize ![/u] ).[/b] Given are three nonconcurrent lines $d_{k}\ ,\ k\in \overline{1,3}$ with the equations $D_{k}(x,y)=a_{k}x+b_{k}y+c_{k}=0$ respectively,
i.e. $|\begin{array}{ccc}a_{1}& b_{1}& c_{1}\\\ a_{2}& b_{2}& c_{2}\\\ a_{3}& b_{3}& c_{3}\end{array}|\ne 0\ .$ Consider the triangle $A_{1}A_{2}A_{3}$, where $A_{1}\in d_{2}\cap d_{3}$, $A_{2}\in d_{3}\cap d_{1}$, $A_{3}\in d_{1}\cap d_{2}\ .$

Ascertain the equation of the (interior or exterior) bisector of the angle with the vertex $A_{k}$, where $k\in \overline{1,3}\ .$

[b]Answer.[/b] The equation of the (interior) bisector of the angle $\widehat{A_{i}A_{k}A_{j}}$, where $k\not\in \{i,j\}$ and $i=\ne j$ is \[\boxed{\ \mathrm{sign\ }|\begin{array}{cc}a_{i}& b_{i}\\ \ a_{k}& b_{k}\end{array}|\cdot \frac{D_{i}(x,y)}{\sqrt{a^{2}_{i}+b^{2}_{i}}}+\mathrm{sign\ }|\begin{array}{cc}a_{j}& b_{j}\\ \ a_{k}& b_{k}\end{array}|\cdot\frac{D_{j}(x,y)}{\sqrt{a^{2}_{j}+b^{2}_{j}}}=0\ }\ .\] [b]Remark.[/b] In the our case, the equations of the lines  $AB\ ,\ BC\ ,\ CA$ are $2x-y=0\ ,\ 3x+y=0\ ,\ x+2y-5=0$ respectively and the equation

of the (interior) bisector of the angle $\widehat{ABC}$ is $\mathrm{sign\ }|\begin{array}{cc}2 &-1\\\ 1 & 2\end{array}|\cdot\frac{2x-y}{\sqrt 5}+\mathrm{sign\ }|\begin{array}{cc}3 & 1\\\ 1 & 2\end{array}|\cdot\frac{3x+y}{\sqrt{10}}=0$, i.e. $(2x-y)\sqrt 2+(3x+y)=0$ a.s.o.

[b]Notation.[/b] $\mathrm{sign\ (x)}=\{\begin{array}{ccc}-1 & \mathrm{if}& x<0\\\ 0 & \mathrm{if}& x=0\\\ 1 & \mathrm{if}& x>0\end{array}\ .$[/color]


This is a problem I made up some time ago and i don't know how to solve.

Let $ ABC$ be a triangle. Then pick any point $ O$ in a triangle.

Expres the angle $ \angle BOC$ with the following anglels:

$ \angle CAB$
$ \angle CBA$
$ \angle OAB$
$ \angle OBA$

Is it possible to do it (without trigonometric functions)?

I believe you need trig Ceva, it can't be done without it.
We can easily find $ \angle ACB$, we know $ \angle ACO$ and $ \angle OCB$ add up to it, and using trig Ceva we know the ratio of their sines. So we solve for them and then $ \angle BOC$.


During a college interview, many times you are asked this kindn of question:

"Many qualified students apply to this college. What characteristics single you out from the others?"

How exactly would you answer this question? Should you talk about your accomplishments, personality, etc? I was wondering exactly what someone would say without comming off as arrogant if you know what I mean...

Answer it truthfully... :maybe:

Don't try to be something you're not.  It's likely the interviewer will pick up on it.

If the school is your top pick, dream school I would say something like 'I don't know about everyone else applying, but this is the school I've dreamed about going to.  I feel that going to this school would give me the opportunity to thrive in a place filled with students with similar goals and pursuits as myself, professors willing and knowledgeable enough to guide me along my college journey, and the inspiration to accomplish great things.'  or something of that sort.

It doesn't necessarily have to be something that most people would consider "great" or an "accomplishment." Just say something that makes you unique and interesting. Mention a non-academic hobby, maybe. You could always answer with something silly to buy some time to think of a serious answer, if you think your interviewer has a good sense of humor.

[quote="Elemennop"]joml's answer is actually pretty cool

I'd just be like, "I'm really aloof" or something along the sorts.[/quote]

yea i'm trying to memorize that answer word for word :wink: jk...well, no not really....

Oh geez, that question's always super-easy for me to answer. "I'm a math geek, but I want to be a vet. I like science, but I like working in the theatre too. In my spare time, I like to take photographs and do stupid tricks on my bike. Each of these things by themselves may not seem that unique, but together, they make for a pretty unaverage person."

[quote="joml88"]If the school is your top pick, dream school I would say something like 'I don't know about everyone else applying, but this is the school I've dreamed about going to.  I feel that going to this school would give me the opportunity to thrive in a place filled with students with similar goals and pursuits as myself, professors willing and knowledgeable enough to guide me along my college journey, and the inspiration to accomplish great things.'  or something of that sort.[/quote]

Generic and boring?

I would try to say something more like bluemoose so that the interviewers think I actually have character; joml's idea could probably be a nice conclusion.

Sorry - can't resist it - If explicitly asked this question, i'd just smile and say..
[quote]Oh I am special, very special, in fact unique, ..
[hide]...just like everybody else :) [/hide][/quote]

[quote="blahblahblah"]Generic and boring?[/quote]

No no, I didn't say anything about Canada.

A spoof answer if you DON'T want to get in is,"I'm schizophrenic!"

The truth is if you don't have an answer say,"I'm hard working!"

But if you worked all your life for an answer say, "I'm an athlete and a math nerd!" (but make sure you trying to get into stanford....JK)

So my answer is talk about your best thing(usually personality.)
If you've had any great accomplishments you should REALLY list these.

[quote="blahblahblah"]A comment about where I live would be more funny if you didn't live in a state ending in 'Carolina' (the worse one, at that!).

just kidding, just kidding :)[/quote]

That's actually historically inaccurate.  South Carolina is not a state in the US.

And actually I don't really live in South Carolina anymore...so I'll go change that :P

[hide="Edit"]I'm jk (in this post and the last), of course :D  ...just making sure I don't totally offend anyone... :maybe: [/hide]

[quote="bleumoose"][quote="K81o7"]start reciting digits of pi[/quote]

yeah, because [i]that's[/i] really sure to impress them![/i][/quote]

start reciting random digits and say that they are digits of pi...

prove that you're unique by assuming that there are two distinct verions of you, reaching a contradiction, and then concluding that you are unique

haha, just kidding. i would mention a really good personality trait, like i have a very positive outlook on the future and i'm willing to work to achieve my goals.

drunner2007 wins, hands down.  Although to take it home, you'd have to tell us what your clever proof is.  It also works better if the inteviewer specifically uses the word "unique."

