geometry circumcircle calculus integration AMC USAMO

Let $ \mathcal{P}$ be a convex polygon with $ n$ sides, $ n\ge3$. Any set of $ n \minus{} 3$ diagonals of $ \mathcal{P}$ that do not intersect in the interior of the polygon determine a [i]triangulation[/i] of $ \mathcal{P}$ into $ n \minus{} 2$ triangles. If $ \mathcal{P}$ is regular and there is a triangulation of $ \mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $ n$.

[hide="My Outline"] Consider the part of the triangulation that contains the center. Then, it's not hard to see that the minor arcs (on the circumcircle) above the three sides must have 1 less than a power of two number of points. So let them be $ 2^{a}\minus{}1,2^{a}\minus{}1,2^{b}\minus{}1$, since it's isosceles. Adding gives $ 2^{a\plus{}1}\plus{}2^{b}$ points in total, or $ 2^{x}\plus{}2^{y}$, where $ x,y$ are not both 0.

To show that there exists such a triangulation, first consider $ 2^{x}\plus{}1$, for positive integral $ x$. It's not too hard to get the construction. Then, if $ 2k$ can be done, $ k$ can be done, so everything follows.[/hide]

Hmm weird is this the same answer?

I had all powers  of 2 greater than 4 (4,8,16....)

and then the recursively defined sequence (3,5,9,...,) where $ a_{m\plus{}1} \equal{} 2a_{m} \minus{} 1$ where a1=3 and its for n>=1

for base cases and it can be any of those odd bases times a integral power of 2

So I  had like (3,4,5,6,8,9,10,12,16,17...)

it seems like the same answer but what if i totally worded it differently

I'm glad I could complete this one!  A friend of mine at school also qualified for USAMO and said he could do this problem too.

I'm not certain how many points I'll get for it though, since I beat around the bush at first when I proved that $ n\equal{}4k\plus{}3$ where integer $ k>0$ doesn't fit the criteria.  Later, I managed to prove that if $ n$ is odd, all $ 2^a\plus{}1$ work and are the only solutions, where $ a>0$.

[quote="frodo"]I'm glad I could complete this one!  A friend of mine at school also qualified for USAMO and said he could do this problem too.

I'm not certain how many points I'll get for it though, since I beat around the bush at first when I proved that $ n \equal{} 4k \plus{} 3$ where integer $ k > 0$ doesn't fit the criteria.  Later, I managed to prove that if $ n$ is odd, all $ 2^a \plus{} 1$ work and are the only solutions, where $ a > 0$.[/quote]

uhh i think 5 works...
which is 2^2+1

anyways, dang i seem to have gotten the right answer but i wrote it in such a bad format i hope they wont count off too much...

[quote="the future"]Hmm weird is this the same answer?

I had all powers  of 2 greater than 4 (4,8,16....)

and then the recursively defined sequence (3,5,9,...,) where $ a_{m \plus{} 1} \equal{} 2a_{m} \minus{} 1$ where a1=3 and its for n>=1

for base cases and it can be any of those odd bases times a integral power of 2

So I  had like (3,4,5,6,8,9,10,12,16,17...)

it seems like the same answer but what if i totally worded it differently[/quote]

That's what I got.  And I semi-proved that those were the only ones by proving that the only odd numbers that existed were one more than a power of 2.

Do you guys think that only a rigorous solution like t0rajir0u's with nice notation and stuff will get full points, or will a solution that just sorta states everything (e.g. if n is odd, n-1 must be a power of 2 because as you keep on dividing the small polygons into smaller ones, they must have an odd number of sides) also get 6 or 7 points?

ARGG I got the right solution but I don't think my solution is rigorous enough

I loosely defined an [i]trapezoid-like[/i] polygon and showed that each regular n-gon must be divided into these trapezoid-like polygons if all the sides were to be used in isosceles triangles.

In the end my solution was much like an essay :(

I divided the conclusion into three parts:
1. $ 2^p$
2. $ 2^p \plus{} 2^q$
3. $ 2^p \plus{} 1$

It's easy to prove that they work (actually you only need to prove 1, since the other two are just its application), but it's hard to prove that others don't. I used a lot of weird notations and wordy paragraphs to prove the converse. All I can remember is this point (I incoporated the converse into the proof of part 3):

($ n$ is not in form of power-of-2 or sum of non-zero-power of 2) Labelling the vertices $ 0,1,...,n\minus{}1$. WLOG let $ 0$ be connected to $ x$, if $ x$ is not a power-of-2, then $ x\geq \frac {n \plus{} 1}{2}$, and $ n \minus{} x$ is a power-of-2. Then I basically proved $ n \minus{} x$ must be $ 1$ by some shaky argument and then the above "lemma". Yeah, I know I'm due for deduction on lack of rigor, but hopefully let it be light...

[quote="Math Geek"]

That's what I got.  And I semi-proved that those were the only ones by proving that the only odd numbers that existed were one more than a power of 2.[/quote]

How do you prove the even ones that don't work?

So this was a bit similar to IMO 2006 number 2. Pretty nice question.

[quote="t0rajir0u"][hide="Solution"] Let $ v_1, v_2, ... v_n$ be the vertices of a regular $ n$-gon.  Define $ P(q, n)$ to return $ 1$ if the polygon $ v_1, v_2, ... v_q$ of a regular $ n$-gon has an isosceles triangulation, and $ 0$ otherwise.  We claim that $ P(n, n) \equal{} 1$ if and only if $ n \equal{} 2^a(2^b \plus{} 1), a, b \ge 0$.

We set $ P(1, n) \equal{} P(2, n) \equal{} 1$.

[b]Lemma:[/b]  $ P(2n, 2n) \equal{} P(n, n)$

Proof:  The side lengths can only be part of an isosceles triangle if they are paired up with an adjacent side.  This gives a regular $ n$-gon.

[b]Lemma:[/b]  $ P(2n \plus{} 1, 2n \plus{} 1) \equal{} P(n \plus{} 1, 2n \plus{} 1)$

Proof:  Exactly one side length must be part of an isosceles triangle that contains the center; the other $ n$ sides must be paired up.  This divides the $ 2n \plus{} 1$-gon into the polygons $ v_1 v_2 ... v_{n \plus{} 1}$ and $ v_{n \plus{} 1} v_{n \plus{} 2} ... v_{2n \plus{} 1}$, which either both do or both do not have a triangulation.

[b]Lemma:[/b]  $ P(2k \plus{} 1, n) \equal{} P(k \plus{} 1, n), n > 4k \plus{} 2$

Proof:  Same as above.

[b]Lemma:[/b]  $ P(2k, n) \equal{} 0, n > 4k$

Proof:  The longest side $ v_1 v_{2k}$ cannot be paired with a vertex opposite it to make an isosceles triangle.

The result then follows by strong induction. [/hide][/quote]

Tiny flaw, though. If a = 0, and b = 0, we are left with a degenerate 2-sided polygon. Ah well, that's what I got too (except rectified the a,b both = 0)

[quote="Xantos C. Guin"][quote="t0rajir0u"]We claim that $ P(n, n) \equal{} 1$ if and only if $ n \equal{} 2^a(2^b \plus{} 1), a, b \ge 0$.[/quote]

minor technicality that I discovered during the test: $ (a,b) \equal{} (0,0)$ yields $ n \equal{} 2$ which violates $ n \ge 3$. But if we allow degenerate polygons, then $ n \equal{} 2$ works.[/quote]

If I remember correctly, i think they stated that $ n > 2$ in the beginning

They stated that $ n\ge 3$, yeah.

Just to be safe, I wrote at the end that they couldn't both be equal to 0.

[quote="timwu"]

How do you prove the even ones that don't work?[/quote]

umm for even ones i said that if you could divide the even ones some amount of time by 2 such that it will be come odd (so if the even number is a*2^k, you can divide by 2^k) and if a is a solution then so is a*2^k but if a wasnt, the a*2^k isnt either...thats where ill probably lose points...

i manage to get the right answer and prove that it works for those ones but i can really prove the other ones dont work...

why is this required :(

The answer is $n=2^a(2^b+1)$ for non-negative $a,b$.

[b][color=#f00]Claim: [/color][/b]$2n$ works if and only if $n$ works.

[i]Proof: [/i]It's easy to see that if $n$ works, just triangulate the border of the $2n$-gon with small isosceles triangles, which will make a regular $n$-gon in the interior. Conversely, pick a side on the $2n$-gon. Note that the only way to triangulate is if the side is on an isosceles triangle with two sides on the $n$-gon. Repeating this argument, we find that the only possible way to triangulate the $2n$-gon is as described, hence the goal for $2n$ is only possible if it is also possible for $n$.

[b][color=#f00]Claim: [/color][/b]If $n$ is odd, $n$ works iff $n = 2^k+1$ for some $k$.

[i]Proof: [/i]Let a triangle be called [i]long[/i] if it passes through the center of the $n$-gon. Note that there is exactly one [i]long[/i] triangle. Now this triangle divides the $n$-gon into two equal polygons. It's obvious that when each equal polygon has $2^{k-1}+1$ vertices for some $k$, we can triangulate it. Conversely, note that the outer perimeter of the equal polygons concurrent with the side of the $n$-gon must be triangulated with small isosceles triangles, and etc. Thus it is only possible when $\frac{n-1}{2}$ is a power of $2$.

Wowza problem. I was stuck on the harder direction for quite some time :(

The answer is all $n \ge 3$ expressible in the form $n = \boxed{2^a+2^b}$ for $a$, $b$ nonnegative integers.

Call $n$ [i]triangular[/i] if it satisfies the conditions of the problem. We start with this claim:

[b]Claim.[/b] If $n$ is [i]triangular[/i] then $2n$ is also.
[i]Proof.[/i] Start with a $2n$-gon. Assume $n$ is [i]triangular[/i]. Consider the isosceles triangles formed with two adjacent sides of the polygon. Remove a set of $n$ of these disjoint triangles - then we will be left with a regular $n$-gon, which we know is [i]triangular[/i]. This means $2n$ is as well. $\square$

In light of this, it suffices to show that odd $n$ are [i]triangular[/i] if and only if they are $1$ more than a power of $2$.

First, we show that these work. Choose a vertex, and create the isosceles triangle formed with this vertex and the side opposite to this vertex as the base. Now two congruent non-regular $2^{x-1}$-gons are left. Consider the following example for $n=17$:
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This can be extended for higher $n$ - the pattern should be evident.

Now we show that these are the only ones that work. Choose two vertices $A$ and $B$, draw diagonal $AB$, and consider the polygon of the resulting two that doesn't contain the center. Define the [i]boundary[/i] length as the number of sides between $A$ and $B$.

[b]Claim.[/b] This polygon can only be triangulated into isosceles triangles if the [i]boundary[/i] length is a power of $2$.
[i]Proof.[/i] Let the [i]boundary[/i] length be $k$. Since the polygon does not contain the center, it follows that the only isosceles triangles that can be constructed with vertices of the polygon are ones whose legs are sides of the polygon. Since every vertex must be included in a triangle, we must be able to draw all these border isosceles triangles. This implies that $k$ should be even. Now removing these triangles leaves us with another polygon, whose [i]boundary[/i] length is $\frac{k}{2}$. Clearly its circumcenter is not in the polygon, so we repeat the process. Iterating, we find that $k$ must be a power of $2$, as needed. $\square$

Now go back to the original $n$-gon. The very first isosceles triangle divides the $n$-gon into four polygons - the triangle, and three polygons whose [i]boundary[/i] lengths are $a_1$, $a_2$, $a_3$ such that $a_1+a_2+a_3 = n$. Clearly two of these should be equal - WLOG $a_2 = a_3$, which means that $a_1$ must be the [i]boundary[/i] length of the polygon touching the base of the isosceles triangle. By our second claim, in order for these to be triangulated into isosceles triangles, all of $a_1$, $a_2$, $a_3$ must be powers of $2$. But since $n$ is odd, by definition $a_1$ should also be odd. This means $a_1 = 1$. Since $a_2+a_3 = 2 \cdot 2^m$ for some $m$, it follows that $n = 2^{m+1}+1$, completing the direction.

