a and b are positive integers, a is a multiple of 19, b is a multiple of 47 and
1000 = a + b. Find the positvie defference between a and b.


linear algebra
matrix
linear algebra unsolved

Two graphs that are isomorphic have the same vertices and edges. You can obviously draw them differently if you liked by just rearranging the vertices. But how can you prove that two isomorphic graphs can be drawn identically?

By isomorphism, if a drawing is a model of one graph, it is a model of the other. That's really the only meaningful question I can see here, and it's trivial.

Suppose you have a graph on $n$ vertices. We can associate to the graph a symmetric matrix $A$, with $1$ in the $(j,k)$ location if there's an edge from the $j$th vertex to the $k$th vertex.

"Redrawing" the graph could involve renumbering the vertices. If we do that, then we have a permutation $P$ such that the newly represented graph has matrix $A'=P^TAP.$

That would seem to be an appropriate algebraic criterion for graph isomorphism - that the two matrices are similar by a permutation.


inequalities
trigonometry
geometry proposed
geometry

[color=darkblue]In any triangle $ABC$ exists the inequality : $8Rr\le\boxed{\frac{a^{3}+b^{3}+c^{3}+abc}{a+b+c}\le 4R^{2}}\ ,$ where $2p=a+b+c$ and $R$- the its circumradius.

[b]Remark.[/b] I like it ! The LHS is evidently. It belongs to the "background". [/color]

ok I have two solution for $LHS$ !Actually $LHS$ is very hard!
The first: trigonometric method
By law of sin we can prove that
$\sum\sin A\ge\sum\sin^{3}A\prod\sin A$
$\Leftrightarrow\sum\sin A\ge\frac{1}{4}(3\sum\sin A-\sum\sin 3A)+\prod\sin A\Leftrightarrow\frac{1}{4}\sum\sin A+\frac{1}{4}\sum\sin 3A\ge\prod\sin A\Leftrightarrow\prod\cos\frac{A}{2}-\prod\cos\frac{3A}{2}\ge\prod\sin A \Leftrightarrow\prod\cos\frac{A}{2}-\prod(4\cos^{3}\frac{A}{2}-3\cos\frac{A}{2})\ge\prod(2\sin\frac{A}{2}\cos\frac{A}{2}) \Leftrightarrow\prod\cos\frac{A}{2}-\prod\cos\frac{A}{2}.\prod(4\cos^{2}\frac{A}{2}-3)\ge 8\prod\sin\frac{A}{2}\cos\frac{A}{2}\Leftrightarrow 1-\prod(2(1+\cos A)-3)\ge8\prod\sin\frac{A}{2}\Leftrightarrow 1+\rod(1-2\cos A)\ge 2(\sum\cos A-1) \Leftrightarrow 3+1-2\sum\cos A+4\sum\cos B\cos C-8\prod\cosh\ge\sum\cos A \Leftrightarrow 4-4\sum\cos A+4\sum\cos B\cos C-4\prod\cosh\ge 4\prod\cos A \Leftrightarrow \prod(1-\cos A)\ge\prod\cos A(1)$
if triangle $ABC$ isn't acute $\Rightarrow \prod\cos A\le 0\le\prod(1-\cos A)$ wich is true.
If triangle $ABC$ acute
$(1)\Leftrightarrow \prod(1-\cos^{2}A)\ge\prod\cos A\prod(1+\cos A) \Leftrightarrow\prod\sin^{2}A\ge\prod\cos A\prod(2\cos^{2}\frac{A}{2}) \Leftrightarrow\prod\tan A\ge\prod\cot \frac{A}{2}$
which is true because
$\sum\tan A=\sum\frac{\tan B+\tan C}{2}\ge\sum\tan\frac{B+C}{2}=\sum\cot\frac{A}{2}$
The second solution and solution for $RHS$ I will post later, I think $RHS$ is more easlier than $LHS$  :)

I don't think the $LHS$ is hard. Here's my approach: \[8Rr(a+b+c) \leq a^{3}+b^{3}+c^{3}+abc\]  \[\iff 16RS=16R\cdot \frac{abc}{4R}\leq a^{3}+b^{3}+c^{3}+abc\]  \[\iff a^{3}+b^{3}+c^{3}-3abc \geq 0\] So done.

ok Jan sorry I understand wrong between $RHS$ and $LHS$  :D

Here's what I would do with the RHS:

$2R = \frac{abc}{2S}$

So use the formula of Heroon, and we have to prove that \[(a^{3}+b^{3}+c^{3}+abc)(a+b-c)(a-b+c)(b+c-a) \leq 4a^{2}b^{2}c^{2}\] Now, use the Ravi substitution, and we have to prove that \[\sum_{cyc}{4x^{4}(y-z)^{2}}+8(\sum_{cyc}x^{3}y^{3}+3x^{2}y^{2}z^{2}-\sum_{sym}x^{3}y^{2}z) \geq 0\] And after substitution $xy=\alpha, xz=\beta, yz=\gamma$ we can use Schur and $f(\alpha)=\alpha$ to prove that \[\sum_{cyc}x^{3}y^{3}+3x^{2}y^{2}z^{2}-\sum_{sym}x^{3}y^{2}z \geq 0\] so we are ready!  :lol:

[color=darkblue]Apply the inequality $\sum a\sin\frac{A}{2}\ge p\mathrm{\ \ (*)\ \ (\ see\ }$ http://www.mathlinks.ro/Forum/viewtopic.php?t=80834 $)$

to the orthic triangle $DEF$ of the triangle $ABC\ ,$ where $D\in BC\ ,\ AD\perp BC$ a.s.o. Preliminary :

$\{\begin{array}{c}abc=4RS=4Rpr\\\\ \sum \sin 2A=4\prod\sin A=\frac{2S}{R^{2}}\end{array}$ $\ldots\ \maltese\ \ldots$ $\{\begin{array}{c}m(\widehat{EDF})=180^{\circ}-2A\\\\ EF=R\sin 2A=a\cos A\end{array}\ .$ Therefore, from the inequality $(*)$ obtain :

$\sum EF\cdot\sin\frac{m(\widehat{EDF})}{2}\ge \frac{1}{2}\cdot\sum EF$ $\Longrightarrow$ $\sum a\cos A\cdot\cos A\ge \frac{R}{2}\cdot\sum \sin 2A$ $\Longrightarrow$ $\sum a(1-\sin^{2}A)\ge \frac{S}{R}\ \ \|\ \bullet\ 4R^{2}$ $\Longrightarrow$

$\sum(4R^{2}a-a^{3})\ge 4RS$ $\Longrightarrow$ $\sum a^{3}\le 4R^{2}\sum a\--abc$ $\Longrightarrow$ $\boxed{\ \frac{a^{3}+b^{3}+c^{3}+abc}{a+b+c}\le 4R^{2}\ }\ .$

[b]Remark.[/b] With LHS (Left Hand Side) I pulled (up) ... the zip ![/color]


geometry
circumcircle
projective geometry
power of a point
radical axis
geometry proposed

[quote="romano"]Let $ ABC$ be a triangle. A line parallel to $ BC$ intersects the lines $ AB$ and $ AC$ at $ D$ and $ E$. Let $ P$ be a point inside the triangle $ ADE$. The lines $ PB$ and $ PC$ intersect the line $ DE$ at $ M$ and $ N$. The circumcircle of triangle $ PDN$ intersects the circumcircle of triangle $ PEM$ at a point $ Q$ (apart from $ P$). Prove that the points $ A$, $ P$, $ Q$ are collinear.[/quote]

Unfortunately, all what I have right now is a solution using projective geometry:

Consider the following theorem which was proven by Grobber in post #2 of http://www.mathlinks.ro/Forum/viewtopic.php?t=5860 :

[b]Theorem 1.[/b] If P is an arbitrary point in the plane of a triangle ABC, and if a line intersects the lines PA, PB, PC at the points K, M, N and the lines BC, CA, AB at the points F, E, D, then the circumcircles of triangles PKF, PME and PND have a common point apart from the point P.

Applying this to your configuration, where the line is your line parallel to BC, we see that the point F, where this line intersects BC, is an infinite point, so that the circumcircle of triangle PKF degenerates into a line, namely the line PK, which is, of course, simply the line PA (since the point K lies on the line PA). Thus, instead of saying that the circumcircles of triangles PKF, PME and PND have a common point apart from the point P, we can say that the line PA and the circumcircles of triangles PME and PND have a common point apart from the point P. But the common point of the circumcircles of triangles PME and PND apart from the point P is the point Q; thus, we can conclude that the point Q lies on the line PA, and thus, the points A, P, Q are collinear. $ \blacksquare$

Unfortunately, proving Theorem 1 is not so easy (see the thread I linked to).

  Darij

The problem is not difficult. Let me give the solution :D
       Your goal is to prove $A$ lies on the radical axis of this two circles. Call $K, L$ the second intersection of $(PME)$ and $AC$, $(PDN)$ and $AB$. Now, it is sufficient to prove that $B, C, K, L$ lie on a circle.
    We have $(BK, KC) = (BK, KP) + (KP, KC) = (BC,CP) + (PB, BC) = (PB, PC)$, here we notice that $B, K, C, P$ lie on a circle since $PKME$ is cyclic and $ME//BC$. Analogously, we obtain $(BL,LC) = (PB, PC)$. Hence we 're done.

Another solution :
 Suppose $ DQ $  :cup:  $ BP $ at $ R $ , $ EQ $  :cup:  $ PC $ at $ S $.
 It's easy to show that : $ RS // DE // BC $
 Applying  Menelaus 's theorem for triangle $ BDR $ ==> $ Q.E.D $

the problem still true if P inside the triangle ABC

[quote="nttu"]$ RS // DE // BC $ [/quote]
$nttu$ can you show me why this is true?
and can anybody tell me what is the meaning by $(BK, KC)$? :(

[quote="zhaobin"][quote="nttu"]$ RS // DE // BC $ [/quote]
$nttu$ can you show me why this is true?[/quote]

Well, here is how I understand Nttu's solution:

Let the lines DQ and BP meet at R, and let the lines EQ and CP meet at S.

We will use directed angles modulo 180°. Since the point Q lies on the circumcircle of triangle PDN, we have < QDN = < QPN; in other words, < (DQ; DE) = < (PQ; CP). Similarly, < (EQ; DE) = < (PQ; BP). Thus,

  < (DQ; EQ) = < (DQ; DE) - < (EQ; DE) = < (PQ; CP) - < (PQ; BP) = < (BP; CP).

In other words, < RQS = < RPS. Thus, the points R, S, Q and P lie on one circle. Consequently, < RSP = < RQP; in other words, < (RS; CP) = < DQP. But since the point Q lies on the circumcircle of triangle PDN, we have < DQP = < DNP, so that < (RS; CP) = < DNP. In other words, < (RS; CP) = < (DE; CP). Consequently, RS || DE. Since we also know that DE || BC, we thus have RS || DE || BC.

Now, from here on, there are two possibilities to complete the solution, i. e. to prove that the points A, P, Q are collinear. Nttu does it somehow using the Menelaos theorem, applied to triangle BDR (I don't know how exactly he does it, but I haven't really tried to understand). Actually, there is a simpler argument: Since RS || DE || BC, the lines RS, DE, BC concur at an infinite point; hence, the triangles RDB and SEC are perspective. Hence, by the Desargues theorem, the points $DB\cap EC$, $BR\cap CS$ and $RD\cap SE$ are collinear (hereby, the abbreviation $g\cap h$ means the point of intersection of two lines g and h). But $DB\cap EC = A$, $BR\cap CS=P$ and $RD\cap SE=Q$. Thus, the points A, P and Q are collinear. Proof complete.

[quote="zhaobin"]and can anybody tell me what is the meaning by $(BK, KC)$?[/quote]

In Treegoner's solution, $\left(BK,KC\right)$ means the directed angle between the lines BK and KC (Treegoner works with directed angles modulo 180°). In my opinion, it is better to use the notation $\measuredangle\left(BK,KC\right)$ (or $\measuredangle\left(BK;\;KC\right)$) instead.

  darij

Lemma 1.
Let X, Y, S, and T be four ciclyc points on a plain. Then if U lies on XS and V lies on YT it holds that UV//ST iff UVYX is ciclyc.

proof

Easy.

Going to the problem, let L = AB^c(PDN) and let K = AC^c(PME). From lemma 1, as DN//BC it results that LPBC is ciclyc and, for the same reason, KPBC is ciclyc too. Let's now define Q=c(PDN)^c(PME) and R=PQ^c(BCP). From lemma1 it follows that BR//DQ and CR//EQ, so the triangles DEQ and BCR are omothetics the center of this omothetie is A=BD^CE^RQ. Then the thesis.

$darij grinberg$,thank you very much.I'll study it when I'm back home.


It is well known that $t^{2}\in${0,$\pm1$}$(mod 5)$ and from Fermat's theorem we have x^5$\equiv$x$(mod 5)$ $\Rightarrow$ $x+y\equiv \alpha+\beta (mpd 5)$ where $\alpha , \beta \in{0,\pm 1}$ $\Rightarrow$ x+y $\in$ {0,1,2} because x,y $\geq$ 0 So we have (x,y) $\in$ {(0,0),(1,0),(1,1),(2,0)} .
 I hope it's right :D

I don't think so- for all integers $t,u$, we can set
$x=t^{2}$, $y=u^{2}$, $a=t^{5}$, $b=u^{5}$
to give a solution.  :maybe:

You only proved that x + y = 0, +/- 1, +/- 2 MOD 5.  That is in fact true of every pair of integers!

Some counterexamples to the result shown include:
(i) x = -y, a = b = 0, so that one of x and y CAN be negative.
(ii) x = y = 2, a = 8, b = 0.
(iii) x = 3, y = 1, a = 10, b = 12.

This is a difficult problem I think.


Solve the equation:
\[2\sqrt{x^{2}+3}-\sqrt{8+2x-x^{2}}=x\]


What are all ordered triples of positive integers (x,y,z) whose product is 4 times their sum, if x < y < z?

