floor function
ratio
geometry
IMO
IMO 1970

Given 100 points on a plane with no three collinear. Show that at most 70% of the triangles, whose vertices (distinct) are chosen from the given points, are acute-angled.

Why 70%?

and this old book contains old IMO :D it is from IMO1970
http://www.kalva.demon.co.uk/imo/isoln/isoln706.html

Let $A(n)$ be the maximum number of acute triangles among $n$ points.  For example, $A(4) = 3$.  The solution mentioned above shows that
\[\frac{A(n)}{{n \choose 3}} \leq \frac{A(n-1)}{{n - 1 \choose 3}}\]
for $n \geq 4$.  From that inequality, we first get $A(5) \leq 7$ and then $\frac{A(n)}{{n \choose 3}} \leq 70\%$ for $n \geq 5$.

Actually, we can do a little better because the numbers are integers. By rearranging the inequality above, we get the recurrence
\[A(n) \leq \left\lfloor \frac{n}{n-3} A(n-1) \right\rfloor\]
for $n \geq 4$.  I had my computer solve that recurrence starting with $A(4) = 3$.  I found that $A(100) \leq 107811$, which means $\frac{A(100)}{{100 \choose 3}} \leq 66.6734694\%$.  It looks like we are getting a constant near $\frac{2}{3}$ for the ratio.

Question: Is $\frac{2}{3}$ the limiting ratio for the recurrence above?

Sorry for the double post, but I figured out that the limiting ratio of the recurrence above is $\frac{2}{3}$.  Namely, the exact solution to the recurrence is
\[A(n) \leq \frac{2}{3} {n \choose 3}
                    + \frac{1}{3} \left\lfloor \frac{n}{3} \right\rfloor\]
for $n \geq 4$.

That solution means that the limiting ratio for the original geometric problem is at most $\frac{2}{3}$, though it could be less.  By considering regular polygons, we can get a lower bound of $\frac{1}{4}$.  Does anyone have better bounds?

Can someone explain why the following bound holds?
\[\frac{A(n)}{{n \choose 3}} \leq \frac{A(n-1)}{{n - 1 \choose 3}}\]

 Consider $n>4$. Notice that the maximum number of acute triangles is simply equal to $A(n-1)$ multiplied by the amount of $(n-1)$ element subsets of the $n$ points, divided the number of ways of choosing $n-1$ points including a given $3$ points (representing an acute triangle). Thus, we have
$$A(n) \le \frac{A(n-1)}{{n-3 \choose n-4}} {n \choose n-1} = \frac{n}{n-3}\cdot A(n-1)$$Dividing by ${n \choose 3}$ we see that this is equivalent to $\frac{A(n)}{{n \choose 3}} \le \frac{A(n-1)}{{n-1\choose 3}}$.

[quote]Notice that the maximum number of acute triangles is simply equal to $A(n-1)$ multiplied by the amount of $(n-1)$ element subsets of the $n$ points, divided the number of ways of choosing $n-1$ points including a given $3$ points (representing an acute triangle).[/quote]

Sorry I'm still not getting... can you elaborate a little further I don't understand...

See that $A(n)$ is the maximum amount of acute triangles. So $A(n-1)$ multiplied by the amount of $(n-1)$ subsets gives us the number of acute triangles in each subset. This is $\binom{n}{n-1}$ But we overcounted. There are many subsets of $(n-1)$ points that contain any given acute triangle. We can count this as $\binom{n-3}{(n-1)-3}$ subsets cover this acute triangle. Thus divide the number of acute triangles to get the number of distinct acute triangle.

We can do way better than $70\%$.

Polish MO somewhen between 1970 and 1974

Full solution showing that we can get arbitrarily close to $2/3$:

Let $f(n)$ be the maximum number of acute triangles in $n$ points. We will show that $f(n)/\binom{n}{3}<2/3+\epsilon$ for sufficiently large $n$, for any fixed $\epsilon>0$.

Consider $n$ points, and let $T$ be the sum over all $n-1$ subsets of the points of the number of acute triangles in that subset. We see that $T\le \binom{n}{n-1}f(n-1)$. However, each triangle is counted $\binom{n-3}{n-4}$ times, so we have that the number of acute triangles is at most $\frac{n}{n-3}f(n-1)$. Thus, $f(n)\le\frac{n}{n-3}f(n-1)$. In fact, we must have
\[f(n)\le\left\lfloor\frac{n}{n-3}f(n-1)\right\rfloor.\]
Note that with $4$ points, there must always be at least one obtuse triangle, so $f(4)=3$. Let $a(n)$ be the function defined by $a(4)=3$ and $a(n)=\left\lfloor\frac{n}{n-3}a(n-1)\right\rfloor$. Clearly, we have that $f(n)\le a(n)$. We claim that in fact, $a(n)=\frac{2}{3}\binom{n}{3}+\frac{1}{3}\left\lfloor n/3\right\rfloor$. The base case of $n=4$ is true. It is easy to see that this quantity is an integer, so we see that
\[\frac{2}{3}\binom{n}{3}+\frac{1}{3}\left\lfloor n/3\right\rfloor=\left\lfloor\frac{2}{3}\binom{n}{3}+\frac{n}{n-3}\frac{1}{3}\left\lfloor\frac{n-1}{3}\right\rfloor\right\rfloor\iff 0=\left\lfloor\frac{n}{3(n-3)}\left\lfloor\frac{n-1}{3}\right\rfloor-\frac{1}{3}\left\lfloor\frac{n}{3}\right\rfloor\right\rfloor,\]
which can be checked by casework on $n\pmod{3}$. Therefore by induction, we have that $a(n)=\frac{2}{3}\binom{n}{3}+\frac{1}{3}\left\lfloor n/3\right\rfloor$, so $f(n)/\binom{n}{3}\le\frac{2}{3}+\epsilon(n)$ where $\epsilon(n)\to 0$ as $n\to\infty$. Thus, by taking $n$ sufficiently large, we are done.

We will prove a stricter bound. Let $a_n$ be the maximum number of acute triangles for $n$ points. We know that $a_3 =1$ and $a_4 = 3$. We claim that $a_n \le \frac{2}{3} \binom{n}{3} + \frac{1}{3} \lfloor \frac{n}{3} \rfloor$, and we can prove this with induction. The base case is already done. Now, suppose the claim is true for $n=k$.  In a set of $k+1$ points, there are at most $(k+1) a_k$ acute triangles in total in the subsets with exactly $k$ points.  Each of these triangles is counted $\binom{k-2}{k-3} = k-2$ times. Therefore,
\[ a_{k+1} \le \frac{k+1}{k-2} a_k \implies a_{k+1} \le \left\lfloor \frac{k+1}{k-2} a_k \right\rfloor  \le \left\lfloor \frac23 \binom{k+1}{3} + \frac13 \frac{k+1}{k-2} \left\lfloor \frac{k}{3} \right\rfloor \right\rfloor \]
We note that this is equal to $\frac23 \binom{k+1}{3} + \frac13 \left\lfloor \frac{k+1}{3} \right\rfloor$ because we have
\[ \left\lfloor \frac{k+1}{3(k-2)} \left\lfloor \frac{k}{3} \right\rfloor - \frac{1}{3} \left\lfloor \frac{k+1}{3} \right\rfloor \right \rfloor = 0 \]
 for $n >4$. This completes the induction, and we see that the fraction of acute triangles is at most $\frac{2}{3} + \epsilon$ for sufficiently large $n$.

@above @2above what's the motivation that the geometry is [i]mostly irrelevant[/i], and that the bound + optimizations is somehow enough to give a 2/3 bound? I tried this for a long time but tried to get more geometric statements (i.e. looking at convex hulls to bound the number of obtuse angles), but i couldn't see how to proceed without mass algebra bash + a bunch of hours wasted.

Solved with [b]nukelauncher[/b], [b]Th3Numb3rThr33[/b].

I'll prove that for $n$ large, the upper bound for fraction of acute triangles approaches $\tfrac23$. The key is the following claim:

[color=red][b]Claim:[/b][/color]      If for any \(n\) points (in general position), at most \(c\) triangles are acute, then for any \(n+1\) points (in general position), at most \(c\cdot\frac{n+1}{n-2}\) triangles are acute.

[i]Proof.[/i]      Select any point \(v\) among the \(n+1\) points. Then there are at most \(c\) acute triangles not containing \(v\). Summing over all \(v\), we count each acute triangle \(n-2\) times, so the number of acute triangles is at most \(c\cdot\frac{n+1}{n-2}\). \(\blacksquare\)

So if \(c_n\) denotes the maximum number of acute triangles for any \(n\) points in general position, then \[c_{n+1}\le\left\lfloor\frac{n+1}{n-2}\cdot c_n\right\rfloor.\]

We can now verify via induction that \[c_n\le\frac23\binom n3+\frac13\left\lfloor\frac n3\right\rfloor,\] and the result follows.

Let $f(n)$ be the maximum number of acute triangles in a configuration of $n$ points. Note that clearly $f(4)\leq 3$ by considering the convex hull.

$\textbf{Claim: }$
\[f(n)\leq \frac{n}{n-3}f(n-1)\]

$\textbf{Proof: }$
We double count the $X$, which is the number of pairs of (acute $\triangle,S$) where $S$ is a set of $n-1$ points which contain the $\triangle$.
\[(n-3)\cdot f(n)\leq X\leq n\cdot f(n-1)\]
Thus, $f(n)\leq \frac{n}{n-3}f(n-1)$.$\square$.

$\textbf{Claim: }$
\[f(n)\leq \frac{1}{3}\left(2\binom{n}{3}+\lfloor \frac{n}{3} \rfloor \right)\]
$\textbf{Proof: }$
We induct. The base case is clear, $f(4)\leq 3 = \frac{1}{3}\left(2\binom{4}{3}+\lfloor 3/3\rfloor \right)$. For the inductive step,
\[f(k) \leq \frac{k}{k-3}f(k-1) \leq \frac{1}{3}\left( \frac{k}{k-3}\cdot 2\binom{k-1}{3}+ \frac{k}{k-3}\lfloor\frac{k-1}{3}\rfloor\right)\]
which equals
\[= \frac{1}{3}\left(2\binom{k}{3}+ \frac{k}{k-3}\lfloor \frac{k-1}{3} \rfloor\right)\]
Now since $f(n)$ must be an integer, it suffices to show that
\[1+ \frac{1}{3}\lfloor\frac{k}{3}\rfloor > \frac{1}{3} \cdot \frac{k}{k-3}\cdot \lfloor \frac{k-1}{3} \rfloor\]%6: 1/3(40+2)
Which is clearly true since
\[\frac{1}{3}\left(\lfloor \frac{k-1}{3} \rfloor+ \frac{3}{k-3} \lfloor \frac{k-1}{3}\rfloor\right)< \frac{1}{3}\left(\lfloor \frac{k}{3}\rfloor+ 3\right)\]$\square$
From this claim, it is clearly true that we can get a bound that is arbitrarily close to $\frac{2}{3}$ of the triangles, and we're done. $\blacksquare$.

$\textbf{Irrelevant Observation: }$
For all $n$ that are multiples of 3, $\frac{n}{n-3}f(n-1)$ is an integer.

Let $f(n)$ denote the maximum number of acute triangles over $n$ points.

Notice that $f(4) = 3$.

Claim: For any positive integer $n\ge 3$, we have \[ f(n) \le \frac{n}{n-3} \cdot f(n-1)\]
Proof: In any set of $n$ points, there are $n$ different groups of $n-1$ points. Each of these groups has at most $f(n-1)$ triangles, and each triangle is counted $n-3$ times in these groups. Therefore, $f(n)\le \frac{n\cdot f(n-1)}{n-3}$, as desired. $\square$

Claim: $f(n) \le \frac{n(n-1)(n-2)}{9} + \frac{ \left \lfloor \frac{n}{3} \right\rfloor }{3}$
Proof: We induct on $n$. Suppose the result was true for $n-1$ for some $n > 4$. Then we have \begin{align*}
f(n) \le \frac{n}{n-3} \cdot f(n-1)  \le \frac{n(n-1)(n-2)}{9} + \frac{n}{3(n-3)} \cdot \left \lfloor \frac{n-1}{3}\right  \rfloor  \\
\le \frac{n(n-1)(n-2)}{9} + \frac{1}{3} \cdot \left \lfloor \frac{n-1}{3}\right  \rfloor + \frac{1}{n-3}  \cdot \left \lfloor \frac{n-1}{3}\right  \rfloor \\
< \frac{n(n-1)(n-2)}{9} + \frac{ \left \lfloor \frac{n}{3} \right \rfloor }{3} + 1 ,\\
\end{align*}where the last step holds because $\left \lfloor \frac{n-1}{3}\right  \rfloor  \le \left \lfloor \frac{n}{3} \right \rfloor $ and $\left \lfloor \frac{n-1}{3}\right  \rfloor  \le \frac{n-1}{3}$. Since $f(n)$ is an integer, and so is $\frac{n(n-1)(n-2)}{9} + \frac{ \left \lfloor \frac{n}{3}  \right \rfloor }{3} $, so the claim must hold true. $\square$

The fraction of triangles that are acute is at most \[ \frac{f(n)}{ \binom{n}{3} } \le \frac{2}{3} + \frac{ 6\left \lfloor \frac{n}{3} \right \rfloor}{n(n-1)(n-2)} \]

The second term on the RHS tends to zero for large $n$, which means that an upper bound for the fraction of triangles approaches $\frac{2}{3}$ as $n\to \infty$. Therefore, for each positive integer $\frac{2}{3} < C \le 1 $, there exists some $M$ such that all positive integers $n > M$ satisfy the fraction of triangles that are acute over $n$ points is at most $C n $.

