geometry
circumcircle
quadratics
parallelogram
algebra
geometry proposed

Let $M$, $N$ and $P$ be three non-collinear points. Construct using straight edge and compass a triangle for which $M$ and $N$ are the midpoints of two of its sides, and $P$ is its orthocenter.

Let M, N be midpionts of AB, AC and (O) circumcircle of $\triangle ABC.$ Let the A-altitude $AP \perp MN$  meet MN at X, BC at D and (O) again at Q. Then DP = DQ. Denote h = DA, p = XP, m = XM, n = XN. Power of D to (O) is $4 mn = DB \cdot DC = DA \cdot DQ = h(h/2+p).$ This quadratic equation for h or x = h/2 is easy to solve: $x^{2}+px-2mn = 0,$ $x =-p/2 \pm \sqrt{p^{2}/4+2mn}.$ Let (K) be circle with diameter MN, Y one intersection of (K) with XP and XYZU a square. Then $XZ = \sqrt{2mn}.$ Circle (X) with this radius meets MN at Z'. Let X' be midpoint of XP. Then $X'Z' = \sqrt{p^{2}/4+2mn}.$ Circle (X') with this radius meets XP at D, D', the 2 possible A-altitude feet. DB = D'B' = 2XM, DC = D'C' = 2XN on parallels to MN through D, D'. BM meet CN at A on XP and B'M meet C'N at A' also on XP.

[quote="Jutaro"][color=brown]Let $ ABC$ a triangle, $ M$, $ N$ the midpoints of $ AB,AC$ and $ H$ the orthocenter.
Given the points $ M,N,H$ construct the triangle[/color][/quote]

If the triangle $ MHN$ is right then $ \triangle ABC$ is also right, which means that $ H$ coincides with some vertice
(if it is right at $ H$ then $ H\equiv A$, if it is right at $ M$ then $ H\equiv B$, if it is right at $ N$ then $ H\equiv C$)

So suppose that $ \triangle MHN$ is not right

Let $ w$ the circumcircle of $ \triangle ABC$. Let $ F$ be the anti-diametric point of $ B$.
It is $ FA \perp BA \Rightarrow FA \parallel HC$ and
$ FC \perp BC \Rightarrow  FC\parallel AH$.
[i](notice that noone of the above segments is zero, because $ \triangle ABC$ is not right)[/i]
So $ AHCF$ is a parallelogram. This means that $ N$ is the midpoint of $ FH$
So the points $ H,F$ are symmetric wrt $ N$

[color=blue][size=134]Construction[/size][/color]
Let $ \varepsilon$ be the line peprendicular to $ MN$ passing through $ H$. Clearly $ A\in \varepsilon$
Let $ F$ be the symmetric point of $ H$ wrt $ N$. The circle with diameter $ MF$ intersects the line $ \varepsilon$ at two points $ A_{1},A_{2}$

For $ A\in\{A_{1},A_{2}\}$ let $ B,C$ be the symmetric points of $ A$ wrt $ M,N$ respectively

[color=blue][size=134]Proof[/size][/color]
From the construction it is $ AF\perp AB$. But $ AHCF$ is a parallelogram (because $ N$ is the midpoint of $ AC$ and $ HF$)
So $ CH\perp AB$. On the other hand it is $ AH\perp MN \Rightarrow  AH\perp BC$, hence $ H$ is the orthocenter of $ \triangle ABC$


In triangle $ ABC$, an interior point $ P$ is such that $ \angle PAB\equal{}\angle PBC\equal{}\angle PCA\equal{}\theta$.

Prove that $ \frac{1}{\sin^2\theta}\equal{}\frac{1}{\sin^2A}\plus{}\frac{1}{\sin^2B}\plus{}\frac{1}{\sin^2C}$

Let $ D$, $ E$, $ F$ be  the intersection points of $ BC$, $ CA$, $ AB$ and $ AP$, $ BP$, $ CP$ respectively.
The triangles $ ABD$ and $ BPD$ are similar.
$ \frac{\sin\theta}{\sin B}\equal{}\frac{BD}{AD}\equal{}\frac{PD}{BD}$

$ \implies \frac{\sin^2\theta}{\sin^2 B}\equal{}\frac{BD}{AD}\frac{PD}{BD}\equal{}\frac{PD}{AD}$

On the same way:
$ \frac{\sin^2\theta}{\sin^2 A}\equal{}\frac{PF}{CF}$

$ \frac{\sin^2\theta}{\sin^2 C}\equal{}\frac{PE}{BE}$

But:
$ \frac{PD}{AD}\plus{}\frac{PE}{BE}\plus{}\frac{PF}{CF}$

$ \equal{}\frac{[BCP]}{[ABC]}\plus{}\frac{[ACP]}{[ABC]}\plus{}\frac{[ABP]}{[ABC]}\equal{}1$

$ \implies \frac{\sin^2\theta}{\sin^2 C}\plus{}\frac{\sin^2\theta}{\sin^2 A}\plus{}\frac{\sin^2\theta}{\sin^2 B}\equal{}1$

$ \implies \frac{1}{\sin^2\theta}\equal{}\frac{1}{\sin^2A}\plus{}\frac{1}{\sin^2B}\plus{}\frac{1}{\sin^2C}$

In the "JOURNAL DE MATHÉMATIQUES ÉLÉMENTAIRES", 1888, p.222; Boutin  gave the property of Brocard point: $ ctg\theta \equal{}ctg A\plus{}ctg B \plus{}ctg C.$

[quote="jbmorgan"] [color=darkred]In $ \triangle ABC$, an interior point $ P$ is such that $ m\left(\widehat {PAB}\right)\equal{}m\left(\widehat {PBC}\right)\equal{}m\left(\widehat {PCA}\right) \equal{} \theta$.

Prove that $ \frac {1}{\sin^2\theta} \equal{} \frac {1}{\sin^2A} \plus{} \frac {1}{\sin^2B} \plus{} \frac {1}{\sin^2C}$ and $ \cot\theta \equal{}\cot A\plus{}\cot B\plus{}\cot C$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Apply the [b]Sinus' theorem[/b] in the triangles :

$ \left\|\ \begin{array}{cccc}
\triangle\ APC : & \frac {PA}{\sin\theta}\equal{}\frac {b}{\sin A} & \implies & PA\equal{}\frac {b\sin\theta}{\sin A}\\\\
\triangle\ BPA : & \frac {PB}{\sin\theta}\equal{}\frac {c}{\sin B} & \implies & PB\equal{}\frac {c\sin\theta}{\sin B}\\\\
\triangle\ CPB : & \frac {PC}{\sin\theta}\equal{}\frac {a}{\sin C} & \implies & PC\equal{}\frac {a\sin\theta}{\sin C}\end{array}\ \right\|$ . Thus, $ [APC]\plus{}[BPA]\plus{}[CPB]\equal{}2S$ $ \implies$

$ b\sin\theta\cdot PC\plus{}c\sin\theta\cdot PA\plus{}a\sin\theta\cdot PB\equal{}2S$ $ \implies$ $ \sum \left(b\sin\theta\cdot\frac {a\sin \theta}{\sin C}\right)\equal{}2S$ $ \implies$

$ \sum \left(\frac {2S}{\sin^2C}\cdot\sin^2\theta\right)\equal{}2S$ $ \implies$ $ \boxed {\ \frac {1}{\sin^2\theta} \equal{} \frac {1}{\sin^2A} \plus{} \frac {1}{\sin^2B} \plus{} \frac {1}{\sin^2C}\ }$ . I used $ ab\equal{}\frac {2S}{\sin C}$ .

The relations $ \frac {1}{\sin^2x}\equal{}1\plus{}\cot^2x$ and $ \sum\left(\cot B\cot C\right)\equal{}1$ are well-known. Therefore,

$ \left(\sum\cot A\right)^2\equal{}\sum\cot^2A\plus{}2\cdot\sum\left(\cot B\cot C\right)\equal{}$ $ \sum\left(\frac {1}{\sin^2A}\minus{}1\right)\plus{}2\equal{}\frac {1}{\sin^2\theta}\minus{}1\equal{}\cot^2\theta$ .

In conclusion, $ \left|\sum\cot A\right|\equal{}\left|\cot\theta\right|$ . [b]If $ \triangle ABC$ is acute, then $ \boxed {\ \cot\theta \equal{}\cot A\plus{}\cot B\plus{}\cot C\ }$[/b] .[/color]


FIe $a,b,c$ numere pozitive cu suma $1$. Demonstrati ca:
   $\frac{\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{b}}{1-\sqrt{b}}+\frac{\sqrt{c}}{1-\sqrt{c}}\geq \frac{3+3\sqrt{3}}2$

cika $f(x)=\frac{1}{\sqrt{x}(1-\sqrt{x})}$ e convexa,nu mai calculez acu $f^{''}$ ca ma ia cu ameteala.

si aplicam jensen ponderat:

$a\cdot f(a)+b\cdot f(b)+c\cdot f(c)\geq f(a^{2}+b^{2}+c^{2})$

$\Longleftrightarrow \sum{\frac{\sqrt{a}}{1-\sqrt{a}}}\geq  \frac{1}{(\sqrt{a^{2}+b^{2}+c^{2}})(1-(\sqrt{a^{2}+b^{2}+c^{2}}))}$

deci ne ramane:
$\frac{1}{(\sqrt{a^{2}+b^{2}+c^{2}})(1-(\sqrt{a^{2}+b^{2}+c^{2}}))}\geq \frac{3+3\sqrt{3}}{2}$

si d`aici notezi $\sqrt{a^{2}+b^{2}+c^{2}}=x$ si ai o inegalitate cu ecuatia de gradu 2, dar chiar nu mai fac nici un calcul :|


Just checking...

If someone were to rate the questions they got as hard and the ones they missed as easy, wouldn't that increase their score? By doing this on hundreds of questions, it would make a difference. The person would be definitely unmoral, but with over 5,800 users, it's bound to happen...

Edit: I guess I was unclear. I was talking about the Alcumus score, not the points. r15s11z55y89w21 adequately answered my question.

It's very simple.

They're only hurting themselves.

How the user rates problems is not considered in determining points for a problem.

yeah. the point value is determined only from their Alcumus Score. Read more topics in this forum and all of your answers will be answered.


search
combinatorics proposed
combinatorics

Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one or more distinct names written on it, although different cards may share some name. For each name, we define a [i]shuffle[/i] by moving every card that has the name written on it to the opposite stack. Prove that it is always possible to end up with more cards in the right stack by picking several distinct names, and doing in turn the shuffle corresponding to each name.


Posted before : http://www.mathlinks.ro/viewtopic.php?p=848070#848070


AH is the altitude in triangle ABC (AB<AC). HD and HE are the altitudes in triangles AHB, AHC, respectively. DE intersects BC at F. BE intersects AH at I. M is the midpoint of BC. Prove that IM is perpendicular to AF.

Could you help me, please? I'm stuck.   :?

Maybe missing somthing but my computer says they are not perpendicular

My computer says it's only true if AB=AC.  :P  :D

[quote="JavaMan"]What software are you guys using?[/quote]

I'm using Geometer's Sketchpad. And my previous post is incorrect. This construction can't exist if AB=AC, because point F wouldn't exist. But when AB and AC are very close, it is very close to being perpendicular. Also when angle ABC is close to being a right angle, it is close to being perpendicular. Although, the right angle would mean triangle ABH is degenerate, or something (B and H are the same point).  :?

That problem is my creation. Sorry that it isn't true for all cases. However, it makes me to think of another problem: when is IM perpendicular to AF? There must be other suppositions. Try to find it is an effective way to learn math, because find the problem is more valuable than trying to solve it.  :P

PS: It seems that learning math on the computer is a very good way.  :!:


The [b]increasing[/b] sequence $ 1,3,4,9,10,12,13,...$ consists of all those positive integers that are powers of 3 or sums of distinct powers of 3. Find the 100th term of the sequence.

[hide="My answer"]
$ 3^6\plus{}3^4\plus{}3^3\plus{}3^1$[/hide]

[quote="mathgirl_fractals"]I believe the answer is 3^6+3^5+3^2. Is that correct?[/quote]

You are absolutely right.
I did a really stupid mistake. The answer was $ T_{64}\plus{}T_{36}\plus{}T_4$
But i calculated $ T_{64}\plus{}T_{26}\plus{}T_4$  :(
I should really go back to nursery school and learn to add. :(

You should consider a bijection using base 2 and base 3.

[quote="JBL"]No, that's the 90th term of the sequence.  This has appeared many times on the forum before, but I'm not sure what the best thing to search for is.[/quote]
I'm fairly certain this is an old AIME problem.

It also appeared in several German math competitions last year.