Once x gets to 4, we can divide by 4 to get:
yz=4+y+z
y and z are going to have to be greater than x,
so we think of using 5 and 6.  From that, we see
that xyz will be greater than 4(x+y+z), and as
we go to bigger values of y and z we see
that xyz continues to get bigger and bigger
compared to 4(x+y+z) (when x is equal to 4).
Another way to see this is if we try 4 in the place of
x, getting 4yz=4(4+y+z).  The smallest possible value
of y is 5, so trying this gives us that z=36/16=9/4<3.
Thus, we see that x=4 is not a possible case, as z
is automatically smaller than y in the conditions given.
When x is greater than 4, xyz is always going
to be great than 4(x+y+z)(for simiilar reasons
as when x=4), so we really only have 3 cases:
[i]x=1[/i]: From this, we have that
yz=4(1+y+z). We see that y has to be greater
than 4, so we try it for 5, 6, 7...
For y=6 we get [u]{1,6,14}[/u]
For y=5 we get: [u]{1,5,24}[/u]
For y=8 we get [u]{1,8,9}[/u]
trying y=9 gives us z=8.  At this point, since z<y,
we know that every greater value of y will give
a lesser value of z, so we are done with this case
[i]x=2[/i]: Here we get 2 values, using
a similar process as case 1:
[u]{2,3,10}[/u] and
[u]{2,4,6}[/u]
[i]x=3[/i]: (No cases; trying y=4 gives
z=7/2<4, so 3 is not a possible case).

To eliminate some of the brute force I'd suggest to use a trick called Simon's Favorite Factoring Trick

[url]http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick[/url].

For case $ x \equal{} 1$

$ yz \equal{} 4 \plus{} y \plus{} z \rightarrow yz \minus{} 4y \minus{} 4z \minus{} 4 \equal{} 0$
Using Simon's Favorite Factoring Trick $ (y \minus{} 4)(z \minus{} 4) \equal{} yz \minus{} 4y \minus{} 4z \plus{} 16$
$ (y \minus{} 4)(z \minus{} 4) \equal{} 20$

Listing factors of 20
$ (1, 20)$, $ (2, 10)$, $ (4, 5)$

To yield 3 solutions for case $ x \equal{} 1$

$ (1, 5, 24)$, $ (1, 6, 14)$, and $ (1, 8, 9)$

Same technique can be applied for cases $ x \equal{} 2, 3$

I think this was an old mathleague problem?

I posted the same question before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194288


limit
function
real analysis
real analysis unsolved

Let $ y(x)$ is a solution of differential equation $ xy''\plus{}y'\plus{}xy\equal{}0$ with conditions $ y(0)\equal{}1,\;y'(0)\equal{}0.$
Find $ \lim_{x\to 0}\frac{y(x)\minus{}1\plus{}\frac{x^2}{4}}{x^4}$.

Just brutal force applies to this problem, but there is also an elegant way to solve this if you have some knowledge about special functions. This is because $ y \equal{} J_0(x)$, where $ J_0$ is [url=http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1]the Bessel function of the 1st kind of order 0[/url]. Therefore the answer is $ 1/64$.

hello Yuriy, i think the limit is $ 1/64$.
Sonnhard.


integration
calculus
derivative
trigonometry
calculus computations

Is this right?

$ \displaystyle \frac{d \int_{0}^{x^{2}}}{dx} \frac{\sin(t)}{t} dt \equal{} \frac{\sin(x^{2}) \cdot \cos(x^{2}) \cdot 2x }{x^{2} \cdot 2x } \equal{} \frac{ \sin(x^{2})\cos(x^{2} )}{x^{2}}$

No. The derivative of $ \int_a^{h(x)}f(t)\,dt$ is $ f(h(x))h'(x)$.

No, it's supposed to be:

$ \frac{2 \sin(x^2)}{x}$

If $ F'(x)\equal{}\frac{\sin(x)}{x}$ then you have:

$ (F(x^2)\minus{}F(0))'\equal{}2xF'(x^2)\equal{}2x \frac{\sin(x^2)}{x^2}\equal{}\frac{2 \sin(x^2)}{x}$

let me re-phrase the original question a bit... just adding in an $ x^{2}$ in the statement gives:

$ \displaystyle \frac {d x^{2} \int_{0}^{x^{2}}}{dx} \frac {\sin(t)}{t} dt$

I found the derivative to be $ 2x \left( \int_{0}^{x^{2}} \frac{\sin(t)}{t}dt \right) \plus{} x^{2} \left ( \frac{2x \sin(x^{2}) }{x^{2}} \right)$ (just use product rule)

Is it okay to leave the integral $ \int_{0}^{x^{2}} \frac{\sin(t)}{t} dt$ as it is, or can it be simplified somehow?

[quote="Prime factorization"]
Is it okay to leave the integral $ \int_{0}^{x^{2}} \frac {\sin(t)}{t} dt$ as it is, or can it be simplified somehow?[/quote]
Unless you'd want to give it a series representation.

Bah, this is annoying me too.

$ \frac{d  \int_{\sin(\theta)}^{\cos(\theta)}}{d\theta} \frac{1}{1\minus{}x^{2}} dx$ I'm not 100% sure what to do here because there's a variable present in both the upper and lower bound of the integral.

The best idea I have so far is to write $ \frac{1}{1\minus{}x^{2}}$ as $ \left(\frac{1}{\sqrt{1\minus{}x^{2}}}\right)^{2}$ (so it's anti-derivative is $ \sin^{\minus{}1}(x)$  ), but that still leaves me kinda stuck.

$ \displaystyle\frac {d}{dx}\int_{a(x)}^{b(x)}f(t)dt \equal{} b'(x)f(b(x)) \minus{} a'(x)f(a(x))$

This is basically a generalization of what mlok said in his post. You are just using the following

$ \int_a^bf(x)dx \minus{} \int_a^cf(x)dx \equal{} \int_c^bf(x)dx$

Yeah, I just figured that out right before I saw your post. I did:

$ \frac{ d \int_{\sin(\theta)}^{\cos(\theta)} }{d\theta} \frac{1}{1\minus{}x^{2}} dx \equal{}$

$ \frac{d \; F( \; \cos(\theta) \; ) \minus{} F( \; \sin(\theta) \; )}{d \theta} \equal{} f(\cos(\theta)) \cdot (\cos(\theta))' \minus{} f(\sin(\theta)) \cdot (\sin(\theta))'$


find the minimum of:

(x + y + z)/(root(x-1)+root(y-1)+root(z-1))

given x,y,z>0, 1/x + 1/y + 1/z = 1

probably 3/root(2), need proof.

Latex doesnt like my computer, sorry. here root(x) = x^0.5

[b]Q: find minimum of f(x,y,z) = (x+y+z)/((x-1)^(1/2) + (y-1)^(1/2) + (z-1)^(1/2)) with condition x,y, z > 0 and 1/x + 1/y + 1/z = 1.
A: The answer is 3/sqrt(2).
It is not that hard to see Jensen's pattern here due to the cyclicity of the expression. So quite natural we try Jensen's for h(x) = (x-1)^(1/2) to get h''(x) = -1/4(x-1)^(-3/2) < 0 for x > 1. So Jensen's inequality gives:
f(x,y,z) >= [(x+y+z)/3]/[((x+y+z)/3 - 1)^(1/2)]. But (x+y+z)(1/x + 1/y + 1/z) >= 9 which implies (x+y+z) >= 9 and hence t = (x+y+z)/3 >= 3.
So g(t) = t/((t-1)^(1/2)) has g'(t) = (t-2)/(2(t-1)^(3/2)) > 0 for t >= 3 and therefore f(x,y,z) >= g(t) >= g(3) = 3/sqrt(2) as claimed and that
"=" occurs when t = 3 which yields x=y=z=3.
and we're done.
____________________________________________________________[/b]


Noah's favorite flower shop sells 5 different types of flowers. How many different ways are there to buy 5 flowers if there are no restrictions on how many of each flower may be purchased?

You can solve this using the hockey stick identity.

Let's pretend that Noah organizes the plants by putting the same type of flower on the same table, but no others on that table. Let l represent the tables, and let o represent the flowers. We'll use a diagram to solve this problem. The flowers that are to the left of a table are considered the flowers of that type.

For example, if we had the diagram ollloolool, this would mean that there are two tables that don't have any flowers on it, one table that has one flower on it, and two tables that have two flowers on it.

Now we know that there are a total of $ 5$ l's. There are also a total of $ 5$ o's. We know that one of the tables has to go to the very right, because otherwise, that would mean that there is a flower that doesn't have a table.

So we have $ 5$ possible tables for the left most spot. Now there are nine symbols left, but the o's are the only ones that are indistinguishable, so we divide $ 9!$ by $ 5!$, and then multiply that by $ 5$ to get the answer.

You can factor out a $ 5!$ from the $ 9!$ to cancel out with the $ 5!$ that is dividing $ 9!$, and you are left with $ 6$, $ 7$, $ 8$, and $ 9$ multiplying each other. We also have to multiply this by five. The product turns out to be;

$ 15120$

@Wickedestjr - Refer to the other thread for the correct answer and solution. Also, in the future, try not to post in a double-posted thread to avoid having a split discussion, as is the case here.

[color=red]@mods - Please lock/delete this thread.[/color]


See here http://www.mathlinks.ro/Forum/viewtopic.php?p=469045#469045 and please use English in your post  :wink:


limit
calculus
integration
advanced fields
advanced fields unsolved

Let $ \lambda_i \;(i=1,2,...)$ be a sequence of distinct positive numbers tending to infinity. Consider the set of all numbers representable in the form \[ \mu= \sum_{i=1}^{\infty}n_i\lambda_i ,\] where $ n_i \geq 0$ are integers and all but finitely many $ n_i$ are $ 0$. Let \[ L(x)= \sum _{\lambda_i \leq x} 1 \;\textrm{and}\ \;M(x)= \sum _{\mu \leq x} 1 \ .\] (In the latter sum, each $ \mu$ occurs as many times as its number of representations in the above form.) Prove that if \[ \lim_{x\rightarrow \infty} \frac{L(x+1)}{L(x)}=1,\] then \[ \lim_{x\rightarrow \infty} \frac{M(x+1)}{M(x)}=1.\]

[i]G. Halasz[/i]

Solution in PDF
We will use upper Darboux integrals see here: http://mathworld.wolfram.com/DarbouxIntegral.html


Each of the four students on each of 16 MATHCOUNTS teams wins a math book valued at $ \$125$. What is the total value, in dollars, of all of the math books received by the students?

$ 4\cdot 16\cdot 125$
$ 4\cdot 16\cdot 5\cdot 25$
$ 4\cdot 4\cdot 20\cdot 25$
$ 4\cdot 20\cdot 100$
8000

associative + commutative properties of multiplication to 'make' a 100. $ n\cdot 25$ is just $ \frac {n}{4}\cdot 100$ it's obvious why but most people don't realize how useful it is.

Edit: I fail

Since there are $ 4$ students on $ 16$ teams, we have $ 4(16)(125)\equal{}64(125)\equal{}\boxed{8000}$.


Dear friends:

Here is the official website for IMO 2007.

http://imo2007.edu.vn

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=131283  :rotfl:

Hi friends,

Welcome to IMO at Vietnam, I'm hopefully u will have a nice time and medal here for all of you.

Today I took an examination that to choose guides for IMO participants. We have to take English test and I took Japanese test too.
There are many students to join this selection. Hope I will pass. Pray for me, ah aha ah.

But anyway, we are preparing for this special festival and we want to send you the message of WELCOME and contact us if you need anything.

Want to become friends with u all.
Enjoy your days.

toato, you have been studying Japanese, gan-batte-kudasai. :)

kunny

Great thanks, hope I can become the Japanese guide.  :P
So happy, see u then in Hanoi


geometry
quadratics
number theory unsolved
number theory

Let $ a, b, c, d$ be positive integers such that $ ab \equal{} cd$ and $ a\plus{}b \equal{} c \minus{} d.$ Prove that there exists a right-angled triangle the measure of whose sides (in some unit) are integers and whose area measure is $ ab$ square units.

[hide]
We have $ ab \equal{} cd$ and $ a \plus{} b \equal{} c \minus{} d$. Substituting gives $ ab \equal{} d(a \plus{} b \plus{} d)$, or $ d^2 \plus{} (a \plus{} b)d \minus{} ab \equal{} 0$. For this quadratic in $ d$ to be solvable over the integers, we must have the discriminant $ (a \plus{} b)^2 \minus{} 4( \minus{} ab)$ be an integer. So

$ a^2 \plus{} 6ab \plus{} b^2 \equal{} k^2$
$ (a \plus{} 3b)^2 \minus{} 8b^2 \equal{} k^2$
$ (a \plus{} 3b)^2 \minus{} k^2 \equal{} 8b^2$
$ (a \plus{} 3b \plus{} k)(a \plus{} 3b \minus{} k) \equal{} 8b^2$

Those two factors are of the same parity, so they must be even.

$ \left(\frac {a \plus{} 3b \plus{} k}{2}\right)\left(\frac {a \plus{} 3b \minus{} k}{2}\right) \equal{} 2b^2$

Now if we let those two factors be $ p$ and $ q$, we have $ p \plus{} q \equal{} a \plus{} 3b$ and $ pq \equal{} 2b^2$. Then we must have $ p \equal{} 2rs^2$ and $ q \equal{} rt^2$ for some integers $ r,s,t$. So now $ b \equal{} rst$ and $ a \equal{} p \plus{} q \minus{} 3b \equal{} 2rs^2 \plus{} rt^2 \minus{} 3rst \equal{} r(2s \minus{} t)(s \minus{} t)$. Then

$ ab \equal{} r^2 st(2s \minus{} t)(s \minus{} t)$

Now set $ m \equal{} s$ and $ n \equal{} s \minus{} t$.

$ ab \equal{} r^2 m(m \minus{} n)(m \plus{} n)n \equal{} r^2 mn(m^2 \minus{} n^2)$

Now if $ x \equal{} r(m^2 \minus{} n^2)$ and $ y \equal{} 2mn$, we'll have $ ab \equal{} \frac {1}{2}xy$, so $ ab$ will be the area of the triangles whose legs are $ x$ and $ y$. Then the hypotenuse is an integer too by the classic identity

$ r^2(m^2 \minus{} n^2)^2 \plus{} r^2(2mn)^2 \equal{} r^2(m^2 \plus{} n^2)^2$

So the triangle with legs $ x$ and $ y$ is the triangle we want.