We conclude that all $n = 2^x+1$ are [i]triangular[/i], so by the first claim, all $n \ge 3$ in the form $n = \boxed{2^a+2^b}$ are [i]triangular[/i]. $\blacksquare$

\begin{proof}
We claim that the solution set is $$S = \{a\mid a = 2^p(2^q+1),\;  \text{$p$, $q$ nonnegative integers not both $0$}\}$$ Then let $f(n)$ denote the existence of a triangulation of a regular $n$-gon into isosceles triangles, where $f(n) = 1$ if there exist one, and $f(n) = 0$ otherwise.\\
Consider the following lemma.
\begin{lemma}
By writing an even positive integer $n$ as $2^k\cdot m$ where $m$ is odd, we have $$f(n) =1 \iff f(m) = 1$$
\end{lemma}
\begin{proof}
Notice that for each side of this polygon, it must be a part of an isosceles triangle. However since $n$ is even, each side can only form an isosceles triangle with the its two adjacent sides. Then we connect the vertices such that we have $n$ isosceles triangles on the side of the polygon and another regular $\frac{n}{2}$-gon inside it. For example, consider the regular $8$-gon. For each side of it, the only isosceles triangle it can be a part of are the four small ones up to rotation. Hence after connecting the segments, we are left to deal with the square inside.
Then we can continue this process on the regular $\frac{n}{2}$-gon, where if $\frac{n}{2}$ is even, we draw in the $\frac{n}{4}$ isosceles triangles and continue with the $\frac{n}{4}$-gon inside. Therefore this implies that the value of $f(n)$ is equivalent to that of $m$, as desired.
\end{proof}
Now we can focus on the value of $f(n)$ for odd $n$s. Since there are an odd number of sides, there has to be an odd number of sides forming a triangle that doesn't have two sides on the polygon. It is not hard to see that the only isosceles triangles with sides on the polygon for $n$ odd is are either having one side on the polygon or having two sides on the polygon. For the case where it only has one side on the polygon, its vertices are two adjacent ones and another that is directly opposite the side connecting the two adjacent vertices, where for brevity we shall name this type of triangles $\Delta Z$. In particular, for a triangulation, there can be at most one isosceles triangle like such. Thus we can first draw in this triangle, and deal with the rest $n-1$ sides. For convenience, all the reasoning that follows are WLOG referring to the right longer side of $\Delta Z$. \\
For the two longer sides of $\Delta Z$, it is evident that there can be no other isosceles triangle in the polygon with this side as one of its sides of equal length. Thus we may draw in the perpendicular-bisector of this side of $\Delta Z$ in hope to hit a vertex. Since the diagonals of the polygon must not intersect in the interior, this bisector must hit an vertex on the right side of the polygon. However this bisector hits an vertex if and only if $\frac{n-1}{2}$ is even, where if it is, we may draw in the diagonals connecting the two right vertices of $\Delta Z$ with this 'midpoint' of the right $n-1$ sides of the polygon to create an isosceles triangle. This connection forms another two smaller section on the right side, each with $\frac{n-1}{4}$ sides. By continuing this process, we notice that if at some point $\frac{n-1}{2^k}$ is $2$, we have found a valid triangulation. However if $\frac{n-1}{2^k}$ is odd, by previous reasoning it is not hard to see that there cannot exist a valid triangulation. In particular, we also know that $n=3$ can obvious work. Therefore $$f(n) = 1 \iff \frac{n-1}{2^k}=2\Rightarrow n = 2^{k+1}+1$$ for odd $n$ and nonneg integer $k$. \\
Lastly, we have shown that the solution set for odd numbers are $2^a+1$ for positive integer $a$. Moreover, by doing the previous process, it is not hard to see that a such triangulation exists for all numbers of the form $2^b(2^a+1)$ for nonnegative integer $b$, and we are done.

Hm, was not aware this was USAMO when I first solved it. This proof is awful (perhaps in a legit writeup I'd include diagrams) but oh well!

The answer is all $n$ expressible in the form $2^a(2^b+1)$ where $a,b$ are nonnegative integers. First notice that if $n$ works, then $2n$ works: we can connect every other vertex in the $2n$-gon, which forms $n$ isosceles triangles and a regular $n$-gon. If we can isosceles triangulate the $n$-gon, we can also isosceles triangulate the $2n$-gon. Thus it suffices to only consider odd numbers and $4$.
We will take care of $4$ first, since it's trivial: just draw any diagonal and you get an isosceles triangle.
Now we look at odd numbers. It suffices to show that any odd $n$ must be in the form $2^a+1$. Suppose $n=2k+1$ for some $k$. Let the vertices of the $n$-gon be $A_1,A_2,\ldots,A_{2k+1}$, where indices are taken mod $n$ (so $A_{2k+2}=A_1$ for instance). Define a "big triangle" to be a triangle of the form $A_{i}A_{k+i}A_{k+i+1}$. Also define a "small triangle" to be of the form $A_{i}A_{i+1}A_{i+2}$. We prove the following lemma:

[b]Lemma 1:[/b] Every isosceles triangulation of a $2k+1$-gon must contain exactly one big triangle and $k$ small triangles.
[b]Proof:[/b] We will first show that there is at least one big triangle. Consider the triangle containing $A_{k+i}A_{k+i+1}$ for an arbitrary $i$. For it to be isosceles, it must either be a big triangle or a small triangle. Now suppose that all of the triangles containing an edge were small triangles. Then we would need an even number of sides, since each small triangle contains two edges. But this is impossible since $2k+1$ is odd. So there must be at least one big triangle. Now observe that if there were multiple big triangles, there would be intersecting diagonals, which is not allowed. So there must be exactly one big triangle. This occupies 1 edge. Hence the remaining $2k$ edges are occupied by $k$ small triangles, as desired. $\blacksquare$

Now suppose that we draw in the big triangle $A_1A_{k+1}A_{k+2}$. This splits the polygon into an isosceles triangle and two "half-polygons" (kind of), each with $k+1$ sides: $k$ of equal length and one "long side". We will focus on the half-polygon $A_1A_2\ldots A_{k+1}$. Then the "long side" is $A_1A_{k+1}$. Consider the triangle which contains the "long side" as a side. Since no other sides or diagonals of the half-polygon are that long, we conclude that the long side must be the base of the isosceles triangle. This means that $k$ must be even, otherwise there would be no isosceles triangles. If $k$ is even, then we draw the triangle $A_1A_{k+1}A_{k/2 + 1}$, which is the only isosceles triangle containing the long side. But this creates even smaller $k/2$-gons (quarter-polygons, maybe). Applying the same argument again and again, we see that $k$ must be a power of $2$, otherwise we will eventually reach an odd number when dividing by $2$ and be unable to further triangulate. Conversely, if $k$ is a power of $2$ then this "splitting" algorithm will eventually produce an isosceles triangulation: it will terminate when everything becomes a triangle and it won't produce any non-isosceles triangles. This implies that an isosceles triangulation exists for odd $n$ if and only if $n=2^a+1$ for some $a$, which implies the desired result. $\blacksquare$

Suppose a polygon is good if it satisfies the condition.

Lemma 1: $n$ good$\implies 2n$ good.

Proof: Just draw half of the shortest diagonals in a regular $2n$-gon to obtain a regular $n-gon$ and $n$ isosceles triangles. Obviously this means $2n$ is good if $n$ is good.

Hence, it suffices to check odd $n$. Let an ear of a triangulation be a triangle sharing two sides with the polygon, and let the span of a diagonal be $\min{|S_1|, |S_2|}$, where $S_1$ and $S_2$ are the sets of sides of the polygon on either side of it. We proceed with casework.

Case $1$: There exists at least one $\triangle ABC$ in the triangulation with $AB = AC$ and $BC$ a side of the polygon. Obviously $A$ is the midpoint of the longer walk from $B$ to $C$. Now consider the smaller portion of the polygon cut off by side $AB$. Clearly every diagonal of this smaller polygon is shorter than $AB$, so $AB$ must be a base of the other isosceles triangle with it is a side. Hence, the span of $AB$ is even. But by repeating the same argument over and over, we get that the span of $AB$ must be a power of $2$. In particular, $n$ is one more than a power of  $2$. Note that the construction for $n = 2^k + 1$ is evident in the above explanation for this case.

Case $2$: There exists no such triangle as described in case $1$. Let $\triangle ABC$ be such that $AB = AC$ and none of $AB$, $BC$, and $AC$ are sides of the polygon. By the same reasoning as before, the spans of $AB$ and $AC$ are powers of $2$. If $\angle BAC$ is obtuse, the span of $BC$ must also be a power of $2$. But then $n$ is even since none of $AB$, $BC$, and $AC$ are sides, but we assumed $n$ is odd -- contradiction. Hence, $\angle BAC$ is obtuse. If $BC$ is the base of the other isosceles triangle with it as a side, then the length of the longer walk from $B$ to $C$ must be a power of $2$ by the same reasoning as before and again we get that $n$ is even. Hence, $BC$ is not the base of that triangle. Let the triangle be $BCD$ with $BC = BD$. Then clearly $\angle CBD$ is acute, so the span of $CD$ is a power of two by the same reasoning again. But again this implies that $n$ is even, so we have exhausted all cases.

Finally, we get that $n$ must be of the form $2^a(2^b + 1)$ for nonnegative integers $a$ and $b$ (and all such $n\neq 2$ work).

Sorry if this is poorly written and/or has typoes, I am really tired while writing this.

[hide]The solution set is $2^x+2^y$ where $x,y$ are nonnegative integers (and the quantity is greater than $2$).

We first show that if $n$ works, then $2n$ also works. To do this, take a $2n$gon and label the vertices $A_1,A_2...A_{2n}$. Then connect $A_1$ to $A_3$ to $A_5$... to $A_{2n-1}$. Then we get a regular $n$gon in the interior (can be proved by angle chasing to be concrete), and thus we have proved the claim.

Thus, it suffices to solve the problem over odd $n$. We will show only $2^x+1$ works where $x$ is a nonnegative. Consider an $n$gon that works and label its vertices $A_1,A_2...A_n$ (let $A_k$=$A_{t*n+k}$ for integer $t$). Now each side of this $n$gon, we will let it be $A_xA_{x+1}$, must be part of some isosceles triangle, and this is only possible if $A_xA_{x+1}A_{x+2}$ or $A_{x-1}A_xA_{x+1}$ or $A_xA_{x+1}A_{x+\frac{n+1}{2}}$ is the triangle the side is part of. Due to parity reasons, we must have at least one side in the last form, WLOG let it be $A_1A_2$, and because no intersections can be possible, the rest must be of the other two forms.

Let $b=\frac{n+3}{2}$. Now the triangle $A_1A_2A_{b}$ splits the polygon into two smaller polygons (excluding the triangle) each with $\frac{n+1}{2}$ vertices. WLOG we will focus on $A_2,A_3...A_{b}$, and call this polygon $P$. But now note that $A_1A_{b}$ is the longest possible diagonal, so it must be a base of an isosceles triangle in $P$. But this is only possible if we connect it to the 'middle' vertex of the polygon or $A_{\frac{b}{2}+1}$, and this makes two more polygons with equal size. We can then make this same argument on these smaller polygons, and repeat this process until we get triangles that contain $3$ consecutive vertices (since this value of $n$ works by our assumption).