[hide="Solution"]If $ f(i)$ denotes the $ i$-th integer of that form, then $ n$ assigns to each nnatural number with the representation $ n \equal{} \sum_{i\in\mathbb{N}} a_i2^i\quad\left(a_i\in\{0,1\}\right)$ the number $ \sum_{i\in\mathbb{N}} a_i3^i$. Very easy to show by bijection/induction/anything.[/hide]


trigonometry
inequalities
inequalities unsolved

[quote="payman_pm"]$x,y$ are positive and we have
$x^2+y^3 \ge x^3+y^4$ prove that
$x^3+y^3\le 2$[/quote]

$\frac{2x^3+1}{3}+y^3\ge x^2+y^3\ge x^3+y^4\ge x^3+\frac{4y^3-1}{3}$

What do you want to talk,Werwulff??
Here my proof :
Sine [tex]x^2+y^3\geq\ x^3+y^4[/tex]
It's not hard to check following inequalities:
[tex]x^3+y^3\leq\ x^2+y^2\leq\ x+y\leq\ 2[/tex]

you can look here  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1012]www.mathlinks.ro/Forum/viewtopic.php?t=1012[/url]

[quote="keira_khtn"]What do you want to talk,Werwulff??[/quote]
AM-GM  implies
$\frac{2x^3+1}{3}\ge x^2$ and $\frac{3y^4+1}{4}\ge y^3\Rightarrow y^4\ge \frac{4y^3-1}{3}$
Thus,
$\frac{2x^3+1}{3}+y^3\ge x^2+y^3\ge x^3+y^4\ge x^3+\frac{4y^3-1}{3}$
$\Rightarrow \frac{2}{3}\ge\frac{x^3}{3}+\frac{y^3}{3}$


Joe has an odd number of dollars, and Mary has the next greater odd number of dollars. They each paid the same price for a T-shirt. To buy the T-shirt, Joe spends $ \$1$ less than $ \frac{2}{5}$ of his money, and Mary spends $ \frac{1}{3}$ of her money. What was the total amount of money Joe and Mary originally had altogether?

Hello
Let Joe had $ x$ dollars and Mary had $ x\plus{}2$ dollars.
$ \frac{2}{5}x\minus{}1\equal{}\frac{1}{3}(x\plus{}2)$
$ x\equal{}25$
Thus the total money=$ 25\plus{}27$=$ 52$
Thank u.


geometry
circumcircle
incenter
trapezoid
cyclic quadrilateral
geometry proposed

Point $ M$ is taken on side $ BC$ of $ \triangle ABC$ such that the centroid $ T_c$ of triangle $ ABM$ lies on the circumcircle of $ \triangle ACM$, and the centroid $ T_b$ of $ \triangle ACM$ lies on the circumcircle of $ \triangle ABM$. Prove that the medians of the triangles $ ABM,ACM$ from $ M$ are of the same length.

Which centroid.circumcenter , incenter or orthocenter ??????

The centroid is the intersection of the three medians, which are the three lines joining each vertex to the midpoint of the opposite side.  :)

[size=134][color=darkblue][b]Very nice problem ! I'll present at first a metrical proof for it.[/b][/color][/size]

[quote="easternlatincup"][color=darkred]Point $ M$ is taken on the side $ [BC]$ of $ \triangle ABC$ such that the centroid $ G_1$ of $ \triangle ABM$ belongs to the circumcircle of $ \triangle ACM$ and the centroid $ G_2$ of $ \triangle ACM$ belongs to the circumcircle of $ \triangle ABM$. Prove that the $ M$ - medians of the triangles $ ABM$ , $ ACM$ are equally.[/color][/quote]

[b][u]Proof I.[/u][/b] Denote $ \left\|\begin{array}{c} M_1\in (BM)\ ,\ M_1B \equal{} M_1M \equal{} x \\
\ M_2\in (CM)\ ,\ M_2C \equal{} M_2M \equal{} y \end{array}\right\|$ . Observe that $ \boxed {\ x \plus{} y \equal{} \frac a2\ }\ (1)$ ,

$ \left\|\begin{array}{c} G_1\in AM_1\ ,\ AM_1 \equal{} 3\cdot G_1M_1 \\
 \\
G_2\in AM_2\ ,\ AM_2 \equal{} 3\cdot G_2M_2\end{array}\right\|$ . Apply the [b]Stewart's relation[/b] to the rays $ (AM_1$ , $ (AM_2$ in $ \triangle ABC$ :

$ \left\|\begin{array}{c} a\cdot AM_1^2 \plus{} ax(x \plus{} 2y) \equal{} c^2(x \plus{} 2y) \plus{} b^2x \\
 \\
a\cdot AM_2^2 \plus{} ay(2x \plus{} y) \equal{} c^2y \plus{} b^2(2x \plus{} y)\end{array}\right\|$ $ \implies$ $ \boxed {\ 2\cdot\left(AM_1^2 \minus{} AM_2^2\right) \equal{} (c^2 \minus{} b^2) \plus{} a(x \minus{} y)\ }\ (2)$ .

$ \blacktriangleright\ AG_1MC$ , $ AG_2MB$ are cyclically $ \Longleftrightarrow$ $ \left\|\begin{array}{c} M_1G_1\cdot M_1A \equal{} M_1M\cdot M_1C\ \Longleftrightarrow\ AM_1^2 \equal{} 3x(x \plus{} 2y) \\
 \\
M_2G_2\cdot M_2A \equal{} M_2M\cdot M_2B\ \Longleftrightarrow\ AM_2^2 \equal{} 3y(2x \plus{} y)\end{array}\right\|$ $ \Longleftrightarrow$

$ 2\cdot\left(AM_1^2 \minus{} AM_2^2\right) \equal{} 3a(x \minus{} y)$ $ \stackrel {(2)}{\ \Longleftrightarrow\ }$ $ 4a(x \minus{} y) \equal{} c^2 \minus{} b^2$ $ \stackrel {(1)}{\ \Longleftrightarrow\ }$ $ x \equal{} \frac {2a^2 \plus{} c^2 \minus{} b^2}{8a}$ $ \Longleftrightarrow$ $ \boxed {\ MB \equal{} \frac {2a^2 \plus{} c^2 \minus{} b^2}{4a}\ }$.

$ \blacktriangleright$ Denote the midpoints $ U$ , $ V$ , $ W$ , $ L$ of the segments $ [AB]$ , $ [AC]$ , $ [BC]$ and $ [UV]$ respectively and the projections

$ D$ , $ T$ of the points $ A$ , $ L$ to $ BC$ respectively. Since $ L\in AW$ , i.e. $ TD\equal{}TW$ , [b]the $ M$ - medians of $ \triangle AMB\ ,\ \triangle AMC$

are equally[/b] $ \Longleftrightarrow$ $ M\equiv T$ $ \Longleftrightarrow$ $ BD \plus{} BW \equal{} 2\cdot BM$ $ \Longleftrightarrow$ $ 2\cdot BM \equal{} \frac {a^2 \plus{} c^2 \minus{} b^2}{2a} \plus{} \frac a2$ $ \Longleftrightarrow$ $ \boxed {\ MB \equal{} \frac {2a^2 \plus{} c^2 \minus{} b^2}{4a}\ }$ .

[u]In conclusion[/u],  $ AG_1MC$ , $ AG_2MB$ are cyclically $ \Longleftrightarrow$ $ MU \equal{} MV$ $ \Longleftrightarrow$ $ M\equiv T$ $ \Longleftrightarrow$ $ \frac {MB}{MC} \equal{} \frac {2a^2 \plus{} c^2 \minus{} b^2}{2a^2 \plus{} b^2 \minus{} c^2}$ .

[hide="Solution"]
Let $ P$ and $ Q$ be the midpoints of $ AB$ and $ AC$ respectively and let $ R$ where $ PQ$ meets $ AM$.  It follows that $ PQ$ is a midline of $ \triangle ABC$, so $ PR\parallel BM$ and $ AP \equal{} PB$, which means that $ AR \equal{} RM$, so $ R$ is the midpoint of $ AM$.  Now, notice that $ \angle AQP \equal{} \angle ACB \equal{} \angle AT_cP$ because of the cyclic quadrilateral $ AT_cMC$.  Hence, $ AQT_cP$ is cyclic.  Similarly, $ APT_bQ$ is cyclic, so $ APT_cT_bQ$ is cyclic, which means that $ PT_cT_bQ$ is cyclic.  Now, notice that $ \frac {MT_c}{T_cP} \equal{} 2 \equal{} \frac {MT_b}{T_bQ}$, so $ T_cT_b\parallel PQ$.  Hence, $ T_cT_bQP$ is a cyclic trapezoid, which implies that $ T_cP \equal{} T_bQ\implies \frac {MP}{3} \equal{} \frac {MQ}{3}\implies MP \equal{} MQ$, as desired. [/hide]

I have just one question about Virgil`s solution. Probably I am doing some elementary mistake, but I am constantly obtaining $ BD\equal{}\frac{b^{2}\minus{}a^{2}\minus{}c^{2}}{2a}$, instead of $ \frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{2a}$. The problem probably appears because my triangle is obtuse in $ B$. However, if this is the case, then it should be proven, that triangle is acute in $ B$, or not?


Hi everyone

I'm sorry I couldn't find a more appropriate place to post this, guess it'll have to go in this topic ...

I read in http://mathworld.wolfram.com/CauchySequence.html that Cauchy sequences in the rationals do not necessarily converge, but they do in case of reals.

Can someone please give me an example of a Cauchy sequence in the rationals that does not converge?

Thanks a lot, and once again I apologize for posting this in a potentially inappropriate forum ...

Rik

1 1-1/3 1-1/3+1/5 1-1/3+1/5-1/7.... converges to pi/4
Even easier, 3 3.1 3.14 ... where it's the first n decimal digits of pi.

Ya, it doesn't converge in the rationals, that's the whole point of a cauchy sequence.
Because pi/4 and pi are not rationals, that's what they mean
If it diverges then the limit doesn't exist, so the cauchy sequence is meaningless.

Try 1,2,3,... if you want a sequence that does not converge

Ok, thanks a lot for your response. I should have framed my question a bit differently, I believe. Let me rephrase by asking - can we find a Cauchy sequence of rational numbers (i.e. all the elements of the sequence are rationals) that does not converge? Bear in mind that the sequence HAS to be Cauchy, so that (1, 2, 3 ...) is ruled out, and it must not be convergent, so that the sequence (a_n = the first n digits of pi) is also ruled out, as it eventually converges to pi.

Thanks a lot everyone.

Rik

You're not getting the point here. A cauchy sequence will always converge to a real number. However, when viewing the rationals as a metric space on their own, a sequence of rational numbers converging to pi when viewed in the real numbers, does NOT converge in the metric space of the rationals.

Now I get the point.  :blush:

Sorry not to have understood that earlier.  :oops:

Thanks a lot,

Rik


Let's look at this sequence...
1/[1*2] + 1/[2*3] + 1/[3*4] + ....

Get the pattern?

Yah its...
1/[n(n+1)] = n/[n+1]

So we have...
1/[1*2] + 1/[2*3] + 1/[3*4] + .... + 1/[n(n+1)] = n/[n+1]

I'm supposed to show that the statement holds for all positive integers n.

This will require the use of mathematical induction.

I haven't really learned this yet so be patient. I know I have to show that when n=1 the statement is true.

So we have...

n=1 means that we just do the first term (left side), which is 1/(1*2) = 1/2 or 0.5. The right hand side is 1/(1+1) = 1/2 = 0.5 SAME YAY so far so good!

But then it wants me to show that they n=k and then it wants me to show that n=k+1.

Can someone walk me through these steps?

You have already shown the base case, so I will just show the inductive step.

The inductive step:

Suppose that some $ n=k$ satisfies the statement.  We wish to show that the statement holds for $ n=k+1$.

By our supposition, we have $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} = \frac{k}{k+1}$.  We wish to show that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{(k+1)((k+1)+1)} = \frac{k+1}{k+2}$.

We have:

\begin{align*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} &= \frac{k}{k+1}
\\ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} & = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}
\\ & = \frac{k(k+2)+1}{(k+1)(k+2)}
\\ & = \frac{k^2+2k+1}{(k+1)(k+2)}
\\ & = \frac{(k+1)^2}{(k+1)(k+2)}
\\ &=\frac{k+1}{k+2},
\end{align*}

as desired.

Thus, by induction, we have shown that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1}$ for all $ n \in \mathbb{N}$.

Or , we have :
 $ \frac {1}{k} \minus{} \frac {1}{k \plus{} 1} \equal{} \frac {1}{k(k \plus{} 1)}$ using this later we'll have :
$ \frac {1}{1.2} \plus{} \frac {1}{2.3} \plus{} . . . \plus{} \frac {1}{k(k \plus{} 1)} \equal{} \frac {1}{1} \minus{} \frac {1}{2} \plus{} \frac {1}{2} \minus{} \frac {1}{3} \plus{} . . . \plus{} \frac {1}{k} \minus{} \frac {1}{k \plus{} 1}$
$ \equal{} 1 \minus{} \frac {1}{k \plus{} 1} \equal{} \frac {k}{k \plus{} 1}$

Redwane


AMC
AIME
geometry
inradius
area of a triangle
Heron\u0027s formula
AoPS Books

15.  The triangle ABC has AB = 1995, BC = 1993, CA = 1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX.

How in the world do you do this problem?

