Let $G$ be a $k$-connected graph with $n$ vertices and $m$ edges (that is, for any two vertices there are $k$ disjoint paths joining them), and let $A \subset V(G)$ be an arbitrary set of $k$ vertices of $G$. Intitially, there are $2m-kn+1$ frogs, each staying on some vertex of $G$ (there may be more than one frogs on the same vertex). Each second, the following happens: from some vertex  $v \in G$  with at least $d(v)-k+1$ frogs on it, $d(v)-k+1$ of these frogs jump to some of the $d(v)$ neighbors of $v$, each on a different neighbor. Show that, at some point, some vertex in $A$ will be visited by a frog.

--Vesselin

NB: It is [i]not[/i] necessary that [i]all[/i] frogs who can make a move will jump; it is only essential that there is [i]at least one[/i] move every second. Thus, even the special case $k=1$ is a generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24406&highlight=]this[/url].

I didn't understand you, Vess.
Color the vertices red if a frog is [b]allowed[/b] to jump in there and green otherwise. We must prove that there are at most $k$ green vertices. Now if we choose the vertices s.t. any red vertex has at most $k-1$ green neighbours, then we can construct a counter-example to your problem:
From every red vertex $v$ there are at least $d(v)-k+1$ red neighbours so the frogs can jumpl [b]only in red vertices[/b]. Now for $k>1$ we can construct such a graph, isn't it?


Solve equation:
\[ n! + 1 = p^m \]
where $n$ and $m$ are integers, and $p$ is prime number.

Maybe no replies because it doesn't have finite solutions? For example, 11!+1 is a prime. Larger values of m as well most likely.

Maybe you are right, but additional condition of the problem is $m > 1$.

[quote="dblues"]As far as I can remember, the problem of $n!+1 = p^2$ is an open problem, with $(n,p) = (5,11), (7,71)$ as the only known solutions so far.

The general problem of $n!+1 = p^m$, $m>1$ is then even harder.[/quote]
$n!+1=p^2$ had solutions (4,5)(5,11),(7,71).
By effective abc hipothesis: $a+b=c,(a,b)=1 \Longrightarrow rad(abc)>c^{3/5}$ easy to prove, that equation
$n!+1=x^m,m>1$ had only these solutions.

No I 'm wrong. Liouville is about (p-1)!+1=p^m  :blush:

[quote="dblues"]Oops! I missed the case of $(4,5)$.

Anyway, isn't the ABC Conjecture still unsolved?[/quote]

[url=http://mathworld.wolfram.com/abcConjecture.html]not sure is it is unsolved[/url]

[quote="e^pi"][quote="José"][quote="Beat"]Here is a similar problem that i found. Solve in natural numbers the equation:
$1!+2!+\cdot\cdot +x!=y^z$[/quote]

Is there any general formula for $1!+2!+3!+...+n!$?????[/quote]

http://mathworld.wolfram.com/FactorialSums.html[/quote]

Thanks


Hi
Solve this difficult (I think) problem :
Two points A , B are fixed and M is variable on the circle C(O,R) . Find locus foot into the middle AM perpendicular to the line BM .
[url=http://www.iimmgg.com/image/532c900764ac32338d37ee9d57c322f2][img]http://i.iimmgg.com/images/th/0a2a5d99b7082697c31427b49a70cdb1.png[/img][/url]
I'v tried to solve it for a long time , but I couldn't solve it in any way .
Please help me as soon as you can !  :help:
Thanks

i think that your problem is this ;
Let $ N$ be the midpoint of $ AM$. then denote by $ H$ the projection of $ N$ on $ BM$.now find the locus of $ H$ as $ M$ varies on the circle.
Let $ B'\equal{}BO \cap (O)$.since $ B'M \perp BM$ and also $ NH \perp BM$ we get $ B'M \parallel{} NH$.so ,since $ N$ is the midpoint of $ AM$, $ NH$ bisects $ AB'$.so $ NH$ passes thru a fixed point ,call it $ S$.now it's obvious that $ H$ is on the circle with diameter $ BS$.thus,the locus is found.

But dear [b]LJK[/b], how many times have you posted the same problem in this Forum?

Kostas Vittas.

PS. The same problem has been deleted one time and another time has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=322896][color=Blue]Here[/color][/url].


Stupid question, but I'm only a beginner. How do you signify a turn of the page?

\pagebreak forces the text to start a new page (I don't [i]think[/i] you need any special packages for it), but $\LaTeX$ automatically goes to a new page at the end of the previous one.


geometry
circumcircle
inequalities
geometric transformation
geometry solved

Point $K, L, M$ are respectively on sides $BC, AC, AB$ of $\triangle ABC$. Prove, that circumcircles of $\triangle ALM, \triangle BMK, \triangle CKL$ meet in one point.

[i]Alternative formulation:[/i] Given a triangle $ABC$ and three points $K$, $L$, and $M$ such that $K\in BC$, $L\in CA$, and $M\in AB$. Denote the circumcircles of triangles $ALM$, $BMK$, and $CKL$ by $k_{a}$, $k_{b}$, and $k_{c}$, respectively. Prove that these circles $k_{a}$, $k_{b}$, $k_{c}$ concur at one point.

This is the Miquel pivot theorem and proven in

http://cut-the-knot.com/proofs/Pivot.shtml Statement B;
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=248620#p248620]http://www.mathlinks.ro/Forum/viewtopic.php?t=39473 post #2[/url] (where I prove it collaterally: "the point P lies on the circumcircles of triangles AYZ, BZX and CXY").

Or try proving it for yourself, it is really one of the simplest theorems in geometry...

  Darij

[quote="dondigo"]Point $K, L, M$ are respectively on segments :$BC, AC, AB$ of $\triangle ABC$. Prove, that circumcircles of $\triangle ALM, \triangle BMK, \triangle CKL$ meet in one point[/quote]
Yes, this is quite easy. :)
[hide]Let $P$ be the intersection point of the circumcircles of $\triangle BMK$ and $\triangle CLK$. Then $\angle MPL=2\pi -\angle MPK-\angle LPK=2\pi - (\pi-\angle B)-(\pi -\angle C)=\pi -\angle A$. Therefore, $A,M,L,P$ are concyclic.[/hide]

[quote="leepakhin"]P.S. Darij, I think you have posted the wrong link; I was linked to a geometric ineqaulity which seems to have nothing to do with this problem.[/quote]

In the proof of this inequality I showed the Miquel theorem as a lemma. I was just too lazy to copy the right part of the proof:

[quote="darij grinberg"][...] Let ABC be a triangle, and let X, Y, Z be three points on the segments BC, CA, AB, respectively. [...]

[We have to show that the circumcircles of triangles AYZ, BZX and CXY have a common point.]

Let the circumcircles of triangles AYZ and BZX meet at a point P (apart from Z). Then, the quadrilaterals AYPZ and BZPX are cyclic, so that $\measuredangle AYP=180^{\circ}-\measuredangle AZP$ and $\measuredangle BZP=180^{\circ}-\measuredangle BXP$. In other words, < AYP = < BZP and < BZP = < CXP. Hence, < CXP = < AYP; equivalently, $\measuredangle CXP=180^{\circ}-\measuredangle CYP$. Hence, the quadrilateral CXPY is cyclic, so that the point P must lie on the circumcircle of triangle CXY.

So thus we have obtained that the point P lies on the circumcircles of triangles AYZ, BZX and CXY. [Thus, these circumcircles have a common point, qed.] [...][/quote]

[b]PS.[/b] Oh, as I see you've just noticed it too...

  Darij

Yes, I JUST noticed it before you made your reply. :D

I've done it just after I posted the problem, it's really easy, but the next problem is:
Prove, that triangle, which vertexes are centres of this circles is similar to $\triangle ABC$

I think you mean "similar".
(A guess without thinking would be that the two triangles are related by a spiral similarity with centre being the intersection point of the three circles. Am I correct? :? )

[quote="dondigo"]Prove, that triangle, which vertexes are centres of this circles is congruent to $\triangle ABC$[/quote]

It's similar, not congruent. See http://www.mathlinks.ro/Forum/viewtopic.php?t=56923 .

[b]PS.[/b] Another simultaneous post...  :rotfl:

  darij

[quote="dondigo"][i]Alternative formulation:[/i] Given a triangle $ABC$ and three points $K$, $L$, and $M$ such that $K\in BC$, $L\in CA$, and $M\in AB$. Denote the circumcircles of triangles $ALM$, $BMK$, and $CKL$ by $k_{a}$, $k_{b}$, and $k_{c}$, respectively. Prove that these circles $k_{a}$, $k_{b}$, $k_{c}$ concur at one point.[/quote]

[hide="hint"]Denote $K$ is common point of $k_{a}$ and $k_{b}$ , where $K\not = M$. You need prove $K\in k_{c}$.[/hide]


Two small talks separated by a distance $ d$ emit sounds of frequency $ \nu$ and equal intensity, keeping a constant phase difference between them. An observer moves along a straight line $ A$ separated by a distance $ D$ from the system of talks.

a) When the observer stands on the point $ C$, equidistant from the two talks, he hears no sound. Find the phase difference between the sounds emitted by the talks.

b) Find the distance $ L$ the observer has to walk along the line $ A$ in order to find the first maximum of sound intensity.

Data: $ \nu \equal{} 3400\ \text{Hz}$, $ D \equal{} 10\ \text{m}$, $ d \equal{} 0.5\ \text{m}$. The speed of sound is $ 340\ \text{m/s}$.

[img]http://img294.imageshack.us/img294/9568/prob2oc5.jpg[/img]

Solution

(a) If the observer hears no sound, then the sound signals from the two talks are destructively interfering. This means that at $ C$, the crests of the waves from one source coincides with the troughs of the other. The phase difference at $ C$ is hence $ \pi$ radians. Since $ C$ is equidistant from the two sources, then the phase difference of the two sources is $ \pi$ radians.

(b) The wavelength of the waves are $ \lambda =\frac{340}{3400}= 0.1\,m$

For there to be a maximum of sound intensity, constructive interference must occur so that path difference between the sources must be a whole number of half wavelengths. For the first maximum, the path difference is half a wavelength or $ 0.05\,m$. Lets say the position of this maximum is "$ B$". Without loss of generality, let $ B$ be right of $ C$. Call the talk on the left "$ F$" and the talk on the right "$ G$". Draw an altitude through $ B$, and call the point that it intersects the line joining the two talks "$ E$". Let $ FE = f$ and $ EG = g$. If $ E$ is between $ F$ and $ G$ then $ f+g = 0.5$. If $ E$ is right of $ G$ then $ f-g = 0.5$. Thus: $ f\pm g = 0.5$

Let $ GB = x$. Since the path difference is $ 0.05\m$ then $ FB = x+0.05$.

Using Pythagoras' theorem on $ \triangle FEB$ gives $ f =\sqrt{((x+0.05)^{2}-100)}$. Similarly $ g =\sqrt{(x^{2}-100)}$

$ \Rightarrow\sqrt{((x+0.05)^{2}-100)}\pm\sqrt{(x^{2}-100)}= 0.5$

$ \Rightarrow\pm\sqrt{(x^{2}-100)}= 0.5-\sqrt{((x+0.05)^{2}-100)}$

$ \Rightarrow x^{2}-100 = 0.25-\sqrt{((x+0.05)^{2}-100)}+(x+0.05)^{2}-100$

$ \Rightarrow x^{2}= 0.25-\sqrt{((x+0.05)^{2}-100)}+x^{2}+0.1x+0.0025$

$ \Rightarrow\sqrt{((x+0.05)^{2}-100)}= 0.1x+0.2525$

$ \Rightarrow (x+0.05)^{2}-100 = 0.01x^{2}+0.0505x+0.0637563$

$ \Rightarrow x^{2}+0.1x+0.0025-100 = 0.01x^{2}+0.0505x+0.0637563$

$ \Rightarrow 0.99x^{2}+0.0495x-100.061 = 0$

$ \Rightarrow x =\frac{-0.0495\pm\sqrt{396.244}}{1.98}$

Since $ x > 0$ then: $ x = 10.0285$

This satisfies the initial equation where a minus is used. Thus $ E$ is right of $ G$.

$ L = 0.25+g = 0.25+\sqrt{(10.0285^{2}-100)}= 0.25+0.7555 = 1.006\, m$


Euler
analytic geometry
geometry proposed
geometry

Four lines that intersect in pairs form four trinagles .
Prove that if one of these  lines is parallel to the Euler line of the triangle which is formed vy the other three lines , then the same holds for the other three lines .

the only sol. that i know uses coordinates with suitable choice of origin.its due to isharygin.


In convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel.  Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD.
Prove that ABCD is a concyclic quadrilateral if and only if the triangles ABP and CDP have equal areas.

Pls explain to me with steps.. THZ A LOT!!

Again go to the contest section of the forum and you can find all the past IMO questions and solutions.


geometry
rectangle
trigonometry
circumcircle
area of a triangle
trig identities
Law of Sines

Ethan is trying to find the area of a triangle, but he only has its side lengths and does not know Heron's formula.  Mistakenly, he decides to multiply the side lengths together (similar to how he would find the area of a rectangle) to find its area.  Miraculously, Ethan computes the correct area!

Find all triangles for which this is possible.

[hide="Large hint"] The correct triangle area formula you'll want to keep in mind is $ [ABC] \equal{} \frac{1}{2} ab \sin C$. [/hide]

$ A\equal{}abc\equal{}\frac{abc}{4R}\iff R\equal{}\frac{1}{4}$ is obviously necessary and sufficient.