Remark. We had $ a \equal{} n(m \plus{} n) \equal{} mn \plus{} n^2$ and $ b \equal{} m(m \minus{} n) \equal{} m^2 \minus{} mn$. We can also find
$ c \equal{} m(m \plus{} n) \equal{} m^2 \plus{} mn$ and $ d \equal{} n(m \minus{} n) \equal{} mn \minus{} n^2$. Then we have $ ab \equal{} cd$ and $ a \plus{} b \equal{} c \minus{} d$.
[/hide]

[hide="Solution"]
By the Four Numbers Theorem, we can say that $ a \equal{} wx$, $ b \equal{} yz$, $ c \equal{} xy$, $ d \equal{} wz$.  Now, since $ a \plus{} b \equal{} c \minus{} d$, we have that $ a \plus{} d \equal{} c \minus{} b$, so $ w(x \plus{} z) \equal{} y(x \minus{} z)$.  By the Four Numbers Theorem, we can say that $ w \equal{} kl$, $ x \plus{} z \equal{} mn$, $ y \equal{} lm$, and $ x \minus{} z \equal{} nk$.  Then, we get that $ x \equal{} \frac {nk \plus{} mn}{2}$ and $ z \equal{} \frac {mn \minus{} nk}{2}$.  It follows that\[ ab \equal{} xyzw \equal{} \frac {(nk \plus{} mn)(mn \minus{} nk)(lm)(kl)}{4} \equal{} \frac {n^2l^2mk(m \plus{} k)(m \minus{} k)}{4}\]Letting $ \frac {m^2 \minus{} k^2}{2} \equal{} t$ and $ u \equal{} mk$, so $ u$ and $ t$ are the legs of a right triangle.  Scaling everything up by a factor of $ nl$, we get that $ t' \equal{} \frac {nl(m^2 \minus{} k^2)}{2}$ and $ u' \equal{} mkln$, so $ u'$ and $ t'$ are still the legs of a right triangle with integer sidelengths.  Then, $ \frac {t'u'}{2} \equal{} \frac {mkl^2n^2(m^2 \minus{} k^2)}{4} \equal{} ab$, so the area of the triangle is equal to $ ab$, as desired. [/hide]


1) It was alredy disscused. Try a search about  "condensation test".


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

$ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 200 \qquad \textbf{(C)}\ 300 \qquad \textbf{(D)}\ 400 \qquad \textbf{(E)}\ 500$

[hide="Click for solution"]
When Dave buys seven windows separately he'll purchase six, and get one for free, for $ \$600$.  When Doug buys eight windows separately, he'll purchase seven and get one for free, for a total cost of $ \$700$.  The total cost for them purchasing separately is $ \$1300$, whereas if they purchase fifteen together, they need to purchase $ 12$ windows for $ \$1200$.  The savings is $ \$100$ or $ \boxed{\textbf{(A)}}$.
[/hide]


geometry
3D geometry
factorial
function
algorithm
symmetry
calculus

um...wow. i'll take them one at a time

(1) [hide]let a:le:b:le:c. Then we can factor out an a! to get a!+b!+c!=a!(1+(a+1)(a+2)...b+(a+1)(a+2)...c). Well, if this is to be a power of 2, then both factors must be powers of 2. So, if a! is a power of 2, a is either 0, 1, or 2. Alright. Note that a!+b!+c!:ge:3, so n cannot be 0 or 1. The only way n can be 2 is if a=b=0 or 1 and c=2. Okay, now mod 24, all 2^n, n:ge:3, are :equiv:8 or 16. a!:equiv:1 or 2, and b! and c!:equiv:1, 2, 6, or 0.  The only possible solution here is a=2, b=3, and c:ge:4. c!=8(2^(n-3)-1). This means c! has exactly 3 factors of 2, i.e. 4:le:c:le:5. Thus, the only possible quadruples (a,b,c,n) are (0,0,2,2), (0,1,2,2), (1,1,2,2), (2,3,4,5), (2,3,5,7) and permutations of the first 3 variables. correct?[/hide]

I won't pass up on an opportunity to use Simon's favorite factoring trick:

3) I assume this should be solved in positive integers. (a+1)(b+1)(c+1)=1001. Assume a<=b<=c. a+1=7, b+1=11, c+1=13, so (a,b,c)=(6,10,12) (and its rotations). If it's supposed to be nonnegative integers, one would have to deal with the cases with a+1=1 and b+1=1. If it includes negative integers, there are several more cases, but the idea is the same.

(2) [hide]still working on it[/hide]

(3) [hide]factor as (a+1)(b+1)(c+1)=7*11*13. just factor and subract one to get all possibilities...[/hide]

(4) [hide]if a:ge:2, we can write it as c:^2:-3^b=2^a. A mod 4 analysis shows us that b is even, i.e. b=2x. Then (c-3^x)(c+3^x)=2^a. Then, if c-3^x=2^m and c+3^x=2^n, 3^x=(2^n-2^m)/2=2^(m-1)(2^(n-m)-1). This will only be a power of 3 if m=1 and n-m=1 or 2, i.e. if n=2 or 3. Thus, the only solutions in this case are (a,b,c)=(3,0,:pm:3), (4,2,:pm:5).

Now, if a=0, we have (c-1)(c+1)=3^b, meaning (a,b,c) is (0,1,:pm:2). If a=1, we have 2+3^b=c:^2:. If b>0, mod 3 gives us c:^2::equiv:2 mod 3, which is impossible. b=0 gives us a nonsquare. So there are no solutions in this case. Therefore, all solutions are (3,0,:pm:3), (4,2,:pm:5), and (0,1,:pm:2) Correct?[/hide]

hmmm...(5)[hide](a+b):^2:+1=ab(2+c) => c=(a+b):^2:/ab - 2. If c is an integer, then ab|(a+b):^2:. Therefore a|(a+b):^2: and b|(a+b):^2: => a|b:^2: and b|a:^2: not sure where to go from there[/hide]

(6)[hide]mod 5 analysis shows x:equiv:0, 3, 4 mod 5. Hmmm...what else can I say...[/hide]

(7) [hide]So, the sum of the squares from 1 to n is n(n+1)(2n+1)/6 = m:^2:n

So (n+1)(2n+1)/6 = m:^2:

So 6|(n+1)(2n+1).  This means either (6|n+1) or (6|2n+1) or (3|2n+1 and 2|n+1).

Case 1: 6|n+1

Now,  (n+1) and (2n+1) are going to be relatively prime, which means that (n+1)/6 and (2n+1) would have to be perfect squares.  However, if 6|n+1, 2n+1 :equiv: 5 (mod 6) which is never a perfect square.  Throw out this case.

Case 2: 6|2n+1

This never works.  That was easy.

Case 3: 3|2n+1 and 2|n+1.  This gives us that n :equiv: 1 (mod 6).

n = 6j + 1.   m:^2: = (6j + 2)(12j + 3)/6 = (3j + 1)(4j + 1).  These two numbers are relatively prime to each other (since they are relatively prime to their common difference), so we have that 3j + 1 and 4j + 1 are both perfect squares.  Hmm.  What can I do with this?  I'm sure someone else knows how to finish this off.[/hide]

(8) m:^2: = n(n + p) where p is an odd prime

[hide]

4m:^2: + p:^2: = 4n:^2: + 4np + p:^2:

p:^2: + (2m):^2: = (2n + p):^2:

Now, we resort to the parametrized form of Pythagorean triples, which looks roughly like d(u:^2: - v:^2:), 2uvd, and d(u:^2: + v:^2:) for u, v relatively prime positive integers, u > v.  Now, p is an odd prime and one of the legs, so p = 2uvd (which fails on odd=even) or

p = d(u:^2: - v:^2:), which factors as p = d(u+v)(u-v).  This implies

 u + v = p, u - v = 1, d = 1.  Then

p = 2v + 1

2m = 2v(v+1)

m = v:^2: + v

2n + p = v:^2: + (v+1):^2:

2n + 2v + 1 = 2v:^2: + 2v + 1

n = v:^2:.

Thus, all solutions are (m, n, p) = (v:^2: + v, v:^2:, 2v + 1) provided that 2v + 1 is prime.[/hide]

(9) [hide]Let s = 6^(1/3).  Okay, so, ignore the right side for a moment and just look for the square root of 49 + 20s.  Guess that it has the form a + bs + cs:^2:, where a, b, and c are relatively nice looking numbers.  I don't know if you can prove that this must work.  I don't think you can.  But anyhow, guess that it works.  Square it out.  Get the equations

b:^2: + 2ac = 0

a:^2: + 12bc = 49

3c:^2: + ab = 10

All right.  So this is totally non-legit.  I guessed that bc = 4, because that would make me very happy because it would give me a nice solid a = :pm:1.  And shazaam!  it works.  We get solutions (a, b, c) = :pm:(-1, 2, 2).  (This is not unique.  There are more.  But that's okay.  They aren't nice-looking.)  Now, take the member where a = -1, not positive 1, because we need m and n to be positive.  This gives me the sums of the cube roots of m and n is equal to twice the cube root of 6 plus twice the cube root of 36.  This should give you that {m, n} = {48, 288}.  I would be willing to wager that this is unique, but I definitely can't prove it.[/hide]

7. [hide]n(n+1)(2n+1)/6=m:^2:n, as n:ge:1, (n+1)(2n+1)=6m:^2: as n is odd, we replace n by 2a-1 to get a(4a-1)=3m:^2: a:equiv:0 or 1 mod 3. If a:equiv:1 mod 3, we can write it as (3k+1)(4k+1)=m:^2:. well, gcd(k,3k+1)=1, so gcd(3k+1,4k+1)=1, so 3k+1 and 4k+1 must both be perfect squares. ok, so if 3k+1=x:^2:, then 4k+1=(4x:^2:-1)/3=(2x-1)(2x+1)/3 => if x=3y+1, then(6y+1)(2y+1) is a perfect square blah this isn't going anywhere.[/hide]

8. [hide]either gcd(n,p) is 1 or gcd(n,p) is not 1. (duh) If gcd(n,p) is 1, then gcd(n,n+p) is also 1. Then n and n+p must both be perfect squares. So n+p=b:^2: and n=a:^2: => p=(b-a)(b+a) => a=(p-1)/2 => n=(p-1):^2:/4.  This makes n+p=(p+1):^2:/4 and m=(p-1)(p+1)/4. This will be an integer, as p is odd so both p-1 and p+1 will have a factor of 2. So, one solution will be (m,n,p)=((p:^2:-1)/4,(p-1):^2:/4,p), where p can be any odd prime.

Now, if gcd(n,p) is not 1, n=kp. That means m:^2:=kp(k+1)p=k(k+1)p:^2:. This means k(k+1) is a perfect square. However, k(k+1)=k:^2:+k is always between k:^2: and (k+1):^2:=k:^2:+2k+1, for positive k, so k(k+1) cannot be a perfect square. So no solutions in this case. Thus, all solutions are (m,n,p)=((p:^2:-1)/4,(p-1):^2:/4,p), where p is any odd prime. Correct?[/hide]

Seeing all those factorials reminded me of this cool little problem:

Prove that there are infinitely many solutions (positive integers) for a!*b!=c!. I've struggled with it for a long time ( :D  :oops: believe it or not), but it's REALLy easy.

There's only one ``nontrivial'' solution as far as we know: 10!=7!6!. You can get a lot more by taking (n!)!=(n!-1)!*n!. I think there's another class that generates infinitely many, but I can't remember what it is.

1!*n! = n! ?  Or are you thinking of something slightly less trivial?

Hehe ... some comments.

(1) Taken from the dutch 'Pythagoras' contest 2003.

Solved by Mysticterminator.

(2) British Mathematical Olympiad 2003 (First round).

No solution not yet.

Hint : [hide]Consider the cases a < b and a = b.[/hide]

(3) Taken from the dutch 'Pythagoras' contest 2002.

Easy problem, solved by Complexzeta & Mysticterminator.

(4) British Mathematical Olympiad 1996.

Solved by Mysticterminator.

http://www.kalva.demon.co.uk/bmo/bsoln/bsol961.html

(5) I found this on the mathlinks forum. It's quite difficult.

No solution not yet.

Hint : [hide]First prove that c must equal 3.[/hide]

(6) Aha ... Interesting. I will mention the source later

No solution not yet.

Hint : [hide]Modular arithmetic. If p is a prime then (p - 1)/2 is interesting ...[/hide]

(7) Adapted from British olympiad 1994.

Partial solution by JBL.

Mysticterminator is wrong here since n = 337 works.

(8) British olympiad 1992.

Solved by JBL and Mysticterminator.

(9) British olympiad 1999, adapted.

An attempt by JBL.

A hint : [hide]Compare rational and irrational parts of both sides.[/hide]

(0) I will mention my source later

No solution not yet.

MysticTerminator -- your solution to number 7 is wrong because when you substituted n = 2a -1, you replaced (n + 1) with (2a - 1) instead of 2a.

for those you say you will mention the source later, they had better not be IMOs. ;) (well, at least not the recent ones)

wurg i knew 7 must have been too easy.  :(

Arne wrote:Solve in integers :
(9) 49 + 20(61/3) = (m1/3 + n1/3 - 1)2 ;
(0) x2 + 4y = s2, y2 + 4x = t2.

(9)[hide]m and n can be written as a:^3:*b and c:^3:*d:^2:. I'm pretty sure that's the only way it would work. Then multiplying everything out gives you:

49+20*6^(1/3)=a:^2:b^(2/3)+c:^2:dd^(1/3)+1

+2acb^(1/3)d^(2/3)-2ab^(1/3)-2cd^(2/3). All the things with something to the (2/3) has to cancel, and we equate the things to the (1/3) and the real numbers:

49=1. okay, this isn't right. this means that there is another term that produces a real number. this must be the 2acb^(1/3)d^(2/3) term, meaning that b=:pm:d. So we replace all the d's by :pm:b's, and now we set the real and irrational terms equal:

49=1+2abc (we don't need to worry about :pm: since the d term is squared)

a:^2:=2c (again we don't need to worry about :pm:)

20*6^(1/3)=b^(1/3)*(bc:^2:-2a) (again, we don't need to worry about :pm: since the d term is to the 4/3 power, which is even, sort of)

okay, since there is only one term to the 1/3 power in the RHS of the last equation, b must equal :pm:6. Now we get:

:pm:4=ac

a:^2:=2c

:pm:20=:pm:6c:^2:-2a

wow lot of :pm:s, but we can solve from here. I think all possible solutions are as follows (a,b,c)=(:pm:2,:pm:6,2) Now we remember that d=:pm:b=:pm:6 and get our original solutions:

(m,n)=(:pm:48,288). correct?[/hide]

edited: cool i followed your hint even though i didn't look at it. and i got the same answer as JBL, but i shouldn't have had the :pm:. Hmm...why did i have it then? my steps seem legit...