This problem itself can be solved simply by noting that \[ \frac{f(n-1)}{ \binom{n-1 }{3} }  = \frac{6f(n-1)}{(n-1)(n-2)(n-3)} \ge \frac{6f(n)}{n(n-1)(n-2) } = \frac{f(n)}{\binom{n}{3}},\] so the fraction of acute triangles is nondecreasing. Since $f(5) \le \frac{15}{2}$, we get $f(5) \le 7$, so $\frac{f(100}{ \binom{100}{3}} \le \frac{ f(5) }{ 10} \le 0.7 $, as desired.

We will prove that this ratio approaches $\frac{2}{3}$

Denote $f(n)$ to be the maximum number of acute triangles for $n$ points. Take a configuration of $n+1$ points and note for each point there is at most $f(n)$ acute triangles not containing that point. Now summing over all points and dividing for overcount we get that $f(n+1) \le \frac{n+1}{n-2} \cdot f(n)$. Let $f(n) = \frac{2}{3} \cdot {n \choose 3} + \epsilon$. Now since $f(n+1)$ is an integer we receive $$f(n+1) \le \left \lfloor \frac{n+1}{n-2} \cdot f(n)\right \rfloor = f(n) + \left \lfloor \frac{3 \cdot f(n)}{n-2} \right \rfloor$$ Simply substituting we recieve that $$f(n+1) = \frac{2}{3} \cdot {{n+1} \choose 3} + \epsilon + \left \lfloor \frac{3 \cdot \epsilon}{n-2} \right \rfloor$$ which finishes the problem by just taking large $n$.

Very nice problem!

We solve this by recursion: Let $a_n$ denote the max number of acute triangles in n points; note that $a_4\le 3$ (since otherwise all triangles are acute, in a convex shape the sum of angles <360, contradiction; in a nonconvex quad with WLOG D inside ABC, letting $AD\cap BC=E$ implies ADC>AEC,ADB>AEB implies ADC+ADB>AEC+AEB=180, so one of them are obtuse). Then, there are at most $a_n$ triangles not containing an arbitrary vertex v; summing over all v, note that any triangle is counted n-2 times (once in each of the other vertices), we have $a_{n+1}\le\lfloor a_n\frac{n+1}{n-2}\rfloor\implies a_n\le\frac23\binom n3+\frac13\left\lfloor\frac n3\right\rfloor$, which is accurate as n gets arbitrarily large and not a multiple of 3.

Let $T_m$ denote the maximum number of acute triangles given $m$ points. Then we will show that the ratio $\frac{T_n}{\binom{n}{3}}$ approaches $\frac{2}{3}$ as $n$ grows large.
We will first show the bound:
\begin{align*}
T_m \leq \frac{T_{m-1} \cdot m}{m-3}.
\end{align*}

To see this consider the $m+1$ sets of $m$ points, within a set of $m+1$ points. Each of these sets has at most $T_m$ acute triangles. Note that for any acute triangle it appears in $\binom{n-2}{n-3}$ such sets, so we must divide due to the overcount. Thus this gives $T_{m+1} \leq \frac{T_m \cdot (m+1)}{m-2}$. Re-indexing we find the desired relation.

Once again we bound upon the remainders. We claim that,
\begin{align*}
T_m \leq \frac{1}{3} \cdot \left( 2\binom{m}{3} + \left\lfloor \frac{m}{3} \right\rfloor \right)
\end{align*}
The proof is through induction. The base case of $T_4$ is clearly true. Now inducting assume the result holds for $m = k - 1$. We will show it holds for $m = k$. Indeed note,
\begin{align*}
\left\lfloor \frac{T_{k-1} \cdot k}{k-3} \right\rfloor &\leq \frac{1}{3} \cdot \left( 2 \binom{k}{3} + \left\lfloor \frac{k}{3} \right\rfloor \right)\\
\iff  \left\lfloor \frac{1}{3} \cdot \left( 2 \binom{k-1}{3} + \left\lfloor \frac{k-1}{3} \right\rfloor \right) \cdot \frac{k}{k-3} \right\rfloor &\leq \frac{1}{3} \cdot \left( 2 \binom{k}{3} + \left\lfloor \frac{k}{3} \right\rfloor \right)\\
\iff \left\lfloor  \frac{k(k-1)(k-2)}{3} + \frac{k}{k-3} \cdot \left\lfloor \frac{k-1}{3} \right\rfloor \right\rfloor &\leq \frac{k(k-1)(k-2)}{3} + \left\lfloor \frac{k}{3} \right\rfloor\\
\iff \left\lfloor \frac{k}{k-3} \cdot \left\lfloor \frac{k-1}{3} \right\rfloor \right\rfloor &\leq \left\lfloor \frac{k}{3} \right\rfloor\\
\iff \frac{k}{k-3} \cdot \left\lfloor \frac{k-1}{3} \right\rfloor &< \left\lfloor \frac{k}{3} \right\rfloor + 1
\end{align*}
However this is clearly true by taking cases on the remainder of $k \pmod{3}$. Now due to our claim we can see as $m$ grows large we have,
\begin{align*}
\frac{T_m}{\binom{m}{3}} \approx \frac{2}{3}
\end{align*}
and we are (finally) done.

Can someone explain to me the motive behind getting to the 2/3rd's of m choose 3 + floor m/3?

Let $a_n$ be the greatest possible number of acute triangles across all possible configurations of $n$ points.

Suppose we start with $k$ points. If we remove 1 these $k$ points, the remaining points form at most $a_{k-1}$ acute triangles, so we have at most a total of $ka_{k-1}$ acute triangles. However, each triangle is counted $k-3$ times. Thus our bound is
\[a_k \leq \left\lfloor \frac{k}{k-3} \cdot a_{k-1} \right\rfloor = \left\lfloor \frac{\binom k3}{\binom{k-1}{3}} \cdot a_{k-1}\right\rfloor,\]
from which induction tells us
\[a_k \leq \frac23\binom k3+\frac13\left\lfloor\frac k3\right\rfloor.\]
Hence the limit for our maximum goes to $\frac 23$. $\blacksquare$


calculus
geometry
geometric transformation
reflection
Support
complex analysis

Hi I am a rising senior in high school who is a bit nervous about college next year, and it would be AWESOME if you guys could suggestion some colleges for me. Even some words of wisdom would be amazing and inspiring.
I am looking for a decent math college and I am aspiring to become a math professor.

A brief preview of my application would be:
3.5 UWGPA
3.7 UCGPA
2000-SAT (800M,650CR,550W)
SATII-800M2, 800Chem, 790 Physics
One AP class this year but took 2 AP tests-both 5s
By the application deadline I would have completed single and multivariable calculus, linear alg., modern alg, and real analysis. I am also planning to complete diffy eq, complex analysis by the end of my senior year and will be taking 4 APs at school along with those classes.

I know my grades are absolutely terrible but I hope maybe my higher level math classes could overshadow it? and I plan on retaking the SAT and will be aiming for a 2150.

Anyways, what colleges would you recommend for me? What are my reaches?match?safeties?Which UCs should I apply to?
And any other bits of advice would just be amazing.
Thanks for your time and if you guys ever need something from me, feel free to ask.

[quote="polarbears"]Which UCs should I apply to?[/quote]
The fall 2010 admissions cycle for California public universities will be something outside the experiences of anyone you've ever talked to. The utter collapse of the California state budget has not spared the universities. I know the CSU system a lot better than I know the UC system but to give you an idea: the state contribution to the budget of the CSU system (all 23 campuses and 450,000 students) was \$3.0 billion for 07-08. It looks like it's going to be \$1.6 billion for 09-10. We've had many university budget crises in the past, but we've never had a budget crisis that looked like this one.

Something's going to have to give. More to the point, [i]everything[/i] is going to have to give. Student fees will go up by a lot, maybe a 25%-30% one-year change. We'll still be fairly cheap by national standards, and because we'll still be fairly cheap, that won't cover the massive drop in state support. We're also going to have to get smaller - we're going to have to admit many, many fewer students than we have been admitting.

As I said, I don't know nearly as much about the UC's, but they're affected by the same budgetary pressures. Whether they can cushion it any better than we can in terms of student admissions is something I don't know. UC fees are substantially higher than CSU fees, but those fees still don't cover the cost of the university.

CSU admissions is very mechanical and formulaic. We ask that you cover the (a)-(g) subjects in high school (the same list as for the UC), and we compute an index by the following formula: 800*GPA + SAT-M + SAT-V. The minimum for CSU admission is 2900 which is, well, very low. (Most people anywhere close to 2900 have severe trouble meeting the (a)-(g) requirements.) Your index seems to be about 4410. (Your beyond-HS courses wouldn't really make any difference in our system, at least for admissions.)

Here's the situation for 2009 admissions: Each campus has a "local guarantee area" in its neighborhood from which it must admit qualified students with a 2900 index, but but can set higher standards for non-local students. The CSULB minimum non-local index for 2009 was something like 3920 or 3950.

But that's the 2009 information. All I can tell you about 2010 is that it won't look exactly like that. I can't tell you what it will look like.

What I'm telling you about UC admissions is this: if someone says to you, "I had __ qualifications in 2008 or 2009 and I got into __," that may not be that accurate a reflection on 2010. There are many things you don't know, in some cases because no one knows.

Edit: I'm probably projecting more of the CSU situation onto the UC than is warranted. Between the higher student fees and the much greater income from grants, contracts, and gifts, the UC is probably considerably less sensitive to the state budget than we are.

[quote="Kent Merryfield"][quote="polarbears"]Which UCs should I apply to?[/quote]
The fall 2010 admissions cycle for California public universities will be something outside the experiences of anyone you've ever talked to. The utter collapse of the California state budget has not spared the universities. I know the CSU system a lot better than I know the UC system but to give you an idea: the state contribution to the budget of the CSU system (all 23 campuses and 450,000 students) was \$3.0 billion for 07-08. It looks like it's going to be \$1.6 billion for 09-10. We've had many university budget crises in the past, but we've never had a budget crisis that looked like this one.
[/quote]

$ 3.0billion to$billion!? WOW that's like halved.
Are private universities going to be impacted too?

The most famous (and richest) private universities have very large endowments, meaning large investment portfolios. Those investment portfolios lost significant value in the financial panic of 2008 and early 2009, and that has caused some budget changes for them. To a large extent this has already happened and the effects have already been seen, unlike the CSU/UC situation, which is still developing. But all private universities charge tuition and fees that is several times what UC charges, which is in turn something like twice what CSU charges - so their tuition income is important to their operation costs. Financial pressure on rich private universities might manifest itself in partial retreats from "need blind" admissions policies. There are other private colleges and universities, like some of the less well known private liberal arts colleges, that may well be primarily funding themselves from tuition.

Part of that dramatic drop in direct state support for the CSU system comes from a one-time application of about \$600 million in federal stimulus money, using it to replace state money rather than supplement it.

Oh wow, thanks for enlightening me to what's going on.

Would you be able to suggest some colleges I should apply for?
I know the situation has changed but lets pretend I was applying to school 2 years ago, which schools would I be able to get into?

As none of us know you, none of us can answer that question.  Speak to your college adviser.


trigonometry
geometry
geometric sequence

In $ \triangle ABC$, $ AB \equal{} BC$, and $ BD$ is an altitude. Point $ E$ is on the extension of $ \overline{AC}$ such that $ BE \equal{} 10$. The values of $ \tan CBE$, $ \tan DBE$, and $ \tan ABE$ form a geometric progression, and the values of $ \cot DBE$, $ \cot CBE$, $ \cot DBC$ form an arithmetic progression. What is the area of $ \triangle ABC$?
[asy]unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair D=(0,0), C=(3,0), A=(-3,0), B=(0, 8), Ep=(6,0);

draw(A--B--Ep--cycle);
draw(D--B--C);

label("$A$",A,S);
label("$D$",D,S);
label("$C$",C,S);
label("$E$",Ep,S);
label("$B$",B,N);[/asy]$ \textbf{(A)}\ 16 \qquad \textbf{(B)}\ \frac {50}{3} \qquad \textbf{(C)}\ 10\sqrt3 \qquad \textbf{(D)}\ 8\sqrt5 \qquad \textbf{(E)}\ 18$

Let $ tan^2 DBE \equal{} a$, $ tan^2x \equal{} b$. The first condition is equivalent to $ tan(DBE \plus{} x)tan(DBE \minus{} x) \equal{} tan^2DBE$, which is equivalent to
$ \frac {a \minus{} b}{1 \minus{} ab} \equal{} a$, or $ a \equal{} 1$, or $ tan DBE \equal{} 1$, or $ DBE \equal{} 45$. So $ BD \equal{} DE \equal{} 5 \sqrt {2}$. From the second condition, write cot as the recipriocal of tans, now we know $ tan DBE \equal{} 1$, so it simpifies (after a bit of algebra) to $ \frac {1 \minus{} b}{b \plus{} \sqrt {b}} \equal{} 2$, so $ b \equal{} \frac {1}{9}$ (the root 1 is extraneous), so $ tan x \equal{} 1/3$, so $ DC \equal{} AC \equal{} \frac {5\sqrt {2}}{3}$ and the answer is $ \frac {50}{3}$

[quote="cheeseyicecream"]How would you find that $\cot (45 - \alpha) = \frac {\cot \alpha + 1}{\cot \alpha - 1}$ during the test? Is this a really important identity that I should memorize? It is just that I would easily be able to derive the left side from the right, but I wouldn't be able derive the right from the left.[/quote]

You can use the tangent difference of angle identity.
[hide]
The identity is:

$\tan(x-y) = \frac{\tan{x} - \tan{y}}{1+\tan{x}\tan{y}}$

So, we know that

$\cot(x-y) = \frac{\tan{x} - \tan{y}}{1+\tan{x}\tan{y}}$

so, just plugging in values:

$\cot(45-a) = \frac{\tan{45}-\tan(a)}{1+\tan{45}\tan{a}}$

$\cot(45-a) = \frac{1 - \tan{a}}{1+\tan{a}}$

From this onwards we just need to substitute and simplify using the fact $\cot(x) = \frac{1}{\tan{x}}$ and $\tan{x} = \frac{1}{\cot{x}}$ to get what is aimed for. Please respond if you have any questions or you want me to complete the solution.
[/hide]

[quote="azjps"][hide]Let $ \alpha \equal{} DBC$. Then the first condition tells us that
\[ \tan^2 DBE \equal{} \tan(DBE \minus{} \alpha)\tan(DBE \plus{} \alpha) \equal{} \frac {\tan^2 DBE \minus{} \tan^2 \alpha}{1 \minus{} \tan ^2 DBE \tan^2 \alpha},
\]and multiplying out gives us $ (\tan^4 DBE \minus{} 1) \tan^2 \alpha \equal{} 0$. Since $ \tan\alpha \neq 0$, we have $ \tan^4 DBE \equal{} 1 \Longrightarrow \angle DBE \equal{} 45^{\circ}$.