I don't understand your "solution", Altheman. What is $ R$?

Ethan only knows the triangle's side lengths. He does not know the angles. How would we use trigonometry?

$ R$ is the radius of the circumscribed circle about the triangle.

$ A\equal{}\dfrac{abc}{4R}$ is good to remember.

Altheman's solution makes this problem quite straightforward.  But according to your hint, would using that fact make the problem easier to solve?  Or is the area formula in your hint a motivation for deriving the area formula for the inscribed triangle in terms of the circumradius?  :maybe:

$ K \equal{} \frac{abc}{4R}$ is a consequence of the area formula I gave together with the [url=http://mathworld.wolfram.com/LawofSines.html]extended Law of Sines[/url] (which is what I was actually thinking of).

I know the derivation of the area formula in terms of the circumradius, but it does not implement the extended law of sines $ \frac {a}{\sin A} \equal{} 2R.$
However, I suspect the derivation of the extended law of sines is analogous to that area formula in terms of three sides and the circumradius; it is as follows:

first consider the case where the triangle is acute (the case where the triangle is obtuse is similar); then use the fact that any of the three angles in the triangle is half the measure of the central angle subtended by its corresponding arc.  The sine of that angle (wlog, call it A) in the triangle is simply $ a/(2R)$ where a is the side opposite angle A and R is the circumradius.  Then we have the area of the triangle as $ 1/2 bc a/(2R).$  The rest follows.

Is that the same derivation (or proof) that you (& mathworld) have in mind?


function
search
calculus
derivative
integration
real analysis
advanced fields

Why do we need to introduce Holder spaces $C^{k,\alpha}(\overline\Omega)$ (what are "heuristic considerations" leading to the definition of these spaces)?

We need function spaces to solve partial differential equations. Useful as they are, continuity and differentiability are [b]qualitative[/b] concepts: a function is either continuous or not. To work with PDE, we need [b]quantitative[/b] versions of such concepts, e.g. Holder continuity. Krylov's book [url=http://www.fetchbook.info/compare.do?search=9780821805695]Lectures on Elliptic and Parabolic Equations in Holder Spaces[/url] explains this in detail. A simple example: the Poisson equation $\Delta u=f$ with $f\in C$ does not necessarily have a $C^{2}$ solution, but if $f\in C^{\alpha}$ with $\alpha>0$, then there is a solution $u\in C^{2,\alpha}$.

Thank you, mlok! Unfortunately, the book is not free and I don't have it on my bookshelf.  :(

The intuition behind $W_{0}^{k,p}(\Omega)$ spaces in simple. E.g. we want to allow vibrating string take angular forms, hence admit non-differentiable functions. So if we have a sequence $(u_{n})$ of $C_{0}^\infty(\Omega)$ solutions to the Dirichlet problem for our PDE, and if $(u_{n})$ converges together with all its derivatives $D^\alpha u$, $|\alpha| \le k$ in $L^{p}(\Omega)$, then the limit function $u$ should also be considered as a solution of our PDE. And it is now easy to understand why one of the norms of $W^{k,p}(\Omega)$ is $\|u\|_{W_{0}^{k,p}(\Omega)}= \sum_{|\alpha| \le k}\|D^\alpha u\|_{L^{p}(\Omega)}$, since, in fact, in the example above we were considering the completion of $C_{0}^\infty(\Omega)$ in this norm.

The situation with $C^{k,\alpha}(\overline\Omega)$ is different. I don't have intuition behind the following things: 1) why do we demand all the $D^\beta u$ with $|\beta|=k$ to be Holder continuous; 2) why do we endow $C^{k,\alpha}(\overline\Omega)$ with its very norm.

Maybe, introduction of Holder spaces in intimately related with the fact that fundamental solution of the Laplace equation is
\[\mathcal E(x-y) = \mathcal E(|x-y|) = \frac 1{n(2-n)\omega_{n}}|x-y|^{2-n}\]
for $n \ge 3$?

The PDE books by Ladyzhenskaya and Uraltseva, or by Gilbarg and Trudinger may be easier to find in Russia. You are right about the connection with $\mathcal E$ (in 2 or more dimensions). Let $u$ be the Newtonian potential of $f$, namely $u(x)=\int \mathcal E(x-y)f(y)\,dy$. As long as $f$ is bounded and integrable, we can differentiate $u$ once and obtain $D_{i}u(x)=\int D_{i}\mathcal E(x-y)f(y)\,dy$. However, it's not so easy to differentiate $u$ twice. For this we use the usual trick $u(x)=\int \mathcal E(x-y)(f(y)-f(x))\,dy+f(x)\int \mathcal E(x-y)\,dy$. Here the second term is not too hard to deal with. But to differentiate the first term twice, we need the integral $\int D_{ij}\mathcal E(x-y)(f(y)-f(x))\,dy$ to be convergent. Since $|D_{ij}\mathcal E(x-y)|\le \frac{C}{|x-y|^{n}}$, the Holder condition $|f(y)-f(x)|\le C|y-x|^{\alpha}$, $\alpha>0$, is exactly what we need.

Well, a weaker condition would work too: $|f(y)-f(x)|\le \omega(|y-x|)$ with $\int_{0}^{1}t^{-1}\omega(t)\,dt<\infty$. (Dini continuity).


This is really weird. Follow the rules.

Check out the following exercise, guaranteed to freak you out. There's no trick or surprise, but, trust me: just follow these
instructions, and answer the questions one at a time and as quickly as you can! Again, as quickly as you can but don't advance
until you've done each of them... really.

What is:

    1+5 ?
    2+4 ?
    3+3 ?
    4+2 ?
    5+1 ?

Now repeat saying the number 6 to yourself as fast as you can for 15 seconds. Then

[hide="Click Here!"]
QUICK!!! THINK OF A VEGETABLE! Then

[hide="Click Here"]
Are you thinking of a carrot???
[/hide]
[/hide]

I saw this on some website, and I had to post it here.  It read my mind, that was pretty awesome!

[hide="comment hidden on request of mysmartmouth"]
it didn't work for me
[/hide]

[hide="WOOHOOHEREYOUGOHIDETAGWOOOO"]When I saw that hide tag, I was like HUH then pictured a lettuce in my head[/hide]

it didnt work for me...the first time i did it i got brocoli. second time, even after it said carrot, i tried it again, and i got brocoli.

****, I was like [hide]tomato[/hide] and didn't realize it's a fruit...

[hide]I thought carrot.  :blush:  :oops:[/hide]

[hide]
hmm... six rhymes with sticks as in "carrot sticks"

[quote="kstan013"][hide]Do you think there's anything to do with the fact that there are 6 letters in carrot? The other common answer was tomato; also with 6 letters.[/hide][/quote]

I thought of potato which has six letters.

[/hide]

[quote="kstan013"][hide]Do you think there's anything to do with the fact that there are 6 letters in carrot? The other common answer was tomato; also with 6 letters.[/hide][/quote]
[hide]what about the fact that 6 with the shift key is a caret?[/hide]

yay for brocoli :D we got 3 ppl with brocoli now

[quote="bubka"]Oops, I also thought tomato.
Actually I thought I was supposed to think of something rhyming with 6, but found none. So I thought tomato. :oops:[/quote]

There was actually a [url=http://en.wikipedia.org/wiki/Nix_v._Hedden]Supreme Court decision[/url] about whether a tomato was a fruit or a vegetable. :roll:

ProtestanT, it's "broccoli"

[quote="ra5249"][hide]Tomato. =)[/hide]
But [hide]tomatoes[/hide] are actually classified as fruits.[/quote]
yay! 4 for tomato

[hide="I thought of a"]carrot.[/hide]

Where did you find this???

A carrot is defined to be a fruit in European Community law


algebra
polynomial
Vieta
algebra unsolved

A quadric trinomial $f(x)$= $x^{2}$ + $ax$ + $b$ is given.The equation
$f(f(x)$=0 has four real roots,two of which sum up to $-1$.
Prove that $b$ $\leq$ $\frac{-1}{4}$


[quote="Farazy"]Let P be a polynomial function such that, for all real $x$:
$P(x^2-1)=X^4+5x^2+3$

What is $P(x^2+1)$?[/quote]

[hide]Since the $x^4$ coefficient is $1$, the coefficient of $x^2$ in $P(x)$ is $1$. $(x^2-1)^2=x^4-2x^2+1$. To get $5x^2$, we must add $7x^2$, so the $x$ coefficient in $P(x)$ is $7$. $(x^2-1)^2+7(x^2-1)$, and we need to add $9$ to get $6$. So $P(x)=x^2+7x+9$. $P(x^2+1)=\boxed{x^4+9x^2+17}$.[/hide]


Solve  0 < |3x + 1| < 1/3.

I'm having trouble solving these absolute value inequalities.

0< 3x+1 < 1/3 (3x+1 is positive because it states|3x + 1| >0)
-1< 3x < -2/3
-1/3 < x < -2/9 (My answer, which is incorrect)

Can somebody write me a very detailed solution (please include steps you think are obvious)

Also, can someone please give me a good explanation on how to solve absolute value inequality problems, or point me to a good *free* resource.

Any help greatly appreciated.

[quote=".Student"]Solve  0 < |3x + 1| < 1/3.

I'm having solving these inequalities.

0< 3x+1 < 1/3 (3x+1 is positive because it states|3x + 1| >0)
[/quote]

This part is wrong. :(

You need to do case work, 1) $ 3x \plus{} 1\geq 0$, 2) $ 3x \plus{} 1 < 0$. :)

What makes it wrong? And how can I fix it?

So when dealing with absolute value, you always have to look at it case by case? What do you make of |x|>0? Is this supposed to be an obvious statement? Something that is true no matter what? So for my problem it would not make a difference if they stated that or not?

All right now reattempting my problem:

First I take note of the Absolute Value definition:

|x|= { x if x≥0 , -x if x<0 }

Next I apply it to my problem:

Case 1:
If 3x + 1≥0, then:

-1/3 < 3x+1 < 1/3

-4/3 < 3x < -2/3

-4/9 < x < -2/9

Case 2:
If 3x + 1<0, then:

-1/3 < -(3x+1) < 1/3

-1/3 < -3x-1 < 1/3

2/3 < -3x < 4/3

-2/9 > x > -4/9 ( :huh:  why is this the same as above?)

So is -4/9 < x < -2/9 the answer?

[quote=".Student"]So when dealing with absolute value, you always have to look at it case by case? What do you make of |x|>0? Is this supposed to be an obvious statement? [/quote]

Yes. For any real number $ x$, $ |x|\geq 0$.

[quote=".Student"]

Next I apply it to my problem:

Case 1:
If 3x + 1≥0, then:

-1/3 < 3x+1 < 1/3

[/quote]

You should do such this: If $ 3x \plus{} 1\geq 0$, then: the given inequality become $ 3x \plus{} 1 < \frac {1}{3}$, yielding $ x < \minus{} \frac {2}{9}$,

Taking $ 3x \plus{} 1\geq 0$ into consideration, we get $ \minus{} \frac {1}{3}\leq x < \minus{} \frac {2}{9}$.

Analogously Case 2 will do.

Comment:

The inequality can be solved in 3 ways, but you should acquire the basic way. Good luck!

I don't understand...

What is the basic way?

Sorry, I am not intend to teach rapid way now.

At first, from basic method :roll:, [color=blue]which is effective, rather than, useful for any case[/color]...

I would just like to learn how to solve these kind of problems. Sorry but I am not understanding your solutions :(

See here, and study for yourself.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1258672&search_id=1129197060#1258672[/url]

I appreciate your efforts in trying to help me :)  Thanks for the link, I will go check that out.


geometry
cyclic quadrilateral
geometry proposed

let $ O$ be the intersection of $ AC,BD$ in cyclic quadrilateral $ ABCD$.$ M,N$ are midpoints of $ AD,BC$ and $ P,Q$ are the projection of $ O$ on $ AB,CD$ prove that $ PQ$ is perpendicular to $ NM$.

It has been posted many times before, last one with an excellent proof of Jean-Louis.

Best regards,
sunken rock

we must prove that $ MP\equal{}MQ$ and $ NP\equal{}NQ$ .i will to use this easy lemma.
lemma.let triangles $ PAB$ and $ QAC$ on triangle $ ABC$ ($ P,Q$ arent in $ ABC$) such that $ APB\equal{}AQC\equal{}90$ and $ PAB\equal{}QAC$.let $ M$ be the midpoint of $ BC$ then $ QM\equal{}PM$ and $ PMQ\equal{}2(PBA)$.this lemma can prove easily.let $ T,N$ be the midpoints $ AB,AC$ and use that $ PT\equal{}TA\equal{}TB$ and for $ Q$ simlliary.


Let a triangle ABC where A(2;1), B(10;1) and C(8;5). Find the equation of the geometric place of the intersection point of the diagonals of all the inscribed rectangles in the triangle ABC, so that one of the sides is always on the side AB


Let n be a number in N and a,b,c in R  so
[an]+[bn]=[cn]  for every x in R.
prove that  : a or b are  in N.
thanks.

I think you mean "for all n in N".

Since lim [an]/n=a we have a+b=c. So [an]+[bn]=[an+bn] ==> {an}+{bn}={an+bn}.
Let u=a-[a]={a}, v=b-[b]={b}.