(0)[hide](x-y)(x+y-4)=(s-t)(s+t)

(x+2)^2+(y+2)^2=s^2+t^2+8

x&s have same parity

y&t have same parity [/hide]

(5) a:^2: + b:^2: = abc - 1

[hide]c = (a:^2: + b:^2: + 1)/ab  is an integer.  Thus

b | a:^2: + 1

a | b:^2: + 1.

Either: wlog a>b or a=b.

If a=b, 2a:^2: = ca:^2:-1

1 = (c - 2)a:^2:

a = b = 1, c = 3.

If a>b,

a:^2: > b:^2:

a:^2: >= b:^2: + 2b + 1 > b:^2: + 1

k = (b:^2: + 1)/a < a.

Let us now examine the ordered triple (k, b, c).  I claim that this ordered triple also satisfies the given equation.

Proof:

b:^4: + 2b:^2: + 1 + a:^2:b:^2: = (a:^2: + b:^2: + 1)(b:^2: + 1) - a:^2:   by multiplication and other fun stuff.

(b:^2: + 1):^2: + a:^2:b:^2: = abc(b:^2: + 1) - a:^2:, since (a, b, c) satisfies our given relation.

((b:^2: + 1)/a):^2: + b:^2: = bc((b:^2: + 1)/a) - 1  by division

k:^2: + b:^2: = kbc - 1

Which is what we wanted.

So.  That means that from any ordered pair, we can get an ordered pair with a smaller value for one of the coefficients, provided they are not equal.  Note that this process is exactly reversible -- if we use the process we used to generate (k, b, c) from (a, b, c) on (k, b, c), we get right back to (a, b, c).  (That is, the function is its own inverse.)  Now, start from any ordered pair with a > b.  Use our process to generate a pair with a smaller value a' for a.  Now, either a' > b, a' < b, or a' = b.  In the latter case, we must now have the ordered triple (1, 1, 3).  In the former cases, we can continue to repeat the algorithm to generate smaller triples.  Since all members are integers, we must end up, no matter what, with the ordered triple (1, 1, 3).  But, as noted above, the process is reversible.  Thus, every solution can be found by using the inverse process starting with (1, 1, 3) and building up.  Thus, our first few triples are (1, 1, 3), (2, 1, 3), (5, 2, 3), (13, 5, 3), (34, 13, 3), etc.

The pattern is easy to see (think Fibonacci numbers) and can be proven with an inductive argument that should present no problem.

Note:  if the problem read had a plus one instead of a minus, we would have solutions using the alternate pairs of Fibonacci numbers such as (8, 21, 3), although I don't know offhand if those are unique.[/hide]

You know what -- for number 9, I knew I could say that, and then I thought, "Wait, but what if the cube root of m is the sum of an integer and a rational multiple of the cube root of six," because I forgot that it was a diophantine equation!  Bother!  I think if I had tossed that in, it would have essentially rigorized my solution, or at least half of it.

For (9) I did the following. The given relation is equivalent with

48 + 20(61/3) = m2/3 + n2/3 + 2(mn1/3) - m1/3 - n1/3.

Consider the following cases :

(I) Both m and n are perfect cubes.
Contradiction since the LHS is irrational and the RHS is rational.

(II) (WLOG) m is a perfect cube but n is not, write m = k3.
Then 48 + 20(61/3) = k2 + n2/3 + 2k(n1/3) - k - n1/3.
Comparing rational and irrational parts we get :
k2 - k = 48
... = 20(61/3)
but there does not exist an integer k such that k2 - k = 48.
So we have a contradiction.

(III) None of m and n is a perfect cube.
Then mn MUST be a perfect cube. (See Joel's remark :)).
Comparing rational and irrational parts we get :
48 = 2(mn1/3)
20(61/3) = m2/3 + n2/3 - m1/3 - n1/3.
Now we have mn = 2933.
Write m = 2a3b, n = 2c3d.
Now only routine work remains. It's easy to check all cases.
Then (m, n) = (288, 48) easily comes out as the only answer.

PS. The problems for which I didn't mention the source are certainly not IMO problems ...

Still unsolved :

(2)
(6)
(7) -> Partial solution
(0)

PS. Nice work on (5), JBL !

Thanks, Arne.  The argument took me a long time playing around to get it to work -- then, when I was looking back at my notes, I couldn't even figure out how I had realized that b and (b:^2: + 1)/a would be a solution in the first place.

I really like how the structure of the solution is, though -- the reductio argument plus a reversible algorithm . . . I thought it was very pretty.

(0) x:^2: + 4y = s:^2:, y:^2: + 4x = t:^2:.

[hide]So, I would like to claim that there are no solutions.  As MysticTerminator said, we can see from the given equations that the parities of s and x must be the same, ditto for y and t.  Also, being diophantine equations, x, y > 0, which implies that s > x,

t > y.  These two conditions give us that s = x + 2c, t = y + 2d.  Substituting these values into the given equations, we get

x:^2: + 4y = (x + 2c):^2:

x:^2: + 4y = x:^2: + 4cx + 4c:^2:

y = cx + c:^2: > x.

But by symmetry, we get that x > y.  This is a contradiction.  Thus, there are no positive integral solutions (x, y, s, t) to the given system.[/hide]

For problem (0), x and y can be negative or zero !

Darn.  I thought that might be the case.

For (6), a hint : [hide]2(5) + 1 = ?[/hide]

(Remember my previous hint for (6) :

[hide]If p is prime then (p - 1)/2 is interesting[/hide].

Maybe that hint was a bit unfair  )

To conclude, worth remembering :

[hide]If 2k + 1 is prime, then it can be useful to look at ak (mod 2k + 1).[/hide]

Arne wrote:Solve in integers :
(2) a! b! = a! + b! + c!
Happy solving

(2) [hide]let us assume, wlog, that a:le:b. then we see that a:le:c:le:b, since a!|LHS and a!|(a!+b!); also, b!|LHS and the 2nd term of the RHS, but not the 1st term, so b cannot divide c!. okay, so we divide everything by a! we get b!=1+(a+1)*...*b+(a+1)*...*c. Well, let us assume for the moment that b:ge:2. Then the LHS is even. That means the sum of the last two terms of the RHS is odd. So one term is odd and one term is even. Well, if the 3rd term is even, the 2nd term is definitely even, so the 3rd term is odd and the 2nd term is even. ok, so the 3rd term will be odd if c=a or if c=a+1 and a is even.

okay, if c=a+1, then b:ge:a+2 (else the 2nd term of the RHS would also be odd). Plugging in c=a+1 gives us b!=(a+2)+(a+1)*...*b . Cancelling an (a+2) gives us 1*2*...*(a+1)*(a+3)*....*b=1+(a+1)*(a+3)*...*b. The LHS is even, so the 2nd term of the RHS must be odd. Remember, a is even, so this means b=a+2 or a+3. b=(a+2) gives the solution a=0, and b=a+3 gives no solutions (the term a+3 has to divide every term, as it divides the LHS and the 2nd term of the RHS, but it obviously does not divide 1 for a:ge:0).

Okay, now we go back and look at the case c=a. this gives us 2=[(a+1)*...*b](a!-1). the second factor in the RHS can only divide 2 if a=2. This gives 4=b!. darn no solutions.

Okay, now we go back and check the cases b=0, 1. This gives us 1=1+1+c!. Okay, no solutions.

It could be cleaned up from there, but is that the general idea, or have I made a fatal mistake (which I suspect I did), like in #7? [/hide]

Here is my solution to (6) y2 = x5 - 4, following Arne's generous hint.

 [hide]To see why there are no integer solutions to the equation, take mod 11 of both sides.  First, I claim that x5  is equivalent to -1, 0, or 1 mod 11.  We can show that either by trying all x from -5 to 5, or by using Fermat's Little Theorem, which implies that 11 divides x11 - x = x(x5-1)(x5+1).  Therefore the right-hand side of the original equation is equivalent to 6, 7, or 8 mod 11.  Second, I claim that the left-hand side, y2, is equivalent to 0, 1, 3, 4, 5, or 9 mod 11, which we can see by trying all y from -5 to 5.  Thus the two sides of the equation are different mod 11, so they can't be equal.    [/hide]

Mysticterminator, you did something wrong for (2).

Verify that (3, 3, 4) is a solution ! It is the only solution in fact.

Hint : [hide]If a < b then there it's quite easy to prove that there are no solutions at all. The case a = b is more difficult.[/hide]

(6) isn't IMO at all, it is problem 4 of the Balkan Olympiad 2000.
There's a solution at [url]http://www.kalva.demon.co.uk[/url].
That's why I didn't mention the source :)

The kalva site says that the problem is from Balkan 1998, not 2000.

OK, Should I post solutions for the remaining problems ?
I think no one is really interested anymore ...

wurg no don't. really. i still haven't given in and looked at your hints.

Thanks. Still unsolved :

(2) C'mon guys ! It's not that difficult.
(7) Partial solution.
(0) Some preliminary work is done.

A hint for (0) :

The case where x, y > 0 is done by JBL.
The case x < 0 < y is not so difficult.
It is more difficult to deal with the case x, y < 0.

A hint for (2) :

[i]If a and b are distinct then WLOG a > b.
So a! does not divide b!.
Hence ... does not divide a!b! - a! - b!.
Hence ... does not divide ... (from the given relation).
Hence ... > ... .
So a!b! = a! + b! + c < ... or (...)(...) < ... .
This gives a few easy cases to check.
...
So there are no solutions for distinct a and b.
Consider now the case a = b.[/i]

OK ?

I'll post solutions soon since no one seems to be still interested.

For (2) :

Suppose a > b. Then a! does not divide b!.
Hence c! = a!b! - a! - b! is not divisible by a!.
Hence a > c. So a!b! = a! + b! + c! < 2a! + b!.
Hence (a! - 1)(b! - 2) < 2.
It is easy to check (routine) that this gives no solutions.

So we must have a = b. Then a! = 2a! + c!.
Hence a! divides c! and therefore a < c.
So we write c = a + k where k > 0 is an integer.
Division by a gives a! = 2 + (a + 1)(a + 2)...(a + k).
Obviously RHS > 2 so a! > 2 => a > 2 => 3|a!.
So we must have (a + 1)(a + 2)...(a + k) = 1 (mod 3).
Hence k < 3. (If k > 2 then 3|(a + 1)(a + 2)...(a + k).)
If k = 1 then we get a! = a + 3. This gives a = 3.
If k = 2 then we get a! = a + 3a + 4.
Then a((a - 1)! - a - 3) = 4 => a|4. So a = 1, a = 2 or a = 4.
It is a routine job to check that these cases fail.

Therefore the only solution is (a, b, c) = (3, 3, 4).

For (0) :

A solution can be found at the kalva site :

http://www.kalva.demon.co.uk/apmo/asoln/asol994.html

It is APMO 1999/4.

(7) is straightforward (but ugly) using Pell's equation.

So this topic is finally closed. What did you think of it ?

DANGIT YOU USE MOD 3 I WAS TOO STUCK ON MOD 2 TO USE MOD 3 RGGGGG!

wurg i thought of using modulo 3, but I thought that might give too much away. sigh . . .

the argument for no solutions w/ a>b was much much much nicer than what I came up with though.

topic = fun; #2 = :evil:

I liked them.  The topic is maybe not my favorite.  It was a good set of questions, though.


geometry
rectangle
probability
search
combinatorics unsolved
combinatorics

1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?

2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.

3. 25 people sit around a circular table. Three of them are chosen randomly. What is the probability that at least two of the three are sitting next to one another?

Searching for "birch trees" gets
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=277174
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=66513
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=135810
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=126224
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=87431
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1518516264&t=48556

Exercise: repeat this experiment to find your other questions.  (Hint: you're posting questions from a particular source.  Many people include the sources when they post questions.  Incidentally, this is something you could do, too, especially when the questions come from some established source like a collection of mathematical problems or a national olympiad.)


prove that for all $n\geq 2$
$5^{5n+2}+ 3^{5n-1} - 7 \equiv 0\pmod{11}$

Firstly notice that:
$5^5\equiv{1}\pmod{11}$
$5^{5n}\equiv{1}\pmod{11}$
$5^{5n+2}\equiv{3}\pmod{11}$
$5^{5n+2}-7\equiv{-4}\pmod{11}$
And now:
We have to prove, that:$5^{5n+2}+ 3^{5n-1} - 7 \equiv 0\pmod{11}$
I will use induction.
So first I check if $1$ satisfy this, it's true.
Now I will prove, that if $n$ satisfies this, so $n+1$ will also satisfy.
For $n+1$ our theorem looks like this:
$5^{5(n+1)+2}+3^{5(n+1)-1}- 7 \equiv 0\pmod{11}$
From our assumption we can conclude, that :$3^{5n-1}\equiv{4}\pmod{11}$.
${5^{5(n+1)+2}-7}\equiv{-4}\pmod{11}$ for all $n$ (we proved it at the beginning)
So we have to prove, that if $3^{5n-1}\equiv{4}\pmod{11}$ then $3^{5(n+1)-1}\equiv{4}\pmod{11}$
$3^{5n-1}\equiv{4}\pmod{11}$
$3^{5n-1}\cdot{3^5}\equiv{4\cdot{3^5}\pmod{11}}$
$3^{5(n+1)-1}\equiv{4\cdot{3^5}\pmod{11}}$
$3^{5(n+1)-1}\equiv{4}\pmod{11}$

[hide]since n is an integer, it is odd or even, so we have $n=2k+1$ and $n=2m+2$
note that $\phi(11)=10$, so we can eliminate the k and m for each case, and just add up the results, and they are in fact, 0 mod 11[/hide]

[quote="dondigo"][quote="Andreas"]$(5^5)^n \cdot 5^2 + (3^5)^n : 3 - 7 \equiv 1 \cdot 3 + 1 : 3 - 7 \equiv 0$
$3 + \frac{1}{3} - 7 \equiv 0$
$-\frac{11}{3} \equiv 0$
$-11 \equiv 0$ which is true.
[/quote]
Andreas, you cannot use fractions in congruences.[/quote]

dondigo, note that $11$ is prime. :)


modular arithmetic
superior algebra
superior algebra unsolved

Consider a prime p and let $ f(x) \in \mathbb{Z}[x] - {0}$ have degree $ \leq p-2$. Then find $ \sum_{i=0}^{p-1}f(i) \mod p$

let $ g$ be a primitive root modulo $ p$. then $ 1^k\plus{}2^k\plus{}\cdots\plus{}(p\minus{}1)^k\equal{}g^k\plus{}g^{2k}\plus{}\cdots\plus{}g^{(p\minus{}1)k}\equal{}g^k\cdot \frac{g^{k(p\minus{}1)}\minus{}1}{g^k\minus{}1}$. the numerator is divisible by $ p$ because $ g^{k(p\minus{}1)}\equiv 1\pmod{p}$, and if $ k\leq p\minus{}2$, the denominator is not divisible by $ p$ by the fact that $ g$ is a primitive root. so $ \sum_{i\equal{}0}^{p\minus{}1}f(i)\equiv 0 \pmod{p}$.


