linear algebra
matrix
geometry
incenter
absolute value
national olympiad

PROB4:AT the end of a soccer tournament in wich each pair of teams played exactly once and there were no games ending in a draw, it was observed that for any three teams A,B  and C if A beat B and B  beat C  then  A  beat C. each team calculated the psitive difference between the number of games they won and the number of games they lost. the sum of all these differences was 5000. ¿how many teams were in the tournament? find all possible solutions.

prob5. let A  and B be 2 circunferendces such that the center of B is on A .  let  C and D  be the intersections between the circunferences. take A'  on  A  and  B'  on  B  such that A'C  is tangent to B on C  and B'C is tangent to A  in the same point C.  the segment A'B'  meets B  again at E  and A  at F . the linne  CE  meets  A again G and  CF  cuts GD at  H. PROVE that point of intersection of GO and EH is the center of the circle passing through D, E , and F.

PROB6. ¿WHAT IS THE MAXIMUM NUMBER OF DIRECTION CHANGES IN A STROLL THROUGH THE LINES OF A 2004 X 2004 DIVIDED INTO 2004X 2004 UNIT SQUARES WITH LINES PARRALEL TO THE FOUR SIDES OF THE SQUARE IF THE STROLL DOESNT PASS THROUGH A SAME POINT TWICE?

 SO HERE IS THE SECOND DAY ¿ANY NICE SOLUTION FOR PROBLEM 6? THE OTHER ARE PRETTY STRAIGHT FORWARD IN MY OPINION.

[quote="marko avila"]PROB6. ¿WHAT IS THE MAXIMUM NUMBER OF DIRECTION CHANGES IN A STROLL THROUGH THE LINES OF A 2004 X 2004 DIVIDED INTO 2004X 2004 UNIT SQUARES WITH LINES PARRALEL TO THE FOUR SIDES OF THE SQUARE IF THE STROLL DOESNT PASS THROUGH A SAME POINT TWICE?
[/quote]
Sorry Marko, I am an spanish and I can't understand the problem, can you traduce to Spanish, thanks (gracias)

RX TCM

[quote="marko avila"]OKS AI TE VA EN ESPAÑOL:
¿CUAL ES EL MAYOR NUMERO POSIBLE DE CAMBIOS DE DIRECCION EN UN RCORRIDO SOBRE LAS LINEAS DE UNA CUADR?CULA DE 2004X2004 CASILLAS, SI EL RECORRIDO NO PASA DOS VECES POR EL MISMO LUGAR? :D[/quote]
JAJAJA, thanks I supposed that   :blush:  :blush: Well, it's easy to prove that in a $n\times n$ square, the maximum walk you get in a snake walk, so the maximum number of edges where the snake passes is $(n-1)^2-1$, but we don't  have the maximum of changes, now if we get the maximum of changes with same length we complete the problem. But, now if we try to turen in every moviment we can see that in every row the number of changes in every posibility of maximum we try to get is the same (try to prove it) so the strategy is the following:

START: if we consider the  "cruzes" as a elemnt of a $(n-1)\times (n-1)$  matrix then I start in Left of (1,1), and go in the following way:
\[
\to (1,1) \to (2,1) \to (2,2)\to (1,2)\to \cdots (1,k)\to (1,k+1)\to (2,k+1)\to \cdots,
\]
as so far as is posible to continue, if I conclude in $(1,n)$  go to $(2,n)$, so I don't add a new change of direction [$2n-3$ changes] go up two positions and do the same to the oposite direction, in this case we need $2(n-2)$ changes, again up two positions until I can't go up two positions, in this case I only can go until the  end straight on and turn left in the last posible moviment, so we get a zig-zag movimient which is optimal in moviments and changes, so the solution depends of the parity of $n$ , if $n$ is even $(n-2)^2+3$ and if $n$ odd $n^2-3n+4$, so taking $n=2004$ then the maximum you need is $2002^2+3=4004007$.

Best regards.

RX TCM

I don't know if this works, but it could be a nice approach, without having to go through a constructive solution, showing that this or that paths are "the best".

I assumed that your large square was divides in $n^2=2004^2$ small squares. This means we have $(n+1)^2=2005^2$ points in our grid. Assume you make $2004\cdot 2005$ direction changes. Since $n$ is even, it shouldn't be too hard to show that there can't be a change of direction on all the $n+1$ points of some line, so on each of the $n+1$ linea/columns, we must have exactly $n$ changes of direction.

Consider, however, the terminal vertex of the path, and assume it was reached through a horizontal segment of the path. It means it was paired off with one of the vertices where we have a change of direction. We are now left with $n-1$ vertices with change of direction on that line, and, because $n-1$ is odd, it's easy to get a contradiction (we get the same sort of contradiction when we show that we can't have $n+1$ vertices with change of direction on one line): each vertex with change of direction adds another horizontal segment, which connects it with an adjacent vertex with change of direction. This means we can eliminate the vertices two by two, and we are left with one vertex, which cannot have a horizontal segment of the path attached to it, so it can't have a change of direction, contradiction.

Sorry if it's sort of unclear.

Edit:

Forgot to add this: the above was intended to show that we can't have more than $n(n+1)-1$ changes of direction when $n$ is even. Conversely, it's not hard to find paths with exactly $n(n+1)-1$ changes of direction, so the bound this is indeed the answer for the problem.

[quote="grobber"]
Edit:

Forgot to add this: the above was intended to show that we can't have more than $n(n+1)-1$ changes of direction when $n$ is even. Conversely, it's not hard to find paths with exactly $n(n+1)$ changes of direction, so the bound this is indeed the answer for the problem.[/quote]
Well, I think the exact value if $n$ is even is $(n-2)^2+3$.

RX TCM

I don't know if your $n$ is the same as my $n$. For me, the $n$ is the number of squares on each side of the large square. This means that we have $(n+1)^2$ points in the grid on which we are moving. For this definition of $n$, I can construct an example for $n(n+1)-1$ changes of direction, and that's much more that the value you wrote. However, if your definition of $n$ is another one, who knows? maybe we're both right :).

[quote="grobber"]
Edit:

Forgot to add this: the above was intended to show that we can't have more than $n(n+1)-1$ changes of direction when $n$ is even. Conversely, it's not hard to find paths with exactly $n(n+1)$ changes of direction, so the bound this is indeed the answer for the problem.[/quote]

MY answer is WRONG, I didn-t understood the problem  :?  :blush: :blush:   :no:

WILL SOMEONE PLEASE SOLVE THE OTHER PROBLEMS???
UM I THINK I WAS TOLD TO CREATE POSTS FOR EACH OF THE PROBLEMS FOR SOLUTIONS BUT I DONT KNOW HOW TO DO THAT¡¡¡ :blush:  HOW DO I DO IT??' :blush:  :(

I'll sketch a proof for the second one, but there are too many properties to prove (it's easy to prove them; basicly, it's all angle chasing).

First show that $CE=CF$, then $GE=GD$, and then show that $\angle CA'F=\angle GA'D\Rightarrow CF=GD$. From here we find that $FC=CE=EG=GD$. We thus find $HC=HG$, so $HE$, being ammedian in $GHC$, it's also a bisector. It's easy to see that $GO$ is the bisector of $\angle CGH$, so their point of intersection is the incenter of $CGH$. Call this result $(*)$. Since $CE=CF,HF=HD,GD=GE$, it means that $(DEF)$ is the incenter of $CGH$. Together with $(*)$, this gives the result.

For the first one:

Because of the property $A\to B,B\to C\Rightarrow A\to C$, we can define a total order on the set of vertices of the graph (teams, if you prefer :)) by putting $A\le B$ iff $A=B$ or there's an arrow from $A$ to $B$. We can now arrange the $n$ vertices in an increasing order. For each point, we take the absolute value of the number of points before it and after it, and sum all of these up, getting $5000$.

If $n$ is even, then that sum is $2(1+3+\ldots+n-1)=\frac{n^2}2=5000$, so $n=100$. If $n$ is odd, we get $2(2+4+\ldots+n-1)=\frac{(n-1)(n+1)}2=5000$, which does not have a solution. I suppose the only solution is $n=100$ (actually, for it to be complete, we would also have to construct an example with $100$ teams :)).

Is it correct, marko? The fact that you wrote "find all possible solutions" gives me some doubts... :?

yes it is correct¡¡¡¡ your solution is new to me so many thanks¡¡¡ :(

[quote="marko avila"]yes it is correct¡¡¡¡ your solution is new to me so many thanks¡¡¡  :)

[quote="Valentin Vornicu"]Marko please stop using CAPS LOCK in your messages / titles of the topics. It is not good for the eye to see that many capitalized letters :)[/quote]

very sorry¡¡¿ :blush:


Is it good to understand balls and urns for mc?

Well, anything is good to learn.

However, I think there are a few distinguishability problems in MATHCOUNTS, so memorizing the formula and how to derive it will do you good  :) .


this question was posted 4 d sake of a junior of mine who had serious allegations about d forum statin dat questions posted here were not of their syllabi

now for those interested add 13 to rhs of question and try again


$ i_{1}$ has 3 different possibilities
$ i_{2}$ and $ i_{3}$ have 5 possibilities
$ i_{4}$ has 6 possibilities
$ i_{5}$ and $ i_{6}$ have 4 possibilities

You multiply that together and you get $ 3*5*5*6*4*4\equal{} 7200$

We count the configurations:

4xx1xx = 4!- ( (3!)4-2(2!) ) = 4

4xxx1x = 4xxxx1 = 5xx1xx = 6xx1xx = 4! - 2(3!) = 12

5xxx1x = 6xxx1x = 5xxxx1 = 6xxxx1 = 4! - 3! = 18

Number of required permutations = 4 + 4(4) + 18(4) = 92


[hide]
she has 3 quaters, 1 dime, 3 pennies.
That means she has [b]1 dime[/b][/hide]

[hide]3 quarters and 3 pennies makes 6 coins with 78cents with 1 coin left over. That would be a dime. $1$ $dime$[/hide]

[quote="joejia"]Molly has seven U.S. coins with a total value of 88 cents. She does not have any half-dollars. How many dimes does Molly have?[/quote]

[hide="Click here"][hide="Click here"][hide="Click here"]3 quarters, 3 pennies, 1 dime.[/hide][/hide][/hide]

lol, that took forever to click.  and my answer disagrees with your answer, someperson01. can you tell me how you did it?

He got carried away with the hide tags  :rotfl:

There is
[hide]1 dime. (grammar gave it away  :P )[/hide]

[quote="someperson01"][hide][hide][hide]3 dimes[/hide][/hide][/hide][/hide][/quote]

[quote="236Factorial"]He got carried away with the hide tags[/quote]

Funny. Well, how did you get your answer?

Anyway, mine had 15 hide tags.

[quote="236factorial"]Me? same way as the others.

Clearly, 3 dimes can't be the answer because you can't have 4 coins that total 58 cents. You'll need a 55-cent coin, which doesn't exist.[/quote]
i think he was talking to someperson1 becuase he/she got a different answer then most of us.