Suppose a and b aren't integer numbers ==> 0<u,v<1.

It is clear {un}+{vn}={un+vn} ==> {un}+{vn} < 1 for all n.

Let
u=0,u0u1u2...,
v=0,v0v1v2...
are binary representations.

1) Suppose uk=vk=1 for some k then {2kuk}+{2kvk}=0,1... + 0,1... \geq 1 - contradiction.
2) Suppose uk+vk<2 for all k. We can find ui=1 and vj=1 ==> uj=0 and vi=0. Consider n=2i+2j ==> {un}=0,1... and {vn}=0,1... ==> {un}+{vn} \geq 1 - contradiction.


number theory
prime numbers
number theory proposed

Find all prime numbers $ p\geq3$ such that numbers $ 1\plus{}k(p\minus{}1)$ is prime for all $ k\in\{1,2,...\frac{p\minus{}1}{2}\}$.

[i]Adrian Stoica[/i]

Note that $ k\equal{}1, l\equal{}2$ is an unique solution to the equation $ 3^k \plus{} 1 \equal{} 2^l$. Indeed, looking $ \mod{3}$ we can conclude than $ l\equal{}2a$ and then it factorizes as $ (2^a \minus{} 1)(2^a \plus{} 1) \equal{} 3^k$. Now it's simple to finish.

Suppose that $ 3 \leq \frac{p\minus{}1}{2}$, i.e. $ p \geq 7$.Let $ q$ be the any prime less or equal to $ \frac{p\minus{}1}{2}$. If $ q$ does not divide $ p\minus{}1$ then $ (q, p\minus{}1)\equal{}1$ and there exist $ 0<k<q$ such that $ q|1\plus{}k(p\minus{}1)$, which is a clear contradiction since $ k<q \leq \frac{p\minus{}1}{2}$ and $ 1\plus{}k(p\minus{}1)>q$, so it's not prime. It means that $ p\minus{}1$ is divisible by all primes less or equal to $ \frac{p\minus{}1}{2}$, in particular $ 6|p\minus{}1$. We have $ 2|p\minus{}3$ as well, so $ p\minus{}3$ is not prime and prime divisors of this number are less then $ \frac{p\minus{}1}{2}$. But $ (p\minus{}3, p\minus{}1)\equal{}2$ and $ p\minus{}1$ is divisible by all prime numbers less than $ \frac{p\minus{}1}{2}$, so it follows that $ p\minus{}3\equal{}2^l$ for certain $ l$. Analogously $ p\minus{}4\equal{}3^k$ for some $ k$. From the observation at the beggining we conclude that $ p\equal{}7$. We check by hand that $ p\equal{}3$, $ p\equal{}7$ work and $ p\equal{}5$ does not. So $ p\equal{}3$ or $ p\equal{}7$.


integration
calculus
real analysis
real analysis theorems

Assume h is Riemann integrable on [c,d] and if $x \in [c,d]$ let $g(x) = \int_{a}^{x}{h(t)dt}$where a is a fixed point in [c,d]. Then if f is Riemann integrable on g([c,d]) the integral $\int_{c}^{d}{f[g(t)]h(t)dt}$exists and

$\int_{g(c)}^{g(d)}{f(t)dt}= \int_{c}^{d}{f[g(t)]h(t)dt}$

Can anyone point out the proof of this theorem?

EDIT: Made a correction to the integrand

I made a correction to the bottom integral since it read h(t)dt instead of f(t)dt.

Does no one know this? I think this is the most general criterion for change of variables in Riemann integration and surely that must make it known/solvable?

That is the Change of Variables Theorem. If you have a library closer to where you are you may want to check out the proof in this text: A friendly Introduction to Analysis: Single and Multivariate 2nd edition. By Witold A.J.Kosmala. The proof is on page 261. I am too lazy to reproduce it here.

If anyone has the book could you perhaps scan/take a picture of the above proof for me? My university library does not have the book.

I believe this book: Advanced Calculus: A course in Mathematical Analysis by Patrick M.Fitzpatrick  is in many libraries because it is kinda outdated but very interesting. Check page 153 of this book.

The reason why I am lazy to type the proof is that it will be longer because I will then have to explain a couple of lemmas or so that are cited in the proof. I am sure you have learned the Fundamental  Theorem of Calculus and its consequences. If this book is not in your library either, then just go to your library and pick three different books on basic Analysis and check the index for the keyword such as Change of Variables or Substitution Formula or Theorem, I am sure you will find it in one of the three books you will pick.

I am familiar with all of the 'standard' theorems on Riemann Stieltjes integrals available in a book like Rudin or Apostol. But this is a general theorem while those variations covered in textbooks usually require f to be continuous. Here f merely needs to be integrable. This level of abstraction is not treated even in Rudin (Rudin even requires monotonicity in his proof).

You can just state the lemmas without justification while giving the proof and I will fill in the details on my own.

Hello,
Disregard my earlier suggestions, but go and hunt for this text it is one of those yellow texts. It is called Real Mathematical Analysis by Charles Chapman Pugh. I just read the proof (stronger version of substitution Theorem) you are looking for in this text. Page 177 -78


inequalities
calculus
inequalities unsolved

[quote="Yosh..."]Let $a,b,c$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}=a+b+c$

Find the highest real number $r$ such that $a^{r}+b^{r}+c^{r}\leq 3$[/quote]

Set a = b = c = 1. We get $r\le 1$.
Now lets prove for $r = 1:$
\[a+b+c\le 3 \]
If we multiply that with
\[a+b+c=a^{2}+b^{2}+c^{2}\]
we get
\[a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\le 3a^{2}+3b^{2}+3c^{2}\iff ab+bc+ac \le a^{2}+b^{2}+c^{2}, \]
which is clearly true.

[b]EDIT: [/b]
I mistaken. I proved that the inequality holds for r = 1, and then it clearly holds for [b]r=2[/b], because of $a+b+c=a^{2}+b^{2}+c^{2}$. But I didn't prove that 2 is the highest value.

[quote="A. S. K."][quote="Yosh..."]Let $a,b,c$ be non-negative real numbers such that $a^{2}+b^{2}+c^{2}=a+b+c$

Find the highest real number $r$ such that $a^{r}+b^{r}+c^{r}\leq 3$[/quote]

Set a = b = c = 1. We get $r\le 1$.
Now lets prove for $r = 1:$
\[a+b+c\le 3 \]
If we multiply that with
\[a+b+c=a^{2}+b^{2}+c^{2}\]
we get
\[a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\le 3a^{2}+3b^{2}+3c^{2}\iff ab+bc+ac \le a^{2}+b^{2}+c^{2}, \]
which is clearly true.[/quote]

I don't know the answer, but if a=b=c=1 , then r can be any nonzero number. ??

[quote="mathisfun1"]

I don't know the answer, but if a=b=c=1 , then r can be any nonzero number. ??[/quote]

You're right, my mistake. I proved that the inequality holds for r = 1, and then it clearly holds for [b]r=2[/b], because of $a+b+c=a^{2}+b^{2}+c^{2}$. But I didn't prove that 2 is the highest value.

By a somewhat lengthy calculation I got that also $r=3$ is OK.
(I hope I didn't make any mistakes.)
Again by calculus small perturbations of $a,b,c = 1$ (such that the condition still holds)
show, that $r>3$ does not work. (Again I hope I didn't make any mistakes.)

But all this is really messy, and I do hope there is a better proof for this nice problem.


geometry
geometric transformation
combinatorics unsolved
combinatorics

There is n points on a plane.
Question:这 个顶点的度数能否分别为
4,5,6,....,n-4,n-3,n-2,n-2,n-2,n-1,n-1,n-1

It's very difficult, I can't even understand the question. :D  Can I have the English translation please?

these points lead to other points,can these lead to 4,5,6,....,n-4,n-3,n-2,n-2,n-2,n-1,n-1,n-1


The famous conjecture of Goldbach is the assertion that every even integer greater than $2$ is the sum of two primes. Except $2$, $4$, and $6$, every even integer is a sum of two positive composite integers: $n=4+(n-4)$. What is the largest positive even integer that is not a sum of two odd composite integers?

If $ a < n\minus{}3$ is a composite odd integer that is congruent to $ n$ modulo $ 3$, then $ n\minus{}a \equal{} 3k$ is also a composite integer.

Now let's have a look here: http://www.research.att.com/~njas/sequences/A000040.

[i]Case 1.[/i] $ n\equiv 0\left(\bmod\, 3\right)$
$ a \equal{} 9$ works, so all $ n > 12$ are good.

[i]Case 2.[/i] $ n\equiv 1\left(\bmod\, 3\right)$
$ a \equal{} 25$ works, so all $ n > 28$ are good.

[i]Case 3.[/i] $ n\equiv 2\left(\bmod\, 3\right)$
$ a \equal{} 35$ works, so all $ n > 38$ are good.

The answer is $ \boxed{38}$ (if my computations are correct).


No twists this time!!!

rules here.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225132

me is joining

me hopes this post is not too short

Chair 8.

Now isn't that a really short message.

I'm still learning the rules, and I'll probably die first, but I take chair 6

I don't know if egglv has to be the one who initiates rounds, but if so, that holds true next round. The music has stopped. I've got chair 6. Somebody else post the people table thing, becuase I'm knew and I do't know exactly how to do it.

I'm pretty sure egglv has to initiate the round.

1
jjx1
2
abcak
3
auroraak
4
powerofpi
5
joe10112
6
algebra2000

PAITEINCE!

But I seriously wonder WHY no one is joining...

I asked if we were starting because you posted a list of current participants.

no, evidently, in spite of the huge amount of people that use this forum :huuh:  :yup:  :whistling:

So we say a chair number?

[b]CHAIR 4[/b]

EDIT: Oh, so I say 4 and I edited my post. This is my first time playing.

[quote="PowerOfPi"]So we say a chair number?

[b]CHAIR 2[/b][/quote]

say one close to where you are

HOW DARE YOU STEAL MY CHAIr!

[b]chair 3[/b]

1
jjx1
2
3 abcak
auroraak
4 powerofpi
5
joe10112
6
algebra2000


Let $ a,b\ge 0$. Find the best k constant such that:
\[ a^2(b \plus{} 2k) \plus{} 2a( \minus{} k \minus{} 3) \plus{} (kb^2 \minus{} kb \minus{} 3b \plus{} k^3)\ge 0
\]

\[ a^2k \plus{} a( \minus{} 3 \minus{} k) \plus{} (kb^2 \minus{} kb \minus{} 3b \plus{} k^3)\ge 0
\]
Hope that [b] ji chen, kunny or argady[/b] will solve it  :wink:

@sorry, I have made mistake in typing  :roll: I have edited it :D

[quote="zaizai-hoang"]Let $ a,b\ge 0$. Find the best k constant such that:
\[ a^2(b \plus{} 2k) \plus{} 2a( \minus{} k \minus{} 3) \plus{} (kx^2 \minus{} kx \minus{} 3x \plus{} k^3)\ge 0
\]

\[ a^2k \plus{} a( \minus{} 3 \minus{} k) \plus{} (kb^2 \minus{} kb \minus{} 3b \plus{} k^3)\ge 0
\]
Hope that [b] ji chen, kunny or argady[/b] will solve it  :wink:[/quote]

What is x?  :?:

Noone solve it :D Please spend your time to solving above inequality, ji chen :D

[quote="zaizai-hoang"]Let $ a,b\ge 0$. Find the best constant $ k$ such that

$ (b \plus{} 2k)a^2 \minus{} 2(k \plus{} 3)a \plus{} kb^2 \minus{} (k \plus{} 3)b \plus{} k^3\ge 0$.[/quote]
Put $ a \equal{} b \equal{} 1$, we have $ k^3 \minus{} 8\geq0\Longrightarrow k\geq 2$. On the other hand,

$ (b \plus{} 2k)a^2 \minus{} 2(k \plus{} 3)a \plus{} kb^2 \minus{} (k \plus{} 3)b \plus{} k^3$

$ \equal{} (k \minus{} 2)^3 \plus{} 6(k \minus{} 2)^2 \plus{} \left[(a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \plus{} a^2 \plus{} b \plus{} 10\right](k \minus{} 2)$

$ \plus{} \frac {(2b \plus{} 7)(a \minus{} 1)^2 \plus{} (a \plus{} 2b \minus{} 3)^2}{2}\geq0$,

which is clearly true for $ k \geq 2$.

Very nice, ji chen  :wink: Can you solve the second?  :wink:

[quote="zaizai-hoang"]Let $ a,b\ge 0$. Find the best constant $ k$ such that

$ ka^2 \minus{} (k \plus{} 3)a \plus{} kb^2 \minus{} (k \plus{} 3)b \plus{} k^3\ge 0$.[/quote]
Put $ a \equal{} b \equal{} 1$, we have $ k^3 \minus{} 6\geq0\Longrightarrow k > 0$.

