limit
real analysis
real analysis unsolved

Let $x_{n},y_{n}$ convergent seqvences, with positive limit.

Prove that $\lim_{n\to \infty}\frac{n}{\sum_{i=1}^{n}{\frac{1}{x_{i}y_{n+1-i}}}}=\lim_{n\to \infty}x_{n}y_{n}$

[quote="ffrogg"]It is well known that if $x_{n}\to x,y_{n}\to y$,then $\frac{x_{1}y_{n}+\cdot+\cdot+\cdot+x_{n}y_{1}}{n}\to xy$,which leads to the desired result.[/quote]

Can you argue a little bit? I dont understand the conclusion:)

see http://www.mathlinks.ro/Forum/viewtopic.php?t=132026  :wink:


integration
real analysis
real analysis unsolved

Prove that if $P\in R[X]$ have no real root then $P+P'+P''+...+P^{(n)}$, where $n= deg P$ have no real root too.

A small extension. We can show for Q = P + a*P' + ... a^n*P^n, where a is real, not zero.
We write Q - a*Q' = P.[1].
We suppose the contrary, so Q has at least one real root.

A. If Q has 1 real root with the order of multiplicity 1, and no other real root, then Q has an odd degree, so from [1] P has an odd degree, so P has at least one real root, contradiction.
B. If Q has 1 real root with the order of multiplicity 2 or higher, from [1] results that P has this root also, contradiction.
C. If Q has at least 2 real distinct roots, let's denote p < q those 2 real consecutive roots, so we can write: Q(x) = (x - p)*(x - q)*R(x).
We can suppose that p and q have the order of multiplicity 1, otherwise we use case B.
Using this and because p and q are consecutive, R(p)*R(q) > 0.[2].
Then Q'(x) = (x-q)*R(x) + (x-p)*R(x) + (x-p)*(x-q)*R'(x), so Q'(p) = (p-q)*R(p), Q'(q) = (q-p)*R(q) and using [1]:
P(p)*P(q) = a^2 * Q'(p) * Q'(q) = a^2 * (p-q)^2 * R(p) * R(q) * (-1) < 0, due to [2].
So P has at least one real root in (p, q), contradiction.

Clearly $P$ has even degree and WLOG is $>0$ everywhere.  Put $D(t)=\sum_i P^{[i]}(t)$.  Then we can write $D(t)-D'(t)=P(t) \ (*)$.

Say that $D$ has a real root.  Then either there are (at least) two distinct roots or (at least) a root of multiplicity two.  If there is a root of multiplicity two, then $D'(t)$ vanishes also there, and by $(*)$ we see that $P(t)$ also vanishes there which is absurd.  So there are two distinct roots, say, at $t_1$ and $t_2$, which we can choose so that $D(t)\not=0$ for $t\in(t_1,t_2)$.  By Rolle's theorem we see that $D'(t_3)=0$ for some $t_3\in (t_1,t_2)$.  So $D(t_3)=P(t_3)<0$ which contradicts the hypotheses on $P$.

I think everything essential is there, but it looks very awkwardly worded to me - maybe I'll fix it tomorrow.

[quote="blahblahblah"] we can choose so that $D(t)\not=0$ for $t\in(t_1,t_2)$. [/quote]

since D have even degree you can chose them so that  $D(t) <0$ (the fact you use later in the proof).  I understood your demonstration you do not have to fix it any more ;)


If $ y \equal{} f(x) \equal{} \frac {x \plus{} 2}{x \minus{} 1}$, then it is incorrect to say:

$ \textbf{(A)}\ x \equal{} \frac {y \plus{} 2}{y \minus{} 1} \qquad\textbf{(B)}\ f(0) \equal{} \minus{} 2 \qquad\textbf{(C)}\ f(1) \equal{} 0 \qquad\textbf{(D)}\ f( \minus{} 2) \equal{} 0$
$ \textbf{(E)}\ f(y) \equal{} x$

[hide="Click for solution"]
The value $ x\equal{}1$ makes the denominator zero.  Thus, $ f(1)$ is undefined, so $ \boxed{\textbf{(C)}}$ is incorrect.
[/hide]


[hide="**ANSWER**"]If they purchase the windows together, they will need 15 windows.  The deal is basically 5 windows for $\$ 400$.  That means that they will need $400\cdot\frac{15}{5}=400\cdot3=\$ 1200$.  If they purchase them separately, Dave will need $400+200=\$ 600$.  Doug will need $400+300=\$ 700$.  That means that they save $600+700-1200=\$ 100$ by buying them together.  Answer A is 100, so the answer is $\boxed{A}$[/hide]

[hide]
Buying seperately Dave would buy 4 for 400 get one free and buy another 2. Total of 600. Doug would buy 4 for 400 get one free and buy another 3 for 300. Total of 700. Together, they would buy 4, get one free, 4 more, get one free, four more, get one free. Thats a total of 1200. Seperately costs 1300 so they save 100. The answer is A.[/hide]

[quote="Silverfalcon"]A store normally sells windows at $\$ 100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How much will they save if they purchase the windows together than rather separately?

A. 100
B. 200
C. 300
D. 400
E. 500[/quote]

[hide="answer"]
one. hundred.
[/hide]


number theory
greatest common divisor
number theory proposed

Well if the gcd of the two starting terms is one, it just settles down into a fibbanoichi like recurrence, as can be checked by induction. you can also see that if the gcd is not 1, then the next term in the sequence will be less than the largest of the two before it. So, looking at it like this, we get the terms will decrease, so eventually the gcd becomes 1, (if it naturally becomes 1, or if the terms is forced to become one) which will then settle back down to a fibanoichi recurrance.

Not sure if this really answers your question, but then again your question was very vague! :P

[quote="seamusoboyle"] ... so eventually the gcd becomes 1 ...[/quote]
This isn't necessarily true. For example, the sequence $a_i=2$ (for every $i$) works, and the GCD is always 2.

Sorry, your right. I forgot to include the case when they're equal. Then the next number is 2, and we proceed as before, except if the number before was 2, which keeps giving 2, as said.


geometry
geometric transformation
homothety
angle bisector
geometry solved

In triangle $ABC$ let $M$ and $N$ are midpoints of $AB$ and $BC$. The incircle meets $BC$ and $AC$ in $D$ and $E$. There is a suggestion that $X=MN\bigcap DE$ lies on the bisector of $\angle BAC$.

On my drawings it works. I tried to prove it by proving that distance from $X$ to $AB$ and $AC$ are equal, writing the distance from $X$ to $AB$ as combination of distances from $D$ and $E$ to $AB$, but it fails...
Help to prove or confirm it fails.

It does seem to be true, but I haven't completed the computations. They are, however, rather easy. Just put $P=AI\cap MN,\ P'=DE\cap MN$, where $I$ is the incenter. Now use $MP=AM$ and $\frac{NP'}{CE}=\frac{ND}{DC}$. That way you can compute $MP,NP'$ in terms of the sides of $ABC$, and what we should show is that $\frac {AC}2+NP'=\frac{AB}2$.

But grobber, all the lengths you need are obvious!

$MP = AM = \frac{c}{2}$, $P'N = ND = \left|\frac{b-c}{2}\right|$.  Now you just have to consider whether $D$ is between $B$ and $N$ or between $N$ and $C$.

[quote="Valiowk"]But grobber, all the lengths you need are obvious!
[/quote]

I didn't say they weren't, did I? :) I just don't like such "computations", may they be trivial or not :).

This problem was proposed at 20th Tournament of Towns, spring-1999, difficult level, juniors, problem № 5.

[quote="Feuerbach"]In triangle $ABC$ let $M$ and $N$ are midpoints of $AB$ and $BC$. The incircle meets $BC$ and $AC$ in $D$ and $E$. There is a suggestion that $X=MN\bigcap DE$ lies on the bisector of $\angle BAC$.[/quote]

I wrote up a solution for this one more than a month ago, but iexplorer crashed down before I could post it, so I had to retype it again...

As it's often easier to prove more than less, I will prove a stronger assertion:

[color=blue][b]Extended problem:[/b] In a triangle ABC, let M and N be the midpoints of the sides AB and BC, and let the incircle of triangle ABC meet the sides BC and CA at the points D and E. Let X be the orthogonal projection of the point B on the angle bisector of the angle CAB. Then, this point X lies on the lines MN and DE.[/color]

[i]Solution of the extended problem.[/i] Let the line BX meet the line CA at a point Y. Then, since the point X is the orthogonal projection of the point B on the angle bisector of the angle CAB, we have < BAX = < YAX on the one hand (because of the angle bisector) and < AXB = < AXY = 90° on the other hand (because of the orthogonal projection). Also, clearly, AX = AX. Thus, the triangles AXB and AXY are congruent (from ASA). It follows that BX = YX. In other words, the point X is the midpoint of the segment BY. Equivalently, the point X is the image of the point Y in the homothety with center B and factor $\frac12$. On the other hand, the points M and N, being the midpoints of the segments AB and BC, are the images of the points A and C in this homothety with center B and factor $\frac12$. Altogether, the points X, M and N are the images of the points Y, A and C in this homothety. Therefore, since the points Y, A and C are collinear, their images X, M and N must be collinear, too; thus, the point X lies on the line MN.

Remains to prove that the point X lies on the line DE.

Let I be the center of the incircle of triangle ABC. Then, the line AI is the angle bisector of the angle CAB and thus passes through the point X, and we have < BXI = 90° (since the point X is the orthogonal projection of the point B on the angle bisector of the angle CAB). On the other hand, since the point I is the center of the incircle of triangle ABC, while the point D is the point of tangency of this incircle with the side BC, we have $ID\perp BC$ and thus < BDI = 90°.

From < BXI = 90° and < BDI = 90°, it follows that the points X and D lie on the circle with diameter BI. Thus, the quadrilateral BDXI is cyclic, and it follows that < BDX = 180° - < BIX. Hence,

  < XDC = 180° - < BDX = 180° - (180° - < BIX) = < BIX = 180° - < AIB.

But by the sum of angles in triangle AIB, we have

  $180^{\circ}-\measuredangle AIB=\measuredangle IAB+\measuredangle IBA=\frac{A}{2}+\frac{B}{2}=\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$

(since A + B = 180° - C by the sum of angles in triangle ABC). Thus,

  $\measuredangle XDC=\frac{180^{\circ}-C}{2}$.

On the other hand, since the two tangents from a point to a circle are equal in length, we have CD = CE, so that the triangle DCE is isosceles, and we get

  $\measuredangle EDC=\frac{180^{\circ}-\measuredangle DCE}{2}=\frac{180^{\circ}-C}{2}$.

Consequently, < XDC = < EDC. It follows that the point X lies on the line DE. This completes the solution of the extended problem.

  Darij


geometry
incenter
circumcircle
geometric transformation
reflection
geometry proposed

Hello!

Let $ ABCD$ be a quadrilateral inscribed in a circle $ \left(O\right)$, $ \left\{V\right\}\equal{}AC\cap BD$ and $ M$, $ N$ are the midpoints
of the arcs $ \widehat{ACB}$, $ \widehat{CBD}$ respectivelly. If $ \left\{K\right\}\equal{}VO\cap MN$ then prove that $ \frac{KM}{KN}\equal{}\frac{AB}{CD}$.

Best regards, Petrisor Neagoe :)

Denote $ I_1, I_2,I_3,I_4$ the incenters of triangles $ AVB, BVC,CVD,DVA$ respectively.
From topic [url=http://www.mathlinks.ro/viewtopic.php?t=315105]here[/url] we get $ MI_1\cap NI_2\cap OV\equal{}\{Q\}$.
On the other side, $ \angle DI_4V\equal{}90^o\plus{}1/2\angle DAV\equal{}90^o\plus{}1/2\angle CBV\equal{}\angle CI_2V$ then $ MI_2\equal{}MI_4$. Similarly, $ NI_2\equal{}NI_4$.
Therefore $ MN\perp I_2I_4 \Rightarrow MN// I_1I_3$.
Applying Thales's theorem we have $ \frac{KM}{KN}\equal{}\frac{I_1V}{I_3V}\equal{}\frac{AB}{CD}$ (QED)

My solution:

First, let's prove that the bisector of $ \angle AVB$ and $ MN$ are parallel.
Construct $ NX\parallel{}AC$ and $ NY\parallel{}BD$, $ X,Y\in(O)\Rightarrow$
$ \Rightarrow$ the bisectors of the angles $ \angle AVB$ and $ \angle XNY$ are parallel   (1)
$ NX\parallel{}AC\Rightarrow \widehat{AN}=\widehat{CX}$
$ NY\parallel{}BD\Rightarrow \widehat{BN}=\widehat{DY}$
$ \widehat{AN}+\widehat{DY}=\widehat{BN}+\widehat{CX}\Rightarrow\widehat{ND}+\widehat{AY}=\widehat{NC}+\widehat{BX}\Rightarrow\widehat{AY}=\widehat{BX}\Rightarrow$
$ \Rightarrow$ the bisector of $ \angle XNY$ is the bisector of $ \angle ANB$   (2)
(1),(2)$ \Rightarrow$ the bisector of $ \angle AVB$ and $ MN$ are parallel.

$ (I)$ is the circumcircle of $ \Delta AVB\Rightarrow$ the points $ M$, $ O$, $ I$ are collinear.
$ (J)$ is the circumcircle of $ \Delta CVD\Rightarrow$ the points $ N$, $ O$, $ J$ are collinear.
$ \{P\}=MI\cap(I)$, $ P\notin\widehat{AVB}$ and $ \{Q\}=NJ\cap(J)$, $ Q\notin\widehat{CVD}$
$ \angle AVP=\angle ABP=\angle BAP=\angle BVP\Rightarrow(VP$ is the bisector of $ \angle AVB$   (3)
$ \angle CVQ=\angle CDQ=\angle DCQ=\angle DVQ\Rightarrow(VQ$ is the bisector of $ \angle CVD$   (4)
(3),(4)$ \Rightarrow$ the points $ P$, $ V$, $ Q$ are collinear.