A similar type of response would be list enough trivial characteristics that are likely to statistically distinguish you from every other person on earth, because frankly, the question is stupid, and I would never ask it.  The only point of it would be to figure out how good the student is at interviewing.  Or maybe I'm just still sore because I was unprepared for that question when someone asked me.  ;)

[quote="yenlee"]drunner2007 wins, hands down.  Although to take it home, you'd have to tell us what your clever proof is.  It also works better if the inteviewer specifically uses the word "unique."
[/quote]

yes...
if there is someone like me then he will have to visit the same places the same time with me so I would have seen him/her. Well, I didn't. QED

this is really silly but maybe(?) clever

Proof: Suppose that there are two distinct versions of me, Mark A and Mark B. By distinct we mean that Mark A and Mark B are different physical beings, but they have done exactly the same things throughout life. However, a particular friend knows only one Mark, contradicting the fact that both Mark A and Mark B are friends of that friend (since everything they have done is identical). So Mark A and Mark B are the same individual.

[quote="drunner2007"] a particular friend knows only one Mark[/quote]

how do you know that a friend with this attribute exists?

[quote="Pocket Sand"]How can one be distinct, and at the same time have done everything the other has? I believe that itself is a contradiction.[/quote]

i guess by considering them as distinct physical entities. i know that would imply that they coincide, but I guess thoeretically that doesn't matter(?)
not sure. maybe prove that its impossible to give a valid proof of that. or maybe not.


calculus
integration
function
logarithms
calculus computations

(1) A function is defined $ f(x) \equal{} \ln (x \plus{} \sqrt {1 \plus{} x^2})$ for $ x\geq 0$. Find $ f'(x)$.

(2)  Find the arc length of the part $ 0\leq \theta \leq \pi$ for the curve defined by the polar equation: $ r \equal{} \theta\ (\theta \geq 0)$.

Remark:

[color=blue]You may not directly use the integral formula of[/color] $ \frac {1}{\sqrt {1 \plus{} x^2}},\ \sqrt{1 \plus{} x^2}$ here.

hello, using the chaine rule we get
$f'(x)=\frac{1}{1+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)=\frac{1}{\sqrt{1+x^2}}.$
Sonnhard.

The use of the chain rule was perfect, however the mistake crept in at the denominator when x was forgotten, thus the solution was supposed tho start the given part but now the denominator need to contain x instead of the given 1. Hence the solution will change as well...

I think Sonnhard might have made a small typo in the first fraction. The result is the same, however.
\[\frac{1}{x + \sqrt{1 + x^2}}\times\left(1 + \frac{x}{\sqrt{1 + x^2}}\right) = \frac{1}{\sqrt{1 + x^2}},\] which is the same as the answer above.


[hide="A doubt about B"]Using my lemma [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250283]here[/url](post#3) (Sorry if it has been discovered before  :blush: ) we have that Your 2nd inequality is true. But since the quantity in the LHS must be real, so $ x^2 \plus{} y^2 \plus{} kyx\geq 0\Leftarrow k\geq \minus{} 2$ is a necessary condition.
Please rectify me if I am wrong. :( [/hide]


Let $S$ be the set of all triangles $ABC$ for which \[ 5 \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},  \] where $r$ is the inradius and $P, Q, R$ are the points of tangency of the incircle with sides $AB, BC, CA,$ respectively. Prove that all triangles in $S$ are isosceles  and similar to one another.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

this problem is very nice, my solution is as follows.

let $x,y,z$ be $AP,BQ,CR$ respectively and suppose wlog $x\geq y\geq z$.
now we have that $a=y+z, b=x+z, c=y+z$, so we have $x=s-a, y=s-b, z=s-c$.

from the area formulas we have that
$\displaystyle r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt\frac{xyz}{x+y+z}$ or
$6\sqrt{\frac{x+y+z}{xyz}}=\frac{6}{r}$

returning to the problem we have that $\displaystyle\frac{5}{x}+\frac{5}{y}+\frac{2}{z}=6\sqrt{\frac{x+y+z}{xyz}}$, squaring both sides and multiplying by $(xyz)^2$ gives $4x^2y^2+25y^2z^2+25x^2z^2-16xy^2z-16x^2yz+14xyz^2=0$.

we may see the equation above as a quadratic in $z$.
so it turns into $z^2(25x^2+25y^2+14xy)-z(16xy(x+y))+4x^2y^2=0$, using the fact that the minimum of a cuadratic is $\displaystyle\frac{-\Delta}{4a}$, where $\Delta$ is the discriminant of the equation we have that $\displaystyle 0=z^2(25x^2+25y^2+14xy)-z(16xy(x+y))+4x^2y^2\geq\frac{-(256x^2y^2(x+y)^2-4(25x^2+25y^2+14xy)(4x^2y^2))}{4(25x^2+25y^2+14xy)}=\frac{36(x^2y-xy^2)^2}{25x^2+25y^2+14xy}\geq0$

$\Rightarrow 36(x^2y-xy^2)^2=0 \Rightarrow x=y$.

we've proved that all of them are isosceles. now, substituting $x=y$ in the equation above gives $64x^2z^2-32x^3z+4x^4=0 \Rightarrow 4x^2(x-4z)^2=0 \Rightarrow x=y=4z$

$\Rightarrow a=y+z=4k+k=5k, b=x+z=4k+k=5k, c=x+y=4k+4k=8k$

$\Rightarrow (a,b,c)=(5k,5k,8k)$
and finally we're done.

Possibly faster finish from

$ 4x^{2}y^{2}\plus{}25y^{2}z^{2}\plus{}25x^{2}\minus{}16xy^{2}z\minus{}16x^{2}yz\plus{}14xyz^{2}\equal{}0$. We will show that the left hand side is $ \ge0$. Let $ a\equal{}xz,b\equal{}xy,c\equal{}yz$. It turns into $ 4b^{2}\plus{}25c^{2}\plus{}25a^{2}\minus{}16bc\minus{}16ab\plus{}14ac\ge0$. Consider smoothing $ a$ and $ c$. Say that $ b$ and $ a\plus{}c$ are fixed, and if $ a\neq c$, consider replacing $ a$ and $ c$ with $ \dfrac{a\plus{}c}{2},\dfrac{a\plus{}c}{2}$. We don't care about the $ 4b^{2}\minus{}16bc\minus{}16ab\equal{}4b^{2}\minus{}16b(a\plus{}c)$ because this is fixed. Note that $ 25c^{2}\plus{}25a^{2}\plus{}14ac\equal{}25(a\plus{}c)^{2}\minus{}64ac$, so when we move $ a$ and $ c$ together so they are equal, $ ac$ goes up, and the whole sum goes down (strictly). Thus we only need to consider $ a\equal{}c$, in which the inequality turns into $ (b\minus{}4c)^{2}\ge0$. Thus, the LHS is at least 0, with equality if and only if $ a\equal{}c,b\equal{}4c$. The result follows.