[quote="mysmartmouth"]What was that different answer?[/quote]

3 dimes.


analytic geometry
linear algebra
matrix
topology
linear algebra unsolved

I want to evaluate the determinant of the matrix $ (a_{ij})_{1 \leq i,j \leq n}$ with entries

$ a_{ii} \equal{} \left\|y\right\|_2^{\minus{}4}(\left\|y\right\|_2^2 \minus{} 2 y_i^2) , a_{ij} \equal{} \left\|y\right\|_2^{\minus{}4} (\minus{} 2 y_i y_j)$ for $ i \neq j$

(where $ y \in \mathbb{R}^n \backslash \{0\}$ is fixed; the purpose is to show that $ S^n$ is an oriented manifold; I chose the sterographic projection so that the change of coordinates is $ y \mapsto \left\|y\right\|_2^{\minus{}2} y$, and $ a$ is its derivative).

but I don't know how to compute this determinant; examples show that it seems to be $ \minus{}\left\|y\right\|_2^{\minus{}2n}$.

You can solve this problem directly using row-reduction.  To connect back to the geometric setup it is helpful to observe that the determinant is of the form
\[ det(M\plus{}\kappa I)\] where $ M \equal{} |y \rangle \langle y|$ is a projection matrix satisfying
$ M |y \rangle   \equal{} |y|^2 |y \rangle .$  It is straightforward to check that $ M$ has $ |y \rangle$ as an eigenvector with eigenvalue $ |y|^2$ and the rest of $ M's$ eigenvalues are 0 (multiplicity $ dim M \minus{} 1$).

The eigenvalues of $ M \plus{} \kappa I$ are $ |y|^2 \plus{} \kappa$ and $ \kappa$ with multiplicity $ n\minus{}1.$  The desired determinant follows since the determinant of a matrix equals the product of its eigenvalues.

sorry I don't understand your notation, you write $ A\equal{}M \plus{} \kappa I$ with which $ M$ and $ \kappa$?

[quote="-oo-"]I want to evaluate the determinant of the matrix $ (a_{ij})_{1 \leq i,j \leq n}$ with entries

$ a_{ii} \equal{} \left\|y\right\|_2^{ \minus{} 4}(\left\|y\right\|_2^2 \minus{} 2 y_i^2) , a_{ij} \equal{} \left\|y\right\|_2^{ \minus{} 4} ( \minus{} 2 y_i y_j)$ for $ i \neq j$

(where $ y \in \mathbb{R}^n \backslash \{0\}$ is fixed; the purpose is to show that $ S^n$ is an oriented manifold; I chose the sterographic projection so that the change of coordinates is $ y \mapsto \left\|y\right\|_2^{ \minus{} 2} y$, and $ a$ is its derivative).

but I don't know how to compute this determinant; examples show that it seems to be $ \minus{} \left\|y\right\|_2^{ \minus{} 2n}$.[/quote]
In fact,You just want to calculate the determinant of the matrix $ A \equal{} \left\|y\right\|_2^2I_n \minus{} 2yy'$
where $ I_n$ is the identity of the n square,and $ y \equal{} (y_1,y_2,\cdots,y_n)'$
then
$ \det(A) \equal{} \left\|y\right\|_2^{2n} \det(I_n \minus{} \frac {2}{\left\|y\right\|_2^2}yy') \equal{} \left\|y\right\|_2^{2n}\det(1 \minus{} \frac {2}{\left\|y\right\|_2^2}y'y) \equal{} \minus{} \left\|y\right\|_2^{2n}$

So we get the your answer  $ \minus{} \left\|y\right\|_2^{ \minus{} 2n}$.

the critical fomular here is:
$ \det(I_n \minus{} AB) \equal{} \det(I_m \minus{} BA)$. :)
You can also use geometric to find n eigenvalues.


trigonometry
function
calculus
integration
geometry
rate problems

1. A voltmeter reads 88 V when connected across the terminals of a sinusoidal power source with frequency 888 Hz. The equation for instantaneous voltage (in volts) provided by the source can be given as

A. $ 88 \sin 888t$
B. $ 88 \sin 5580t$
C. $ 125 \sin 5580t$
D. $ 125 \sin 888t$

2. A mini-power station provides 10 MW of power to a small town. The resistance of the transmission cable is $ 10.0 \Omega$ and the power is transmitted at a voltage of $ 50.0 \textup{ kV}$. The potential drop across the cable in $ \textup{kV}$ os

A. 1.0
B. 2.0
C. 10.0
D. 50.0

3. A sinusoidal current of peak value $ I_0$ is passed through a resistor $ R$ with a diode connected in series with it. The average rate of heat dissipated in $ R$ is

A. $ 0.25 I_0^2 R$
B. $ 0.50 I_0^2 R$
C. $ 0.71 I_0^2 R$
D. $ I_0^2 R$

Solutions steps would be appreciated  :)

These are AC current questions.  These should be in the Advanced Physics forum.

The answer to the first question is C.  Remember that the voltmeter actually reads the rms voltage, which is the voltage amplitude divided by $ \sqrt {2}$.  Also, the coefficient of $ t$ is the angular frequency (the frequency multiplied by $ 2\pi$), not the frequency itself.

I don't understand question 2.

As for the third question, I'm not getting an answer that's on the list.  With a diode, the graph of the current through the resistor will look like a sinusoidal function with the negative parts cut off; you can determine the average value of the current by integrating (which isn't an introductory physics topic), which I think is either $ \frac {2}{\pi}I_0$ or $ \frac {I_0}{\pi}$, but that doesn't give any of the answers on your list.

Question 1

The 88 volts is the root mean squre voltage so $ V_{peak} \equal{} \sqrt {2} V_{rms} \equal{} \sqrt {2} \times 88 \approx 125 \, V$

The angular frequency is $ \omega \equal{} 2\pi f \equal{} 2\pi \times 888 \approx 5580 \, rad \, s^{ \minus{} 1}$

So the answer we require is C.

Question 2.

Write all quantities in standard form.

$ P \equal{} 1.0 \times 10^7 W$
$ R \equal{} 1.00 \times 10^1 \Omega$
$ V \equal{} 5.00 \times 10^4 V$

Now we can calculate the current: $ P \equal{} IV \Rightarrow I \equal{} \frac {P}{V} \equal{} \frac {1.0 \times 10^7}{5.00 \times 10^4} \equal{} 2.0 \times 10^2 \, A$

$ \Rightarrow V \equal{} IR \equal{} 2.0 \times 10^2 \times 1.00 \times 10^1 \equal{} 2 \times 10^3 \, V \equal{} 2 \, kV$

So the answer is B

Question 3

I would like to stress to genericme that the square of the mean is not the mean of the square for continuous random variables. Otherwise there would be no such thing a variance or standard deviation, which depends on the difference of such quantities. I highlight this in the following example.

Take an electrical DC circuit. Say there is a resistor of 1 Ohm attached in parallel with the power supply. By changing the voltage at the supply a current of 1 Ampere flows through the resistor for 1 second and then a current of 2 Amperes flows through the resistor for the following second. Now lets calculate the energy transfered to the resistor during this time:

For the 1st second:

$ P \equal{} I^2R \equal{} 1^2 \times 1 \equal{} 1 \, W$ so one Joule of energy is transferred.

For the 2nd second:

$ P \equal{} I^2R \equal{} 2^2 \times 1 \equal{} 4 \, W$ so four Joules of energy is transferred.

Thus total energy transferred, I hope you will agree is 5 Joules, and hence the mean power is $ P \equal{} \frac {5}{2} \equal{} 2.5 \, W$.

Now lets average the current over the time interval. It can be seen immediately that this is 1.5 Amperes. You can check this trivial result by using integration (Note for the first two seconds, the area under the current time graph is 3 which may be done with basic geometry). Without further ado, we can find the mean power calulated by this method as:

$ P \equal{} I^2R \equal{} 1.5^2 \times 1 \equal{} 2.25 \, W$

This is not in agreement with the actual mean power. Note this is just a physical argument that if you average 1 and 2 then square, then this is different to if you square both then average.

This whole difficulty has been simplified by the concept of root mean square voltage. Now contrary to popular belief in school students (and some teachers!) the root mean square voltage/current, is not the mean magnitude of voltage/current over a time interval. You can check this by performing the integral $ \frac {\int_{t \equal{} 0}^{t \equal{} \frac {2\pi}{\omega}} |V_0 \sin(\omega t \plus{} \phi)| dt}{\frac {2\pi}{\omega}}$ where $ \phi$ is just a phase angle, which may be removed with appropriate redefining of the integration limits. So this begs the question, what is root mean square voltage? Well it is the steady DC voltage that if applied to a circuit component such as a resistor or bulb, will on average give the same power to the component as a particular sinusoidal AC voltage does. There is however a small problem with this. If we choose the frequency of the AC supply to be low, then we will visually see a bulb becoming bright then dimming and then becoming bright then dimming... and so on. So the power is not constant during a single oscillation of the supply voltage. However, each oscillation must deliver the same total energy to the circuit component since each is identical, so we can define the power as the energy transferred by one oscillation, divided by the time taken for such an oscillation (we could actually take each "half" oscillation instead since each idelivers identical energy - energy transfer is independent of the direction of the voltage!). Now the definition of root mean square current follows naturally: it is the steady DC current that if applied to a circuit component such as a resistor or bulb, will on average give the same power to the component as a particular sinusoidal AC current does. Now note that even for variable voltages, Ohm's law must apply at any instant in time. So we write them as functions of time to make this obvious:

$ R \equal{} \frac {V(t)}{I(t)}$

and so it follows $ R \equal{} \frac {V_{rms}}{I_{rms}} \Rightarrow V_{rms} \equal{} I_{rms} R$.

So Ohm's Law agrees if we replace variable voltages/currents with their root mean squares. Now to find these "rms" quantities.

Take a sinusoidal current with peak $ I_0$ and angular frequency $ \omega$. For convenience, take the zero of time to be when the current is 0, but increasing. Then:

$ I(t) \equal{} I_0 \sin{\omega t}$
$ P(t) \equal{} (I(t))^2R \equal{} I_0^2 R \sin^2{\omega t}$ - here we SQUARE.

The period of an oscillation is $ T \equal{} \frac {2\pi}{\omega}$. So we need to integrate our power from $ t \equal{} 0$ to $ t \equal{} \frac {2\pi}{\omega}$ to find the energy transferred.