$ k^3 \plus{} (a^2 \plus{} b^2 \minus{} a \minus{} b)k \minus{} 3(a \plus{} b)$

$ \equal{} \frac {k(a \minus{} b)^2}{2} \plus{} \frac {(ka \plus{} kb \minus{} k \minus{} 3)^2}{2k} \plus{} \frac {2k^4 \minus{} k^2 \minus{} 6k \minus{} 9}{2k}$

$ \equal{} \left(2\sqrt {2}k \minus{} 1 \minus{} \sqrt {1 \plus{} 12\sqrt {2}}\right)\left(2\sqrt {2}k \minus{} 1 \plus{} \sqrt {1 \plus{} 12\sqrt {2}}\right)\frac {\sqrt {2}k^2 \plus{} k \plus{} 3}{8\sqrt {2}k}$

$ \plus{} \frac {(ka \plus{} kb \minus{} k \minus{} 3)^2}{2k} \plus{} \frac {k(a \minus{} b)^2}{2}\geq 0$

holds for all nonnegative real numbers $ a,b$ if and only if $ k\geq\frac {1 \plus{} \sqrt {1 \plus{} 12\sqrt {2}} }{2\sqrt {2}} \equal{} 1.85\cdots$.
[quote="isomorphism"][quote="zaizai-hoang"]Let $ a,b\ge 0$. Find the best constant $ k$ such that

$ a^2(b \plus{} 2k) \minus{} 2(k \plus{} 3)a \plus{} kx^2 \minus{} (k \plus{} 3)x \plus{} k^3\ge 0$.[/quote]What is $ x$? [/quote]
$ x$ is a real number.

I am really curious why you can factor expression like that  :o It is not easy to factor like that  :roll: Can you explain something about this  :wink:

[quote="zaizai-hoang"]Let $ a,b\ge 0$. Find the best constant $ k$ such that

$ ka^2 - (k + 3)a + kb^2 - (k + 3)b + k^3\ge 0$.[/quote]

$ ka^2 - (k + 3)a + kb^2 - (k + 3)b + k^3\ge 0$

$ \Longleftrightarrow k(a^2 + b^2) - (k + 3)(a + b) + k^3\geq 0$

$ \Longleftrightarrow k(1^2 + 1^2)(a^2 + b^2) - 2(k + 3)(a + b) + 2k^3\geq 0$ for $ a + b\geq 0.\ \cdots (*)$

Case 1: $ k = 0$
We have $ - 6(a + b\)geq 0$ and $ a + b\geq 0$, yielding $ a + b = 0.$

Case 2: $ k > 0$
Let $ a + b = t\ (t\geq 0),\ (*)$ is $ kt^2 - 2(k + 3)t + 2k^3\geq 0$ for $ t\geq 0.$

If $ t = 0,$ then $ 2k^3\geq 0$ which is valid.

If $ t > 0,$ then for $ f(t): = kt^2 - 2(k + 3)t + 2k^3 = k\left(t - \frac {k + 3}{k}\right)^2 + \frac {2k^4 - (k + 3)^2}{k},$ the axis $ t = \frac {k + 3}{k} > 1,$ so

$ f_{min} = f\left(\frac {k + 3}{k}\right) = \frac {2k^4 - (k + 3)^2}{k}\geq 0$ with $ k > 0\Longleftrightarrow \sqrt {2}k^2\geq k + 3$ and $ k > 0$, yielding

$ k\geq \frac {1 + \sqrt {1 + 12\sqrt {2}}}{2\sqrt {2}}$

If $ k < 0,$ then $ f(0) = 2k^3 < 0$ so this case is invalid. Therefore the best constant of $ k$ is $ \boxed{k\geq \frac {1 + \sqrt {1 + 12\sqrt {2}}}{2\sqrt {2}}}$


integration
calculus
derivative
calculus computations

Hello
Try to find extreme value of $ h\left( x \right) \equal{} {\left( {\int\limits_0^x {{e^{ \minus{} {t^2}}}dt} } \right)^2} \plus{} \frac{\pi }{4}{e^{ \minus{} {x^2}}} \minus{} \frac{\pi }{4},x \in R$

thx

hello, i have found $ h(x)=\sqrt{\pi}\text{erf(x)}e^{-x^2}-\frac{\pi}{2}x\cdot e^{-x^2}$ and the derivative is $ h^'(x)=\frac{1}{2}e^{-x^2}(2\sqrt{\pi}\text{erf(x)}-x\pi)$ by a numerical way we get
$ h'(x)=0 \Leftrightarrow x_1\approx 0.8981847897794521200$ or $ x_2 \approx-.89818478977945212$
Sonnhard.

Though how much is the maximum value? Can we say that it is 0 :maybe: ?

hello, you must insert $ x_1$ or $ x_2$ in $ h(x)$.
Sonnhard.

Yup, but i cant calculate the values since i dont have a programme like mathematica or maple or something like that
Thx though

hello, we have $ h_{max}\approx 0.062770120$
Sonnhard.

Hhmmm... not the 0 that i was expected...
Could it be 0 (maybe cause of the approximation of arithetic number that maximum value appears we have small mistake about maximum value)?
Tanx for your reply :lol:

hello, no, $ x\equal{}0$ gives the global minimum of $ h(x)$.
Sonnhard.


$\frac{2 \sqrt 6}{\sqrt 2 + \sqrt 3 + \sqrt 5}$ equals

A. $\sqrt 2 + \sqrt 3 - \sqrt 5$
B. $4 - \sqrt 2 - \sqrt 3$
C. $\sqrt 2 + \sqrt 3 + \sqrt 6 - 5$
D. $\frac{1}{2} (\sqrt 2 + \sqrt 5 - \sqrt 3)$
E. $\frac{1}{3} (\sqrt 3 + \sqrt 5 - \sqrt 2)$

[quote="JesusFreak197"]How did you know which values to choose to multiply top and bottom by?[/quote]

Well, he got lucky there but that's also what I did.

Let's say you have $\sqrt a + \sqrt b + \sqrt c$. I recommend you multiplying $(\sqrt a + \sqrt b) - \sqrt c$ than $\sqrt a - \sqrt b - \sqrt c$ because the first way ensure that you will get some cancellations no matter what $a,b,c$ are.

[hide]Group the denominator to get $(\sqrt2+\sqrt3)+\sqrt5$. This technique was illustrated in AoPS book vol. 1. Multiply the numerator and denominator by $(\sqrt2+\sqrt3)-\sqrt5$ to get

$\frac{2\sqrt6(\sqrt2+\sqrt3-\sqrt5)}{3+2-5+2\sqrt6}=\sqrt2+\sqrt3-\sqrt5$. [/hide]


Running at constant speed, Andy can run around a circular track once in 40 seconds. Bob, running in the opposite direction, slaps hands with Andy every time they run past each other. If they only slap hands every 15 seconds, how many seconds does it take Bob to run around the track once?

How did you make the equation? Can you please explain your solution more thoroughly?

Thanks,
pi man

This is a (maybe different) solution.

Suppose that Andy and Bob are starting at the same spot but going opposite ways.  Then, we see that after $15$ seconds, they will meet.  Let the track have length $1$ mile (this is actually WLOG), then we see that Andy and Bob's speeds must sum to $\frac{1}{15}$ (they are going towards each other, so we add speeds).  We see that Andy's speed is $\frac{1}{40}$ (because he takes $40$ seconds to run around the track once), so Bob's speed is $\frac{1}{15} - \frac{1}{40} = \frac{25}{40 \cdot 15} = \frac{1}{24}$, so Bob takes $\frac{1}{\frac{1}{24}} = \boxed{24}$ seconds to run around the track once.

Every 15 seconds Bob and Andy meet. That means the sum of their speeds will take 15 seconds to go around the track, because their distance traveled all together is equivalent to the entire track. Let's say the track is 1. So $\frac{1}{40}$ is Andy's speed and $\frac{1}{15}$ is the sum of the speeds. So, then Bob's speed is the reciprocal of $\frac{1}{15}-\frac{1}{40}$ which equals $\frac{1}{24}$ so Bob's speed is 24 seconds to go around the track.

Andy runs around the track in 40 seconds, so in 15 seconds, he goes 3/8 of the way around the track.  That means Bob has to go 5/8 of the track in 15 seconds.  15*8/5 = 24.

[quote=math154]This means that Andy can run $ \frac1{40}$ of the track every second. If Bob takes $ x$ seconds to run the track once, then his speed is $ \frac1x$ per second. Thus, we have $ \frac{15}{40}+\frac\frac{15}x=1$ ($ d=rt$), and $ x=\boxed{24}$.[/quote]

check ur LaTeX


[quote="Karth"]Hmm... I don't understand the solution. Could you explain it to me?  :maybe:[/quote]
1) Square both sides.
2) Use $|a|+|b|\geq |a+b|$ and $|z|^{2}=z\cdot \bar{z}$.

Ah I understand  :) Nice solution! What was the motivation?

It's the difinition of Complex number which is $a+bi\ (a,\ b\in\mathbb{R})$.


The arithmetic mean of these five expressions is 24. What is the
value of $ x$?
\[ x \plus{} 8 \hspace{.5cm} 15 \hspace{.5cm} 2x \hspace{.5cm} 13 \hspace{.5cm} 2x \plus{} 4\]

If the mean of these five expressions is 24, then the sum is $ 24 \cdot 5 \equal{} 120$. So we can write the following equation:

$ x\plus{}8\plus{}15\plus{}13\plus{}2x\plus{}2x\plus{}4\equal{}120$

Solving for $ x$, we find that $ x \equal{} \boxed{16}$.

Alternatively, we find the "distance" from each one to 24 (term-24) and we get

x-16,-9,2x-24,-11,2x-20

These must all sum to 0 (can u see why?)

x-16-9+2x-24-11+2x-20=5x-80=0 so x=16.


I hope you like it:
$a,b,c$ are positive real numbers such that $abc=1$. Prove the inequality
$\frac{a^5}{2bc-1}+\frac{b^5}{2ac-1}+\frac{c^5}{2ab-1}\ge 3$

I have edited it. Hope this time is correct  :|
Solution
[hide]
$\frac{a^5}{2bc-1}+\frac{b^5}{2ac-1}+\frac{c^5}{2ab-1}\underbrace{\ge}_{Chebishov}\frac{1}{3}(a+b+c)\left(\frac{a^4}{2bc-1}+\frac{b^4}{2ac-1}+\frac{c^4}{2ab-1}\right)\underbrace{\ge}_{We \cdot use\cdot that(a^2-1)^2\ge 0\cdot etc.}\frac{1}{3}(a+b+c)\left(\frac{2a^2-1}{2bc-1}+\frac{2b^2-1}{2ac-1}+\frac{2c^2-1}{2ab-1}\right)\underbrace{\ge}_{Cauchy\cdot in\cdot Engel\cdot form}\frac{1}{3}(a+b+c)\frac{3(2(a^2+b^2+c^2)-3)}{2(ab+bc+ca)-3}\underbrace{\ge}_{a^2+b^2+c^2\ge ab+bc+ca}(a+b+c)\underbrace{\ge}_{AM-GM}3\sqrt[3]{abc}=3$
[/hide]

I think there must be another condition for $2ab,2bc,2ca > 1$, :)
here is another proof:
$(6-a-b-c)(\frac{a^5}{2bc-1}+\frac{b^5}{2ac-1}+\frac{c^5}{2ab-1})=(2abc-a+2abc-b+2abc-c)(\frac{a^5}{2bc-1}+\frac{b^5}{2ac-1}+\frac{c^5}{2ab-1}) \underbrace{\ge}_{Cauchy} (a^3+b^3+c^3)^2$
then we should only to prove:
$(a^3+b^3+c^3)^2 \ge 3(6-a-b-c)$
this is so easy :)

[quote="Beat"]Very nice . But how will you prove the last inequality (the so easy part)
 :blush:  :icecream:[/quote]
Sorry,It must be:$(a^3+b^3+c^3)^2 \ge 3(6-a-b-c)$
proof:
$a+b+c \ge 3,a^3+b^3+c^3 \ge 3$ :lol:


search
USAMTS
summer program
PROMYS
geometry
3D geometry
tetrahedron

IMTS is International Mathematical Talent Search which is good source you can use to study for USAMTS. But sadly, the solution to the problems are not available. So, here is the idea. Why not make the solution thread of it so people can look at it and study for USAMTS?

I'll majorly post the problems. Not that I know better than others but that I have more time :P .

Anyway, here is one problem:

Posting problem, please source:

Year, Problem Number, and original wording. Thanks.

Round 41
Problem 1

Determine the unique positive two-digit numbers $m$ and $n$ for which the approximation $\frac {m}{n} = 0.2328767$ is accurate to the seven decimals; i.e. $0.2328767 \leq \frac {m}{n} < 0.2328768$.

[hide="My_Sol"]

First, my answer is m = 17 and n = 73. I'm pretty sure this is the answer because it worked when I punched in the calculator.

But I think the answer is the second most important thing in this problem. The most important thing is "how" to get the answer.

Since we know that $0.2328767 \text {close to} 0.23$ or $0.24$, we can start from either one. However, it is more close to $0.23$ or $\frac {23}{100}$. Thus, here's what I did.

Changing the numerator and denominator, I began to find the close form of the number.

$\frac {22}{95}, \frac {21}{90}, \frac {20}{86}, \frac {19}{82}, \frac {18}{78}$ and then I got my answer $\frac {17}{73}$.

But I'm pretty confident that there are non-guess-and-check ways because if this problem was reworded like asking you to find the answer, with proof, then I would've got into trouble because my guess-and-check doesn't prove anything.[/hide]

How would you do this problem? Also, is my answer correct? :)

Round 5
Problem 1

The set S consists of five integers. If pairs of distinct elements of S are added, the following ten sums are obtained: 1967, 1972, 1973, 1974, 1975, 1980, 1983, 1984, 1989, 1991. What are the elements of S?