$ \angle VAP=\angle VAB+\angle PAB=\angle VDC+\angle PVB=\angle VDC+\angle QVC=$
$ =\angle VDC+\angle QDC=\angle VDQ\Rightarrow\angle VAP=\angle VDQ$
$ \frac{VP}{VQ}=\frac{\frac{VA\cdot sin(\angle VAP)}{sin(\angle VPA)}}{\frac{VD\cdot sin(\angle VDQ)}{sin(\angle VQD)}}=\frac{VA}{VD}\cdot\frac{sin(\angle VQD)}{sin(\angle VPA)}=\frac{AB}{CD}\cdot\frac{sin(\angle VCD)}{sin(\angle VBA)}=\frac{AB}{CD}\Rightarrow$
$ \Rightarrow\frac{VP}{VQ}=\frac{AB}{CD}$   (5)
$ \frac{VP}{KM}=\frac{OV}{OK}=\frac{VQ}{KN}\Rightarrow\frac{VP}{VQ}=\frac{KM}{KN}$   (6)
(5),(6)$ \Rightarrow\frac{KM}{KN}=\frac{AB}{CD}$

Best regards, Petrisor Neagoe :)

let M' is reflection of point M wrt O and N' of N wrt O. and OV intersect M'N' at K' easy to see that K'N/K'M = KN/KM. easy to see that OM' is perpendicular to AB and ON' is perpendicular to CD let VH is perpendicular on AB and OH' on CD and M" is midpoint of AB and N" of DC easy to see that S(OM'V)/S(ON'V) = (M"H*OM'/2)/(N"H'*ON'/2) = K'N/K'M = M"H/N"H' and easy to see thet M"H/N"H' = AB/DC


induction
quadratics
modular arithmetic
number theory unsolved
number theory

find the number of $ n(2^{2009}>n \ge 0)$ such that $ n \equiv k^2(mod 2^{2009})$ for some integer $ k$

If $ 0\leq n < 2^{2009}$ then $ n \equiv k^2 \mod 2^{2009}$ means that $ n\equal{}k^2$ and answer is $ [\sqrt{2^{2009}}]\plus{}1$.

[quote="Yuriy Solovyov"] then $ n \equiv k^2 \mod 2^{2009}$ means that $ n \equal{} k^2$.[/quote]
Why is that?

Maybe I don't understand something, but $ a\equiv a \mod N$ if $ a < N$, because $ a \equal{} 0\cdot N \plus{} b$ means $ a \equal{} b$ and we can't replace $ 0$ by $ 1$ because $ a < N$.

Yes, but there's no such restriction on $ k$.

yeah $ k$ is can be arbitrary integer...

Use induction and the fact that for odd $ n$. $ n$ is a quadratic residue modulo a power of $ 2$ iff $ n\equiv 1\pmod{8}$


A triangle has sides of length 24, 10, and 26. What is the radius of the inscribed circle ?

(A)26 (B) 4 (C) 3 (D) 8 (E) none of these

We can find the inradius of a such right triangle can be found by first dividing the right triangle into 3 simliar triangles, with the inradii as altitudes. This is a bit hard to explain without a picture... :/  But i'll try.. So, let's say i have right triangle with sides $ a$, $ b$ and $ c$, with $ a$ and $ b$ are the legs. And a right triangle with vertex $ A$ is the right angle. Then, with what i stated before, we can find the area of the three simliar figures. So, we can solve the formula
$ \frac {A*B}{2} \equal{} \frac {r*a}{2} \plus{} \frac {r*b}{2} \plus{} \frac {r*c}{2} \equal{} \frac {r(a \plus{} b \plus{} c)}{2}$, which is equal to the area of the triangle relative to the inradii. So, then, with the previous formula, we can solve for $ r$ and get $ r \equal{} \frac {a*b}{a \plus{} b \plus{} c}$, then we can solve this to get an answer of $ \frac {10*24}{10 \plus{} 24 \plus{} 26}$ $ \implies$ $ \frac {240}{60} \equal{} \boxed 4$, so the answer would be $ B$, or $ 4$. :)

EDIT: i added a picture to help you understand what i'm saying. :)

I just use the formula $ a\equal{}rs$ where $ a$ is the area of the triangle, $ r$ is the radius of the inscribed triangle and $ s$ is the semi perimeter:

$ 30r \equal{} 120$
$ r \equal{} 4$

[quote=origamiguy777]oh.. :blush:
I suppose that works too... xD
(personally, i think that your way is better than mine..)
i just wanted to help... :)

$ The$ $ Origami$ $ Guy$[/quote]

I used your way until I found out I could use 2@bove method

[hide=1 second solve](a+b-c)/2=4[/hide]
also 3above im pretty sure its A=rs not a=rs


Functional Analysis
real analysis
real analysis unsolved

Let $E$ be finite dimensional $\mathbb{R}$-vectorial space. Prove that $E$ cannot be represented as a countable union of it's subspaces, different from $E$.

That's false: let $E$ be the space of all  sequences of real numbers $\{x_n\}$ such that $x_n$ is $0$ for all sufficiently large $n$. Let $E_k$ be the subspace of all sequences such that $x_n=0$ for $n\ge k$. Clearly, $E_k\ne E$ but $\bigcup_k E_k=E$.

[quote="3X.lich"]Perhaps a countable intersection of its subspaces...[/quote]
Then it is true but not very interesting: the intersection of subspaces is contained in every of them, so, of course, it cannot be $E$.

Maybe Eugene forgot to say that $dim(E) < \infty$ ?

Yes, i'm sorry-as Alekk said i forgot to say that $dimE < \infty$

Take a norm on $E$ (for instance, take the norm of $\sum a_ie_i$ to be $\sum|a_i|$, where $(e_i)_i$ is any fixed basis). This makes $E$ a Banach space, so by Baire's Theorem, $E$ cannot be a countable union of nowhere dense subsets. Every subspace of $E$ with dimension less than that of $E$ is such a nowhere dense subset, so the conclusion follows.


man mikham bedoonam behine sazi chie age kasi midoone chie bege

salam
fekr mikonam injoor matalebo mishe too topice pih post kard.be harhal behine sazi yani
inke az chizayii ke dar ekhtiaretan be nafve ahsan estefade koni.

taa oonjayi ke man beyad daram, in mozoo roo too ghesmat haye avaliye ye hesaban, baraye peyda kardan e max, va min tavaabe hast...

Bebinid behineh sazi yek masalye kheili kolli va mohemmiye ke dar Systemhaye dinamiki, kontrol va reshtehaye mohandesi kheili karbord dare. Masaleh ine ke ma ye system darim va mikhahim oon raftari ke delemoon mikhad baraye ma dashte bashe, Yaani mikhahim ye vijegi ra tooye oon system Behine konim. Ye halate khassesh hamoone ke amirhtlusa goft.

Ye mesal mizanam:

Farz konid ye radio az ferestande yek seri amvaj migire ke too oonha noise oftade. Inke radio ba tavajoh be amvaje daryaft shode che joori sedayi ta hadde momken nazdik be sedaye vaghei pakhsh kone va noise ha ra begire ta sedye nahanjari be gooshe ma narese ye masaleye bahine sazi too kontrole.

bebakhshid shayad enghadr khoob nabashe ke soal konam vali hala in behine sazi dar ehtemal chikar mikone?


I'm an international student looking for Colleges in the US.
I know the usual common ones like MIT, Caltech, Stanford, Princeton etc.

However I wanted to know if there are other good colleges that have a strong mathematics program and provide financial aid to international students.

Thanks!

Williams and Middlebury Colleges are need blind for internationals, but I don't know about the strength of their Math and Science programs. Could anyone comment?

U Chicago is a good one for pure studies.  I have heard U Michigan has a good math department, but it definitely doesn't line up with the ones you mentioned.

[url=http://www.ams.org/employment/group_i.html]These[/url] are some good departments. Besides their reputation, find out about the workload for teaching assistants. (Which tends to be heavier at public universities.)

Our (myself and the OP) problem is that we will be very hard put even with an EFC of $ \$$5000. Thus we have to consider only those schools which offer full or nearly full fin. support. I know that we are in an extremely competitive pool but we want to consider our options. Thus most public schools etc. are almost out of question.

Could anyone tell me which [b]need-blind [/b]colleges are easier to get into (higher acceptance rates)  for majoring in math? I don't have any awards in sports, club leaderships and I have never been selected for the IMO (in fact, no international contest). Hence, it will be difficult to get into highly selective colleges.

All states will treat me as out-of-state (since I have lived in India for the last 16 years) and hence, I cannot apply to public universities.


Your solution is wrong!

The Chineese Remainder Theorem could no be applied directly to (a1,..,an) as they are not pairwise coprime

[quote="mto"]Let P(x) be a polynomial with integer coefficients. Suppose that there exist n positive integers a1,..., an satisfying:

1. GCD(a1,...,an)=1

2. For each natural number k, there exists an index j such that  aj is a divisor of P(k)

Show that we can find  j such that P(k) is divisible by aj for all k[/quote]
Not necessary condition i)
For each p we consider number $ a_i$ such that $ v_p(a_i)\equal{}b$ is maximum
From the condition we can chose a natural number $ k_i$ such that $ p^b|P(k)$
Consider the Equation
$ x\equiv k_i (\mod p_i^{a_i}$
From chinese remainder theorem exist x satisfy condition.
So $ p_i^{a_i}|P(x)$
Because $ v_{p_i}(a_i)$ is chosen maximum so  $ a_j|P(x_0),j\equal{}1,..,n$

[quote="aviateurpilot"]we suppose that $ \forall j\in\{1,2,...,n\},\exists k_j\in\mathbb{Z}: \ a_j\not|P(k_j)$

we know that $ \exists n\in\mathbb{Z},\forall j\in\{1,2,...,n\}: \ n\equiv k_j\ (mod\ a_j)$
then $ \forall j\in\{1,2,...,n\}: \ P(n)\equiv P(k_j)\ (mod\ a_j)\not\equiv 0\ (mod\ a_j)$
so $ \forall j\in\{1,2,...,n\}: \ a_j\not|P(n)$ (impossible).

then $ \exists j\in\{1,2,...,n\},\forall k\in\mathbb{Z}: \ a_j|P(k)$[/quote]
sorry here,
the first solution is true just if $ gcd(a_i,a_j) \equal{} 1 \forall i,j$ :blush:

[u]solution[/u]
if $ a_j \equal{} 1$ for some $ j \equal{} 1....n$ then $ a_j \equal{} 1|P(k) \forall k$
if $ min(a_1,a_2,...,a_n) > 1$

let the sequence $ (u_f)_{f\in\mathbb{N}}$ such that $ u_0 \equal{} (P,a_1,a_2,a_3,....,a_n)$
if $ u_n \equal{} (Q,x_1,x_2,....,x_n)$ where $ Q,x_1,x_2,....,x_n$ cheks the conditions of the problem with the polynom $ Q$
and $ min(x_1,x_2,...,x_n) > 1$

[quote]we take $ S \equal{} \{p\ prime: \ p|\prod_{k \equal{} 1}^{n}x_k\} \equal{} \{p_1,p_2,...,p_h\}$

$ i)$ $ \forall i,j\in\{1,2,..,h\}\ s.t\ i\neq j: \ gcd(p_i,p_j) \equal{} 1$
$ ii)$ $ \forall k\in\mathbb{Z},\exists i\in\{1,2,..,h\},\ p_i|Q(k)$ (because $ p_i|x_l$ for some $ l \equal{} 1.2...n$)

$ i)$ and $ ii)$ gives $ \exists j\in\{1,2,...,h\},\forall k\in\mathbb{Z}: \ p_j|P(k)$ (see my first solution).[/quote]
so $ u_{n \plus{} 1} \equal{} (\frac {Q}{p_j},x'_1,x'_2,....,x'_n)$ where $ x'_i \equal{} \frac {x_i}{gcd(p_j,x_i}$

finally, $ \exists d\in\mathbb{N}$ such that $ u_d \equal{} (\frac {P}{A},y_1,y_2,....,y_n)$
and $ y_j \equal{} 1$ for some $ j \equal{} 1....n$ and it's evident that $ a_j|A$
so we have $ a_j|A|P(k)$, $ \forall k\in\mathbb{Z}$
and $ gcd(a_1,a_2,...,a_n)\equal{}1$ is not necessary.

i hope that i havn't make another mistake :oops:


Look at the generalization of the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=5722
\[\underline {\overline {\left| \triangle \ ABC,\ x,y>0,\ x+y=2\Longrightarrow R^2+xa^2+yb^2\ge xyc^2\right| }}.\]When is it equality ?

\[\underline {\overline {\left| \ \clubsuit\ \blacksquare\ \Longrightarrow\ \bigstar\ \Longleftarrow\ \blacksquare\ \clubsuit\ \right| }}\] \[\emptyset\emptyset\emptyset\emptyset\emptyset\] \[\emptyset\emptyset\emptyset\] \[\emptyset\]
\[\underline {\overline {\left| \ Farewell,\ my\ dear\ Mathlinks.ro\ !\ \right| }}\]


topology
parameterization
geometry
geometric transformation
analytic geometry
advanced fields
advanced fields theorems

Circle is an one-dimensional manifold and cartesian product of two circles gives a two dimensional manifold torus. Defineitely the cartesian product is not in $ R^2$, I mean, in $ R^2$ a circle is represented by two coordinates. So, a torus $ T\equal{}S^1\times S^1$ should be represented by four coordiates if it is a cartesian product. Isnt it?

I mean I think I understand the whole business of torus, but I dont understand what they mean by "cartesian product" in this case?

The Cartesian product of two sets $ A \times B$ is the set of all ordered pairs $ (a, b)$ such that $ a \in A, b \in B$.  In this case, the reason a torus is described as the Cartesian product of two circles is that every point on a torus can be located by describing its corresponding locations on two circles:  one circle giving the "width" of the torus, and the other circle giving its "length," as per the following diagram:

[img]http://upload.wikimedia.org/wikipedia/commons/5/54/Torus_cycles.png[/img]

The most natural parametrization of the torus is in $ \mathbb{C}^2$: $ (e^{i\theta_1},e^{i\theta_2})$. The $ \mathbb{R}^3$ version is homeomorphic, but loses some of the geometric symmetry.

The easiest form to work with is $ T^2\equal{}\mathbb{R}^2/\mathbb{Z}^2$: the Cartesian plane, with the orbits under translation by integers identified. We can't embed the torus in a plane, but we can make it a quotient of a plane.