The second condition tells us that $ 2\cot (45 \minus{} \alpha) \equal{} 1 \plus{} \cot \alpha$. Expanding, we have $ 1 \plus{} \cot \alpha \equal{} 2\left[\frac {\cot \alpha \plus{} 1}{\cot \alpha \minus{} 1}\right] \Longrightarrow (\cot \alpha \minus{} 3)(\cot \alpha \plus{} 1) \equal{} 0$. Evidently $ \cot \alpha \neq \minus{} 1$, so we get $ \cot \alpha \equal{} 3$.

Now $ BD \equal{} 5\sqrt {2}$ and $ AC \equal{} \frac{2BD }{\cot \alpha} \equal{} \frac {10\sqrt {2}}{3}$. Thus, $ [ABC] \equal{} \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} \equal{} \frac {50}{3}\ \mathbf{(B)}$.[/hide][/quote]

The second condition tells us that $ 2\cot (45 \minus{} \alpha) \equal{} 1 \plus{} \cot \alpha$.

...
can someone go into detail into this? thanks

[quote=azjps] $ 2\cot (45 \minus{} \alpha) \equal{} 1 \plus{} \cot \alpha$. Expanding, we have $ 1 \plus{} \cot \alpha \equal{} 2\left[\frac {\cot \alpha \plus{} 1}{\cot \alpha \minus{} 1}\right] \Longrightarrow (\cot \alpha \minus{} 3)(\cot \alpha \plus{} 1) \equal{} 0$. [/quote]

Why is $\cot(45-\alpha)=\frac{\cot \alpha + 1}{\cot \alpha -1}$?

Because $\cot{45}=1$ and you just do the tangent difference of two angles trick

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

Wait what! This is an AOPS class problem.

(Specifically precalculus)

Edit:
Nvm, the problem asks for an angle and not the area.

@below,
I realized that.
Also, people have posted problems in the past and they could easily gone unnoticed. Just thought I would see if that was a case.

$\cot (45-\alpha)=\frac{1}{\tan (45-\alpha)}=\frac{1}{\frac{\tan 45-\tan \alpha}{1+\tan 45\tan \alpha}}=\frac{1+\tan\alpha}{1-\tan\alpha}=\frac{1+\frac{1}{\cot \alpha}}{1-\frac{1}{\cot \alpha}}=\frac{\cot \alpha +1}{\cot \alpha-1}$. Is that the fastest way to do it?

In that case azjps must have left out a lot of steps :|

[quote=hinna]Wait what! This is an AOPS class problem. Lock please[/quote]

...dude this is 2004 AMC12B #24 calm the heck down...I'm not even taking an AoPS class. And read the dates please.

Sorry to revive, but is it necessary that one sets up the equation as DBE + alpha and DBE - alpha? I personally attempted this problem initially with angle ABD = angle DBC = X and angle CBE =Y, but could not reach the equation derived above. I suspect this is because in my case angle DBE, which is the nice angle, is in terms of two other angles, making it difficult to compute.
Nevertheless, I don't see a particular reason why one would attempt this problem with the angles labeled as in the solution above, rather than the way I did, so perhaps there is a way to calculate it my way..?


Show that the product of $k$ consecutive positive integers can't be the kth power of an integer


[hide="answer"]
the surface tension is $ T\equal{}\frac{\minus{}3P_{0}\Delta V}{4\Delta A}$
 where $ \Delta V$=change in volume
$ \Delta A$=change in area[/hide]

Perfect ans. I actually thought you were wrong initially but you've just given it as notation and not in terms of r1, r2, r3.

Answer:
$ T \equal{} \frac {P_0 (r^3 \minus{} r_1^3 \minus{} r_2^3)}{4(r_1^2 \plus{} r_2^2 \minus{} r^2)}$


A baseball is cut in half through the center. If the area of the circular region formed by the cut would be increased by 800%, then the volume of the baseball would be increased by K%. Compute K.

Start with 2 cups: Cup A, with 100mL or tea and Cup B with 100mL of cofee. Take 2 mL from Cup A, and mix it into Cup B. Then take 3 mL from Cup B and mix it into Cup A. Finally, take 1mL from Cup A, and mix it into Cup B. If in the end Cup A has $x$ mL of tea, and Cup B has $y$ mL of cofee, compute $x-y$

1.
[hide]I'll call the radius of the baseball $r$.  The area of the circle is ${r^2}{\pi}$, and increased by 800% is ${9r^2}{\pi}$.  That makes the radius of the increased baseball $3r$, and the volume of it $\frac{108r^3\pi}{3}$, compared to $\frac{4r^3\pi}{3}$ for the original baseball.  So the smaller baseball is increased by 2600% to get the area of the larger baseball.[/hide]

[quote]Start with 2 cups: Cup A, with 100mL of tea and Cup B with 100mL of coffee.
Take 2 mL from Cup A, and mix it into Cup B.
Then take 3 mL from Cup B and mix it into Cup A.
Finally, take 1 mL from Cup A, and mix it into Cup B.
If in the end Cup A has $x$ mL of tea, and Cup B has $y$ mL of cofee, compute $x-y.$[/quote][hide]
The classic "Water and Wine" problem.

No matter how much is transferred each time and how many transfers are made,
if the cups have their original volumes of liquids, the problem is stunningly simple.

At the end, Cup A has $x$ mL of tea.

Where is the missing $100-x$ mL of tea? . . . in Cup B, of course.

This means that Cup B has $100-x$ mL of tea and $x$ mL of coffee.

Get it? . . . There is as much tea in the tea cup as there is coffee in the coffee cup.

Therefore: $x = y\;\;\;\Longrightarrow\;\;\;x - y = 0.$[/hide]


Let Pn denote the number of regions formed when n lines are drawn in the Euclidean plane in such a way that no three are concurrent and no two are parallel. Show that Pn+1 = Pn + (n+1).

hey I just showed that in another topic some minutes ago, your problem inside a circle is an IMO longlisted prolem :)

Anyway, it's quite easy: since there are only n lines, the (n+1)'th can only intersect n lines, thus only create n+1 extra areas.

And there are enough easy methods that allow to continue an infinite amount of times.

Induct as follows:

Basis step: 1 line divides the infinite plane into 2 regions, and 2 lines divide it into 4 regions, so $P_2=P_1+(1+1)$.

Induction step: The (n+1)st intersects all n of the other lines, and thus creates n+1 more regions.  Therefore, $P_{n+1}=P_n+(n+1)$, and the induction is complete.

[quote="firefox"]Thanks, that is quite easy. How can you show it can continue infinitely?[/quote]

You can easily find a line that intersects all lines ;)

Any line intersecting them all works ;)


A chemical producer plans to sell 50,000 liters (to the nearest 1,000 liters) of a particular chemical at a price of Rs.10.00/= (to the nearest Rupee) per liter.

The cost of material used to produce the chemicals is expected to be Rs.100, 000/=, but depend on the wastage levels this is subjected to an error ± 5%. Other costs are estimated to be Rs.300, 000/= (± 10%)

What can be the maximum error occurring when you are calculating the total cost?

what?!? can you explain the problem again in a simpler way so I can understand? (sry...thanks!)

i don't know what maximum error is. :blush:

maximum error is the difference when you use the estimations to find the amount and when you use the actual original numbers to find the amount


Hi, I want to show that this function satisfies the Mean Value Thereom however, I am stuck when I try to obtain c. The function is f (x) = x/(x+2) on the interval [1,4].

So far I have:

That because it is a rational fuction, f is continuous on [a,b] and differentiable on (a,b).

f(4) - f(1) = f'(c)(4-1)
1/3 = f ' (c)*3
f '(c) = 1/9

f'(x) = 2/[(x+2)^2]

however when i try to set f'(x) = 1/9, i do not obtain a value within [1,4]

Thank you so much for your help!

Everything is correct until [quote="AquaGlass88"] i do not obtain a value within [1,4][/quote]  Or, rather, that statement may be true, but if so it's because you are solving the equation $ f'(c) \equal{} \frac{1}{9}$ incorrectly :)


A house was appraised at 75,000 dollars one year and 87,000 dollars the next year. How much more must the value of the house increase in order to show a 20% increase in value from the original value?

ANSWER: 0.44

Thanks,
Burnorth

A house originally cost 75000 dollars one year, and the next year it cost 87000 dollars. We want to find how much more percent this new price must increase so that the newest price is 20% more than the original cost. In other words, what percent does 87000 need to increase by in order for it to reach 90000? This is clearly 3000 dollars, which means there is an error in the problem.


An infinite sequence $(x_{n})$ is defined by its first term $x_{1}>1$, which is a rational number, and the relation $x_{n+1}=x_{n}+\frac{1}{\lfloor x_{n}\rfloor}$ for all positive integers $n$. Prove that this sequence contains an integer.
[i]A. Golovanov[/i]

If $x_{1}$ irrational, then $x_{n}$ irrational for all n.

But problem said that $x_{1}\in\mathbb{Q}$.

Well, Rust, I don't really get your solution. From the first line (what is $ x_{m_2}$)?

Can you clarify, or can anyone else post a new solution?

$ x_{m_i}$ denote member $ x_k$, suth that $ \{x_j\}>\{x_{m_i}\} \ \forall j<m_i$.

look at the fractional part of x{n} = r{n}
say x{n} floor is m
then we add (1/m) till the fractional part crosses 1
after that we add 1/(m+1) and so on
so , the new fractional part that we get r{n+1} is either = r{n} or r{n} - 1/m
and if we look at the new fractional part r{n+1]} as p/q , then its numerator is less than the numerator of r{n}

this is just a rough thing. but i think its correct. hope u guys think so.

[quote="Rust"]Let $ k_{1} = [[x_{1}](1 - \{x_{1}\})] + 1$, then ${ x_{m_{2}} = x_{k_{1} + 1} = x_{1} + \frac {k_{1}}{[x_{1}]}}$  [/quote]

Why is this?

[quote="gagan_goku"]
so , the new fractional part that we get r{n+1} is either = r{n} or r{n} - 1/m
[/quote]

Are you sure this is true?

Please, anybody give a "readable" solution to this one...  :wink:

Hey RUST.I think that your solution is correct.But I can't understand it now.Can you explain obvious for me.

This problem essentially relies on the existence of [url=http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html]Egyptian Fractions[/url]

First note that the set $\{\lfloor x_n \rfloor \}_{n=1}^{\infty} = \{m,m+1,m+2,...\}$ for some $m\in \mathbb{N}$. So define $k_j = \min\{ k  \, : \, k \in \mathbb{N}\, \text{and} \, \lfloor x_k \rfloor = j\}$ which exists for all $j\ge m$.

The sequence is increasing by increments of $\frac{1}{j}$ between terms $x_{k_j}$ and $x_{k_{j+1}}$. So we certainly have $\{x_{k_j}\} < \textstyle\frac{1}{j-1}$ and then $\{x_{k_{j+1}}\} <  \{\textstyle\frac{1}{j-1} + \frac{j-1}{j} \} = \frac{1}{j^2-j}$.

Choose $r>j$ so that $\frac{1}{r} \le \{x_{k_j} \} < \frac{1}{r-1}$, then $x_{k_j+i} = x_{k_j} + \frac{i}{j}$ for $1 \le i \le j$.

Specifically for $i=j$ we have $x_{k_{j+1}}=x_{k_j} + 1$, and repeating the process we end up with $x_{k_r} = x_{k_j} + (r-j)$.

Now it follows that $x_{k_{r+1}} = x_{k_r} + \frac{r-1}{r}$ and hence $\{x_{k_{r+1}}\} = \{x_{k_r} +1 - \frac{1}{r}\} = \{x_{k_j} - \frac{1}{r}\} < \frac{1}{r(r-1)}$.

Repeating this process implies that all terms $\{x_{k_{\ell}}\}$ will have the form $\{x_{k_j} - \textstyle\frac{1}{r_1}-\textstyle\frac{1}{r_2}-...-\textstyle\frac{1}{r_t}\}$ for some $t$ and distinct positive integers $r_i$.

Since $\{x_{k_j}\} \in \mathbb{Q}$ the process will eventually contain a term such that $\{x_{k_{\ell}}\} = 0$ because all rational numbers have an Egyptian fraction representation. So the proof is complete $\square$

In fact, you should be able to determine $\min\{ k \, : \, x_k \in \mathbb{N}\}$ by studying the Egyptian fraction representation of $\{x_1\}$.

Let the $\textit{entrance}$ value for $n$ be the value of $ x_k - \left \lfloor{x_k}\right \rfloor$ such that $k$ is the smallest integer such that $x_k > n$. We want to show that this is eventually zero.