Now if the size of a polygon we are focusing on is $x$ and then we apply the process above on this polygon, the new number of sides becomes $\frac{x+1}{2}$. This means that since $3$ works (trivially), then $5$ works, and then $9$ works, and if $x$ works then $2x-1$ works too, but since $3=2^1+1$, then all the (odd) numbers that work are one more than a power of $2$, and this easily gives the answer we claimed above, so we are done. [/hide]

The answer is all $n = 2^i+2^j$ for nonnegative integers $i, j$.

First we show that these numbers work. We claim the following:

[b][color=#f00]Claim.[/b][/color] If a regular polygon with $n$ sides can be triangulated subject to the given conditions, then a polygon with $2n$ sides can also be triangulated in that way.

Draw the $n$ shortest diagonals in the $2n$-agon such that no two diagonals overlap. Hence we are left with an $n$-agon and valid isosceles triangles, so a polygon of $2n$ sides can be triangulated. $\blacksquare$

Therefore we only need to prove existence for $2^i+1$-agons for $i$ nonnegative. Before doing so we show the following lemma as well:

[b][color=#ff9a00]Lemma.[/color][/b] A polygon with one long side and $k$ shorter sides of equal length can be triangulated if and only if $k$ is a power of two.

It suffices to show that such polygons with an odd number of equal sides cannot be triangulated. Observe that in a valid triangulation of any form, there must exist a triangle with one of its sides being the bottom side, and thus two of its vertices being the bottom two vertices. But there does not exist any vertex which we can form an isosceles triangle with the bottom two vertices, and hence the construction is impossible for odd $n$.

To note that all powers of 2 work, simply induct upwards -- if $k=i$ can be triangulated, then $k=2i$ can also be triangulated by connecting the topmost vertex with the bottom two vertices, thus leaving two polygons with $k=i$. All other numbers do not work since at every step, we must triangulate the bottom two vertices with the topmost vertex, and thus at some point we will be left with such polygons with $k$ odd remaining, which are impossible to triangulate as we have shown. $\blacksquare$

With this, we first present our construction for $n = 2^i+1$. Triangulate the topmost vertex with its two opposite vertices. This leaves us with two polygons with $2^{i-1}$ equal sides and one longer bottom side. By the Lemma, each of these can be triangulated, so the polygon can also be triangulated.

Now, if $n \neq 2^i+1$, notice first that we must triangulate the top vertex with the bottommost two vertices. If we do not, then we will be left with a polygon as described in the Lemma with an odd number of equal sides at the ``bottom" of the polygon, which cannot be triangulated validly. However, then we are left with two polygons in the Lemma that do not have $2^n$ equal sides. Thus, they cannot be triangulated validly, and the $n$-agon can never be triangulated validly, as desired.

The anser is all numbers of the form $(2^a)(2^b+1)$. Let the polygon be defined by $A_1A_2\ldots A_n$. We begin by showing that all powers of $2$ work.
		[b]Claim:[/b]
			Even $n$ work if and only if $\frac{n}{2}$ works.

		[i]Proof of the claim.[/i]
			From the definition of triangulation, it follows that each side $A_{2i}A_{2i+1}$ must be part of a triangle. If $n$ is even, it is easy to that this triangle must in fact be $\triangle A_{2i}{A_{2i+1}}A_{2i+2}$, we will call it "small". So after drawing all the lines $A_{2i}A_{2i+2}$ in (we obtain $\frac{n}{2}$ triangles), we obtain a regular $\frac{n}{2}$-gon. From here the result is immediate.  $\square$
		Now we look at numbers of the form $n2^kr$, where $r$ is odd. Because of parity reasons, there exists a side $A_iA_{i+1}$ which cannot be part of a small triangle, say $A_1A_2$. The only vertex which can make this triangle be isoceles is the point $A_{1/2(n+3)}$. Now each of the polygons $A_1A_{n}A_{n-1}\ldots A_{1/2(n+3)}$ and $A_2A_3\ldots A_{1/2(n+3)}$ must be further disected. Using the small triangle argument it is easy to see that this can only happen when their number of sides is each a power of $2$, which must be the same power of $2$ as they have equal sides. Hence, $n$ must be of the form $2^b+1$ as desired.

badly written sol

The answer is all numbers n of the form 2^a+2^b, a,b nonnegative. Notice 2n works if n works because you can take A_iA_{i+2} as a side which makes isosceles triangle A_iA_{i+1}A_{i+2}, which we'll classify as [i]short[/i] triangles. It's evident that doing this for the entire polygon reduces each two segments into one, effectively halving the sides.

On the other hand, we show that a polygon with one long side and l shorter sides can be triangulated iff l=2^k (we would think of this by drawing any diagonal which motivates it by reading this proof backwards). From the 2n<->n footnote, we only need to consider odd # of sides polygon not of the form 2^k+1 and show they don't work (since this will effectively cover all evens as well), and also show 2^k+1 works. WLOG orient the longest side to be at the bottom. Since 2^k+1 is odd, for the longest side to form an isosceles triangle, it needs to connect to the top most vertex. But this reduces the problem to (2^k+1-1)/2=2^{k-1}, which we know works. On the other hand, if it's not 2^k+1 sides, then by connecting the same vertices we reduce to a problem that is the form $2^ij\forall j=2m+1$, which by induction hypothesis we know cannot be triangulated. Hence all numbers of the form $2^a(2^b+1)$ work.

[hide=horribly written]We claim that the answer is all $n=2^i+2^j$ for $i,j$ being non-negative integers. Before showing that this satisfies the given conditions, we first set up the following claims.\\

$\underline{Claim:}$ If we can triangulate in such a manner for $n$, we can also triangulate for $2n$.\\

$\underline{Proof:}$ We just form a construction based on parity. Label the parities from $1$ to $2n$, and connect $1$ and $3$, $2$ and $4$, an so on. After this, we are left with $n$ isosceles triangles and $n$-regular polygon, as desired.\\

$\underline{Claim:}$ A polygon with $1$ long side and $k$ shorter sides can be triangulated in such a manner if and only if $k$ is a power of $2$. \\

$\underline{Proof:}$ To prove the if $k$ is a power a $2$ statement we proceed with induction.\\

For the base case, we consider $k=2$, which is already an isosceles triangle, so it works.

For the inductive step, assume this is true for $k=2^i$. For $k=2^{i+1}$, proceed by taking the center of the polygon, $2^{i}+1$ and connecting that to $1$ and $2^{i+1}$. This results in an isosceles triangle and two additional congruent polygons that have $k=2^i$ shorter sides and one long side, as desired.

To prove the converse, we proceed in a similar manner. Using a similar algorithm, we take the center of the polygon and connect it to the edges resulting in two smaller polygons with $k$ as half as much and similar to the original one. However, this algorithm stops when $k$ eventually becomes odd.\\

Also, in order to become triangulated in the desired manner, the center must be connected in the above manner, as the longer side is not the same as any of the shorter sides, as desired.\\

$\underline{Main Proof:}$

From the second claim we can conclude that $n=2^i+1$ will works for any non-negative $i$. This is because we can connect a vertex to its opposite two vertices leaving use with two congruent polygons with one long side and $k=2^{i-1}$ shorter sides, and an isosceles triangle.\\

Connecting this with the first claim, we have proven that $n=2^i+2^j$ works for all non-negative $i$ and $j$.\\

We know prove that these are the only ones, or that everything else does not work. It suffices to show that this will not work when $n$ is not $1$ modulo a power of $2$ and is still odd.\\

By the second claim, we have shown that when $k$ is not a power of $2$, it cannot be triangulated for that specific polygon. However, if $n$ is not $1$ modulo a power of $2$, we must take an isosceles triangle with a vertex and its two opposite vertices. Otherwise, we must end up with $3$ of the polygons described in the second claim with at least one having an odd number of sides, which does not work by the second claim.\\

However, this also does not work, as the two remaining polygons will not be able to triangulate validly by the second claim, finishing our proof.
[/hide]

We claim that all the numbers of the form $2^m+2^n$ for nonnegative $m,n$ work.

Label the vertices of a regular n-gon from $0$ to $n-1$. Notice that all edges must be part of some triangle, so edge $(k,k+1)$, vertices taken mod $n$, must be a part of one of the following three triangles
\begin{align*}
    &(k,k+1) - (k+1, k+2) - (k, k+2)\\
    &(k-1,k) - (k, k+1) - (k-1,k+1)\\
    &(k,k+1) - (k+1, k+1 + \frac{n-1}{2}) - (k+1 + \frac{n-1}{2}, k)\\
\end{align*}
When $n$ is even, we must use either option 1 or 2 repeatedly for all the edges, so we see that $n$ has a triangulation if and only if $n/2$ has a triangulation.

However, when $n$ is odd, we must use option 3 exactly once. WLOG, $(n-1,0)$ is the edge that uses option 3, so we can't have vertex $\frac{n-1}{2}$ be blocked by $(\frac{n-3}{2},\frac{n+1}{2})$, so we need $2\mid \frac{n-1}{2}$, giving $n=4c+1$. Now, the polygon is split into two halves, and for each half, we must be able to connect consecutive edges, because we can't use option 3 more than once. From this, we can arrive at $n=2^t+1$ for some nonnegative $t$.

Using the previous two cases, we can have $n=2^t(2^u+1)$ or $n=2^t$, which is equivalent to what we claimed.

[hide]
The answer is $\boxed{\text{all }n\mid s_2(n)\le2}$, i.e., $\tfrac{n}{2^{v_2(n)}}=2^k+1$ for some $k\in\mathbb{Z}_{\ge0}$.
Observe that for even $n$, no side can be the base of an isosceles triangle in any triangulation, so each side must be a leg.
We are accordingly forced to connect every other vertex, which produces $\tfrac{n}{2}$ outer isosceles triangles and an inner $\tfrac{n}{2}$-gon.
Hence, an even $n$ satisfies the condition iff $\tfrac{n}{2}$ does, so it suffices to prove that of the odds, only $n=2^k+1$ work. \bigskip

\noindent In fact, we shall prove said result for all \textit{$n$-semigons}, an $n$-sided polygon with $n-1$ congruent sides forming $n-2$ congruent angles and an $n$th arbitrarily long ``oddball" side.
Clearly even $n$ fail, as the oddball must be the base of an isosceles triangle; and for odd $n$, connecting said isosceles triangle produces two smaller $\tfrac{n-1}{2}$-gons.
Applying the selfsame logic, we find that an $n$-semigon is triangulable in the desired fashion iff all of $n,\tfrac{n-1}{2},\tfrac{(n-1)/2-1}{2},\dots$ are odd, which is precisely equivalent to having $n=2^k+1$, as claimed. \qed \\

The answer is $n=\boxed{2^a(2^b+1)}$ for nonnegative integers $a,b$.

For the sake of definitions, begin with a regular $(2n+1)$-gon with points $A_1, A_2, \dots, A_{2n+1}$ in that order. Define a \textit{small} triangle to be a triangle likewise to $A_1A_2A_3$ and define a \textit{large} triangle to be a triangle likewise to $A_1A_{n+1}A_{2n+1}$.

-----

[color=#f00][b]Claim:[/b][/color] If a $n$-gon can be triangulated, so can a $2n$-gon.

[i]Proof:[/i] Notice that filling out the $2n$-gon with small triangles will result in an $n$-gon, which proves the claim. $\square$

-----

Now, consider the aforementioned regular $(2n+1)$-gon. Evidently, there must be at least one large triangle. Upon further inspection, we see that each of the large triangles contains the center and thus, there are only $1$ such triangle.

The large triangle splits the polygon into two $n+1$ sided polygons with $n$ equal sides. Then, it is clear that $n$ must be a power of two, so the only $k$-gons with an odd number of sides must have $k=2^b+1$ for a nonnegative integer $b$. Then, we can multiply this by any factor of $2$, so our answer results.