Thanks everyone. It is making more sense now.


conics parabola LaTeX calculus derivative

[quote="indialover"]Find the shortest distance between the parabola, y2= 4x and circle
 x2 + y2 -24y +128 = 0[/quote]
Latex: Find the shortest distance between the parabola $ y^2 \equal{} 4x$ and the circle $ x^2 \plus{} y^2 \minus{} 24y \plus{} 128 \equal{} 0$

[hide]The expression of the circle becomes $ (x)^2 \plus{} (y \minus{} 12)^2 \equal{} 16$. A point on the parabola can be represented as $ (x,2\sqrt {x})$. The center of the circle is $ (0,12)$ and we want to minimize the distance between the center and the point. We have the equation:
$ d^2 \equal{} x^2 \plus{} (12 \minus{} 2\sqrt {x})^2$
$ d^2 \equal{} x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x$
$ d \equal{} \sqrt {x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x}$
Taking the derivative of this and setting it equal to 0 results in:
$ \frac {x\plus{}12x^{ \minus{} \frac {3}{2}} \plus{} 2}{\sqrt {x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x}} \equal{} 0$
$ x \plus{} \frac {12}{\sqrt {x^3}} \plus{} 2 \equal{} 0$
$ x \plus{} 2 \equal{} \minus{}\frac {12}{\sqrt {x^3}}$
$ x^2 \plus{} 4x \plus{} 4 \equal{} \frac {144}{x^3}$
$ x^5 \plus{} 4x^4 \plus{} 4x^3 \equal{} 144$
$ x^5 \plus{} 4x^4 \plus{} 4x^3 \minus{} 144 \equal{} 0$
And....this doesn't really factor nicely...:(
So after plugging this into mathematica...the root is approximately 2.06
This gives a distance of approximately 9.4... and this gives the minimum distance of 5.4...which is still greater than 5.05...hmmmmmmmm
 [/hide]
I really think your problems should go in the calculus forum  :wink:

[quote="ckck"][quote="indialover"]Find the shortest distance between the parabola, y2= 4x and circle
 x2 + y2 -24y +128 = 0[/quote]
Latex: Find the shortest distance between the parabola $ y^2 \equal{} 4x$ and the circle $ x^2 \plus{} y^2 \minus{} 24y \plus{} 128 \equal{} 0$

[hide]The expression of the circle becomes $ (x)^2 \plus{} (y \minus{} 12)^2 \equal{} 16$. A point on the parabola can be represented as $ (x,2\sqrt {x})$. The center of the circle is $ (0,12)$ and we want to minimize the distance between the center and the point. We have the equation:
$ d^2 \equal{} x^2 \plus{} (12 \minus{} 2\sqrt {x})^2$
$ d^2 \equal{} x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x$
$ d \equal{} \sqrt {x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x}$
Taking the derivative of this and setting it equal to 0 results in:
$ \frac {x \minus{} 12x^{ \minus{} \frac {1}{2}} \plus{} 2}{\sqrt {x^2 \plus{} 144 \minus{} 48\sqrt {x} \plus{} 4x}} \equal{} 0$
$ x \minus{} \frac {12}{\sqrt {x}} \plus{} 2 \equal{} 0$
$ x \plus{} 2 \equal{} \frac {12}{\sqrt {x}}$
$ x^2 \plus{} 4x \plus{} 4 \equal{} \frac {144}{x}$
$ x^3 \plus{} 4x^2 \plus{} 4x \equal{} 144$
$ x^3 \plus{} 4x^2 \plus{} 4x \minus{} 144 \equal{} 0$
$ (x \minus{} 4)(x^2 \plus{} 8x \plus{} 36) \equal{} 0$
$ x \equal{} 4$
Thus the minimum distance is $ \sqrt {16 \plus{} 144 \minus{} 48 \plus{} 16} \equal{} \sqrt {128} \equal{} 8\sqrt {2}$
But since we're looking for the distance between the parabola and the circle subtract the radius to get a final answer of $ 8\sqrt {2} \minus{} 4$. [/hide]
I really think your problems should go in the calculus forum  :wink:[/quote]

ckck,

your method must have a wrong assumption/flaw in it.
I sketched the curves together.

As a counterexample to your shortest distance, [$ 8\sqrt {2} \minus{} 4) \sim 7.31$],
I picked two specific points, one from each curve, and I computed the
distance between them.

Let's look at a point from the lower half of the circle and a point from the
upper branch of the parabola.

Circle:
Suppose $ x \equal{} 2$.  Solve for the lower $ y$-value.

$ x^2 \plus{} (y \minus{} 12)^2 \equal{} 16$
$ (2)^2 \plus{} (y \minus{} 12)^2 \equal{} 16$
$ 4 \plus{} (y \minus{} 12)^2 \equal{} 16$
$ (y \minus{} 12) \equal{} 12$

Lower part of the curve:
$ y \minus{} 12 \equal{} \minus{} \sqrt {12}$
$ y \equal{} 12 \minus{} \sqrt {12}$

The point on the lower half of the circle is $ (2, 12 \minus{} \sqrt {12})$.
--------------------------------------------------------------------------

Parabola:
Suppose $ x \equal{} 5$.  Solve for the $ y$-value.

$ y^2 \equal{} 4x$
$ y^2 \equal{} 4(5)$
$ y^2 \equal{} 20$

Upper part of the curve:
$ y \equal{} \sqrt {20}$

The point on the upper part of the parabola is $ (5, \sqrt {20})$.
--------------------------------------------------------------------------

$ d \equal{} \sqrt {(5 \minus{} 2)^2 \plus{} [\sqrt {20} \minus{} (12 \minus{} \sqrt {12})]^2}$

$ d \equal{} \sqrt {9 \minus{} (\sqrt {20} \minus{} 12 \plus{} \sqrt {12})^2}$

$ d \sim \sqrt {25.514}$

$ d \sim 5.05 < (8\sqrt {2} \minus{} 4)$, that you worked it to.

Oh I see why...I took the derivative incorrectly...jeeez and I call myself an AP Calc BC student....heh I'll edit that  :blush:


There were 3 people and they went to hotel.  They decide to share a room which they think costs 30 dollars.  So they split it three ways, each paying 10 dollars.  The casheir person realizes that they payed 30 while the room costed 25 dollars.  The casheir gives a billboy 5 dollars to give bacjk to the 3 people.  On the way, the billboy pockets 2 dollars for himself.  There are 3 dollars left ans gives 1 back to each person.  This means they only payed 9 dollars each.  9*3+2=30??? how is this possible?

Bah, it's an easy paradox. If you draw it out with a flow chart, it's simple, but I'm too lazy :P .

[hide]The bellboy's 2 dollars are [i]part[/i] of the 27 dollars the three people payed, and they got 3 dollars back total, which comes to 30 dollars.  This is probably as old as the one with the goat/goose, fox/wolf/dog, grain, and you can only hold one in the boat.  However, it is a good problem for those who haven't seen it.[/hide]

[quote="dwx314"][hide]I think that your equation's wrong.
It should be 9*3-2=25
Since each person essentially paid 9 dollars, but the billman *pocketed* (not added) 2 dollars from that sum, making the total 25 dollars.[/hide]

EDIT: Hmm...Don't know about Treething's explanation though...[/quote]

no...each person paid a total of 9 dolars after.  this is 27 dolars.

the bellboy took 2 so thats 29.

Im not sure I'm getting what your saying

maybe I should try this sometime soon in a real situation. That one dollar seemed to disappear!!!!

[hide]

This is a trick question.

The total money paid is 9 * 3 = 27.

The total money received is 25 for the room and 2 to the bellboy = 27.

There simply is no extra dollar. The 2 to the belloy is included in the 27 the three men paid. By adding it to the 27 it is being counted twice.

[/hide]

[quote="4everwise"][hide]I got 29 as the answer. I hope I understood this correctly[/hide][/quote]that's why I posted the problem

i dont get the question...it makes sense to me...maybe your equation is wrong...i didnt read the equation...[hide]30-3=27...27-2=25...no extra dollar here[/hide]

oh i get the question now!

good explanation...

[quote="236factorial"]let's think of it this way

after paying the 30 dollars, assume the men have 0 dollars total. Then the 5 dollars extra is given to the billboy, and then he gives 3 dollars to each of the men. That's 3 dollars total for the men. the bill boy has 2 dollars, and the hotel cashier has 25 dollars. That's 3 + 2 + 25 = 30.[/quote]

yeah...if you think of it that way, it makes sence but...

when the bell boy gives 3 dollars back to each of the three people, they each technically only pay 9 dollars.  in total they all pay 27 dollars.  the bellboy takes two.  2+27=29...

yes but they paid 27 dollars in this manner: 2 to the billboy and 25 for the room. That gives 27 dollars which is what they paid :D

yes, and that other three dollars they have now, right?

easy but tricky paradox,
take it this way...........
step by step....
1st event 3 give 30 bellboy has 5 of those  and manager has 25 of them.
so now when he gives 3 they has given 27 and he still owes them 2 which he had pocketed so they have 25 that means 25=25 which is correct and there is no room for 1 more dollar.

this is a traditional brain teaser.
 :D

is there a brain teaser section in this forum? i have plenty to post....

[quote="Barnacle"]The men forked over 27 dollars in total, 25 went to the hotel, 2 went to the bellboy.  the equation at the end of the story is just bogus and meant to confuse you  :lol:[/quote]
Or 25+3+2=30

30 - 3 = 25 + 2 fits the problem scenario most i think, but thinking about that 3 is just a waste of time lol


geometry
parallelogram
rectangle
trigonometry
Pythagorean Theorem
cyclic quadrilateral

P is an interior point of a paralleogram whose distances from the corners are 1,4,7 and 8. What is the largest possible area of paralleogram?

i know it's a rectangle, but how did you get 18sqrt(5)???

how did you get that? that's even higher than the maximum that i believe i got right....

I thought the diagonal was $9$ so I just kinda did trial and error

thats impossible, a diagonal of length 9 would be too long for this quadrilateral. trial and error... you trust trial and error?

[quote="Palytoxin"][quote="pkothari13"]the british flag theorem states that for any point P in [b]space [/b]relative to rectangle ABCD, AP^2 + CP^2 = BP^2 + DP^2.

1 + 64 = 16 + 49, so you know it's a rectangle.  is there some sort of pythagorean theorem thing you can apply to this?[/quote]
its true if you assume that it is a rectangle then $AP^2 + CP^2 = BP^2 + DP^2$, but is the converse true as well?[/quote]

yep, it sure is  :) and if you wanna be sure of it, try other quadrilaterals to convince yourself

Let P be an interior point of a parallelogram ABCD. Translate the point P into a point $P_1$ in the direction of the parallel lines $BA \parallel CD$ by the distance BA = CD. Similarly, translate the point P into a point $P_2$ in the direction of the parallel lines $DA \parallel CB$ by the distance DA = BC. The quadrilaterals $APDP_1 \cong PAP_2B$ are congruent, because the triangles $\triangle BPC \cong \triangle AP_1D \cong \triangle PP_2B$ are congruent and the triangles $\triangle CPD \cong \triangle BP_2A \cong \triangle PDP_1$ are also congruent. Moreover, the area of the parallelogram ABCD is equal to the sum of the areas of the quadrilaterals $APDP_1 \cong PAP_2B$, hence

$|ABCD| = \frac{|APDP_1|}{2} = \frac{|PAP_2B|}{2}$

and the area |ABCD| is maximum iff the areas $|APDP_1| = |PAP_2B|$ are maximum. The quadrilateral $APDP_1$ has given sides

$AP = a = 8,\ \ PD = d = 1,\ \ DP_1 = CP = c = 4,\ \ P_1A = PB = b = 7$

The order obviously does not matter. Let $s = \frac{a + d + c + b}{2} = 10$ be the quadrilateral semiperimeter. For the maximum area, this quadrilateral has to be cyclic and its area is then given by Brahmagupta's formula for the area of cyclic quadrilaterals:

$|APDP_1| = \sqrt{(s - a)(s - d)(s - c)(s - b)} = \sqrt{2 \cdot 9 \cdot 6 \cdot 3} = 18$

As a result, the maximum area of the defined parallelogram ABCD is $|ABCD| = 2 |APDP_1| = 36$. In this particular case, the diagonals $AD, PP_1$ happen to be perpendicular to each other, $AD \perp PP_1$ and since the lines $CD \parallel PP_1$ are parallel, $AD \perp CD$, which means that the parallelogram ABCD is a rectangle. But this is not necessary in a general case - cyclic quadrilaterals with non-perpendicular diagonals do exist.
____________________________________________________________________________________

Here is how you create the problem: Pick an arbitrary full square with at least 4 different factors $S^2 = q_1q_2q_3q_4$ and solve the set of 5 equations for 5 unknowns s, a, b, c, d:

$s - a = q_1,\ \ s - b = q_2,\ \ s - c = q_3,\ \ s - d = q_4,\ \ s = \frac{a + b + c + d}{2}$

Substituting the first 4 equations into the last one,

$s = \frac{4s - (q_1 + q_2 + q_3 + q_4)}{2},\ \ s =  \frac{q_1 + q_2 + q_3 + q_4}{2},\ \ a = s - q_1,\ ...$, etc.