[hide="Solution"]

First and most of all, this was pretty tough problem for #1. But again, this is INTERNATIONAL so I assume it has to be hard.

Let's call each five integers $a,b,c,d,e$. Notice that every sum that is found there are in form of $a+b, b+c,d+e,$ and so on.

So, all numbers are in form of:
$a+b,a+c,a+d,a+e,b+c,b+d,b+e,c+d,c+e,d+e.$
Hence, adding them will give us $4a+4b+4c+4d+4e = 4(a+b+c+d+e) = \text {sum of 10 resulted sums}$.

To find $a+b+c+d+e$, we divide 4 by both sides and get that it equals to 4947.

Now, here is the hard part. Looking at the unit digit, they are:
7,2,3,4,5,0,3,4,9,1.

And since one digit is 1, 2 out of 5 digits must be 0 and 1. Now, rest of 3 can be done by finding the repeated ones. So, the unit digits are:
1,0,4,3,9.

Since 1991 is equal to 0+1+one number+another number, we get that if we subtract 1991-1 = 1990, we find that another number has to be 1000. If the first number was over 1000, the calculation wouldn't work. So, we find that first two numbers are 991 and 1000.

Using 991, we find the two other numbers. For the last one, however, we use the earlier fact that a+b+c+d+e = 4947 and get the fourth number is 984.

Therefore, all five numbers in set S are:

991, 1000, 983, 984, 989.[/hide]

Do you guys have any other ideas on doing this problem?

Better solution than mine!

Thanks to pilot.

[hide="Alt"]

$a<b<c<d<e$ since it's distinct. So, it became obvious that if we rewrite each smaller terms in a,b,c,d, and e, it's easy to find what they are.[/hide]

This method is MUCH better than mine. Mine was long-extra-working method while this method was easy. :)

[quote]Now, here is the hard part. Looking at the unit digit, they are:
7,2,3,4,5,0,3,4,9,1.

And since one digit is 1, 2 out of 5 digits must be 0 and 1.[/quote]

This isn't necessarily true - two digits could be 5 and 6, for example.  You got lucky in that 0 and 1 were actually the digits.

As an alternate method, you can find one of the five original numbers as follows: let a < b < c < d < e.  Then a+b will be the smallest member of the group of pairwise sums, and d+e will be the greatest.  So you know that $c=(a+b+c+d+e) - (a+b) - (d+e) = 4947 - 1967 - 1991 = 989$.  Using this information, you should be able to find the other four much more easily.

Oh, yeah. I got lucky :D  :D

Yeah, your method is way easier than my guess-and-check.

I'm not sure how I'll prove for this one. This is bit obvious but I'm not sure where to prove this.

Round 21
Problem 2

Find the smallest positive integer that appears in each of the arithmetic progressions given below, and prove that there are infinitely many positive integers that appear in all three of the sequences.

5,16,27,38,49,60,71...
7,20,33,46,59,72,85....
8,22,36,50,64,78,92...

[hide="Um"]

I got to the part on defining each sequence as:

In order:

$5 + 11k\\
7 + 13k\\
8 + 14k$

But I'm not sure how to find the first term and proving there is infinitely many solutions. I KNOW there is infinite number of solutions but I'm not sure how to prove this.[/hide]

Help! :(

[hide]
So you have 11k+5, 13k+7, and 14k+8.

Do it by pairs. Find the first common term in 11k+5 and 13k+7. This is 49. Then make a new sequence 49+143k  which describes the numbers that satisfy the first two. Find the first common terms among 49+143k and 14k+8. This is something big... just call it x for now (you can find it using promys number theory, but its not fun, its better just to use a computer or something). Then all numbers of the form x+11*13*14k are in those 3 sequences so there are infinite.                                                                           [/hide]

Oh.

Thanks.

Yeah. If it was real problem, I might say like since $k$ can be any real number, it means that there are infinitely many solutions.

[quote="Silverfalcon"]

Determine the unique positive two-digit numbers $m$ and $n$ for which the approximation $\frac {m}{n} = 0.2328767$ is accurate to the seven decimals; i.e. $0.2328767 \leq \frac {m}{n} < 0.2328768$.
[/quote]
One  standard/easy way to do this problem .

[hide]Convert this into continued fraction:  [Take reciporcal,  note down the integerpart and take reciprocal of the fractional part, and repeat this)

You will get something like: ${\frac {1}{4+\frac {1}{3+\frac {1}{2+\frac {1}{2}}}}}$

And "best"  approximations of the above would be $\frac {1}{4} , \frac  {3}{13} , \frac{7}{30} , \frac {17}{73} $ etc..

[/hide]

Cool way, Gyan! :D

Here is another problem that I think may have more than one solution. I'm not quite sure of the "wording."

Round 32
Problem 1

Exhibit a 13-digit integer N that is an integer multiple of $2^{13}$ and whose digits consist of only 8s and 9s.

[hide="Two_Answers"]
Maybe there are more but here are two I found:

9888888888888
8989898989898[/hide]

What does word "exhibit" means? Does that mean list them? :?

[quote="Silverfalcon"]Here is another problem that I think may have more than one solution. I'm not quite sure of the "wording."

Round 32
Problem 1

Exhibit a 13-digit integer N that is an integer multiple of $2^{13}$ and whose digits consist of only 8s and 9s.

hide=Two_Answers
Maybe there are more but here are two I found:

9888888888888
8989898989898
/hide

What does word "exhibit" means? Does that mean list them? :?[/quote]
9888888888888 = 8 * 1236111111111
8989898989898 = 2 * 4494949494949
Neither are divisible by $2^{13}$.

Verifying solutions is an important part of a USAMTS answer (and the IMTS had the same standards).  if you had done that, you would not have posted erroneous solutions.

Describing your method of finding the solution is another important part of a USAMTS answer. I have no idea who you created your erroneous solutions, though repetition of digits seems to be a part of it.

"Exhibit" is a synonym for "show". It means show such a number N. If you find more than one such number, you need give only one. However, finding them all and proving you found them all could lead to a commendation.

If you want a strong hint:
[hide]Since N is divisible by $2^{13}$, it is even, so its rightmost digit has to be an even number. So the rightmost digit is 8.

N is divisible by 4. Any multiple of 100 is divisible by 4, so we need check only the rightmost two digits of N to discover whether it is divisible by 4. Since the rightmost digit is 8, the rightmost two digits are 88 or 98. 88 is divisible by 4, but 98 is not. So the rightmost two digits are 88.

N is divisible by 8. Any multiple of 1000 is divisible by 8,...

Since this is only a hint, I am stopping here.[/hide]

Since your goal is to create a list of answers for all the IMTS problems, perhaps you should not jump from round to round so randomly. This thread would be more useful if you went through the rounds in order (or reverse order, because the later problems have more polished wording).

Erin Schram
USAMTS grader

Ooh. :blush:

Yes, I'll start to do from Year 1 to current.

Although I can't do right now since I"m busy little bit. :lol:

$8898989989888=2^{13}(1086302489)$

Really simple:
If divisible by $2^n$ the number formed by the last n digits is divisible by $2^n$.
xxxxxxxxxxxx8
xxxxxxxxxxx88
xxxxxxxxxx888
xxxxxxxxx9888
xxxxxxxx89888
xxxxxxx989888
xxxxxx9989888
xxxxx89989888
xxxx989989888
xxx8989989888
xx98989989888
x898989989888
$8898989989888$

Try it with an 8 every time.  If it divides evenly, then use an 8. If not, use a 9.

Round 1
Problem 1

For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases:

(i) $10 \leq n \leq 99$
(ii) $100 \leq n \leq 999$
(iii) $1000 \leq n \leq 9999$
(iv) $10000 \leq n \leq 99999$

Since IMTS problems are so long and stuffs, I'll put one problem per thread or two. Here is my answer:

I'm pretty sure I'm wrong but it is pretty LONG.

[hide="Answer"]

(i) Rewrite the expression:

$\frac {n}{s(n)} \Rightarrow \frac {10a+b}{a+b}$

Thus, the value is minimum when they are close to each other. Further they're apart, the greater value is because $n > s(n)$.

[b]Lemma 1[/b] $n>s(n)$

Assume $n<s(n)$. If we plug in 1 and 0 for $a$ and $b$ respectively, 10+0 = 10 for $n$ and 1+0 = 1 for $s(n)$. But 10>1 so this contradicts first assumption, therefore, $n>s(n)$.

For $10a+b$, the number will get extremely large as $a$ increases, so the minimum would be when $a=1$.

$\frac {10+b}{1+b}$

I can prove the case for $b$ but since there are only 10 possible choices for $b$, it's easier to simply test each value.

$\frac {10}{s(10)} \Rightarrow 10\\
\frac {11}{s(11)} \Rightarrow 5.5\\
\frac {12}{s(12)} \Rightarrow 4\\
\frac {13}{s(13)} \Rightarrow 3.25\\
\frac {14}{s(14)} \Rightarrow 2.8\\
\frac {15}{s(15)} \Rightarrow 2.5\\
\frac {16}{s(16)} \Rightarrow \frac {16}{7}\\
\frac {17}{s(17)} \Rightarrow 2.125\\
\frac {18}{s(18)} \Rightarrow 2\\
\frac {19}{s(19)} \Rightarrow 1.9$

It is true that other small numbers may occur even when 99 where $99/(9+9) = 5.5$. But the MINIMUM value is when number is 19. Therefore, the minimum value is $\fbox {1.9}$.

(ii). $\frac {n}{s(n)} \Rightarrow \frac {100a+10b+c}{a+b+c}$

Just like above, if we apply the smallest restriction for $a$, which is 1 and then see that when $b=c=9$, the minimum occur so $\frac {n}{s(n)} \Rightarrow \frac {199}{1+9+9} \Rightarrow \text {Minimum is \fbox {199/19}}$.

I'll skip explanation for iii. and iv.

iii. $\fbox {1999/28}$
iv. $\fbox {19999/37}$.[/hide]

Is this wrong? :(

Rest of Round 1

Problem 2
Find all pairs of integers, $n$ and $k$, $2<k<n$, such that binomial coeffcients:

$\binom {n}{k-1}, \binom {n}{k}, \binom {n}{k+1}$

form an increasing arithmetic series.

Problem 3
On an 8 x 8 board we place $n$ dominoes, each covering two adjacent squares, so that no more dominoes can be placed on the remainding squares. What is the smallest value of $n$ for which the above statement is true?

Problem 4
Show that an arbitary acute triangle can be dissected by straight line segments into three parts in three different ways so that each part has a line of symmetry.

Problem 5
Show that it is possible to dissect an arbitary tetrahedron into six parts by planes or portions thereof so that each of the parts has a plane of symmetry.

[hide="Number_2"]

I thought this had "no solution" but I found one by brute forcing.

n=14 and k=5.

This form 1001,2002,3003.

I wonder if there's an algebraic way, (which there should be), on this problem.

And also, is this only answer?[/hide]

Rest of the three problems, I couldn't do it. I didn't even get what "arbitary" meant in 4-5.

Please give help on getting solutions to these problems.

Thank you! :)

For number 4, "arbitrary acute triangle" just means show it can work for any acute triangle.

[hide="Big hint for number 4."]All of the line segments that dissect the triangle pass through a special point belonging to the triangle.
[hide="This point is..."]The incenter.[/hide][/hide]

[quote="Silverfalcon"]Well, can you show whole solution though?

I'm trying to get whole solution to all problems, not just hint ;)[/quote]

[hide="Here it is."]Consider the incircle of the triangle, and its center.  Drop perpendiculars from the incenter to each side (they hit the sides at the points of tangency).  Those perpenduculars are the desired segments.  Each of the four sided regions that the triangle is divided into is a kite, so they have an axis of symmetry (the axes of symmetry are the segments connecting the incenter to the vertices of the triangle).[/hide]

For number 2:
[hide]you know that nCk = n!/[(n-k)!k!], so if you let nCk = a, and since it's in an arithmetic sequence, so nC(k-1) = a - b, and nC(k+1) = a+b. (a-b)+(a+b) = 2a, so divide by 2, so put it in the form and isolate one variable and you'll find out your answer is just one of many[/hide]

Let me clear this up little bit more.

So, do you mean:

$\binom {n}{k} = a\\
\binom {n}{k-1} = a-b\\
\binom {n}{k+1} = a+b$

And that:

$\binom {n}{k-1} + \binom {n}{k+1} = 2 \binom {n}{k}$

But the problem that I have (maybe you posted but I'm not reading it well), is how to find $n$ and $k$ from here...

:-(

[hide]kinda ugly cuz i know no tex
n!/[(n-k-1)!(k+1)!] + n!/[(n-k+1)!(k-1)!] = 2n!/[(n-k)!k!]
1/[(n-k-1)!(k+1)!] + 1/[(n-k+1)!(k-1)!] = 2/[(n-k)!k!]
[(n-k)(n-k+1) + k(k+1)]/[(n-k+1)!(k+1)!] = 2/[(n-k)!k!]
[(n-k)(n-k+1) + k(k+1)]/[(n-k+1)(k+1)] = 2
(n^2 - nk + n - nk + k^2 - k + k^2 + k) = 2(nk + n - k^2 - k - k + 1)
n^2 - 2nk + n + 2k^2 = 2nk + 2n - 2k^2 - 4k + 2
n^2 - 4nk + 4k^2 = n + 2
(n - 2k)^2 = n+2
n - 2k = sqrt(n+2)
2k = n - sqrt(n+2)
k = [n - sqrt(n+2)]/2

Basically, since n has to be greater than k. Take the square root of n+2 and subtract from n and divide by 2, so it'll basically always work as long as n+2 is a positive perfect square greater than 4.