[hide]Here's how I did it: to find out $CX$, we have to find out area of triangle $ABC$ by Heron's Formula: $A=\sqrt{s(s-a)(s-b)(s-c)}$. We see $A=1,721,673.225$, thus $\frac{1995 \cdot CX}{2}=1,721,673.225$ and $CX$ is approximately $1726$, now to find out $AX$ and $XB$ we use Pythagorus: $1994^2=AX^2+1726^2 \Longrightarrow AX=998.5$ and $1993^2=BX^2+1726^2 \Longrightarrow BX=996.5$. Now to find the radii of each of the two triangles, we use $rs=A \Longrightarrow r \cdot 2,359.25=861,705.5 \Longrightarrow r$ is approximately $365.25$ and $r \cdot 2,357.75=859,979.5 \Longrightarrow r$ is approximately $364.75$ and the distance betwen their heights is the same as the distance where they touch $CX$, so the answer is $\boxed{\frac{1}{2}}$. This isn't $\text{exactly}$ right, so does anyone have solutions?[/hide]

[quote="JesusFreak197"]I'm not entirely sure how to do it in a reasonable amount of time without a calculator, but I would hazard a guess that your answer isn't right because you keep rounding.[/quote]

exactly, I $\text{had}$ to use a calculator since the numbers were huge, and I had to round since there were too many decimals.

i got $\frac{332}{665}$, one hint that i can give you is to make the sides n,n+1,n+2, and subtracting the inradii gives you the answer

It might be more useful to try $n-1$, $n$, and $n+1$, but I haven't finished the problem.

[hide="hint"]also something in the aops books is that if you have a right triangle with sides a,b, c being the hypotenuse, then the inradius is (a+b-c)/2, and also when you subtract the radii, CX cancels out so you do not need to find it[/hide]

Wow, this problem is very easy for a aime 15.  Besides the seemingly inevitable squaring of these huge numbers...

there is no problems with large numbers if you do it the way i have been suggesting

I tried it your way, and it does simplify it a lot, but I still end up having to square one of those numbers when you plug it back in for n.

Would you mind posting your solution, Al?
I simplified the distance between the two points of tangency as 998 - x, where x is XB, 1995-x is XA. Any hints?

[size=150]
SOLUTION:[/size]

http://www.mathlinks.ro/Forum/files/1

[hide]

Ok...it looks long, but it does not take much to write on paper.

Let the $n=1993$, so we have that $BC=n$, $AB=n+2$, and $AC=n+1$
note, anything in the aops books, i will not bother to prove...

if we want to find the distance between the 2 points of tangency, all we need to do is subtract the inradii of the circles because the inradii are perpendicular to AB and CX is perpendicular to AB, thus CX and the inradii are parallel, and we can move over the lengths onto CX and get our answer

Since AXC and BXC are right triangles, the inradius for them respectively are $\frac{AX+CX-AC}{2}$ and $\frac{BX+CX-BC}{2}$ so subtracting these, we get the distance = $\frac{AX-BX+BC-AC}{2}$
let $f=AX-BX$, so we find f and we are almost done,

We know that $AX+BX=n+2$, so we get that $AX=\frac{n+2+f}{2}$ and $BX=\frac{n+2-f}{2}$

Using the pythagorean theorem on ACX and BCX, (i am omitting that CX=CX step)

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

expanding and collecting terms gives us

$f=\frac{2n+1}{n+2}$

using distance = $\frac{AX-BX+BC-AC}{2}$, we have

$d=\frac{ \frac{2n+1}{n+2}+n-(n+1) }{2}$
$d=\frac{n-1}{2(n+2)}$ knowing that $n=1993$, it follows that the answer is

$\boxed{\frac{332}{665}}$

[/hide]

Thank you Altheman, I think your solution was very good :)

[quote=Iversonfan2005]15.  The triangle ABC has AB = 1995, BC = 1993, CA = 1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX.
[/quote]

I was wondering: is there any way to solve this question for valid values of AB, BC, and CA? So, for instance, solving it for AB = x, BC = y, and CA = z? Obviously, there would be some restrictions on those values, due to Triangle Inequality and the like. In other words, can a problem of this type be generalized?

Thanks!

SNPIXELMATHLETE88


a small army of 120 soldiers are holding a fort.  they have enough food for only 6 days.  at the start of every day, 4 soldiers are killed.  how many days will they be able to survive off the food?

changed to 6 days

Am I understanding the problem wrong, or is this what it means?

Every day, 4 soldiers die, which leaves more food left for the next day. Thus, in day $n$, $\frac{13}{14}-\frac{1}{14}\left( \frac{116}{120}\right)^{n-1}$ is left. But this leaves a negative number for $n$. Is there a flaw in my solution?

[quote="jli"][quote="asdff"]So at the start of the first day-4 people are killed or is it they all start with 120 people on day 1, so day 2 has 116 people.[/quote]

yup...they die before they eat :D[/quote]

Yea I know I get x^2-61x+3000=0 giving me an imaginary number meaning they dont finish the food..just find the number of days they survive.  which is [hide]31 just using arithmetic sequence..[/hide]

[quote="asdff"][hide]31 just using arithmetic sequence..[/hide][/quote]
Won't they all be dead by then?

oops sorry guys...change it to 6 days :blush:

[hide]They have $120\cdot6=720$ "units" of food.

On the first day, 116 units are used up.
On the second, 112, and so on.

There is still enough food for only 6 days because the units of food that were supposed to be eaten by the soldiers that are dead don't add up to an extra day of food. On the 7th day, 8 soldiers (of the remaining 92) die of hunger. :([/hide]


geometry
angle bisector
exterior angle
geometry solved

Let [i]O[/i] be the center of circle[i] W[/i]. Two equal chords [i]AB[/i] and [i]CD [/i]of[i] W [/i]intersect at [i]L [/i]such that [i]AL>LB [/i]and [i]DL>LC[/i]. Let [i]M [/i]and[i] N [/i]be points on [i]AL[/i] and [i]DL[/i] respectively such that ([i]ALC[/i])=2*([i]MON[/i]). Prove that the chord of [i]W[/i] passing through [i]M [/i]and [i]N[/i] is equal to [i]AB[/i] and [i]CD[/i].

[quote="Armo"]$(XYZ)=$(area of $XYZ$)??  or what?[/quote]
ooops...I'm sorry! (ALC),(MON)-angles

[quote="Armo"]Heh... That's why it was so difficult (I thought them areas).[/quote]
OK! :P ...And now what do you think about it :D

[color=blue][b]Problem.[/b] Let W be a circle, and AB and CD two equal chords of this circle. Let these two chords AB and CD intersect at a point L such that AL > LB and DL > LC. Finally, let M and N be two points on the segments AL and DL, respectively, such that < ALC = 2 < MON, where O is the center of the circle W. Prove that the chord of the circle W passing through the points M and N is equal to the chords AB and CD.[/color]

[i]Solution.[/i] Let's start with a lemma:

[b]Lemma 1.[/b] If ABC is a triangle, and P is a point in its plane such that:

- this point P lies on the angle bisector of the angle CAB;
- we have $\measuredangle BPC=\frac{180^{\circ}-\measuredangle CAB}{2}$;
- the points A and P lie in different halfplanes with respect to the line BC,

then the point P must be the A-excenter of the triangle ABC (i. e., the excenter opposite to the vertex A).

[hide="Proof of Lemma 1"]

[i]Proof of Lemma 1.[/i] Let X be the A-excenter of the triangle ABC. We want to show that P = X.

Well, at first, it's clear that the point X lies on the angle bisector of the angle CAB (since the point X is the A-excenter of the triangle ABC). It is also clear that the points A and X lie in different halfplanes with respect to the line BC. Moreover, denoting by A = < CAB, B = < ABC and C = < BCA the angles of triangle ABC, we have $\measuredangle XBC=90^{\circ}-\frac{B}{2}$ and $\measuredangle XCB=90^{\circ}-\frac{C}{2}$ (since the point X is the A-excenter of the triangle ABC and thus lies on the exterior angle bisectors of the angles ABC and BCA), and thus, by the sum of angles in triangle BXC, we obtain

$\measuredangle BXC=180^{\circ}-\measuredangle XBC-\measuredangle XCB=180^{\circ}-\left(90^{\circ}-\frac{B}{2}\right)-\left(90^{\circ}-\frac{C}{2}\right)$
$=\frac{B}{2}+\frac{C}{2}=\frac{B+C}{2}=\frac{180^{\circ}-A}{2}$,

where we have used the fact that B + C = 180° - A (after the sum of angles in triangle ABC); in other words, $\measuredangle BXC=\frac{180^{\circ}-\measuredangle CAB}{2}$.

Now, the locus of all points T such that $\measuredangle BTC=\frac{180^{\circ}-\measuredangle CAB}{2}$ and such that the points A and T lie in different halfplanes with respect to the line BC must be a circular arc with endpoints B and C; it is clear that the angle bisector of the angle CAB can intersect this circular arc at only one point. Hence, there exists only one point T lying on the angle bisector of the angle CAB such that $\measuredangle BTC=\frac{180^{\circ}-\measuredangle CAB}{2}$ and such that the points A and T lie in different halfplanes with respect to the line BC. But we know that both points P and X are such points (both of them lie on the angle bisector of the angle CAB, both of them satisfy $\measuredangle BPC=\frac{180^{\circ}-\measuredangle CAB}{2}$ and $\measuredangle BXC=\frac{180^{\circ}-\measuredangle CAB}{2}$, and both of them are such that the points A and P or, respectively, X lie in different halfplanes with respect to the line BC). Hence, these two points P and X must coincide. Thus, P = X, and Lemma 1 is proven.[/hide]

Now let's solve the problem itself:

Let the line MN intersect the circle W at two points Q and R. Then, the chord QR of the circle W is the chord passing through the points M and N. And we have to prove that QR = AB = CD.

Since < ALC = 2 < MON, but < ALC = 180° - < NLM, we have 2 < MON = 180° - < NLM, so that $\measuredangle MON=\frac{180^{\circ}-\measuredangle NLM}{2}$. Moreover, it is clear that the points L and O lie in different halfplanes with respect to the line MN.

Two chords of a circle are equal if and only if the center of the circle has equal distances to these two chords. Hence, as AB and CD are two equal chords in the circle W, the center O of the circle W has equal distances from these two chords AB and CD; in other words, the point O has equal distances from the lines ML and NL, and consequently, this point O lies on the angle bisector of the angle NLM.

So let's combine our results:

- The point O lies on the angle bisector of the angle NLM.
- We have $\measuredangle MON=\frac{180^{\circ}-\measuredangle NLM}{2}$.
- The points L and O lie in different halfplanes with respect to the line MN.

Hence, by Lemma 1, applied to the triangle LMN, it follows that the point O is the L-excenter of the triangle LMN. Thus, the point O has equal distances from the sidelines MN, NL and LM of this triangle. In other words, the point O has equal distances from the chords QR, CD and AB of the circle W. But remembering again that two chords of a circle are equal if and only if the center of the circle has equal distances to these two chords, we see that, since the point O is the center of the circle W, the equality of the distances of the point O from the chords QR, CD and AB yields the equality of these chords. In other words, we have QR = AB = CD. This solves the problem.

  Darij

We start from the converse of the statement: Prove that if the chord passing through M,N equals to AB and CD, then $\angle ALC=2\angle MON$.
Let X,Y,Z be the feet of the perpendicular from O to lines AB,CD,MN respectively. Then OX=OY=OZ, which implies $\angle XOM=\angle ZOM,\angle YON=\angle ZON$. By simple angle chasing, we can conclude that $\angle ALC=2\angle MON$.
Now we go back to the original problem. We first choose the position of M. Then clearly the position of N is fixed, which solves the problem.

Let $O$ be the center of circle $ W$. Two equal chords $AB$ and $CD $ of $ W $ intersect at $L $ such that $AL>LB $ and $DL>LC$. Let $M $ and $ N $ be points on $AL$ and $DL$ respectively such that $\angle ALC=2 \angle MON$. Prove that the chord of $W$ passing through $M $and $N$ is equal to $AB$ and $CD$.


An evil wizard has imprisoned 64 math geeks. He decides to play with them, violently. Each geek will have a white or black hat placed on his/her head. He will not know what his hat color is. They stand in a line, facing the person in front of them. Each math geek can only see the colors of the hat in front of him. Each geek has to say white or black. If he says the color of his hat, he gets a cookie. If he doesn't he gets killed.

Before this, the math geeks made a ordeal to help them survive. What's the highest number of mathgeeks that can be garenteed to survive?

Err, what do the hats have to do with anything?  I think you left part of the problem out.

I don't think that any can be gauranteed, given the problem you proposed. If all but one of the hats were black, how would they be able to predict? Maybe if there must be 32 black hats/32 white hats.. hmm

Although I think I've heard a riddle like this before, so there's probably some trick, but iunno.

whoops sorry double post... read the next one. bleh. sorry about this:( :oops:

[quote="BornForMath"]Two questions:
Does each geek have to tell the truth?
What should the first person say (he can't see anyone)?[/quote]

The first person asked is the person in the back, if I remember correctly. I know the answer is $63$ will survive, but I've totally  forgotton the trick :blush:

EDIT: I believe it has something to do with whether there is an even or odd amount of black hats or whatnot. Like, if there are an odd number of black hats, the first dude would say black. If there there are an even number of black hats, (s)he would say white. Let's say there's an odd number and the first person says black. Then, the second person would see how many black hats there were. If there were an odd number, then his or her hat must be white. But! If he or she sees an even number, then his or her hat must be the one that made it odd... It's something like that.

Yeah, but there were no restrictions on what types or how many or if there was an order of white/black hats, so we really can't assume it.

It'll be interesting if there's a solution to this problem, but I doubt it.