For example, $S^2 = 36 = 1 \cdot 2 \cdot 3 \cdot 6$ leads to s = 6, a = 5, b = 4, c = 3, d = 0, which is no good. Let $S^2 = 144 = 1 \cdot 3 \cdot 6 \cdot 8 = 2 \cdot 3 \cdot 4 \cdot 6$. The 1st possibility leads to s = 9, a = 8, b = 6, c = 3, d = 1, while the 2nd one to

$s = \frac{15}{2},\ \ a = \frac{11}{2},\ \ b = \frac{9}{2},\ \ c = \frac{7}{2},\ \ d = \frac{3}{2}$

Or let $S^2 = 324 = 2 \cdot 3 \cdot 6 \cdot 9 = 1 \cdot 3 \cdot 9 \cdot 12$. The 1st possibility leads to s = 10, a = 8, b = 7, c = 4, d = 1 (the original problem), while the 2nd one to

$s = \frac{25}{2},\ \ a = \frac{23}{2},\ \ b = \frac{19}{2},\ \ c = \frac{7}{2},\ \ d = \frac{1}{2}$
____________________________________________________________________________________

Consider the original problem with s = 10, a = 7, b = 8, c = 4, d = 1 (the order does not matter). Denote $\alpha = \angle PAP_1, \delta = \angle PDP_1$. Since $\alpha + \delta = 180^o,\ \cos \delta = -\cos \alpha$. Using the cosine theorem for the triangles $\triangle PAP_1, \triangle PDP_1$ with the common side $PP_1 = BA = CD$,

$a^2 + b^2  - 2ab \cos \alpha = PP_1^2 = c^2 + d^2 - 2cd \cos \delta = c^2 + d^2 + 2cd \cos \alpha$

$\cos \alpha =  \frac{(a^2 + b^2) - (c^2 + d^2)}{2(ab + cd)} = \frac{(49 + 64) - (16 + 1)}{2(56 + 4)} = \frac{96}{120} = \frac 4 5$

$PP_1^2 = a^2 + b^2 - 2ab\ \frac{(a^2 + b^2) - (c^2 + d^2)}{2(ab + cd)} = \frac{ab(c^2 + d^2) + cd(a^2 + b^2)}{ab + cd} =$

$= \frac{56 \cdot 17 + 4 \cdot 113}{56 + 4} = \frac{1404}{60} = \frac{117}{5}$

$BA = CD = PP_1 = \sqrt{\frac{117}{5}} = 3 \sqrt{\frac{13}{5}}  = \frac{3 \sqrt{65}}{5}$

Denote $\beta = \angle AP_1D, \gamma = \angle APD$. In an entirely similar way, we can find the length of the other diagonal $AD$ of the quadrilateral $APDP_1$:

$\cos \beta =  \frac{(b^2 + c^2) - (d^2 + a^2)}{2(bc + da)} = \frac{(64 + 16) - (1 + 49)}{2(32 + 7)} = \frac{30}{78} = \frac{5}{13}$

$AD^2 = \frac{bc(d^2 + a^2) + da(b^2 + c^2)}{bc + da} = \frac{32 \cdot 50 + 7 \cdot 80}{32 + 7} = \frac{2160}{39} = \frac{720}{13}$

$BC = AD = \sqrt{\frac{720}{13}} = 12 \sqrt{\frac{5}{13}}  = \frac{12 \sqrt{65}}{13}$

$AD \cdot CD =  \frac{3 \sqrt{65}}{5} \cdot \frac{12 \sqrt{65}}{13} = 36 = |ABCD|$

and it follows that in this case, the defined parallelogram ABCD with the maximum area happens to be a rectangle.
____________________________________________________________________________________

Consider the same problem with s = 9, a = 6, b = 8, c = 3, d = 1. The area of the cyclic quadrilateral $APDP_1$ is now

$|APDP_1| = \sqrt{(s - a)(s - d)(s - c)(s - b)} = \sqrt{3 \cdot 1 \cdot 6 \cdot 8} = 12$

and the parallelogram area is $|ABCD| = 2|APDP_1| = 24$. In exactly the same way as above, we get

$\cos \alpha =  \frac{(a^2 + b^2) - (c^2 + d^2)}{2(ab + cd)} = \frac{(36 + 64) - (9 + 1)}{2(48 + 3)} = \frac{90}{102} = \frac{15}{17}$

$\cos \beta =  \frac{(b^2 + c^2) - (d^2 + a^2)}{2(bc + da)} = \frac{(64 + 9) - (1 + 36)}{2(24 + 6)} = \frac{36}{60} = \frac{3}{5}$

$PP_1^2 = \frac{ab(c^2 + d^2) + cd(a^2 + b^2)}{ab + cd} = \frac{48 \cdot 10 + 3 \cdot 100}{48 + 3} = \frac{780}{51} = \frac{260}{17}$

$AD^2 = \frac{bc(d^2 + a^2) + da(b^2 + c^2)}{bc + da} = \frac{24 \cdot 37 + 6 \cdot 73}{24 + 6} = \frac{1326}{30} = \frac{221}{5}$

$BA = CD = PP_1 = \sqrt{\frac{260}{17}} = \frac{2 \sqrt{5 \cdot 13 \cdot 17}}{17}$

$BC = AD = \sqrt{\frac{221}{5}} = \frac{\sqrt{5 \cdot 13 \cdot 17}}{5}$

$AD \cdot CD = \frac{2 \sqrt{5 \cdot 13 \cdot 17}}{17} \cdot \frac{\sqrt{5 \cdot 13 \cdot 17}}{5} = 26 \neq 24 = |ABCD|$

and it follows that in this case, the defined parallelogram ABCD with the maximum area is [b]not[/b] a rectangle.

.................................

 :surrender:

wow. yetti sure owned that problem. O_O

Tx. If I knew the British flag theorem (honestly, I did not :blush:), I could save myself some work. In the 1st case with s = 10, a = 7, b = 8, c = 4, d = 1:

$a^2 + c^2 = 49 + 16 = 65 = 64 + 1 = b^2 + d^2$

so that the parallelogram ABCD is a rectangle, while in the 2nd case with s = 9, a = 6, b = 8, c = 3, d = 1:

$a^2 + c^2 = 36 + 9 = 45 \neq 65 = 64 + 1 = b^2 + d^2$

so that the parallelogram ABCD is not a rectangle. On the other hand, from what I did, I can get radii of the circumcircles. In the 1st case,

$\sin \alpha = \sqrt{1 - \cos^2{\alpha}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac 3 5$

$|\triangle PDP_1| = \frac{cd}{2} \sin \alpha = \frac{12}{10} = \frac 6 5$

$c = 4,\ \ d = 1,\ \ PP_1 = \frac {3 \sqrt{65}}{5}$

$R = \frac{cd \cdot PP_1}{4|\triangle PDP_1|} = \frac{12 \sqrt{65}}{24} = \frac{\sqrt{65}}{2}$

while in the 2nd case

$\sin \alpha = \sqrt{1 - \cos^2{\alpha}} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}$

$|\triangle PDP_1| = \frac{cd}{2} \sin \alpha = \frac{24}{34} = \frac{12}{17}$

$c = 3,\ \ d = 1,\ \ PP_1 = \frac {2 \sqrt{5 \cdot 13 \cdot 17}}{17}$

$R = \frac{cd \cdot PP_1}{4|\triangle PDP_1|} = \frac{6 \sqrt{5 \cdot 13 \cdot 17}}{48} = \frac{\sqrt{5 \cdot 13 \cdot 17}}{8}$

Ha !!


geometry
3D geometry
frustum
pyramid
similar triangles

The diagram show a box that has six faces. The top and bottom faces are squares.The other faces are all identical.
what is the volume of the box.(ans 312)

Is that a frustrum? I don't really know what it is.

Looks like a frustum to me. By similar triangles (or enlargements), the "missing" pyramid, on top of the square of side 4, must have a slant height of 4. Then we can find the volume of the frustum by subtacting the volume of this little pyramid from the volume of the whole pyramid (given frustum + little pyramid), using the formula "one third the area of the base times the perpendicular height".

Hope this helps you, joshisunny16

You cannot tell from the diagram if 6 cm. is the slant height or the
altitude, but it turns out that the label of "6 cm" is the altitude.

Let $ B_1$ = area of the lower base
Let $ B_2$ = area of the upper base
Let $ h$ = the altitude

Volume = $ \frac13h(B_1 \plus{} B_2 \plus{} \sqrt {B_1B_2})$

Volume = $ \frac13(6)[100 \plus{} 16 \plus{} \sqrt {(100)(16)}]$

Volume = $ 2(116 \plus{} \sqrt {1600})$

Volume = $ 2(116 \plus{} 40)$

Volume = $ 2(156)$

Volume = $ 312$

Volume = $ 312 cm.^3$

thanks sir for the answer ,can anyone pls tell that the formula used in above post is a general formula or a derived one, pls give a detail

Sorry, joshisunny16;  I misread the diagram. Had I worked through it, I would have realised my mistake as I would not have got $ 321$.

Anyway, we can derive Arrange your tan's formula as follows (probably not the best way of doing it, but at least you'll see why it works):

Using Arrange your tan's notation and using $ g$ for the height of the "missing" little pyramid on top, similar triangles give

$ \frac{g}{\frac{\sqrt{B_2}}{2}} \equal{} \frac{g\plus{}h}{\frac{\sqrt{B_1}}{2}}$.

Cross-multiplying, we get

$ g\equal{}\frac{h\sqrt{B_2}}{\sqrt{B_1}\minus{}\sqrt{B_2}}$.

Then using the formula for the volume of a pyramid, we have

$ V \equal{} \frac{1}{3}B_{1}h \plus{} \frac{1}{3}g(B_1\minus{}B_2)$.

$ \equal{} \frac{1}{3}B_{1}h \plus{} \frac{1}{3}\frac{h\sqrt{B_{2}}}{\sqrt{B_{1}} \minus{} \sqrt{B_{2}}}(B_{1}\minus{}B_{2})$

$ \equal{}\frac{1}{3}B_{1}h \plus{} \frac{1}{3}h\sqrt{B_{2}}(\sqrt{B_{1}} \plus{} \sqrt{B_{2}})$

$ \equal{}\frac{1}{3}h(B_{1} \plus{} B_{2} \plus{} \sqrt{B_{1}B_{2}})$.

I hope that this is of some use to you.


probability trigonometry LaTeX quadratics algebra polynomial

$ 1.$ Let $ a$, $ b$, and $ c$ be the sides of a triangle. Find the number of triples $ (a, b, c)$ such that:
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} ab^2 \minus{} a^2c \minus{} ac^2 \minus{} b^2c \minus{} bc^2 \plus{} 3abc > 0$

----------------------------------------------------------------------------------------

$ 2.$ Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where $ a_2$, $ a_1$, $ a_0$ are real constants and $ |a_i| < \equal{}  2$ for $ i \equal{} 0, 1, 2$. Let $ r$ be the largest positive real number which satisfies at least one of these equations. Then $ r$ lies on which interval?

$ \textbf{(A)}\ [1,\frac {3}{2})\qquad\textbf{(B)}\ [\frac {3}{2}, 2)\qquad\textbf{(C)}\ [2,\frac {5}{2})\qquad\textbf{(D)}\ [\frac {5}{2}, 3)\qquad\textbf{(E)}\ [3,\frac {7}{2})$

----------------------------------------------------------------------------------------

$ 3.$ In the right triangle $ YUO$, $ < YUO$ is trisected and the angle trisectors
intersect $ YO$ at $ T_1$ and $ T_2$. If the order of points in $ YO$ is $ Y, T_1, T_2, O$ and it is given that $ YU \equal{} 54$ and $ OU \equal{} 47$, find the remainder when $ (YU \plus{} T_2U)(T_1U \plus{} OU)$ is divided by $ 35$.

----------------------------------------------------------------------------------------

$ 4.$ Let $ S$ be the set of all $ (x, y)$ ordered pairs such that $ x, y$ are positive integers that satisfy:

$ x \equal{} \frac{6 \minus{} x}{y^2 \minus{} x}$

What is the probability that randomly chosen ordered pair in set $ S$ have $ x > y$?

#4 isn't so interesting. it's law of cosines bash. i'll leave the grueling calculation to you

Oh, whoops. It's a very annoying problem.

I'll replace it with something else.

[hide=" Arguably Cheap Solution for 1"]By schur's $ \sum{x^{3}} + \sum{xyz} - \sum{x^{2}y + xy^{2}} \ge{0}$. Equality holds iff $ x = y = z$. So all non-equilateral triangles work.[/hide]

[hide="Intended Solution"]We can rewrite as $ \sum_\text{cyc}a(b^2+c^2-a^2)<{3abc}\implies{2abc(\cos{A}+\cos{B}+\cos{C})}<3abc$

$ \implies{\cos{A}+\cos{B}+\cos{C}<\frac{3}{2}}$ which is always true unless it is an equilateral triangle in which case we have equality. [/hide]

[quote="veezbo"]$ 1.$ Let $ a$, $ b$, and $ c$ be the sides of a triangle. Find the number of triples $ (a, b, c)$ such that:
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} ab^2 \minus{} a^2c \minus{} ac^2 \minus{} b^2c \minus{} bc^2 \plus{} 3abc < 0$

----------------------------------------------------------------------------------------

$ 2.$ Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where $ a_2$, $ a_1$, $ a_0$ are real constants and $ |a_i| < \equal{}  2$ for $ i \equal{} 0, 1, 2$. Let $ r$ be the largest positive real number which satisfies at least one of these equations. Then $ r$ lies on which interval?

$ \textbf{(A)}\ [1,\frac {3}{2})\qquad\textbf{(B)}\ [\frac {3}{2}, 2)\qquad\textbf{(C)}\ [2,\frac {5}{2})\qquad\textbf{(D)}\ [\frac {5}{2}, 3)\qquad\textbf{(E)}\ [3,\frac {7}{2})$

----------------------------------------------------------------------------------------

$ 3.$ In the right triangle $ YUO$, $ < YUO$ is trisected and the angle trisectors
intersect $ YO$ at $ T_1$ and $ T_2$. If the order of points in $ YO$ is $ Y, T_1, T_2, O$ and it is given that $ YU \equal{} 54$ and $ OU \equal{} 47$, find the remainder when $ (YU \plus{} T_2U)(T_1U \plus{} OU)$ is divided by $ 35$.

----------------------------------------------------------------------------------------

$ 4.$ Let $ S$ be the set of all $ (x, y)$ ordered pairs such that $ x, y$ are positive integers that satisfy:

$ x \equal{} \frac {6 \minus{} x}{y^2 \minus{} x}$

What is the probability that randomly chosen ordered pair in set $ S$ have $ x > y$?[/quote]

----------------------------------------------------------------------------------------

EDIT: ARGH. So apparently the mock had a mistake. For #1, the sign should be a less-than sign. It's fixed now.

Oh... the solution is still pretty much the same.