$ E \equal{} \int_{t \equal{} 0}^{t \equal{} \frac {2 \pi}{\omega}} P(t) \, dt$
$ \equal{} I_0^2 R \int_{t \equal{} 0}^{t \equal{} \frac {2 \pi}{\omega}} \sin^2{\omega t} dt$
$ \equal{} \frac {I_0^2 R}{2} \int_{t \equal{} 0}^{t \equal{} \frac {2 \pi}{\omega}}(1 \minus{} \cos(2\omega t)) dt$
$ \equal{} \frac {I_0^2 R}{2} \left[ t \minus{} \frac {1}{2 \omega} \sin(2 \omega t) \right]^{\frac {2 \pi}{\omega}}_{t \equal{} 0}$
$ \equal{} \frac {I_0^2 R}{2} \frac {2 \pi}{\omega}$

Now divide by the time interval $ T \equal{} \frac {2 \pi}{\omega}$ to obtain the power - here we MEAN

$ P \equal{} \frac {I_0^2 R}{2}$

Now by definition $ P \equal{} I_{rms}^2 R$

$ \Rightarrow I_{rms}^2 R \equal{} \frac {I_0^2 R}{2}$

$ \Rightarrow I_{rms}^2 \equal{} \frac {I_0^2}{2}$

$ \Rightarrow I_{rms} \equal{} \frac {I_0}{\sqrt {2}}$ - here we ROOT

So we have just proved the relationship between rms and peak (currents) - hence it transfers by Ohm's law to voltages. Also I have hopefully demonstrated through the use of capitals, why it is root mean square. (You may ask, why if you square then mean then root, is it not called square mean root....the answer is mathematical convention! Say we have three functions $ f(x), g(x), h(x)$. If we want to take an $ x$ then let $ f$ act on $ x$ first, then let $ g$ act on the result, then let $ h$ act on this result, in that order. Then we write the answer as $ h(g(f(x)))$, or simply if you consider functions as elements in a "group" (a set of elements, with a binary relation satisfying certain conditions) $ hgf \,(x)$)

Now much all of the above was just for clarification, but we can use the fact that root mean square current gives the same power in the problem:

The diode only allows current to flow in one direction. But as remarked obove, the rms current agrees for mean power outputs of half oscillations. So now to calculate the energy transferred in one oscillation of the supply voltage, with period "T" say. Then the energy transfered while the current is allowed through on the first half of the oscillation is

$ E \equal{} Pt \equal{} I_{rms}^2 R \frac {T}{2} \equal{} \frac {I_0^2}{4} RT$

In the second half of the oscillation, no energy is transferred. So total energy transferred by one oscillation of supply voltage is $ E \equal{} \frac {I_0^2}{4} RT$ and dividing this by the period of  an oscillation gives the mean power as $ P \equal{} \frac {I_0^2}{4} R \equal{} 0.25I_0^2R$

Answer is A.

(as another note to genericme, dimensional analysis would clearly make your answer suspect).

I studied this at school - perhaps with less mathematical rigour - when I was 16. It is not advanced physics, since we are effectively approximating with D.C instead of using complex phasors to model oscillations.

Cool, I see where I went wrong, namely using the average value of the current rather than the rms value.

*brain fart*

I can't believe I misread question 2.

[quote="sludgethrower"](as another note to genericme, dimensional analysis would clearly make your answer suspect).
[/quote]

Are you referring to my factor of $ \frac {2}{\pi}$?  That's dimensionless - it's just the average value of the sine function over one half-period.

[quote="sludgethrower"]I studied this at school - perhaps with less mathematical rigour - when I was 16. It is not advanced physics, since we are effectively approximating with D.C instead of using complex phasors to model oscillations.
[/quote]

It's preferred that you don't use calculus in the introductory physics forum, though.  I actually think phasors are a neat and not-too-complicated trick, and even easier to understand for someone at the introductory level than your treatment above.  I also happen to be 16, btw.  But I think the regular physics education system in U.S. high schools is too crappy to cover this material.

Apologies, genericme, I thought that your answer was refering to the power of the system, but I look back, and see that you were giving the mean current.

I understand what you mean. The above may be a little swamped in formal derivation. A simpler, (non-calculus) way of deriving the relation between peaks and r.m.s is as follows.

Consider a sinusoidal oscillation on a current-time graph. Say the peak current is $ I_0$. We need to find the mean value of $ I^2$ so we can use it in our calulation of $ P \equal{} I^2R$. Well square our signal. The graph obtained will be another sinusoidal wave, which ranges from 0 to $ I_0^2$, instead of $ \minus{} I_0$ to $ I_0$ as before. Secondly the mean value is going to be the middle, so at $ \frac {I_0^2}{2}$. Thus $ I_{rms}^2 \equal{} \frac {I_0^2}{2}$

Thanks all.. I'm quite sure these belong to introductory physics as they are 'A' Level problems  :)

Yes, they are "Higher" problems in Scotland. Higher is slightly more demanding than AS level...but not by much.


induction
combinatorics proposed
combinatorics

Let $n$ be a positive integer. Does there exist a positive integer $m$ s.t. every tounrament with $m$ players contains a transitive subtournament with $n$ players?

(a transitive tounrament is one in which whenever $A$ beats $B$ and $B$ beats $C,\ A$ beats $C$)

Yes;
We may choose $m = 2^n.$

We prove this assertion by induction on $n$.
For $n=1,2$ the result is trivial.
Now, assume that it holds for $n-1.$
Let $T$ be any tournament with $2^n$ people. Let $x$ be any of them.
Let $W$ (resp. $L$) be the set of all people $x$ has beaten (resp. who have beaten $x$). We have $|W|+|L| = 2^n - 1$ so that one of these two sets has at least $2^{n-1}$ elements. Wlog, we will assume that $|W| \geq 2^{n-1}$. By the induction hypothesis, we may find a transitive sub-tournament in $W$ with cardinality $n-1.$ Now, add $x$ and we have a transitive tournament with cardinality $n$. Hence result.

Pierre.

Hmm...I think $m = 2^{n-1}$ also works in exactly the same way.
Does anybody know about the best bounds?

Pierre.

Yes. Let $ T(n) $ be the order of the smallest tournament which will always contain a transitive subtournament of $ n $ vertices. We'll show by induction that $ T(n) $ is finite for all $ n $.

It's easily checked that $ T(3) = 4 $. Suppose that $ T(i) = k $. Consider a tournament of $ 2k $ vertices. Choose any vertex; it must either beat $ k $ other vertices, or be beaten by $ k $ other vertices. Within these $ k $ vertices, we can, by the induction hypothesis, find a transitive subtournament of $ i $ vertices, so together with this vertex, we have a transitive subtournament of $ i + 1 $ vertices, as required, and hence $ T(i+1) \leq 2k $.

The bound obtained from this proof, $ T(n) \leq 2^{n-1} $ is quite poor tho.

Edit: Oops, I posted just after pbornsztein

Edit: The arguments in the Cases are incorrect.

On furthur thought, I think I can show that $ T(n+1) = 2T(n) $, and consequently $ T(n) = 2^{n-1} $. For, if $ T(n) = k $, there must exist some tournament $ T $ of $ k-1 $ vertices that does not have a transitive subtournament of $ n $ vertices. Let the vertices of $ T $ be $ v_1, \ldots, v_k $. Construct a tournament $ T' $ isomorphic to but disjoint from $ T $, and let it's vertices be $ v'_1, \ldots, v'_k $, where $ v_i $ corresponds to $ v_i' $ under the isomorphism. Now, construct the tournament $ G = T \cup T' \cup \{ w \} $, where $ w $ is an additional vertex, and the missing arcs of $ G $ are given by:

$ v_i \rightarrow v'_i $
$ v_i \rightarrow v'_j $ iff $ v_j \rightarrow v_i $, where $ i, j $ are distinct.
$ w \rightarrow v_i, v'_i \rightarrow w $ for all i.

I claim that $ G $ contains no transitive subtournament of $ n + 1 $ vertices. For if $ n + 1 $ such vertices can be chosen, we have two cases:

Case 1: $ w $ is chosen.
Then no pair $ v_i, v'_i $ can also be chosen, otherwise $ w \rightarrow v_i \rightarrow v'_i \rightarrow w $ is a cycle. So all chosen indices are distinct. We cannot have chosen three vertices $ v_i \rightarrow v'_j \rightarrow v_k $, or $ v'_i \rightarrow v_j \rightarrow v'_k $, otherwise we would also have $ v_k \rightarrow v_i $ or $ v'_k \rightarrow v'_i $, by the transitivity of the chosen vertices. So must have either every chosen vertex in $ T $ beating every chosen vertex in $ T' $, or the converse. If it is the latter case, then by taking the 'images $ T^* $  in $ T $ of the chosen vertices in $ T' $, every vertex in $ T^* $ would beat every chosen vertex in $ T $, so $ T^* $, together with the chosen vertices in $ T $, would induce a contradicting transitive subtournament of $ T $. The former case is similar.

Case 2: $ w $ is not chosen.
Then some pair $ v_i, v'_i $ must be chosen. No other pair $ v_j, v'_j $ may be chosen as well, otherwise one of $ v_i, v'_i, v_j $ or $ v_i, v_j, v'_j $ must form a cycle. Let $ N^-(v_i) $ and $ N^+(v'_i) $ be the vertices in $ T $ that beat $ v_i $, and the vertices in $ T' $ that are beaten by $ v'_i $ respectively. Then any vertex in $ N^-(v_i) $ or $ N^+(v'_i) $ would form a cycle with $ v_i, v'_i $, so none of these can be chosen. By again taking the images of the chosen vertices in $ T $, and arguing as in Case 1, we arrive at a contradiction as well.

So we must have $ T(n+1) > 2T(n) - 1 $, and hence $ T(n+1) = 2T(n) $. I guess the bound wasn't so poor after all...


I also have a converges as question:

Let X1 X2...be iid with a symmetric distribution. Then
Sum_n [ (Xn)/n] converges almost surely iff E(|X1|) is finite.

I am not sure how Law of Large Numbers applies here.

As the summation is over [(Xn)/n]..
but not [SUM(Xn)]/n
or they mean the same?

:blush:  you are right, I have read it too fast.


One cubic foot is what fraction of a cubic yard? Express your
answer as a common fraction.

We have $ \left (\frac{1}{3} \right )^3\equal{}\boxed{\frac{1}{27}}$.


find a hundred digit natural number with non zero digits such that it is divisible by the sum of its digits

[hide="hidden by mod"]the divisibility rule for 128 is that if the last 7 digits are divisible by 128 then the whole number is divisible by 128. The first 7 digit number divisible by 128 is 1111168. There are 93 other numbers after. If we assume all those digits are 1s, then sum of all the digits is 112. We can split the remaining 16 among the 93 digits allowing it to be divisible by 128. Obviously there are many numbers that fit the criteria.[/hide]

[hide="A somewhat simpler answer"]Using $ 202$ as the sum, $ 222\ldots2223232$.[/hide]

What's the divisibility test for [hide="hidden by mod"]101?[/hide]

Another possibility:

Sum is 150. If it ends in 00, and the sum of the first 98 digits is 150, then it works.

1. You missed the nonzero digits part. (which I did too at first)

2. I wasn't really thinking of divisibility tests. More the fact that [hide]$ 2222$ is a multiple of $ 202$, so you could cull four digit blocks of $ 2$s until you got to $ 3232 \equal{} 16(202)$.[/hide]

Oh well I failed. Another go:

[hide]
The sum is 144. This means that the sum has to be a multiple of 9, and the last 4 digits must be divisible by 16. Clearly, the first condition is true. For something divisible by 16, we can do 2128. So we have

______________2128 where in the ______ there are 96 digits, which add up to 131.[/hide]

[quote="Mewto55555"]What's the divisibility test for [hide="hidden by mod"]101?[/hide][/quote]
[hide]It's analogous to the test for 11... Start from the right, and group the digits into groups of 2, then do alternating subtracting and adding.
Example: 2222 is divisible by 101 since 22-22 is 0, which is divisible by 101. 12222 is not, since 1-22+22=1 is not divisible by 101.
Note that this works because $ 100\equiv \minus{}1 \pmod{101}$.[/hide]

[hide="Here is another one"]The sum for my number is 125. The number ends in 625 and after that, the rest of the 97 digits just choose those digits whose sum is 112. This works because $ 125|1000$.[/hide]


geometry
geometric transformation
rotation
homothety
ratio
trapezoid
quadratics

Find all the complex numbers z such that the points with affixes $z$, $z^2$, $z^3$, $z^4$ are on a circle.