Hint

[hide]It has something to do with parity and something the first person says.
The answer [i]is[/i] that 63 can be garenteed to survive. What's the trick though?
Remember, parity and what the first person says. You guys are really close.[/hide]

[quote="break314dancer"]There doesn't need to be an order or a certain amount. Let's say the first person, the one in the back, says that there is an odd number of black hats, but the second person only sees an even number. Therefore, his or her hat must  be black. Likewise, if he or she sees an odd number of black hats also, then his or her hat must be white, so as not to disrupt the order. If one person screws up, then the rest of the people also screw up.[/quote]

Ah ok, yeah I've heard this before, but never remember it.

[quote="A+MATH"][hide]Have the first geek as black if there is an even number of black hats and white if there is an odd number of black hats. The rest can deduce his hat color from what the first person said and the first has a half-half probabilty.[/hide][/quote]

Good work!  :D

But the problem only works if this condition is satisfied (translation: this is what was on the sheet that they gave us):

The wizard has to start from the back and go to the front. If he happens to be mean and start from the front, no one is guaranteed to survive.

i think this was a 2005 BAMM problem

part of the problem is left out, the geeks start saying their colors from the back.


Show that $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{(a)(c+a)}\geq \frac{27}{(a+b+c)^{2}}$

EDIT- nvm, the problem is still up.

Oops... foolish mistake. The problem is re-unsolved.

For $x,y,z > 0$ it holds I think. \[2(a+b+c)\left(\sum_{cyc}{\frac{1}{b(a+b)}}\right) \geq \left(\sum_{cyc}\frac{1}{\sqrt{a}}\right)^{2}\] So now all we have to prove is \[\left(\sum_{cyc}\frac{1}{\sqrt{a}}\right)^{2}\geq \frac{27}{a+b+c}\] Which is equivalent to power mean: \[\frac{x+y+z}{3}\geq \sqrt{\frac{3}{x^{2}+y^{2}+z^{2}}}\] Done  :)

[b]OR[/b]

Jensen:

Because $\frac{1}{\sqrt{x}}$ is convex on $\mathbb{R}^{+}_{0}$, we can apply Jensen.

First Cauchy, then Jensen: \[\left(\sum_{cyc}\frac{1}{b(a+b)}\right)(a+b+c) \geq \left(\sum_{cyc}{\frac{1}{\sqrt{a+b}}\right)^{2}\geq \left(\frac{3}{\sqrt{\frac{2(a+b+c)}{3}}}\right)^{2}=\frac{27}{2(a+b+c)}}\]


What online chess server do you use?
If you play ICC tell me your handle.

FICS... what's so good about ICC that you would pay money to subscribe to it?

I use Yahoo, when I rarely play Online Chess.


Two of your friends live 4 miles apart and on the same "east-west" street, and you live halfway between them. You are all talking on a three-way phone call when you hear an explosion. Six seconds later your friend to the east hears the explosion, and your friend to the north hears it 8 seconds after you do. Find the equations of two hyperbolas that would locate the explosion. Sound travels at a rate of 1100 ft/sec.

I don't understand what "east-west street" means.

A street that runs from east to west, i.e. the $x$-axis.  :?:

so how is a friend to the north and a friend to the east possible?

Then how is the problem possible? The branches of each hyperbola won't intersect.

Straight forward:(All units in feet)
4 miles =  21120 ft,
6 sound-seconds = 6600 ft
8 sound seconds = 8800 ft,

first curve (Hyperbola) is given by: $\sqrt{(x+21120)^{2}+y^{2}}-\sqrt{x^{2}+y^{2}}= 6600$
Second curve: $\sqrt{(x-21120)^{2}+y^{2}}-\sqrt{x^{2}+y^{2}}=8800$

(We are assumng, one friend is 4 mile west, and the other is  4 mile east)

(If the  friends are  one in north, one on east, then one of the equation would look some thing like:
$\sqrt{x^{2}+(y-21120)^{2}}-\sqrt{x^{2}+y^{2}}=8800$

You may like to simplify a little, if you like.
(Hint: Both equation can be  written in a little simpler form)

BTW:  Important fact: You still can't locate where the explosion ..(No there are more than 2(or 4)  possible points, in fact There are INFINITE points where this can take place, See if any of you can guess why?

[quote="iamagenius"]i'm confused...how did you get that and why isn't it in standard form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$? can't you just find the focus, then $a$, then use $c^{2}=a^{2}+b^{2}$ to find $b$, and then plug it into the formula there?[/quote]
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is a hyperbola but is very specific  (that is the axes are x-y and center is the origin etc) and NOT a general or standard form ... (For example another nice form  for a hyperbola is  $xy=1$)

How did I get the equation I got?  it: Just the difference between the distances is constant (equal to distance travelled by sound - in this case )  - this is actually how a hyperbola is defined.

Any way what I posted above,  trust me,  be patient,   square and simplify, (You may have to do it twice) I promise that everything will get simplified, and you will get something of the form -
$\frac{(x-c)^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$?

In general hyperbola looks like:
$ax^{2}+bxy+cy^{2}+dx+ey+f=0$ as long as  $b^{2}>4ac$


topology
advanced fields
advanced fields theorems

Does anyone have this book?
my e-mail is sisioyus@hotmail.com.
thank you!!!

Kelley's book? I suppose china-pub has it?


Find an O(n) time algorithm that decides if there is a path connecting two given vertices of a graph. (n is the number of vertices in the graph.)

The book indicates that the reachability problem is different from the connectedness problem--deciding if the graph is connected, but I can't figure out how to solve one problem without the other.

Well, BFS would be O(m) where m is the number of edges. I wonder if O(n) can even be done...


Show that $n(n+1)(n+2)$ is not a perfect square for positive integer values of $n$.

[hide]
Suppose it is a square. Since $gcd(n,n+1) = 1$ and $gcd(n+1,n+2) = 1$, we must have $n+1 = a^{2}$. Then $n(n+2) = (a^{2}-1)(a^{2}+1) = a^{4}-1$ which is $1$ less than a square. However, two squares differ by one iff the smaller is $0$ so $a^{4}-1 = 0 \Rightarrow a = 1 \Rightarrow n = 0$, which is not a positive integer. Contradiction, so the result is proven. [/hide]


Let $ \underline{A} \ \underline{B} \ \underline{C}$ denote a three-digit base 10 number whose digits are $ A,B,$ and $ C$ with the condition $ A \geq 1.$  Fnd the smallest prime number that can be attained from $ \underline{A} \ \underline{B} \ \underline{C} \minus{} \left(A^2 \plus{} B^2 \plus{} C^2\right)$

Level: AIME #1-#2

There's no taker in this problem and one after so I'm going to give hint (I want to have these proposal problems solved so anyone in future can take a look at it):

[hide="Hint"]

The problem is actually very simple. Just solve the problem as if you're just looking for smallest number. Then, start going up until you hit a prime number. Bingo! [/hide]

Consider the number A0C. If we make the middle digit greater than 0, it can only add to the final answer. So we want C as large as possible, A and B as small as possible. 108 gives 43. So if there is a smaller number, it must end in 9.  129 also gives 43. Thus, 43 must be the smallest number.


hi !  view the solution in  doc format!!!!!!!!!!!!!!!!!


geometry
trigonometry
circumcircle
angle bisector

$\triangle ABC$ is an isosceles right triangle with $\angle A=90^\circ$. The angle bisector of $\angle B$ meets $AC$ at $D$. Let $E$ be the point on $\overleftrightarrow{BD}$ such that $BE\perp CE$. If $CE=5$, find the area of $\triangle BCD$.

draw a perpendicular line through D, intersect BC at F. since ABC is isosceles so ∠ACB=45 °, so DCF is isosceles. let CF=x, so DF=x, CD=√ 2x. Since BD is the angle bisector of ∠ABC and ∠A=90  °,so AD=DF=x,    AC=x+√ 2x, BC=2x+√ 2x, BF= x+√ 2x.∵ABD≌BDF, AB=BF and ABD∽DEC, so AD/DE = AB/EC, which is X/DE=
(x+√ 2x)/5, DE= 5 √ 2-5. DE²+EC²=DC²,  DC²=100-50√ 2=2x², so x²=50-25√ 2. Area of BCD= BC*DF/2=(2x+√ 2x)*x/2=(2x²+√ 2x²)/2, sub x²=50-25√ 2 into it, then we can get area is 25

That's correct. :)

Does anyone have other solution(s)?

$\frac{S_{ABC}}{S_{BCD}}=\frac{AC}{DC}=\frac{AD+DC}{DC}=\frac{AD}{DC}+1=\frac{AB}{BC}+1=1+\cos 45$ so we have $S_{BCD}=\frac{S_{ABC}}{1+\cos 45}=\frac{AC^2}{2(1+ \cos 45)}$
as $\angle BEC=90$ E is on the circumcircle $\angle AEC=180-45=135$ and $CE=AE=5$
by Cosine theoreme $AC^2=AE^2+BE^2-2AE*BE*\cos 135$ or
$AC^2=25(2(1+\cos 45))$ so $S_{BCD}=25$

in the same way one can prove that if $\angle B=\beta$ and $CE=d$ then    $S_{BCD}=d^2 \cot \beta$

[img]http://i5.photobucket.com/albums/y194/jelly119/ae5513ea.jpg[/img]

\begin{eqnarray*}[\triangle BCD]&=&\frac52\times{BD} \\ &=&\frac52\times{AB}\sec22.5\\ &=&\frac52\times\frac1{\sqrt2}BC\times\sec22.5\\ &=&\frac52\times\frac1{\sqrt2}\times5\csc22.5\times\sec22.5\\ &=&\frac52\times\frac1{\sqrt2}\times5\times\frac1{\sin22.5\cos22.5}\\ &=&\frac{25}{2\sqrt2}\times\frac2{\sin45}\\ &=&25 \end{eqnarray*}


Have any of you ever tried to create your own olympiad-level problem? Any tips on how to make one?

i have a couple of times, but unfortuanately they weren't anywhere near olympiad. :P

I made some (a couple got into the QEDMO and are or will be posted on the forum).

Tips:
- take some not so trivial problem and take a special case
- take some identity and use it in some way
- just take a random formula/sequence/drawing and try to find something

To me, the most difficult part of writing a hard question is making sure that (a) there is a solution, (b) the solution isn't terribly long and ugly, and (c) you don't need some special, obscure bit of knowledge in order to find a nice solution.
I think a good way to "practice" is to create very hard problems and see what "problem-creation techniques" tend to result in the existence of nice solutions and which ones create just ugliness.

Examples of my recent attempts with inequalities:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102894]I think it's ok[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103268]Not good because trivial[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103251]Not good because ugly[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103249]This one ok but not as good as the first[/url]

If you really want I can explain how I made those.


function
induction
algebra proposed
algebra

Define the sequence $ (a_n)$ by $ a_0 \equal{} 0$ and $ a_n \equal{} a_{n\minus{}1} \plus{} \frac{3^{r\plus{}1}\minus{}1}{2}$ if $ n \equal{}3^r(3k \plus{} 1)$ and $ a_n \equal{} a_{n\minus{}1} \plus{} \frac{3^{r\plus{}1}\plus{}1}{2}$ if $ n \equal{} 3^r(3k \plus{} 2) (r; k\geq0)$. Prove that every integer appears exactly once in the sequence.

Obviosly $ a_n$ is strongly monotonic, therefore your statement  is not true, because $ a_n\minus{}a_{n\minus{}1}>1$ for some n.

[quote="tdl"]Define the sequence $ (a_n)$ by $ a_0 \equal{} 0$ and $ a_n \equal{} a_{n \minus{} 1} \plus{} \frac {3^{r \plus{} 1} \minus{} 1}{2}$ if $ n \equal{} 3^r(3k \plus{} 1)$ and $ a_n \equal{} a_{n \minus{} 1} \plus{} \frac {3^{r \plus{} 1} \plus{} 1}{2}$ if $ n \equal{} 3^r(3k \plus{} 2) (r; k\geq0)$. Prove that every integer appears exactly once in the sequence.[/quote]
the second relation must be:

$ a_n\equal{}a_{n\minus{}1}\minus{}\frac{3^{r\plus{}1}\plus{}1}2$

[quote]if we want say that:
$ a_{n} \equal{} a_{n \minus{} 1} \plus{} ( \minus{} 1)^{i \plus{} 1}\frac {3^{r \plus{} 1} \plus{} ( \minus{} 1)^i}{2}\ (if\ n \equal{} 3^{r}(3k \plus{} i),i\in\{1,2\})$[/quote]

we have $ 2a_n \equal{} \minus{} 1 \plus{} 2a_{n \minus{} 1} \plus{} ( \minus{} 1)^{i \plus{} 1}3^{r \plus{} 1}\ (if\ n \equal{} 3^{r}(3k \plus{} i),i\in\{1,2\})$
then $ 2a_n \equal{} \minus{} n \plus{} \sum_{r \equal{} 0}^{\plus{}\infty}3^{r \plus{} 1}(E\left(\frac {E(\frac {n}{3^r}) \plus{} 2}{3}\right) \minus{} E\left(\frac {E(\frac {n}{3^r}) \plus{} 1}{3}\right))$
so we can see that if $ n\equal{}\overline{b_hb_{h\minus{}1}..b_2b_1b_0}^{(base\ 3)}$ then $ 2a_n\equal{}\sum_{r \equal{} 0}^{h}3^{r}(3g(b_r)\minus{}b_r)$ where $ g(0)\equal{}g(2)\equal{}0$ and $ g(1)\equal{}1$

finally if $ n\equal{}\overline{b_hb_{h\minus{}1}..b_2b_1b_0}^{(base\ 3)}$ then $ a_n\equal{}\sum_{r \equal{} 0}^{h}3^{r}f(b_r)$ where $ f(0)\equal{}0,f(1)\equal{}1,f(2)\equal{}\minus{}1$
the rest is easy :wink:

let $ f: \{0,1,2\}\to\{ - 1,0, + 1\}$ be a function such that $ f(2) = - 1$,$ f(1) = 1$ and $ f(0) = 0$.