[hide="1"]We can rewrite as $ \sum_\text{cyc}a(b^2 + c^2 - a^2) > {3abc}\implies{2abc(\cos{A} + \cos{B} + \cos{C})} > 3abc$

$ \implies{\cos{A} + \cos{B} + \cos{C} > \frac {3}{2}}$ which is impossible if $ A + B + C = 180^{\circ}$ so there are $ \boxed{0}$ solutions. [/hide]

[hide][quote="veezbo"]$ 1.$ Let $ a$, $ b$, and $ c$ be the sides of a triangle. Find the number of triples $ (a, b, c)$ such that:
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} ab^2 \minus{} a^2c \minus{} ac^2 \minus{} b^2c \minus{} bc^2 \plus{} 3abc > 0$

----------------------------------------------------------------------------------------

$ 2.$ Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where $ a_2$, $ a_1$, $ a_0$ are real constants and $ |a_i| < \equal{}  2$ for $ i \equal{} 0, 1, 2$. Let $ r$ be the largest positive real number which satisfies at least one of these equations. Then $ r$ lies on which interval?

$ \textbf{(A)}\ [1,\frac {3}{2})\qquad\textbf{(B)}\ [\frac {3}{2}, 2)\qquad\textbf{(C)}\ [2,\frac {5}{2})\qquad\textbf{(D)}\ [\frac {5}{2}, 3)\qquad\textbf{(E)}\ [3,\frac {7}{2})$

----------------------------------------------------------------------------------------

$ 3.$ In the right triangle $ YUO$, $ < YUO$ is trisected and the angle trisectors
intersect $ YO$ at $ T_1$ and $ T_2$. If the order of points in $ YO$ is $ Y, T_1, T_2, O$ and it is given that $ YU \equal{} 54$ and $ OU \equal{} 47$, find the remainder when $ (YU \plus{} T_2U)(T_1U \plus{} OU)$ is divided by $ 35$.

----------------------------------------------------------------------------------------

$ 4.$ Let $ S$ be the set of all $ (x, y)$ ordered pairs such that $ x, y$ are positive integers that satisfy:

$ x \equal{} \frac {6 \minus{} x}{y^2 \minus{} x}$

What is the probability that randomly chosen ordered pair in set $ S$ have $ x > y$?[/quote][/hide]
[hide="LaTeX hints"]
2. for less than or equal to, try $ \ge$ (\ge)
Anyways, what is that paragraph sign doing there?
to make the brackets cover the fractions, use \left and \right as follows
$ \left[ 1, \frac{3}{2} \right)$
3. I prefer $ \angle$ (\angle) to $ <$
[/hide]

[hide="number 2"]
as r increases and becomes 'more positive', the quadratic, linear, and constant terms must compensate more (by being 'more negative')
Therefore, our largest r appears when $ a_2\equal{}a_1\equal{}a_0\equal{}\minus{}2$
Let the polynomial be $ P(x)$
$ P(3)\equal{}1, P(2.5)\equal{}\minus{}3.875$
This tells us that r is in the interval in choice D
[/hide]
[hide="number 4"]
arg i have to go, but the fact that x and y are positive integers tells is that $ 6\minus{}x \ge y^2\minus{}x \implies 6 \ge y^2 \implies y\equal{}1, 2$
Then we have casework
[/hide]


For $ x>0$, the perimeter of the triangle with vertices $ (0,0)$, $ (x,0)$ and $ (x,5)$ is 30 units. What is the value of $ x$?

The triangle's base is $ x$, and its height is $ 5$, so $ \frac{1}{2}bh\equal{}\frac{5}{2}x\equal{}30\Rightarrow x\equal{}12$.

Although Brut3forc3 got the right answer, it wasn't the right way to do the problem. He was calculating the area in stead and coincidentially got the same correct answer.

you have points (0,0),(x,0),(x,5). It forms a right triangle. And, you can see that the lengths of the base and height(leg) are x and 5, respectively. The only pythagorean triple containing 5 is 5-12-13. So, x=12.


In my humble opinion it would be a good idea to add the possibility to search not only in subforums but also in the "favorites" selection. Or is it too hard to implement? Or, maybe, it is already possible and I just do not know how  :?

It's not but I see no point. You cannot possibly have THAT many favourite topics :D

[quote="Valentin Vornicu"]You cannot possibly have THAT many favourite topics :D[/quote]
Well, as a matter of fact my intention was to add to favorites everything I like and leave it there. One thing I need it for is to create a collection of nice problems for Putnam team coaching and it is quite tiresome to copy the links to a file and then scratch your head recalling what the post #12345 was about. It accumulates at the rate of about 1 thread a day, so in a year I'll have about 20 pages. It will still be feasible for a manual search but quite time consuming. Perhaps it is a misuse of the "favorite" section, but indeed, what other options can you offer for creating your own "mini-forum" consisting of the posts you like that you can manage efficiently?

[quote="fedja"]What other options can you offer for creating your own "mini-forum" consisting of the posts you like that you can manage efficiently?[/quote]None except viewing favorites, watched topics (will be released soon) and another function (also in tests right now). It is not in our main focus at this point. As you said above you can easily maintain a collection of links, I've specially created easy to copy-paste shortcuts of the topics and posts.

[quote="Valentin Vornicu"]None except viewing favorites, watched topics (will be released soon) and another function (also in tests right now).[/quote]
Thanks! I'll impatiently wait for their release and will really appreciate if one of them comes with a search function.  :)  :P  As to collection of links, that's what I am doing now too but the only way to search in them I know is to view them one by one... :(

Well, thanks for your time anyway, [b]Valentin[/b].


Let $ x,y,z > 0$ , such that $ x \plus{} y \plus{} z \equal{} xyz$. Prove that
$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \leq 6 \sqrt{3} \minus{} 10$

The condition $ x \plus{} y \plus{} z \equal{} xyz$ implies $ x \equal{} \frac {y \plus{} z}{yz \minus{} 1}$, therefore $ yz > 1$. Hence at least two of the numbers $ x,y,z > 1$, therefore we may assume that every number is greater than $ 1$.
Denote $ a \equal{} x \minus{} 1>0$, $ b \equal{} y \minus{} 1>0$, $ c \equal{} z \minus{} 1>0$. Hence $ x \plus{} y \plus{} z \equal{} xyz \Leftrightarrow 2 \equal{} abc \plus{} ab \plus{} bc \plus{} ca$ and it's left to prove that $ abc \le 6\sqrt3 \minus{} 10$.
From the condition $ 2 \equal{} abc \plus{} ab \plus{} bc \plus{} ca$ we get
$ 2 \minus{} abc \equal{} ab \plus{} bc \plus{} ca \ge 3\sqrt [3]{a^2b^2c^2} \Leftrightarrow (2 \minus{} abc)^3 \ge 27a^2b^2c^2 \Leftrightarrow (abc \plus{} 1)( \minus{} a^2b^2c^2 \minus{} 20abc \plus{} 8) \ge 0$.
Therefore, $ (abc)^2 \plus{} 20abc \minus{} 8 \le 0 \Leftrightarrow (abc \minus{} ( \minus{} 10 \plus{} 6\sqrt3))(abc \minus{} ( \minus{} 10 \minus{} 6\sqrt3)) \le 0$. Hence $ abc \minus{} (6\sqrt3 \minus{} 10) \le 0 \Leftrightarrow abc \le 6\sqrt3 \minus{} 10$ as desired.


Why is it that $ 12^4\equiv 8^2\equiv \minus{}4\pmod{34}$, and $ 12^{32}\equiv(\minus{}16)^2\equiv \minus{}16 \pmod{34}$ and $ 12^{64}\equiv (\minus{}16)^2\equiv \minus{}16\pmod{34}$? Sometimes you are squaring and building of previous powers of 12 (as with $ 12^{32}$) and $ 12^{64}$, but something more is used.

Thanks!

Notice that any positive power of $ 12$ is $ 0\mod{2}$.  Now notice that $ \minus{}16 \equiv 1\mod{17}$.  Now think Chinese Remainder Theorem.  In other words, it all depends on the residue of $ 12^n \mod{17}$.


In 1772 $Euler$ discovered the curious fact that
$n^{2}+n+41$
is prime when$n = 0, 1, 2, 3, ..., 39$. Show that there exist $40$consecutive integer values of $n$ for which this polynomial is NOT prime.


[quote="vinskman"][quote="enndb0x"][quote="vinskman"]Prove that $ 1010\cdots 101$ is a composite number, where the integer mentioned contains $ 1998 \text { 0's},1999 \text{ 1's}$.

 :roll:[/quote]

We can write 'your' number  as :

                        $ 10^{3996} \plus{} 10^{3994} \plus{} 10^{3392} \plus{} 10^{3990} \plus{} ... \plus{} 10^{2} \plus{} 10^{0} \equal{} 10^{3994}(10^{2} \plus{} 1) \plus{} 10^{3990}(10^{2} \plus{} 1) \plus{} ...10^{2} \plus{} 1 \equal{} (10^{2} \plus{} 1)(10^{3994} \plus{} 10^{3990} \plus{} ...1)$
               so it is composite

[i]I don't know why LaTeX looks like this [/i]  :mad:[/quote]

Sorry, from $ 0$ to $ 3996$, there are $ 1 \plus{} 1998 \equal{} 1999$ terms, how can you join perfect pairs at all in

$ 10^{3994}(10^{2} \plus{} 1) \plus{} 10^{3990}(10^{2} \plus{} 1) \plus{} ...10^{2} \plus{} 1$?[/quote]

Anybody has idea?

A brief check shows that 1999 is prime, thus we cannot split
\[ 10^{3996}\plus{}10^{3994}\plus{}\dots\plus{}10^{2}\plus{}10^{0}
\]
into $ m$ groups of $ n$ terms, where $ m,n<1999$.

So it must be shown another way.

Ah I completely missed that particular type of factorization :P

So in fact:
$ 10^{3996}\plus{}10^{3994}\plus{}\dots\plus{}10^2\plus{}1\\
\equal{}(10^{1998}\plus{}10^{1997}\plus{}\dots\plus{}10\plus{}1)(10^{1998}\minus{}10^{1997}\plus{}\dots\minus{}10\plus{}1)$

Since both terms are clearly less than the original value and greater than 1, they must be two non-trivial factors of the number, and therefore it is composite. :D

[quote="vinskman"]Prove that $ 1010\cdots 101$ is a composite number, where the integer mentioned contains $ 1998 \text { 0's},1999 \text{ 1's}$.

 :roll:[/quote]

[color=brown]Another solution:[/color]

[hide]
Let  $ K\equal{}\underbrace{1010\cdots 101}_{1999 \text{ 1's} ,1998 \text{ 0's}}$

Then $ 10K\plus{}K\equal{}\underbrace{111\cdots 111}_{3998\text{ 1's}}$

$ \implies 11K\equal{}\frac{1}{9}\underbrace{999\cdots 999}_{3998\text{ 9's}}$

$ \implies 11K\equal{}\frac{1}{9}(10^{3998}\minus{}1)$

$ \implies K\equal{}\frac{(10^{1999}\plus{}1)}{11} \cdot \frac{(10^{1999}\minus{}1)}{9}$

$ \implies K\equal{}mn ,\text{integer } m\equal{}\frac{(10^{1999}\plus{}1)}{11} ,\text{integer } n\equal{}\frac{(10^{1999}\minus{}1)}{9}$

$ \because (10^{1999}\plus{}1)$ is divisible by $ 11$ , and $ (10^{1999}\minus{}1)$ is divisible by $ 9$

[/hide] :bruce:

[quote="enndb0x"]sorry vinskman but I still think my solution is correct : $ 101010...01$ , $ 1998 0's$ and $ 1999 1's$ we can write as
 $ 100...0 ({3996} \minus{} 0's) \plus{} 100...0 ({3994} \minus{} 0's) \plus{} 100...0 ({3992} \minus{} 0's) \plus{} ... \plus{} 100 \plus{} 1$
and this is equal to : $ 10^{3996} \plus{} 10^{3994} \plus{} 10^{3992} \plus{} 10^{3990} \plus{} ... \plus{} 10^{2} \plus{} 1$
 We split into group with 2 elements : $ (10^{3996} \plus{} 10^{3994}) \plus{} (10^{3992} \plus{} 10^{3990}) \plus{} ... \plus{} (10^{2} \plus{} 1)$
by factorization we see  :   $ (10^{3994}(10^{2} \plus{} 1)) \plus{} (10^{3992}(10^{2} \plus{} 1)) \plus{} ...(10^{2} \plus{} 1)$
                                      $ (10^{2} \plus{} 1)(10^{3994} \plus{} 10{3992} \plus{} 10^{3990} \plus{} 10^{3988} \plus{} ... \plus{} 1)$ so it is composite .

Here : $ 100...0 ({3996} \minus{} 0's) \plus{} 100...0 ({3994} \minus{} 0's) \plus{} 100...0 ({3992} \minus{} 0's) \plus{} ... \plus{} 100 \plus{} 1$ the  $ (3996 \minus{} 0's)$ tells that the number $ 100...0$ have $ 3996$ zeroes.

P.S- Sorry for bad english ! :wink:[/quote]

Dear enndb0x,

when we consider the sequence of the powers$ \left\{0,2,4,6,8,\cdots,3994,3996\right\}$ .It is an A.P.,so
$ 3996 \equal{} 0 \plus{} 2(n \minus{} 1)\implies n \equal{} \frac {3996}{2} \plus{} 1 \equal{} 1999\cdots$ an odd number.

i.e. there are 1999 terms from $ 10^0$ to $ 10^{3996}$

How can you group all terms in the series evenly pairwised without missing some term?

Your grouped  terms stated in the above may be a joke !!
:lol:  :D

[quote="enndb0x"]example : Prove that the number $ 1010101$ is composite ?