Notice that in between every interval of $\[ n, n+1 \]$, you either add $\frac{1}{n}$ to $x_k$ $n$ or $n-1$ times. This depends on the value of $entrance$. If the value of entrance is less than $\frac{1}{n-1}$, $\frac{1}{n}$ is added $n$ times. If not, it is added $n-1$ times. Now, lets see the effect of this in the fractional part of $x_k$. When $\frac{1}{n}$ is added $n$ times, the fractional part does not change. If added, $n-1$ times however, it is equivalent to subtracting $\frac{1}{n}$ from the fractional part.

Therefore, the effect of this recursion on the fractional part can be summarized as follows. If the entrance value of $n$ is less than $\frac{1}{n}$, subtract $\frac{1}{n}$ from the fractional part. If not, the fractional part is unaltered.

We show that this is possible by showing how each time we subtract a fraction from our fraction parts, the numerator decreases. This is true because we subtract $\frac{1}{k}$ from some fraction $\frac{a}{b}$ iff $\frac{1}{k} < \frac{a}{b} < \frac{1}{k-1}$. So, the new numerator is $ka - b$ which is less than $a$, supported by how $ \frac{a}{b} > \frac{1}{k-1}$.

Let $f(a)$ be the minimum value $x_k$ obtains when $\lfloor x_k \rfloor =a$.

Check that $f(a)=f(a+1)$ if $f(a) < \frac 1a$ and $f(a+1)=f(a)-\frac 1a$ otherwise.

Since $\frac 1a$ is divergent, it follows that $f(a)<\frac 1a$ for some $a$. Now let $f(a)=\frac xy$. The key insight is that $x$ strictly decreases if the fraction decreases. To see this, say $\frac 1b > \frac xy > \frac{1}{b+1}$ then $f(b+1)=f(b)-\frac{1}{b+1} = \frac{x(b+1)-y}{y(b+1)}$. Note $x(b+1)-y > x(b+1)- xb = x$. Therefore at one point the numerator must be 1, the end.


calculus
integration
trigonometry
function
calculus computations


geometry
ratio
angle bisector
number theory
relatively prime

In triangle $ ABC,$ angle bisector $ AD$ and $ BE$ intersect at $ P$. If $ a \equal{} 3, b \equal{} 5, c \equal{} 7, BP \equal{} x,$ and $ PE \equal{} y,$ compute the ratio $ x: y,$ where $ x$ and $ y$ are relatively prime integers.

I don't usually answer my own question but I'm writing this solution because a) I want to know if what I did here is right  :D and b) just to present another solution. Feel free to post any other solutions to this problem!

[hide="Mass Point Geometry"]

Construct another angle bisector, namely $ CF$. Then, find the length of sides as follows from Angle Bisector Theorem:

$ AF \equal{} \frac{35}{8}, BF \equal{} \frac{21}{8}, BD \equal{} \frac{7}{4}, CD \equal{} \frac{5}{4}, CE \equal{} \frac{3}{2}, AE \equal{} \frac{7}{2}$

Now, knowing the sides, I can assign mass (which works out really nicely here). 3 to $ A$, 5 to $ B$, and 7 to $ C$. Then, I can see that $ E$ will be assigned with mass of 10. So, considering $ BE,$ I know that $ BP \equal{} 10$ and $ PE \equal{} 5$ so the ratio is 2:1 as desired. [/hide]

I also used mass points and obtained the same answer. You can also use menelaus' theorem, which is guessable since the associativity of mass points is due to menelaus' theorem.  :D

Hmm, I wonder if there is any pure geometry solution to this? Or is that not possible (I'm not good at constructing extra lines)?  :maybe:

You can use similar triangles, but again, that's just equivalent to Menelaus' theorem.  :)


Find positive integers such that:
$(m/n)^{m}=(mn)^{n}$

[hide]Evidently $n|m$. Let $m=dn$. Then we have $d^{d-1}=n^{2}$. If $d$ is odd then $n=d^{\frac{d-1}{2}}, \ \ m=d^{\frac{d+1}{2}}$
If $d$ is even then also $n=d^{\frac{d-1}{2}}, \ \ m=d^{\frac{d+1}{2}}$ but $d$ perfect square.[/hide]


geometry
rotation
analytic geometry
graphing lines
slope
AMC

A point $ (x,y)$ in the plane is called a lattice point if both $ x$ and $ y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to

$ \textbf{(A)}\ 4.0\qquad
\textbf{(B)}\ 4.2\qquad
\textbf{(C)}\ 4.5\qquad
\textbf{(D)}\ 5.0\qquad
\textbf{(E)}\ 5.6$

[hide="Not very thorough, but still"]
The 3 points have to be as close together as possible. Either they are in a row, or form an L shape. Through testing it is impossible to have them in a row, so they are an L shape. The square would be rotated 45 degrees with respect to the origin, and would contain 3 points (example: (1,1) (1,2) (2,1)) and thus the largest square would be tangent to (2,2) (1,0) and (0,1). The perpendicular distance between (2,2) and the segment joining (0,1) and (1,0) is $\frac{3\sqrt{2}}2$, which makes the area of the square $\frac92$, the answer is C
[/hide]

according to the official solutions the actual answer is D
will someone care to explain why it is D though?

Suppose the 3 points are the origin, (0, 1), and (1,0). Then the optimal square has 2 sides with slope 2 and 2 sides with slope 1/2; it has (1,1) and (0, -1) on a pair of opposite sides and (-1,0) and (1,-1) on the other pair. The side length is $ \sqrt{5}$ so its area is $ 5$.

[quote="alkjash"]Then the optimal square has 2 sides with slope 2 and 2 sides with slope 1/2.[/quote]
I don't think this is correct - the slopes must have opposite signs (in order to be perpendicular)

it was probably a typo, cuz the two lines mentioned do have slopes of 2 and -1/2

EDIT: fail, i typoed the word typo...its fixed now, though.

[quote="AdithyaGanesh"]Is there a way to use Pick's theorem for this kind of problem?   :huh:[/quote]

plugging the given info into pick's theorem we get 4, which is not correct :huh:

pick's won't work, cuz the sides of the square don't have to be lattice points


find all function injective such:
$f: \mathbb{N}\to\mathbb{N}\\ f(f(n))\le \frac{f(n)+n}{2}$
i know that it was posted before.but i want an other solution.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=17637#17637]this topic[/url].
[quote="fermat3"]I know that it was posted before.but i want an other solution.[/quote]
If you know it was posted before, then why did you post it again?
You could have posted your (somewhat vague) request in that topic. Why? Because it is undesirable to have solutions of a problem spread around the forum, especially when that happens "on purpose".

It seems only logical to lock this topic and continue any discussion there.

If $f(n)\not=n$ for some $n\in\mathbb{N}$, then $f^{(k)}(n)=\underbrace{f(f(\cdots f(n)\cdots))}_{k\text{ times}}<\max\{n,f(n) \}$ for all $k\in\mathbb{N},k\geq 2$ by induction.
But $\{1,2,\ldots,\max\{n,f(n) \}\}$ is finite, so $f^{(k)}(n)=f^{(l)}(n)$ for some $2\leq k<l$ with $l-k\geq 2$.
But as $f$ is injective, we get
$n=f^{(l-k)}(n)<\max\{n,f(n) \}$ and $f(n)=f^{(l-k+1)}(n)<\max\{n,f(n) \}$.
Contradiction!

EDIT Ok, same solution. :oops: Just checked.

good olorin.for perfect-radio. i will poste my soution such exercice:
1: soit $u_{n}$ a positive sequence such $u_{n+2}\le{\frac{u_{n+1}+u_{n}}{2}}$.
prove that $u_{n}$ converge.
2:use this fact to prove that the only function $f$ is $f(n)=n$. :wink:


geometry
trigonometry
circumcircle
incenter
angle bisector
geometry unsolved

Let's take the triangle $\ ABC$ whith $\ AH\bot\ BC$ and $\ BE$ the interior bisector of the angle$\ ABC$.If $\ m(\angle BEA)=45$, find $\ m(\angle EHC).$

What is $H$? Is that orthocenter, or a feet of altitude from $A$ or just some point? (the last one has no sense!)

1) $\ H$ is the feet of the altitude from $\ A$.
2) Yes, you're right.The answer is $\ 45$.But please post a full solution.

Since $\measuredangle AHB = 2\measuredangle AEB$, then $H$ is the center of the circle passing through $A, E, B$. Since $\measuredangle AHB = 90^{\circ}$, then $\measuredangle EBA = \frac{45}{2}^{\circ}$ and $\measuredangle AHE = 45^{\circ}$. Hence $\measuredangle EHC = 45^{\circ}$.

[quote="Andreas"]Since $\measuredangle AHB = 2\measuredangle AEB$, then $H$ is the center of the circle passing through $A, E, B$.[/quote]

This is not correct! Suppose that is, then $HA=HB$ so $\angle ABC=45^{o}$ which doesn't have to be true at all.

Here is the correct solution:

We want to prove that $HE$ is an angle bisector of $CHA$, so we want to prove that $\frac{CE}{EA}=\frac{CH}{AH}$. $BE$ is an angle bisector of $\angle ABC$ so $\frac{CE}{EA}=\frac{CB}{AB}$. We have that $\angle C+\angle \frac{B}2 = 45^{o}$ so angle $A>90^{o}$ so feet of altitude from $B$ (let's say $D$) to $AC$ isn't on side $AC$. Now simple angle calculation gives that triangles $CAH$ and $BAD$ are similar, so $CH\cdot AB = AC\cdot BD=2\cdot Area=AH\cdot CB$, so from that $\frac{AB}{CB}=\frac{AH}{CH}$ and $\frac{AB}{CB}=\frac{EA}{CE}$  so $\frac{AH}{CH}=\frac{AE}{EC}$ and that is the end of the solution.

Bye!!

[color=darkblue][b]I. The first (trigonometrical) solution.[/b] Denote $m(\widehat{ABE})=m(\widehat{HBE})=x$ and $m(\widehat{EHC})=y\ .$
Thus, in the quadrilateral $ABHE$ there is the relation : $\sin \widehat{BAH}\cdot\sin\widehat{HBE}\cdot\sin \widehat{EHA}\cdot\sin\widehat{AEB}=$ $\sin\widehat{ABE}\cdot\sin\widehat{BHA}\cdot\sin\widehat{HEB}\cdot\sin\widehat{EAH}\ .$
Therefore, $\cos 2x\sin x\cos y\sin 45^{\circ}=\sin (45^{\circ}+x)\sin x\sin 90^{\circ}\sin (y-x)$ $\Longrightarrow$ $\cos 2x\cos y=2\cos 45^{\circ}\sin (45^{\circ}+x)\sin (y-x)$ $\Longrightarrow$ $\left(\cos^{2}x-\sin^{2}x\right)\cos y=(\cos x+\sin x)\sin (y-x)$ $\Longrightarrow$ $(\cos x-\sin x)\cos y=\sin y\cos x-\sin x\cos y$ $\Longrightarrow$ $\cos y=\sin y$ $\Longrightarrow$
$y=45^{\circ}$ $\Longrightarrow$ $m(\widehat{EHC})=45^{\circ}\ .$

[b]II. The second (metrical) solution (from [u]SpongeBob[/u]).[/b] Denote the projection $D$ of the point $B$ to the line $AC$. We have $A\in (CD)$ because $A=90^{\circ}+C$. Also, $\triangle ABD\sim\triangle ACH$. Thus, $\frac{AB}{AC}=\frac{BD}{CH}$ $\Longrightarrow$ $\underline{AC\cdot BD=AB\cdot CH}$. But $AC\cdot BD=[ABC]=BC\cdot AH$ $\Longrightarrow$ $\underline{AC\cdot BD=BC\cdot AH}$. Thus, $BC\cdot AH=AB\cdot CH$, i.e. $\frac{BA}{BC}=\frac{HA}{HC}$. But $\frac{BA}{BC}=\frac{EA}{EC}$. Therefore, $\frac{EA}{EC}=\frac{HA}{HC}$, i.e. the ray $[HE$ is the bisector of the angle $\widehat{AHC}$ $\Longrightarrow$ $m(\widehat{EC})=45^{\circ}\ .$

[b]III. The third (pure syntetical) solution.[/b] Let $w=C(I)$ be the incircle of the triangle $ABH$ and denote the second intersection $F$ between the line $BE$ and the circumcircle of the triangle $ABH$. Then $FA=FH=FI$ and $m(\widehat{IHA})=m(\widehat{IEA})=45^{\circ}$, i.e. the quadrilateral $AIHE$ is inscribed in a circle with the center $F$. Therefore, $FI=FE$ $\Longrightarrow$ the point $E$ is the $B$ - exincenter for the triangle $ABH$ $\Longrightarrow$ the ray $[HE$ is the bisector of the angle $\widehat{AHC}$ $\Longrightarrow$ $m(\widehat{EHC})=45^{\circ}\ .$

[b]IV. The fourth (pure syntetical) solution.[/b] Denote the point $F\in BC$ for which $AF\perp AC$. Thus, $A=C+90^{\circ}$ $\Longrightarrow$ $\widehat{BAF}\equiv\widehat{BCA}\equiv\widehat{HAF}$, i.e. the point $I\in BE\cap AF$ is the incenter and the point $E$ is $B$ - exincenter for the triangle $ABH$ a.s.o.