This is equivalent yet less ugly, even though it uses Lagrange multipliers. There's probably a simple noncalculus way to do the inequality; I'm just not good enough at inequalities.

Let $ A$, $ B$, and $ C$ be the angles of the triangle; let $ a \equal{} \tan\frac {A}{2} \equal{} \frac {r}{AP}$, $ b \equal{} \tan\frac {B}{2} \equal{} \frac {r}{BQ}$, and $ c \equal{} \tan\frac {C}{2} \equal{} \frac {r}{CR}$. If we WLOG let $ A\le B\le C$ and therefore $ a\le b\le c$, $ AP\ge BQ\ge CR$. Our condition for $ S$ is now $ 5a \plus{} 5b \plus{} 2c \equal{} 6$. $ \frac {A}{2} \plus{} \frac {B}{2} \plus{} \frac {C}{2} \equal{} \frac {\pi}{2}$, so $ \Re((1 \plus{} ai)(1 \plus{} bi)(1 \plus{} ci)) \equal{} 0$; this is equivalent to $ ab \plus{} bc \plus{} ca \equal{} 1$. We wish to minimize $ f(a,b,c) \equal{} 5a \plus{} 5b \plus{} 2c$ under the constraint $ ab \plus{} bc \plus{} ca \equal{} 1$. This gives $ F(a,b,c,\lambda) \equal{} 5a \plus{} 5b \plus{} 2c \minus{} \lambda(ab \plus{} bc \plus{} ca \minus{} 1)$.
\[ \frac {\partial F}{\partial a} \equal{} 5 \minus{} \lambda(b \plus{} c) \equal{} 0 \\
\frac {\partial F}{\partial b} \equal{} 5 \minus{} \lambda(a \plus{} c) \equal{} 0 \\
\frac {\partial F}{\partial c} \equal{} 2 \minus{} \lambda(a \plus{} b) \equal{} 0
\]So $ a \equal{} b$ at the minimum by the first two and by the third $ c \equal{} 4a$. Substituting into the constraint, we have $ a^2 \plus{} 4a^2 \plus{} 4a^2 \equal{} 1$, or $ a \equal{} \frac13$. This leaves $ b \equal{} \frac13$ and $ c \equal{} \frac43$ with a minimum of $ 6$ for $ 5a \plus{} 5b \plus{} 2c$. Since $ \frac {A}{2}$ ranges from $ 0$ to $ \frac {\pi}{2}$, its tangent uniquely determines it. Because $ a$ is fixed across all triangles in $ S$, as there is only one $ a$ at which the minimum occurs, $ A$ is fixed as well, as are $ B$ and $ C$. All the triangles are thus similar. Since $ A \equal{} B$, they are all isosceles too.

[hide="Solution"]
Without loss of generality, let $ AP \le BQ \le CR,$ and let $ \tan \frac {A}{2} = a, \tan \frac {B}{2} = b,$ and $ \tan \frac {C}{2} = c$.  Then from the given equation, we have
\begin{eqnarray*} 5 \left(\frac {1}{AP} + \frac {1}{BQ} + \frac {1}{CR} \right) - \frac {3}{AP}} & = & \frac {6}{r} \\
2 \cdot \frac {r}{AP} + 5 \cdot \frac {r}{BQ} + 5 \cdot \frac {r}{CR} & = & 6. \end{eqnarray*}
Because triangles $ IAP$, $ IBQ$, and $ ICR$ are right triangles, we have $ \frac {r}{AP} = \frac {IP}{AP} = \tan \angle PAI = \tan \frac {A}{2} = a$.  Similarly, $ \frac {r}{BQ} = b$ and $ \frac {r}{CR} = c$.  Thus,
\[ 2a + 5b + 5c = 6.\]
Since $ \tan \theta \tan(\frac {\pi}{2} - \theta) = 1$ for $ 0 < \theta < \frac {\pi}{2}$, we have
\[ \tan \frac {A}{2} = \frac {1}{\tan(\frac {B}{2} + \frac {C}{2})} = \frac {1 - \tan \frac {B}{2} \tan \frac {C}{2}}{\tan \frac {B}{2} + \tan \frac {C}{2}}.\]
Thus, $ a = \frac {1 - bc}{b + c}$.

Substituting this in to the original equation, we obtain
\[ 2 \cdot \frac {1 - bc}{b + c} + 5b + 5c = 6,\]
which can be rearranged as
\[ (2b + c - 1)^2 + (b + 2c - 1)^2 = 0.\]
As a sum of squares can only be equal to zero when each square is zero, we must have $ b = c = \frac {1}{3}$.
This uniquely determines the angles $ B$ and $ C$.
Thus, all triangles whose lengths satisfy the original equation are isosceles and are similar to one another.[/hide]

Isn't there a solution with a more geometric idea ?

I don't think there's a solution with a more geometrical idea, because of the super-contrived condition making it clear that one must find an equality case in an inequality. Similar to 2003 USAMO 2.

[hide=solution]
We start with a well-known trig identity.

Lemma: For triangle $ABC$, $\sum_{cyc}\tan\frac{A}{2}\tan\frac{B}{2}=1$.
Proof: By tangent sum identity,
\[ 1-\tan\frac{A}{2}\frac{B}{2}=\frac{\tan\frac{A}{2}+\tan\frac{B}{2}}{\tan(\frac{A}{2}+\frac{B}{2})} \]
Obviously $\tan(\frac{A}{2}+\frac{B}{2})=\tan(90-\frac{C}{2})=\cot\frac{C}{2}$, so
\[ \tan\frac{A}{2}\tan\frac{B}{2}=1-\tan\frac{C}{2}(\tan\frac{A}{2}+\tan\frac{B}{2}) \]
which easily rearranges to the desired.

$AP=\frac{r}{\tan\frac{A}{2}}$ and similarly for $BQ, CR$. WLOG, $\min\{\tan\frac{A}{2}, \tan\frac{B}{2}, \tan\frac{C}{2}\}=\tan\frac{A}{2}$.
Then, the hypothesis becomes $6=2\tan\frac{A}{2}+5\tan\frac{B}{2}+5\tan\frac{C}{2}$. For simplicity, $\tan\frac{A}{2}=x, \tan\frac{B}{2}=y, \tan\frac{C}{2}=z$, so we have $5y+5z+2x=6$ and $xy+yz+zx=1$ from our Lemma.

The first equation, after replacing $x$, becomes $5y^2+y(8z-6)+(5z^2-6z+2)=0$ when viewing as a quadratic in $y$. $\triangle ABC$ exists so
\[ 4(4z-3)^2-20(5z^2-6z+2)\ge0 \implies 9z^2-6z+1\le0 \]
In light of this we have $z=\frac{1}{3}$ and symmetrically $y=\frac{1}{3}$. We can proceed to find $x=\frac{4}{3}$.