Our answer is $n = 2^x (2^y + 1).$

We will first need the following:

[b]Claim:[/b] $2n$ works if and only if $n$ works.
[i]Proof:[/i] If $n$ works, then $2n$ works because we can connect two corners that are not connected but closest possible with an edge. This gives $n$ isosceles triangles (which we will call $1-1-2$ triangles) on the side. Now, if $2n$ works, we note that every edge is contained in some triangle. Suppose that two adjacent vertices of a $2n$-gon are $A$ and $B,$, where $A$ and $B$ are in the same triangle. There is no point diametrically opposite an edge on a $2n$-gon, so to make an isosceles triangle, so the third point of this triangle $C$ must be adjacent to either $A$ or $B,$ creating a $1-1-2$ triangle. This repeats for every edge, giving a regular $n$-gon in the middle. Since $2n$ works, that forces $n$ to work, as claimed.

Since $n = 4$ clearly works, all powers of $2$ work, at which point we can note that $2^k = 2^{k-1}(2^0 + 1).$ Therefore, we only need to consider odd $n.$

There are two parts to the problem from here: proving that odd numbers of the form $2^k + 1$ work, and proving that the rest don't.

[b]Part 1:[/b] Proving that $2^k + 1$ work. We attach a construction for $k = 5;$ it is obvious how to generalize.

[b]Part 2:[/b] Proving that the rest don't work. Suppose that we have adjacent vertices $A$ and $B$ in our $(2k + 1)$-gon. Then they must be part of the same triangle. There are $2$ ways that this triangle is isosceles: the first is if $A$ and $B$ are in a $1-1-2$ triangle, and the second is if this triangle includes the point diametrically opposite segment $AB,$ which we will call $C.$ Call a triangle $ABC$ [i]very long.[/i] Note that every $1-1-2$ triangle covers $2$ edges of the polygon, but there are $2k + 1$ edges. Therefore, there is exactly one very long triangle in such a triangulation.

This very long triangle splits the polygon into $2$ congruent regions, each with $k$ edges facing outward and $1$ edge being inside the polygon. Note that $k$ is even, otherwise we cannot include the outward edges inside our triangulation (triangles containing outward edges are forced to be $1-1-2$). Similarly, $\frac{k}{2}$ is even, and so is $\frac{k}{4},$ and so on. This forces $k$ to be a power of $2,$ so $2k + 1 = 2^{x} + 1.$

Therefore, the only odd $n$ that work are $n$ of the form $2^x + 1.$ As a result, the only $n$ that work are $n$ of the form $2^x (2^y + 1),$ as claimed at the beginning.

The key intuition is to consider the "small sides" of each shape we are triangulating. We use casework based on the value of $n$.
[list]
[*] $n$ is even: Each side must be the leg of an isosceles triangle rather than the base. Consequently, we are forced to pair the sides off at every other vertex, creating an interior $\frac n2$-gon. Repeating this, we see that an $n$-gon is successfully triangulated $\iff$ an $\frac{n}{2^{v_2(n)}}$-gon can be triangulated.

The only special case to consider if when $n$ is a power of two, but we note these always work as we can use this algorithm to obtain a square.

[*] $n$ is odd: Let $n=2n_1+1$. We pair off the sides again, but this time 1 is left out. We are forced to use this side as the base with the vertex being the opposite vertex of the $n$-gon. Hence if $\frac{n_1}{2}<1$, we're done, but otherwise it must be an integer.

Then let $n_1=2n_2$. After the previous move, we are left to triangulate two congruent polygons with $n_2$ small sides and one long side. This long side is longer than any diagonal in this new polygon, so it must be used as the base. Hence, if $\frac{n_2}{2}<1$, we're done, but otherwise it must be an integer.

We continue, concluding that we require $n=2^x+1$ for a positive integer $x$.
[/list]

Hence our solutions for $n$ are $\boxed{2^x, 2^x+1, 2^x\left(2^y+1\right) \forall x,y \in \mathbb{Z}^+}$. $\blacksquare$


Let ABC be a triangle with <A=60 degree and area of ABC triangle is 72√3 .Take point D on [AB] side and take point E on [AC] side.If area of ADE triangle is equal area of DECB quadrilateral what is the minumum length of [DE] line segment

It is pretty easy problem.
We have that $S_{ADE}=36\*\sqrt{3}$ so $AD\cdot{AE}=144$.Then
by the cosines rule ${DE}^{2}={AD}^{2}+{AE}^{2}-AD\cdot{AE}\geq{AD}\cdot{AE}=144$ and  the equality holds iff $AD=AE=12$ thus the minimum of $DE$ is $DE_{min}=12$.


Yes, I'm hoping to have a Mock AIME up by Friday, provided I don't get drowned in schoolwork. I have the first half done... the easy half, that is. Anyway, if anyone wants to take it, you can probably take it starting from around Friday, 9 PM ET, which is when I hope to get it up. I'll post answers sometime Sunday and you guys can post solutions. The test will probably appear in this thread as an attached PDF file because I want to TeX it =).

That would be great. :)

Since you're going to make mock test (AIME or AMC, it doesn't matter), here is the secret code only given to the ones who made them :P

[hide="Secret"]
If you didn't make any of them, you're not eligible to look at this.
[hide="Real"]
[hide="Not_Kidding"]

 :first:  :rotfl:  :cool: --> My favorite icons![/hide][/hide][/hide]

Will the problems on the mock AIME be, or taken from past contests?

[quote]
1234567890 says:
Could you not make it like mildorf's (cough:too hard)  :)
[/quote]

but could you not make it like silverfalcon's (cough:easier than i thoguht) :)
thank u for your hard work, jeffrey
and since i will be taking my wireless laptop, ill see you posting a test
friday in Vancouver!

Yeah, I'm trying to make it a bit harder than silverfalcon's and a bit easier than mildorf's (maybe more than a bit on this side) but it's hard to gauge the difficulty of questions that I write, so it might end up one way or another. I'll try hard, thoguh.

These are all my own problems, though some are inspired by past AIME problems. There isn't anything TOO similar yet I don't think.


[quote="Shahzada"]-4^x=x-3

x=?[/quote]
[hide="uhhh"]Lambert?[/hide]

$\ ln$ is definited in $\ [0,+\infty]$
then there isn't $\ ln(-4)$

[quote="binomial_4eva"]$-4^{x}=x-3$

$ln(-4)^{x}=ln(x-3)$

and i dont know how to get x from $ln(x-3)$[/quote]

i also don t know so i asked you. :)

[hide="There no integer solutions."][/hide]
[hide="Proof"]
[u]Two cases:[/u]

$1) \text{Even Integer}$
$-4^{x}= x-3$
$4^{x}= x-3$
Plug in $x = 0$: we get $1 =-3$.
As $x$ increases, the LHS goes to infinity rapidly while the RHS goes slowly, so they will not overlap.
As $x$ decreases, the LHS will approach a limit of $0$ which is $>-3$.
Therefore, there are no even integer solutions.

$2) \text{Odd Integer}$
$-4^{x}= x-3$
$(-4)(4^{x-1}) = x-3$
$4^{x-1}=-\frac{x-3}{4}$
Plug in $x = 1$.  We get $1 = \frac{1}{2}$
As $x$ increases, the LHS goes to infinity rapidly while the RHS goes slowly, so they will not overlap.
As $x$ decreases, the RHS goes to infinity slowly while the LHS goes rapidly, so they will not overlap.
Therefore, there are no odd integer solutions.

Therefore, there are no integer solutions.
[/hide]

The Product-Log function aka. "Lambert function" $W(x)$ is the inverse of $xe^{x}$; that is, $W(xe^{x})=x$.

$-4^{x}= x-3 \implies-64 (4^{x-3}) = x-3 \implies \frac{64}{4^{3-x}}= 3-x$

$\implies 64=4^{3-x}(3-x) \implies 64 = e^{\ln 4 (3-x)}(3-x)$

$\implies (64 \ln 4) = e^{\ln 4 (3-x)}[\ln 4 (3-x)]$

Here's our opportunity to use $W$:

$W(64 \ln 4) = W(e^{\ln 4 (3-x)}[\ln 4 (3-x)]) = \ln 4 (3-x)$

$\implies 3-x = \frac{W(64 \ln 4)}{\ln 4}\implies \boxed{x=3-\frac{W(64 \ln 4)}{\ln 4}}$


$ S(n) \equal{} 1 \plus{} \frac {1} {2}  \plus{} \frac {1} {3} \plus{} .......... \plus{} \frac {1} {n}$

prouve that: $ \forall n\in \aleph$   ,  $ S(n)$ is not in $ \aleph$


geometry
geometric transformation
reflection
angle bisector
geometry unsolved

(5) Let ABC be a triangle with circum-circle T. Let M be the point of intersection of this tangent to T at B and C. Suppose N is a point choosen on the line AC such that MN is parallel to AB . prove that NA =NB.

It is equivalent to this problem : In triangle $ ABC$, let $ P,Q,R$ be the point of tangencies of the incircle with the sides $ BC, CA, AB$ respectively. Let $ I$ be incenter. Let $ CI$ intersect $ RQ$ at $ Y$, then $ BY$ is perpendicular to $ YC$

Let $ QR$ intersect $ CB$ at $ Y$ .It is well known that $ (XBPC)$ is harmonic (since $ AP, BQ$ and $ CR$ concur)
Also observe that $ Q$ is the reflection of $ P$ about $ CI$ (or $ CY$). Hence, $ \angle PYC \equal{} \angle QYC$, that is, $ YC$ is the external angle bisector of triangle $ XYP$. Hence the result

[img]http://img4.imageshack.us/img4/5629/imagewa.png[/img]

Since $ AB \parallel MN$, $ \angle ABM \plus{} \angle BMN \equal{} 180$, or $ \angle B \plus{} \angle A \plus{} \angle BMN \equal{} 180$. Thus $ \angle BMN \equal{} \angle C$, so $ BMCN$ is cyclic. So $ \angle ABN \equal{} \angle BNM \equal{} \angle BCM \equal{} \angle A$, so $ \triangle BAN$ is isoceles.

Lajanugen's problem is from a different direction. Since $ Y$ is on the bisector of $ \angle C$ it is the point on $ RQ$ equidistant from $ P$ and $ Q$. If we can show that $ PQ \parallel BY$, then we are done. But $ PQ \perp CY$, so if we show $ BY \perp CY$ we are done.


Από ένα σημείο Γ ενός κύκλου φέρουμε κάθετη ΓΔ σε μία χορδή
ΑΒ = σταθερό .
 Από το Γ φέρουμε μεταβλητή χορδή , η οποία τέμνει την ΑΒ στο Ε και την περιφέρεια στο Ζ .
α) Να βρεθεί ο γ.τ των κέντρων των περιγεγραμμένων κύκλων στα τρίγωνα ΑΖΕ και ΒΖΕ
β) Αν Ο1 , Ο2 είναι τα κέντρα των πιο πάνω κύκλων , να δείξετε  ότι το άθροισμα των εμβαδών ΑΟ1Γ  και ΓΟ2Β είναι σταθερό .

Σε τι χρειάζεται η κάθετη ΓΔ ??  :maybe:

Έκ δεδομένου σημείου Γ περιφερείας άγεται κάθετος ΓΔ έπί δοθείσαν χορδήν ΑΒ . Περί το Γ στρέφεται χορδή , τέμνουσα την ΑΒ είς το Ε και την περιφέρειαν είς το Ζ .Ζητείται : α) Να εύρεθή ο γεωμετρικός τόπος των κέντρων των περιφερειών των περιγεγραμμένων είς τα τρίγωνα ΑΖΕ και ΒΖΕ
β) Έάν Ο1 , Ο2 είναι τα κέντρα των έν λόγω περιφερειών , να δειχθή ότι το άθροισμα των έμβαδών ΑΟ1Γ  και ΓΟ2Β είναι σταθερόν .