$ 1010101 \equal{} 100000 \plus{} 10000 \plus{} 100 \plus{} 1 \equal{} 10^6 \plus{} 10^4 \plus{} 10^2 \plus{} 1 \equal{} (10^6 \plus{} 10^4) \plus{} (10^2 \plus{} 1) \equal{} (10^{4 \plus{} 2} \plus{} 10^4) \plus{} (10^2 \plus{} 1) \equal{} (10^4 \cdot 10^2 \plus{} 10^4) \plus{} (10^2 \plus{} 1) \equal{} 10^4(10^2 \plus{} 1) \plus{} (10^2 \plus{} 1) \equal{} (10^2 \plus{} 1)(10^4 \plus{} 1)$

I don't know what is the problem here  :o  , and I am not joking , I really think this is the solution  or ...maybe I missunderstood the problem$ ^1$ , or you are not understanding the solution $ ^2$ (probably $ ^1$)[/quote]

If the numbers are in the following types:

$ 101010101\equal{}10^8\plus{} 10^6 \plus{} 10^4 \plus{} 10^2 \plus{} 1$

$ 1010101010101\equal{}10^{12}\plus{}10^{10}\plus{}10^{8}\plus{}10^6 \plus{} 10^4 \plus{} 10^2 \plus{} 1$

How can you use such ways?

hmm ...It seems the way I used is OK  just for odd number of digits.Anyway ...1998+1999=3997 is odd   :wink:

[quote="Flame"]enndb0x, you are trying to divide odd number by 2 and get integer which is clearly impossible. Let me explain: as vinskman already proved, there are $ 1999$ terms in your sequence ($ 10^{3996} \plus{} 10^{3994} \plus{} 10^{3992} \plus{} 10^{3990} \plus{} ... \plus{} 10^{2} \plus{} 1$) so you cannot split it in any number of pairs. Let's prove this by contradiction: let $ x$ be the number of pairs (which must be a positive integer), then total number of terms in sequence is $ 2x$, hence $ 2x \equal{} 1999 \implies x \equal{} \frac {1999}{2}$ but this is contradicting the fact that $ x$ must be a positive integer.[/quote]

Yes, it is a joke to get odd no. of terms making each pair in 2 members.

Haha...

I think it is quite weak in basic concept in Maths.

[quote="cdymdcool"]is there any ways to factor x^7+x^5+x^3+x^1+1 and x^6+x^4+x^2+1?[/quote]

The first polynomial, $ x^7+x^5+x^3+x^1+x^0$ is in no way a "pretty" polynomial to factor, considering the fact that the the exponents in the last two terms differ by 1, not 2.

The second polynomial, $ x^6+x^4+x^2+1$, has a pretty simple factoring by grouping.
\begin{align*}
x^6+x^4+x^2+1 &= x^4(x^2+1)+1(x^2+1) \\
&= (x^4+1)(x^2+1)
\end{align*}

btw, yongyi, you can factor further:

$ x^4 \plus{}1 \equal{} x^4\plus{}1 \plus{}2x^2 \minus{} 2x^2 \equal{} (x^2\plus{}1)^2 \minus{} 2x^2 \equal{} (x^2 \minus{} \sqrt{2}x\plus{}1)(x^2 \plus{} \sqrt{2}x\plus{}1)$

[quote="facis"]btw, yongyi, you can factor further:

$ x^4 \plus{} 1 \equal{} x^4 \plus{} 1 \plus{} 2x^2 \minus{} 2x^2 \equal{} (x^2 \plus{} 1)^2 \minus{} 2x^2 \equal{} (x^2 \minus{} \sqrt {2}x \plus{} 1)(x^2 \plus{} \sqrt {2}x \plus{} 1)$[/quote]

Right, but that kind of factorization will not help in determining whether, for example, $ 10^4\plus{}1$ is prime (since $ x^2\pm\sqrt 2 x \plus{} 1$ isn't an integer anyway).

yeah I wasn't thinking about the context of the discussion when I posted, just got caught up with factoring.

[quote="enndb0x"]thnx Flame , and I was not joking vinskman ...I was just drunk  :lol:
sorry to all ![/quote]

Sorry to enndb0x, believe you not in joking. In my mind , the  main purposes  to write much more on the discussing board is that

(1) sharing the perfectness in Mathematical world together.
(2) reasoning discuss to understand what is the reality .
(3) enrich our insight from  other personal views.

Do you agree?

 :)


[quote="kohjhsd"]Find the value of $\lfloor 100x \rfloor$ if
\[\left\lfloor x+\frac{19}{100}\right\rfloor+\left\lfloor x+\frac{20}{100}\right\rfloor+\left\lfloor x+\frac{21}{100}\right\rfloor+\cdots+\left\lfloor x+\frac{54}{100}\right\rfloor=321. \]
[/quote]

All the numbers can be just 1 integer or they can be n and n+1.  You just have to find how many n's and n+1's.

To finish out the problem

There are 54-19+1 terms so 36 terms.  Divide 321 by 36 and be get a fractional number between 8 and 9, meaning it must be seperated into n and n+1 (remember that it must be n and n+1 because (54-19)/100 is less than 1).  Well n has to be 8 and n+1 is 9 becuase when we divided, the numbers were between there.  Ok there are 36 terms

So a+b=36
8a+9b=321

solve the system of equation

8a+8b=288
8a+9b=321

-b=-33, b=33 a=3

ok we start counting up
floored{x+21/100} must equal 8 because there are 3 terms that equal 8
however floored{x+22/100} must equal 9 becuase the next term is 9
so if floored{x+21/100}=8 and the next term by simply adding 1/100 is 9, we assume that x+21/100=somewehre between (8+99/100) and (9).  So x=somewehre between (8+78/100) and (8+79/100).
Well, it doesnt matter what it is because once be do floor{100x} to it it must be 878, so no there are no multiple solutions

[quote="kohjhsd"]Find the value of $\lfloor 100x \rfloor$ if
\[\left\lfloor x+\frac{19}{100}\right\rfloor+\left\lfloor x+\frac{20}{100}\right\rfloor+\left\lfloor x+\frac{21}{100}\right\rfloor+\cdots+\left\lfloor x+\frac{54}{100}\right\rfloor=321. \]
[/quote]
This is like an old AIME problem, isn't it?
[hide]
There are 36 terms.
x must be between 8 and 9.
The value of all but three of these must be 8.
So $x+\frac{52}{100}=9$
$100x=848$[/hide]

[quote="diophantient"][quote="kohjhsd"]Find the value of $\lfloor 100x \rfloor$ if
\[\left\lfloor x+\frac{19}{100}\right\rfloor+\left\lfloor x+\frac{20}{100}\right\rfloor+\left\lfloor x+\frac{21}{100}\right\rfloor+\cdots+\left\lfloor x+\frac{54}{100}\right\rfloor=321. \]
[/quote]
This is like an old AIME problem, isn't it?

There are 36 terms.
x must be between 8 and 9.
The value of all but three of these must be 8.
So $x+\frac{52}{100}=9$
$100x=848$[/quote]

Wait why is $x+\frac{52}{100}=9$, I thought it was $x+\frac{22}{100}=9$
So it would be $878$


Let $ S$ be the set of all pairs of real numbers $ (\alpha;\beta)$ such that equation $ x^3 \minus{} 6x^2 \plus{} \alpha x \minus{} \beta \equal{} 0$ has three real roots(not necessarily disctint) and they are all greater than $ 1$.Find the maximum value of $ T \equal{} 8\alpha \minus{} 3\beta$ for $ (\alpha;\beta) \in S$

Let $ p,\ q,\ r$be three roots of the equation, we can write $ f(x): \equal{} x^3 \minus{} 6x^2 \plus{} \alpha x \minus{} \beta \equiv (x \minus{} p)(x \minus{} q)(x \minus{} r)$.

$ \therefore f'(x) \equal{} 3x^2 \minus{} 12x \plus{} \alpha \equal{} (x \minus{} q)(x \minus{} r) \plus{} (x \minus{} p)(x \minus{} r) \plus{} (x \minus{} p)(x \minus{} q)$.

By Vieta's formula, $ p \plus{} q \plus{} r \equal{} 6,\ pq \plus{} qr \plus{} rp \equal{} \alpha ,\ pqr \equal{} \beta$.

$ p > 1,\ q > 1,\ r > 1\Longleftrightarrow \left\{ \begin{array}{ll} (p \minus{} 1) \plus{} (q \minus{} 1) \plus{} (r \minus{} 1) > 0 & \quad \\

(p \minus{} 1)(q \minus{} 1) \plus{} (q \minus{} 1)(r \minus{} 1) \plus{} (r \minus{} 1)(p \minus{} 1) > 0 & \quad \\

(p \minus{} 1)(q \minus{} 1)(r \minus{} 1) > 0 & \quad \end{array} \right.\Longleftrightarrow \left\{ \begin{array}{ll} p \plus{} q \plus{} r > 3 & \quad \\

f'(1) > 0 & \quad \\

f(1) < 0 & \quad \end{array} \right.\Longleftrightarrow 6 > 3,\ \alpha \minus{} 9 > 0,\ \beta > \alpha \minus{} 5\ \cdots (*)$.

$ \therefore T \equal{} 8\alpha \minus{} 3\beta < 8\alpha \minus{} 3(\alpha \minus{} 5) \equal{} 5(\alpha \plus{} 3)$

$ \equal{}5(pq\plus{}qr\plus{}rp\plus{}3)\leq 5\cdot \frac{(p\plus{}q\plus{}r)^2}{3}\plus{}15\equal{}\boxed{75}$.

The equality hods when $ p\plus{}q\plus{}r\equal{}6, p\equal{}q\equal{}r\Longleftrightarrow p\equal{}q\equal{}r\equal{}2\Longleftrightarrow \alpha \equal{}12,\ \beta \equal{}8$,

which satisfy the condition $ (*)$.

If p,q and r are the roots, then $ p\plus{}q\plus{}r \equal{} 6; \sum pq \equal{} \alpha; pqr \equal{} \minus{}\beta$

It is given that $ p,q,r \ge 1 >0$

We have $ (p\plus{}q\plus{}r)^2 \ge 3 \sum pq \Rightarrow \sum pq \le 12$

Also $ \sqrt [3] {pqr} \le \frac{p\plus{}q\plus{}r}{3} \equal{} 2 \Rightarrow pqr \le 8$

Hence $ \alpha \le 12$ and $ \minus{}\beta \le 8$

Hence $ T \equal{} 8 \alpha \minus{} 3 \beta \le \boxed {120}$

I think there is a mistake in your solution kunny.
if $ \alpha\equal{}12;\beta\equal{}8$ then $ T\equal{}12.8\minus{}8.3\equal{}72$.
 $ \beta<\alpha\minus{}5$ so the equality can't hold. :D

[b]dulldrake[/b] —

$ \minus{}3\beta < \minus{}3(\alpha \minus{} 5) \Longleftrightarrow \beta > \alpha \minus{} 5$. Remember that you're dividing by a negative number.

oh..I'm sorry :blush:
but the equality can't hold
sorry for my bad english


[color=blue]There're $N$ tortoises, each one is on the point of the regular polygon with $N$ edges. The length of one edge is $a$
Each tortoise is chasing the one next to him with the velocity $V$.
Find the time $\Delta T$for the tortoise  to catch the one next to him.
Here is the answer
[hide]$\large\Delta T=\frac{a}{2V\sin^2\frac{\pi}{N}}$[/hide]
[/color]

Haha, bugs, ants, dogs, and now turtles.   :lol:

Anyway, the trick is
[hide]
realize the radial component of the turtle's velocity is constant
[/hide]

[color=blue]It's a solution. There's another solution to this problem. Pay attention that the velocity of each turtle to his next one is also a constant.[/color]

i could never understand this problem.
can someone explain it in a detailed manner. thanks.

Taking the relative velocity of approach is the BEST method for the fastes solution...

additional questions...

What is they are moving with constant acceleration?
Find the distance travelled by each turtle?
Find the equation of the locus each turtle follows!?


geometry
3D geometry
tetrahedron
trigonometry
inradius
incenter
sphere

[b]i.)[/b]  Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tedrahedron, each of whose edges has length $ s,$ is circumscribed around the balls. Find the value of $ s.$

[b]ii.)[/b] Suppose that $ ABCD$ and $ EFGH$ are opposite faces of a retangular solid, with $ \angle DHC \equal{} 45^{\circ}$ and $ \angle FHB \equal{} 60^{\circ}.$ Find the cosine of $ \angle BHD.$

Let r be the ball radius. (This is better than r = 1, because equations can be checked for dimensions.) Let $\sigma$ be a horizontal plane tangent to the top ball from the bottom. The top tetrahedron portion cut off by the plane $\sigma$ is still a regular tetrahedron with one insphere of radius r. Let $s_1, h_1$ be the edge and height of this regular tetrahedron, $h_1 = s_1 \sqrt{\frac 2 3}$ (which is simple to calculate by a double use of Pythagorean theorem). Let $F_1, S_1, V_1$, be the face area, surface and volume of this regular tetrahedron, $F_1 = \frac{s_1^2 \sqrt 3}{4}$, $S_1 = 4 F_1$ and $V_1 = \frac{h_1 F_1}{3}$. The inradius is given by $r = \frac{3V_1}{S_1} = \frac{h_1 F_1}{4 F_1} = \frac{h_1}{4}$, just like for a triangle. Thus we have $h_1 = 4r$ for the height of this regular tetrahedron. The height of the top ball center above the plane $\sigma$ is equal to r. The centers of the 4 balls also form a regular tetrahedron with the edge $s_C = 2r$ and height $h_C = s_C \sqrt{\frac 2 3} = 2r \sqrt{\frac 2 3}$. The centers of the bottom 3 balls are at the height r above the floor. Thus the total height of the entire regular tetrahedron is $h = r + h_C + h_1 - r = h_C + h_1 = 2r \sqrt{\frac 2 3} + 4r$ and its edge $s = h \sqrt{\frac 3 2} = \left(2r \sqrt{\frac 2 3} + 4r\right) \sqrt{\frac 3 2} = 2r \left(1 + \sqrt 6\right)$.

[quote="yetti"]The inradius is given by $ r \equal{} \frac {3V_1}{S_1} \equal{} \frac {h_1 F_1}{4 F_1} \equal{} \frac {h_1}{4}$, just like for a triangle. [/quote]
How is this formula derived?

If I is incenter and r inradius of a tetrahedron ABCD, its volume V(ABCD) is equal to the sum of 4 tetrahedron volumes

$ V(ABCD) \equal{} V(IABC) \plus{} V(IBCD) \plus{} V(ICDA) \plus{} V(IDAB) \equal{}$
$ \equal{} \frac{_r}{^3} S(ABC) \plus{} \frac{_r}{^3} S(BCD) \plus{} \frac{_r}{^3} S(CDA) \plus{} \frac{_r}{^3} S(DAB)  \equal{} \frac{_r}{^3} S(ABCD)$

Oh yeah... I should have realized that, but I don't really see your claim for $ h_1\equal{}4r$ because the top vertex of the tetrahedron does not intersect with the highest point of the topmost sphere.