Taking your example:
[14 - sqrt(16)]/2
= (14-4)/2
= 10/2
= 5
[/hide]

Thanks a lot, theone.

So, this suggests that this problem has infinitely many solutions. :D

Good problem.

Not sure so I'm asking this.

Did all problems for the round 1 answered?

I answered 1, theone answered 2, dts answered 4 but no one did #3 and #5.

Can anyone give a hand?

I tried #3 and here's what I got:

2x2 --> 2 dominoes
3x3 --> 4 dominoes
4x4 --> 6 dominoes
5x5 --> 9 dominoes!!!!

It's breaking the step. :(

I haven't tried 8 yet because it looks really nasty.

Any ideas?

:?

Is there any specific method on doing the dominoe problem like this?

just letting you know, your numbers are off

[quote]Find the smallest positive integer that appears in each of the arithmetic progressions given below, and prove that there are infinitely many positive integers that appear in all three of the sequences.

5,16,27,38,49,60,71...
7,20,33,46,59,72,85....
8,22,36,50,64,78,92... [/quote]

[hide]Well, going one term backwards, we find that they are
$-6+11k$
$-6+13k$
$-6+14k$

Since 11, 13, and 14 are relatively prime, the next common term will be $11 \cdot 13 \cdot 14 - 6$.

There will be infinitely many common terms, all of the form $11 \cdot 13 \cdot 14 \cdot k -6$, so we are done.[/hide]

$k$ is assumed to be in set of $[1,2,3....\infty)$, right? :D

Round 3 (I forgot where I put Round 2 papers so let's do this first..)

Problem 1:

Note that if the product of any two distinct members of {1,16,27} is increased by 9, the result is perfect square of an integer. Find the unique positive integer $n$ for which $n+9$, $16n+9$, and $27n+9$ are also perfect squares.

I only have an approach for number one. :? Here is how far I got:

[hide="Approach.."]

Let's say:

$n+9 = x^2\\
16n+9 = y^2\\
27n+9 = z^2$

Since $9(3n+1) = 27n+9$, we can have $3n+1 = u^2$.

So, we have:

$n+9 = x^2\\
16n+9 = y^2\\
3n+1 = u^2$

Only problem is how am I going to work from here? :blush: [/hide]

Problem 2:

Note that 1990 can be "turned into a square" by adding a digit on its right, and some digits on its left; i.e., $419904 = 648^2$. Prove that 1991 can not be turned into a square by the same procedure, i.e. there are no digits $d,x,y...$ such that ....$yx1991d$ is a perfect square.

Problem 3:

Find $k$ if P,Q,R, and S are points on the sides of quadrilateral ABCD (P in AB, Q in BC, R in CD, S in AD) so that:

$\frac {AP}{PB} = \frac {BQ}{QC} = \frac {CR}{RD} = \frac {DS}{SA} = k$,

and the area of quadrilateral PQRS is exactly 52% of the area of quadrilateral ABCD.

Problem 4:

Let $n$ points with integer coordinates be given in the xy-plane. What is the minimum value of $n$ which will ensure that three of the points are the vertices of a triangle with integer (possibly, 0) area?

Problem 5:

Two people, A and B, play the following game with a deck of 32 cards. With A starting, and thereafter the players alternating, each player takes either 1 card or a prime number of cards. Eventually all of the cards are chosen, and the person who has none to pick up is the loser. Who will win the game if they both follow the optimal strategy?

Wow.. I'm proud of myself for finding the answer! :D  :D

It's not that neat but hey, I found the answer.

[hide="Whew!"]

From where my last post stopped, I used basic systems of equation:

$16n + 144 = 16x^2\\
16n + 9 = y^2\\
---------------\\
135 = 16x^2 - y^2\\
135 + y^2 = 16x^2\\
\frac {135 + y^2}{16} = x^2$

Also:

$3n + 27 = 3x^2\\
3n + 1 = u^2\\
----------------\\
26 = 3x^2 - u^2\\
\frac {26 + u^2}{3} = x^2$

Set them equal:

$x^2 = \frac {135 + y^2}{16} = \frac {26 + u^2}{3}$

$u$ values work for every non-3 multiple numbers like 1,2,4,5,7,8.. so on.

$y$ values work for 3,5,11,13,19,21,27 etc.. +2 and then +6 terms

So, here was my approach:

I started from $y$ values so that $x^2$ is perfect square and when put in for the $u^2$ value, it's also perfect square.

This goes like:

29,35,37,43,45,51,53,59,61,67

And for $u^2 = 29^2$, $y^2 = 67^2$, $x^2 = 17^2$, it works.. Making $n = 280$.

If there was more than one pair, I would have hard time proving it.. :blush:

Is there anymore better way? There should be..[/hide]

I recognize #2 and #3 from the AoPS books (volume 2; I think #2 was in the last chapter and I know #3 was in the chapter on
[hide]analytic geometry[/hide]
), so you could perhaps look in the solutions manual for the solutions for those.

As for #4, I think it asks to be solved with the
[hide]shoelace formula, which states that for a triangle with coordinates $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, the area is half the absolute value of the determinant of the following matrix

[code]x_1   y_1   1
x_2   y_2   1
x_3   y_3   1[/code][/hide]

[quote="Silverfalcon"]Mind on sharing them?

Not all of us have AoPS Volume 2 you know..[/quote]

Uhh... I'm not sure if I can post the solutions from the manual.  It might be breaking a copyright law...

Hmm..

I'll try to do it and see how it goes..

According to my knowledge, as long as you don't copy by word by word, there aren't such strict laws on math materials like math problems. Of course, there are exceptions like ARML and Mandelbrot, I think..

You can always imply what to do you know.. Like:

[hide="Example"]

If it was proving on algebraic inequality, then you can say something like:

$a>b$ so this can mention $a^2 > b^2$ :lol: [/hide]

Some problems will be scattered because I don't like all the problems in each round.

[url=http://www.artofproblemsolving.com/Forum/topic-33297.html]Round 33 Number 1[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-27258.html]Click on the thread, it will tell you[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-27256.html]Nice hard problem[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=227378#p227378]Two nice problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=234129#p234129]Hard Number 4 on Cotangents from Round 23[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=235948#p235948]Round 21 Problem 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=237063#p237063]Arrange the numbers. Don't know the number or round[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38279]#2 on Round 23[/url]

I joined here several minutes ago.
This is my first postion.
I am interested in making math solutions, especailly IMTS problems these days.

Below, the pdf file, is my solution for Round 1 Problem 1.

[quote="Silverfalcon"]Round 1
Problem 1

For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases:

(i) $10 \leq n \leq 99$
(ii) $100 \leq n \leq 999$
(iii) $1000 \leq n \leq 9999$
(iv) $10000 \leq n \leq 99999$

Since IMTS problems are so long and stuffs, I'll put one problem per thread or two. Here is my answer:

I'm pretty sure I'm wrong but it is pretty LONG.

[hide="Answer"]

(i) Rewrite the expression:

$\frac{n}{s(n)} \Rightarrow \frac{10a+b}{a+b}$

Thus, the value is minimum when they are close to each other. Further they're apart, the greater value is because $n > s(n)$.

[b]Lemma 1[/b] $n>s(n)$

Assume $n<s(n)$. If we plug in 1 and 0 for $a$ and $b$ respectively, 10+0 = 10 for $n$ and 1+0 = 1 for $s(n)$. But 10>1 so this contradicts first assumption, therefore, $n>s(n)$.

For $10a+b$, the number will get extremely large as $a$ increases, so the minimum would be when $a=1$.

$\frac{10+b}{1+b}$

I can prove the case for $b$ but since there are only 10 possible choices for $b$, it's easier to simply test each value.

$\frac{10}{s(10)} \Rightarrow 10\\ \frac{11}{s(11)} \Rightarrow 5.5\\ \frac{12}{s(12)} \Rightarrow 4\\ \frac{13}{s(13)} \Rightarrow 3.25\\ \frac{14}{s(14)} \Rightarrow 2.8\\ \frac{15}{s(15)} \Rightarrow 2.5\\ \frac{16}{s(16)} \Rightarrow \frac{16}{7}\\ \frac{17}{s(17)} \Rightarrow 2.125\\ \frac{18}{s(18)} \Rightarrow 2\\ \frac{19}{s(19)} \Rightarrow 1.9$

It is true that other small numbers may occur even when 99 where $99/(9+9) = 5.5$. But the MINIMUM value is when number is 19. Therefore, the minimum value is $\fbox {1.9}$.

(ii). $\frac{n}{s(n)} \Rightarrow \frac{100a+10b+c}{a+b+c}$

Just like above, if we apply the smallest restriction for $a$, which is 1 and then see that when $b=c=9$, the minimum occur so $\frac{n}{s(n)} \Rightarrow \frac{199}{1+9+9} \Rightarrow \text {Minimum is \fbox {199/19}}$.

I'll skip explanation for iii. and iv.

iii. $\fbox {1999/28}$
iv. $\fbox {19999/37}$.[/hide]

Is this wrong? :([/quote]

[quote="mathland"]I joined here several minutes ago.
This is my first postion.
I am interested in making math solutions, especailly IMTS problems these days.

Below is my solution for Round 1 Problem 1.

[quote="Silverfalcon"]Round 1
Problem 1

For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases:

(i) $10 \leq n \leq 99$
(ii) $100 \leq n \leq 999$
(iii) $1000 \leq n \leq 9999$
(iv) $10000 \leq n \leq 99999$

Since IMTS problems are so long and stuffs, I'll put one problem per thread or two. Here is my answer:

I'm pretty sure I'm wrong but it is pretty LONG.

<Hidden Answer snipped>

Is this wrong? :([/quote]<IMTSRound1Problem1Solution(Eng).gif not included>[/quote]

Sorry, mathland, but the characters in your gif file have been rendered into small rectangles. I cannot read it.

By the way, Silverfalcon, your answer was wrong for cases iii and iv. For example, 1099/19 < 1999/28.

Erin Schram

[quote="Erin J. Schram"]Sorry, mathland, but the characters in your gif file have been rendered into small rectangles. I cannot read it.[/quote]
[url=http://img89.imageshack.us/img89/721/imtsround1problem1solutioneng9.gif]does this help?[/url]
btw if youre running internet explorer you may have to set an option so that it shows the full picture, and not the scaled-down one (which is what it does by default for big pictures last i used it)

IMTS Round1 Problem2 Solution

Here again...

Round 44 problem 1 can be solved by defining
f(w) is a function from words to integers defined as
f(w) = (number of a's in w) + 3*(number of b's in w)

f('a') = 1
Applying rule 1 to w to get w' gives f(w')=f(w)+3
Appying rule 2 gives f(w')=f(w)
Appying rule 3 gives f(w')=f(w)-9
Appying rule 4 gives f(w')=2*f(w)
Then you can show that f(w') can never be a multiple of 3 if f(w) is not a multiple of 3. Since f('a')=1 is not a multiple of three, no word w can have f(w) = 0 (mod 3).

f('baabaabaa')=15 is a multiple of three, therefore it is not a word.

-someone posted this with my account, some moderator please delete it-


ARML
geometry
search
ratio
power of a point
Pythagorean Theorem
similar triangles

I'm trying to learn power of a point theorem but I need some actual practice problems, not the proofing problems like ones in aops books.  Can anyone post any useful ones for me to do?

I' got one. This was adapted from this year's ARML Super Relay. It combines the Pythagorean Theorem as well as power of the point.

In a circle, there are chords AB and CD, which intersect at E perpendicular to each other. DE is 4, EB is 16, and AD is 5. Determine the length of CE.

[hide]DE=4, AD=5, and its right triangle, so AE=3.

AE*BE=DE*CE, so 48=4CE, so CE is 12.[/hide]

Am I wrong?

I got the same answer as Platypus. Unfortunately I had never heard of power of a point before this discussion. In geometry we learned things like tangent-secant, secant-secant, chord-chord. I didn't realize it was so simple.

[url=http://www.cut-the-knot.org/pythagoras/PPower.shtml]Here[/url] for power of a point. [url=http://www.google.com/search?hl=en&ie=UTF-8&q=power+of+a+point&btnG=Google+Search]Google[/url] is your friend.

Power of a Point is a very powerful theorem. (Excuse the pun.) Click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12578]here[/url] (caution: link leads to intermediate forum) for a question I made up. The solution requires the usage of the said theorem.

[/shameless self-promotion]

[quote="Yasha"]I did do a Google search and I did look at that specific webpage before I posted. (I'm not completely ignorant of people's signatures, you know). I guess that I should have been clearer in my post that I was not asking what power of a point was, just commenting on how it's funny that I haven't heard of it until this discussion.[/quote]
Oh, sorry. No offense meant. I never heard of that term before, either. I knew what the theorem said, though. Sorry if I sounded harsh.

[quote="Duck-Billed Platypus"]Oh, sorry. No offense meant. I never heard of that term before, either. I knew what the theorem said, though. Sorry if I sounded harsh.[/quote]

Sorry if I sounded offended :-). I just posted that so I wouldn't look ignorant :-). It's funny how written things make people sound a lot angrier than they actually are...