let  $ n = \sum_{k = 0}^p x_k3^k$ be the representation of $ n$ in base $ 3$,we claim that:

$ a_n = \sum_{k = 0}^p f(x_k)3^k$

we can prove this statement using induction on $ n$ as follow:

the statement clearly holds for $ n = 0$.now assume that the statement holds for $ n - 1$,we want to prove it for $ n$.

instead of using $ n = \sum_{k = 0}^p x_k3^k$ we use $ n = (x_p,x_{p - 1},\ldots ,x_0)$.

now we have these two cases:

$ \boxed{\textrm{ Case 1. }}$

$ n$ is a natural number and $ k\geq 0$ and $ r\geq 0$ and we have $ n = 3^r(3k + 1)$.in this case we obviously have:

$ n = \left(x_p,\ldots, x_r,1,\underbrace{0,0,\ldots,0}_{\textrm{r times}}\right)$

hence we have:

$ n - 1 = \left(x_p,\ldots ,x_{r + 1},0,2,2,\ldots,2\right)$

now according to the hypothesis of induction we have:

$ a_{n - 1} = \left(f(x_p),\ldots ,f(x_{r + 1}),0, - 1, - 1,\ldots, - 1\right)$

but:

$ \frac {3^{r + 1} - 1}2 = 3^r + \ldots + 3 + 1 = \left(0,0,\ldots,0,1,1,\ldots,1\right)$

thus:

$ a_n = a_{n - 1} + \frac {3^{r + 1} - 1}2 = \left(f(x_p),\ldots ,f(x_{r + 1}),f(1),f(0),f(0),\ldots ,f(0)\right)$

as claimed.

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

$ \boxed{\textrm{ Case 2. }}$

$ n$ is a natural number and $ k\geq 0$ and $ r\geq 0$ and $ n = 3^r(3k + 2)$.in this case we have:

$ n = \left(x_p,\ldots ,x_{r + 1},2,0,0,\ldots ,0\right)$

hence:

$ n - 1 = \left(x_p,\ldots ,x_{r + 1},1,2,2,\ldots ,2\right)$

now according to the hypothesis of induction we have:

$ a_{n - 1} = \left(f(x_p),\ldots ,f(x_{r + 1}),1, - 1, - 1,\ldots , - 1\right)$

but we have:

$ - \frac {3^{r + 1} + 1}2 = - 2\times 3^r + \left(3^{r - 1} + 3^{r - 2} + \ldots + 3 + 1\right) = \left(0,0,\ldots ,0, - 2,1,1,\ldots ,1\right)$

hence:

\begin{eqnarray*}a_n = a_{n - 1} - \frac {3^{r + 1} + 1}2 = \left(f(x_p),\ldots ,f(x_{r + 1}),1, - 1, - 1,\ldots, - 1\right) + \left(0,0,\ldots ,0, - 2,1,1,\ldots ,1\right)\end{eqnarray*}

hence:

$ a_n= \left(f(x_p),\ldots ,f(x_{r + 1}),f(1),f(0),f(0),\ldots ,f(0)\right)$

QED


[quote="secrets"][quote="hophinhan"][quote="secrets"]Give $ a,b,c \ge 0$Prove
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge \frac {4max(a,b,c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]

Very nice ineq but my solution not very interesting  :maybe: .... The idea from a technique of Ba Can  :)

[b]Solution [/b]: (hope I made no mistake)

WLOG : $ b \plus{} c \equal{} 1\ ;\ a \equal{} max\{a,b,c\}\ ;\ bc \equal{} x \le \frac {1}{4}$ . Set : $ a \rightarrow 3a$

Then:

$ inq \Leftrightarrow 1 \plus{} \frac {1}{(3a \plus{} b)^2} \plus{} \frac {1}{(3a \plus{} c)^2} \ge \frac {12a}{(3a \plus{} b)(3a \plus{} c)}$

Because : $ \frac {1}{(3a \plus{} b)^2} \plus{} \frac {1}{(3a \plus{} c)^2} \ge \frac {1}{2}(\frac {1}{(3a \plus{} b)} \plus{} \frac {1}{(3a \plus{} c)})^2 \equal{} \frac {(6a \plus{} b \plus{} c)^2}{2(3a \plus{} b)^2(3a \plus{} c)^2}$

Then we need to prove that :

$ f(x) \equal{} 1 \plus{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} x)^2} \minus{} \frac {12a}{9a^2 \plus{} 3a \plus{} x} \ge 0$

+Case 1: $ a\ge 1$ : $ f(x) \equal{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} x)^2} \plus{} \frac {9a(a \minus{} 1) \plus{} x}{9a^2 \plus{} 3a \plus{} x} \ge 0$

+Case 2: $ a\le 1$: $ f'(x) \equal{} \frac { \minus{} (6a \plus{} 1)^2}{(9a^2 \plus{} 3a \plus{} x)^3} \plus{} \frac {3a}{(9a^2 \plus{} 3a \plus{} x)^2} \equal{} \frac {27a^2(a \minus{} 1) \plus{} 3a(x \minus{} 4) \minus{} 1}{(9a^2 \plus{} 3a \plus{} x)^3} < 0$

$ \Rightarrow f(x) \ge f(\frac {1}{4}) \equal{} 1 \plus{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} \frac {1}{4})^2} \minus{} \frac {12a}{9a^2 \plus{} 3a \plus{} \frac {1}{4}} \equal{} \frac {9(a \minus{} \frac {1}{2})^2}{(3a \plus{} \frac {1}{2})^2} \ge 0$

Equality point :$ \Leftrightarrow a \equal{} b \equal{} c \equal{} \frac {1}{2}$[/quote]
This solution is not true ;),try it[/quote]

Could you plz tell me my mistake ... Note that I have changed $ a$ to $ 3a$ so the Equality point is  :  $ a\equal{}3b\equal{}3c$  :roll:

When c=0,equality occur ;),when c=t>0,equality also occur :D

1. You are taking nonsense

2. If you're true..It not enough to conclude that my solution not true :)

[hide]Hi`,cach giai? cua? anh em cung~ chua check,tai hoi phuc tapj,anh xem lai di a.,dau bang` cua no' hoi khac' chut xiu' :D[/hide]
P/S nonsense la sao a.?

[quote="secrets"]Give $ a,b,c \ge 0$Prove
$ \frac {1}{(a + b)^2} + \frac {1}{(b + c)^2} + \frac {1}{(c + a)^2} \ge \frac {4max(a,b,c)}{(a + b)(b + c)(c + a)}$[/quote]
Hello, secrets. Thank you for your present!
Here is my solution for it:
Assume that ${ a = \max\{a,b,c},$ and set $ b + c = x,c + a = y,a + b = z,$ then we have to prove that
$ \frac {yz}{x} + \frac {zx}{y} + \frac {xy}{z} \ge 2(y + z - x),$
or
$ \fac{yz}{x} + \frac {x(y + z)^2}{yz} \ge 2(y + z),$
which is obviously true by the AM-GM Inequality.
Under the hypothesis ${ a = \max\{a,b,c},$ we have equality when $ x(y + z) = yz,$ or $ (a - b)(a - c) = (b + c)^2.$ :)

@Secrets: Le ra phai la "dedicated" chu khong phai "delicated" em a, truoc anh cung nham chu nay, h anh biet nen da sua lai rui. :)

[hide]Em go lon chut xiu', =)),Tieng anh em te lam' a.,nonsense nghia la` gi` a.?[/hide]

[quote="can_hang2007"][quote="secrets"]Give $ a,b,c \ge 0$Prove
$ \frac {1}{(a + b)^2} + \frac {1}{(b + c)^2} + \frac {1}{(c + a)^2} \ge \frac {4max(a,b,c)}{(a + b)(b + c)(c + a)}$[/quote]
Hello, secrets. Thank you for your present!
Here is my solution for it:
Assume that ${ a = \max\{a,b,c},$ and set $ b + c = x,c + a = y,a + b = z,$ then we have to prove that
$ \frac {yz}{x} + \frac {zx}{y} + \frac {xy}{z} \ge 2(y + z - x),$
or
$ \fac{yz}{x} + \frac {x(y + z)^2}{yz} \ge 2(y + z),$
which is obviously true by the AM-GM Inequality.
Under the hypothesis ${ a = \max\{a,b,c},$ we have equality when $ x(y + z) = yz,$ or $ (a - b)(a - c) = (b + c)^2.$ :)

 [/quote]
Nice solution.My solution is same idea.assume a=max{a,b,c}
$ \frac{1}{(a+b)^2}+ \frac{1}{(b+c)^2}+ \frac{1}{(c+a)^2}= \frac{2(a+c)}{(a+b)(b+c)(c+a)}+ \frac{(a-c)^2}{(a+b)^2(b+c)^2}+ \frac{1}{(a+c)^2} \ge \frac{4a}{(a+b)(b+c)(c+a)}$

[quote="hophinhan"][quote="secrets"]Give $ a,b,c \ge 0$Prove
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge \frac {4max(a,b,c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]

Very nice ineq but my solution not very interesting  :maybe: .... The idea from a technique of Ba Can  :)

[b]Solution [/b]: (hope I made no mistake)

WLOG : $ b \plus{} c \equal{} 1\ ;\ a \equal{} max\{a,b,c\}\ ;\ bc \equal{} x \le \frac {1}{4}$ . Change : $ a \rightarrow 3a$

Then:

$ inq \Leftrightarrow 1 \plus{} \frac {1}{(3a \plus{} b)^2} \plus{} \frac {1}{(3a \plus{} c)^2} \ge \frac {12a}{(3a \plus{} b)(3a \plus{} c)}$

Because : $ \frac {1}{(3a \plus{} b)^2} \plus{} \frac {1}{(3a \plus{} c)^2} \ge \frac {1}{2}(\frac {1}{(3a \plus{} b)} \plus{} \frac {1}{(3a \plus{} c)})^2 \equal{} \frac {(6a \plus{} b \plus{} c)^2}{2(3a \plus{} b)^2(3a \plus{} c)^2}$

Then we need to prove that :

$ f(x) \equal{} 1 \plus{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} x)^2} \minus{} \frac {12a}{9a^2 \plus{} 3a \plus{} x} \ge 0$

+Case 1: $ a\ge 1$ : $ f(x) \equal{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} x)^2} \plus{} \frac {9a(a \minus{} 1) \plus{} x}{9a^2 \plus{} 3a \plus{} x} \ge 0$

+Case 2: $ a\le 1$: $ f'(x) \equal{} \frac { \minus{} (6a \plus{} 1)^2}{(9a^2 \plus{} 3a \plus{} x)^3} \plus{} \frac {3a}{(9a^2 \plus{} 3a \plus{} x)^2} \equal{} \frac {27a^2(a \minus{} 1) \plus{} 3a(x \minus{} 4) \minus{} 1}{(9a^2 \plus{} 3a \plus{} x)^3} < 0$

$ \Rightarrow f(x) \ge f(\frac {1}{4}) \equal{} 1 \plus{} \frac {(6a \plus{} 1)^2}{2(9a^2 \plus{} 3a \plus{} \frac {1}{4})^2} \minus{} \frac {12a}{9a^2 \plus{} 3a \plus{} \frac {1}{4}} \equal{} \frac {9(a \minus{} \frac {1}{2})^2}{(3a \plus{} \frac {1}{2})^2} \ge 0$

Equality point :$ \Leftrightarrow a \equal{} b \equal{} c \equal{} \frac {1}{2}$[/quote]
This solution is totally wrong.


geometry
parallelogram
geometry unsolved

Given a convex quadrilateral $ABCD$ such that for any transversal line $l$ intersecting the lines $AB,AC,AD$ at $L,M,N$ respectively the following equation holds
$\frac{AB}{AL}+\frac{AD}{AN}= \frac{AC}{AM}$
Where $AB,AC,AD,AL,AN,AM$ are directed line segments which may have positive or negative values.
Prove that $ABCD$ is a parallelogram.[IS CONVEXITY NECESSARY?]


geometry
incenter
circumcircle
geometry solved

Let $ABC$ be a triangle, $P$ a point, $A'B'C'$ the pedal triangle
of $P$, and $A''B''C''$ the pedal triangle of $P$ wrt triangle $A'B'C'$.

The Parallel through $P$ to $B''C''$ intersects $AB, AC$ at $A_b,A_c$, resp.

The Parallel through $P$ to $C''A''$ intersects $BC, BA$ at $B_c,B_a$, resp.

The Parallel through $P$ to $A''B''$ intersects $CA, CB$ at $C_a,C_b$, resp.

Which is the locus of $P$ such that

$AA_b + BB_c + CC_a = AA_c + BB_a + CC_b$ ?