$ h_1$ is height of a tetrahedron tangent to one sphere (that's why I put the index 1 there), not to the tetrahedron tangent to 4 spheres. To get height $ h$ of the tetrahedron tangent to 4 spheres, you have to add the height $ h_C$ of a tetrahedron formed by the centers of the 4 spheres, $ h \equal{} h_1 \plus{} h_C.$


If the first ineq is true, then the second ineq is also true.

$\sqrt[3]{\frac{\sum\limits_{cyc}(\frac{a}{a+b})^{3}}{3}}\geq \sqrt{\frac{\sum\limits_{cyc}(\frac{a}{a+b})^{2}}{3}}$

Hence $\sum\limits_{cyc}(\frac{a}{a+b})^{3}\geq 3 \cdot \sqrt{\frac{\sum\limits_{cyc}(\frac{a}{a+b})^{2}}{3}}^{3}\geq 3 \cdot \sqrt{\frac{1}{4}}^{3}=\frac{3}{8}$

So the importance is solving the first inequality, who can solve it? :D

I think you should move this question to the  'inequalities' section...which is more suitable for this problem


conics
ellipse
geometry
calculus
integration
3D geometry
tetrahedron

1. find the circumference of the following figure:  $ 12x^2 \minus{} 300x \plus{} 9y^2 \plus{} 234y \plus{} 3288 \equal{} 0$.
(I don't know how to find the circumference of an ellipse)

2. Find the volume of the figure enclosed by (4,2,-3), (2,5,6,), (-3,1,5), and (3,1,4).
I could use geometry and a lot of right triangles to solve this one, but I get the feeling there is probably a nicer analytical way to do this, especially since I know almost no analytic geo.

Thanks in advance

1)  There is no nice way to find the circumference of an ellipse.  The answer is given by what's called an elliptic integral.  Are you sure this is the intended question?

2)  There is a [url=http://www.mathlinks.ro/Wiki/index.php/Shoelace_Theorem]nice analytic way[/url].

yea, I find it weird too, and I checked that I got the question straight. This test should not require calculus...

is there another way to compute the exact value of the circumference?

No.  The only possible explanation is that the ellipse is degenerate.  Have you actually completed the square?

Yea, I got $ \frac {(y \plus{} 13)^2}{12} \plus{} \frac {(x \minus{} \frac {25}{2})^2}{9} \equal{} 1$ this morning, so it came out quite nicely...

EDIT: Hm, this will help me understand how to do this, but the answer choices given are

A. $ \sqrt {42}\pi$    B. $ 7\pi$  C. $ \sqrt {6}\pi$   D. $ (3 \plus{} 2\sqrt3)\pi$    E. none of the answers

EDIT: for a polyhedron with more than 4 vertices, does the order of the points matter when I do shoelace? Just wondering...
I can't seem to easily find a lot of information on 3D use of shoelace, although I see how it works now.

hm, I think this should be taken off the vast majority of middle school and high school tests. The students that care to do well on those tests often learn this method for efficiency without being able to prove it until several years later (like me, and I still can't prove it..)

In the stupid formula booklet my teacher gave me that I try to never consult, there is a formula for the circumference: $ \sqrt{\frac{a^2\plus{}b^2}{2}}$.

Does this give the exact value or is it just an approximation?

[quote="hurdler"]the answer choices given are

A. $ \sqrt {42}\pi$    B. $ 7\pi$  C. $ \sqrt {6}\pi$   D. $ (3 \plus{} 2\sqrt3)\pi$    E. none of the answers[/quote]
Then the answer has to be none of the above.  The answer is not anywhere near this nice.

[quote="hurdler"]In the stupid formula booklet my teacher gave me that I try to never consult, there is a formula for the circumference: $ \sqrt {\frac {a^2 \plus{} b^2}{2}}$.

Does this give the exact value or is it just an approximation?[/quote]
Approximation.  Elliptic integrals are called that for a reason - they don't evaluate to any nicer function.

EDIT: that should have been $ 2\pi \sqrt{\frac{a^2 + b^2}{2}}$

yea, you're right. Ok, thanks.

How exactly do you do shoelace on 3 dimensions?

There is no direct analogue of the shoelace theorem in three dimensions (sorry, I misread that question).  This is because the [url=http://en.wikipedia.org/wiki/Scalar_triple_product]triple scalar product[/url] doesn't have quite the same properties as the cross product.

Edit:  Fortunately, four points determine a tetrahedron, so once you find three vectors that determine the tetrahedron you can use the triple scalar product instead.

If you are stuck with question 2 you could always do it the long way and find the volume of the pyramid

1. you can find the area of of the plane between the points (4,2,-3),(2,5,6) and (-3,1,5) by using the distance formula to find the length between the points and aplying herons formula. This is the base of the pyramid

2. To find the height of the pryamid we have to find the direction that is perpendicular to the side of the base. So find the vectors between (4,2,-3),(2,5,6) and (4,2,-3),(-3,1,5) and (2,5,6) ,(-3,1,5). Then create three simultaneous equations by multiplying each vector with $ pi \plus{} qj \plus{} rk$ and making it equal zero
(because if a.b=0 then they are perpendicular - dot product)

3. take your perpendicular vector (make sure it is in unit vector form) and apply scalar resolute to find the exact height of the pyramid

4. now you have the area of the base and height so plug it into v=(1/3)A.h


Is this class appropriate for an entering 6th grader?

My daughter is taking MathCounts Problem Series now as an entering 5th grader, but I sit next to her during the class to assist her.  The other kids are a lot quicker than she is, but she gets the basic points.

The MATHCOUNTS problem series doesn't have a pre-test or a post-test.

[quote="MithsApprentice"]which MC class? problem series?

It's probably illogical to estimate your problem solving ability based on what grade you're in. Try to take the Pre-test and Post-test evaluations.[/quote]

Yes, the Mathcounts Problem Series (sorry for leaving that out).  I don't think the problem series classes have pre- and post- tests - do they?  My question is: given the standard body of knowledge one has at the beginning of 6th grade, could he/she handle the class?

The problem series classes do not have a pre- or post- test.

As for the MATHCOUNTS class... The class is currently designed to help students who can already solve most MATHCOUNTS problems.  Some 5th and 6th graders are able to handle the problems well.  Others might need more curriculum to make the most of the class.

We eventually plan on hosting two levels of MATHCOUNTS classes.  The current class will become the harder of the two classes while a more basic class will be added to help students who are not as experienced with harder algebra and geometry problems.  The curriculum will be a bit more basic and will be designed to help a student move smoothly into the harder class.  The more basic MATHCOUNTS class will not be available until at earliest the summer of 2005.


pigeonhole principle
inequalities
absolute value
combinatorics unsolved
combinatorics

Size of chessboard is $2k+1$ x $2k+1$. In each fild we write numbers $-1,0,1$. Lets $a_{1},...a_{2k+1}$ be the sums of numbers in rows and $a_{2k+2},...a_{4k+2}$ be the sums of numbers in columns.  Prove that exist such $k,l$ $k \not = l$ that $a_{l}=a_{k}$

I have a part of solution.

We will check three cases:
1) there is no m such that $a_{m}= 2k+1$ and no n such that $a_{n}=-(2k+1)$

This is the easiest case. Since for all i , $-(2k+1) \leq a_{i}\leq 2k+1$, there are 4k +3 possible values of $a_{i}$. If there is no m such that $a_{m}= 2k+1$ and no n such that $a_{n}=-(2k+1)$, then there are 4k+1 possible values of $a_{i}$. The pigeonhole principle does it's job and we're done.

2) There exists such m, that $a_{m}= 2k+1$, but there doesn't exist such n, that $a_{n}=-(2k+1)$.

Of course we have = $a_{1}+...+a_{2k+1}= a_{2k+2}+...+a_{4k+2}$
Suppose that for every m and l such that $m \not = l$ there is  $a_{m}\not = a_{l}$
Summing up all $a_{i}$, we get then 2k+1. So wew cannot divide all $a_{i}$ into two groups, each with diffrent sum. Contradicction.

3) There exist m and l such that $a_{m}= 2k+1$ and $a_{l}=-(2k+1)$

Well, I didn't figured out this case yet, but I'll try working on it.

It's easy to see that we don't have one even number. If we have chessboard of size $2kx2k$ then exist such numbers

Yeah, I can construct a $2k x 2k$ chessboard which has all the sums in the columns and werses diffrent.

for 2 x 2 we have:

1  0
1(-1)

for 4x4:

1 -1  -1  -1
1  1   0   -1
1  1   -1  -1
1  1   1    0

as we can see, the center is the 2x2 good chessboard. Now we can put the 4x4 in the center of 6x6 chessboard, and build the same frame around it, as we did around 2x2. I mean the frame looks something like this:

1 -1 -1 -1 -1 -1
1 _________-1
1 _________-1
1 _________-1
1 _________-1
1 _________-1
1  1   1  1  1  0

It's hard to draw a frame here, but I hope it's clear.
Building such frames all the time we can have as big 2n x 2n good chessboard as we want.

Yes, it's obvious that the missing number is even, and my cases are preety dumb  :wink:  still, i got stuck and don't know how to solve your problem wojoto. Any tips??

Also, where is this problem from?

I can't help you, because I don't know anything more :((. Really I think that this is hard problem.

If $\pm(2k+1)$ are both there (they have to be), then they're both rows or both columns (else they'd intersect).  WLOG, they're both columns.  Then the max and min of any row are $\pm(2k-1)$.  One of $\pm2k$ exists, though.  WLOG, some column has sum $2k$.  So every row now has bounds $-(2k-3), 2k-1$, except the one with the $0$, which has bounds $\pm(2k-2)$.  Thus, there exists a column with sum $-(2k-1)$.  It looks like this is getting very complicated very quickly, but it might work out...

EDIT: A possible start?

Suppose $2a$ is the removed sum.  Now we know that the columns with sums $2k+1$ and $-(2k+1)$ are parallel.  [i]Remove them[/i]
Now we have a $2k+1 \times 2k-1$ board.  If $a<k$, we must then have two rows with sums $2k,-2k$.  [i]Remove those as well[/i].  And now we again have a $2k-1 \times 2k-1$ board.

Continue this process until we get a $2a+1 \times 2a+1$ board which is missing the sum $2a$ (WLOG $a$ is positive.  We can deal with the case $a=0$ later).  This makes the problem more specific, but i suppose it's no easier...

Dude@ I think I solved it, and I am pretty excited. OK so here goes.

First of all, the rows/columns have all sums between -2k-1 and 2k+1 except for one, so WLOG they're missing a negative sum (or zero). Thus we may split up the rows/columns into two sets--P, those with positive sum and N, those with non-positive sum.

Now let p be the number of 1's such that its row and column are not both in P, and let n be the number of -1's such that its row and column are not both in N. Now consider the 2k+1 rows/columns in P. They intersect each other x(2k+1-x) times, where x is the number of rows in P. However, the sum of their sums (counting some numbers in the grid twice) is 1+2+...+(2k+1) = (2k+2)(2k+1)/2 = (k+1)(2k+1), and furthermore, there are n -1's which appear in at least one element of P. Therefore, since the 1's with both row and column in P can contribute at most 2x(2k+1-x) to the total sum, and we must subtract n for the -1's, we have $p \ge (k+1)(2k+1)-2x(2k+1-x)-n$.

For the non-positive rows/columns, we have an analogous result (except that the absolute value of their sum is k(2k+1) instead of (k+1)(2k+1)). Thus $n \ge k(2k+1)-2x(2k+1-x)-p$.

Adding those two inequalities, we have $4x(2k+1-x) \ge (2k+1)^{2}$, but by AM-GM this implies that x = 2k+1-x, an impossibility.


symmetry
MATHCOUNTS
modular arithmetic
probability

[hide="HINT"]
To be divisible by $ 8$, the last three digits must form a multiple of $ 8$.  That simplifies the problem greatly.
[/hide]

Now, find the number of 3 digit multiple of 8s you can make by rolling a die. Hmm... how do you do that without listing them all out... or can you?

[quote="ghertfort"][quote="AIME15"]HINT
To be divisible by $ 8$, the last three digits must form a multiple of $ 8$.  That simplifies the problem greatly.
[/quote]

Your hint is useless. If you do not know how to solve it, please do not spam.[/quote]

Explain how the hint is useless. Find a different way to solve it. You want to list out about a hundred thousand numbers?

[quote="BOGTRO"][
Explain how the hint is useless. Find a different way to solve it. You want to list out about a hundred thousand numbers?[/quote]

Can you please explain how the hint is useful? Can you please show us how to get the answer to the problem based on this "hint"?

[hide="how that hint is helpful"] A number is divisible by 8 if and only if the three digit number formed by the three rightmost digits is divisible by 8. This is useful, because now you only need to examine the case where three dice are rolled.[/hide]

[hide="rest of solution"] We now can use casework, either on the first digit by itself, or the parity of the first digit. The second option turns out to be better:
Case 1: The hundred's digit is odd: Now, we can just examine when the first digit is 1, since adding 200 (which is divisible by 8) gives the ones for the numbers beginning with 3 and 5. The second two digits can be 12, 36, 44, or 52. That gives 12 for this.
Case 2: The hundred's digit is even: We proceed in a similar fasion. The second two digits can be 16, 24, 32, 56, or 64. That's 15 more from here.

So, in all we have 27, while there are a total of $ 6^3 \equal{} 216$ ways to roll three dice. Our answer is $ \frac {1}{8}$.[/hide]

EDITED after finding a relatively nice solution.

[quote="xpmath"]This is useful, because now you only need to examine the case where three dice are rolled.

[/quote]

I said it was useless because I already know this and knowing this does not help further. I think everyone who works on Nationals should have known this.

[quote="xpmath"]The second two digits can be 12, 36, 44, or 52.
[/quote]

How did you get these numbers? By trying every one? What is the way you used?

[quote="xpmath"] EDITED after finding a relatively nice solution.[/quote]

Can you share with us your nice solution?

The solution I gave [i]was[/i] my so-called nice solution :P. It's better than doing regular casework.

Ok, to find the numbers:

Do you understand that we can just count the numbers that start with either 1 or 2 and then multiply by three?
After that, I divided it into two cases.