Let $ a, b, c, d$ be positive real numbers such that

$ a\plus{}b\plus{}c\plus{}d \equal{} 12$

and

$ abcd\equal{}27\plus{}ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd$.

Find all possible values of $ a, b, c, d$ satisfying these equations.

By AM-GM, $ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\geq 6\sqrt{abcd}$ so $ abcd\equal{}27\plus{}ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd\geq 27\plus{}6\sqrt{abcd}$.  This factorizes into $ (\sqrt{abcd}\minus{}9)(\sqrt{abcd}\plus{}3)\geq 0$ so $ \sqrt{abcd}\geq 9$, or $ 4\sqrt[4]{abcd}\geq 12\equal{}a\plus{}b\plus{}c\plus{}d\geq4\sqrt[4]{abcd}$.  Equality if $ a\equal{}b\equal{}c\equal{}d\equal{}3$.


Έστω το τετράγωνο $ ABCD$ πλευράς $ a$ και $ M$ το μέσο της $ AB$. Αν η $ DB$ τέμνει την $ MC$ στο $ E$, να υπολογίσετε την ακτίνα $ R$ του εγγεγραμμένου κύκλου στο τρίγωνο $ DEC$ συναρτήσει του $ a$. :)

[hide]An einai sostes oi prakseis:

$ r\equal{}\frac{1}{\sqrt{5}\plus{}\sqrt{2}\plus{}\frac{3}{2}}\cdot \alpha$[/hide]

Βγαίνουν  $ DE \equal{} \frac {2a\sqrt {2}}{3}$ ,...$ EC \equal{} \frac {a\sqrt {5}}{3}$, ...$ (DEC) \equal{} \frac {a^2}{3}$

και με δεδομέμο ότι DC = α, και τον τύπο

Ε = τρ προκύπτει το ζητούμενο ...

Ys.  Νίκο νομίζω ότι έχεις φάει ένα 2 στον παρονομαστή του ρίζα 5.

Αλλά μπορεί να έχω κάνει και εγώ λάθος στις πράξεις ( ως συνήθως )
γιατί τις έκανα βιαστικά...

[quote="vangelismak"]Βγαίνουν  $ DE \equal{} \frac {2a\sqrt {2}}{3}$ ,...$ EC \equal{} \frac {a\sqrt {5}}{3}$, ...$ (DEC) \equal{} \frac {a^2}{3}$

και με δεδομέμο ότι DC = 1 και τον τύπο

Ε = τρ προκύπτει το ζητούμενο ...

Ys.  Νίκο νομίζω ότι έχεις φάει ένα 2 στον παρονομαστή του ρίζα 5.

Αλλά μπορεί να έχω κάνει και εγώ λάθος στις πράξεις ( ως συνήθως )
γιατί τις έκανα βιαστικά...[/quote]

me auto ton tropo to ekana ki ego...tora gia tis prakseis-h episthmh shkonei ta xeria psila.

Πραγματικά νομίζω ότι είναι το πιο δυσκολο μέρος της άσκησης  :rotfl:

Mια λύση που δεν είναι βασανιστική ... ( λίγες πράξεις μόνο ... ) :|

[hide]Έστω τα σημεία Α(0,α), Β(α,α), C(α,0) και D(0,0)  που είναι κορυφές τετραγώνου. Τότε θα είναι Μ( $ \frac {a}{2}$,α ).
Eύκολα βρίσκουμε ότι οι εξισώσεις των ευθειών BD και CM είναι
BD: y=x (1) και  CM: y= -2x+2α (2).Λύνοντας το σύστημα των (1) και (2) προκύπτει ότι το σημείο Ε έχει συντεταγμένες E$ \left(\frac {2a}{3},\frac {2a}{3}\right)$. Από τον τύπο Ε = $ \frac {1}{2}|det( \vec{DE},\vec{DC})|$ υπολογίζουμε το εμβαδόν του τριγώνου DEC,
E = $ \frac {a^2}{3}$και από τον τύπο απόστασης δυο σημείων βρίσκουμε ότι DE=$ \frac {2a}{3}\sqrt {3}$.. και CE = $ \frac {a}{3}\sqrt {5}$,..και δεδομένης της απόστασης DC = a, μπορούμε να βρούμε την ημιπερίμετρο του τριγώνου DE,...τ = $ \frac {1}{2}$$ \left(\frac {2a}{3}\sqrt {2} \plus{} \frac {a}{3}\sqrt {5} \plus{} a\right)$

Στη συνέχεια από τον πολύ γνωστό τύπο Ε = τρ προκύπτει το ζητούμενο ...[/hide]


I have been thinking about this a lot lately.

Just recently I have seen proofs (on AoPS website and in my class), that are formal and prove why the theorem works, or prove that something is not true.  For example, if someone makes a statement "x^3-5>0 for all x" and then you can say "what about x=cuberoot(5)?" and that is a contradiction, so the statement is proved false.  However, they are of course way more complicated than this simple example.

However, in math classes my teacher has always given some sort of justification as to why we use a certain theorem or formula.  When you see it you think "Ohhh, so that's why it works!"  This justification is not usually very formal.

Now the first type of proof I was talking about does not really elicit the same reaction (for me at least).  Do I just have to get used to proofs, or will this always be the case?

Here's the best example that I could think of for this:

[tex] g(x)= \sqrt{f(x)} [/tex]

Now, I am going to make the statement "the maximum of g(x) is the same as the maximum of f(x)."

Proof:

[tex]g(x)\ge g(x)_{min}[/tex]
[tex]\sqrt{f(x)} \ge \sqrt{f(x)_{min}}[/tex]
[tex]f(x) \ge f(x)_{min}[/tex]

Justification:

If your highest value for f(x) is one number, then if you take the square root of all of the values of f(x), the square root of the maximum f(x) will still be the greatest, b/c sqrt(bigger #)>sqrt(smaller number), b/c of how square roots work.

I am sure you guys can find lots of fundamental flaws in this, but I am just trying to provide an example.  This is a bad one, I know.  A good example might be the "justification" (dividing out b/c of overcounting, how it works w/multiplication counting principle) and "proof" (more formal, I don't know it, induction maybe) or the combinations and permutations formulas.

I don't quite see how what you are saying is true. Here's a counterexample: Say f(x)=4. Then g(x)=2. The maximum of f(x) is 4, but the maximum of g(x) is 2.

The main reason rigorous proofs are used is so that you are certain the proof is correct. The many false proofs out there show that it is important to be as rigorous as possible.

You are saying that a rigorous proof does not really explain to you why something works, unlike a less formal justification. The best thing to do is to think about what a statement in a rigorous proof really means.

I guess a good example would be if you see:

[tex]x=\frac pq,[/tex] where [tex]p,q\in\mathbb Z,q>0,GCF(p,q)=1[/tex].

You know it's just saying that x is written as a fraction in lowest terms.

I was under the impression she meant that the x at which the maximum for one was achieved was the same.

Julie, I'm not really sure what you are asking/talking about.  Your examples confuse me more, not less.  What do you want to know/have people talk about?

I just want to know the difference between formal proofs and informal justifications.  I don't know why informal justifications make more sense to me than proofs do.  Will formal proofs ever help me understand why something is so it "clicks" when I am using it (a formula, theorem, etc)?

Yes and I realize my examples make no sense.

It depends on the skill and goals of the proof-writer. If the proof-writer's main objective is just to establish that something is true - then they may choose a proof that is quick and simple, but that gives little "insight" into why it is true.

However if the proof-writer's main purpose is to [i]explain[/i] why something is true - then they may choose to provide either a more constructive or intuitive proof of the statement.

You often find this as the main difference between the style of proofs that you find in advanced textbooks or in journals, and the proofs that you find in elementary textbooks. In elementary textbooks, when authors are faced with the choice between styles of proof, they will often choose one that is more instinctive, or constructive, or visual.

In some cases though, early on in math, you are taught theorems that are important, but that you just don't have the mathematical machinery yet to prove. So the only option is to give at least a "motivation" for why the theorem is true.

The ability to come up with intuitive, though maybe not entirely formal, justifications for things is a great skill to have. Along with this goes an intuition as to when things are true or not true (especially the ability to come up with counterexamples). Mathematicians use this type of thinking all the time to come up with conjectures, or to lead them in the right direction. But also they realize that a statement is never really completely safe until a formal proof of it has been provided. Also formal proofs often reaveal the conditions under which a statement is true. That's something that is often hard to do with hand-wavy explainations.

Deep within every proof though, is a good intuitive "reason" why the statement is true. It just depends on how deeply or abstractly you can look the proof to see it. In some cases, proof-writers intentionally obscure and hide their work and thought processes for the sake of a cleaner and more elegant final product. But even then, the ultimate reason behind the truth or the statement is in there.

[quote="JS1527"]
[tex] g(x)= \sqrt{f(x)} [/tex]

Now, I am going to make the statement "the maximum of g(x) is the same as the maximum of f(x)."

[/quote]

You don't need to rely on shaky logic here.

The idea I'm going to use here is that of a bijection between two continuous functions - that is to say, that over the set R+ (the positive reals), the function p(x) = sqrt(x) does two things.

1: It maps R+ onto R+ --> every element in R+ is mapped onto a new element in R+ by p(x).

2: For every element a E R+, the equation p(x) = a has exactly one solution.

So this is nice.  The second thing that we need to note is that sqrt(x) is monotonically increasing - it is increasing EVERYWHERE.

So what the function p(x) = sqrt(x) does is preserve order.

Given a > b, we have that sqrt(a) > sqrt(b).  But we also know that we will not randomly have some number c < a such that sqrt(c) = sqrt(a), because it is a bijection.

So, let t = the value where f(x) assumes a maximum.

Then we know that f(t) >= f(x), x E R+ by hypothesis, correct?

So we now know that sqrt(f(t)) >= sqrt(f(x))  Hence f(t) and sqrt(f(t)) are both maximized for the same value.


Nice attempt at vector art, much better than my first. A bit too contrasted, try lowering it.

edit: Also it's very aliased.

Yeah, she really liked this on Facebook.

I like this one better than the other ones.
Though the shades is a close second.

Woah, this is pretty impressive! Did you just like mess with teh filter on gimp and merged the light saber from Yoda?

Actually, you use a Gaussian blur and colorizing. I was holding a dowel for the picture.


parameterization
number theory unsolved
number theory

[quote="Moonmathpi496"]Prove that the equation $ x^2 \equal{} y^3 \plus{} z^5$ has infinitely many solutions in positive integer.

[hide="Hint"]A magical substitution(I mean constructing solutions by parameters) solves this problem, but I don't really understand what the motivation behind that substitution is... and also how to cast this type of magic...:( [/hide][/quote]
Once you have found the solution $ x\equal{}3, y\equal{}2, z\equal{}1$, I think it is pretty easy to find the rest of the solutions, since there is no restriction, like they should be coprime :)


[quote="lotrgreengrapes7926"][quote="kimby_102"][quote="PenguinIntegral"]If a, b, c are positive numbers such that
$(1+a)(1+b)(1+c) = 8$
prove that abc <= 1.[/quote]
I think the only solution to this problem is $a=b=c=1$.
So $abc=1$..
but i don't understand why there is an inequality $abc\leq 1$..[/quote]
Positive NUMBERS, not only positive integers.[/quote]
sorry, my mistake :blush:

then we apply AM-GM inequality.
$1+a\geq 2\sqrt{a}$
$1+b\geq 2\sqrt{b}$
$1+c\geq 2\sqrt{c}$.
Now multiply thses three inequal. we get:
$(1+a)(1+b)(1+c)\geq 8\sqrt{abc}$.
But since $(1+a)(1+b)(1+c)=8$, $1\geq abc$. :D


number theory
prime numbers
number theory unsolved

Suppose that $ p$  is a prime number .Find all $p$ s.t. the remainder of $p$ to all other prime numbers be the square-free . :huh:  :o

well I have showed that for big primes, Not too big I guess $p>10$ will do, we must have one of the following situations:

$p=3^k+4=2^t+9$, or $p=3^k+4=2^t.7^s+9=5^l.7.3^2+16$ or $p=3^k+4=7.2^t+9$ and some of them can be shown to have no nontrivial solutions.


hmm links don't open in a new window/tab anymore...
I'm on the ML skin/firefox

As required per web 2.0 standards links shouldn't open in new tabs / windows (basically the user should choose to do so by right-click open in a new tab/window). In plus the target attribute will not validate in xhtml.


LaTeX
linear algebra
linear algebra unsolved

Are these ordered triples two distinct bases in Q^3?

(4/5,3/4,2/3) and ( 11/13,7/11,5/7) ?

If they are, how would i go about proving that they span Q^3?

If they aren't-- what's a good example of an ordered triple that is a basis?

No. Each of those triples is [i]one[/i] element of $ \mathbb{Q}^{3}$, and you need three elements for a basis.

How about these?

([1,0,0],[0,1,0],[0,0,1]) and ([4/5,0,0],[0,3/4,0],[0,0,2/3]).

are those two distinct bases in Q^3?

Also, how does someone go about writing tex in these forums?

Yes, those are bases.