This means that $B=C$ so $\triangle ABC$ is isosceles, and all angles $A, B, C$ of $\triangle ABC$ are fixed so all such triangles are similar to each other, as requested.
[/hide]

We will prove the inequality \[ \frac{2}{AP} + \frac{5}{BQ} + \frac{5}{CR} \ge \frac 6r \] with equality when $AP : BQ : CR = 1 : 4 : 4$. This implies the problem statement.

Letting $x= AP$, $y = BQ$, $z = CR$, the inequality becomes \[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \] Squaring both sides and collecting terms gives \[ 	\frac{4}{x^2} + \frac{25}{y^2} + \frac{25}{z^2} + \frac{14}{yz} 	\ge \frac{16}{xy} + \frac{16}{xz}. \] If we replace $x=1/a$, $y=4/b$, $z=4/c$, then it remains to prove the inequality \[ 64a^2 + 25(b+c)^2 \ge 64a(b+c) + 36bc \] where equality holds when $a=b=c$. This follows by two applications of AM-GM: \begin{align*} 	16 \left( 4a^2 + (b+c)^2 \right) &\ge 64a(b+c) \\ 	9(b+c)^2 &\ge 36bc. \end{align*}

[quote=mssmath]Upon trying is problem it can also finished with $ 4b^{2}+25c^{2}+25a^{2}-16bc-16ab+14ac=7(a+b-\frac{c}{2})^2+18(a-\frac{b}{4})^2+18(c-\frac{b}{4})^2\ge0 $ after CatalystOfNostalgia's substitution.[/quote]
Sorry for the revive, but I don't believe this works. However, $4a^2 + 25b^2 + 25c^2 -16ab-16ac + 14bc = 4(a-2b-2c)^2 + 9(b-c)^2 \geq 0$ does.

[hide=My Solution]Let $BC=x+y,AB=x+z,AC=y+z$, so $AP=z, BQ =x, CR = y$. WLOG $x\leq y \leq z$. So, $ \frac{2}{x} + \frac{5}{y} + \frac{5}{z} = \frac{6}{r}$. Now, the area of the triangle is $(x+y+z)r$ and $\sqrt {(x+y+z)xyz}$, thus $r = \sqrt{\frac{xyz}{x+y+z}}$, and $\frac{2}{x} + \frac{5}{y} + \frac{5}{z} = 6\sqrt{\frac{x+y+z}{xyz}}$. Now, I claim $\frac{2}{x} + \frac{5}{y} + \frac{5}{z} \geq 6\sqrt{\frac{x+y+z}{xyz}}$.
Proof: Let $x= \frac{1}{a}, y = \frac{1}{b}, z= \frac{1}{c}$, so our inequality becomes $2a+5b+5c \geq 6\sqrt{ab+bc+ca}$. Squaring both sides and moving to one side, this becomes $4a^2 + 25b^2 + 25c^2 -16ab-16ac + 14bc \geq 0$. However, note the LHS can be written as $4(a-2b-2c)^2 + 9(b-c)^2 \geq 0$, so it is [i]trivially[/i] true.
Equality, which is what we want, occurs only when $b=c$ and $a=2b+2c = 4b = 4c$, so $a:b:c = 4:1:1$, or $x:y:z = 1:4:4$, so any triangle with side lengths in a ratio $5:5:8$ works and only these triangles do, hence the claim. $\blacksquare$. [/hide]

I've been trying to cauchy this for a long time due to the trig problem in USAMO 2002, but apparently quadratic formula trivializes it:

Making the substitutions that campos and v_Enhance did leads to the equation:

$$4a^2 + 25b^2 + 25c^2 -16ab - 16ac +14bc = 0$$

Applying the quadratic formula with respect to a gives:

$$ a = \frac{16b + 16c + \sqrt{(16b + 16c)^2 - 16(25b^2 + 25c^2 + 14 bc)}}{8} = \frac{16b + 16c + \sqrt{-(3b-3c)^2}}{8} = 2b + 2c $$

Subtracting $$4(a-2b-2c)^2$$ from the original equation gives:

$$4a^2 + 25b^2 + 25c^2 -16ab - 16ac +14bc - 4(a-2b-2c)^2 = (3b-3c)^2 = 0$$

So $b = c \rightarrow y = z$.

From this, substituting y = z into
$\[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \]$
Gives 4x = y = z, which gives equality at $AP : BQ : CR = 1 : 4 : 4 \rightarrow AB : BC : AC = 5 : 8 : 5$, as desired.

I am seeing solutions and getting frustrated.....I came till the part...

$\displaystyle\frac{5}{x}+\frac{5}{y}+\frac{2}{z}=6\sqrt{\frac{x+y+z}{xyz}}$

n then, I started believing that MAA wouldn't do something so bad to me as to give me full-length quadratic....

Why am I even alive?

A proof using some (healthy) amount of linear algebra.

Suppose without loss of generality that ${\rm min}(|AP|,|BQ|,|CR|)=|AP|$.
Let the side lengths of this triangle be $m,n,k$ with $u=\frac{m+n+k}{2}$ being the semiperimeter. Then, $|AP|=u-m$, $|BQ|=u-n$, and $|CR|=u-k$ hold. Using now Heron's formula, we obtain
$$
u\cdot r=\sqrt{u(u-m)(u-n)(u-k)}.
$$
Let now $x=u-m$, $y=u-n$, and $z=u-k$. Observe that $u=x+y+z$. With this, we thus obtain
$$
r=\sqrt{\frac{xyz}{x+y+z}}.
$$
Upon taking squares, we now conclude
$$
\frac{4}{x^2}+\frac{25}{y^2}+\frac{25}{z^2}+\frac{14}{xz}=\frac{16}{xy}+\frac{16}{yz}.
$$
Let now $a=x^{-1}$, $b=y^{-1}$, and $c=z^{-1}$. We then obtain that
$$
4a^2+25b^2+25c^2-16ab-16ac+14bc=0.
$$
We now claim
$$
4a^2+25b^2+25c^2-16ab-16ac+14bc\geqslant 0,
$$
with equality iff $a=4b=4c$. This will then yield the desired claim. For this we appeal to a matrix-vector product notation. Note that
$$
4a^2+25b^2+25c^2-16ab-16ac+14bc = \begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}4 & -8 & -8 \\ -8 & 25 & 7 \\ -8 & 7 & 25\end{pmatrix}\begin{pmatrix}a\\ b \\ c\end{pmatrix}\triangleq \xi^T M\xi.
$$
Now, a calculation yields the eigenvalues of $M$ to be $0,18$, and $36$. Since they are all non-negative, $M$ is positive semidefinite; hence $z^TMz\geqslant 0$ for every $z\in\mathbb{R}^3$.