I'm not sure about this right now, but I think it's equivalent to 1,z,z2 and z3 being on a circle, because we can just apply a similitude of z' (z conjugate) to the initial figure (by similitude of z' I mean a rotation composed with a homothety, of angle -(the angle of z) and ratio 1/|z|).

There was a condition for z1,z2,z3,z4 being on a circle. I think it was something like [(z2-z1)/(z4-z1)]/[(z2-z3)/(z4-z3)] is real (not sure).

If we apply that to z1=1, z2=z, z3=z2 and z4=z3 we get |z|=1, but I'm not sure about the order in which we should take them (maybe it's possible for the 4 numbers to be on a circle in another order).

You can see that all z s.t. |z|=1 do check out, because the corresponding quadrilateral is an isosceles trapezoid, which is cyclic. However, I'm not sure that these are the only solutions.

I really wonder why this problem stands in the "Unsolved" section. Actually, Grobber, your answer is completely correct and I really don't see why you are not sure about it!

Indeed, for a complex number z, the points z, $z^2$, $z^3$, $z^4$ lie on one circle if and only if |z| = 1.

[i]Proof.[/i] It is clear that if |z| = 1, then the points z, $z^2$, $z^3$, $z^4$ lie on one circle; in fact, |z| = 1 yields $\left|z\right|^2=1$, $\left|z\right|^3=1$ and $\left|z\right|^4=1$, and thus, the points z, $z^2$, $z^3$, $z^4$ all lie on the unit circle.

Remains to prove the converse: If the points z, $z^2$, $z^3$, $z^4$ lie on one circle, then |z| = 1.

In fact, four arbitrary points $z_1$, $z_2$, $z_3$, $z_4$ lie on one circle or on one line if and only if the cross-ratio $\frac{z_1-z_3}{z_2-z_3}:\frac{z_1-z_4}{z_2-z_4}$ is real. Since the points z, $z^2$, $z^3$, $z^4$ lie on one circle, we thus see that the cross-ratio

$\frac{z-z^3}{z^2-z^3}:\frac{z-z^4}{z^2-z^4}$

is real. But after some algebra,

$\frac{z-z^3}{z^2-z^3}:\frac{z-z^4}{z^2-z^4}=1+\frac{z}{z^2+z+1}$.

Hence, the number $\frac{z}{z^2+z+1}$ must be real. Thus, its inverse, $\frac{z^2+z+1}{z}$ is also real. Thus, $\frac{z^2+z+1}{z}-1=\frac{z^2+1}{z}$ is also real. Let $k=\frac{z^2+1}{z}$, so that k is real. Then, $kz=z^2+1$, so that $z^2-kz+1=0$. Thus, the number z is a solution of the quadratic equation $z^2-kz+1=0$. Solving this equation, we see that either $z=\frac{k}{2}+\frac{\sqrt{k^2-4}}{2}$ or $z=\frac{k}{2}-\frac{\sqrt{k^2-4}}{2}$. The number $\sqrt{k^2-4}$ must be imaginary (else, z would be real, and thus, the numbers z, $z^2$, $z^3$, $z^4$ would be all real and wouldn't lie on a non-degenerate circle). Hence, we have either $z=\frac{k}{2}+\frac{\sqrt{4-k^2}}{2}i$ or $z=\frac{k}{2}-\frac{\sqrt{4-k^2}}{2}i$, with $\sqrt{4-k^2}$ being real. Thus,

$\left|z\right|^2 = \left(\frac{k}{2}\right)^2+\left(\pm \frac{\sqrt{4-k^2}}{2}\right)^2 = \frac{k^2}{4}+\frac{4-k^2}{4} = 1$,

so that |z| = 1, qed..

  Darij


The condition may be $a_{1}a_{2}\dots a_{n}=1$
this is a problem in Russian Olympaid.

[quote="Albanian Eagle"]this is wrong by taking every variable large enough ! :maybe:
please check your condition again![/quote]
i see.i though about this problem 2 hours but i get nothing.so it was wrong,now i am happy. :D


function
floor function
geometry
national olympiad

Here are the problems of the Costarrican math olympiad... but i can bet all of these were taken from other olympiads and some of them are relatively too easy

q1) Consider the set $S=\{1,2,...,n\}$. For every $k\in S$, define $S_{k}=\{X \subseteq S, \ k \notin X, X\neq \emptyset\}$. Determine the value of the sum \[S_{k}^{*}=\sum_{\{i_{1},i_{2},...,i_{r}\}\in S_{k}}\frac{1}{i_{1}\cdot i_{2}\cdot...\cdot i_{r}}\] [hide]in fact, this problem was taken from an austrian-polish[/hide]
http://www.mathlinks.ro/Forum/viewtopic.php?t=103787

q2) let $a,b,c$ be the sides of a triangle. prove that \[\displaystyle \frac{3(a^{4}+b^{4}+c^{4})}{(a^{2}+b^{2}+c^{2})^{2}}+\frac{ab+ac+bc}{a^{2}+b^{2}+c^{2}}\geq 2\] http://www.mathlinks.ro/Forum/viewtopic.php?t=14618 ; http://www.mathlinks.ro/Forum/viewtopic.php?t=1456 .

q3) let $ABC$ be a triangle. let $P,Q,R$ be the midpoints of $BC, CA, AB$ respectively. let $U,V,W$ be the midpoints of $QR, RP, PQ$ respectively. let $x=AU, y=BV, z=CW$. prove that there exist a triangle with sides $x,y,z$

q4) let $f$ be a function that satisfies \[\displaystyle f(x)+2f\left(\frac{x+\frac{2001}{2}}{x-1}\right)=4014-x\] find $f(2004)$

q5) let $n$ be a positive integer, and let $p$ be a prime, such that $n>p$. prove that \[\binom{n}{p}\equiv \left\lfloor\frac{n}{p}\right\rfloor \ (mod \ p)\] q6) It is given a triangle $ABC$ in which $AC+BC=3AB$. Incircle with center $I$ touches $BC$ and $CA$ at respectively $D$ and $E$. Let $K$, $L$ be symmetrical points of $D$, $E$ with respect to $I$. Prove that $A,B,K,L$ lie on one circle.
http://www.mathlinks.ro/Forum/viewtopic.php?t=76458

Just posting (all) the links so that this contest is added to the resources section :

#1 : http://www.mathlinks.ro/Forum/viewtopic.php?p=479410#479410
#2 : http://www.mathlinks.ro/Forum/viewtopic.php?p=4663#4663
#3 : http://www.mathlinks.ro/Forum/viewtopic.php?p=516070#516070
#4 : http://www.mathlinks.ro/Forum/viewtopic.php?p=516073#516073
#5 : http://www.mathlinks.ro/Forum/viewtopic.php?p=516076#516076
#6 : http://www.mathlinks.ro/Forum/viewtopic.php?p=439274#439274


Let $ \overline{AB}$ be a diameter of a circumference with center $ O$. Denote by $ C$ the point of intersection of the tangent to the circumference in $ B$ with the straight line $ \overleftrightarrow{DE}$, where $ D$ and $ E$ are two arbitrary points on the circumference. Let $ F$ and $ G$ be the points of intersection of the line $ \overleftrightarrow{OC}$ with the lines $ DA$ and $ EA$, respectively. Show that $ \left|FO\right| \equal{} \left|GO\right|$.

[quote="behzad"]let $X$ be the point on original circle $C_{1}$ different from $B$ where $CX$ is tangent to $C_{1}$. if we show that $AX\parallel OC$ then since $DEXB$ is harmonic , so $FOG\infty$ is harmonic , so $FO=GO$. but $\angle XOB=2\angle XAB=2\angle COB$ and so $AX\parallel OC$.
so we are done.[/quote]

Please explain one more time , why the pencil $A.FBEX$ with center $A$ and rays $AF, AB, AE,, AX$ is harmonic.Only based on this theorem I can undrestand your nice and short solution. Or I have missed any other theorem ?
  Many Thanks- Babis

[color=darkred][b]Lemma.[/b] Let $w=C(O)$ be the circumcircle of the triangle $ABC$ and $d$ be a line for which $O\in d$
Denote the intersections $X$, $Y$, $B'$, $C'$ of the line $d$ with the lines $AA$, $BC$, $AB$, $AC$ respectively. Then $OB'=OC'\Longleftrightarrow$ $OX=OY\ .$[/color]

See and read with a great atention the second butterfly problem ([b]P.B.2[/b]) from
http://www.mathlinks.ro/Forum/viewtopic.php?t=110196
and you"ll obtain the above lemma as a particular case $A\equiv B$ !!

[color=darkblue][b]Proof of the proposed problem.[/b] Construct the tangent $t$ in the point $A$ to the circle $(t\parallel BC)$. Denote the intersection $L\in OC\cap t$.
Apply the above lemma to the triangle $ADE$ and the line $OC$ : $OC=OL$ (is true) $\Longleftrightarrow$ $OF=OG$.[/color]

For this problem, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=108346 and Theorem 11 in my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i]. I have included Virgil's solution as First Proof of Theorem 11 in this note.

  darij

I proved the next problem:

  Let's consider the triangle ABC and the circumconic C* with the center O
(O is the midpoint of the side BC). Let P be a point on the circumconic C*,
the point Q is the intersection of the line AP with the tangent to the
circumconic C* at the point C, the point F is the intersection of the lines
AB and OQ,  the point G is the intersection of the lines BP and OQ.
Prove that OF=OG.

A=(1:0:0), B=(0:1:0), C=(0:0:1)
O is the midpoint of the side BC  =>  O=(0:1:1)
C*: pyz+qzx+rxy=0
O is the center of circumconic C*  =>  O=( p(q+r-p):q(r+p-q):r(p+q-r) )
From p(q+r-p)=0 and q(r+p-q)=r(p+q-r)    =>  p=q+r
So C*: (q+r)yz+qzx+rxy=0
If P=(u:v:w) is on the circumconic C*  =>  (q+r)vw+qwu+ruv=0
the tangent (to the circumconic C* at the point C)
equation: qx+(q+r)y=0  and  AP: wy-vz=0
the point Q is the intersection of the line AP with the tangent to the
circumconic C* at the point C  =>  Q=( v(q+r):-qv:-qw )
BP: wx-uz=0  and  OQ: (w+u)x+uy-uz=0
the point G is the intersection of the lines BP and OQ  =>  G=(u:-u:w)
AB: z=0  => infinite point of the line AB is (1:-1:0)
CG: x+y=0  =>  infinite point of the line CG is (1:-1:0)
So AB||CG.
Since AB||CG and O is the midpoint of the side BC  =>  OF=OG

Best regards,  :)
Petrisor Neagoe


In a town with a population of $ n$, a person sends $ 2$ letters to two separate persons each of whom is asked to repeat the procedure. Let us denote the $ i$-th person as $ P_i$. Thus for each letter received $ 2$ letters are sent to separate persons in the town. This means taht at the $ i$-$ th$ stage $ 2^i$ letters are sent. Then

$ 1.$ The value of $ P(A_2)$ [ probability that $ P_1$ does not receive a letter at second stage ] is
a) $ \frac{2^2}{2^n}$;   b) $ \frac {n\minus{}2}{n\minus{}1}$;   c) $ (\frac{n\minus{}2}{n\minus{}1})^2$;   d) $ none$ $ of$ $ these$;

$ 2.$ The value of $ P(A_3/A_2)$ [ probability that $ P_1$ does not receive a letter in the third stage ] is
a) $ \frac{2^3}{2^n}$;   b) $ \frac {n\minus{}2}{n\minus{}1}$;   c) $ (\frac{n\minus{}2}{n\minus{}1})^2$;   d) $ none$ $ of$ $ these$;

$ 3.$ Probability that $ P_1$ does not receive a letter at $ m$-th stage is
a) $ \frac{2^{m\minus{}1}}{2^n}$; b) $ \frac{2^m}{2^n}$; c) b) $ \frac {n\minus{}2}{n\minus{}1}$; d) $ none$ $ of$ $ these$;

Thank you... :)

[quote="hzbrl"] This means taht at the $ i$-$ th$ stage $ 2^i$ letters are sent. [/quote]

I'm not sure about this, I think is $ 2^1\plus{}2^2\plus{}2^3\plus{}...\plus{}2^i$

For number 1
At stage $ 2$, two people (neither are P_1) send out $ 2$ letters to [b]different[/b] people.
So the probabiliy that $ P_1$ does not recieve a letter from one of the people is $ \dfrac{\binom{n\minus{}2}{2}}{\binom{n\minus{}1}{2}}$

Therefore the probability that he doesn't get a letter from either is $ \left(\dfrac{n\minus{}3}{n\minus{}1}\right)^2\longrightarrow D$

For number 2 and 3 is that suppose to be the probability that $ P_1$ does not recieve a letter at stage $ i\plus{}1$ given that he did not recieve a letter at stage $ i$?