[b]Generalization.[/b] Given are a triangle $ABC$ for which there is a point $H\in BC$ so that $m(\widehat{AHB})=2\cdot m(\widehat{AEB})$, where the point $E$ is the intersection between the line $AC$ and the bisector ray of the angle $\widehat{ABC}$. Prove that $m(\widehat{EHC})+m(\widehat{AEB})=90^{\circ}\ .$
[b]Proof.[/b] Let $I$ be the incenter of the triangle $ABH$. Then the quadrilateral $AIHE$ is inscribed. Therefore, the point $E$ is the $B$ - exincenter of the triangle $ABH$,
i.e. the ray $[HE$ is the bisector of the angle $\widehat{AHC}$ $\Longrightarrow$ $m(\widehat{EHC})=90^{\circ}-m(\widehat{AEB})\ .$[/color]

Mr Virgil Nicula, can you show your metrical solution?


What is the "Savebox" and how do I use it? What happens if my Inbox gets full?

You can send messages from your inbox to your savebox, a storage for old messages that you want to keep.

How to use it: Simply mark the check boxes of the ones you want to save and click on "Save Marked" in bold at the bottom of the screen. You could also delete some or all of your mail here also so be careful...

If the inbox was full then the oldest message gets deleted.


[quote="yzymat"]I  cannot  install MikTex2.8.

[size=150]
[color=red][b]Could  not  create  a  unique  sub-directory.
Details: C:\DOCUME~1\username\LOCALS~1\Temp\[/b][/color][/size]

 I  cannot   understand   what it  mean

Many thanks![/quote]

You could be more specific... What steps did you take? etc....

Thank  you  very much!
I  romove  2.7  and  some  files  still exist
 when  I     install  2.8,   the   problem  occur

So delete those files, or install in a different directory.

Better yet would be to delete the whole directory.


calculus
integration
logarithms
trigonometry
function
calculus computations

This is confusing me:

lets say { means integral

Say you have the problem

{ (1/x)*(1/(ln(x))) *dx

and using integration by parts:

{u dv = uv - { v du

if 1/x * dx = dv
and 1/(lnx) = u

solve the problem and you end up with

{ (1/x)*(1/(ln(x))) *dx = (lnx/lnx) + { (1/x)*(1/(ln(x))) *dx

in other words, 1-0 if you subtract the original integral from both sides.

Any input?

You are saying as folows,right?

$\int \frac{1}{x\ln x} dx=\int \frac{1}{\ln x}d(\ln x)=\frac{1}{\ln x}\cdot \ln x - \int \ln x\  d\left(\frac{1}{\ln x}\right)$

$=1+\int \ln x \cdot  \frac{1}{x(\ln x)^2}dx=1+\int \frac{1}{x\ln x} dx \Longleftrightarrow 0=1$

(I'll be moving this to "Calculus computations and tutorials.")

If I were to say $\int\sin x\,dx = 1-\cos x$ would you object?

Let's take a look at your problem:

$\int\frac1x\cdot\frac1{\ln x}\,dx$

Let $u=\frac1{\ln x}$ so that $du=-\frac1x\cdot\frac1{(\ln x)^2}\,dx$

Let $dv = \frac1x\,dx$ so that $v=\ln x$

Then $\int\frac1x\cdot\frac1{\ln x}\,dx=\frac{\ln x}{\ln x} + \int\frac1x\cdot\frac1{\ln x}\,dx$

So far, this is correct. Useless, but correct.

But the symbol $\int f(x)\,dx$ does not stand for a unique function. This symbol (for primitive, or antiderivative) represents an infinite family of functions, all differing by a constant. That "$+C$" that you ritually put at the end of your answers to antiderivative problems isn't just there for decoration.

Each use of $\int f(x)\,dx$ comes with its own "$+C$". Employ that in this case to get:

$\int\frac1x\cdot\frac1{\ln x}\,dx+C_1=1 + \int\frac1x\cdot\frac1{\ln x}\,dx+C_2$

Subtract the integrals from both sides to get that

$C_1-C_2=1.$

So we discover that .... 1 is constant. As I said, useless.

Why is there a + C for the first one?  It wasn't in the original equation, and if it was, then wouldn't there be a + Cx somewhere in the integral of the original?

And then, it would mean that 0 = 1+C, meaning that C = -1, but C doesn't have to be that number, C can be any constant, but if you integrate by parts in this case, it says that is must be -1.


[b]1[/b]. Expressed in terms of $ n$, what is the sum of all the terms in this arithmetic sequence:

               $ (n\minus{}7),(n\minus{}2), ................, (n\plus{}428)$

[b]2[/b].  RIght triangle ABC with right angle at vertex C has median AD of length 5 and median BE of length $ 2\sqrt {10}$.   Find the length of the hypotenuse of triangle ABC.  Express your answer in simplest radical form.

huh, I just did a test with the first question on there

[hide="hints"]
1) You can still use the formula for the sum of the first $ k$ terms of an arithmetic sequence even with variables.

2) Use the definition of median and notice a symmetric system of equations.
[/hide]


For some reason, I broke my 4x4 rubik's cube. How do I fix it?

Not that I know very much about them, but:
There are at least two different types of mechanics, so it would be usefull if you could give more information.
This is where my knowledge stops, never disasembled one.

Mine's a traditional Rubik's brand. Its mechanism has a ball-core spindle, not the one like Eastsheen. So how do u assemble that?

Easy.
First get the bottom centers in.
Then do the edges.
Keep working your way up (2nd layer, then 3rd layer)
Last layer:
Centers in first, then 1 edge, then 2 corners, 2 edges, last 2 corners, last edge.

5x5 assembly is same method

NOW EASTSHEEN IS A BEAST TO ASSEMBLE

but i couldn't press in the center pieces...is something wrong? did i try to jam them?

There's like a slot in the middle where you can insert them.

Ok. I finished fixing my cube, except that one of the corners is placed incorrectly. How do I safely disassemble the cube again just to be able to fix that one corner piece?

get an adjacent edge out then the corner out then put it back correctly then put the edge back in

whats eastsheen?
u can pm me

no i mean how do u [b]deliberately[/b] get an edge out without destroying the cube entirely?

eastsheen is another brand of rubik's cube...more durable than traditional rubik's brand.
eastsheen is made so taht screws can be used to tighten or loosen the cube.
It's not like the traditional ball-core spindle mechanism used in the classic Rubik's brand.

orient the cube such that 1 face is facing you
turn the right half up slightly
turn the top layer left (clockwise looking down onto the top face) so that one edge looks delightfully vulnerable to being taken out
take that edge out
blah

try to understand the image... i have horrible wording skills

Take it out just like how you would take out a regular piece from a 3x3 cube.
Turn it 45 degrees, and lift the piece. Then arrange the corner.


number theory
Diophantine equation
number theory proposed

\[ 2m^2\plus{}1\equal{}3^{n\plus{}1}\]

have you got any idea ?

$ n\equal{}2t$ then we can let $ y\equal{}3^t$. We first solve the Diophantine Equation $ 3y^2 \minus{} 2m^2 \equal{} 1$

When $ n\equal{}2t\minus{}1$, denote $ y\equal{}3^{t}$ and solve $ y^2 \minus{} 2m^2 \equal{} 1$ (one solution is $ (3,2)$)


analytic geometry
graphing lines
slope
modular arithmetic

Determine all points on the straight line which joins $(-4,11)$ to $(16,-1)$ and whose coordinates are positive integers.

[hide="Answer?"]The slope of the line is $\frac{-5}{3}$. Therefore the equation of the line (from point-slope form) is $y-11 = \frac{-5}{3} (x+4)$. So since they're both positive integers, we know $x+4 \equiv 0 \pmod3$ then start testing. $x=2$ is the first one that satisfies both conditions, and plugging in yields $y-11 = -10 \Rightarrow y=1$. Trying the next one, $x=5$ gives $y-11 = -15 \Rightarrow y=-4$ so since it's decreasing (negative slope) $(2,1)$ is the only solution.[/hide]

the equation of line should be $3x+5y=43$

So the points are $(1,8),(6,5),(11,2)$

[hide="Answer"]$|m|=\frac{3}{5}$
Therefore, every $5$ over or $3$ down (same thing), we have another point with integer coordinates. Not counting the endpoints, since both of those have negative coordinates, we find $\boxed{(1,8),(6,5),(11,2)}$.[/hide]


Let $ 0 \leq a_i \leq b_i$ for all $ i$ and $ \sum a_i \equal{} \sum b_i < \infty$.  Prove or give a counter example that $ a_i \equal{} b_i$ for all $ i$.

I came across a result like this when using the class equation in algebra.  I was curious if the conclusion holds for infinite sums as well.  My intuition says of course not, but I have yet to find a counter example.

Let $ r_i\equal{}b_i\minus{}a_1$
Because $ \sum a_i\equal{}\sum b_i<\plus{}\infty$ we have
 $ r_i\geq 0$
and  $ \sum r_i\equal{}0$
Suppose have an $ r_k>0$ then $ \sum r_i \geq r_k >0$ (tradition)
So $ a_i\equal{}b_i$
If $ \sum a_i \equal{}\plus{}\infty$ then it is not true.


Determine all continuous functions f:R-->R so that we have:

f(x/( sqrt( 1+x^2))=f(x) for all x real.

I'm thinking of [hide]a trig substitution here ...[/hide] dunno if it will work, though.  Must go do homework ... must go do homework ...

my first guess would be [hide]f(x) = constant because if a0 = k, and an+1 = an/( :sqrt: 1+an^2), the limit as n ->  :inf: an = 0, so by continuity, f(k)=f(0).

[/hide]

f(x)=f(x/sqrt(1+x^2)). Let's concentrate on positive x, because for negative x it's practically the same. For positive x, x^2>0 => 1+x^2 > 1 => sqrt(1+x^2) > 1 => x > x/sqrt(1+x^2). So we see that for any number a, we can construct a decreasing sequence of numbers that has the same value as f(a). All of these sequences converge to 0, so by continuity, we have that f(0)=f(a) for every positive number a. Let us call f(0) = k. Then f(a) = k for all a > 0. Using similar reasoning, we note that x < x/sqrt(1+x^2) when x<0, construct an increasing sequence, and determine that f(a) = k for all real a. All constant functions work because all constant functions satisfy the conditions of the problem.


Let positive integers a,b,c satisfy a + 2b +11c = abc
Find min P=a + 6b + 5c

:D  :D
HINT: lagrange
ithis pro is mine :D


modular arithmetic
number theory
least common multiple
AMC 8
AMC

Can you help me with this AMC problem from 2006?

#23. A box contains gold coins. If the coins are equally divided among 6 people, 4 coins are left over. If the coins are equally divided by among 5 people, 3 coins are left over. If the box holds the smallest no. of coins that meets these 2 conditions, how many coins are left when equally divided among 7 people?

Thank you in advance. :)

[hide]Let the # be $ x$.
$ x\equiv4\pmod{6}$
$ x\equiv3\pmod{5}$
So $ x \plus{} 2 \equal{} 6\cdot{5\cdot{j}}$ for some number j. Since we want the smallest, j=1. So there are 28 coins, and the answer is 0.[/hide]

shaggy75-where can you find these problems on the internet?

xpmath-is it posible to do [hide="this"]
$ x\equal{}4  \pmod 6$
$ x\equal{}3 \pmod 5$
$ x\plus{}2\equal{}6\equal{}0 \pmod 6$
$ x\plus{}2\equal{}5\equal{}0 \pmod 5$
$ x\equal{}\minus{}2\equal{}28 \pmod {30}$
$ 28\equal{}0 \pmod 7$[/hide]
or does it only work in this case

I don't know, but I think it's only possible if (m,n)=1 in $ \pmod{m}$ and $ \pmod{n}$.
But I like my way better.

Thanks guys.

Abacadaea- I got some from [url=http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/2007-AMC8Bro.pdf]this site[/url] and i got some more by searching google

[quote="abacadaea"]$ x \plus{} 2 \equal{} 6 \equal{} 0 \pmod 6$
$ x \plus{} 2 \equal{} 5 \equal{} 0 \pmod 5$
$ x \equal{} \minus{} 2 \equal{} 28 \pmod {30}$[/quote]

You're saying that if $ 5 | x \plus{} 2, 6 | x \plus{} 2$ then $ 30 | x \plus{} 2$.  In general, for moduli $ a, b$, this implication is valid if and only if $ \gcd(a, b) \equal{} 1$.

This method of adding $ 2$ and applying the above result would fail if the problem were instead, say,

$ x \equiv 2 \bmod 6$
$ x \equiv 3 \bmod 5$

in which case you would either test out values of $ x$ (say, starting with the first equation: $ x \equal{} 2, 8, 14, ...$) until one of them satisfied both equations or use the [url=http://en.wikipedia.org/wiki/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].

[quote="t0rajir0u"]

$ x \equiv 2 \bmod 6$
$ x \equiv 3 \bmod 5$

[\quote]
[hide="My solution, since I dont know  the Chinese Remainder Theorem"].
So, $ x \equal{} 6j \plus{} 2$ and $ x \equal{} 5k \plus{} 3$ for some k and j.
Equating, $ 6j \plus{} 2 \equal{} 5k \plus{} 3$, and taking modulo 5, $ j \plus{} 2\equiv3\pmod{5}$, or $ j\equiv1\pmod{5}$. Now, if instead you wanted k, modulo 6:
$ 5k\equiv \minus{} 1\pmod{6}$, so $ x\equiv1\pmod{6}$.
The smallest value of x is 8.[/hide]
Would that work anywhere, or just here?

That works, and is essentially the computation that the Chinese Remainder Theorem generalizes.

this was from the 2007 AMC 8, which I bombed! :D

if the problem-writers give such a shortcut (e.g. just find 2 less than lcm) than you could just go about solving it the old-fashioned with of solving a pair of linear congruences.

Hehe, what I always do in a problems like this (as in, 6-4=2 and 5-3=2) is take the LCM and subtract 2, like mihail said. But it doesn't hurt to know the congruences.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A \in M_{3}(C)$ so that $|\det(A+xI_{3})| \leq 1$ for any complex number x, with $|x|=1$. Prove that $\det (A+xI_{3}) = x^3$.