To write $ \text{\LaTeX}$ math mode here, just put your code between dollar signs.


function
calculus
derivative
inequalities
inequalities unsolved

Basicly I don't quite have the energy to post my solution but it uses derivatives. You set z=const and you substitute x=1-y-z and you make a function of y. You can set y to be between 0 and 1/3 ( basicly the smallest of the 3, without loss of generality cuz it is a cyclic function). Next you should take the first derivative and see that one of the roots ( the maximum of the funcion) is not between those two numbers which means that one of them ( or two as the case may be) must be the value for which f(y) is maximum. Next by substituting y=0 you will see that x=2/3 and z=1/3 (easy again with derivative of the function (x^2)y since all other monomials are equal to 0 and that would lead you to one solution (0,2/3,1/3) ) and the second y=1/3 leads to x=1/3 z=1/3 again with derivative of a function.
          It is certainly not an easy way to do this and not the most pretty but certainly all BEAUTIFUL inequalities like AM, GM, Cauchy or anything else would lead you nowhere cuz equality is reached basicly when x=y=z and that is not the only case we are seeing in this problem so I don't see any way around derivatives or Lagrange. Hope you find an easier way. But at least I hope I saved everyone some time in checking for an elegant solutions with the above inequalities. THEY WILL NOT WORK!!!
         So there are two answers: (1/3;1/3;1/3) AND (0;1/3;2/3) and their permutations of course.

Hello. Here is a simple solution:

WLOG suppose that $ x$ is the middle variable (i.e. $ y\ge x\ge z$ or $ z\ge x\ge y$). Then clearly, $ (x\minus{}y)(x\minus{}z)\le 0$. So $ \minus{}y(x\minus{}y)(x\minus{}z)\ge 0$.

Now let $ F[x,y,z]\equal{}x^2 y \plus{} y^2 z \plus{} z^2 x \plus{} x y z$. It is easy to verify that

\[ F[x, 0, y \plus{} z] \minus{} F[x, y, z]\equal{}\minus{}(x \minus{} y) y (x \minus{} z)\ge 0\]

Let $ w\equal{}y\plus{}z$. Then $ x\plus{}w\le 1$, and we need to prove that $ w^2x\le \frac{4}{27}$.

But this is trivial by AM-GM because \[ w^2x\equal{}4\cdot\frac{w}{2}\frac{w}{2}x\le 4\cdot \left(\frac{ \frac{w}{2}\plus{}\frac{w}{2}\plus{}x}{3}\right)^3\equal{}\frac{4}{27}\cdot (x\plus{}w)^3\le \frac{4}{27}\]

Ugh, Altheman beat me to it but i'm pretty sure i have the nicer proof anyway.

Let $ x\geq z \geq y$, and clearly we may assume $ x\plus{}y\plus{}z\equal{}1$. Then, $ 0\geq (x\minus{}z)(y\minus{}z)\iff z(x\plus{}y)\geq xy\plus{}z^2\qquad (*)$.

Now,
\[ yz(x\plus{}y)\plus{}x(xy\plus{}z^2) \leq z(x\plus{}y)^2\qquad \text{ by }(*)\]
And, by Am-Gm we have \[ \frac13\equal{}\frac{\frac{x\plus{}y}{2}\plus{}\frac{x\plus{}y}{2}\plus{}z}{3}\geq \sqrt[3]{\frac{z(x\plus{}y)^2}{4}}\iff z(x\plus{}y)^2\leq \frac{4}{27}\]
QED
 :D

It seems to me that our proofs are equivalent...you should be careful because you cannot assume $ x\ge z\ge y$ since the inequality is cyclic by not symmetric.

Yes, that was sort of the joke... and good poitn about the cyclicity, I should have caught that mistake.
Anyway, I was thinking about the whole "motivation" behind comparing $ xy\minus{}z^2$ to $ z(x\plus{}y)$ and realized that when you substitute t+a,t-a = x,y we have $ |t\minus{}z|\leq a$ as the square root of $ xy\minus{}z^2\leq z(x\plus{}y)$, so it's totally a topological statement

[quote="jumbogaba"]If x, y, z > 0 and $ x \plus{} y \plus{} z \le 1$ prove that:

$ x^2y \plus{} y^2z \plus{} z^2x \plus{} xyz \le 4/27$[/quote]

$ x^2y \plus{} y^2z \plus{} z^2x \plus{} xyz \le \frac {4}{27}(x \plus{} y \plus{} z)^3 \le \frac {4}{27}$  :lol:


Given $ a, b, c > 0$ satisfy $ (4a\plus{}5b)(4b\plus{}5c)(4c\plus{}5a)\equal{}729$.

Prove that: $ abc(a^2\plus{}bc\plus{}ca)(b^2\plus{}ca\plus{}ab)(c^2\plus{}ab\plus{}bc) \le 27$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ satisfy $ (4a \plus{} 5b)(4b \plus{} 5c)(4c \plus{} 5a) \equal{} 729$.

Prove that: $ abc(a^2 \plus{} bc \plus{} ca)(b^2 \plus{} ca \plus{} ab)(c^2 \plus{} ab \plus{} bc) \le 27$
 :)[/quote]
We can solve this problem by "ony Am-Gm"  :)

[quote="nguoivn"]Given $ a, b, c > 0$ satisfy $ (4a \plus{} 5b)(4b \plus{} 5c)(4c \plus{} 5a) \equal{} 729$.

Prove that: $ abc(a^2 \plus{} bc \plus{} ca)(b^2 \plus{} ca \plus{} ab)(c^2 \plus{} ab \plus{} bc) \le 27$
 :)[/quote]
See it here: http://can-hang2007.blogspot.com/2010/01/inequality-67-t-q-anh.html
(Note: Visitors need to use Firefox in order to see the Math Formulas in my blog.)

It also exists a shorter solution by AM-GM  :)


function
calculus
integration
LaTeX
real analysis
real analysis unsolved

Hi all, just studying for my analysis exam, I am stuck on this problem:

Suppose mu is a Borel measure on X, f:X -> [0,infinity) is measurable and g:[0,infinity) -> [0,infinity) is an increasing C-one function st. g(0) = 0

Show the integral over X of g(f(x)) d(mu) = the integral from 0 to infinity of g'(t)*mu({x:f(x)>t}) dt

Thanks for the help, sorry I don't know Latex, I intend to learn it over the summer.

You have

\[ \int^\infty_0 \mu(\{f>t\})g'(t)dt \equal{} \int^\infty_0 \int_X g'(t) \chi_{\{f>t\}}d\mu dt
\]

Now use Fubini.

Ah yes of course, thanks very much, I appreciate the help. I am stuck on another problem now, about Fourier coefficients of measures on the unit circle:
Prove that, if mu^(n) -> 0 as n->+infinity, then for any bounded Borel function f on S^1, the integral over S^1 of
exp(-2Pi*i*n*t)f(t)dmu(t) ->0 as n->+infinity

*Here mu^(n) means the nth Fourier coefficient of mu, namely the integral over S^1 of exp(-2Pi*i*n*t)dmu(t)

Any advice would be appreciated.


How high would the column be if you were to put all the millimeter cubes contained in a cubic meter,on top of each other.

a cubic meter contains 1 billion milimeter cubes. So if you put those on top of eacother you get something 1 billion milimeters heigh, that is 1 million meters, or 1.000 kilometers.
again, far too easy.  what was the point?


How does one go about finding the first few digits of a mammoth number by just using a calculator and logs?.  For instance, the newest discovered prime is 2^25,964,951-1.  How can the first 3 digits be found by just using the methods above?.

Thanks,
Cody

This was fun.  I came up with this method just now, then checked it to make sure I wasn't crazy.

$log_2(2^{25,964,951}-1)$ is, for our purposes, about equal to ${log_2(2^{25,964,951}})=25,964,951$.  Therefore, $log_{10}(2^{25,964,951}-1)$ is about $25,964,951*log_{10}2=7816229.0869454841986461649283156$.  If we subtract an integer from that value, we are essentially just moving the decimal place over 1 point in our behometh number.  Therefore, we take $10^.0869454841986461649283156$ to get $1.2216463006127794810775378964564$.  Those should be the first couple of digits of the number.

Edit: Those first few digits match up with what they posted on mathworld.  It matched up to 9810, then mathworld stop posting the number.  Presumably, the rest are correct as well.

Thanks. I was close. I got as far as 25964951*log2=7816229.88695, but I didn't multiply 10 times the decimal part. Clever. Thanks

Cody


$ xy|x^{2}\plus{}y^{2}$. Proof that $ x\equal{}y$

[quote="inny"]$ xy|x^{2} \plus{} y^{2}$. Proof that $ x \equal{} y$[/quote]
$ x^2 \minus{} kxy \plus{} y^2 \equal{} 0$, where $ k\in\mathbb{N}$. Divide the equation by $ y^2$ and note $ t \equal{} \dfrac{x}y$. So: $ t^2 \minus{} kt \plus{} 1 \equal{} 0$. In order to admit rational solution its discriminant must be rational. So $ k^2 \minus{} 4 \equal{} a^2$ or $ 4 \equal{} (k \minus{} a)(a \plus{} k)$ So $ k \equal{} 2, a \equal{} 0$ which implies $ x \equal{} y$


Let $ S(n)$ denote the sum of all digits of a positive integer $ n$. Do there exist positive integers $ a,b$ such that for any positive integer $ n$, $ S(an)<S(bn)$?

No, it is not. Consider $ n\equal{}[\frac{10^k}{b}\plus{}1]$.

[quote="Rust"]No, it is not. Consider $ n \equal{} [\frac {10^k}{b} \plus{} 1]$.[/quote]

Please write more clearly, Rust!  :mad:


There are 16 coins in a bank.  If the coins are all nickels and dimes and they total $1.05 how many nickels are there?

[hide] Simple problem.
11 nickels.  If you view each dime as two nickels, that's 21 nickels.  Squash the five extra nickels into five other nickels to form five dimes, and you have 11 nickels and 5 dimes.[/hide]

[hide]with all dimes it's 1.60
so every time a nickel replaces a dime it goes down .05
1.05 is down 11 so 11 nickels[/hide]


Four circles of radius $ 1$ are each tangent to two sides of a square and externally tangent to a circle of radius $ 2$, as shown. What is the area of the square?
[asy]unitsize(5mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

real h=3*sqrt(2)/2;
pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h);
pair X=O0+2*dir(30), Y=O2+dir(45);

draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle);
draw(Circle(O0,2));
draw(Circle(O1,1));
draw(Circle(O2,1));
draw(Circle(O3,1));
draw(Circle(O4,1));

draw(O0--X);
draw(O2--Y);
label("$2$",midpoint(O0--X),NW);
label("$1$",midpoint(O2--Y),SE);[/asy]$ \textbf{(A)}\ 32 \qquad \textbf{(B)}\ 22 \plus{} 12\sqrt {2}\qquad \textbf{(C)}\ 16 \plus{} 16\sqrt {3}\qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 36 \plus{} 16\sqrt {2}$

[hide]The side length of the square is $\left(\sqrt{2}+1+2\right)\cdot 2=2+3\sqrt{2},$ the desired area is $(2+3\sqrt{2})^{2}=22+12\sqrt{2}. (B).$[/hide]

That's how I did it, except I misread that 3 as a 4 and squared the quantity 2+4rt(2) which led me to choice E, which I didn't bother checking since it was listed as an answer :oops:

why is a side of the square is [img]http://www.mathlinks.ro/Forum/latexrender/pictures/9a7e15804a9ba062f5a6e5f8382a0527.gif[/img]?

although can make a reasonable guess by

diagonal is little bigger than 8 ( 4 + 2 + 2),

so 64 = 2x^2
32 = x^2

22+12root2 is 38.9, so it can be answer

16+16root3 is too big, 43.7 means square of diagonal is 87.4, bigger than 9

but i dun get why it's [img]http://www.mathlinks.ro/Forum/latexrender/pictures/9a7e15804a9ba062f5a6e5f8382a0527.gif[/img]

Could you show me how you got the side as well?

[hide="Solution"]
Draw the diagonal connecting the center of the 2 small circles and the center of the big one.  We know this length is $6$ and if you connect the bottom right circle with the bottom left, and the bottom left with the top left, you get an isosceles right triangle with side $x$.

Therefore: $2x^{2}=6^{2}$ so $x=\sqrt{18}$

The length of the side of the square is just going to be $x+2$, so the area will be $(2+3\sqrt{2})^{2}=22+12\sqrt{2}$
$\boxed{B}$
[/hide]

This can trick many poeple on the test, like me :oops:

Heres what I did:

you can see that if you connect the radius of the large circle to one of the small circle you will get 3.

You can visualize a 1x1 square at the top left, this would form a 45 45 90 right triangle allowing you to figure the the hypotenuse is sqrt(2).

So the distance from the middle of the large circle to the corners of the square is 3 + sqrt(2), you can multiply this by two to get the diagnols of the square.

The area of a rhombus can be expressed as (d1 * d2)/2 (half of diagnol 1 * diagnol 2).

So the answer is: (6 + 2sqrt(2))^2 / 2 = 22 + 12sqr(2)

Yeah, a rhombus is a square, so you can just multiply the diagonals and divide by 2.

How I did it:

We can see the diagonal of the square is $d = 6+2\sqrt{2}$ so the side length is $s = \frac{d\sqrt{2}}{2} = 2+3\sqrt{2}$

$s^2 = 4+12\sqrt{2}+18 = \boxed{22+12\sqrt{2}}$

[quote=Calculus123]Can someone make a diagram? For some reason, I cannot visualize this.[/quote]

It's in the OP...

The diagram of the "diagonals connecting the two smaller circles with the larger circles." It was in a solution.

[hide=Solution]
The diagonal is $2(1+\sqrt2)+4=6+2\sqrt2$, so the area = $\frac{1}{2}(6+2\sqrt2)^2 = \boxed{\textbf{(B)}22+12\sqrt2}$.
[/hide]


LaTeX
geometry
geometric transformation
Gauss
graph theory
national olympiad

Here are problems:http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=15&Id=3671&page=14.
But  I do not translate.
I hope it is translated into English.

I can not download that file, therefore I can not help you.

This is file.I really thank you if you can help me

I really hope people continue to translate it

[quote="math10"]Why people do not translate to?[/quote]

Why don't you do anything but only ask for things? How about formatting those problems which were already translated above in $ \text{\LaTeX}$ and posting them on the forum and posting the links to them here? And for the other problems you could provide a translation suggestion via google translator!