We now study the equality case. For this, we rely on eigendecomposition of the matrix. A further inspection reveals that the eigenvectors $v_2,v_3$ corresponding respectively to the eigenvalues $18$ and $36$ are
$$
v_2 = \begin{pmatrix}0\\-1 \\ 1\end{pmatrix}\quad\text{and}\quad
v_3 = \begin{pmatrix}-1\\2 \\ 2\end{pmatrix}.
$$
Hence, $M=18v_2v_2^T + 36v_3v_3^T$, and thus
$$
\xi^T M\xi = 18(-b+c)^2+36(-a+2b+2c)^2.
$$
In particular, for equality to hold, it must hold that $b=c$; and $-a+2b+2c=0$, that is, $a=4b=4c$, as claimed earlier.

Is this Titu?

Let $BC=a,CA=b,AB=c$ and $s=(a+b+c)/2$. WLOG $a=\max(a,b,c)$. Then the given is
\[\frac{6}{r}=\frac{2}{s-a}+\frac{5}{s-b}+\frac{5}{s-c}.\]
Let $x=s-a,y=s-b,z=s-c$. By Heron's Formula, we have
\[r=\frac{[ABC]}{s}=\sqrt{\frac{xyz}{x+y+z}}.\]
Thus we can write
\[\frac{6\sqrt{x+y+z}}{\sqrt{xyz}}=2/x+5/y+5/z.\]
By scaling we can WLOG $x=1$. Then $6\sqrt{yz+(y+z)yz}=2yz+5y+5z$. Squaring yields
\[36yz+36(y+z)yz=4y^2z^2+20(y+z)yz+50yz+25y^2+25z^2\implies 0=4y^2z^2-16(y+z)yz+14yz+25y^2+25z^2=\]
\[(2yz-4y-4z)^2+(3y-3z)^2.\]
Thus $y=z,2y^2=8y\implies y=z=4$. Thus $a=8,b=c=5$. Thus all possible triangles $ABC$ are similar to the isosceles $5-5-8$ triangle.

It seems that every problem on an olympiad which asks for solutions to a single real-number equation is an inequality.

Here is a solution that does not rely on finding a sum of squares and expanding a large quadratic.
I will first prove the inequality:
$$\frac{2}{AP}+\frac{5}{BQ}+\frac{5}{CR} \geq \frac{6}{r}.$$
We can see that if $s$ is the semiperimeter and $a,b,c$ are the side lengths as usual, we have $AP=s-c,BQ=s-b,CR=s-a$. Also, we have the well-known formula $A=rs \implies \tfrac{1}{r}=\tfrac{s}{A}$, where $A$ is the area. Further, by Heron's we have $A=\sqrt{s(s-a)(s-b)(s-c)}$. Substsituting $p=s-c,q=s-b,r=s-a$ (where $p,q,r \in \mathbb{R^+}$) and noting that $p+q+r=s$, the inequality becomes:
$$\frac{2}{p}+\frac{5}{q}+\frac{5}{r} \geq 6\sqrt{\frac{p+q+r}{pqr}}.$$
Now, making the substitution $\tfrac{1}{p}=x,\tfrac{1}{q}=y,\tfrac{1}{r}=z$, the inequality becomes:
$$2x+5(y+z)\geq 6\sqrt{x(y+z)+yz},$$
where $x,y,z \in \mathbb{R^+}$.
To prove this inequality, we first perform a bit of smoothing. If we keep $y+z$ constant and vary $y,z$ individually, we can see that the LHS remains constant and the RHS attains its maximum value when $y=z$. As such we can prove the stronger inequality (obtained by putting $z=y$):
$$2x+10y\geq 6\sqrt{2xy+y^2} \iff x+5y \geq 3\sqrt{2xy+y^2}.$$
Squaring and simplifying, this is equivalent to:
$$x^2-8xy+16y^2=(x-4y)^2 \geq 0,$$
which is true. Further, equality holds if and only if $x=4y=4z$, so $\tfrac{1}{x}=\tfrac{1}{4y}=\tfrac{1}{4z} \implies 4p=q=r$. Let $p=k$, so $q=r=4k$. Then we have $s=9k$, from which we can recover $c=8k,a=b=5k$. Observe that this inequality and its equality case implies that the equation holds if and only if $\min\{AP,BQ,CR\}=AP$, and $a:b:c=5:5:8$. Since $a:b:c=5:5:8$ implies that $\min\{AP,BQ,CR\}=AP$ always, we obtain that the equation holds when $AB:BC:CA=5:5:8$ and permutations, so all triangles in $S$ are isosceles and similar to each other. $\blacksquare$

ew

Let $x = s - a$, $y = s - b$, $z = s - c$, and we say that WLOG $x$ is the smallest. Then our equality reduces to \begin{align*}
    \frac 2 x + \frac 5 y + \frac 5 z = 6 \frac{\sqrt{x + y + z}}{\sqrt{xyz}}.
\end{align*} Multiplying, we get \begin{align*}
    2 \sqrt{ \frac{yz}{x}} + 5 \sqrt{\frac{xz}{y}} + 5 \sqrt{\frac{xy}{z}} = 6 \sqrt{x + y + z}.
\end{align*} Noting that both sides are homogeneous, we set $x = 1$, which yields \begin{align*}
    6 \sqrt{1 + y + z} &= 2 \sqrt{yz} + 5 \sqrt{\frac y z} + 5 \sqrt{\frac z y} \\
    &= \frac{2yz + 5y + 5z}{\sqrt{yz}} \implies \\
    (2yz + 5y + 5z)^2 &= 36yz(1 + y + z).
\end{align*}

However, observe that \begin{align*}
    (2yz + 5y + 5z)^2 - 36yz(1 + y + z) &= \\
    4y^2z^2 + 25y^2 + 25z^2 + 14yz - 16yz(y + z) &= \\
    (2yz - 4y - 4z)^2 + (3y - 3z)^2 &\geq 0.
\end{align*} This forces $y = z$ as well as $yz = 2y + 2z$, so our solution is $(x, y, z) = (1, 4, 4)$. If we solve for $a, b, c$ and then dehomogenize, we get $(8k, 5k, 5k)$ as our solution set.

WLOG let $AP\leq BQ\leq CR$. Then the condition (after some simplifying) is equivalent to $$2\tan \frac{A}{2}+5\tan \frac{B}{2}+5\tan \frac{C}{2}=6,$$ and since $$\tan \frac{A}{2}=\frac{1}{\tan \left(\frac{B}{2}+\frac{C}{2}\right)}=\frac{1-\tan \tfrac{B}{2}\tan \tfrac{C}{2}}{\tan \tfrac{B}{2}+\tan \tfrac{C}{2}},$$ the condition is now equivalent to $$\frac{2-2\tan \tfrac{B}{2}\tan \tfrac{C}{2}}{\tan \tfrac{B}{2}+\tan \tfrac{C}{2}}+5\tan \frac{B}{2}+5\tan \frac{C}{2}=6.$$

Set $x=3\tan \frac{B}{2}-1$ and $y=3\tan \frac{C}{2}-1$. We may verify that this condition is equivalent to $$5x^2+8xy+5y^2=0.$$ Now suppose WLOG that $y\neq 0$; then we may divide this by $y^2$ to get $$5\left(\frac{x}{y}\right)^2+8\left(\frac{x}{y}\right)+5=0$$ which has no solutions. Thus $x=0,y=0$, which easily works, so that $\tan \frac{B}{2},\tan \frac{C}{2}=\frac{1}{3}$. From here the rest is easy; the isosceles triangle in question has side lengths in the ratio $AB:BC:CA=5:8:5$ (and permutations!).