The first case was when the first number was odd, and I just counted the ones that began with 1 and multiplied by three. By simple trial and error, the first number that we can create in that category is 112. Now we add 8 at a time and see if we can create those numbers. So, we examine 112, 120, 128, 136, 144, 152, 160 and 168. After that, we'd need at least a 7, which we can not create. Out of these, only 112, 136, 144, and 152 work. For each of these, we can add 200 (which is still divisible by 8) to get the numbers that begin with 3. We add 200 again to get the numbers that begin with 5. That gives 12 for this case.

The second case was when the first number was even, and I just counted the ones that began with 2 and multiplied by three. Once again, the first number in this category is 216. Using the same reasoning as in the first case, we examine 216, 224, 232, 240, 248, 256, and 264. We find 216, 224, 232, 256, and 264 work. That means that 416, 424, 432, 456, 464, 616, 624, 632, 656, and 664 also work. 15 in all.

So we get 27 in all.

If you don't understand something, just respond.

the first time i saw this problem i was like "ok so 1/8 of all numbers are divisible by 8. isn't that obvious?"
i think they should make it so the answer isn't 1/8 so that ppl can't just skip all the hard work and get it right,

Bleh, intuitively, that appears obvious, but I was thinking that you can't create all possible three digit numbers, so that symmetry may be off. It's interesting how it works here. Does it work in all cases similar to these?

i think it's more likely if there are more numbers.
my point in the earlier post was that mathcounts made this problem too guessable. generally you should try to solve the problem to be sure.

try mod 8 of 100a+10b+c

4a+2b+c must divide 8.
where a,b,c are dicerolls.

c={2,4,6}

If c is 2, then 2a+b+1 must divide 4.
If c is 4, then 2a+b+2 must divide 4.
If c is 6, then 2a+b+3 must divide 4.

Therefore all that is left is listing out all mods 2a+b (mod4)
Note that a = {1,2},{3,4},{5,6} preserves 2a+b mod4.
2a = {0,2} mod 4
Also note that adding 2 mod 4 is equivalent to subtracting 2 mod 4.
Therefore we can pair up residues 0 and 2, 1 and 3.
We can obtain every possible residue in every ab-block of a in the set {1,2},{3,4} or {5,6} and b also in the set {1,2},{3,4},{5,6}
i.e. each residue mod 4 has 1/4 chance to be picked.

c to be even: 1/2 probability.
2a+b mod 4 to be of the right mod: 1/4 probability.
They are independent events, so answer is 1/2*1/4=[b]1/8[/b].

This method reduces casework to investigating a grand total of [b]0 cases.[/b]

does that mean it is 1/8 as long as there are as many odd numbers as even numbers? so a 4-sided die and a 8-sided die would also yield 1/8

Yes, I'm pretty sure that's correct.

To experiment:
ab2: 11,23,31,43
ab4: 14,22,34,42

8/64 = 1/8  :D

[quote="math154"]Once we get it to $ 100a \plus{} 10b \plus{} c\equiv4a \plus{} 2b \plus{} c\pmod8$, we can note that only the parities of $ a,b,c$ matter, and since the odd and evens are equally likely, express this as $ \overline{abc}_2$. Clearly, this is only divisible by $ 8$ if $ \overline{abc}_2 \equal{} 000_2$, so the probability is $ \frac1{2^3} \equal{} \boxed{\frac18}$. This also generalizes pretty nicely for $ 8|4a \plus{} 2b \plus{} c$, as we just need to consider the parities, and multiply the probabilities that each digit is even together to get the answer.[/quote]

revive much?  :wink:
but nice solution.


Find all pairs of real number a and b for which the system of equation:
  $\frac{x+y}{x^2+y^2} =a , \frac{x^3+y^3}{x^2+y^2} =b$
with unknowns x and y has a solution in the real domain.

Case 1: Let y=0 in this case 1/a=b=x where a&b can take any real value except 0.

Case 2: Let y != 0
Consider the product of the equations

(x+y)(x3 + y3)/(x2 + y2)2  = ab

Converting the equation into homogeous form in x/y=z we have
(1+z)(1+z3)/(1+z2)2 = ab
rearranging the terms & substituting z/(1+z2 )=Z we have 1+Z-2Z2 = ab. Evaluating for D >= 0 we have ab >= 9/8
for all a,b  :varepsilon: R.


calculus
derivative
real analysis
real analysis unsolved

How to show that if $f: \mathbb{R}^{n}\to \mathbb{R}$ is lipschitz, then $f$ has partial derivatives almost everywhere?

I like the exposition of Rademacher's theorem in [url=http://www.math.jyu.fi/research/reports/rep100.pdf]Lectures on Lipschitz Analysis[/url] by Heinonen.

I think this might work.

Let consider the case when $n=1$.
Then $f$ Lipschitz implies $f$ is absolutely continuous from which it follows that $f$ is differentiable almost everywhere.

Didi


Find all integers $ n\ge 2$ with the following property:

There exists three distinct primes $p,q,r$ such that
whenever $ a_1,a_2,a_3,\cdots,a_n$ are $ n$ distinct positive integers with the property that at least one of $ p,q,r$ divides $ a_j - a_k \ \forall 1\le j\le k\le n$,
one of $ p,q,r$ divides all of these differences.

Hmm ,really I don't understand this problem very much .Can you explain in more details ?

Actually ith_power, you have got to use 3 relations :
$ p(r\minus{}q) \plus{} q(r\plus{}p)\equal{} r(p\plus{}q)$
$ p(q\minus{}r)\plus{} r(p\plus{}q) \equal{} q(r\plus{}p)$
$ q(r\minus{}p) \plus{} r(p\plus{}q) \equal{} p(q\plus{}r)$.
And then it is from now on very very simple to finish.

[quote="ith_power"]it asks to find n such that if you fix n, three fix primes p,q,r are there (depending on n only) such that for any sequence a_1,a_2,...,a_n, the given property holds with that p,q,r.
[hide]
Its pretty easy. Use
$ (r\minus{}q)p\plus{}(r\plus{}p)q\equal{}r(p\plus{}q)$[/hide][/quote]

I really think that you hv made a typo in the question .. it seems absurd :D

So, for $n=3$, take the three numbers $a_1 = p(q+r)$,  $a_2 = q(r+p)$, $a_3 = r(p+q)$. Each difference is divided by one of $p,q,r$, bot none of $p,q,r$ divides all differences.

Then, for $n\geq 4$, take the first three as above, and all others  $a_m = a_1 + k_m qr$, for $4\leq m\leq n$ and arbitrary $k_m$. Each difference  is divided by one of $p,q,r$, bot none of $p,q,r$ divides all differences.

Therefore no $n$ satisfies.


Since there are so much topics for only one game, I have decided to make a topic where you can discuss you're favorite game.

First you say what you're favorite game is, and then what kind of game it is e.g. racing/ first person shooter (FPS) and then tell us what the story is about (If you want to).

Later, when there is enough posts, I will add a poll.


inam az marathone hendese!! khashi jooon komak kon to in marathon.faghat omidvaram in marathon mesle tarkybiat bijoon nabashe :)

$ABCD$ chaharzeli mohaddab ast va $N$ vasate $BC$ ast . hamchenin , $\angle AND=135$ .sabet konid :
$AB+CD+\frac{1}{\sqrt{2}}BC \geq AD$

[quote="sepehr ahmadzadeh"]$ABCD$ chaharzeli mohaddab ast va $M$ vasate $BC$ ast . hamchenin , $\angle AMD=135$ .sabet konid :
$AB+CD+\frac{1}{\sqrt{2}}BC \geq AD$[/quote]

farz konid $BA$ , $CD$ ra dar dar emtedade an az tarafe $D$ ghat konad.

darim $\alpha+\angle AMD=\angle MDC<45^{\circ}\longrightarrow \alpha<45^{\circ}\longrightarrow \cos\alpha > \frac{\sqrt 2}{2}$

va hamchenin darim:
$AD=AB\cos\beta+BC\cos\alpha+CD\cos \gamma< AB+BC\frac{\sqrt 2}{2}+CD$

halate tasavi be ezaye $BA\parallel CD$ etefagh mioftad, ke khodetoon peidash konid.

[img]5855[/img]

[quote="sepehr ahmadzadeh"]makan hendesie hameye noghtahye manande $P$ ra daroone lozie $ABCD$ tori biaibid ke $\angle APD+\angle BPC = 180$[/quote]

Just for fun:

darim $\angle APB=180^{\circ}-\angle CPD$ pas shoa'e dayerehaye mohitiye $\triangle APB$ va $\triangle PCD$ barabar hastand.
az $P$ khati movazi ba $AC$ rasm mikonim ,darim$\triangle CP'D\sim \triangle APB$ banabar in $\angle ABP=\angle CDP'=\angle CPP'=\angle ACP$ pas$\triangle APC\equiv \triangle ABP$ pas makanhndesiye $P$ aghtare lozi ast.

[img]5856[/img]

[quote="sepehr ahmadzadeh"]amir merC az hallet :wink: soal bezar too in marathone nazar inam mesle tarkybiat kam ronagh beshe[/quote]

Fagaht be khatere shoma :wink:

[b][color=blue]A nice problem:[/color] [/b]

[color=darkblue]A quadrilateral ABCD with AB>AC circumscribes about a circle with center O.Let E and F be the points of intersection of BD with the circle.The line OH passing through O cuts orthogonally AC at H. Prove that $\widehat{BHE}=\widehat{DHF}$[/color]

[img]3194[/img]

chera kasi in soalha ro hal nemikone?:maybe:

jaye afrade khafani chooon [b]Ashegh[/b] khaliye ke betooonand in soal ha ro hal konand. :roll:

tebghe ghazieye pascal  AC,MK,NI va mahalle barkhord e (NK,MI) ham khat hastand ⇒ AC ghotbie P ast ⇒ OP bar AC amood ast
(sabet mikonim HT nimsaz har do zavieye FHE , DHB ast)

Amma chon AC, (EN,FM) , (FN,ME) ham khat and(tebghe ghazieye pascal) ⇒ P,F,T,E  hamsaaz and
va be tarighe moshabeh (ba pascal) saabet mishavad  P,D,T,B niz hamsaaz and ⇒ chon HP bar HT amood ast ⇒ hokm sabet mishavad.

Khob albate soalesh asoon bood. :D

inham ye soale asoone dige:
dar mosalase $ABC$ ,$H$ markaze ertefaei ast,az $H$ do khate amood bar hame $HX$ va $HY$ ra mikeshim($X\in AB\ \ ,\ Y\in AC$).
az $B$ ca $C$ 2 khat be movazate $HY$ va $HX$ beartib mikeshim ta yek digarro dar $Z$ ghat konand .

sabet konid $X,Y,Z$ ham khatand.

ABC va noghteye delkhahe P daroon an mafroozand:
C1 dayereye delkhahi ast ke az P migozarad

E mahalle barkhord e BP ba C1 ast va  J mahalle barkhord e dayereye mohotie APC ba C1  ast... ( baghieye noghte ha be shekle moshabeh rooye shekl moshakhas and)

sabet konid EJ,GH,IF  hamresand  :idea:


Ok, this is kind of embarrasing but, since most people on here have very high SAT scores (especially Math), this seemed the best place to turn.  In regards to conventional classroom tests(math), I know the material front and back, and Im usually one of the few people who "get it" after the first lecture.  When the test comes around, everything changes, I cant remember important things and my mind comes crashing down in flames.  Ive been told that I stress way too much over tests, and that this hinders my ability to do well. Does anyone have any advice to help become a better test taker.  I know that SAT is different from conventional tests, but I am convinced that if I improve my test taking abilities, the SAT will be no problem.

Thanks.

[quote="now a ranger"]i would try relaxing and just knowing this is just a test, you can retake it again if necessary. To sum it up, do your best and don't get the jitters. Remember, the SAT (pertaining to the math section) contains stuff that has most likely been taught in school.

I don't know about the reading section, since my score isn't so high. But I would just do all the ones I know and know that I can improve. but hey, im an 8th grader, took sat last year just for some stupid thing, and i didn't get nervous. My scores werent that good since I hadnt learned much but i have siblings who have done well when they took it older by relaxing.[/quote]

Agreed.  I took the SAT again after I didn't care about it anymore (already scored 1550 out of 1600), didn't study, left my calculator at home on purpose, and did better.  It might be hard to relax on purpose, so it might help to take a lot of practice tests to get used to the environment as much as possible.

[quote="haoye"]
Agreed.  I took the SAT again after I didn't care about it anymore (already scored 1550 out of 1600)...[/quote]

Eh, why?  :maybe:

You might be different, but I must have confidence to relax and do well. Otherwise, I freak out, especially on standardized tests.

this is what I do at school
-enrolled in some of the hardest classes I can
-always try paying attention. Focus is key.
-if I get a lame "fill in the blank" guided reading worksheet, I read the section/chapter it's about; then I try to complete the work from memory. This is great reading practice.
-I complete all tests as fast as I can without going "too" fast. This is mainly to work on pacing and focus.

out of school
-I read. Actual books.
-take practice tests in realistic time settings, etc. This is probably the best thing you can do. For example, if you take a random practice test once a week, you will be so ready for anything.

weird things that probably won't help you...
-to me, knowledge is power. For every single test I take, I must know what sections are on it, what material they cover, how long each one is, what are possible orders, etc.
-I also like to know where the test will be given, what time it begins, if I can bring food/drink, etc.
-I wear an outfit I like, and a sweatshirt I like. Comfort is a must also.
-I have more, but I don't want you guys to think I have OCD

Anyway, I know many people that are experiencing the same thing as you. They're usually bitter about it and lazy, but I really think you can get better. Good luck!

Thanks guys, there is no doubt that I will try my absolute hardest to do my best on these tests.


probability AMC USA(J)MO USAMO function \/closed

Hey this may seem really n00bish but I dont care: What are the best problems and/or chapters in Intermediate Counting and Probability for USAMO prep?

Chapters 5, 8, and 15 cover essential tactics for proof problems that you may not have much experience with.  The rest of the book covers all sorts of counting tactics that may be helpful on the test.  I don't think there are any chapters in the book that I'd describe as "can be safely ignored", though some are more important than others.  (Generating functions, for example, are not frequently useful.  But when they are, they're *very* useful.)