Since $\det(A+xI_3)$ is a monic polynomial of third degree, we can use the following result:

Let $P\left( x\right) =x^{n}+a_{1}x^{n-1}+\ldots+a_{n}$ be a polynomial with complex coefficients such that $\left\vert P\left( z\right) \right\vert \leq1$ for all $z\in C$ with $\left\vert z\right\vert =1.$ Then $a_{1} =a_{2}=\ldots=a_{n}=0.$

We prove first that $a_{n}=0.$ Let $\varepsilon_{0},\varepsilon_{1} ,\ldots,\varepsilon_{n-1}$ be the $n$-th roots of the unity. We have
\[ P\left( \varepsilon_{0}\right) +P\left( \varepsilon_{1}\right) +\ldots+P\left( \varepsilon_{n-1}\right) =n+na_{n},  \]
since
\[ \varepsilon_{0}^{k}+\varepsilon_{1}^{k}+\ldots+\varepsilon_{n-1}^{k}=0,  \]
for all positive integers $k,$ $1\leq k\leq n-1.$ But
\[ \left\vert P\left( \varepsilon_{0}\right) +P\left( \varepsilon_{1}\right) +\ldots+P\left( \varepsilon_{n-1}\right) \right\vert \leq\left\vert P\left( \varepsilon_{0}\right) \right\vert +\left\vert P\left( \varepsilon _{1}\right) \right\vert +\ldots+\left\vert P\left( \varepsilon_{n-1}\right) \right\vert =n,  \]
hence
\[ \left\vert 1+a_{n}\right\vert \leq1.  \]
Similarly, considering $\varepsilon_{0}^{\prime},\varepsilon_{1}^{\prime },\ldots,\varepsilon_{n-1}^{\prime}$ the $n$-th roots of $-1,$ we obtain
\[ \left\vert -1+a_{n}\right\vert \leq1.  \]
This implies $a_{n}=0$ (because the distances from the point in the complex plane corresponding to $a_{n}$ to the points $-1$ and $1$ are both less than or equal to $1$).

Now, observe that if $\left\vert z\right\vert =1$ then
\[ \left\vert P\left( z\right) \right\vert =\left\vert z^{n}+a_{1} z^{n-1}+\ldots+a_{n-1}z\right\vert =\left\vert z^{n-1}+a_{1}z^{n-2} +\ldots+a_{n-1}\right\vert ,  \]
hence the polinomial $P_{1}\left( x\right) =x^{n-1}+a_{1}x^{n-2} +\ldots+a_{n-1}$ has the same property: $\left\vert P_{1}\left( z\right) \right\vert \leq1$ for all $\left\vert z\right\vert =1.$ It follows that $a_{n-1}=0,$ etc. Eventually, we obtain $P\left( x\right) =x^{n},$ as desired.


limit
real analysis
real analysis unsolved

Hello all.

Here's another one I don't quite see:

Prove: if $f(x)>0\mbox{ }\forall \ x\mbox{ then }\lim\limits_{x\rightarrow x_o}f(x) \geq0\mbox{ for any }x_o$

I'm assuming the best way to prove this is by contradiction. Therefore,

Assume $\lim\limits_{x\rightarrow x_o}f(x) =A\mbox{ with }A<0$.

Then I'd need to show:

$\forall \ \epsilon>0\mbox{ }\exists \ \delta>0\mbox{ such that}$

$0<|x-x_o|<\delta\mbox{ implies } |f(x)-A|<\epsilon$

I'm not quite sure how to find an appropriate $\epsilon$ to show the contradiction.

Bailey

Er...No...not that.

Assume $\lim\limits_{x\rightarrow x_o}f(x) =A<0$ for some $x_0$.
Let $\epsilon = -\dfrac{A}{2}>0$, then there exists $\delta>0$ such that
$|f(x)-A|<-\dfrac{A}{2}$ for $x\in(x_0-\delta,x_0+\delta)$.
and it is $\dfrac{3}{2}A<f(x)<\dfrac{A}{2}<0$,
which contradicts with $f\geq 0$ for all $x$

Thanks...I should have seen that....excuse me while I go bang my head against the wall.

Bailey


trigonometry
geometry
incenter
trig identities
Law of Sines
perpendicular bisector
geometry unsolved

Triangle $ABC$ has the following property -- there is an interior point $P$ such that $\angle PAB=10^{\circ}$, $\angle PBA = 20^{\circ}$, $\angle PCA = 30^{\circ}$, and $\angle PAC = 40^{\circ}$. Prove that triangle $ABC$ is isosceles.

This problem was posted before at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56291]url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56291[/url].

Here is my solution, which uses primarily Law of Sines...I want to see if anyone has a purely geometrical approach to this.

[hide]

Let , , , , , and . By angle chasing, we can find that , , , and . Let  and , so we have . By Law of Sines, we have

from triangles , , , and , respectively.

From the LoS expression for  and , we have  and . Multiplying together and solving yields

(*)

From the LoS expression again on  and , we have  and . Multiplying again, using the  double angle formula, and solving for  yields

Substituting our expression for  back into what we found in (*) yields . (**)

By LoS on , we know that . Substituting (**) into this for  and simplifying using double angle formulas yields

By LoS on , we have . Substituting what we found for , followed by simplifying through double angle formulas yields

However, we know by  that , so solving for one in terms of the other and plugging in gets rid of all the lengths, leaving us with

Substituting , and simplifying with the identities  yields

Make the substitution , then use double angle formula for  to simplify down to

Recall the identity . Using this on the LHS, we have

Using the identity , we have

or . Since , then we have , so the only solution on that interval to  is , yielding .

Now we have , so . However, we are given that , so  is isosceles, as desired.

[/hide]

We know the angle $\angle A = \angle PAB + \angle PAC = 10^\circ + 40^\circ = 50^\circ$. Let M be the midpoint of the side AC and let CP meet the perpendicular bisector of the side CA at a point O. The triangle $\triangle AOC$ is isosceles, hence $\angle OAC = \angle OCA \equiv \angle PCA = 30^\circ$ and $\angle OAB = \angle A - \angle OAC = 50^\circ - 30^\circ = 20^\circ$. Since $\angle PAB = 10^\circ$, AP bisects the angle $\angle OAB$. Let the perpendicular bisector OM of the side AC meet the line AB at a point B'. From the isosceles triangle $\triangle AOC$ with

$\angle AOC = 180^\circ - (\angle OAC + \angle OCA) = 120^\circ$

we get $\angle AOP = \angle OAC + \angle OCA = 60^\circ$ and $\angle AOB' = 180^\circ - \frac{\angle AOC}{2} = 120^\circ$. Therefore, $CP \equiv OP$ bisects the angle $\angle AOB'$. Since we already showed that AP bisects the angle $\angle OAB' \equiv \angle OAB = 20^\circ$, P is the incenter of the triangle $\triangle AOB'$. Hence, B'P bisects the remaining angle $\angle AB'O$ of this triangle, which is equal to

$\angle AB'O = 180^\circ - (\angle OAB' + \angle AOB') = 40^\circ$

Consequently, $\angle AB'P = \frac{\angle AB'O}{2} = 20^\circ$, which means that the points $B' \equiv B$ are identical. As a result, the vertex B lies on the perpendicular bisector OM of the side AC, which means that the triangle $\triangle ABC$ is isosceles with AB = BC.


geometry
trigonometry
parallelogram
incenter
geometry proposed

Let $ABC$ be a triangle.

Let $R$ be the point on $AB$ so that $BR=BC$ and $B$ is between $A$ and $R$.
Let $S$ be the point on $AC$ so that $CS=CB$ and $C$ is between $A$ and $S$.
Let $A'$ be the point where the diagonals of $BRSC$ intersect.

Define $B'$ and $C'$ analougsly.

Prove that $[AC'BA'CB'] = [ABC] + [A'B'C']$.

An easy angle-chasing shows that $\angle {ABC'}=\angle {ACB'}=\frac{\alpha}2$, so $\angle {C'BC}+\angle {BCB'}=\pi \longrightarrow BC'//CB'$.

But $BC'=\frac{c\cdot \sin{\frac{\beta}2}}{\cos{\frac{\gamma}2}}=\frac{b\cdot \sin {\frac{\gamma}2}}{\cos{\frac{\beta}2}}=B'C$

and so $BCB'C'$ is a parallelogram; in particular $B'C'=BC$.

In the same way we show that $A'B'=AB, C'A'=CA$, thus $ABC$ and $A'B'C'$ are congruent.

So we must show that $[ABC']+[AB'C]+[A'BC]=[ABC]$

If $I$ is the incenter of $ABC$, it is obvious that $[ABI]=[ABC']$ and cyclic, so we have done.


geometry
geometric transformation
reflection
trigonometry
circumcircle

In acute triangle $ABC,\, \angle A < 45^\circ.$ Point $D$ lies in the interior of triangle $ABC$ such that $BD = CD$ and $\angle BDC = 4\angle A.$ Point $E$ is the reflection of $C$ across line $AB,$ and point $F$ is the reflection of $B$ across line $AC.$ Prove that $AD\perp EF.$

we have to show that $AF^{2}-AE^{2}=DF^{2}-DE^{2}$

but we know that $AF=AB,AE=AC$ so we get that:

$\boxed{AF^{2}-AE^{2}=c^{2}-b^{2}}(I)$

now note that $\angle ACF=\angle C$ and $CF=a$ and also in the isosceles triangle $\triangle BDC$ we have $\angle DCB=90-2A$ so we get that $FCD=C-(90-2A)+C=2A+2C-90=270-2B$ so we get that $\cos FCD=-\cos (90-2B)=-\sin 2B$

also according to $\sin$law in $\triangle BDC$ we get that:

$\frac{\cos 2A}{DC}=\frac{\sin 4A}{a}$ so we get that $DC=\frac{a\cos 2A}{2\sin 2A\cos 2A}=\frac{a}{2\sin 2A}$

now by $\cos$law in $\triangle DFC$ we get that:

$DF^{2}=CF^{2}+CD^{2}-2CF.CD.\cos{\angle DCF}$

$\Rightarrow DF^{2}=a^{2}+\frac{a^{2}}{4\sin^{2}2A}+\frac{a^{2}}{\sin 2A}.\sin 2B$

similary we get that $DE^{2}=a^{2}+\frac{a^{2}}{4\sin^{2}2A}+\frac{a^{2}}{\sin 2A}.\sin 2C$

so we get that

$\boxed{DF^{2}-DE^{2}=\frac{a^{2}}{\sin 2A}(\sin 2B-\sin 2C)}(II)$

so according to $(I),(II)$ its sufficient to show that:

$c^{2}-b^{2}=\frac{a^{2}}{\sin 2A}(\sin 2B-\sin 2C)$

which is true

[hide]

we have to prove that $\frac{c^{2}-b^{2}}{a^{2}}=\frac{\sin 2B-\sin 2C}{\sin 2A}$

so we have to prove that:

$\frac{c^{2}-b^{2}}{a^{2}}=\frac{\sin B\cos B-\sin C\cos C}{\sin A\cos A}$

according to $\sin$law we know that $\frac{\sin B}{\sin A}=\frac{b}{a}$ and similary $\frac{\sin C}{\sin A}=\frac{c}{a}$

so the previews equality becomes:

$\frac{c^{2}-b^{2}}{a^{2}}=\frac{b}{a}.\frac{\cos B}{\cos A}-\frac{c}{a}.\frac{\cos C}{\cos A}$

$\Rightarrow \frac{c^{2}-b^{2}}{a}=\frac{b\cos B-c\cos C}{\cos A}$

now according to $\cos$ law we get that $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$ and similary we get that $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}$ and $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$ so we have to prove that:

$\frac{c^{2}-b^{2}}{a}=\frac{\frac{ba^{2}+bc^{2}-b^{3}}{2ac}-\frac{ca^{2}+cb^{2}-c^{2}}{2ab}}{\frac{b^{2}+c^{2}-a^{2}}{2bc}}$

so we have to prove that:

$\frac{c^{2}-b^{2}}{a}=\frac{\frac{b^{2}a^{2}+b^{2}c^{2}-b^{4}-c^{2}a^{2}-c^{2}b^{2}+c^{4}}{2abc}}{\frac{b^{2}+c^{2}-a^{2}}{2bc}}$

$\frac{c^{2}-b^{2}}{a}=\frac{b^{2}a^{2}-b^{4}-c^{2}a^{2}+c^{4}}{a(b^{2}+c^{2}-a^{2})}$

so we have to prove that:

$(c^{2}-b^{2})(b^{2}+c^{2}-a^{2})=b^{2}a^{2}-b^{4}-c^{2}a^{2}+c^{4}$

which is obviously true after expanding...  :) [/hide]

Another formulation of this problem, has already been posted by [b][size=100]Huyen Vu[/size][/b], at http://www.mathlinks.ro/Forum/viewtopic.php?t=158177, with a solution by [b][size=100]bravado.[/size][/b]

Kostas Vittas.