1966.....1958......
mu fenete oti sto teleyteo 8ema aytu tu topik 8a xriastume metafrasi stin arxea eliniki... :P
brika mia orea lisi (grafika,xoris apodiksi gia to 1o...) me tin zografiki ton windows......

 :D

[quote="Ramanujan"]Σε τι χρειάζεται η κάθετη ΓΔ ??  :maybe:[/quote]

Το πρόβλημα αυτό ήταν το δεύτερο από τα τέσσερα θέματα Γεωμετρίας, που δόθηκαν στις εισαγωγικές εξετάσεις για την Αρχιτεκτονική σχολή του ΑΠΘ, το έτος 1958. Διαβάζοντας τη λύση στο ''ΕΤΗΣΙΟΝ ΔΕΛΤΙΟΝ 1958'' του Α. Πάλλα από τις εκδόσεις Παπαδημητρόπουλου (Κάθε χρόνο, από το 1947 έως το 1980, αν δεν κάνω λάθος, ο Α. Πάλλας έβγαζε ένα τεύχος με τα θέματα που είχαν δοθεί στις εισαγωγικές εξετάσεις, μαζί με τις λύσεις τους), βλέπω ότι η ευθεία $\Gamma\Delta,$ δεν χρειάζεται σε τίποτα για την εκφώνηση, αλλά είναι πολύ χρήσιμη για τη λύση.

Φαίνεται πως ο εξεταστής καθηγητής, θέλησε να βοηθήσει τους εξεταζόμενους, δίνοντάς τους  μία βοηθητική γραμμή, σαν υπόδειξη. Σε αντίθεση με τον αντίστοιχο καθηγητή στη ΣΜΑ (Σχολή Μηχανικών Αεροπορίας), ο οποίος, όπως διαβάζω στο ίδιο δελτίο, έδωσε για πέμπτο θέμα Γεωμετρίας (τα θέματα ήταν πέντε...), το πρόβλημα του Malfatti, ο αθεόφοβος (Να εγγραφούν σε δοσμένο τρίγωνο τρεις κύκλοι, έτσι ώστε ο καθένας να εφάπτεται των δύο άλλων και ταυτόχρονα, δύο πλευρών του τριγώνου).

Ο Πάλλας βέβαια δεν του το συγχώρησε. Γράφει στο σχόλιό του:

''Το θέμα αυτό δεν δίδεται εις εισαγωγικάς εξετάσεις. Ο εξεταστής Καθηγητής είναι δυστυχώς εκτός πραγματικότητος''.

Προσπαθήστε να λύσετε την άσκηση αυτή που δεν έχει ιδιαίτερες δυσκολίες και το σχήμα βοηθάει πολύ στη λύση της (αρκεί να προεκταθεί η $\Gamma\Delta,$ μέχρι να τμήσει τον κύκλο).

Κώστας Βήττας.

'Εχει βρει κανείς λύση για την άσκηση 13;

EDIT

Ooops....!!! καταλάθος το έγραψα αυτό εδώ....κανονικά προοριζόταν για αλλού....

...eixe politexneio i thessaloniki tote? :roll:

το πολυτεχνειο θεσσαλονικης ειναι διαφορετικο απο το Αριστοτελιο? το τελευταιο εχει ιστορια 80 χρονων...

egw mia politexneiki sxoli kserw na exei to apth :P

Δεν έχω καταλάβει γιατί ονομάζεται Πολυτεχνική Σχολή του ΑΠΘ (Αριστοτέλειο Πανεπιστήμιο Θεσσαλονίκης) και όχι Πολυτεχνείο Θεσσαλονίκης. Πιθανόν έχει να κάνει με διοικητικά θέματα και το ίδιο έγινε  με το ΔΠΘ (Δημοκρίτειο Πανεπιστήμιο Θράκης). Δεν ξέρω αν  συμβαίνει το ίδιο στην Πάτρα και την Κρήτη και αν εξακολουθεί το ΕΜΠ (Εθνικό Μετσόβιο Πολυτεχνείο) μόνο, να φέρει επίσημα το τίτλο του Πολυτεχνείου.

Ο όρος ''Σχολή'', παραπέμπει σε ένα μόνο γνωστικό αντικείμενο σπουδών Μηχανικών (Αρχιτέκτονες, Πολιτικοί, ...) και άρα είναι αδόκιμος όταν ένα εκπαιδευτικό ίδρυμα περιλαμβάνει πολλές ''Σχολές'' (Αρχιτεκτονική Σχολή κλπ).

Κώστας Βήττας.

το πολυτεχνειο εκπαιδευει μηχανικους => μετα απο καποιο ετος διαλεγεις τομεα πχ στο ΕΜΠ 9 σχολες--42 τομεις πολυτεχνειο αυτονομο ειναι το ΕΜΠ και της Κρητης στα χανια. υπαρχει σελιδα στο ιντερνετ  του πολυτεχνειου Πατρας ή Θεσσαλονικης?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118744&start=20
4o post...
:rotfl:

Mpori kapios na dosi tinolokliromeni lisi stin askisi....??

[u]@thatin[/u]
[hide]
gt ti exi to 4o post,,,?
a...sto... [url]http://www.blackstate.gr/Brain_game.htm[/url]
36:302!!!!!!!!!!!!!!!!!!!!!!! :winner_first:  :clap:  :wow:  [/hide]

προς thatin αλλα και οποιον αλλο τι σου αρεσει περρισσοτερο στο ΕΜΠ ? :P

i vivliothiki :P

[hide="@kostas"]esi se mas den eisai? lol[/hide]

η βιβλιοθηκη ?.πολυ καλη,και εμενα το ιδιο
 στον Θανο-->ΕΜΠ ειμαι   :)

[quote="mostel"]@RAKAR:

[hide]Πλάκα έχει το παιχνίδι! Πόσες φορές το παίξες για να κάνεις 36'' ?  :maybe: [/hide][/quote]

[u]@mostel[/u]
[hide]neo::37:386!!!!!!!! :wow:  :wow:
x8es mu to ipe o thatin...+ka8e fora pu mpeno net pezo kai ligo..fasi exi..
mia sini8ia ine....<lei oti kai kala ,metrai posa deyterolepta mporun na liturgun taytoxrona ta 2 imisferia tu egefalu.. >:P  :P trixes katsares...etsi ipa aytos pu evgale to pexnidaki gia na kolisis ma ayto...
kalitera kamia askisi me kana anagogo polionimo......
i kamia geometriula...
[/hide]

[u]@thatin[/u]
[hide][quote]thatin:<esi se mas den eisai?[/quote]>
gt re.. toso diaforetiki imaste emis i ap8ites apo esas tus empites...??:maybe:
kalo olus tus ap8ites lipon se sispirosi...enomeni mporume...!! :P
[/hide]
 :D

@RAKAR:

[hide]Άντε και με το καλό να τα ...κατοστίσεις!  :winner_first:
Ναι, όντως έχει πλάκα! Εγώ το παιξα μία φορά και έκανα 9.6  :rotfl:[/hide]

[quote="RAKAR"][

[u]@thatin[/u]
[hide][quote]thatin:<esi se mas den eisai?[/quote]>
gt re.. toso diaforetiki imaste emis i ap8ites apo esas tus empites...??:maybe:
kalo olus tus ap8ites lipon se sispirosi...enomeni mporume...!! :P
[/hide]
 :D[/quote]

[hide="@rakar"]
koita.... oxi aplws diaforetikoi... alla... kathara.... hm... <
looooooooooooooooool
to eipe kai o compiler!
NTUA>AUTH!!
  :P
mexri kai i pascal einai enantion sas (vasika eixame diko mas anthrwpo stin omada...)
kala, den kanete sinolothewria, protasiako logismo ktl?
xtes me rwtouse enas dikos sas ti simainei pVq...!
 :P

koita vasika esi pou eisai mathimatiko den exeis idiaitero provlima... einai ΣHMMY vs ΤHMMY [color=#f8f8f8]kai min tolmisei kaneis na kanei tpt stin pascal![/color]

[/hide]

@kostas:
to kserw!
HMMY είπες?

@thatin  [hide]:dry:  :bruce:  :whistling:  :harhar:[/hide]
@RAKAR [hide]Katafera na kanw tin aspri mpala na kineitai katakoryfa panw katw sti mesi tis othonis kai tin mauri na einai statheri sti mesi tis katw dokou. Dystyxws omws kounithike katalathws to pontiki kai ekana mono 35 sec kai kati psila... :very_very_sad: (<--den yparxei tetoio emoticon) :juggle: [/hide]
@mostel [hide]Den peirazei, egw tin prwti fora ekana 0,6 giati den ixera ti prepei na kanw...Lol! [/hide]
@kostas [hide]Diladi ti to ksexwristo exei i vivliothiki sas pou i dikia mas de to exei? :agent: [/hide]
@fotis [hide]btw  :welcomeani:[/hide]
[hide="Lol"]Auto to egraya gia na dextei ton arithmo twn smiles pou evala parapanw LoOlolOol![/hide]
pros olous [hide]To xeskisame to topic... :starwars: LoOol![/hide]

:spam:  :no:

Btw, αυτό που λέει ο thatin είναι κάτι παραπλήσιο με αυτό που είπε ένας φίλος μου από τα Χανιά:

"Τοπικιστής είναι αυτός που θεωρεί την πόλη του ομορφότερη από τα Χανιά"

 :harhar:

προς stedes--->μα δεν κανω συγκρηση της βιβλιοθηκης του ΑΠΘ με την δικη μου (δεν μου αρεσει να συγκρινω ετσι κι αλλιως διαφοερετικα πραγματα) απλα λεω τι μου αρεσει στο στο ΕΜΠ.ειναι το μοναδικο μερος που μπορω να διαβασω μαθηματικα  :)
προς ολους---> ΚΑΛΗ ΧΡΟΝΙΑΑΑΑΑ  :D

χρόνια πολλά με πολλές επιτυχίες και μια υπέροχη εικόνα για wallpaper!!!

να και η βιβλιοθηκη που λεγαμε . Θανο ΜΜ ειμαι  :)  ομορφη εικονα εβαλες

lipon.....katarxin i  maxi ine anisi...
parolayta..emis..oi 300 ap8ites :chief:...8a ta vgalume pera...

@nikos
[hide]orea geometria...[/hide]
@thatin
[hide]apagoreuonte ta optikoakustika mesa.... :spam: [/hide]
@stedes (1)
[hide]o fotis dn nomizo na estile s ayto to topic...ligo akiro ine... :P [/hide]
@mostel
[hide]mpa.. stamatisa na to pezo...tin proti fora+ego den iksera ti eprepe na kano kai ekana 1,3!!!!!meta :read: tis odigies....[/hide]
@kostas
[hide]i propagandistiki politiki pu akolu8ite den 8a karpoforisi....
[quote]μα δεν κανω συγκρηση της βιβλιοθηκης του ΑΠΘ με την δικη μου [b][u](δεν μου αρεσει να συγκρινω ετσι κι αλλιως διαφοερετικα πραγματα)[/u][/b][/quote]     :clap2:  :heli:  :clap:  :ewpu:  [/hide]
@kosta+thatin
[hide]ti....mas kanete... :flex: ???siga.....[/hide]
@stedes(2)
[hide]+ego etsi eftasa sta 37.... :rotfl: (ala den kuni8ike to pontiki...pire faltso i mpala...)[/hide]
@all
[hide]happy new year!!welcome...$2^{11}-7^{2}+2^{3}=...$[/hide]

προς RAKAR+STEDES --->δε θεωρω κανενα ΕΜΠ καλυτερο. κινδυνευω να παρεξηγηθω ,δεν ξανα λεω τιποτα για ΕΜΠ οκ.