[quote="math10"]I am very sorry because do not have Chinese font(although I find) So I can not do.[/quote]

I know it is impossible that you also contribute something?!

1: The problems posted by Potla as .doc-format above surely cannot be posted to the forum!

2.) I guess you are using Windows. Then to change or add a new language go to regional and language settings in the control panel and add Mandarin/Chinese. For your convenience I even provide a link from a discussion forum [url=http://www.trap17.com/forums/index.php?s=007fd747797956f55bfeb6b5a2f352dd&showtopic=49038&view=findpost&p=332407]here.[/url]

[quote="Potla"][quote="orl"]Thanks for translating a few of the documents!  :) How comes you speak Mandarin, or did you just use an online translator as [url=http://www.google.ca/language_tools?hl=en]google translator?[/url] Do you mind providing the translatiosn as $ \text{\LaTeX}$ documents/code?[/quote]
I TRIED to use GOOGLE translate but it gave WEIRD TRANSLATIONS........ :rotfl:
So I did the job by COPYING and PASTING small parts of a sentence and putting many senseless pieces together, it was a tough job. I will try to do more, actually I forgot about this topic COMPLETELY, sorry. :oops:  :([/quote]
I am sorry that I do not have the TeXNic Centre or the Miketex... I saw there were download links in the wiki, but I did not install it, nor did I download. :(
Could you TeXify it, orl? :oops:

I do not have too much time.Today i would translate the (G):

1. For what positive integer $ h$, there exists infinitely many positive integer $ n$ such that $ [n\sqrt {h^2\plus{}1}]$ is a perfect square?
$ [x]$ means the Gauss Function.

2.Find maximum positive real number $ c$ such that for any positive integer $ n$, $ \{ \sqrt {7}n \} \ge \frac {c}{\sqrt {7}n}$
$ \{ x \}$ means the fractional part of $ x$

3.Find how many ways are there to express $ 42$ in the form $ 42\equal{}x^3\plus{}y^3\plus{}z^3\minus{}w^2$ with integers $ x,y,z,w$.Prove your conclusion.

I would only translate the words now because you guys can understand math symbols and type them into latex form.

(1) is (A) , (2) is (B) and (3) is (F)

[quote="math10"]it has been translated polta(Test A,B)[/quote]

since he doesn't understand Mandarin, his version may not quite clear to understand.so I translated them again.

oh,I understand why you translate it.
I really hope you continue to work.

the fourth quiz and sixth quiz
(the fifth are in post #13)

[quote="math10"]I feel China TST 2001 quiz is not really interesting as the other year
Some problems are too long and divided into many small ideas
This is not good in the exam as  China TST.[/quote]

I agree with you.

the quiz in 2001 was very strange. there are only two geometry problem in total, but more than five problems about graph theory!

also these problems are quite hard. maybe I could only solve half of them.

I think the quiz in year 2002 are much better, you can try it.


$a$, $b$, $c$, $d$ are positive integers and
\[ad=b^{2}+bc+c^{2}\]
Prove that
\[a^{2}+b^{2}+c^{2}+d^{2}\]
is a composed number.

Set $p = a^{2}+b^{2}+c^{2}+d^{2}.$ Next assume $p$ is a prime. Then

\begin{eqnarray*}p & = & (a^{2}+2ad+d^{2})+(b^{2}+c^{2}-2ad) \\ & = & (a+d)^{2}+b^{2}+c^{2}-2(b^{2}+c^{2}+bc) \\ & = & (a+d)^{2}-(b^{2}+2bc+c^{2}) \\ & = & (a+d)^{2}-(b+c)^{2}\\ & = & (a-b-c+d)(a+b+c+d). \end{eqnarray*}

Since $p$ is a prime, the only possible factorization af $p$ is $1 \cdot p$. Hence $a-b-c+d = 1$ and $a+b+c+d = p$, which implies that $a+d ={\textstyle \frac{p+1}{2}}$ and $b+c ={\textstyle \frac{p-1}{2}}$. Now

${\textstyle \frac{(p+1)^{2}}{16}\; = \; (\frac{a+d}{2})^{2}\geq ad \;=\; b^{2}+c^{2}+bc \; \geq \; \frac{3}{4}(b+c)^{2}\; = \; \frac{3(p-1)^{2}}{16}.}$

This gives the following chain of implications:

\begin{eqnarray*}{\textstyle \frac{(p+1)^{2}}{16}}& \geq &{\textstyle \frac{3(p-1)^{2}}{16}}\\ 3(p-1)^{2}-(p+1)^{2}& \leq & 0 \\ p^{2}-4p+1 & \leq & 0 \\ p & < & 4 \end{eqnarray*}

Therefore

${\textstyle b+c \;=\; \frac{p-1}{2}\; \leq \; \frac{3-1}{2}\;=\; 1.}$

But $b$ and $c$ are positive integers, so $b+c \geq 2.$ This contradiction implies that $p$ is a composite number.

Here is a simplier way of solving it (I mean my solution)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136491[/url]

Hmm so while waiting in line during Black-Friday-at-eight-o-clock shopping I somehow managed to solve this one after also previously having gotten [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=276020]this one[/url].  Guess that says something about how long the line was.

Note that $2ad=2b^2+2bc+2c^2\implies 2ad-b^2-c^2=(b+c)^2$.  Therefore \begin{align*}a^2+b^2+c^2+d^2&= a^2+2ad+d^2+b^2+c^2-2ad\\&=(a+d)^2-(b+c)^2\\&= (a+b+c+d)(a+d-b-c).\end{align*}  Since $a,b,c,d>0$ and $a^2+b^2+c^2+d^2>0$, in order to show that this is composite we must show that $a+d=b+c+1$ is not possible.  Suppose it is possible.  Then squaring yields \[a^2+2ad+d^2=b^2+2bc+c^2+2b+2c+1\implies a^2+ad+d^2=bc+2b+2c+1.\]  Using the fact that $ad=b^2+bc+c^2$ again gives \[a^2+b^2+c^2+d^2+ad+bc=ad+bc+2b+2c+1\implies a^2+(b-1)^2+(c-1)^2+d^2=1.\]  However, since $a^2+d^2\geq 2$ we have a contradiction.  Therefore $a+d-b-c\geq 2$ and $a^2+b^2+c^2+d^2$ is composite, as desired.  $\blacksquare$

Note that
\[(a+d)^2 - (b+c)^2 = a^2+d^2 + 2(ad-bc) - b^2-c^2 = a^2+d^2 + 2(b^2+c^2)-b^2-c^2 = a^2+b^2+c^2+d^2\]
Thus, $a^2+b^2+c^2+d^2 = (a+d-b-c)\cdot (a+d+b+c)$. Thus, it suffices to show that $a+d\geq b+c+2$. AFTSOC that $a+d\leq b+c+1$, then
\[b^2+bc+c^2=ad\leq \frac{(a+d)^2}{4} \leq \frac{(b+c+1)^2}{4}\]
So,
\[4b^2+4bc+4c^2\leq b^2+c^2+2bc+2b+2c+1\]
Then,
\[3b^2+2bc+3c^2\leq 2b+2c+1\]
but since $b,c$ are positive, we have $3b^2>2b, 2bc>1, 3c^2>2c$, so this is absurd. Thus, $(a+d-b-c)\geq 2$, so $a^2+b^2+c^2+d^2$ can be written as the product of two integers $\geq 2$ and we're done.


induction
trigonometry
algebra unsolved
algebra

How to prove the identity

${{\sin^2(n+1)x}\over {\sin x}} = \sin x+\sin 2x +\sin 3x+\ldots +\sin (2n+1)x$

by induction?

I don't think it's correct  :? (so we can't prove it by induction when assuming consistency of our axioms :P ).
Take $n=1$ to get $\frac{\sin(2x)^2}{\sin(x)} = \sin(x)+\sin(2x)+\sin(3x)$, which is wrong for all $x \neq \frac{n\pi}{2}$.

The correct formula should be
$\sin\alpha+\sin 2\alpha+\sin 3\alpha+\cdots +\sin n\alpha=\frac{\sin\frac{n\alpha}{2}\sin\frac{(n+1)\alpha}{2}}{\sin\frac{\alpha}{2}}$.

Indeed, and your formula can be easily proven using complex numbers ;) (if we don't know the RHS of course, else, induction!)

The correct identity should be:

${{\sin^2(n+1)x}\over {\sin x}}=\sinh+\sin3x +\ldots+\sin (2n+1)x$

For $n=1$:

LHS${= {{\sin^2 2x}\over {\sin x}}={{(2\sin x\cos x)^2}\over {\sin x}} ={{4\sin^2 x\cos^2 x}\over {\sin x}}=4\sin x\cos^2 x}$

RHS$=\sin x+\sin3x =2\sin{{3x+x}\over 2}\cos{{3x-x}\over 2} =2\sin 2x\cos x=4\sin x\cos^2 x=$LHS


1. Find all values of $ x,y \in\mathbb{N}$ that satisfy
$ (x\plus{}y)\plus{}(x\minus{}y)\plus{}xy\plus{}\frac2{y}\equal{}2009$

2. Find the value of $ x$ if
$ x^2\minus{}2\sqrt{x\minus{}1\minus{}\frac1{x}\plus{}\frac1{x^2}}\minus{}x\plus{}\frac2{x}\equal{}1$, $ x\neq 0$

3. It is given that $ x\equal{}\frac1{2\minus{}\sqrt{3}}$. Find the value of
$ x^6\minus{}2\sqrt{3}x^5\minus{}x^4\plus{}x^3\minus{}4x^2\plus{}2x\minus{}\sqrt{3}$

thx

[quote="IW@IT"]1. Find all values of $ x,y \in\mathbb{N}$ that satisfy
$ (x \plus{} y) \plus{} (x \minus{} y) \plus{} xy \plus{} \frac2{y} \equal{} 2009$
[/quote]

[hide="1."]
$ \because y\in\mathbb{N}$ then $ y\in \left\{1,2\right\}$

If $ y\equal{}1$ then $ 2x\plus{}x\plus{}2\equal{}2009 \implies x\equal{}667$

If $ y\equal{}2$ then $ 2x\plus{}2x\plus{}1\equal{}2009 \implies x\equal{}502$

$ \therefore (x,y)\in \left\{(667,1),(502,2)\right\}$
[/hide]

[quote="ffao"][quote="IW@IT"]
$ x^6 \minus{} 2\sqrt {3}x^5 \minus{} x^4 \plus{} x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3}$
[/quote]

Ninja'd by vinskman! ;)

[hide="3"]
$ x \equal{} 2 \plus{} \sqrt {3}$

We have $ x^5(x \minus{} 2\sqrt {3}) \equal{} x^6 \minus{} 2\sqrt {3}x^5 \equal{} x^4$.

Therefore the expression is equivalent to $ x^4 \minus{} x^4 \plus{} x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3} \equal{} x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3}$

This is not too hard to calculate, it's $ 26 \plus{} 15\sqrt {3} \minus{} 28 \minus{} 16\sqrt {3} \plus{} 4 \plus{} 2\sqrt {3} \minus{} \sqrt {3} \equal{} 2$.
[/hide][/quote]

Hello , my guy :lol:  what means "Ninja'd by vinskman"? would you tell me?

[quote="ffao"]It means that you posted exactly what I would post, right before I posted.[/quote]

Thanks.

No worries.You are welcome. :D  :P

[quote="ffao"]
Therefore the expression is equivalent to $ x^4 \minus{} x^4 \plus{} x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3} \equal{} x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3}$
[/quote]
[hide]$ 2\plus{}\sqrt{3}$ is a root of $ x^2\minus{}4x\plus{}1\equal{}0$ so  $ x^3 \minus{} 4x^2 \plus{} 2x \minus{} \sqrt {3}\equal{}x(x^2\minus{}4x\plus{}1)\plus{}x\minus{}\sqrt{3}\equal{}x\minus{}\sqrt{3}\equal{}2$[/hide]


Find all prime numbers $ p$  such that $ 1\plus{}p\plus{}p^2\plus{}...\plus{}p^{p\minus{}1}$ and $ 1\minus{}p\plus{}p^2\minus{}p^3\plus{}....\plus{}p^{p\minus{}1}$ are prime numbers


geometry
incenter
circumcircle
Euler
geometric transformation
reflection

Dear Mathlinkers,
let ABC be a triangle,
1 the incircle , I the incenter, X the pointy of contact of 1 with BC, X' the antipode of X wrt 1, A"B"C" the I-circumtriangle, U the point of intersection of B"C" and AI, and Fe the Feuerbach' s point of ABC.
Prove that Fe, U and X' are collinear.
Sincerely
Jean-Louis

Dear Jean-Louis
What is "$ I$-circumtriangle"?

Dear Zhang Fangyu,
the I-curcumtriangle is the triangle forwicg the vertex are the second intersection od the cevians AI, BI, CI with the circumcircle of ABC.
Sincerely
Jean-Louis

A1,A2 are the projections of A onto BI,CI
△AA1A2∽△ABC  A1A2∥BC
From here we (http://www.mathlinks.ro/viewtopic.php?t=219598) we know X’Fe is the Euler Line of △AA1A2
So now our goal is to prove U is on Euler Line of △AA1A2
Note that AA’’⊥B’’C’’   =>  U,A,C’’,A2 are concyclic
∠UAA2=UA2A=∠CC’’B’’=∠AC’’B’’=1/2∠B
We have △AA2I is a right triangle  , thus U is midpoint of IA    =>  U is the circumcenter of AA2IA1  =>  U lies on the Euler Line of △AA1A2   =>  X’, U , Fe are collinear[/url]

[quote]A1,A2 are the projections of A onto BI,CI
△AA1A2∽△ABC A1A2∥BC
From here we (http://www.mathlinks.ro/viewtopic.php?t=219598) we know X’Fe is the Euler Line of △AA1A2
So now our goal is to prove U is on Euler Line of △AA1A2
Note that AA’’⊥B’’C’’ => U,A,C’’,A2 are concyclic
∠UAA2=UA2A=∠CC’’B’’=∠AC’’B’’=1/2∠B
We have △AA2I is a right triangle , thus U is midpoint of IA => U is the circumcenter of AA2IA1 => U lies on the Euler Line of △AA1A2 => X’, U , Fe are collinear.[/quote]

Indeed plane geometry, the ideas from previous messages keep coming back.
Your short and good solution is even still a bit too long, because $ U$ can be immediately proven to be the desired circumcenter:

Since $ I$ is the orthocenter of $ \Delta A''B''C''$, it follows that $ U$ is the midpoint of $ A$ and $ I$, and hence the circumcenter of $ \Delta AA_1A_2$. Since $ X'$ is on the Euler line of this triangle, the problem now follows from Hatzipolakis' theorem.