Until  went to the ARML trip and started discussing math, I knew Power of a Point as the "Chord-Chord Theorem" or something like that, and it still remains one of my favorite theorems.

Power of a Point is actually much more powerful than just the two-chord case.  The two -chord case reduces easily to similar triangles, using fairly simple circle properties.  (This is the easy way to remember it inside a circle, by the way -- when you have crossed chords, power of a point is just similar triangles.)  However, the theorem also works for points outside a circle.

Here are three problems, all of which are related in some way to Power of a Point.

What do you mean "The Power of point E relative to the circle"?[/quote]

Here are the answers...I think

1.[hide]4[/hide]

2.[hide]JG=3, JI=18[hide]

3.[hide]r=8[/hide][/hide][/hide]

Please tell me what i got wrong and why.

In AoPS book one, there is a chapter for this theorem and a lot of proof question on it.  I tried to understand them but I just don't seem to get it.  If anyone can read those and maybe give explanation, I'll be really greatful.

The way I understood Power of a Point is that you take a point P and a circle O. P can be inside or outside the circle. Then you draw a random line through P which intersects O at points A and B. Note that if the line is tangent to the circle A and B are the same point. Then you multiply (PA)(PB). That value is the power of point P relative to circle O. The cool thing is no matter what other lines you draw through point P intersecting circle O, that product will always be the same.

Could you please explain each case to us, JBL, or some equally knowledgeable person?

Power of a Point is one very great theorem in geometry.  You have a point P, a circle O, and two lines through p and each intersects the circle twice.  One line intersects the circle at A and B, the other at C and D.  Power of a point says that (PA)(PB) = (PC)(PD).  This can be broken up into different cases.  What if both lines only touch the circle once (they are tangent to the circle)?  Then A and B would be in the same place, and C and D would be in the same place.  T
(PA)(PB) = (PA)^2
(PC)(PD) = PC^2

(PA)^2 = (PC)^2
PA = PC

Thus, two tangents to a circle from the same point are equal.  If there is one tangent line and one secant line (a line that touches the circle twice), A = B but C is not in the same place as D.  Again by the power of a point,

(PA)^2 = (PC)(PD).

And lastly, for two secants,

(PA)(PB) = (PC)(PD).

Responding to your BobofBob's answers:

It seems you almost got number 1. Just remember that [hide]the ratio of the areas of two similar triangles is the square of the ratios of the corresponding sides.[/hide]

For number 2, [hide](EF)(EG) is the power of point E relative to the circle, which you know to be 18. Write EG as the sum of EF, FJ, and JG and then solve for JG. Also note that although the power of point P is 18, you don't know the power of point J. You can figure it out by multiplying (JF)(JG).[/hide]

So,

1.[hide]16[/hide]

2.[hide]JG=1, but I can't figure how get HJ or JI.[/hide]

3.[hide]r=31/6[/hide]?

BobofBobs, #1 is correct, and in #2, the problem tells you that HJ=1.

I think that's right.

A hint for #3: [hide]What is important about C in the problem?[/hide]

BobofBobs wrote:Is [hide]C important because it is an extension of the radius to P?[/hide]

[hide]CP is indeed an extension of the radius to P. However, the only way you'd know that O, C, and P are collinear is because it says that C is the closest point on the circle to P. You can use this fact by extending CP through O to the other side of the circle.[/hide]

It seems like everything here is under control.  I've seen two good explanations of power of a point so far.  It happens that the power of a point with respect to a circle can be expressed by a very simple formula relating the distance of the point to the center and the radius of the circle.  Anyone want to find it?

If you ever want an explanation of something, it's always worthwhile to try the [url=http://mathworld.wolfram.com/]Mathworld[/url] website, a truly excellent resource.


Woohoo, having fun with my new Rapidographs (.5 and .25). I should probably get around to cleaning my long-clogged .30 one for more line consistency (And just a better selection.)

First one is just a warm-up. I've been a bit rusty and getting back into drawing. Second one I quite like. Might keep him and use him for something...


function
parameterization
parametric equation

Well, I'm not sure which forum this goes in, but...

How do you find the equation of the line between two points, in 3D space? eg (x1,y1,z1) and (x2,y2,z2)? Thanks.

The standard parametric equation is found by specifying that $ t \equal{} 0$ corresponds to the first point and $ t \equal{} 1$ to the second point; letting $ \mathbf{v}_1 \equal{} (x_1, y_1, z_1), \mathbf{v}_2 \equal{} (x_2, y_2, v_2)$ we have the equation

$ \mathbf{L} \equal{} \mathbf{v}_1 \plus{} t (\mathbf{v}_2 \minus{} \mathbf{v}_1)$.

In component terms, this is

$ x \equal{} x_1 \plus{} t(x_2 \minus{} x_1)$
$ y \equal{} y_1 \plus{} t(y_2 \minus{} y_1)$
$ z \equal{} z_1 \plus{} t(z_2 \minus{} z_1)$.

There is no 'equation of the line' in $ \mathbb{R}^3$, which can be shown in linear algebra. But, you can have a set of linear parametric equations.

In $ \mathbb{R}^2$, the equation of the line between $ (x_0,y_0)$ and $ (x_1,y_1)$ is $ (y\minus{}y_0)(x_1\minus{}x_0)\equal{}(y_1\minus{}y_0)(x\minus{}x_0)$. Parametrizing,

\[ \begin{aligned}x&\equal{}x_0\plus{}(x_1\minus{}x_0)t\\y&\equal{}y_0\plus{}(y_1\minus{}y_0)t\end{aligned}\]

So suppose we want to find such a set of parametric equations of a line passing through $ (x_0,y_0,z_0)$ and $ (x_1,y_1,z_1)$. Using the same analogy as in $ \mathbb{R}^2$, we have

\[ \begin{aligned}x&\equal{}x_0\plus{}(x_1\minus{}x_0)t\\y&\equal{}y_0\plus{}(y_1\minus{}y_0)t\\z&\equal{}z_0\plus{}(z_1\minus{}z_0)t\end{aligned}\]

Note that if $ z\equal{}0$, like 'crunching' the entire z-axis, we are led back to $ \mathbb{R}^2$, which is not surprising.

There is the parametric form $ \vec{r}(t)\equal{}x(t)\vec{\imath} \plus{}y(t)\vec{\jmath} \plus{}z(t)\vec{k}$ too. You would substitute the equations for $ x(t),y(t),z(t)$ as given above and upon evaluation for some $ t$, $ \vec{r}(t)$ would give you an ordered triple.

Another form of representation for a line (the former is for a general curve) is with symmetrical equations. If we solve for $ t$ as a function of $ x,y,z$ then set them all equal to each other, \[ \frac{x\minus{}x_o}{x_f\minus{}x_o}\equal{}\frac{y\minus{}y_o}{y_f\minus{}y_o}\equal{}\frac{z\minus{}z_o}{z_f\minus{}z_o}\]

There would be exceptions to your case, igiul. For instance, lines parallel to one of the axes fail to be parametrized in that form stated because the denominator is going to be zero.


Find all natural numbers n for which $\frac{n}{1!}+\frac{n}{2!}+\cdot \cdot \cdot +\frac{n}{n!}$ is an integer.
(E. Kovac)

If $n \sum_{j = 1}^n \frac{1}{j!}$ is an integer then $n \sum_{j = 1}^n \frac{1}{j!} = k$ for some integer $k$. Equivalently $\sum_{j = 1}^n \frac{1}{j!} = \frac{k}{n} .$
Multipliy both sides by $n!$ to obtain $\sum^n_{j = 1} \frac{n!}{j!} = \sum^{n - 2}_{j = 1} \frac{n!}{j!} + n + 1 = ( n - 1 ) ! \cdot k$
Remark that $( n - 1 ) | \sum^{n - 2}_{j = 1} \frac{n!}{j!}$ for all $n \geqslant 2$. (for $n = 2$, the empty sum is 0 by definition)
Hence $n - 1 | \sum^{n - 2}_{j = 1} \frac{n!}{j!} + n + 1$ is equivalent to $n - 1| n + 1$.
$n - 1| n + 1$ is equivalent to $( n - 1, n + 1 ) = n - 1$.
$( n - 1, n + 1 )$ can be equal only to $1$ or $2$. So $n = 2$ or $n = 3$.
For $n \geqslant 2$ we check that for $n = 2$ and $n = 3$, $n \sum_{j = 1}^n \frac{1}{j!}$ is an integer
For $n = 1$ we check that  $n \sum_{j = 1}^n \frac{1}{j!}$ is also an integer


How many sequences of lenght $ n$ consisting of $ 0,1,2,3$ contain even number of $ 0$ and odd number of $ 1$?

Depends what the other entries can be.  If the sequence can contain rational numbers, then infinitely many.

The sequence consists of
[quote="Sage!"]$ 0, \ 1, \ 2, \ 3$[/quote].

I've found close formula using generating functions. The answer is [hide]$ 4^{n\minus{}1}$[/hide]

[hide="Click to see the solution"]
Using the exponential generating function we have
$ f(x) \equal{} \left( \frac {x^0}{0!} \plus{} \frac {x^2}{2!} \plus{} \frac {x^4}{4!} \plus{} \ldots\right)\left( \frac {x^1}{1!} \plus{} \frac {x^3}{3!} \plus{} \frac {x^5}{5!} \ldots \right)\left(\frac {x^0}{0!} \plus{} \frac {x^1}{1!} \plus{} \frac {x^2}{2!} \ldots\right)^2 \equal{} \\
 \\
f(x) \equal{} \frac {1}{4}(e^{x} \plus{} e^{ \minus{} x})(e^{x} \minus{} e^{x})(e^{x})^2 \equal{} \frac {1}{4}e^{4x} \minus{} \frac {1}{4}$

We can leave $ \frac {1}{4}$ and expand $ \frac {1}{4}e^{x}$

$ \frac {1}{4}e^{4x} \equal{} \frac {1}{4} \sum_{k \equal{} 0}^{\infty} \frac {4^nx^n}{n!}$ so $ \mbox{coeff}[\frac {x^n}{n!}] \equal{} \frac {1}{4} \cdot 4^n \equal{} 4^{n \minus{} 1}$

So the answer is $ 4^{n \minus{} 1}$.
[/hide]

Exponential generating functions are way overkill.  How about a nice recursive method?

[hide]Let $ EO(n)$ be the number we are looking for, and similarly $ EE(n), OE(n)$ and $ OO(n)$.  Then by examining the possible choices for the first entry of a given string we have $ EO(n \plus{} 1) \equal{} 2EO(n) \plus{} OO(n) \plus{} EE(n)$.  It's also clear that $ EO(n) \equal{} OE(n)$ for all $ n$ (switch every 0 to a 1 and vice-versa), so we can write this as $ EO(n \plus{} 1) \equal{} EO(n) \plus{} OE(n) \plus{} OO(n) \plus{} EE(n) \equal{} 4^n$.  Thus for $ n \geq 1$, $ EO(n) \equal{} 4^{n \minus{} 1}$.[/hide]

Here's a bijective proof that uses a very similar idea but doesn't require any recursive equations:
[hide]As in the previous method, $ EO(n) \equal{} OE(n)$ is easy for all $ n$.  For $ n \geq 1$, consider the following involution on strings in $ \{0,1,2,3\}^n$: add 2 to the first entry of the string (modulo 4).  This changes every string counted by $ EO(n) \plus{} OE(n)$ into a string counted by $ EE(n) \plus{} OO(n)$ and vice-versa.  Thus, these two sets have the same size, so $ EO(n) \equal{} \frac {EO(n) \plus{} OE(n)}{2} \equal{} \frac {\frac {4^n}{2}}{2} \equal{} 4^{n \minus{} 1}$ for $ n \geq 1$.[/hide]


[quote="Quevvy"]A right circular cylinder is just large enough to hold $3$ tennis balls. The diameter of a tennis ball is $2.5$ inches. What is the volume of the cylinder measured to the nearest tenth of a cubic inch? Give your answer as a decimal.[/quote]
[hide]
$\frac{25\pi(7.5)}{4}$ is about $150$. multiply it out by hand :)
the answer is actually 147.3 (with the calculator)
[/hide]

[quote="Quevvy"][quote="ckck"][quote="Quevvy"]A right circular cylinder is just large enough to hold $3$ tennis balls. The diameter of a tennis ball is $2.5$ inches. What is the volume of the cylinder measured to the nearest tenth of a cubic inch? Give your answer as a decimal.[/quote]
[hide]So the height would be the sum of all three diameters or 7.5. The radius of the cylinder is the radius of one tennis ball or 1.25. So solving for the volume of a cylinder $\pi r^{2}h=(3.14)(1.5625)(7.5)=36.796875\approx36.8$
I hope I didn't multiply wrong.[/hide][/quote]
I forgot to multiply by pi when we did this problem! :wallbash:[/quote]
I occasionally forget to on some tests. I just kept reminded myself as soon as I saw the problem put 3.14 next to it so I won't forget.

[quote="archimedes1"]:huh: I thought sprints had only 30 problems...[/quote]
This was from 1988. That was when they did 40 problems in 40 minutes. They thought it was too hard and they changed it to 30 problems in 40 minutes.


Using each of the digits {1,2,3,4} exactly once and using zero or more "+" signs, how many different totals can be obtained? No digit can be used as a power, root, etc. Only addition can be done. For example, four values to include are $ 1234$, $ 55 \equal{} 12 \plus{} 43$, $ 127\equal{} 123\plus{}4$, and $ 37 \equal{} 31\plus{}2\plus{}4$.