Καλή χρονιά σε όλους!  :)  Εύχομαι να πετύχετε όλοι τους δικούς σας στόχους το νέο έτος!  :wink:

Κώστα ό,τι είπαμε το είπαμε για πλάκα, λόγω μιας κόντρας που έχουμε με το Θάνο σχετικά με τα πανεπιστήμια. :P  Πες ό,τι θες για το ΕΜΠ, σιγά μη σε παρεξηγήσουμε.  :)

p.s.[hide="@RAKAR"] Έστειλε μήνυμα και σε αυτό το topic ,καταλάθως βέβαια ,αλλά έστειλε.  :P[/hide]

[quote="kostas"]προς RAKAR+STEDES --->δε θεωρω κανενα ΕΜΠ καλυτερο.[/quote]

egw mono ena EMP!!!

[hide="@stedes"]apth...! p->l, den mporw na anextw mia tetoia ntropi! egw na p thessaloniki![/hide]

[hide="@RAKAR"]pws eiste 300? olo ta apth 300? giati den to pisteuw?[/hide]

[hide="@kostas"]thanx! kati tetoio thimomoun alla den imoun sigouros... pws kai apo kei? hm... an thimamai kala mas nikisate sto paintball :mad: ? :rotfl: [/hide]

[quote="thatin"]

[hide="@kostas"]thanx! kati tetoio thimomoun alla den imoun sigouros... pws kai apo kei? hm... an thimamai kala mas nikisate sto paintball :mad: ? :rotfl: [/hide][/quote]
εννοεις γιατι να μην παω ΗΜΜΥ ? δεν την εβαλα ουτε στο μηχανογραφικο  :)

απλώς επειδή οι περισσότεροι που έχουν σχέση με μαθ έρχονται ημμυ... :)


Find $lim [xlnx(1+2/x)]$ x tend to infinity

Just a hint.. denote $\displaystyle y = \frac{1}{x}$ and the rest is simple...

[quote=Archeron]Hello
.
.
How are you?[/quote]

why? in case you're new, don't post on threads that no one has posted on in a long time.


Se considera numerele naturale x si y, astfel incat 2x+1/6y+2 si 6y+7/2x+6.Aflati x si y.

Sorry, am corectat.Acum nu cred ca mai e chiar asa simpla. :D

Poate nu inteleg eu , dar ..astfel incat....ce?

sper sa nu gresesc

2x+1 | 6y+2 deci 2x+1 | 3y+1
6y+7 |2x +6 deci 6y+7 | X+3

dar x si y erau naturale deci 2x+1 <sau= 3y+1 , iar 6y+7 <sau= x+3
de unde x <sau= 3y/2  si x>sau = 6y+4

deci 6y+4 < 3y/2
9y<-8, contradictie cu y natural, deci pb nu are sol

Asta cred ca era o problema frumoasa de-a V-a, insa nivel de locala, poate nivel mai mic.

$\ 2x+1|6y+2$, deci $\ 2x+1\leq\ 6y+2$, adica $\ 2x\leq\ 6y+1$
$\ 6y+7|2x+6$, deci $\ 6y+7\leq\ 2x+6$, adica $\ 6y+1\leq\ 2x$
Aceste doua relatii ne dau $\ 2x=6y+1$, relatie imposibila, caci $\ 2x$ este par, in timp ce $\ 6y+1$ este impar.(lucram in $\mathbb{N}$)


topology
search
superior algebra
superior algebra unsolved

Let $ G$ be an infinite compact topological group with a Hausdorff topology. Prove that $ G$ contains an element $ g \not\equal{} 1$ such that the set of all powers of $ g$ is either everywhere dense in $ G$ or nowhere dense in $ G$.

[i]J. Erdos[/i]


function
combinatorics unsolved
combinatorics

Any help massivly appreciated!  I have been stuck on this problem for months.

A loop to iterate through all combinations of 7 cards

for int c1 = 0; c1 < 52
        for int c2 = c1 + 1; c2 < 52
                for int c3 = c2 + 1; c2 < 52
                        ...
                        permutationIdex++;

 Given an ordered set of 7 integers, EG:

0, 1, 2, 3, 4, 5, 6
4, 5, 6, 10, 11, 40, 51
4, 5, 6, 10, 11, 41, 42

 Convert their values into their respective permutation index.  EG:

The following permutations are the first ones that appears:

Permutation     Index
0,1,2,3,4,5,6   0
0,1,2,3,4,5,7   1
0,1,2,3,4,5,8   2

 The following appears to work correctly for 2 cards

returnValue = 52 - c6 * c6 + ((c6 - 1*c6 - 1) / 2 + c6 - 1 / 2) + (c7 - c6 - 1);

and it doesnt work for 3 cards?

Also, could you copy paste ALL of the code lines you wrote in the program entirely(no "..."), without typing them again here ?

it is possible you wrote the code wrong, and the program doesnt do what you want it to because of that

because i already see what may be a potential problem :

[quote="Gullanian"]Any help massivly appreciated!  I have been stuck on this problem for months.

A loop to iterate through all combinations of 7 cards

for int c1 = 0; c1 < 52
        for int c2 = c1 + 1; c2 < 52
                for int c3 = c2 + 1; [b][color=red]c2 < 52[/color]                 [/b]       ...
                        permutationIdex++;[/quote]

You wrote that the 3rd loop will go until the 2nd card reaches 51
not until the 3rd card reaches 51...

so it works for the first two cards, but the 3rd card it overcounts

Dear Lena,

Thanks for your time!  The problem was just a typo, my program works fine, I just wrote it up without the code handy in a hurry.  Here is the correct full loop:

for int c1 = 0; c1 < 52
for int c2 = c1 + 1; c2 < 52
for int c3 = c2 + 1; c3 < 52
for int c4 = c3 + 1; c4 < 52
for int c5 = c4 + 1; c5 < 52
for int c6 = c5 + 1; c6 < 52
for int c7 = c6 + 1; c7 < 52

permIndex++;

One thing i dont understand

You said "a set of ordered numbers"

but they way you wrote the loop

the first card will always be 0 , becase it never goes up (you neglected to write c1++;   )

the first card may be higher than the 2nd card
the 2nd card may be higher than 3rd
3rd card may be higher than 4th

but sets of this type :

51, 48, 16, 5, 3, 2,1 are not acceptable ?

are sets of this type acceptable ? : 1,1,1,1,1,1,1

After looking at this for a couple of hours, i finally figured it out

Unfortunately, upon testing i found out that my computer cant do more than 3769 things at once :(

so the test was incomplete

but it works fine unti 0 1 2 3 7 21 52 (which is 3768 recursive steps)
upon 0 1 2 3 7 22 23 it gets stuck (its step number 3769)

it works fine for 0 1 2 3 6 22 23 so it should in theory work for the other terms

[code]import java.util.Scanner;
public class Cards
{static int pIndex;
 /**
  * Returns the number of steps it takes to get from 0123456 to 46 47 48 49 50 51 52  */

 public static void main(String[] args)
 { System.out.println("Enter permutation ( 7 numbers)");
     Scanner scan = new Scanner (System.in);
     int[] C = new int [7];
     for (int i=0; i<7; i++)
        C[i]=scan.nextInt();
     for (int i=0; i<7; i++)
       System.out.print(C[i]+" ,");
     front(C);
}//main

public static int rPermutation(int[] Cards, int c1, int c2, int c3, int c4, int c5, int c6, int c7)
    {if ( c1==Cards[0] && c2==Cards[1] && c3==Cards[2] && c4==Cards[3] && c5==Cards[4] && c6==Cards[5] && c7==Cards[6])
        return pIndex;
      else if (c7<52)
                                    {pIndex=1+rPermutation(Cards,c1,c2,c3,c4,c5,c6,c7+1);
                                    //increaces the 7th card until max
                                }
                      else if(c6<51)
                                     {if(c7==52)
                                         pIndex=1+rPermutation(Cards,c1,c2,c3,c4,c5,c6+1,c6+2);
                                       // increases 6th card, when 7th is maxed
                                    }
                                   else if(c5<50)
                                                {if(c6==51)
                                                    pIndex=1+rPermutation(Cards,c1,c2,c3,c4,c5+1,c5+2,c5+3);
                                                //increases 5th card when 6th and 7th are maxed
                                            }
                                                    else if(c4<49)
                                                                {if(c5==50)
                                                                    pIndex=1+rPermutation(Cards,c1,c2,c3,c4+1,c4+2,c4+3,c4+4);
                                                                //increases 4th card when 5,6,7 cards are maxed
                                                            }
                                                                else if(c3<48)
                                                                        {if(c4==49)
                                                                            pIndex=1+rPermutation(Cards,c1,c2,c3+1,c3+2,c3+3,c3+4,c3+5);
                                                                        //increases 3rd card when 4,5,6,7 are maxed
                                                                    }
                                                                        else if(c2<47)
                                                                                {if(c3==48)
                                                                                pIndex=1+rPermutation(Cards,c1,c2+1,c2+2,c2+3,c2+4,c2+5,c2+6);
                                                                                //increases 2nd card when 3,4,5,6,7 are maxed
                                                                            }
                                                                                else if(c1<46)
                                                                                {if(c2==47)
                                                                                pIndex=1+rPermutation(Cards,c1+1,c1+2,c1+3,c1+4,c1+5,c1+6,c1+7);
                                                                                // increases 1st card when 2,3,4,5,6,7 are maxed
                                                                                }
      return pIndex;
      // returns the number of steps it took to get from 0123456 to the desired number
                                                                                }//rPermutation
 public static void front(int[] Cards)
{int c1=0;
    pIndex=0;
  System.out.println("Index "+  rPermutation(Cards,c1,c1+1,c1+2,c1+3,c1+4,c1+5,c1+6));
}//front
                                                                                }//Cards[/code]

If what I understand about your problem is correct, then it is very simple.
Given the sequence $\{a_{i}| 1 \leq i \leq k\}$ which is a subset of $\{i | 0 \leq i < n\}$ where $k \leq n$, find its index in a numerical order of all such sequences.

Let $a_{0}=-1$
Then the index of the sequence is $\sum_{i=1}^{k}{ \sum_{j=a_{i-1}+1}^{a_{i}-1}{ \binom{n-1-j}{k-i}}}$, because for each $a_{0}<x<a_{i}$ there are $\binom{n-1-x}{k-i}$ sequences that begin with the subsequence $\{a_{1}, a_{2}, ... , a_{i-1}, x\}$
For your problem $(n,k)=(52,7)$

By the way, Lena, your code is correct if and only if there are 53 cards (0 to 52) :)

In C++ (You shouldn't use recursion :wink:):
[code]
#include<stdio.h>
long ncr(int n,int r)
{
	long p;
	int a;
	p=1;
	for(a=1;a<=r;a++) p=p*(n+1-a)/a;
	return p;
}
int main()
{
	int n,k;
	int i,j;
	long o;
	int a,s;
	scanf("%d%d",&n,&k);
	o=0;
	j=-1;
	for(a=1;a<=k;a++)
	{
		scanf("%d",&i);
		for(s=j+1;s<i;s++) o+=ncr(n-1-s,k-a);
		j=i;
	}
	printf("%d\n",o);
	return 0;
}
[/code]

Basic code just in case you can't understand C++:
[code]
function ncr(n,r)
	p=1
	for a=1 to r
		p=p*(n+1-a)/a
	next a
	ncr=p
end function

input "n = ",n
input "k = ",k
o=0
j=-1
for a=1 to k
	input i
	for s=j+1 to i-1
		o=o+ncr(n-1-s,k-a)
	next s
	j=i
next a
print "index =";o
[/code]

The C++ code works, so the Basic code should too :P

By a similar method you can also get the sequence from its index.

Thx!