But anyway, your solution is good. I wonder if we can avoid all those overkill far-fetched theorems and find an independent olympiad solution. I bet we can use Monge-d'Alembert with some additional arguments or something. I don't have time right now to investigate further.

Could anyone show me why X’Fe is the Euler Line of △AA1A2   :)
I searched http://www.mathlinks.ro/viewtopic.php?t=219598  but I didn't find a specific proof
Thank you very much!

Dear Mathlinkers,
see for example :
Vonk J., The Feuerbach Point and Reflections of the Euler Line, Forum Geometricorum, Volume 9 (2009) 47–55 ;
http://forumgeom.fau.edu/FG2009volume9/FG200905.pdf
Sincerely
Jean-Louis

Dear jayme，your problem actually is that $ Fe$ is on the $ Line\minus{}OH$ what it the Euler-line of Δ$ DPQ$ .
But your description prompt another way to prove the Problem.

Dear Lym and Mathlinkers,
if I am not wrong, ithink that this way concerns another problem post on Mathlinks recently.
Your observation is nice
Sincerely
Jean-Louis

jayme,I read the proof in your excellent article :) But I find that it used a tough property of $ X_{80}$
It states that $ X_{80}$ is the reflection of incenter wrt feuerbach's.I spent a whole afternoon to solve it,and I create a solution using inversion and a bit long calculation(if necessary I will post here).Do you have or know a syntheic proof for it?I am eager to see it :)

Dear Mathlinkers,
it is not my proof but these of Jan Vonk.
Sincerely
Jean-Louis

Can anyone provide a synthetic proof :)
My solution is very long but I like it very much.

Dear Mathlinkers,
for more explanation, see for example
http://perso.orange.fr/jl.ayme   vol. 4  Symetriques de OI par rapport aux triangles de contact et médian p. 21
Sincerely
Jean-Louis


parameterization function calculus derivative conics ellipse limit

For which real $p$ has system of equations:
$(x-y)^2+z^2=(z-y)^2+x^2=p^2$
$xyz=p^3$
at least two solutions $(x,y,z)$?

Subtracting the first 2 equations immediately yields $y(z - x) = 0$. But if $y = 0$, then from the 2nd equation $p = 0$ and substituting this to the first 2 equations leads to a single solution $x = y = z = 0$. Hence, it must be $z = x$ and we actually have only 2 equations:

$2x^2 - 2xy + y^2 = p^2$
$yx^2 = p^3$

Obviously, this set of 2 equations always has the solution $x = y = p$ for any $p \neq 0$. The second equation represents the function $y = \frac {p^3}{x^2}$, which has derivative at $x = p$ equal to

$y' = -2\frac{p^3}{x^3}, \ y'(p) = -2$

The first equation represents an ellipse centered at the origin $[0, 0]$ (because the equation is symmetrical WRT the exchange $[x, y] \leftrightarrow [-x, -y]$) and passing through the points $[-p, -p], [0, -p], [0, p], [p, p]$. Solving the ellipse equation for $y$ and differentiating:

$y = x \pm \sqrt{p^2 - x^2}$
$y' = 1 \mp \frac {x}\sqrt{p^2 - x^2}$ $, \ \lim_{x\to \pm p}{y'} = \infty$

Hence, this ellipse has vertical tangents at the points $[-p, -p], [p, p]$. This is sufficient to conclude that the ellipse intersects the graph of the function $y = \frac {p^3}{x^2}$ at exactly one more point different from the point $[p, p]$ for any $p \neq 0$.

Edit:
$x = y = -p$ is not a solution, error fixed.
$p\ \#\ 0$ replaced by $p \neq 0$.
Thanks.

...

correct me if i'm wrong, but it seems to be that $x=y=z=-p$ is not a solution?

You are right, I will edit out the error. Tx.

yetti you can use \neq (not equal) command to make $p \neq 0$ if you want ;)


[quote="Altheman"](***)
59.  Let $ x,y$ be distinct positive reals. Determine the minimum value of
\[ \frac {1}{(x \minus{} y)^2} \plus{} \frac {16}{xy} \plus{} 6(x \plus{} y)
\]
and determine when that minimum occurs.

[hide="Comment:"] A proof exists with only Cauchy and AM-GM.[/hide][/quote]
Let $ x \ge y$ then : $ LHS \ge (\frac {1}{(x \minus{} y)^2}\plus{}(x\minus{}y)\plus{}(x\minus{}y))\plus{}(4x\plus{}8y\plus{}\frac {16}{xy} ) \ge (3)\plus{}(24)\equal{}27$;( by AM-GM )
equality is for $ x\equal{}2,y\equal{}1;$ :)

Hey, I like your solution, it is very nice!!

Here is my solution:

$ \frac{1}{(x\minus{}y)^2}\plus{}\frac{64}{4xy}\plus{}6(x\plus{}y)\ge \frac{(1\plus{}8)^2}{(x\minus{}y)^2 \plus{} 4xy}\plus{}6(x\plus{}y)$

$ \equal{}\frac{81}{(x\plus{}y)^2}\plus{}3(x\plus{}y)\plus{}3(x\plus{}y)\ge 3\sqrt[3]{81*3*3}\equal{}27$.

Equality occurs iff $ \frac{1}{(x\minus{}y)^2}\equal{}\frac{8}{4xy}$ and $ \frac{81}{(x\plus{}y)^2}\equal{}3(x\plus{}y)$. The first one factors as $ (x\minus{}2y)(2x\minus{}y)\equal{}0$. The second equation is $ x\plus{}y\equal{}3$. So $ (x,y)\equal{}(1,2),(2,1)$.


What is the maximum number of bishops that can be placed on any $8\times 8$ chessboard such that at most three bishops lie on any diagonal?

Here are my thoughts. Can they be made more rigorous?
[hide="Approach"]Consider all white squares. There are two 1 square diagonals, two 2 square diagonals, ... , and two 7 square diagonals. There is also one 8 square diagonal, for a total of 15 diagonals.

Each bishop is contained in exactly two diagonals, and each diagonal (except the 1 and 2 squares) can contain 3 bishops.

Hence, the bishops can theoretically occupy $3(15)-4-2=39$ spots on these 15 diagonals (subtract 4 and 2 since there are two diagonals of length 1 and two of length 2), where one bishop occupies "two spots" (one on each diagonal). However, since one bishop occupies two spots at a time, there can only be $38$ occupied, so $19$ bishops on white squares.

Similarly, $19$ bishops on black squares, for a total of $38.$
[/hide]
If this method works, I believe this generalization will folllow
[hide="Generalization?"]For each color, $3(2n-1)-4-2$ spots. Only $6n-10$ can be used, so $3n-5$ bishops on each color.

In total $6n-10$ bishops.[/hide]
Not sure how to show that the maximum value is attainable though.


Prove that two medians in a triangle are congruent iff the sides which they bisect are congruent.

[hide]I don't have a diagram, but...

Since two sides of a triangle are congruent, the triangle, say ABC,  is isosceles, and the medians are coming from the 2 base angles B and C. The medians intersect sides AB and AC at M and N, respectively.

Since $ AB \equal{} AC$, $ MB \equal{} NC$, and since  $ \angle B \equal{} \angle C$ and $ BC \equal{} BC$, $ \triangle BMC\cong\triangle CNB$.
Thus, $ CM \equal{} BN$

Now, for the "only if" part:

Consider triangle ABC, with medians CM and BN, with CM=BN. Denote the intersection of CM and BN as P. Because the intersection of the medians divide the medians into a ratio of 2:1, and since CM=BN, PM=PN and PB=PC. Since $ \angle MPB \equal{} \angle NPC$, $ \triangle MPB\cong\triangle NPC$.

Thus, $ \angle ABN \equal{} \angle ACM$. Since $ \angle A \equal{} \angle A$, and $ BN \equal{} MC$, $ \triangle AMC\cong\triangle ANB$, and thus AB=AC.

QED[/hide]

EDIT: Yeah, didn't notice it sad "iff", but added it now.

Let $ A$ be the origin, and let $ 2y$, $ 2z$ be vectors that denote points $ B$ and $ C$, respectively. Let $ N$ and $ P$ be the midpoints of $ AC$ and $ AB$, respectively. Then \[ BN\equal{}CP\iff BN^2\equal{}CP^2\iff (2y\minus{}z)\cdot (2y\minus{}z)\equal{}(2z\minus{}y)\cdot (2z\minus{}y)\]
\[ \iff 2y\cdot 2y\minus{}4y\cdot z\plus{}z\cdot z\equal{}2z\cdot 2z\minus{}4z\cdot y\plus{}y\cdot y\equal{}y\cdot y\equal{}z\cdot z\iff AB\equal{}AC\]

yay

Let $ AA'$ and $ BB'$ be two medians in the triangle $ ABC$
$ AA'\equal{}BB'$,$ AG\equal{}BG$, $ AB'A'B$ is a trapezoid therefore $ AB'\equal{}BA'$ $ \implies$ $ BC\equal{}AC$

[hide] Let ABC be an isosceles triangle with AB=AC

and BD and CE are the medians

clearly AD=AE and <A is common and AB=AC
so triangles ABD and ACE are congruent

so BD=CE QED [/hide]


In an isosceles trapezium the length of the smaller parallel sides is equal to it's height,and the other is equal to a diagonal.Find the ratio of the length of parallel sides.

[hide="solution"]
Let $ ABCD$ be the trapezium, $ AB\parallel CD,\ AB<CD,\ AB\equal{}a,CD\equal{}b$.
Let $ AE\perp CD,E\in CD$.
We have $ AE\equal{}a,AC\equal{}b,EC\equal{}\frac{a\plus{}b}{2}$
In the right triangle $ AEC$ we have $ AC^{2}\equal{}AE^{2}\plus{}EC^{2}$.
Then $  b^{2}\equal{}a^{2}\plus{}\left(\frac{a\plus{}b}{2}\right)^{2}\Rightarrow 5a^{2}\plus{}2ab\minus{}3b^{2}\equal{}0$
Divide by $ b^{2}$ and note $ \frac{a}{b}\equal{}t$.
$ 5t^{2}\plus{}2t\minus{}3\equal{}0\Rightarrow t_{1}\equal{}\minus{}1,t_{2}\equal{}\frac{3}{5}$.
So $ \frac{a}{b}\equal{}\frac{3}{5}$
[/hide]


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

Hello everyone,

I am having some trouble with the following word problems and would appreciate some help. Please note that the diagrams are my own; they are not given by the questions.

Thank you.

---

[b]1. Two concentric circles have centre [i]C[/i]. Two points [i]A[/i] and [i]B[/i] are located on the smaller circle, with radius 10, so that $ \angle ABC \equal{} 52^{\circ}$. Point [i]M[/i] is located on the larger circle so that $ \angle CAM \equal{} 150^{\circ}$ and $ \angle CBM \equal{} 135^{\circ}$. Find the radius of the larger circle.[/b]

My diagram: [img]http://img440.imageshack.us/img440/1125/p22514kf4.jpg[/img]

I made CR the radius of the larger circle, but I do not know how I can calculate more information to get the length of BR.

[b]2. The quadrilateral [i]ABCD[/i], whose sides [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i], is inscribed in a circle. Determine the cosine of angle [i]C[/i] in terms of the lengths of the sides of the quadrilateral.[/b]

My diagram: [img]http://img167.imageshack.us/img167/8496/p22515ki7.jpg[/img]

$ c^2 \equal{} a^2 \plus{} d^2 \minus{} 2ad\cos B$

$ \cos B \equal{} \cos (180 \minus{} c) \equal{} \ \minus{} cos C$

$ \ \minus{} cos C \equal{} \frac { \minus{} c^2 \plus{} a^2 \plus{} d^2}{2ad}$

$ \cos C \equal{} \frac {c^2 \minus{} a^2 \minus{} d^2}{2ad}$

However, this is not the answer given by the textbook. The provided answer is: [hide] $ \frac {b^2 \plus{} c^2 \minus{} a^2 \minus{} d^2}{2ad \plus{} 2bc}$.
[/hide]

[hide="hint for 1"]
CM is a radius of the big circle too, I think
And don't forget you have Law of Sines and Law of Cosines at your disposal.

[/hide]

Thank you CircleSquared for your help; I overlooked CM as the radius for the big circle.

[hide="hint #2"]how did you get the sides to the quadrilateral ABCD to have side the length of only a and d,  it says that the lenghts are a,b,c, and d[/hide]

Thanks for your reply jbano.

I attempted the problem again and I got a slightly different answer. Could someone please see where I may have erred?

---

Answer in textbook = [hide]$ \frac{b^2 \plus{} c^2 \minus{} a^2 \minus{} d^2}{2ad \plus{} 2bc}$[/hide]

[u]My Work:[/u]

$ AD^2 \equal{} c^2 \plus{} a^2 \minus{} 2ac\cos B \equal{} c^2 \plus{} a^2 \plus{} 2ac\cos C$

$ AD^2 \equal{} d^2 \plus{} b^2 \minus{} 2db\cos C$

Therefore:

$ c^2 \plus{} a^2 \plus{} 2ac\cos C \equal{} d^2 \plus{} b^2 \minus{} 2db\cos C$

$ c^2 \plus{} a^2 \plus{} 2ac\cos C \equal{} \cos C(\minus{}2ac \minus{} 2db)$

$ \cos C \equal{} \frac{c^2 \plus{} a^2 \minus{} d^2 \minus{} b^2}{\minus{}(2ac \plus{} 2db}$

$ \cos C \equal{} \frac{\minus{} c^2 \minus{} a^2 \plus{} d^2 \plus{} b^2}{2ac \plus{} 2db}$

[hide="hint 2"] try rotating the lables of the side, since you have all the right sings and conefeents[/hide]

Dear jbano,

Thanks for your reply! I got it!