[quote="hzbrl"]I think what the question means is at the $ i$-th stage $ 2^i$ letters are sent. :)
What you did is the total number of letters sent after $ i$-th stage. :)

please give another try...
thank you... :)[/quote]

Thanks for the correction


geometry
circumcircle
incenter
geometric transformation
rotation
geometry solved

[quote="ashrafmod"]what is the
centre of a triangle[/quote]

Huh?

A generic triangle has lots of different "centers" (i. e. remarkable points defined in a way symmetric with respect to the vertices of the triangle): the circumcenter, the incenter, the centroid, the orthocenter, the symmedian point, etc.. You can't just say "the center of the triangle", because this notion has no defined meaning.

Only if a triangle is equilateral, you can speak of "the center of the triangle". This center is the point P such that the rotation with center P over the angle 120° maps the triangle to itself. This center P is simultaneously the centroid, the orthocenter, the circumcenter and the incenter of the equilateral triangle.

  darij


induction
combinatorics solved
combinatorics

Suppose $F$ is a family of finite subsets of $\mathbb{N}$ and for any 2 sets $A,B \in F$ we have $A \cap B \not= \O$.

(a) Is it true that there is a finite subset $Y$ of $\mathbb{N}$ such that for any $A,B \in F$ we have $A\cap B\cap Y \not= \O$?
(b) Is the above true if we assume that all members of $F$ have the same size?

[hide="just The Ansewr"] for (a) it's No .
for (b) it's Yes
I try to prove it   [/hide]

[quote="Peter VDD"]Suppose $F$ is a family of finite subsets of $\mathbb{N}$ and for any 2 sets $A,B \in F$ we have $A \cap B = \O$.

(a) Is it true that there is a finite subset $Y$ of $\mathbb{N}$ such that for any $A,B \in F$ we have $A\cap B\cap Y = \O$?
(b) Is the above true if we assume that all members of $F$ have the same size?[/quote]
It should be $A\cap B \neq \O$ and $A\cap B\cap Y \neq \O$ ;)

[quote="ZetaX"]It should be $A\cap B \neq \O$ and $A\cap B\cap Y \neq \O$ ;)[/quote]I'm sorry. Has been corrected. :)

(a)

The answer is negative. We'll construct finite sets $A_{1},\ldots,A_{n},\ldots$ inductively, such that $A_{n},A_{n+1}$ have no points in common inside $[n]=\{1,\ldots,n\}$, and $A_{m}$ has exactly one point in common with $A_{n}$ for every $m\ne n$.

First take $A_{1}$ to be an arbitrary set with at least two elements. Now suppose $A_{1},\ldots,A_{n}$ have been constructed so that they have the properties stated above, and we want to construct $A_{n+1}$. Take an element $x_{1}$ of $A_{1}$ which is not contained in $A_{n}$ and put it in $A_{n+1}$. Now do the same for $A_{k}$ with smallest $k$ such that $x_{1}\not\in A_{k}$, and so on. In the end, we'll have a set which has exactly one element in common with each of $A_{1},\ldots,A_{n-1}$. Finally, put an element of $A_{n}\setminus[n]$ in $A_{n+1}$, and also some positive integer greater than $n+1$ which does not belong to any of the sets $A_{1},\ldots,A_{n}$.

(b)

Here, the answer is 'yes'. We'll prove this by induction on the common size $r$ of our sets. It's obvious for $r=1,2$, so assume now that it's proven for sizes $<r$ ($r\ge 3$). Of course, we're only concerned with the case when $\mathcal F$ is infinite.

Suppose we can find $A_{n},B_{n}\in\mathcal F$ such that $A_{n}\cap B_{n}\cap[n]=\emptyset$. Let $C\in\mathcal F$ be another set in our family, and take $n_{0}$ large enough for us to have $C\subset[n_{0}]$. From now on we work only with $n\ge n_{0}$.

There is an element $x\in C$ such that infinitely many of the $A_{n}$ contain $x$. Assume WLOG that all of them contain $x$, in order to keep things simple. By Ramsey's Theorem, either we can find infinitely many $A_{n}$ no two of which have any other point in common except for $x$, or we can find infinitely many such that each two have at least one other point in common. If the former holds, then it's clear that in fact, all sets in $\mathcal F$ contain $x$, and we get a contradiction; this means that the latter holds. Again, for the sake of simplicity, we assume each two $A_{n}$'s have at least two elements in common. Repeating for the $B_{n}$'s, we may assume that there is a $y\in C,\ y\ne x$ such that each $B_{n}$ contains $y$ and each two $B_{n}$'s have at least one other common point. The family $\mathcal F'=\{A_{n}'=A_{n}\setminus x\}\cup\{B_{n}'=B_{n}\setminus y\}$ now satisfies the same conditions as $\mathcal F$, except that all its elements have size $r-1$. The induction hypothesis contradicts the existence of $A_{n}',B_{n}'$, so we have our contradiction.


erm lets say a+b= 5
a+c= 10
b+c=15

The problem is something like this, how do i find out each unknown's number

[hide="Hint"]
Add the equations!
[/hide]

[hide="Solution"]
\[ a \plus{} b \equal{} 5
\]
\[ a \plus{} c \equal{} 10
\]
\[ b \plus{} c \equal{} 15
\]

\[ 2a \plus{} 2b \plus{} 2c \equal{} 30
\]

\[ a \plus{} b \plus{} c \equal{} 15
\]
Now since the sum of all three variables is $ 15$, and the sum of the first two variables is $ 5$ (equation 1), $ \boxed{c \equal{} 10}$. Now, since $ c \equal{} 10$, substituting into the third equation, we get $ \boxed{b \equal{} 5}$, and then $ \boxed{a \equal{} 0}$.
[/hide]

[hide="My Solution"]

$ a\plus{}b \equal{} 5$

$ a \plus{} c \equal{} 10$

$ b \plus{} c \equal{} 15$

Let's add our equations together.

$ 2a \plus{} 2b \plus{} 2c \equal{} 30$

We can divide by two.

$ \frac{2a \plus{} 2b \plus{} 2c}{2} \equal{} \frac{30}{2}$

$ a \plus{} b \plus{} c \equal{} 15$

You may notice that we already know what $ a \plus{} b$ is -- it's 5! We can substitute this into our equation.

$ 5 \plus{} c \equal{} 15$

$ 5 \plus{} c \minus{} 5 \equal{} 15 \minus{} 5$

$ c \equal{} 10$

So now we know that $ c$ is 10! We also know that $ b \plus{} c \equal{} 15$, so we can substitute our value for $ c$ into that equation to find what $ b$ is.

$ b \plus{} 10 \equal{} 15$

$ b \plus{} 10 \minus{} 10 \equal{} 15 \minus{} 10$

$ b \equal{} 5$

And we also know that $ a \plus{} b \equal{} 5$, so let's find what $ a$ is.

$ a \plus{} 5 \equal{} 5$

$ a \plus{} 5 \minus{} 5 \equal{} 5 \minus{} 5$

$ a \equal{} 0$

So $ a$ is $ 0$, $ b$ is $ 5$, and $ c$ is 10.

[/hide]

I'm almost sure this is correct.

[hide]First lets just combine some of the equations.

Basically stating:  A+B=5
                         A+C=10
                          B+C=15
So lets solve for the A's

A+C=10
A+B=5
--------
C-B=5  or -B+C=5

Then we solve our next system.

 B+C=15
-B+C=5
----------
2C=20
C=10

So now that we've found 1 variable we can solve for the next variables.

Plugging C back into the equation C+B=15

We get 10+B=15
           ---------
               B=5

Then we plug B into the equation A+B=5

We get A+5=5
           ------
             A=0

Thus the variables are:

               A=0
               B=5
               C=10

I hope this helps you understand the concept of these problems.
[/hide]

:play_ball: My solution:
[hide]Subtract the first equation from the second or...

(A+C)-(A+B)= 5
C-B= 5
C= B+5

Now using the Substitution Property of Equality and working from there, you can get A=0; B=5; C=10; [/hide]

[quote="ilovepi"][hide]Just add the equations , then you could plug in the A+B=5[/hide][/quote]

How is different from my solution  :dry: ?

:arrow: :ddr:  :welcomeani:
This is how TheHedgehog thinks how you should to this problem.

[hide]
We have to find out a way which a fits in all of the equations. I like to start from the lowest number possible, in this case 0. If we let a=0 then in 0+b=5 b is forced to be 5. In 0+c=10 c this time is forced to be 10. b+c=15, replace the variables with the numbers we just found, and for sure, 5+10=15. Problem Done![/hide]

Brute Force, as useful as it is, can get very rigorous when dealing with large numbers, or more complicated like values.

So, even though I agree that this problem can be Brute Forced easily, it is always a good idea to practice algebraic solutions, too.

Um, just a question. How did you guys come up with '2a + 2b + 2c = 30'? I see were it came from, but how did that work? Is there a rule that says we have to do that?

There's no rule saying you have to combine all the equations when you solve something. It just helps here.

Also, you are allowed to add or subtract equations without changing their truthfulness.

given that,
                  a + b = 5 ---eq.1
                  b + c = 15 --eq.2
                  a + c = 10 --eq.3

     let us subtract eq.1 from eq.2
  then we get c - a = 10 ---eq.4

     let us add eq.3 and eq.4
   then we get
                  2c = 20
             then 'c' = 10
by substituing value of 'c' in eq.2 and eq.3
               we get a = 0
                         b = 5
                   and c = 10


$ \frac {1 + \tan A}{\sin A} - \frac {1 - \cot A} { \cos A } = 2 \csc A $

[quote="Jan"][hide]
\[\frac {1 + \tan A}{\sin A} - \frac {1 - \cot A} { \cos A } = 2 \csc A \]
$Equals$
\[\frac {\frac{\cos A + \sin A}{\cos A}}{\sin A} - \frac{\frac{\sin A - \cos A}{\sin A}}{\cos A}\]
$And that again equals$
\[\frac{2\cos A}{\sin A \cos A} = 2 \csc A\]

P.S.: Does anybody know why my text in LateX is written as one word?
[/hide][/quote]

You can just dump sentences into $\LaTeX$.  The best thing to do is to not put the sentences into $\LaTeX$.  Otherwise you could use the \text{} command.

how do you jump from

[hide]

$\frac {\frac{\cos A + \sin A}{\cos A}}{\sin A} - \frac{\frac{\sin A - \cos A}{\sin A}}{\cos A} = 2 \csc A$

to- >

$\frac{2\cos A}{\sin A \cos A} = 2 \csc A$

??