The number of cards is fixable

I used Java, and not C++, i dont know C++ language so i used what i do know ;)

For some reason thougn, i cant check the program for 0 1 2 3 7 21 52

too many recursion steps

even when i wrote a program to do the following

for int i=1 to 4000 println i

it wouldnt do it

and i just wonder why... ?

Thank you so so so much for your help, it is massivly appreciated, and I am extremely grateful for the volountary time you have both offered.

I will experiment with your solutions now to see if they work.

However, for my project speed is of uptmost importance, this software is going to be used for heavy statistical analysis.  When using the code I expect it to work, but I am 99% sure an algebraic solution will be massivly faster than using loops... is an algebraic solution to this problem possible?

[quote="weiquan"]If what I understand about your problem is correct, then it is very simple.
Given the sequence $\{a_{i}| 1 \leq i \leq k\}$ which is a subset of $\{i | 0 \leq i < n\}$ where $k \leq n$, find its index in a numerical order of all such sequences.

Let $a_{0}=-1$
Then the index of the sequence is $\sum_{i=1}^{k}{ \sum_{j=a_{i-1}+1}^{a_{i}-1}{ \binom{n-1-j}{k-i}}}$, because for each $a_{0}<x<a_{i}$ there are $\binom{n-1-x}{k-i}$ sequences that begin with the subsequence $\{a_{1}, a_{2}, ... , a_{i-1}, x\}$
For your problem $(n,k)=(52,7)$[/quote]

according to this, this is the algebric solution

but i dont understand it yet, as i've just begun to study it

i'm only on my 3rd semester in university, and i cant figure out what this means

$\sum_{k=0}^{n}3^{k}*k*\binom{n}{k}$

so....

[quote="Gullanian"]Weiquan, your solution only works for 2 cards, correct?  It needs to work with 7 which is the problem.[/quote]

Erm... I think I said in my earlier post that it works for $k$ cards chosen from $n$ cards. So it works for $(n,k)=(52,7)$.

[quote="Gullanian"]Also, is it possible to do this algebraicly without the need of loops anywhere?  Speed is of uptmost importance![/quote]

This is already very fast. It is $O(nk^{2})$ if you want to know, which in your case of 7 from 52 cards it should take less than 0.0001s :lol:

By the way, Lena, if you still want to use recursion, you should do it the following way, even though it will be much slower:
[code]
#include<stdio.h>
int n,k;
int i[1024];
long c;
void test(int p,int x,int l)
{
	int m;
	int a,s;
	if( p==k ) { c++; return; }
	if( l==0 ) m=n-1;
	else m=i[p];
	for(a=x+1;a<=m;a++)
	{
//		for(s=0;s<p;s++) printf(" "); printf("%d\n",a);
		if( l==1 && a==i[p] ) test(p+1,a,1);
		else test(p+1,a,0);
	}
}
int main()
{
	int a;
	scanf("%d%d",&n,&k);
	for(a=0;a<k;a++) scanf("%d",&i[a]);
	c=-1;
	test(0,-1,1);
	printf("%d\n",c);
	return 0;
}
[/code]
scanf("%d%d",&n,&k); means input integers $n$ and $k$
printf("%d\n",c); means print integer $c$ and a new-line
// means line-comment, which in this case is to show you the recursive calls and their depths if un-commented
The rest is almost the same as Java.
(And C++ is good and fast, try Dev C++ :wink: )

Dear Weiquan,

Thank you so, so, much!  It works perfectly!  Many many thanks for your time!

Tom

[quote="Gullanian"]One thing though, I am running it, and it is quite slow (for what I am doing).  I'm going to attempt to speed it up by precomputing the choose function answers, but is it possible to simplify this solution into something algebraic?[/quote]

Slow?!? Then you must be doing something on the order of millions before this computation can make a difference :P. Anyway if you really need absolute speed, of course memoization will work, and will give you amortized constant time per call to the function ncr(). But no, I don't think there is a way to simplify this solution.

Dear Weiquan,

Thanks for your help!  It was slow for what I was doing (in the order of billions!) but I've gotten it up to speed now, thanks  :lol:  I've managed to reduce it to a single line formula.

Thanks again, you were a great help!

Tom

[quote="Gullanian"]Thanks for your help!  It was slow for what I was doing (in the order of billions!) but I've gotten it up to speed now, thanks  :lol:  I've managed to reduce it to a single line formula.[/quote]

Oh! What's the formula? And how did you get it?


zpv ibwf opx tpmwfe nz qvaamf uijt pof xbt tjnqmf cvu dpohsbuvmbujpot bozxbz

:spam: the jedi will get you for that :starwars: and the moderaters

I'm thinking this is an actual post.  It's probably one of those crypotgram thingies, where another letter is substituted for another, and you have to figure out what the thing means.  For example, , if x=t, c=h, and v=e, then xcv = the.  This should be moved to Other Problem Solving Topics: Puzzles and Brainteasers though.

[hide]You have now solved my puzzle this one was simple but congratulations anyway[/hide] :)

[quote="jhollenbeck"]Spam the jedi will get you for that and the moderaters[/quote]
Why? I know it's not a standard post, but cryptography and such puzzles seemed to me as a subset of computer science. You see that in the other forums they have chemistry or physics maratons, so why not having a cipher one here?

[quote="#include<math>"][hide]You have now solved my puzzle this one was simple but congratulations anyway[/hide][/quote]
You got it, of course;-)

[quote="kubus"][quote="jhollenbeck"]Spam the jedi will get you for that and the moderaters[/quote]
Why? I know it's not a standard post, but cryptography and such puzzles seemed to me as a subset of computer science. You see that in the other forums they have chemistry or physics maratons, so why not having a cipher one here?[/quote]

Well, it probably would have been helpful to explain what you were doing, to people who couldn't catch on right away.


Well i came across this question that require you to find out the number of possible words of 3 letters(all of which go from A-Z), with the condition that it should consist of atleast 1 A. the answer is 1951

To try solving this I just multiplied 26 *26* 1 because only one A is required the other possible words could be accounted by multiplying the product with 3
giving 2028

but i thought about it and realized obviously  I overcounted many situations like when you have 3 AAAs..

and i solution uses casework to solve this question
by using the number of times A has occured
1, 2, and 3 times

and then doing the subcases of  of the position if A in each of the cases(which can be omitted by just multiplying by 3)

but though this list of possible words checks for words that do not overcount as in the case of A's  it does overlook everyother overcounting situtaion

for example the solution does not overcount AAB two times but it overcounts ABB  twice or CAC or ZZA

The solution for each sub case is  done somethins like 1(for A) * 25(for letrters without A)* 25(for letrters without A) = 625 for the very first subcase something like A-- wher - is not a

the solution given above(1951) will obviouly count the same words again right??

instead a product like 1*25*24 should be done because this will give possible different words

but that is not required by the question it only ststes possible words; hence there method should be right?

Then why isn't my initial 3*(26*26*1) make sense because we don't have to have different words

Can somebody explain some crucial fact I am oevrlooking

Thank you in advance for your time and effort :blush:

[quote="modularmarc101"]Complementary counting would be a good idea. Count how many words have [b]no[/b] A's and subtract it from the total amount of words.

No A's: $ 25 \cdot 25 \cdot 25 \equal{} 15625$

All words: $ 26 \cdot 26 \cdot 26 \equal{} 17576$

$ 17576 \minus{} 15625 \equal{} \boxed{1951}$

Now if you want me to explain how to count it the normal way, tell me.[/quote]

Yes I would like to you to explain it I already know the complementary way.
it would be helpful if you give an example if possible

Thank you

[quote="modularmarc101"]For the different cases, it becomes a lot more complicated and tedious since there lots of times that you overcount since there is a repeated letter and you have to divide the problem into more problems.

Do you mean an example of complementary counting?[/quote]

No actually while doing casework exaples that go along with the steps

[quote="JBL"][quote="Pythogris"]for example the solution does not overcount AAB two times but it overcounts ABB  twice or CAC or ZZA[/quote]  This is just not true: $ 25 \times 25$ counts every ordered pair of letters other than A exactly once, so (for example) it counts BC once, CB once, and BB once.  Thus $ 1 \times 25 \times 25$ counts ABB exactly one time.  If you want to see how this works for yourself, pretend the alphabet contains only the letters A, B, C -- in this case, you can actually write out all the words and see which terms count which words.[/quote]

Yes, thank you for some reason i kept confusing over the fact that when you add 25*25*3 you are cosidering three exclusive cases :P .

That makes much more sense :lol:

Here is how to apply casework to this problem:

Case 1: There exists exactly one $ A$.

There are $ 25 \times 25$ choices for the two other letters, and the $ A$ can go in exactly $ 3$ locations, so there are $ 25\times 25\times 3 \equal{} 1875$ ways to have exactly one $ A$.

Case 2: There exist two occurrences of the letter $ A$.

There are $ 25$ choices for the third letter, and the letter that isn't $ A$ can appear in one of $ 3$ locations, so there are $ 25\times 3 \equal{} 75$  ways to have exactly two occurrences of the letter $ A$.

Case 3: There exist three occurrences of the letter $ A$.

Trivially, there is only one way for this to occur.

We have $ 1875\plus{}75\plus{}1\equal{}\boxed{1951}$ ways to have at least one $ A$.


Let $ u_n \equal{} \frac {1}{\sqrt [4]{n \plus{} 1}}$ , $ n\in N$ and  $ S_n \equal{} \sum_{k \equal{} 1}^{n} u_k$
Find $ S_{2008}$?

I don't think there's a nice way to compute that sum (without a computer). the best we can do is estimate it.

hello, or he has forgotten the floor-function in the problem.
Sonnhard.


Form quadratic equation with roots $ a^{229}$ and $ a^{1004}$ if $ a$ and $ b$ are roots of $ x^2 \plus{} x \plus{} 1 \equal{} 0$

a is a third root of unity, hence $ a^{229} \equal{} a$ and $ a^{1004} \equal{} a^2 \equal{} b$ since $ a^2$ is a root of the equation, and it can't equal $ a$, since then $ a^2 \equal{} a$ <--> $ a(a \minus{} 1) \equal{} 0$ But $ a$ can't be 0 or 1, since neither is a root of $ x^2 \plus{} x \plus{} 1 \equal{} 0$, so therefore, $ a^2 \equal{}$ the other root $ \equal{} b$

Hence the quadratic is just $ x^2 \plus{} x \plus{} 1$


αν α θετικός αριθμός,
να δειχθεί ότι η εναλλασουσα σειρά

a-a^1/2+a^1/3-a^1/4+...  συγκλίνει    (εντάξει αυτό ειναι εύκολο)

αλλά!!!
οι παρακάτω σειρές συγκλίνουν και οι 2

 (a-a^1/2)+(a^1/3-a^1/4)+...

  a-(a^1/2-a^1/3)-(a^1/4-a^1/5)-...

Η πρώτη σειρά που είπα πως αποδεικνύεται εύκολα ότι "συγκλίνει", ήθελα να πω πως "αποκλίνει" βέβαια!

Το πρόβλημα έχει σχέση με το άπειρο...
Αν πάρεις τα αθροίσματα για πεπερασμένο πλήθος τότε πράγματι έχουν το ίδιο άθροισμα.Όταν όμως μπαίνει στη μέση το άπειρο τα πράγματα αλλάζουν.Οι δύο σειρές δεν δημιουργούνται από την ίδια ακολουθία οπότε δεν προσθέτουμε για κάθε $ n$ τα ίδια πράγματα οπότε μπορεί να μην συγκλίνουν και οι δύο.
Υπάρχουν κριτήρια που σου λένε πότε μία αναδιάταξη μιας σειράς συγκλίνει...