My method was...weird. Can anyone help?

My way...

i'd use casework, starting from the lowest sum (3 +s) to the highest (no +,1234):

[hide="My Solution"][hide="Case 1: +++"]This is simple; the only one is 1+2+3+4=10, which is the lowest.

$ \boxed{1}$[/hide]

[hide="Case 2: ++"]Think about this from left to right. Just for example, let the leftmost digit be 1. If we have exactly 2 +'s, exactly one number has 2 digits. Let this be the two leftmost digits. However, the sums 12+3+4,13+2+4,14+2+3 are equal. Since the first digit can be either 1,2,3 or 4, there are 4 totals.

$ \boxed{4}$[/hide]

[hide="Case 3: +"]Now it's either a three digit number plus a one digit number or two 2-digits. Here, if the one's digits of both are interchanged, the sum remains the sum. Thus, we only need to think about the left most digits. There are 4*3=12 ways for this.

For 2 2-digit numbers, the ones are again interchangeable. However, we can also switch the numbers around (i.e. commutative  :P ). This is gives (4C2)/2=3.

$ \boxed{15}$[/hide]

[hide="Case 4:"]This is just 4!.

$ \boxed{24}$[/hide]

Total: The total is then $ 1\plus{}4\plus{}15\plus{}24\equal{}\boxed{44}$.[/hide]

Problem with the above method:
[hide]
13 + 24 = 34 + 1 + 2.
[/hide]
Here's my own line of thought:
[hide]
Imagine four bins: ones, tens, hundreds, and thousands. We have to put the four balls 1, 2, 3, 4 into the bins, under the condition that:
no. of balls in the ones bin $ \geq$ no. of balls in the tens bin $ \geq$ no. of balls in the hundreds bin $ \geq$ no. of balls in the thousands bin.
This is since a number can't have a hundreds digit without a tens digit, and so on.

In other words, say, 3 + 124 becomes putting 1 in the hundreds bin, 2 in the tens bin, and 3 and 4 in the ones bin.
So use casework on the possible distributions of balls, but first note that 1+2+3+4 = 10, which is just barely enough to carry from the ones digit to the tens digit, and means no carrying anywhere else could possibly occur.
[hide="1,1,1,1"]
24 ways. Each larger than 1000. No carrying is possible from the hundreds bin, so these cannot possibly overlap with the other ones.
[/hide]
[hide="0,1,1,2"]
12 ways, each larger than 100. Again, no carrying is possible, and any combination has a unique hundreds digit and tens digit, so no repeats.
[/hide]
[hide="0,0,2,2+0,0,1,3"]
0,0,2,2: 6 ways. However ()()(14)(23) = ()()(23)(14) (for instance, 12 + 43 = 21 + 34 = 55) so only 5 ways exist. Note that I think the above solution erred here. 13+24,12+34,12+43,21+34,21+43,31+42.
0,0,1,3: 4 ways.
However the tens digits could repeat, as 1+2=3 and 1+3=4.
i.e. 31+2+4 = 13+24, 41+2+3 = 12+34, so two ways were counted double.
7 ways for this method.
[/hide]
[hide="0,0,0,4"]
1 way, 10 exactly, which is so small it cannot repeat.
[/hide]
24+12+7+1=44
[/hide]
which, somewhat annoyingly, gives the same answer, since the two mistakes cancel out. Which makes me suspect if there are any more mistakes.

bbgun's solution is impressive.

Im pretty sure the only efficient way to do this problem is casework.

I used the same method as bbgun, but I though it was a little tedious. Thanks guys!


probability ratio parameterization function conics hyperbola geometry

Had to edit, maybe this is better:

Earlier this year I was talking to a math professor who gave me this problem.  I'm not really sure where to start.  The only hint he gave me was that he used jacobians to solve the problem:

Given that xy=1, and all (x,y) are equally likely to be chosen, what is the probability that x <1/10?

Any suggestions?

That question seems to be lacking in information -- unless I make some rather large assumptions, I don't see any way to approach it without me knowing other things about x and y and how we choose them.

well, I didn't have any other information but it's possible that the professor left something out.  Obviously, any point satisfies xy=1 and thus is on the curve y=1/x.  Beyond this I can think to try and evaluate ratios of arc length, but that doesn't seem to yeild anything promising.  Maybe I misinterpreted the conditions for this problem.  If so, has anyone seen any related problem?

The problem is indeed incomplete.  We need to know the original probability distribution on (x, y).

Arc length doesn't lead anywhere, since both sections are infinite ... my personal inclination would just to be to give the answer as 1/10, but that obviously isn't what is meant.  I don't think your restatement (all pairs being equally probably) leads anywhere, either -- doesn't that just reduce it to arclength?  Can you e-mail him and ask him to give a full statement of the question?

Yeah I'll see if I can get a hold of him.  If he can't remember the exact parameters of the problem, I'll at least try to get the proper method and then work backwards from there.

Jacobians are important to integrating a function such as xy=1 because they allow you to transform the curve to an easily integrable function.  Given the equation, it doesnt' surprise me that he used a Jacobian.

Sorry everyone, I got an e-mail back from the professor; I really butchered the problem.  Anyway, for anyone still interested, here is the true problem:

Suppose that two real numbers x and y are chosen independently with uniform distribution in the interval 0<x<1 and 0<y<1.  Suppose that you also know that the product xy=1/1000.  Find the probability that x>1/2.

Once again sorry about the errors in the original post.

Ok, now its just arclength of the hyperbola inside the unit square.

The condition that xy=1/1000 comes after the fact that x and y are chosen in the unit iterval (0<x<1 and 0<y<1).  Think of a function that generates the product of two random numbers.  You don't know the random numbers, but you see that the product is 1/1000.  The problem then is asking the likelyhood that one of those random numbers is in the open interval (1/2,1).

Yes, rcv, the problem is now well-defined (although written a little informally).  There are two probability distributions here.  The first is a uniform distribution on the unit square.  The second is the first distribution conditioned by the event xy = 1/1000.  The problem is asking about the second distribution.  Or, to phrase it in your terminology, the question is asking for p(x > 1/2 | xy = 1/1000).

I can't explain the subtleties of conditional probability (especially for continuous distributions) here, but I trust you can look up the topic.  The problem is analogous to the following: I roll two dice.  Given that their sum is 4, what is the probability that the first die is a 1?  It's a perfectly well-defined problem about conditional probability.

As zableman pointed out, the problem boils down to arc length.  What fraction of the hyperbola xy = 1/1000 in the unit square satisfies x > 1/2?

Actually, zabelman and ravi B, though I also thought the problem boiled down to arclength ratios but this is not the case.  I'll give you a hint in spoiler:[hide]try looking at the areas of lxy-1/1000l <a and 1/2<x<1, andlxy-1/1000l <a and 0<x<1 respectively and then take the limit of the ratio of these areas as a approaches 0.  This is also the reason why arclength doesn't work: because the bandwidth for a given value of a, is not uniform whereas arclength is uniform along the curve.  Jacobians come into play by making the substitutions x=u and xy=v.  Anyway, see if you can take it from there.[/hide]

amd -- we know (x, y) is in some set S, with all elements of S being equally likely to be chosen.  We then find out that (x, y) is also in some subest H of S.  I would argue that all elements of H are equally likely to have been chosen, thus reducing the problem to an arclength problem.  Why is that wrong?

JBL, it is important to remember that both x and y are chosen randomly and independently over the previously specified interval.  In other words the product xy is random over the unit square.  The fact that xy=1/1000 doesn't change how the numbers x and y were chosen.  One way to see that the arclength model doesn't work is to look at the "fattening" of the graph xy=k/1000 for k=0,1,2,...,1000.  This fattening of the bandwidth still leaves some gaps and overlaps and thus doesn't cover the entire unit square.  However, the bands lxy-k/1000l<a for a=1/2000 and k=0,1,2,...,1000 fill up the square exactly.  From a geometric standpoint, if the product xy=1/1000 then that means we are in the bandwidth
lxy-1/1000l<a for some roundoff error a.  Also arclength is the limit of a band of uniform width around the curve.  This is not the case for this problem.  Anyway, I hope I explained myself a little better this time.


Is there a shortcut of some sort to find the sum of the divisors of a whole number, or do you just have to list them out and add?

yeah, there's a short cut. here's how to do it.

Say that your number is 360. The prime factorization of 360 is $ 2^3*3^2*5$. To find the sum of the factors, you have to take the first number in the prime factorization (2, in this case) and add up the powers of 2, starting at $ 2^0$ and going up to $ 2^3$. This is $ 1\plus{}2\plus{}4\plus{}8\equal{}15$. We now have to do the same with each other prime factor. $ 3^2$ gives us $ 1\plus{}3\plus{}9\equal{}13$, and $ 5$ gives us $ 1\plus{}5\equal{}6$. We now have to multiply these numbers together to get the sum of the factors. This gives us $ 15*13*6\equal{}1170$

In general, if your number's prime factorization is $ a^w*b^x*c^y*d^z$, then the sum of it's factors are $ (a^0\plus{}a^1\plus{}a^2\plus{}...\plus{}a^w)*(b^0\plus{}b^1\plus{}b^2\plus{}...\plus{}b^x)*(c^0\plus{}c^1\plus{}c^2\plus{}...\plus{}c^y)*(d^0\plus{}d^1\plus{}d^2\plus{}...\plus{}d^z)$

oh wow...learned that like 2 years ago and forgot it...


The sum of Angie’s age and John’s age equals Michael’s age. Angie’s age is a prime number, but John’s age and Michael’s age are not prime numbers. The sum of John’s age and Michael’s age is a prime number, and Angie’s age minus John’s age is a prime number. One more than half of John’s age is a prime number. Angie is the only teenager, and she is older than John. How many years old is John?

[hide]Let their names be A, J, and M.

A+J=M
A (prime) - J,M (not prime)
J+M (prime)
A-J (prime)
$1+\frac{J}{2}$ (prime)
A>J, 12<A<20

(those were notes for me only)

J is definitely even, and the only possible values for J are 12, 8, and 4.
If J=12, A could be 19 or 17---If J=8, A could be 19 or 13---If J=4, then A could be 17.
J can't be 12 because then M would be prime.
2J+A must be prime, so for (J,A), and (8,13) is the only one that works.

Therefore, the answer is $8$.[/hide]

[hide]

OK, so Anglie can be 13, 17, or 19.

John must be a multiple of 4.
So John can be 4, 8, or 12.

Lets try 12. So now Angie can be 17 or 19.

And we have
A+J=M
J+M is prime
M is prime

12 plus 17 and 19 is prime both times.
8- Angie is a prime number of years older, so 13 or 19 will work here.
8 + 13 is 21, so that works.
We have John as 8, Angie as 13, and Mike as 21.

Answer is 8.[/hide]

[hide]
Guess and check is the best way to solve the problem.
We know that Angie is 13,17,19 and John is not prime and the difference between their ages is prime.
Therefore, John's age has to be even. Since 1/2 of his age + 1 is odd, then his age is a multiple of 4.
So it can be 4,8,12.
Here is where you guess and check.
If Angie is 13 and John is 8, then the difference is prime, 1+1/2 of John's age is prime, and John+Angie's Age is not prime and >19.
Also, John+Michael's age is prime.
So John is 8,Angie is 13, and Michael is 21.
$\boxed{8}$[/hide]

[hide]We are given A+J=M. A is between 13 and 19 and prime, so she must be either 13, 17, or 19. A-J and J+M are prime. .5J+1 is prime, so J must equal 4, 8, or 12.
Possibilities of A and J are,
13,8
17,12
17,4
19,12
19,8
We can say possibilities for Michael's age are 21, 29 and 31.
Michael's age is 21, since the rest are prime.
So, Angie's age is 13, and John's, 8.
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Angie is a teen with a prime age, so she is 13,17, or 19 years old.
John's is not prime and one more than half of his age is prime, so he is 4,8,or 12.
Michael's age is greater than 19 and is composite, so it can be 21,25, or 27.

With a lot of guess and check, you get 8,13,21 so it is 8.


Find all real numbers $a$ such that the equations

$x(x^4 - a)^2 - x^3(1-a)^2 + x^2 - 4 = 0$,
$x^3 + x^2 + x + a = 0$

have a common root

Say the solution $s$ satisfies both equations, and, also, to make the problem work a little easier, we make the substitution $1-a=b$.
So, since $s$ is a solution of both we know:

$s(s^4-1+1-a)^2 - s^3(1-a)^2 + s^2 - 4 = 0$    (1)

$s^3 + s^2 + s + 1 = 1 - a$    (2)

Now, if we make our substitution in (2) and multiply by $s-1$, we get $s^4 - 1 = b(s-1)$. Bringing this back up into (1), we now have $s(b(s-1)+b)^2 - s^3(b)^2 + x^2 - 4 = 0$. Simplifying, we get $s(bs)^2 - s^3(b)^2 + s^2 - 4 = s^2 - 4 \Rightarrow s = \pm 2$ Now that we know the possible values of $s$, we put them into $s^4 - 1 = b(s-1)$ and get $15=16-1=b(\pm 2 - 1)=b, -3b \Rightarrow b = 15, -5$. From $1-a=b=15, -5$, we get $-a = 14, -6$ and finally $\boxed{a=6,-14}$.