[/hide]

I don't see a problem in that step...

$\frac {\frac{\cos A + \sin A}{\cos A}}{\sin A} - \frac{\frac{\sin A - \cos A}{\sin A}}{\cos A}=$
\[\frac{\cos A + \sin A - \sin A + \cos A}{\sin A \cos A}\]
$\text{Which is equal to:}$
\[\frac{2\cos A}{\sin A \cos A}\]

yip everythings correct, just wanted to see the step :>>


On the AMC 12B this year, I found that familiarity with certain types of questions helped me immensely; questions 23 and 24, for example, were fairly easy for anyone with a lot of experience in Number Theory. So practicing all types of problems is the best way to improve yourself here.

In general, however, AMC problems are pretty novel and unique. To improve yourself in this field, I would suggest familiarizing yourself with the "basics" of each subject. For example, questions 20 and 22 on the 12B this year were pretty weird. But, because I had a good fundamental geometry background (I'm not even that good at geometry). All you needed to know for 20 was how to find the area of a triangle (well, basically) and for 22 you just needed to know some basic triangle geometry again.

Overall, I would go for the second paragraph's suggestions more so than the first. If you've thoroughly studied AoPS 1 and 2 there should be no reason you can't get a 150 on the AMC besides the time limit. So, pick up AoPS 1 again (AoPS 2 isn't quite as useful for the AMC) and make sure you know every topic like the back of your hand. Once you have familiarized yourself with these, applying the techniques to problems should be fairly easy.

[quote="1/(ln x)"]Looking at the AMC scores for AoPS members, they seem to be pretty high (120-130). How do you guys get such high scores? I've done AoPS 1 and 2 and a bunch of old tests, and I only got 102 on A and 103.5 on the B.[/quote]

OMG I totally agree...

I don't have AoPS 1, but I read through (most of) AoPS 2 and did most of the problems.  For me, that was enough to boost my AMC 12 score from just over 100 (last year's 12A) to about 120 (this year's 12B) despite the change in scoring.  I don't know how typical that is, but that's my experience.

I have to say the doing old tests helps more than anything on the AMCs. On the AIMEs... AoPS II and other books start becoming more and more useful.


Armed with just a compass  no straightedge  draw two circles that intersect at right angles; that is, construct overlapping circles in the same plane, having perpendicular tangents at the two points where they meet.

Armed with just a compass-Is math problems solving like a war :D

Given two points, using the compasses we can draw a third point s.t. the three points form an equilateral triangle: Draw circles centered at $A,B$ with radius $AB$, and take one of their intersection pts.

In order to solve our problem, just draw a regular hexagon (by drawing equilateral triangles glued to each-other) $A_1A_2A_3A_4A_5A_6$. Now draw the circles centered at $A_1,A_4$, passing through $A_2,A_5$. These are orthogonal because the angles $\angle A_1A_2A_4$ and $\angle A_1A_5A_4$ are right angles.


Hey guys, how do we check for these errors? I mean when you press CTRL + F7, you get like a scan thing. And they show you how much of each of these you have. I know what errors is, but what are warnings and bad boxes? And is there a way to make LATEX highlight the places so we can fix them?

Also, I am writing a math paper and I need to put (last name, page number) at the top right corner of the pages, how do I do that?

That's all, thanks for reading and helping me out!

It should mention which line the error/warning etc occurs in and if you are using TeXnic Center then double clicking on the warning/error should take you to it.


I have the following:

a-b=X

I need one single formula (that uses only addition, substraction, multiplication or division) where if X is a positive number the result will be 1 and if X is a negative number the result will be -1 or 0

Is this possible?

Thanx in advance!
nik :D

No this is not possible:
every function you can reach by this is continous, where as your wished result is not.

ZetaX, thanx for your quick response man :) I will look for alternatives
nik


Let $ \left(d_{1},d_{2},\cdots,d_{n}\right)$ be the sequence of degrees of graph $ G$ and $ k>1$ be a positive integer such that:

$ \sum^{n}_{i\equal{}1}\binom{d_{i}}{2}>\left(k\minus{}1\right)\binom{n}{2}$

Prove that there exist two vertices $ u$ and $ v$ in $ G$ that have at least $ k$ common neighbors.

Double-counting of the number $ N$ of the pairs $ (A, \{B,C\})$ such that $ AB$ and $ AC$ are edges of the graph :
- If $ A$ is given with degree $ d$ then there are $ \binom d 2$ such pairs. Suming over $ A$ leads to $ N \equal{} \sum^{n}_{i\equal{}1}\binom{d_{i}}{2}$.
- Now, assume that there are no two vertices $ u$ and $ v$ with at least $ k$ common neighbours. Then, since there are $ \binom n 2$ choices for $ \{B,C \}$ and, for each, no more than $ k\minus{}1$ choices for $ A$ such that $ AB$ and $ AC$ are edges of the graph, we deduce that $ N \leq (k\minus{}1)\binom n 2$. A contradiction.

Pierre.


calculus
integration
trigonometry
calculus computations

$ \int {{{\frac {\ 3ctg ^2 x - {\sqrt{\sin x}} cos x } {\sin^2 x}} dx}}$

Lets write it as:

$ \int 3 \frac{\cos^2 x}{\sin^4 x} \, dx \minus{} \int \frac{\cos x}{\sin^{\frac32} x}  \, dx$
$ \equal{} \minus{}\int 3 \left(\frac{\cos x}{\sin x}\right)^2 \, d\left(\frac{\cos x}{\sin x} \right) \minus{} \int \frac{d(\sin x)}{\sin^{\frac32} x}  \, dx$
$ \equal{} \minus{}\left(\frac{\cos x}{\sin x}\right)^3 \plus{}2 \frac1{\sqrt{\sin x}}\plus{}C$


I wrote a test with solutions that were used at the 2005 Vestavia Hills High School Math Team Tournament to help select students who will participate at ARML 2006 for the Alabama team.  Here are the test and solutions for others who would like to work the problems.

A quick method for #11 was to notice that if you rotate the original figure three times 90 degrees each about the point where the fourth vertex of the rectangle would be, you get a large square of side $17$ and a small square of side $12$, but your original figure is a fourth of this figure (since you made 3 extra copies), so the area is $\frac{17^2-12^2}{4}=\frac{145}{4}$.


ceiling function
LaTeX
combinatorics proposed
combinatorics

$ 8$ students took part in exam that contains $ 8$ questions. If it is known that each question was solved by at least $ 5$ students, prove that we can always find $ 2$ students such that each of questions was solved by at least one of them.

At least 40 solutions were given, and so at least one student solved at least 5 questions. Suppose that the best score is $ 8\minus{}n$ for $ 0\le n\le3$. Then on the $ n$ questions which the best student did not solve, there were $ 5n$ solutions and at most 7 students who solved any of them. So, at least one student must have solved at least $ \lceil\frac{5n}{7}\rceil$ of them. But, for $ 0\le n\le3$, $ \lceil\frac{5n}{7}\rceil\equal{}n$, and so one of the other students must have solved all of the other problems.

Let $ a_i$ be the number of students ($ s_i$) that solved problem $ i$. We need to prove there is a pair of students that solved $ 8$ problems between the two of them. The sum of all different problems solved by each pair of students is: $ 7\sum_{k \equal{} 1}^{8}a_i \minus{} \sum_{k \equal{} 1}^{8}{{a_i\choose 2}} \equal{} \sum_{k \equal{} 1}^{8}\frac {(15 \minus{} a_i)(a_i)}{2}\geq 25\cdot 8 \equal{} 200$ but on the other hand if the problem wasn't true, then each pair of students would have solved at most $ 7$ different problems. Total: $ {8\choose 2}\cdot 8 \equal{} 196$ so we have the conclusion.

There are 40 solved problems, so one person must have solved at least 5 by pigeonhole. We can assume wlog that person 1 solved questions 1,2,3,4, and 5, and that question 6 was solved by persons 2,3,4,5, and 6. To minimize the possibility of persons 2,3,4,5, and 6 of solving both problems 7 and 8, in which case the problem is over, let persons 7 and 8 both solve problem 7 and 8. But problems 7 and 8 must both be solved by three other people than 1,7, and 8, since there are only five other people, 2,3,4,5, and 6, one must have solved both 7 and 8, which completes the proof. The problem is obvious if person 1 solved 6,7, or 8 problems.

A diagram helps:
O1 2 3 4 5 6  7 8
1X
2X
3X
4X
5X
6O X X X X X
7O O O O O OX X
8O O O O O OX X

And you see that one of 2,3,4,5,6 must have solved both 7 and 8. (I put O's in to fill the space that latex deletes).

For each pair of students, assume there is an unanswered question between them. Since each question has 5 correct solutions, each Q can only have 3 such pairs of students between which neither solved the question.

There are $\binom{8}{2}=28$ pairs of students, but 24 possible pairs where both didn't solve a particular problem. Hence there must exist some pair of students such that each question was solved by at least one of them.


algebra
function
domain
THIS IS A RANDOM TAG

Where is a trustable service for free where you could upload your files and, images and all, it posts it on the World Wide Web?

I don't care about ads or domains. I want a site/service now.

Thank you.

For photos, http://www.photobucket.com is quite popular. Although, if you were just looking for an album or something, Yahoo! photos works wonders. Also, look into ImageStation.

Der...if you just want free web space, put that in Google and it should give you quite a few satisfactory results.

[quote="anirudh"]on your own computer.  :) Download apache webserver download v 1.3.3 (its easy to install-leave aoll features default. for the screeen taht asks for your domain, type in your ip address) and then you just put your photos at c:/programfiles/apache group/htdocs. then it will be http://your ip address/your photo.jpg[/quote]

...and if you turn off your computer? ;)

um lol...that'd be kinda hard to accomplish...comps need turning off at least every couple months...

and yeah photobucket  :D

[quote="mwahahahaha"]um lol...that'd be kinda hard to accomplish...comps need turning off at least every couple months...[/quote]

*nods* Quite so. Lol. Just Google it. "free webspace" would probably give you what you're looking for.

of course. It is obvious that every webthingie turns off their computer every 3 days for every month. That's called *downtime*, but it's unavoidable. OK. whooops, i forgot that all people don't have pd, or dual core. dual core doesn't require you to turn it off, and i also read osmewhhere else, that it uses up 1/2 the amount of power that another normal computer would. but i don't understand it.  :dry:

i use Geocities as a sort of junk storage yard for everything i have to upload (all hail the omni-useful free Yahoo acct). according to its user policy you aren't *technically* supposed to use geocities as a dump site like im doing, but i at least made a *cough* decent site with some of the stuff i uploaded, so it still abides by the rules  :lol:

it's selfevident. look from the begin. it says that they must NOT turn of their computer. what you did was spam.  :spam:  :roll:

[quote=#H34N1]on your own computer.  :) Download apache webserver download v 1.3.3 (its easy to install-leave aoll features default. for the screeen taht asks for your domain, type in your ip address) and then you just put your photos at c:/programfiles/apache group/htdocs. then it will be http://your ip address/your photo.jpg[/quote]

#awesomeavatardude


inequalities
probability
expected value
probability and stats

Hi,

This one has me stomped for a while.  Any help is greatly appreciated.