[quote="Ali.Kh"]I found a good result:
$AP$ is perpendicular to $A_bA_c$
$BP$ is perpendicular to $B_aB_c$
$CP$ is perpendicular to $C_bC_a$
because of this lemma:
if $XYZ$ is pedal triangle of $P$ with respect to $A''B''C''$, $XYZ$ is homothetic to $ABC$.[/quote]

This is wrong. Triangle XYZ is just similar to triangle ABC, not homothetic.

Anyway, don't expect anything nice from the locus. It is easy to see that the locus contains the incenter $X_1$ and the circumcenter $X_3$ of triangle ABC, but dynamic sketches show that it doesn't contain the centroid $X_2$, the orthocenter $X_4$, the nine-point center $X_5$, the symmedian point $X_6$, the de Longchamps point $X_{20}$ and the Bevan point $X_{40}$. These were all points with halfways interesting pedal triangles I could think about (but probably I have forgotten some, as I haven't really done triangle geometry for a long time now).

[b]EDIT:[/b] According to Antreas P. Hatzipolakis in [url=http://groups.yahoo.com/group/Hyacinthos/message/11575]Hyacinthos message #11575[/url], the locus is the union of an isogonal cubic with a curve of degree 10.

  Darij


vector trigonometry ratio inequalities limit

Two vector $\vec{e_1}$, $\vec{e_2}$ satisfy: $|\vec{e_1}|=2$, $|\vec{e_2}|=1$, the angle between the two vectors is $60^{o}$. The angle between $2t \vec{e_1} + 7\vec{e_2}$ and $\vec{e_1}+t\vec{e_2}$ belongs to the interval of $(90^{o},180^{o})$.
Find the range of $t$.

I go to school in Sweden and im 16... and i dont get a wooden knickle about vectors.
How old are you when you go to highschool?

When do you learn vector mathematics?? I know vector physics a bit but Im only $14$ and mathematical vectors are not taught till Im $16$, I dont think vector mathematics should be here.

Of course vector math should be here.  If you haven't learned it, read aops or go to mathworld (although that is not the best site to learn new things from).  Frankly, my school doesn't teach vector math at all besides adding and subtracting vectors.  This is a good problem for intermediate level.

[hide="hint"]
Remember that if $a$ and $b$ are 2d vectors,
$a\cdot b = |a| |b| \cos \theta$ where theta is the angle between them
[/hide]

we learn vectors here in singapore at the age of 15 or 16..

just sub in the values of the 2 vectors respectively to get

cos(theta) = ([b]a[/b].[b]b[/b])/(|[b]a[/b]|.|[b]b[/b]|) ------ (1)

      -1  <      put eqn (1) and sub values into it.             < 1

solve, get ratio of t and that's about it.

hmmm.. am i right..? lazy to solve tho >.<

Chrnox, your (1) is valid, although I do not understand your method of solving.

To solve, it helped me to rename the vectors and write them out using their x and y counterparts as such
$\vec{e_1} = \vec x = \left<a, b\right>$
$\vec{e_2} = \vec y = \left<c, d\right>$
$2t \vec{e_1} + 7\vec{e_2} = \vec p$
$\vec{e_1}+t\vec{e_2} = \vec q$

Using the dot product theorem (or whatever it is called), we see $\vec x \cdot \vec y = 1$, or
$ac + bd = 1$
Also, using the magnitudes we are given, we get
$a^2 + b^2 = 4$
$c^2 + d^2 = 1$

Using the dot product theorem with $\vec p$ and $\vec q$, we get $\vec p \cdot \vec q = |\vec p||\vec q| \cos \theta$, where we know $90 < \theta < 180$
Since $\cos$ of all possible values of $\theta$ is negative, and the magnitude of a vector is always positive, we can set up the inequality
$\vec p \cdot \vec q < 0$
The LHS can be written as
$(2t \vec{e_1} + 7\vec{e_2}) \cdot (\vec{e_1}+t\vec{e_2})$
$(2ta + 7c)(a + tc) + (2tb + 7d)(b + td)$
$t^2(2ac + 2bd) + t(2a^2 + 2b^2 + 7c^2 + 7d^2) + 7ac + 7bd$
Using the equalities listed above, this is nicely simplified
$2t^2 + 15t + 7 < 0$
Since $\frac{-b}{2a} = \frac{-15}{4}$ is negative, and the graph of the inequality increases, there are values of $t$ that satisfy the inequality.  All values of $t$ between the zeroes satisfy it.
$(2t + 1)(t + 7) = 0$
$-7 < t < \frac{-1}{2}$

I hope it is right.  I really liked this problem.

Your answer is right,deej21. :)

Here is my solution.

As you mentioned, the codition satisfying the given sentence is

$(2t\overrightarrow{e_1}+7\overrightarrow{e_2})\cdot(\overrightarrow{e_1}+t\overrightarrow{e_2})<0$

$\Longleftrightarrow 2t|\overrightarrow{e_1}|^2+(2t^2+7)\overrightarrow{e_1}\cdot\overrightarrow{e_2}+7t|\overrightarrow{e_2}|^2<0$

$\Longleftrightarrow 2t\cdot 2^2+(2t^2+7)\cdot 1+7t\cdot 1^2<0$

$\Longleftrightarrow 2t^2+15t+7=(t+7)(2t+1)<0$,yielding [b]deej21[/b]'s answer.

kunny

[quote="shobber"]That's also my answer. But the final answer the paper says: $(-7, -\frac{\sqrt{14}}{2}) \cup (-\frac{\sqrt{14}}{2}, -\frac{1}{2})$.
How can it be?[/quote]

$-\frac{\sqrt{14}}{2}$ must be excluded because, when $t$ equals it, $\vec p$ and $\vec q$ become parallel ($\vec p$ is $-\sqrt{14}$ times $\vec q$), which means that $\vec p \cdot \vec q = 0$.

Where you would come up with this value, however, is still a mystery to me.

Mathfanatic, $\vec p \cdot \vec q = 0$ means $\vec p$ and $\vec q$ are perpendicular, not parallel.  They are perpendicular at $t=-7, \frac {-1}2$, as established by the bounds.  They will be parallel at $\theta = 180$
$\vec p \cdot \vec q = -|\vec p| |\vec q|$
Using the formula established in a recent thread, $|\vec i + \vec j| = |\vec i|^2 + |\vec j|^2 + 2( \vec i \cdot \vec j)$, we get
$|\vec p| = 16t^2 + 28t + 49$
$|\vec q| = t^2 + 2t + 4$
$2t^2 + 15t + 7 + (16t^2 + 28t + 49)(t^2 + 2t + 4) = 0$

Which is probably wrong since I did it pretty fast.  Since I didn't feel like solving that by hand, I graphed it and got no solutions, which makes sense because I don't think it can ever EQUAL 180 degrees.  Perhaps $\lim_{\theta \rightarrow 180}$ of the equation yields $t = \frac {\sqrt{14}} 2$  Or something.  I don't know.

Also kunny, could you explain how you worked out the original dot product you had?  The reason I used the x and y parts of each vector was because I did not know a way to do the dot product of sums of vectors.  At first I thought perhaps yours is the same as mine, except knowing the outcome of using the x and y parts and skipping that part, but that does not seem the case.  Is it just coincidence that all the x and y parts simplified here, or will that always happen?

[quote="deej21"]Mathfanatic, $\vec p \cdot \vec q = 0$ means $\vec p$ and $\vec q$ are perpendicular, not parallel.  They are perpendicular at $t=-7, \frac {-1}2$, as established by the bounds.  They will be parallel at $\theta = 180$
$\vec p \cdot \vec q = -|\vec p| |\vec q|$
Using the formula established in a recent thread, $|\vec i + \vec j| = |\vec i|^2 + |\vec j|^2 + 2( \vec i \cdot \vec j)$, we get
$|\vec p| = 16t^2 + 28t + 49$
$|\vec q| = t^2 + 2t + 4$
$2t^2 + 15t + 7 + (16t^2 + 28t + 49)(t^2 + 2t + 4) = 0$
[/quote]

Oops.  I mix the two up a lot...

[quote="shobber"]Which part don't you understand??
Since $\vec{p}\cdot\vec{q}=|\vec{p}|\cdot|\vec{q}| \cos{\theta}$. If $\theta > 90^{o}$, then $\cos{\theta}<0$ so the dot product is negative.
Is that what confuse you?
[/quote]

No.  I solved the problem... I just didn't understand kunny's solution.  Specifically here.

[quote="kunny"]
$(2t\overrightarrow{e_1}+7\overrightarrow{e_2})\cdot(\overrightarrow{e_1}+t\overrightarrow{e_2})<0$

$\Longleftrightarrow 2t|\overrightarrow{e_1}|^2+(2t^2+7)\overrightarrow{e_1}\cdot\overrightarrow{e_2}+7t|\overrightarrow{e_1}|^2<0$
[/quote]

Sorry for that I made a typo.I have just edited. :blush:

kunny

Still, we don't know how the $\sqrt{14}$ stuff comes out... :(  :?

Yes kunny that typo confused me but I thought that you meant $\vec{e_2}$ there anyway.  I still do not understand how you got from line 1 to line 2 in the part I quoted in my previous post.

The $\frac{-\sqrt{14}}{2}$ exception doesn't make sense.  I plugged that in for $t$, then solved for $\theta$, and got $~94.07782847...$, which clearly falls into the boundary we set for $\theta$

You can expand kunny's equation and  you will understand why.

Also, read our posts carefully, you will find that if $t=-\frac{\sqrt{14}}{2}$, then $\vec{p}= -\sqrt{14} \vec{q}$, which means they are collinear.


geometry
ratio
analytic geometry
graphing lines
slope
search
algebra

[quote="#H34N1"][img]http://i268.photobucket.com/albums/jj11/Mersenne3731/1996_Nationals_Sprint_18.gif[/img][/quote]

there is a question very similar to this in HSB right now

[hide]Clearly, P is the centroid. [AMC]=12, since the bases area in a ratio of 1:1. Now, PC/PM=2/3, so it's 8.[/hide]

Well, I'm sure there's a method with more finesse than mine, but I have a cheap trick method.

P is the intersection of the two given medians.  Now pretend that AB is part of the y-axis of the coordinate plane, BC is part of the x-axis of the coordinate plane, and B is the origin.  Therefore, P's coordinates are the solution to the system of equations:
y=-(3/8)x+3  <--Line MC
y=-(3/2)x+6  <--Line AN

The slopes are found because we know the change in y and x for both equations; we know the lengths of AB and BN (change in y and x for line AN) and we know the lengths of MB and BC (change in y and x for line MC).
The y-intercepts are found even more easily; the coordinates of A and the coordinates of M.

Solving the system of equations:  -(3/2)x+6+(3/8)x-3=0 (subtract the first equation from the second)  -(9/8)x+3=0 (simplify)  9x=24 (simplify)  x=24/9 (divide).  Plugging the x back into either equation, y is obtained; y=2.

So P's coordinates are (24/9, 2).  A's coordinates are (0,6), and C's coordinates are (8,0).  You may be wondering where I am going with this.  I'll tell you in three words... the shoelace method!

I assume that you know what the shoelace method is, if not, well... you should be able to find a detailed description of it by means of google or other search engines online.

I'll grace you with my use of the shoelace method:

0 6
24/9 2
8 0
0 6

(1/2)(abs (0+0+48)-(16+16+0))=(1/2)(16)=8.  The area of the triangle is therefore 8 by the shoelace method.

You might be looking at this method of solving the problem, and you might also be thinking that it's too long and inefficient in a Nationals competition setting.  Well, I am able to do all of this in about a minute to a minute and a half because of practice, and I've seen how good you are at FTW.  I believe you can probably perform this entire process in less than a minute.  So, that's that and the problem is solved!

Use xpmath's method... Centroid. So the triangle is 1/3 of total area. 8.

[quote="#H34N1"][img]http://i268.photobucket.com/albums/jj11/Mersenne3731/1996_Nationals_Sprint_18.gif[/img][/quote]
That's simple. medians divide a triangle into 6 triangles with equal area. since one median is missing, the area of APC is one-third of the area of ABC.
[b]8.[/b]


Βρίσκομαι στην Κυπαρισσία για διακοπές και μόλις βρήκα ένα kafe με δίκτυο. Λοιπόν , εύχομαι στην Ελληνική ομάδα Καλά Αποτελέσματα και καλή επιστροφή !

 Εδώ μπορείτε να κάνετε τα σχόλιά σας με εναλλακτικές λύσεις και εκτιμήσεις για τα μετάλλια  !

 Μπάμπης

Καλησπερα, μολις ειδα τα προβληματα και μου φανηκε οτι δωθηκε μεγαλη σημασια σε Γεωμετρια/Συνδιαστικη/Αλγεβρα, καθως δεν υπηρχε καποια δυσκολη Θεωρια Αριθμων ουτε Ανισοτητα.