[quote="April"]I[color=darkred]n acute triangle $ ABC,\, \angle A < 45^\circ.$ Point $ D$ lies in the interior of triangle $ ABC$ such that $ BD = CD$ and $ \angle BDC = 4\angle A.$ Point $ E$ is the reflection of $ C$ across line $ AB,$ and point $ F$ is the reflection of $ B$ across line $ AC.$ Prove that $ AD\perp EF.$[/color][/quote]
[color=darkblue][b]Leonhard Euler's proof.[/b] First, we will prove a Lemma.[/color]
[quote][color=blue][b]Lemma.[/b] Let be given a triangle $ ABC.$ Extend segments $ AB,\,AC$ through $ B,\,C$ to $ E,\,F,$ respectively such that $ EB=BC=CF.$ Let $ O,\,I_{a}$ be the circumcenter and excenter opposite vertex $ A$ of triangle $ ABC.$
Prove that $ OI_{a}\perp EF$[/color][/quote]
[color=darkblue][b]Proof.[/b] Similar to solution at the topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=147540

Come back with our problem.
Denote $ S=BE\cap CF$
Consider triangle $ SBC,$ we have $ A$ is the excenter opposite vertex $ S$ of triangle $ SBC,$ and $ E,\,F$ are points lie on extend segments $ SB,\,SC$ such that $ EB=BC=CF$ and easy to show that $ D$ is the circumcenter of triangle $ SBC.$ Hence $ AD\perp EF\quad\mathbb{QED}$[/color]

[color=darkblue][b]Remark.[/b]Let $ ABC$ be a triangle and let $ D$ be a point so that $ BD = CD$.
Denote the reflection $ E$ of ther point $ C$ w.r.t. the line $ AB$ and the reflection $ F$ of the point $ B$ w.r.t. the line $ AC$. Prove that :
$ 1\blacktriangleright$ If $ 0<\mathrm A<45^{\circ}$, then $ AD\perp EF\Longleftrightarrow$ the line $ BC\cap [AD]=\emptyset$ and $ m(\widehat{BDC})=4\cdot\mathrm A$.
$ 2\blacktriangleright$ If $ \mathrm A=45^{\circ}$, then $ AD\perp EF\Longleftrightarrow$ $ D\in BC$ (interesting !).
$ 3\blacktriangleright$ If $ \mathrm A>45^{\circ}$, then the line $ BC\cap [AD]\ne\emptyset$ and $ m(\widehat{BDC})=360^{\circ}-4\cdot\mathrm A$.[/color]


They agreed that they should duel!The rules  of their fight:Each one brings a gun with only [b]one[/b] bullet.They will shoot in a certain order.

We know that King1 and king2 are masters in shooting, they never miss  :ninja: .We also know that King3 is such a poooor shooter.He never hits his target,even i am better than him  :D

Who is the most likely to survive?

Sorry for my grammar  :oops: .

[hide="case 1"]The stragety for King 1 is to shoot King 2. Once he does, King 3 must shoot. He won't hit King 1, so therefore Kings one and three survive.[/hide]

[hide="case 2"]King 1 shoots King 3. Now King 2 will shoot King 1. Therefore, King 2 will survive.[/hide]

[hide="solution"]Since King 1 has 1/2 chance of surviving, (because he has 1/2 chance of shooting King 2, which lets him survive)(he also has 1/2 chance of shooting King 3, then King 2 will shoot King 1.)  and King 3 has 1/2 chance of surviving,(look at the cases again) and King 2 also has 1/2 chance of surviving, each has euqal probability of surviving. Fair game!!![/hide]

[hide]  There are 6 cases.

123: 1 shoots 2, 3 misses 1, 1 and 3 survive

132: 1 shoots 2, 3 misses 1, 1 and 3 survive

213: 2 shoots 1, 3 misses 2, 2 and 3 survive

231: 2 shoots 1, 3 misses 2, 2 and 3 survive

312: 3 misses one of them, 1 shoots 2, 1 and 3 survive

321: 3 misses one of them, 2 shoots 1, 2 and 3 survive

1 and 2 survive half the time, 3 always survives.

[/hide]

[hide]I think King3 because the other two have to try to shoot each other.[/hide]

[quote="alex__ib"]Case 2=Incorrect! Why would you shoot somebody who cannot shoot well, if the other guy is going to kill you?

Hope you understood.... :P[/quote]

You never said the kings weren't stupid :wacko:

[hide]king 3 will live because nobody's going to bother with shooting him.  they only each have 1 bullet.[/hide]

[hide] Well, if they hate each other, that makes it much more interesting! :huh: [/hide]
[hide]Anyways, it's pointless for king1 and king 2 to kill king3 because he posesses no threat to them. If king 3 shoots, he'll miss. now you have king 1 and king2. king1 and king 2 are the only threats to each other.  king 1 and king 2 will likely kill each other, and king3 will win. [/hide]

How would it change if king three would always miss the person that he was shooting at, and instead hit the other person, accidentally? I haven't thought about this at all, so it might be an impossible/stupid question.

so even though king3 is the worst at shooting he stays alive :rotfl:

[hide]kings 1 and 2 can't kill each other! Once one of them is dead they can no longer shoot. They could probably take their gun and bullet, though. So whoever goes first between 1 and 2 will live.[/hide]

i think i have a solution....

[hide]King 1 melees King 2 hard in the back, so he dies. King 1 will wait for King 3 to shoot. Once he does, King 1 will simply shoot King 3.[/hide]


Solve the system of equations:
$\left\{\begin{array}{c}{1}x^{2}+y^{2}=1\ 125y^{5}-125y^{3}+6\sqrt{15}=0\end{array}\right.$

[hide="Solution (not very elegant though)"]Substitute $y={3z\over\sqrt{15}}$ into the second equation to get

$3z^{5}-5z^{3}+2=0$

Obviously $z-1$ divides it:

$(z-1)(3z^{4}+3z^{3}-2z^{2}-2z-2)=0$

The second factor is also divisible by $z-1$:

$(z-1)^{2}(3z^{3}+6z^{2}+4z+2)=0$

Putting $f(z)=3z^{3}+6z^{2}+4z+2$ we get $f'(z)=9z^{2}+12z+4=(3z+2)^{2}, f''(z)=6(3z+2), f'(-{2\over 3})=f''(-{2\over 3})=0$ and $f'(z)>0$ for $z\neq-{2\over 3}$

Hence $z=-{2\over 3}$ is an inflexion point and $f(z)$ is increasing, thus has exactly one real zero. However, from the first equation we have $|y|\leqslant 1\implies |z|\leqslant \sqrt{{5\over 3}}$. For $z=-\sqrt{{5\over 3}}$ we have $f(z)=3(4-\sqrt{15})>0$, therefore the remaining zero can't give a solution to the system (because it's less than $-\sqrt{{5\over 3}}$).

So, $z=1\implies y={3\over\sqrt{15}}=\sqrt{{3\over 5}}\implies (x,y)=\left(\pm\sqrt{{2\over 5}},\sqrt{{3\over 5}}\right)$[/hide]

These two equations imply that $125y^5-125y^3(x^2+y^2)+6\sqrt{15}=0$, which gives $x^2y^3=\frac{6\sqrt{15}}{125}$.
By AM-GM Inequality,
\[\begin{aligned}
1=x^2+y^2=2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}\geq 5\sqrt[5]{\frac{x^4}{4}\cdot\frac{y^6}{27}}=5\sqrt[5]{\frac{(x^2y^3)^2}{108}}
\end{aligned}\]
which gives $x^2y^3\leq\frac{6\sqrt{15}}{125}$. It follows that $\frac{x^2}{2}=\frac{y^2}{3}=k$, where $k$ is a nonnegative real number, and using the condition $x^2+y^2=1$, we get $k=\frac{1}{5}$ and $(x^2,y^2)=\left(\frac{2}{5},\frac{3}{5}\right)$.

Note that with $f(y)=125y^5-125y^3+6\sqrt{15}$, we have $f\left(-\sqrt{\frac{3}{5}}\right)=12\sqrt{15}>0$, so the solutions are $(x,y)=\left(\pm\sqrt{\frac{2}{5}},\sqrt{\frac{3}{5}}\right)$.


let ABCDEF be a hexagon inscribed in a circle of radius R. if AD, BE, and CF are concurrent,
prove that at least one of AB, CD, EF has a length less than or equal to R

Try posting in the Olympiad Geometry section.


I don't get what the problem is asking... Can anybody help?
Ten squares, each with sides one inch, are arranged in plane to form a figure so that no square has all 4 edges on the perimeter of the figure, no 2 squares overlap, and where two squares share a point they share an edge.

The answer is a number, and I don't see the problem.... Please help!

Yea, thats what I got, but the answer says 12... Does anybody have the Solution Manual to the Handbook? HELP!

Oh, and another question....

The question (from Mathcounts Handbook 1994-1995 Warm Up 3 Problem 5):

Peng Chu and Li Juan are each saving to buy the same two CD's. Peng Chu is lacking \$22 and Li Juan is lacking \$3. When they combine their money, they still do not have enough money to buy the two CD's. What is the greatest number of whole dollars the two CD's could cost.

My reasoning, well if they are missing \$25 dollars together, there's no way they can buy the two CD's whatever the cost is. Is there a typo in the problem? I think "When they combine their money, they still do not have enough money to buy the two CD's" should be "When they combine their money, they still do not have enough money to buy one CD".

Thanks!

Edit: NVM, Search button is your friend. (And I found the first problem with the search button) but nobody knows). This question is directed at people that have the Solution to the problem, and can post what the Solution says.


factorial
number theory unsolved
number theory

Let C(n) denote the concatenation of the first n natural numbers, e.g. K(5) = 12345 and K(13) = 12345678910111213.

a.) Is there a square of the form C(n) ?

b.) Is there a factorial of the form C(n) ?

c.) Is there a prime of the form C(n) ?

d.) Are there natural numbers n,m such that K(n)|K(m) ?

We claim C(1) = 1 is the only factorial number.

C(m) = n! with n<=10 only for C(1)=1!.  So assume n>5.

The number of digits in n! is ceiling(log(n!)).  However, log(n!)=log 1+log 2+...+log n = 1/ln10 (ln 1+...+ln n) <= 1/ln10 integral(ln x dx, x=1..n+1)
=1/ln10 ((n+1)ln (n+1) - n)<(n+1)ln (n+1) - n, so ceiling(log(n!))<=ceiling((n+1)ln (n+1) - n)<=(n+1)ln (n+1) - n+1 (1).

However, n! ends with at least floor(n/5) zeros (since it has at least floor(n/5) 5's in its prime factorization).  However, the first time floor(n/5) consecutive zeros appear in C(n) is in C(10^floor(n/5)) which has 9*1+90*2+900*3+...+9*(10^(floor(n/5)-1)(floor(n/5))+floor(n/5) digits.  But we can easily verify that 9*1+...+9*10^k=(k+8/9)10^(k+1)+1/9 (just consider the difference between S=9*1+...+9*10^k and 10S).  Therefore, the number of digits exceeds (floor(n/5)-1/9)*10^(floor(n/5))+floor(n/5)>(n/5-1)*10^(floor(n/5)).  This quantity obviously exceeds the upper estimate derived in (1) for n>10, giving a contradiction.

This proves the claim.

Oh this is problem interesting .I like . IS easy or hard :D  :?

b.) and d.) are not too difficult, but a.) and c.) are rather challenging


algebra
polynomial
number theory
number theory proposed

Let $q$ be a prime power and $\mathbb{F}_q$ the finite field of $q$ elements. Let $d \in \mathbb{N}$. Prove that if an element $\alpha \in \mathbb{F}_q$ is expressible as the sum of $d$-th powers [of elements of $\mathbb{F}_q$], then it is expressible as the sum of at most $d$ such powers.

This is really a nice problem - following solution (which originally dealt with $Z_p$ but it actually doesn't change anything I guess) does not belong to me - it can be found in a book of Narkiewicz "Teoria Liczb":

Denote by $A_i$ set of non-zero  elements of $\mathbb{F}_q$ which are sums of at most $i$ $n$-th powers elements of this field. We have $A_1 \subset A_2 \subset ...$ and finiteness of $\mathbb{F}_q$ implies that for some $t$ we have $A_t=A_{t+1}=...$. Denote by $r$ the smallest such number - we'll prove that $r \leq n$. Consider group $G(q)^n$ of $n$-th powers of elements of $G(q)$ (by $G(N)$ I mean group of units modulo $N$). $G(q)$ is cyclic and so quotient group $G(q)/G(p)^n$ is cyclic; because every element $g$ of the latter satisfies $g^n=1$, thus it has at most $n$ elements. Remark that every set $A_i$ is a set sum of some number of cosets of $G(q)$ with respect to $G(q)^n$. Indeed, if $x \mod q$ lies in $A_i$ and $y \mod q$ lies in the same coset as $x \mod q$, then we have $x \equiv x_1 ^n + ...+ x_i ^n \mod q$ and on the other hand $y \equiv xz^n \mod q$ for some $z \not \equiv 0 \mod q$, hence:
$y \equiv (x_1z)^n+...+(x_iz)^n \mod q$
This implies that $r$ cannot be greater than number of cosets and so $r\leq n$.

I tried different approach though without success:
Consider polynomial equation $x_1^n+...+x_r^n = \alpha$ over $\mathbb{F}_q$. Number of solutions (call it $N$) to it is given by: $N = q^{r-1}+\sum \chi _1 \chi _2 ...\chi _r(\alpha) J(\chi _1,...,\chi _r)$ (see "A Classical Introduction to Modern Number Theory" by Ireland and Rosen) where the sum is over all characters of degree $n$. Now as $r>n$ (we can suppose it as otherwise we obtain the thesis) some of the characters will repeat itself so my idea was to prove that there's some constant $C$ such that:
$\sum \chi _1 \chi _2 ...\chi _r(\alpha) J(\chi _1,...,\chi _r)=C \sum \chi _1 \chi _2 ...\chi _n(\alpha) J(\chi _1,...,\chi _n) \pm k$, where $k$ is something unimportant.

Vess, could you show your solution? (if you have a different approach)

Actually, this is precisely the solution I knew; notice that the argument gives a stronger result, namely, every element expressible as the sum of $d$-th powers is expressible as the sum of at most $(d,p-1)$ such powers. Moreover, this (latter) result is sharp; that is, for any $d \mid p-1$ there is an $\alpha \in \mathbb{F}_q$ that is the sum of $d$ $d$-th powers and not the sum of $(d-1)$ $d$-th powers. Can you see this?