Let X be an integrable random variable and m be a median.  That is P(X < m ) <= 0.5 and P(X > m) <= 0.5.

Prove that   min E|X-c| = E|X-m|  (c real number).

Regards.

Use the inequality
\[ \left| X \minus{} c \right| \ge \left| X \minus{} m \right| \plus{} (m \minus{} c) \text{sgn}(X \minus{} m).
\]

Ravi, this problem is hitting me back again :)

I am not sure what to do with that inequality as I have fiddled around with many similar ones.

Haha, I guess you are referring to my signature.

Take the expected value of both sides of my inequality.  The expected value of the final term will be zero, because $ m$ is a median.

It will be zero if the random variable has a continuous cumulative distribution.  In that case P(X > m) = P(X < m) = 0.5 and indeed the expectation of your last term will be zero.

But in general it is (m-c)P(X>m) + (c-m)P(X<m), right?  I am not given a continuous r.v.

Good point.  To fix my argument, redefine $ \text{sgn}(0)$ to be $ \Pr[X < m] \minus{} \Pr[X > m]$.


1.What is Euclidean geometry? Can you give me examples that we learn in high school?
2.If x is a number from column x, y is a number from column y, Column x: -1, 0,1  Column y: a,0,a. and a is not equal to 0, which of the following statements are true.
I.xy always equals a value of y.  Answer says this statement is true because even though xy is not equal to y, every xy is a value of y.  I don’t understand. I think it’s wrong beause -1*a= -a which is not a, so they are different. So why it’s true？
THanks for help.

Euclidean geometry is geometry in which the parallel postulate holds (look it up if you don't know what it is).  Flat surfaces are examples of Euclidean geometry.  The surface of a sphere is not, since you can draw an infinite number of lines parallel to a prime meridian.

JRav, on a sphere there are no parallel lines.

Euclidean geometry is based on 5 postulates, the fifth of which says that given a line and a point not on the line, there is only one line through the given point parallel to the given line.  This is the parallel postulate, and basically is the assumption that the plane we are working in is flat.

There are two other major types of geometry: elliptical (in which the typical plane is a sphere, a line is a great circle of the sphere, and there are thus no parallel lines) or hyperbolic (in which we end up with 2 parallel lines to any given line and then lines which are ultraparallel or something strange like that).

2) are you sure your choices for y aren't -a, 0 and a?

[quote="facis"]

2) are you sure your choices for y aren't -a, 0 and a?[/quote]

I am 100% sure that it's a,0,a, but maybe there is a misprint? but since my book stated every xy is a value of y, so if -1*a=-a, it's still a value of a???

sounds like a misprint - they probably intended your choices for $ y$ to be $ \minus{}a$, $ 0$, and $ a$.  Anyway, don't worry about that problem.

I guess this was a misprint, but thanks.


Can anyone tell me where we can download Latex ??

There is a link on the AoPS page.  I m not sure if you can see it on mathlinks(if that is what you are using) but here it is anyways.
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

Once you go there it will be under the downloads tab.


trigonometry
inequalities proposed
inequalities

let $ x_i$ ($ 1\leq i \leq n$) be some reals satisfying

$ \prod tg(x_i)\equal{}1$

Find max of $ \prod sin(x_i)$

It is not very hard.
$ \cot^2(x) \equal{} \frac {1}{\sin^2(x) }\minus{}1$

[quote="FOURRIER"]Using this method I found $ (\frac {1}{\sqrt {2}})^n$ but I wonder if it's true because it depends on $ n$  :maybe: .[/quote]
your answer is true.


MATHCOUNTS AMC USA(J)MO USAMO AIME ARML HMMT

Especially Nats competitors, if you expected to place higher, and fell short, especially if this year was your last chance, how do you deal with the disappointment? Any advice for those in a similar boat?

[quote="nikeballa96"]mathcounts doesn't matter.

well, it does. but evan o'dorney, from california, got fifth in california mathcounts in eighth grade. the next year, he made IMO.

don't worry about mathcounts. it's a cool competition, but it means virtually nothing. no offense intended to those who have succeeded in mathcounts.[/quote]

Middle school doesn't matter.

well, it does. but Bill Gates, from Washington, got owned by the school officials in eighth grade. now, he's rich.

don't worry about middle school. it's a cool time in your life, but it means virtually nothing. no offense intended to those who have succeeded in middle school.

point: you are pointing out the extreme case of this. i'm pretty sure that MATHCOUNTS gives you scholarships that yes, give you benifits and put you ahead of the rest of the people who did not obtain scholarships. And it's not like you won't gain recognition from colleges. Also 5th at state is still pretty good, especially in CA where 45 at some chapters don't make state. i'm sure if someone said the first 3 paragraphs to you, you would probably try to make something similar up for them.

[quote="nikeballa96"]mathcounts doesn't matter.

well, it does. but evan o'dorney, from california, got fifth in california mathcounts in eighth grade. the next year, he made IMO.

don't worry about mathcounts. it's a cool competition, but it means virtually nothing. no offense intended to those who have succeeded in mathcounts.[/quote]

Two years ago, Evan O'Dorney got 5th at CA States.  The next year, he was a USAMO winner.

Last year, David Yang got 5th at CA States.  The next year, he was a USAMO winner.

This year, I got 5th at CA States.  Next year, :D

Anyway, to mathlearner: Just get on with life.  As stevenmeow said, there are more important things in life than MATHCOUNTS.

I never said that
nikeballe96 said that in a more extreme way
I agree with that, but not in such an extreme way as nikeballa96 does.

let's edit my earlier post...

Made state and got disqualified on some stupid techincality. I coped by getting ready to pwn everyone next year.

I failed Florida state, i got 29th. but i have another chance next year, i'll try harder

Be disappointed and learn from it. Then just suck it up and move on to your future opportunities. For me, I just need to get better at stupid mistakes. If I performed the way I did at chapter and states in nationals, I would have gotten top 30, on a good day with little stupid mistakes. Unfortunately, I had the worst day ever and missed 13 points on adding wrong, plus many more for just a bad "thinking day". For people who made nats, you just got to think about how much you did to get there. And, be thankful that was MC nats, not the SAT!!!

This is rather random, but I care much more about MC Nats than the SAT.

Anyway, when I was in 6th grade I got 5th at MC states, so I used that as motivation for seventh grade. Now, I'm slightly disappointed over my ranking in my final MathCounts competition, so I'm using that plus my USAMO failure to force myself to study more.

CASE #1:   You got first in the world in every competition you entered. You got a $ \$1000$ scholarship.
RESPONSE: So what.

CASE #2:   You got did good on your college entrance exam, got a degeree in your favorite field, and you are know doing your favorite job. You make enough money to buy whatever you want, have a good house and car, and get to live in your dream location.
RESPONSE:  Life is good.

Which case was harder to achieve? Which one do you like better? Which one is more important?
Think about it.

[quote="fooblah"]Well, I can definitely relate to phailing mathcounts...at chapter this year I lost on a tiebreaker and got 11th place so didn't make the cd, even though I fully had 43 points! I am using this as motivation to work REALLY hard on math so I can go to nats next year...[/quote]
But remember, MC nats isn't the only thing in the world.

I had many experiences, personally, where I felt short of my expected MATHCOUNTS goals. This not only happened at Nats this year, but also at chapter in 6th grade, state in 7th grade, chapter in 8th grade, and state in 8th grade. All of these disappointments were due to very stupid mistakes. All I really could do to cope with the disappointment is just keep my mind open for anything I could possibly do, simply just to keep thinking to myself that EVERY POINT COUNTS. If you fall short by a point, you fall short by many places. In sixth grade, I missed the cutoff for state because of just 1 point on the chapter contest. If I had just gotten 1 more point, I would have made state, which unfortunately I haven't. :( Also, in 7th grade, I got 4th place at chapter, which I kinda thought was a little bit more embarrassing because I missed a target question which I should have gotten right if I hadn't misread the question. If only should I have that question right, I would have gotten a better place, possibly first if I could only remember what the top 4 scores at chapter was....Oh well it didn't really matter anymore for me, bleh. But I at least made state, so it didn't really matter to me anymore. Besides, chapter contest is trivial so why does it matter? Basically I just pushed my mind ahead to prepare for states. At states, unfortunately, I felt sick, so I didn't really do as well as I expected to, so I just had to get over it. There wasn't really any way that I had to really cope with the disappointment, even though again, I missed the cutoff for nats by a single question. 6th place at state. But Mathcounts was over for me in my 7th grade year, so realizing that fact just made me want to study even harder over the summer. What I did was just forget about the sadness, try to do problems faster than the given time limit, and just avoid the stupid mistakes that crawl over my head every once in a while. However, in 8th grade chapter contest, I somehow tied for 1st, but tiebreakers cut me down to 3rd. I then realized that there's no matter of being disappointed because chapter is over and I get to go to state, meaning I still have a shot for making Nats, which was my ultimate goal at the beginning of 8th grade. Later, at State, I was 2nd place, falling short of the first place person by 1 point. There we go again. Missing a goal by a single point. Therefore, the moral of the story is that [b]every point counts[/b] in any Mathcounts competition. If you fall short by a point, you fall short by your goals rather quickly, and this is not a surprise, because there are often many, MANY tied scores in most Mathcounts competitions, regardless of whether it's at chapter, state, or nationals. (maybe perhaps Nationals has the most ties, making it such an intense competition and creating discussions of problem triviality) So at Nats, I came in 20th and fell short for countdown qualification just by 1 point, or actually 1 question (target, *sob*).

In general, what I really did to cope with Mathcounts disappointments is just think to myself, "Does Mathcounts matter?" Well, maybe not for 6th and 7th grade, but probably just now, since I'm done with Mathcounts. I probably was MUCH more disappointed with falling short of major goals in 6th and 7th grade just because I really enjoyed the benefits of Mathcounts and wanting to get all of those cool, shiny trophies. But now, I'm done. If you're done with a competition, and you're disappointed about your final placing in such an event, just get over it quickly. There's no reason to lag forever and ever saying that you lost or you failed or whatsoever. Just keep on pressing forward to look ahead to high school competitions, such as the AMC, AIME, USAMO, Mandelbrot, ARML, HMMT, and many others that you can possibly think of. Also, there's one thing that you should always remember in any competition: EVERY POINT COUNTS. Don't just think that you feel okay just because you lost to a person by one point. Losing by one point can mean a lot, especially when your score is very close to getting a major award.

I of all people should not feel disappointed (I placed a good 20 places higher than I expected, and I actually scored ABOVE my test average at the competition), but I still can't help but think that if I could get that 100x4 problem, I would be up 3-1 and pretty much go on to the finals. As such, disappointment.

I'm just coping by looking at the stats. I am the #2 written in America, and #3 cd. That is really good.

Gauss, you came in the top 57 (won't say what place!) out of ALL THE MIDDLE SCHOOL STUDENTS IN AMERICA.