Η εκφωνηση και η λυση μου για το 1ο προβλημα:

Αν $ a_1, a_2, ... a_k \in \{1,2,...,n\}, k \geq 2$ και $ n|a_1(a_2 \minus{} 1), a_2(a_3 \minus{} 1), ... a_{k\minus{}1}(a_k \minus{} 1)$ νδο ο $ n$ δεν διαιρει τον $ a_k(a_1 \minus{} 1)$

Λυση:

Εστω οτι ισχυει η υποθεση, υπαρχουν $ a,b, ab \equal{} n, a|a_{1}, b|(a_{2} \minus{} 1)$, θα δειξουμε με επαγωγη οτι $ a|a_i, b|a_{i \plus{} 1} \minus{} 1, \forall 1 \leq i \leq k, a_{k \plus{} 1} \equiv a_1$
Για $ i \equal{} 1$ προφανως ισχυει
Εστω οτι ισχυει για $ i \equal{} j$, τοτε $ a|a_j, a_j(a_{j \plus{} 1} \minus{} 1), a_{j \minus{} 1}(a_{j} \minus{} 1)$, ομως $ gcd(a_j, a_{j} \minus{} 1) \equal{} 1$ αρα $ a|a_{j \minus{} 1}$, αρα $ a|a_i$ για καθε $ i$ και εντελως αναλογα (με επαγωγη προς την αντιθετη φορα) δειχνουμε οτι και $ b|a_i \minus{} 1$ για ολα τα $ i$ (Οι περιπτωσεις $ a \equal{} 1$ η $ b \equal{} 1$ ειναι τετριμενες).
Ειναι τωρα $ n \equal{} ab| a_m(a_k \minus{} 1), a_k(a_m \minus{} 1) \Rightarrow n| |a_ma_k \minus{} a_m \minus{} a_ma_k \plus{} a_k| \equal{} |a_k \minus{} a_m|$ για καποιο $ m$
το τελευταιο ομως ειναι ατοπο διοτι $ 0 < |a_m \minus{} a_k| \leq n \minus{} 1$.

Ελπιζω να μην εχω κανει καμια χαζομαρα

Καλα αποτελεσματα σε ολους :)

Βαζω και την εκφωνηση-λυση του 2ου προβληματος της 1ης μερας:

Εστω τριγωνο ΑΒΓ, Μ,Ν σημεια των ΑΒ, ΑΓ αντιστοιχα, ετσι ωστε ο περιγεγραμενος κυκλος του τριγωνου ΚΛΣ με Κ,Λ,Σ μεσα των ΜΒ, ΝΓ, ΜΝ αντιστοιχα να εφαπτεται στο τμημα ΜΝ, νδο τα Μ,Ν ισαπεχουν απο το περικεντρο του ΑΒΓ

Λυση:

Επηδη η γωνια χορδης-εφαπτομενης ισουτε με την αντιστοιχη εγεγραμμενη γωνια εχουμε  <ΣΚΛ = <ΜΣΛ = <ΝΜΑ (εντος εναλλαξ στις παραλληλες ευθειες ΑΒ,ΣΛ). Ομοια και <ΑΝΜ = <ΝΣΚ = <ΣΛΚ. Αρα τα ΑΜΝ, ΣΛΚ ειναι ομοια τριγωνα. Αρα ΣΚ/ΣΛ = ΑΜ/ΑΝ, και επηδη ΣΚ = 1/2ΝΓ, ΣΛ = 1/2ΜΒ, ειναι ΣΚ/ΣΛ = ΑΜ/ΑΝ = ΝΓ/ΜΒ => ΑΜ*ΜΒ = ΝΓ*ΑΝ, αρα τα Μ,Ν εχουν ισες δυναμεις ως προς τον περιγεγραμμενο κυκλο του τριγωνου ΑΒΓ και αρα ισαπεχουν απο το περικεντρο του.

Τα αποτελεσματα [url=http://www.imo-official.org/year_country_r.aspx?year=2009]εδω[/url]. Συγχαρητηρια σε ολους και ιδιαιτερα στους χαλκινους :clap2:  :clap:

Πολλα σγχαρητηρια στα παιδια για τα 3 χαλκινα και τις 3 εφημες μνειες!!!

Παιδιά, πολλά συγχαρητήρια !Καλές σπουδές σε όσους πάνε για ΑΕΙ και .... χρυσά στους μικρότερους για του χρόνου  !

 Μπάμπης

Polla polla sugxarhthria sta paidia!!!kali sunexeia stous mikroterous tis omadas!

Συγχαρητήρια και από εμένα σε όλα τα παιδιά....... :)

Συγχαρητήρια σε όλα τα παιδιά της ομάδας που μας έκαναν υπερήφανους. :lol:

Συγχαρητήρια σε όλους για τις εξαιρετικές επιδόσεις!!!

Σε όλα τα παιδιά συγχαρητήρια !!! Μπράβο σας , κατατάξατε για 3η συνεχόμενη χρονιά τη χώρα αρκετά ψηλά στη λίστα.

Σε αυτό το σημείο νιώθω ότι είμαστε υποχρεωμένοι να δώσουμε πολλά συγχαρητήρια στον Χρήστο Αναστασιάδη από την ομάδα της αδελφής Κύπρου, ο οποίος βραβεύθηκε με αργυρό μετάλλιο (25 πόντοι)!!!

[quote="Ipparchos"]Πρώτα απ' όλα καλώς μας ήρθες στο φορουμ!
Με τα ζητήματα τα οποία έθεσες (ελληνικές επιδόσεις στην ΙΜΟ, μαθηματικά στο Ελλαδιστάν) ασχολούμαστε συνήθως στο section Γενικά Θέματα- Κουβεντούλα.
Πάντως, ένα πρώτο σχόλιο: για τη συντριπτική πλεοψηφία των νεοελλήνων το ότι στην Ελλάδα γεννήθηκε η μαθηματική επιστήμη δεν σημαίνει απολύτως τίποτα. Για τους περισσότερους τα μαθηματικά είναι μία ενασχόληση για βαρεμένους φύτουλες, οπότε δεν χρειάζεται το εκπαιδευτικό μας σύστημα να επενδύει κεφάλαια και ανθρώπινο δυναμικό σε αυτά. Το ποιος είναι στην πραγματικότητα "βαρεμένος" είναι εμφανέστατο, νομίζω.[/quote]

Simfono apolita. Kai afou den iparxei megali ependisi sto sigkekrimeno tomea, i proetimasia pou kanoun ta paidia edo stin Ellada einai tipota mprosta se afti twn ksenwn. Ta perissotera themata ton IMO stirizontai kiriws se idees pou mporeis na vreis se diafora vivlia, kai osoi exoun dei para polles tetoies idees mporoun efkola na linoun ta perissotera apo ta themata twn IMO.

Edo i moni proetimasia pou exw dei einai se Geometria-Anisotites-theoria arithmwn, kai se sigkekrimenes morfes sinartisiakwn eksisosewn. Den gnorizw kan pou mporei kapoios na vrei oraia themata algebras san to fetino 3o thema tis 1is meras gia na eksaskithei me afta, opos episeis kai oraia themata diskolis theorias arithmwn opos to persino 3o provlima tis 1is meras... gia themata sindiastikis den to sxoliazw kan...

Κατ' αρχάς, αφού σας ευχαριστήσω θερμά όλους για τα συγχαρητήρια, θα ήθελα να δώσω κι εγώ συγχαρητήρια στα υπόλοιπα μέλη της ομάδας!
Η Ελλάδα κρατήθηκε για άλλη μια χρονιά αρκετά υψηλά στη βαθμολογία και με λίγο διαφορετικά θέματα μπορούσε να πάει ακόμα καλύτερα καθώς όλα τα παιδιά ήταν πολύ καλά προετοιμασμένα :)

Τέλος, εκφράζω μια ελπίδα να πάψουν τα συνεχόμενα άκριτα παράπονα για την όποια θέση της Ελλάδας, καθώς είναι η σημασία που δίνουν οι άλλες χώρες για τα μαθηματικά δεν συγκρίνονται με το ενδιαφέρον της Ελλάδας για τον τομέα αυτό!

Υ.Γ. 5/6 παιδιά της ομάδας συμμετέχουν ενεργά σε αυτό το forum, το οποίο όλοι εκτιμούμε και ευχριστόυμε αφάνταστα!

[quote="Jondrette"]Φίλοι είμαι καινούργιος στο Φορουμ και βρέθηκα τυχαία εδώ ψάχνοντας για τα θέματα της τελευταίας ΔΜΟ. Έχοντας διαβάσει τα διάφορα σχόλια για τα αποτελέσματα της Ελληνικής ομάδα, ένιωσα την ανάγκη να γράψω και εγώ κάτι για το θέμα αυτό.

Η προσπάθεια του κάθε παιδιού είναι αναμφίβολα αξιέπαινη, είτε πήρε μετάλλιο, είτε δεν πήρε. Από εκεί και πέρα η εικόνα της χώρας μας στις ΔΜΟ με θλίβει βαθιά. Ρίξτε μια ματιά στις χώρες που είναι πάνω από μας. Και αν θέλετε να γίνω πιο συγκεκριμένος, δείτε χώρες σαν την Τουρκία, το Ιράν τη ΦΥΡΟΜ. Πριν από χρόνια η Τουρκία είχε αποτελέσματα σαν τα δικά μας. Δείτε τους τώρα...

Εμείς τι βελτίωση είχαμε όλα αυτά τα χρόνια; Το σύστημα διαγωνισμών της ΕΜΕ γίνεται όλο και πιο περίπλοκο. Εκεί που κάποτε υπήρχε μόνο ένας διαγωνισμός για την Γ Λυκείου, τώρα υπάρχουν πολλά στάδια διαγωνισμών για όλες τις τάξεις. Το αποτέλεσμα; τρια χάλκινα μετάλλια... Για τη χώρα που γέννησε τα μαθηματικά...

Μακάρι να κάνω λάθος στην εκτίμησή μου. Περιμένω να ακούσω τις απόψεις σας.

Φιλικά.[/quote]

[color=blue][size=150]Σε καλωσορίζουμε στο φόρουμ των Ολυμπιάδων !

 Ο προβληματισμός σου είναι και δικός μας, αλλά αυτό δεν αρκεί για να αλλάξουμε την κατάσταση. Στην πράξη, όχι μόνο για τα μαθηματικά, αλλά και για πολλά πράγματα που γεννήθηκαν ή αναπτύχθηκαν στην Ελλάδα , οι Έλληνες και το εκπαιδευτικό σύστημα είναι αδιάφοροι. Είμαστε  ως χώρα αυτό που είμαστε και αυτό μας αξίζει ως σύνολο. Όμως θα παλαίψουμε να αλλάξουμε ό,τι μπορούμε .

     Ευτυχώς που υπάρχουν μεμονωμένοι Έλληνες, όπως οι μαθητές μας που συμμετέχουν στους μαθηματικούς ή άλλους διαγωνισμούς , που μόνο με το ταλέντο και τις θυσίες τους προβάλουν συχνά αυτή τη χώρα. Διότι δυστυχώς, αν μας βαθμολογήσει κανείς ως συγκροτημένη Πολιτεία για τη βοήθεια που δίνουμε σε μαθητές ή φοιτητές  που διακρίνονται σε σπουδές και επιδόσεις, μάλλον παίρνουμε μηδέν.

     Αλλά την κουβέντα αυτή μπορούμε από το φθινόπωρο, αν έχουμε μάλιστα και προτάσεις, να την συνεχίσουμε στο ''Κουβεντούλα''. Να ξέρεις όμως ότι εδώ έχουμε αφιερώσει εκατοντάδες ώρες για σκέψεις και προτάσεις για το πώς θα μπορούσαμε να ανέβουμε όχι μόνο ως προς τον αριθμό μεταλλίων, αλλά και ως μαθηματική κοινότητα γενικότερα. Με λίγο ψάξιμο σε παλαιότερα μηνύματα , αυτό θα το διαπιστωσεις και μόνος σου.Για την ώρα ας χαρούμε με τα παιδιά την συγκομιδή των καρπών τους και ...επανερχόμαστε.

  Καλό Καλοκαίρι σε όλους σας - Μπάμπης[/size][/color]

Πολλα συγχαρητηρια σε ολα τα παιδια της εθνικης και απο μενα.Η κριτικη ειναι παντα ευπροσδεκτη αρκει να συνοδευεται απο προτασεις.

      ΚΑΛΟ ΚΑΛΟΚΑΙΡΙ ΣΕ ΟΛΟΥΣ

Υπαρχούν και πολύ πιο αναπτυγμένες χώρες από εμάς που είναι πολύ πιο κάτω ... Για περισσότερη κουβέντα, προτάσεις επί προτάσεων , κριτική και οτιδήποτε άλλο ας συνεχιστεί η συζητήση στο "κοθβεντούλα" ή στο mathematica.gr  :rotfl:

Καταρχήν συγχαρητήρια σε όλα τα παιδιά και ιδιαίτερα σε αυτούς που πήραν τα μετάλλια. Τους εύχομαι καλή πρόοδο.

Σιλουανέ νομίζω ότι είναι καλύτερα να μεταφέρεις εσύ όλα τα μηνύματα από αυτό του Jondrette και κάτω στο «Κουβεντούλα».

Προς αποφυγή παρεξηγήσεων θα ήθελα να πω ότι όσοι εκφράζουν δυσαρέσκεια για τα αποτελέσματα της χώρας μας δεν κατηγορούν τα παιδιά. Όσα τέτοια μηνύματα έχω δει… κανείς δεν κατηγορεί τα παιδιά. Όλοι λίγο πολύ λένε ότι η προετοιμασία τους δεν είναι η κατάλληλη. Το λέω αυτό γιατί τα παιδιά έχουν την τάση να παρεξηγιούνται και να νομίζουν ότι κάποιοι στρέφονται εναντίον τους. Με εξαίρεση κάποια ελάχιστα μηνύματα στο mathematica που λένε ότι δεν επιλέγονται τα κατάλληλα άτομα… οι περισσότεροι πιστεύουν ότι τα συγκεκριμένα άτομα της εθνικής θα μπορούσαν να είχαν πάρει χρυσά αν είχαν υποβληθεί στην προετοιμασία πχ των Τούρκων.