--Vesselin

Well, I cannot see it now but I bet it is highly related to the following fact:

Let $\alpha \in \mathbb{F}_q$. Equation $x^d=\alpha$ has a solution in $\mathbb{F}_q$ iff $\alpha ^{\frac{q-1}{(d,q-1)}}=1$ has a solution in $\mathbb{F}_q$. Moreover then it has precisely $(d,q-1)$ solutions.

Am I right?


Suppose you have a weighted coin in which heads comes up with probability $ \frac34$ and tails with probability $ \frac14$.  If you flip heads, you win 2 dollars, but if you flip tails, you lose 1 dollar.  What is the expected value of a coin flip? Express your answer as a decimal rounded to the nearest hundredth.

By the definition of expected value, $ E\equal{}\frac34\cdot\$2\minus{}\frac14\cdot\$1\equal{}\$\frac54\equal{}\$\boxed{1.25}$.


Discounting a price is different to simply reducing right!

If something is worth 1440 in 2 years time what is its present value if its discounted at 20% per annum. The answer is supposed to be 1000 but it doesn't figure

Could someone please explain

If it's discounted for 1 year it's $1440/1.2$ because 120% = 1.2

If it's discounted for 2 years it's $\frac{1440}{(1.2)^{2}}$ because you divide by 1.2 twice.  and that comes out to 1000.

where do you get 120%?

I am confused as well, if in two years time it is worth 1440, and it has been discounted 20% each year, then shouldnt the current price be > 1440?


If $x,y,z$ are reals satisfying $x+y+z=1$ and $\arctan{x}+\arctan{y}+\arctan{z}=\frac{\pi}{4}$, prove that $x^{2n+1}+y^{2n+1}+z^{2n+1}=1\forall n\in\mathbb{N}$.

[hide="A start"] Well, working backwards from the desired condition, it can only be satisfied if WLOG $x = 1, y =-z$.  So we need to show that this is the only way the first two conditions can be met.

There's probably some kind of convexity argument... [/hide]

That's right!
[hide="hint"]$\tan{ (\arctan{x}+\arctan{y}+\arctan{z})}=1$[/hide]

the second condition is equivalent to $xy+yz+zx=xyz$, and we have $x+y+z=1$, if we eliminate x, and factor, etc, we get $(y+z)(yz-y-z-1)=0$

so two cases...
i) $y=-z$, then $x=1$, the conclusion is obvious

ii) $yz=y+z+1$, so $y=\frac{z+1}{z-1}$, using $1=x+y+z$,
$x=\frac{-(z+1)(z-2)}{z-1}$

uhh....what? i don't think that works...

You factored incorrectly.  It should be $(y+z)(yz-y-z+1) = 0$, and we apply SFFT to finish.


source: KCATM 2005 7-8

A radical middle school principal decides that as the school year progresses and summer approaches, classes should get shorter and passing periods should get longer. The first week of school begins with a standard 8 period day of 45 minutes each with 7 typical 5 minute passing periods. At the start if each new week, the class periods get 1 minute shorter while the passing periods get 20 seconds longer. During what week of school will the time for passing periods overtake the time spent in class?

[hide]Call the number of that week $x$. Therefore, we are looking for a week where $7*(5+\frac{x}{3})>8*(45-x)$. Just solve the inequality and you get $x$ is $31.4$, but $x$ can't be a decimal, so the answer is $32$.[/hide]

[hide]
$8(45-x)=7\left(5+{1\over 3}\cdot x\right)$
$360-8x = 35 + {7\over 3}\cdot x$
$1080-24x = 105 + 7x$
$975 = 31x$
$x=31+{14\over 31}$
So the smallest number of weeks when there will be more passing period is $32$.
[/hide]


function
modular arithmetic
algebra
domain
induction
LaTeX

1. Let $ u_n$ be the $ n$th term of the sequence $ 1,2,4,5,7,9,10,12,14,16,17,\dots$, where one odd number is followed by two evens, then three odds, and so on. If $ u_n \equal{} 2009$, find $ n$.

2. Find all triples of integer solutions $ (x, y, z)$ to $ x^z \plus{} 2009 \equal{} y^z$ where $ z$ is greater than or equal to 2.

3. Find the greatest integer less than $ \sqrt {2\sqrt {3\sqrt {4\cdots\sqrt {2008\sqrt {2009}}}}}$.

4. Prove that there exists no function $ f: \mathbb{N} \rightarrow \mathbb{N}$ such that $ f(f(n)) \equal{} n \plus{} 2009$ for all $ n \in \mathbb{N}$.

My proof
Note that The last number of each frequency is an perfect square number 1,4,9,16...and the number of the numbers in a frequency is 1,2,3,4...
Then $ 44^2\equal{}1936\leq2009$ and from 1937 to 2009 there are 37 numbers.
So n= 1+2+3...+44+37=1027 :lol:

[hide="4"]
Let us work mod 2009. $ \forall a f(a)\not\equiv a\pmod{2009}$, for then $ \exists n$ such that $ f^n(a) \equal{} f(a)$, meaning $ f$ would be a periodic function for $ a$, which is a contradiction since $ g(a,n) \equal{} f^{2n}(a)\equal{}a\plus{}2009n$ and thus diverges.

Thus, $ f$ has period 2 mod 2009 for any number, meaning $ f$ mod 2009 can be expressed purely by disjoint 2-cycles, meaning that $ f$ mod 2009 must have a domain of even order. But $ f$ mod 2009 has a domain of odd order. Contradiction![/hide]

Perhaps the solution is as follows(no 3):
[hide="HINT"]Use strong reverse induction[/hide]
[hide="solution(hopefully)"]
For $ m\geq 2$, we have the problem generalized as:
$ \sqrt {m\sqrt {(m \plus{} 1)\sqrt {.....\sqrt {N}}}} < 3$
Now we use reverse induction. Clearly, for $ m \equal{} N$, $ \sqrt {N} < N \plus{} 1$.
Assume inductively, for $ m < n$,
$ \sqrt {(m \plus{} 1)\sqrt {(m \plus{} 2)\sqrt {...\sqrt {N}}}} < m \plus{} 2$
Then, $ \sqrt {m\sqrt {(m \plus{} 1)\sqrt {...\sqrt {N}}}} < \sqrt {m(m \plus{} 2)} < m \plus{} 1$
Thus, it is proved that  $ \sqrt {2\sqrt {3\sqrt {4\cdots\sqrt {2008\sqrt {2009}}}}} < 3$by plugging $ m \equal{} 2$ :D
Hence$ [ {\sqrt {2\sqrt {3\sqrt {4\cdots\sqrt {2008\sqrt {2009}}}}}} ] \equal{} 2$ [/hide]
Whew!That was a problem!! :) :lol:
[hide="P/S:"]Can anyone tell me how to use the box sign ([*]) in LaTeX which means the greatest integer less than or equal to * ?? :blush:[/hide]

[quote="Potla"]
[hide="P/S:"]Can anyone tell me how to use the box sign ([*]) in LaTeX which means the greatest integer less than or equal to * ?? :blush:[/hide][/quote]
[url=http://www.mathlinks.ro/Wiki/index.php/LaTeX:Symbols]Here[/url]

Hello, I have proceeded quite a bit in problem no 2.
$ y^z \equal{} x^z \plus{} 2009$ implies that $ x < y$ and $ y^z \minus{} x^z \equal{} 2009 \equal{} 49.41$
then we see that both of $ x,y$ cannot be odd or both cannot be even.
One has to be odd, and the other even.
$ (y \minus{} x)(y^{z \minus{} 1} \plus{} y^{z \minus{} 2}x \plus{} ........ \plus{} x^{z \minus{} 1}) \equal{} 7^2.41$
If only IT would have been given that $ x,y,z\in\mathbb{N}$ then I could have taken combinations of the factors of $ y^z \minus{} x^z$. Also 41 is a prime...
[color=red]BUT I DO NOT KNOW HOW TO PROCEED![/color]
Please help me :(


logarithms
function
real analysis
real analysis theorems

[b]
prove that a to the power x[/b] into [b]b to the power x [/b]is equal to [b](ab) whole to the power x[/b]

[b]please do it when x is irrational.[/b]  :?:

$x \log a+x \log b = x \log ab$

This is easily proved by a powerseries argument.

[size=150][color=red][b]But what if i dont go so far as the existance of logs and so on? say i have only the real no.s ;there addition ;multiplication and power defined (acctually in our class the analysis teacher has gone only so far and gave us  this problem in the exam)[/b][/color][/size]

Your approach may depend on how you define $a^{x}$ with $x$ irrational. One way to do this is to define it for $x$ rational, and then prove that $a^{x}$ depends continuously on $x$. Then you already know that both sides of the identity are continuous functions of $x$. Since the identity holds for rational $x$, and rationals are dense in the reals, it must hold for all reals.

If this does not work for you, then please give us the definition of $a^{x}$ that your teacher used. (In black font of normal size, if possible).


Define the sequence $ (a_p)_{p\ge0}$ as follows: $ a_p\equal{}\displaystyle\frac{\binom p0}{2\cdot 4}\minus{}\frac{\binom p1}{3\cdot5}\plus{}\frac{\binom p2}{4\cdot6}\minus{}\ldots\plus{}(\minus{}1)^p\cdot\frac{\binom pp}{(p\plus{}2)(p\plus{}4)}$.
Find $ \lim_{n\to\infty}(a_0\plus{}a_1\plus{}\ldots\plus{}a_n)$.

Using $ f(x) \equal{} \sum_{0}^{p}\frac {C_{p}^{k} x^{k \plus{} 4}}{(k \plus{} 2)(k \plus{} 4)}$ then $ a_p \equal{} f( \minus{} 1)$
I found $ a_p \equal{} \frac {p^2 \plus{} 7p \plus{} 6}{2(p \plus{} 1)(p \plus{} 2)(p \plus{} 3)(p \plus{} 4)} \equal{} \frac {1}{p \plus{} 2} \minus{} \frac {3}{2}\frac {1}{p \plus{} 3} \plus{}\frac{1}{2}\frac {1}{p \plus{} 4}$
$ \sum_{n \equal{} 0}^{ \plus{} \infty}a_n \equal{} \frac {1}{3}$
Please check it :P


I was in awe until I remembered what day it is....

I just wanted to read it....

And Gyan beat you to the first April Fools joke  :P

[quote="hello"]HAHAHAHAHA. This is the beginning of april fool's for AoPS. Wonder how many people will view this topic.[/quote]

He claims he did  :D  :D  :D

[quote="Derek"]You didn't get me. Haha. I am very careful today on trusting people.  :lol:[/quote]

I think I did get you. DeBranges has been disproved I think in at least three ocations, nevertheless, he swears that now he really got the prove.

Best,

Isn't that what happend to that Bierbach conjecture too, he made gaps and then finally proved it. Also, Gyan wrote when it wasn't April 1 here yet. So, I tricked some of u, eh?

Also, I did get you if you looked in this topic, it counts.

What did Gyan do?
EDIT: never mind, it's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31996]this thread[/url] right?

Who knows maybe it is true, International Congress of Mathematician
will be in 2006 in Spain.

Ouch. Now I feel dumb... :blush:

For a second there I thought you were serious. :(

After spending all day reading april foools stories on Slashdot, I thought "this is an april fools joke, isn't it", and by golly I was right.

[quote="pkothari13"]i dont even know who "De Branges" is.. so it could be true for all i know.. *shrug. but i guess you got me? kudos to you....?[/quote]

Louis de Branges de Bourcia is a French mathematician currently working in the University of Purdue (Indiana, USA) who something like twenty years ago proved The Bieberbach Conjecture. He has proposed several times a proof for the Riemann Hypothesis, nevertheless, it has been disproved (someone has find an error for his proof) each single time, and I believe the last proof (probably a modification of the last before that one) is not been disproved yet (not sure), but after a lot of failures, many people do not take him seriously.

[quote="djimenez"][quote="Scrambled"]it would be extremely funny if a mathematician proved a millenium problem shortly before april fools, and held it until then to release it :D[/quote]

Well, it depends on where the mathematician comes from and where is he at that moment. As far as I know, April fools day is a totally American (understanding America as the United States of America) costum. At least it is not used in Central and South America, and I really haven't done a real survey about it, but the few Europeans that I have asked about say that it is not something that they use to celebrate.

Therefore, if the guy is, say, African and he is working in, say, Japan, I really don't think that he would have that many implications that it is a joke.

Best,[/quote]
well there have been so many frauds such as "FLT proved false!", "Reimann hypothesis proved"(This was a prominant joke for quite a few years), and a whole farrago of millenium problems that have circulated through email, but they were only put down some time later as an "april fools joke".

plus, i remember quite a few european mathematicians doing april fools jokes


Sorry if this should be in HSB.
Well, here's the equation (all angles in degrees):
$cos{35}+cos{125}+2sin{185}(sin{130}+sin{140})$

Anyone know how to find the value of this without a calculator? I've simplified it down so that all angles are less than 90 degrees, but I can't do much more from there.

Thanks in advance.

[hide="Identity that might be good to know"] $2 \sin a \sin b = \cos (a-b)-\cos (a+b)$ [/hide]

cos35+cos125+2sin185(sin130+sin140)
=2cos80cos(-45)-2sin5 (2sin135cos(-5))
=2sin10cos45-2sin5cos5(2sin135)
=2sin10cos45-2sin10sin45
=0

We use:
cos a + cos b=2cos[(a+b)/2]cos[(a-b)/2]
sin a + sin b=2sin[(a+b)/2]cos[(a-b)/2]
2sin a cos a=sin2a
to simplify it.

Right, thanks a lot! You guys saved my butt... :P