[hide=Solution]
Let $a=AP$, $b=BQ$, and $c=CR$. Assume without loss of generality $a\leq b\leq c$. Then, the area of $ABC$ is $\sqrt{(a+b+c)abc}$ by Heron's Formula, and since this is equal to $rs=r(a+b+c)$, we have $r=\sqrt{\frac{abc}{a+b+c}}$. Therefore, it suffices to solve the equation
$$\frac2a+\frac{5b+5c}{bc}=6\sqrt{\frac1{bc}+\frac{b+c}{abc}}.$$ Suppose that $b+c=kbc$. Then, since $(b+c)^2\geq4bc$, we have $k^2b^2c^2\geq4bc$, so $bc\geq\frac4{k^2}$. Therefore, we have
\begin{align*}
\frac2a+5k&=6\sqrt{\frac1{bc}+\frac ka}\\
&\leq6\sqrt{\frac{k^2}4+\frac ka}\\
&=3\sqrt{k^2+\frac{4k}a}.\end{align*}
Multiplying by $a$, we have $2+5ak\leq3\sqrt{a^2k^2+4ak}$. Let $x=ak$. Then, $2+5x\leq3\sqrt{x^2+4x}$, so squaring gives $25x^2+20x+4\leq9x^2+36x$, so $16x^2-16x+4\leq0$. This is equivalent to $4(2x-1)^2\leq0$, so $x=\frac12$ and $b=c$. Since $b=c$, all triangles in $S$ are isosceles, and since $x$ is constant, we have $x=ak=\frac{2a}b$ is constant. Therefore, $\frac ab$ is constant, which means that all triangles in $S$ are similar to one another.

This is my solution (pretty similar to ChrisWren's one )

[color=#f00]Lemma:[/color] We know that for a given triangle $\Delta ABC$ if $P,Q,R$ are the tangency points of the incircle with the sides $AB,BC,CA$ respectively,then $AP=s-a,BQ=s-b,CR=s-c$ where $a,b,c$ are side lengths of $AB,BC,CA$ respectively and $s$ is the semi-perimeter of the triangle .
and also $\frac{r}{s-a}=tan(\frac{A}{2}),$ $\frac{r}{s-b}=tan(\frac{B}{2})$,$\frac{r}{s-c}=tan(\frac{C}{2})$

[color=#c00]Claim:[/color][u]If a triangle $\Delta XYZ \in \mathbb{S}$ , then it is an isosceles triangle with sides $(5k,5k,8k) $  (in some order )[/u]

[color=#960000]PROOF:[/color]Consider a triangle $\Delta ABC\in \mathbb{S}$   w.l.o.g, let $a\geq b\geq c$ which directly implies that $s-a \leq s-b \leq s-c $
Now as $\Delta ABC\in \mathbb{S}$,

\[\frac{5}{s-a}+\frac{5}{s-b}+\frac{5}{s-c}-\frac{3}{s-a}=\frac{6}{r}\]
\[\implies \frac{2r}{s-a}+\frac{5r}{s-b}+\frac{5r}{s-c}=6 \]
\[\implies 2   tan \frac{A}{2}+5  tan  \frac{B}{2}+5   tan \frac{C}{2}=6..(*)\]
\[\implies 2 \frac{1}{cot(\frac{\pi}{2}-\frac{B+C}{2})}+5   tan  \frac{B}{2}+5   tan  \frac{C}{2}=6\]
\[\implies \frac{2}{tan({\frac{B}{2}+\frac{C}{2})}}+5 (tan\frac{B}{2}+ tan\frac{C}{2})=6\]
\[\implies \frac{2(1-tan\frac{B}{2}tan\frac{C}{2})}{tan\frac{B}{2}+ tan\frac{C}{2}}+5 (tan\frac{B}{2}+ tan\frac{C}{2})=6\]
\[\implies \frac{2(1-xy)}{(x+y)}+5(x+y)=6\]
\[\implies 2-2xy +5(x+y)^2=6(x+y)\] (where $x= tan \frac{B}{2}$ and $y= tan \frac{C}{2}$)
\[\implies 5x^2+5y^2+8xy+2-6x-6y=0\]
\[\implies 5x^2+x(8y-6)+(5y^2-6y+2)=0 .....(i)\]
\[\implies x=\frac{6-8y \pm \sqrt{36+64y^2-96y-100y^2+120y-40}}{10}\]
\[\implies x=\frac{6-8y \pm \sqrt{-(6y-2)^2}}{10}\]

 Now note that as $(6y-2)^2\geq 0 $ and $x\in \mathbb{R}$  Discriminant of $(i)$ must be $0$ (otherwise $x\notin \mathbb{R}$)
hence \[6y-2=0\implies y= \frac{1}{3} \implies x= \frac{6- \frac{8}{3}}{10}=\frac{1}{3}\implies \boxed{tan\frac{B}{2}=\tan\frac {C}{2}=\frac{1}{3}}....(ii)\]
As $a\geq b\geq c$ ,so,$A\geq B \geq C$ which implies that both $B,C$  is less than $\frac{\pi}{2}$ hence  \[\boxed {B=C}\]
and [u]the triangle $\Delta ABC$ is isosceles triangle.[/u]
Putting the value of $tan \frac{B}{2} $  and $tan{\frac{C}{2}}$ in  $(*)$ we get $tan \frac{A}{2}=\frac{4}{3}$
 This implies that \[\frac{r}{s-a}:\frac{r}{s-b}:\frac{r}{s-c}:=4:1:1\]
\[\frac{1}{s-a}:\frac{1}{s-b}:\frac{1}{s-c}:=4:1:1\]
\[\implies (s-a):(s-b):(s-c)=1:4:4\]
\[\implies s=9k \   \text{for some}  \  k\in \mathbb{R} \implies \boxed{ a=8k,b=c=5k}\]

 Hence our claim is proved to be true.