Πιστεύω να μην σου δημιούργησα περισσότερες απορίες.... :wink:

To pws perneis ena athrisma , tin sira twn orwn tis , otan milame gia apira athrismata pezei poli simantiko rolo...Den kserw ti gnorizeis peri seirwn pantws isxiei to eksis , kathe seira pou den sigklinei apolita tote kathe anadiataksi tis siras sou dinei diaforetiko apotelesma kai malista etsi mporeis mia seira na tin stileis se opoiodipote arithmo theleis....!!!!


Find $ x,y,z,t \in Z$ of this equation: $ 5(x\plus{}y\plus{}z\plus{}t)\plus{}10\equal{}2xyzt$

$ 5(x \plus{} y \plus{} z \plus{} t \plus{} 2) \equal{} 2xyzt$
Does it help?
Lets solve it first in positive integers
WLOG $ x \le y \le z \le t$
$ x \equal{} 0$ possible solutions are $ (x,y,z,t) \equal{} (0,y,z,2 \minus{} y \minus{} z)$ for all $ y,z \in \mathbb {Z}$ or its permutations.

then $ 0 < x \le y \le z \le t$

If $ 3 \le x$ then $ 5(x \plus{} y \plus{} z \plus{} t \plus{} 2) < 2xyzt$

$ 5(x \plus{} y \plus{} z \plus{} t \plus{} 2) \equal{} 70 \plus{} 5(a \plus{} b \plus{} c \plus{} d)$ with $ a,b,c,d \ge 0$
$ 2xyzt \equal{} 2(3 \plus{} a)(3 \plus{} b)(3 \plus{} c)(3 \plus{} d) \equal{} 162 \plus{} 54(a \plus{} b \plus{} c \plus{} d) \plus{} \cdots$

then $ x \equal{} 1$ or $ x \equal{} 2$

same argument to get a constraint for $ y$ and so on...


Putnam
probability
linear algebra
number theory
college contests

Hi all --

What topics (or names of the equivalent college courses) does the putnam cover? I am using this summer to seriously prepare for the putnam, so my plan is to hone my problem solving skills with elementary problems while expaning my knowledge by studying curriculum material so that I can have a hope of doing well (getting one or two correct :P) on the putnam this winter.

Textbook reccomendations, for both the problem solving aspect or the curriculum stuff, would be greatly appreciated. Thanks!

Putnam typically covers calculus/analysis, linear algebra, number theory, combinatorics, probability and some combination of those.  [url=http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml]This[/url] is your best source of problems to work on.


I'm not sure if this has been posted before, but has anyone noticed how when you rate a post "1", its overall rating goes up?

Yes. This has been mentioned.  :)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268896[/url]


calculus
integration
trigonometry
real analysis
real analysis unsolved

The following integral is very nice for me. Compute it: \[\int_{0}^{\frac{\pi }{2}}{\frac{{dx}}{{1+\tan (x)^{\sqrt 2 }}}}\]

I don't think you need to have $\sqrt{2}$ here. For any real $\alpha$, we have:

$\begin{array}{ll}\int_{0}^\frac{\pi}{2}\frac{dx}{1+\tan(x)^{\alpha}}&=\int_{0}^\frac{\pi}{2}\frac{\cos(x)^\alpha}{\sin(x)^\alpha+\cos(x)^\alpha}dx\\\\&=\int_{0}^\frac{\pi}{2}\frac{\cos(\frac{\pi}{2}-x)^\alpha}{\sin(\frac{\pi}{2}-x)^\alpha+\cos(\frac{\pi}{2}-x)^\alpha}dx\\\\&=\int_{0}^\frac{\pi}{2}\frac{\sin(x)^\alpha}{\cos(x)^\alpha+\sin(x)^\alpha}dx\\\\&=\frac{1}{2}\int_{0}^\frac{\pi}{2}\frac{\cos(x)^\alpha+\sin(x)^\alpha}{\cos(x)^\alpha+\sin(x)^\alpha}dx\\\\&=\frac{\pi}{4}.\end{array}$

This result is quite old though - it must have been posted here before.


Who knows what is the Eiffel Towers volume? When it's standing and when it's smelted.
Maybe to figure out the volume of it when it's standing is by using pyramid formula? When it's smelted I've though of taking the weight of different tower ingredients and see how much volume each ingredient uses, and that way see how much volume in all.

What I need is for someone give their ideas of how to calculate the volume of the Eiffel Tower. Both when it's standing and when it's smelted. Would appreciate the answer or just a hint.

[quote="AndrewTom"]I'm not sure how to post a web-link, but here goes:

There is a site called "The official site of the Eiffel tower" at http://www.tour-eiffel.fr/teiffel./uk

This has lots of data which might help you.

Good luck.[/quote]

Yeah, there are some statistic that maybe be used in calculation. But without formula they're useless...


quadratics
calculus
derivative
function
special factorizations

[quote="willydavidjr"]Suppose you have the point for the center of the circle having x as $-\frac{k}{2}$ and y as $\frac{-1-k}{2}$ and a radius of $\sqrt\frac{2k^2+6k+5}{4}$. What would be the value for k? Hydro answer it as $-\frac{3}{2}$ but I am wondering how did he get it...[/quote]

Is the question "The center of a circle is at $(-\frac{k}{2},\frac{-1-k}{2})$ and the radius is $\sqrt\frac{2k^2+6k+5}{4}$. What is $k$?" There should be some other restriction, or else k could be any number, right?

There is no other restriction. What I am looking forward is that how did hydro get $-\frac{3}{2}$? Substituting this value to the radius of the circle and we get $\frac{\sqrt2}{4}$ as the radius of the circle...How did he get k?

This is the original equation: $(x+\frac{k}{2})^2+(y+\frac{1+k}{2})^2=\frac{2k^2+6k+5}{4}$

Well, I'd start with this:

[hide]Rewrite it as $\left(\frac{2x+k}{2}\right)^2+\left(\frac{2y+k+1}{2}\right)^2=\frac{2k^2+6k+5}{4}$
This gives $\frac{(2x+k)^2}{4}+\frac{(2y+k+1)^2}{4}=\frac{2k^2+6k+5}{4}$, so now we know that $4x^2+4xk+4y^2+4yk+4y+2k^2+2k+1=2k^2+6k+5$ and $4x^2+4xk+4y^2+4yk+4y=4k+4 \implies x^2+xk+y^2+yk+y=k+1$...

Not really sure where to go from there.  Is there some quadratic thing that I'm not seeing how to apply it right now since I'm so tired, like completing the square or something?[/hide]

I believe you are referring to the problem that you posted in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85489]this thread[/url].

Basically, hydro found the value of $k$ such that $\frac{2k^2+6k+5}{4}$ is minimized.
Since $2k^2+6k+5=2(k^2+3k)+5=2(k+\frac{3}{2})^2+\frac{1}{2}$, the value of $k$ that minimizes this expression is $\frac{-3}{2}$, which gives $r^2=\frac{0+\frac{1}{2}}{4}=\frac{1}{8}$, or $r=\frac{1}{\sqrt{8}}=\frac{\sqrt{2}}{4}$.

excuse me willydavidjr I didn't see this post before...

to find $k=-\frac{3}{2}$ I just did the derivative of the function $f(k)=\frac{2k^2+6k+5}{4}$ finding that value for minimum. Then, $f\left(-\frac{3}{2}\right)=\frac{1}{8}$, but in the original equation the function represented the square of the radius, so its minimum value is $\frac{1}{\sqrt{8}}$

Is there a way to find that without calculus?

Now I see hydro. I thought you use the way frt do it. What we get is

$2(k+ \frac{3}{2})^2 +\frac{1}{2}$

so the possible values for k from that equation is $(k+ \frac{3}{2})$
will be $k=-\frac{3}{2}$. Am I right Frt?

But now I see that we can use derivatives on arriving the same data.

Yeah, that's the idea.
Since $(k+\frac{3}{2})^2$ cannot be negative, its minimum value is $0$, which is obtained when $k=-\frac{3}{2}$.

BTW, in my previous post, I wrote that "minimum value of $k$ is $-\frac{3}{2}$", but what I meant was "the value of $k$ that minimizes the expression is $-\frac{3}{2}$". (I edited my post now.)

But Frt, is there a time that we need to multiply the value of k or x to their coefficient, like on this thread, the coefficient is 2 ($2(k+\frac{3}{2}$) to get the value of k or x. Or it will be as is just like what we did to get the value of k?

The value of k will be -4 because I multiply it to its coefficient 2. Is this right?

Once we see that $(k+\frac{3}{2})^2$ is minimized when $k=-\frac{3}{2}$, we simply substitute $k=-\frac{3}{2}$ into $\frac{1}{4}\left(2(k+\frac{3}{2})^2+\frac{1}{2}\right)$ to get the minimum value of the original expression ($r^2$).

So minimizing the equation means removing everything even the coefficient "2" to get the value of k?

[quote="frt"]Read my post ;)[/quote]

Oooh.  Oops.

So, on to the matter at hand.  It's not really removing the coefficient, because when $k=-\frac{3}{2}$, it will be multiplied by $0$ anyway...

No, that would be $r=0$.  In this problem, since $r=\sqrt{\frac{2k^2+6k+5}{4}}$, if $k=0$, then $r=\frac{\sqrt{5}}{2}$, and the circle is not degenerate...


what are the largest pythagorean triple in which two numbers are consecutive and each number is  less than 115?

Thanks for posting the problem.

I see you have edited the first problem statement I saw, and now I will ponder how to solve it, perhaps to put the solution in spoiler text tomorrow. The level of this question looks just right for this forum.

[hide]15,112,113

lets

hmm say the middle term is 2*u*v so lets try 112 since its the first possible even number.

it factors 2*2^3*7

2^3^2- or +7^2 looks good[/hide]

Solution:

[hide]

d :^2: + x :^2: = (x+1) :^2:

d :^2: + x :^2: = x :^2: + 2x + 1

So d =  :sqrt: (2x+1)

For d to be an integer, x must half of one less than an odd perfect square. So possible values of x are 4, 12, 24, 40, 60, 84, 112, 144.....

We need x<115, so x = 112. This means that d =  :sqrt: [2(112)+1] = 15. So the triple is 15-112-113.

[/hide]

[hide]i just counted up from those that have two consecutive numbers. 3,4,5  5,12,13   7,24,25  9,40,41  11,60,61  13,84,85  15,112,113

If you had to do it for a larger number, i would've probably used formulas or algebra or some other math to do it.[/hide]

A neat way to find Pythagorean triples in which the last two numbers are consecutive is to take the Nth odd number for the first number (not counting 1), multiply the Nth triangular number by 4, add one to the previous term.

3, 1*4, 1*4 +1
3,4,5

5, 3*4, 3*4+1
5,12,13

So, experiment with that to find what number works.

Triangular numbers: 1, 3, 6, 10, 15, 21, 28

15, 28*4, 28*4 +1
15, 112, 113

I think I said that right.


logarithms
inequalities
induction
algebra proposed
algebra

Let n be a natural number and $ n\geq5$. Prove the inequality:

$ \log_23\plus{}\log_34\plus{}\cdots\plus{}\log_n(n\plus{}1)>n$

[quote="sandu2508"]Let n be a natural number and $ n\geq5$. Prove the inequality:

$ \log_23 \plus{} \log_34 \plus{} \cdots \plus{} \log_n(n \plus{} 1) > n$[/quote]
Induction works here.Notice that $ log_n(n\plus{}1)>1$

[quote="Allnames"][quote="sandu2508"]Let n be a natural number and $ n\geq5$. Prove the inequality:

$ \log_23 \plus{} \log_34 \plus{} \cdots \plus{} \log_n(n \plus{} 1) > n$[/quote]
Induction works here.Notice that $ log_n(n \plus{} 1) > 1$[/quote]

Yes, it's true. My solution was also by induction.