Δεν είναι το ταλέντο αυτό που τους λείπει, ούτε η όρεξη αλλά ούτε και η ποσότητα δουλειάς. Μάλλον είναι η ποιότητα της δουλειάς σε σχέση με τους άλλους. Οι ευθύνες φαίνεται να βαραίνουν την προετοιμασία τους, πράγμα που μεταφράζεται κυρίως σε εξειδίκευση των trainers για τέτοιους διαγωνισμούς αλλά και γενικότερο πρόβλημα παιδείας στη χώρα μας.

Για το δεύτερο δεν μπορεί να γίνουν πολλά σε μικρό χρόνο. Το θέμα του εξειδικευμένου training για τέτοιους διαγωνισμούς είναι αυτό που θα πρέπει να απασχολήσει τους αρμόδιους. Και δε μιλάω για τις γνώσεις των εκπαιδευτών, αλλά για την εξειδίκευση τους για τέτοιους διαγωνισμούς. Δεν αμφιβάλλω ότι οι εκπαιδευτές μπορεί να είναι άριστοι επιστήμονες και με πλούσιο ερευνητικό έργο. Το θέμα είναι καθαρά το αν τα παιδιά μας έχουν υποστεί την ίδια εξειδικευμένη για ΙΜΟ διαγωνισμούς εκπαίδευση με τα παιδιά των χωρών που είναι συστηματικά πάνω από την Ελλάδα… κι ας είναι χώρες χαμηλότερου πνευματικού/πολιτιστικού επιπέδου. Πόσο καλύτερο μπορεί να είναι το εκπαιδευτικό σύστημα της Τουρκίας, της Βραζιλίας και του Περού από το δικό μας; Νομίζω πως το κλειδί της επιτυχίας είναι η εξειδίκευση των trainers για το συγκεκριμένο τύπο εξέτασης.

απευθυνομενος στο αρχικο θεμα του τοπικ αυτου θα ηθελα και εγω να ευχαριστησω ως μελος της ομαδας αυτο το καταπληκτικο φορουμ που αποτελεσε και την αφορμη για την ενασχοληση μου με τους μαθηματικους διαγωνισμους αλλα και την πηγη ερεθισματων και γνωσεων γυρω απο τετοια προβληματα που σε ενα χρονο συντελεσαν καθοριστικα και στο να παω στην ολυμπιαδα και να παρω την ευφημη μνεια εκει.
Καλη τυχη στους επομενους τις χρονιες που ερχονται και να ξερουν οτι οι ολυμπιαδες αυτες ειναι απλα πανεμορφες.  :)

Αγαπητοί φίλοι,

Ευχαριστώ αυτούς που διάβασαν το μήνυμά μου και κατέθεσαν τα σχόλια και την κριτική τους. Κάθε τύπου κριτική είναι ευπρόσδεκτη. Το μήνυμά μου ήταν περισσότερο μια διαπίστωση παρά μια προσπάθεια κριτικής. Διάβασα με μεγάλο ενδιαφέρον τις απαντήσεις και είδα οτι έχω παρόμοιες απόψεις με κάποιους από τους φίλους που απάντησαν.

Το ποστινγκ μου ήταν περισσότερο μια διαπίστωση παρά μια προσπάθεια ασκησης κριτικής. Το τελευταίο πράγμα που θα έκανα ποτέ θα ήταν να κριτικάρω τα παιδιά. 'Εζησα αξέχαστες στιγμές στους διαγωνισμούς αυτούς, έκανα πολύ όμορφες φιλίες και ξέρω από πρώτο χέρι το μέγεθος της προσπάθειας που καταβάλλουν τα παιδιά. Θυμάμαι ότι καθόμασταν πολλές φορές και συζητούσαμε για το τι πρέπει να γίνει ώστε να ανέβει η συγκομιδή μεταλλίων για την Ελληνική ομάδα. Είμασταν όλοι σύμφωνοι και ο gpapargi σχεδόν συνόψισε αυτά που σκεφτόμασταν πολλά χρόνια πριν.

Στο μήνυμά μου δεν άσκησα κριτική σε κανένα. Το βασικό περιεχόμενό του ήταν το αν κανείς θα πρέπει να είναι ευχαριστημένος από τα μετάλλια ή θα πρέπει να περιμένει και να ελπίζει σε κάτι πολύ περισσότερο. Σέβομαι την άποψη των φίλων που μιλούν για επιτυχία αλλά ταπεινά διαφωνώ με αυτή.

Ζητώ συγνώμη για το νέο μήνυμα που απ' ότι έμαθα εκ των υστέρων θα ταίριαζε καλύτερα στον τομέα κουβεντούλα. Ο λόγος που επισκέφτηκα την ιστοσελίδα είναι το μαθηματικό της περιεχόμενο, το οποίο είναι εντυπωσιακό και νομίζω ότι θα εστιάσω την προσοχή μου σε αυτό. Ευχαριστώ τον κο Στεργίου, του οποίου τη δουλειά εκτιμώ ιδιαιτέρως, και τον Ύππαρχο για το καλωσόρισμά τους.

Φιλικά


calculus
derivative
function
calculus computations

How do you prove that if $f(x,y)$ is  function that is continuous at $f(x_{0},y_{0})$ how do you prove that

$f_{xy}(x_{0},y_{0})=f_{yx}(x_{0},y_{0})$?
sorry i dont remember what this theorem is called.

It's called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72477]Clairaut's Theorem[/url] (or Schwarz's Theorem). See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31863]this topic[/url].

This is textbook material.

If I remember correctly, it's not $f$, but its second order derivatives that are supposed to be continuous.

Calculus textbooks usually give an example to show that without the continuity assumption the derivatives could be different, as if this was something important. In fact, $f_{xy}=f_{yx}$ in the sense of distributions without any smoothness assumptions on $f$.


geometry
number theory
number theory proposed

[i]Author of the problem is Melkior Ornik (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 4 (Number Theory) for 11th and 12th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.artofproblemsolving.com/viewtopic.php?t=243614.[/i]

Prove that for every natural number $ k$, there exist infinitely many natural numbers $ n$ such that
$ \frac {n \minus{} d(n^r)}{r} \in \mathbb{Z}$, for every $ r \in \{1,2,\ldots,k\}$.
Here, $ d(x)$ denotes the number of natural divisors of a natural number $ x$, including $ 1$ and $ x$ itself.

[quote="mornik"][i]Author of the problem is Melkior Ornik (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 4 (Number Theory) for 11th and 12th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008.[/i]

Prove that for every natural number $ k$, there exist infinitely many natural numbers $ n$ such that
$ \frac {n \minus{} d(n^r)}{r} \in \mathbb{Z}$, for every $ r \in \{1,2,\ldots,k\}$.
Here, $ d(x)$ denotes the number of natural divisors of a natural number $ x$, including $ 1$ and $ x$ itself.[/quote]
Let $ n\equal{}\prod_{i\equal{}1}^m p_i^{a_i}$ then
\[ d(n^r)\equal{}\prod_{i\equal{}1}^m (ra_i\plus{}1)
\]
Which implies that $ d(n^r)\equiv 1 (\mod r)$. It suggests us to choose $ n\equiv 1 (\mod r),\forall r\equal{}1,...,k$. Obviously $ n\equiv 1 (\mod k!)$ satisfies. Thus there exist infinitely many natural numbers $n$ satisfying the condition.


I got only first 17 problems right........122.........

I tried to do the last 5 pbs....but it was really hard....

and i was kind of frustrated cauze, I solved 22 pbs in 12A when

I took it as a practice.............

Definitely, I think B was harder than A as for the last 5 pbs..

try to keep the discussion in the designated threads


integration
logarithms
Euler
calculus
calculus computations

The DE is:
\[x^{2}y^{\prime\prime}+xy^{\prime}-y=0 \]
Solution: [color=red]The only novel part of this is grouping terms like this[/color]:
\[\left( x^{2}y^{\prime\prime}+2xy^{\prime}\right)-\left( xy^{\prime}+y\right) =0 \]
[color=red]To get this nice product rule form[/color]:
\[\left( x^{2}y^{\prime}\right)^{\prime}-\left( xy\right)^{\prime}=0 \]

\[\int\left( x^{2}y^{\prime}\right)^{\prime}dx-\int\left( xy\right)^{\prime}dx=0 \]

\[x^{2}y^{\prime}-xy= C_{1}\]
divide through by $\frac{1}{x^{2}}$ to get
\[y^{\prime}-\frac{1}{x}y= \frac{C_{1}}{x^{2}}\]
the usual integrating factor is then $u=e^{-\int\frac{1}{x}dx}=e^{-\log |x|}=\frac{1}{|x|}$

so we have
\[\frac{1}{|x|}y^{\prime}-\frac{1}{x|x|}y= \frac{C_{1}}{x^{2}|x|}\]

\[\left(\frac{1}{|x|}y\right)^{\prime}=\frac{C_{1}}{x^{2}|x|}\]

\[\int\left(\frac{1}{|x|}y\right)^{\prime}dx =\int\frac{C_{1}}{x^{2}|x|}dx \]

\[\frac{1}{|x|}y =-\frac{C_{1}}{2x|x|}+C_{2}\]

\[y =-\frac{C_{1}}{2x}+C_{2}|x| \]
or just $y =\frac{c_{1}}{x}+c_{2}|x|$

So how do I conduct a similar creative grouping scheme for a general second order, say
\[y^{\prime\prime}+p(x)y^{\prime}+q(x)y=r(x) ? \]

You're equation had very special form called an Euler Differential Equation. If someone can answer your general question...wow.  :(

More info on Euler D.E's here: http://tutorial.math.lamar.edu/AllBrowsers/3401/EulerEquations.asp


function
algebra
domain
trigonometry
polynomial
algebra unsolved

[color=blue][b]Any solution for this deeply difficult problem?

$ x^x  \minus{} 4^x  \plus{} 6 \equal{} 0(\forall x \in R)$[/b][/color]

Hello
There is only one solution.
x is nearly equal to 1.4806257488028.
Thank u.

hello, taking a graphic calculator and plotting the function $ f(x)\equal{}x^x\minus{}4^x\plus{}6$ we see that there exist two
real solutions of our equation,by a numerical way we find
$ x_1 \approx 1.48062574880283378$
$ x_2 \approx 3.97537367217821648$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, taking a graphic calculator and plotting the function $ f(x) \equal{} x^x \minus{} 4^x \plus{} 6$ we see that there exist two
real solutions of our equation,by a numerical way we find
$ x_1 \approx 1.48062574880283378$
$ x_2 \approx 3.97537367217821648$
Sonnhard.[/quote]
No other means of solving than the calculator???
How did u solve it [color=blue]kabi[/color]?

hello, you can't express the solutions of the equation $ x^x\minus{}4^x\plus{}6\equal{}0$ by the known elementary functions.
Sonnhard.

Hello Dr.
yes there r two solutions.That's the reason i don't believe in my calculator.
Thank u.

hello kabi, what kind of calculator do you have?
Sonnhard.

[quote="Rajiv"]
No other means of solving than the calculator???
How did u solve it [color=blue]kabi[/color]?[/quote]
Hello Rajiv
I too have done it my calculator only.
Hello Dr. actually I was seeing the graph with a domain [-2,2] So I didn't saw the second solution.
It was my mistake.
Thank u.

[quote="Dr Sonnhard Graubner"]hello, this kind of equations can only be solved by a numerical way. The problem is to find all solutions in a given domain.
Sonnhard.[/quote]
Alright!

My graph application too shows the same answers.

[quote="Dr Sonnhard Graubner"]hello, you can't express the solutions of the equation $ x^x \minus{} 4^x \plus{} 6 \equal{} 0$ by the known elementary functions.
Sonnhard.[/quote]
How do you prove that it can't be expressed by the known elementary functions (atleast show how to prove it for trig and polynomial functions)

[color=blue][b]Dear all,
All your solutions and evaluations are very highly appreciated, yet I need a known elementary function solution. May I get it soon. Because you know, using the graphic caculator is not highly feasible and persuasive.
All the best regards,
TUAN KIET[/b][/color]


[b]Pregunta 1:[/b]
[color=blue]¿Alguien sabe en qué hotel nos hospedarían?[/color]

I don't belong to that group, so I can't anwer to that question. :(

será mejor llevar dolares?? euros?? que será lo más fácil de cambiar allá?

en nuestro caso eso no sirve..  en paris reciben euros y en hong kong dolares entonces necesitamos lo que sirva en vietnam...

[quote="R.G.A.M."][color=blue]La verdad que es una buena pregunta…
No sé si en Viet Nam aceptan con facilidad dólares (bromas aparte).
Sugiero llevar el tipo de moneda que mejor se adapte al itinerario del viaje.[/color][/quote]
Buno Rafa, creo q la respuesta es el dolar. En este sitio me indicaron esto => http://www.viajes-vietnam.com/Dinero_precios_Vietnam.php
CHAU.

[color=blue]La moneda es el [b]Dong[/b]. Algo más de 15.000 dongs son un dólar.[/color]