[hide=More over]If $\exists$ a glitch somewhere ,however small (other than latex )  I shall be really obliged if you point it out .
(actually I posted this long solution to find if I am right or not .As cunningham law says ,"The best way to get the right answer on the Internet is not to ask a question; it's to post the wrong answer."[/hide]

Firstly, note that we have $AP = s-a = x, BQ = s-b = y, CR = s-c = z$ for values $x,y,z$. WLOG take $a \leq b \leq c$, so our equality rewrites to  $$5\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) - \frac{3}{z} = \frac{5}{x} + \frac{5}{y} + \frac{2}{z} = \frac{6}{r}.$$ But $r = \frac{\sqrt{sxyz}}{s}$, so we have $$\frac{5}{x} + \frac{5}{y} + \frac{2}{z} = \frac{6s}{\sqrt{sxyz}} = \frac{6\sqrt{s}}{\sqrt{xyz}}.$$ Multiplying by $xyz$ gives $$2xy+5xz+5yz = 6\sqrt{sxyz}.$$ But $s = x+y+z$, so squaring gives $$(2xy+5xz+5yz)^2 = 36xyz(x+y+z).$$ Using the minimum-of-a-quadratic formula it can be shown that $(2xy+5xz+5yz)^2 - 36xyz(x+y+z)$ can only be $0$ when $x = y$. From there we can easily obtain $x = y = 4z$, and upon obtaining $a, b, c$ the similarity follows.

Assume without loss of generality $a=\max(a,b,c)$. We have $\frac2{s-a}+\frac5{s-b}+\frac5{s-c}=\frac6r$. Since $r=\sqrt{\frac{(s-a)(s-b)(s-c)}s}$, we have
$$2(s-b)(s-c)+5(s-a)(s-c)+5(s-a)(s-b)=6\sqrt{s(s-a)(s-b)(s-c)}.$$ Let $x=s-a$, $y=s-b$, and $z=s-c$. Then, $2yz+5xy+5xz=6\sqrt{xyz(x+y+z)}$. Squaring gives $$4y^2z^2+25x^2y^2+25x^2z^2+20xy^2z+20xyz^2+50x^2yz=36xyz(x+y+z),$$ or $$4y^2z^2+25x^2y^2+25x^2z^2-16xy^2z-16xyz^2+14x^2yz=0.$$ This is equivalent to $x^2(25y^2+14yz+25z^2)-16xyz(y+z)+4y^2z^2=0$. The discriminant is $256y^2z^2(y^2+2yz+z^2)-16y^2z^2(25y^2+14yz+25z^2)$. This is nonnegative when $16y^2+32yz+16z^2\geq25y^2+14yz+25z^2$, or $y=z$. Therefore, $64x^2y^2-32xy^3+4y^4=0$, which implies $y=4x$. This means $s=x+y+z=9x$, so $a=8x$, $b=5x$, and $c=5x$. Therefore, the triangle must be isosceles, and it is similar to the triangle with sides $5$, $5$, and $8$.

Let $u,v,w$ be $AP,BQ,CR$, respectively. WLOG assume $\min(u,v,w)=u.$ Since \[\sqrt{\frac{uvw}{u+v+w}}=A=rs=r(u+v+w),\] we have $r=\sqrt{\frac{xyz}{x+y+z}}.$ The equation in the problem becomes:
\[\frac{2}{u}+\frac{5}{v}+\frac{5}{w}=6\sqrt{\frac{u+v+w}{uvw}}.\]
When the equation is squared and multiplied by $(uvw)^2$, we get \[25u^2v^2+25w^2u^2+4w^2v^2+14u^2vw-16uvw^2-16uv^2w=0.\] Now we have $9(uv-wu)^2+4(2uv+2uw-vw)^2=0$ which implies that $uv=wu$, so $v=w$. Another thing it implies is $2uv+2uw=vw\implies 4uv=v^2$ so $v=4u.$ Thus, the lengths $u,v,w$ are in the ratio $4:4:1$ in some order, which makes the sides of the triangle have ratio $8:5:5$ in some order, and we're done.

Let $(AP, BQ, CR) = (x, y, z)$ respectively, and assume $x = \min(x, y, z)$. Then we wish to show that,
\begin{align*}
5 \sum_{cyc} \frac1x = \frac 6r + \frac 3x
\end{align*}
Note that from Herons we have,
\begin{align*}
r &= \frac{[ABC]}{x+y+z}\\
&= \sqrt{\frac{xyz}{x+y+z}}
\end{align*}
Then we wish to find all solutions to,
\begin{align*}
5 \left(\frac{xy+yz+xz}{xyz} \right) &= 6\sqrt{\frac{x+y+z}{xyz}} + \frac 3x\\
5(xy + yz + xz) &= 6 \sqrt{xyz(x+y+z)} + 3yz\\
5xy + 5xz + 3yz &= 6 \sqrt{xyz(x+y+z)}\\
25x^2y^2 + 25x^2z^2 + 9y^2z^2 + 50x^2yz + 20xy^2z + 20xyz^2 &= 36x^2yz + 36xy^2z + 36xyz^2\\
x^2(25y^2 + 25z^2 + 14yz) - 16x(y^2z + yz^2)  + 4y^2z^2 &= 0
\end{align*}
Now from the discriminant we find we must have,
\begin{align*}
16y^2 + 32yz + 16z^2 \geq 25y^2 + 14yz + 25z^2\\
18yz \geq 9y^2 + 9z^2\\
0 \geq (y-z)^2
\end{align*}
whence we must have equality at $y = z$. Substituting this back we find $y= 4x$, from which we conclude $ABC$ has sides of the form $$a:b:c = 8x:5x:5x$$


calculus
integration
real analysis
real analysis unsolved

Hi can anyone solve this?

integral of  ln(sqrt(x)) dx...

Thank You,
Alon

Don't create multiple copies of the same topic. The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98890]other[/url]  one is in the right forum, and has been answered. If you have further questions, ask there. There's not enough difference between the definite and indefinite integrals to justify a new topic.


LaTeX quadratics function calculus integration algebra

How many values of $x\in\left[ 1,3 \right]$ are there, for which $x^2$ has the same decimal part as $x$?

I draw a graph of $f(x)=\{x^2\}$ and $f(x) = \{x\}$, and count the number of intersections, there are seven intersections.

hehe. :D

well if you did that without computer/calculator/anything that's perfectly ok :?

but there are very simple elementary solutions :P

well, of course without computer / calculator...
it is easy to draw the graph, just note that the zeroes of $f(x)=\{x^2\}$ are $1,\sqrt2,\sqrt3,2,\sqrt5,\sqrt6,\sqrt7,\sqrt8,3$  :D

each interval [tex] [\sqrt{n}, \sqrt{n+1}] [/tex] ($\sqrt{n+1}$ is not an integer) contains only intersection  :D

edit: removed my post, since all of the latex formulae appeared corrupted.

anyway liyi, what's your answer? I'm not really sure your solution will end well, can you post your final result?

This solution is pretty simple, but it took a little time.  What I did was to separate the interval [1,3] into 8 sections, each bounded by the square roots of the integers 1-9.  So I ended up with: [1, :sqrt: 2], [ :sqrt: 2, :sqrt: 3].....[ :sqrt: 7, :sqrt: 8], [ :sqrt: 8,3].

For each separate partition, I solved a different quadratic equation, and took the x-value that lay in the interval.  So for [1, :sqrt: 2], I solved
x 2 - 1 = x - 1.  For [ :sqrt: 7, :sqrt: 8], I solved
x 2 - 7 = x - 2.   In general, for every partition [ :sqrt: n,  :sqrt: (n+1)], I solved the equation x 2 - n = x -  [ :sqrt: n], where [ :sqrt: n] is the greatest integer less than or equal to the  :sqrt: n.  After solving all of these equations, you end up with seven different solutions in the interval [1,3].

That was painstaking, but clear, I hope...and most of all, I hope it's correct!

$x^{2}=a+x$ for some $a \in Z$ $\therefore$ we need to solve $x^2-x=a$ for $x \in [1,3]$. Now $x^2-x=x(x-1)$ is an non-decreasing function for $x\geq 1$ (as both $x$ and $x-1$ are non-decreasing) so $a$ can take any integral value between 0 and 6 inclusive so all we do is solve each resulting quadratic.

Therefore the seven solutions for $x\in [1,3]$ are $1,\dfrac {1+\sqrt{5}} {2}, 2,\dfrac {1+\sqrt{13}} {2}, \dfrac {1+\sqrt{17}} {2}, \dfrac {1+\sqrt{21}} {2}, 3$.

Ok so this problem is fairly obscure, but I wanted to note that there is no need to find the solutions themselves. It is sufficient to know that within each interval there are a certain number of intersections, and a shared solution at each end of an $[t,t+1]$ interval. This makes it very easy to solve the problem for, say, $x\in [1,2020]$.


algebra
polynomial
modular arithmetic
function
algebra proposed

Let $ a_{1},...,a_{n}$ be n different numbers, $ n>1$. Prove:

$ \sum_{k\equal{}1}^{n} \frac{1}{\prod_{1 \le j \le n, j \neq k} (a_{k}\minus{}a_{j})} \equal{} 0$

[hide]An elementary way is preferred. It can be solved with some tricks from linear algebra. [/hide]

proof:

let $ f(x) \equal{} (x\minus{}a_{1}) (x\minus{}a_{2} ) \cdots (x\minus{}a_{n})$, we show that we must have the identity:

\[ \frac{1}{f'(a_{1})} \frac{f(x)}{x\minus{}a_{1}} \plus{} \cdots \plus{}\frac{1}{f'(a_{n})} \frac{f(x)}{x\minus{}a_{n}}  \equiv 1
\]

this is becuase the left hand side is a polynomial of degree of at most n-1, and equals to 1 at points $ a_{1}, a_{2}, \cdots, a_{n}$. Hence must be 1 identically. Then compare the coefficient of the term $ x^{n\minus{}1}$, we have the proof.

nice solution!  :D and it even gives us more - identities such as $ \sum_{k\equal{}1}^{n} \frac{a_{k}}{\prod_{1 \le j \le n, j \neq k} (a_{k}\minus{}a_{j})} \equal{} 0$ (if we compare the coefficient of $ x^{n\minus{}2}$).

Exactly
$ \sum_{k \equal{} 1}^{n} \frac {a_{k}^m}{\prod_{1 \le j \le n, j \neq k} (a_{k} \minus{} a_{j})} \equal{} \delta_m$,
were $ \delta_m \equal{} 0$ if $ m < n \minus{} 1$ and $ \delta_{n \minus{} 1} \equal{} 1$.

Nice, Rust. For the sake of completeness, and because of boredom, here's my ugly solution:

$ \sum_{k \equal{} 1}^{n} \frac {1}{\prod_{1 \le j \le n, j \neq k} (a_{k} \minus{} a_{j})} \equal{} 0$ iff (multiply by $ \prod_{i > j} (a_{i} \minus{} a_{j})$):
$ \sum_{k \equal{} 1}^{n} ( \minus{} 1)^{n \minus{} k}\prod_{i > j, i,j \neq k} (a_{i} \minus{} a_{j}) \equal{} 0$ (*)

$ \prod_{i > j, i,j \neq k} (a_{i} \minus{} a_{j})$ is the determinant of a Vandermonde matrix. By Leibniz formula:
$ \prod_{i > j, i,j \neq k} (a_{i} \minus{} a_{j}) \equal{} \sum_{\sigma \in S_{n \minus{} 1}} sgn(\sigma) \, \alpha_1^{\sigma(1) \minus{} 1} \cdots \alpha_{n \minus{} 1}^{\sigma(n \minus{} 1) \minus{} 1}$, where: $ \alpha_i \equal{} a_i$ for $ i < k$, and $ \alpha_i \equal{} a_{i \plus{} 1}$ for $ i \ge k$

Hence, (*) contains only monomials of the form $ \prod_{i \equal{} 1}^{n} a_{i}^{b_{i}}$, where: {$ b_{i} \plus{} 1$ | $ 2 \le i \le n, i \neq k_{1},k_{2}$}={2,..,n-1} and $ b_{k_1} \equal{} b_{k_2} \equal{} 0$, for some k's.
If we take any monomial like this, it appears exactly twice - in $ ( \minus{} 1)^{n \minus{} k_1}\prod_{i > j, i,j \neq k_1} (a_{i} \minus{} a_{j})$ and $ ( \minus{} 1)^{n \minus{} k_2}\prod_{i > j, i,j \neq k_{2}} (a_{i} \minus{} a_{j})$.
I will show that the sum of the coefficients is zero, or: $ ( \minus{} 1)^{n \minus{} k_1} sgn(\sigma_1) \plus{} ( \minus{} 1)^{n \minus{} k_2} sgn(\sigma_2) \equal{} 0$.

$ sgn(\sigma_1)$ is the permutation that sends $ i$ to the $ i$th number in this list: $ b_1 \plus{} 1,....,b_n \plus{} 1$ [b]without[/b] $ b_{k_1} \plus{} 1$
$ sgn(\sigma_2)$ is the permutation that sends $ i$ to the $ i$th number in this list: $ b_1 \plus{} 1,....,b_n \plus{} 1$ [b]without[/b] $ b_{k_2} \plus{} 1$

If WLOG $ k_1 > k_2$, then $ \sigma_1 \equal{} (1 (b_{k_1 \minus{} 1} \plus{} 1))(1 (b_{k_1 \minus{} 2} \plus{} 1))*....*(1 (b_{k_2\plus{}1} \plus{} 1)) \sigma_2$ (representation as transpositions)

Now it is easy to see that $ k_1 \equiv k_2 \pmod 2$ iff $ sgn(\sigma_1) \equal{} \minus{} sgn(\sigma_2)$, and thus the coefficient is 0. (the $ sgn$ function is multiplicative)