Etc.

Anyway, there are two scenarios. You did not do your best on the test as you could have. Take comfort in the fact that, had you been at your best, you would have come in a higher place.

If you did your best, but did not make it to the next level, you still DID YOUR BEST. That is a good life lesson, to do your best udner pressure.

Also, if you failed, just remember, you beat EVERY SINGLE PERSON who did not come to the competition.

Wow, mew. Coming from a person like you (THAT MEANS NOTHING), that's deep.
Well, I failed. But yay. Top 57 (barely...) for me.

How am I coping with it? EPIC AIME PREPARATION =D.

(wow, this post was total spam. akhil0422 is totally gonna quote izzy and tell me how spammy/unnecessary my post was on the next post)

[quote="CEM_College_Champ"]Would doing trigonometry, calculus (whether SV or MV is not important) or complex-valued functions be of any help in Mathcounts preparation?[/quote]
Not a lot of middle-schoolers even KNOW trigonometry (e.g. me) :P

As Mewto55555 said somewhere:

I managed to forget the forumla for the volume of a sphere and rederived it using calculus. (that isn't an exact quote, so I am not going to put quotation marks around it)

well, in response to the actual question of the topic, even though its kinda late, i thought that i would just study a lot harder next year.

But some for some people, that was their last chance. I would just say, try to forget about it. The point of MathCounts is having fun, meeting other people interested in math, and building team skills, not getting first place.

EDIT: I edited it, and added what JBoyd added.

I don't think the point of Mathcounts is just to have fun.  It is to learn something and hopefully open some of the doors to future learnings in mathematics and to have some fun along the way.  You will learn work ethic and how to work as a member of the team.  You will learn that you can accomplish a lot more as a team than you can all by yourself.  Those skills will be useful for the rest of your life in just about everything that you will do.

Mmm... with all due respect - a lot of respect - for Coach Boyd and Coach Boyd's opinion (as posted above this one) - I don't believe MATHCOUNTS is just about "team building" and in fact, I know of number of participants at Nats - several who were even in the top 10%, who made it to Nats STRICTLY as individuals for reasons ranging from that they came from a school with no team, and worse, no interest to start a team!, and/or homeschoolers who again, had no team, and in fact, these are some of the most motivated, self-starting people you'll ever meet - having had the motivation, drive, innate abilities, and sheer determination to succeed without the extreme benefit of having a coach like Coach Boyd (or any coach, in many cases), without having many, or [i]any[/i], MATHCOUNTS practice materials except what they could scrounge on the net and/or afford to purchase (and most Nats tests weren't available in the past) so I guess what I'm saying is perhaps MATHCOUNTS offers different benefits for different people. In some cases, fun IS the biggest benefit, in some cases team-building, and yes, in some cases, just the sheer desire to be the best and to at least have a GOAL of being #1 or in the top 12 or top 57 or whatever someone's goal is, is the main reason for participation. However, I think some angst arises for middle-schoolers if/when their goals are not met, and thus, in some cases, maybe the [u]biggest[/u] benefit is as this post asks - how they ultimately learn to deal with such disappointment.

I didn't know much about MathCounts and I thought the top 8 advance from state, and that was my goal. I got fifth, partially acheiving my goal. I got top 8, but didn't advance. The top 8 went to countdown.

Well trig sometimes helps, but I wouldn't recommend learning it if you are weak in other areas. You should patch those up first.

Case in point: #20-something(27?) on this years sprint was like there is a 15-75-90 triangle with sidelength something opposite the something angle. What is the area? You could split the 75 degree angle into a 45 and 30 and go from there, or remember what sin75 and sin15 are. I did the latter and finished pretty quickly.

...nikeballa96...

how do u think i feel...

76 @ nats!

at least u made top 57...

It happens. The point of mathcounts is to have fun, so if you had fun, there isn't really a problem.

Just work more and try again next year. If this was your last year, move on. Things happen in life. The trick is to come back as if they were nothing.

as Mr VD once said at one of my school assemblies:

[quote="Mr. VD @ 8th grade grad"]
don't look back and say "if only"; if only i did this, if only i made that, etc.
look forward and say "next time";  next time i will do this, now i will do this so that next time i can do that, etc.

[/quote]

if you feel short of your expectations, just look to do better in the future
think about what you can do now, so that you can do better next time

to wrap up, don't keep thinking about what happened
don't cry over spilled milk <-- first time i got to use that =)

However, the dense concentration of high scores near the top calls for a harder test, or less time for the same difficulty.

[quote="maybach"][quote="Mewto55555"][quote="mathlearner"]PowerofPi - I think you make an excellent and valid point, and one that [u]is[/u] sort of topic related in a way.  I think that my own disappointment and the disappointment of the person I'm close to who was really devastated by this year's outcome at Nats had to do with JUST what you had said. In that, they nailed the targets but just couldn't (and still have a hard time) quite comprehending how they can do SO well on Target, USAMO, and even ARML problems, but get about a third of the questions wrong on an "easier" Sprint round - especially a round many people say they felt  was relatively easy this year (note - "ease" is largely opinion - though one can argue that, I guess).

 I also wonder if making "careless errors" or in some cases, "overcomplicating" the problem, and thus making more mistakes on "easy ones" while acing the more diffficult ones, causes the MATHCOUNTS Competition as a whole to sort of "handicap" or "penalize" the more advanced students and allows some participants to make the top 12 who by more difficult tests would NOT have made it, and results in others NOT making the top 12  (or ranking not as high as their skills would lead one to believe) by virtue of careless errors they wouldn't typically make on more advanced problems. (one only needs to review this year's National rankings to support such a conclusion, which by no means applies to all participants) I guess I agree with Coach Boyd in that I'd prefer more difficult tests (at all levels) to separate "the men from the boys" so to speak. (Note - that's my quote - not Coach Boyd's - but, Coach Boyd, I believe you posted about preferring more difficult questions.)[/quote]

Yes well the "handicapping" serves a role in real life preparation. Say you are an engineer, designing an easy bridge to build. No matter how easy, you still can't make a stupid mistake.[/quote]

Well, the "more difficult questions"  isn't really my cup of tea.  The Sprint round questions are Hard enough!  Case in point:  2009 Chapter Sprint Round Problems 28-30, 2009 Chapter State Round Problems23, 28, 29, and 30.  Anyways, I'm not too great at math, and I also make tons of careless mistakes!  After all, you have to really screw up to get 63rd in sttate for Indiana(To make me feel a bit better, I did get first place in chapter, but you can't really call that a huge accomplishment).  Plus, most MC problems ALREADY test hard topics such as Discrete Math, Complex Probability, Combinatrics, Complicated Geometry, as well as those killer algebra problems.  In my opinion, the questions should be made EASIER and we should make there be 40 problems in 30 minutes for the Sprint Round instead of vice versa(like before 1990)[/quote]

Well, MATHCOUNTS is designed to seperate the top people from the really good people.

Most the good contests are like this; USACO Gold, USAMO.

I thought I'd get 10th in State as a 6th grader but Got 76th
Thought:5th in state as a 7th grader
Got:33rd


geometry
3D geometry
rotation
trigonometry

there is a uniform cone of semi vertical angle $\alpha$ .there is a massless spring of natural length $l_{0}$and spring constant $k$ is attached to the vertex of the cone,to(spring) which a block of mass $m$ is attached and which initially is held to the cone and the spring is unextended.now the cone is rotated about it's axis with an constant angular accelaration $\beta$.find

1)the distance $l$ from the top of the cone to the block(assume it as a particle) when the block flies away from the cone?

2)the time $\tau$ after which the block flies away from the cone?

Let it fly off the cone at time t.
Since the mass is moving in a circular path at this instant, we have that the centripetal force is provided by the spring force and the normal force component-

$kx sin{\alpha}-Ncos{\alpha}= m{\beta}^{2}t^{2}({{l_{0}}+x})cosec{\alpha}$
for  the vertical equilibrium
$mg = kx cos{\alpha}+Nsin{\alpha}$

When it just flies of put $N = 0.$
Then u have two linear equations in t and x
solving we get  $t = k sin{\alpha}/{\beta}$$\sqrt{x cos{\alpha}/m({l_{0}}k cos{\alpha}+mg)}$
the required length $l$ is = ${l_{0}}+x$(find x out from the above 2 equations...elimination seems a bit messy)

[quote="ashwinkrishnan"]
$kx sin{\alpha}-Ncos{\alpha}= m{\beta}^{2}t^{2}({{l_{0}}+x})cosec{\alpha}$
[/quote]
thanks a lot :lol:

i had messed up there with the term $cosec \alpha$ i had done with $\sin \alpha$. but that turned out in some way to be helpful becuse i came up with an entirely new interesting problem now u have assumed while proving(which even i did) that there is no slipping between the surfaces and there is just an tangential component along the horizontal circle .then can u comment on how the block moves down the surface and why there is no friction necessary and if $k$ is small enough what friction will be necessary

[quote="ashwinkrishnan"]I didn't understand ur point! could u elaborarte[/quote]
let's analyse the problem.
now it's beauty is it's simpplicity but yet complex nature. now the block  is in rest with respect to the point of contact in cone assuming block to be a point object now
surely friction has to act to avoid relative slipping between the two surfaces. now here comes the trouble. the cone 'tries' to slip tangentially to move in a circular path and the block 'tries' to slip along the cone...now  there has to be two 'componets' of friction to prevent both 'slippings'.now the velocity of the block at each point is along the tangent yet it moves downward along the cone even if the friction acts to prevent it from slipping..
now in our proof we statred with an assumption that there is just tangential friction and no friction along the cone. now if i assume the surfaces are quite rough and the spring constant is quite small then can u solve the problem becuse in that case the force due to eart may  overcome that due to spring. [b]note[/b] the motion of the block is just like an electron in a magnetic field

i think instead of this the solution should be like this though it yields the same result...

we consider friction acting along the slant surface of the cone say $f$

then we

$(f+kx)\sin \alpha-N \cos \alpha$ $=$ $m\omega^{2}l\ cosec \alpha$ and similarly for the vertical component ..

now as the block tends to fly away assuming continious break down

$f\to \ 0$ and $N\to \ 0$ .

so although  this yields the same result i suupose this holds for any situation.
now what i am missing is the motion of the block along the slant of the cone .can u comment on it's motion...


geometry
rectangle
ratio
geometric transformation
homothety
circumcircle

[quote="phuchung"]Let $ABC$ be a triangle. On the sides $AB$, $BC$, $CA$ of the triangle $ABC$, we externally erect three squares $ABDE$, $BCFG$, $CAHI$. Prove that the perpendicular bisectors of the segments $DG$, $FI$, $HE$ are concurrent.[/quote]

It's been posted before and it received several solutions, but I have no idea who posted it or what the topic was called... :(

[i]Moderator edit.[/i]

This problem is from the e-book [url=http://www-math.mit.edu/~kedlaya/geometryunbound/]Kiran Kedlaya's "Geometry Unbound"[/url] (problem 2 for section 5.3). According to this e-book, the source of this problem is Germany 1996, but actually this is not quite right. The true Germany 1996 problem (the problem 3 of the BWM 1996 round 2) was more general: Instead of "squares $ABDE$, $BCFG$, $CAHI$", we have "arbitrary rectangles $ABDE$, $BCFG$, $CAHI$"; hereby, the rectangles needn't even be similar to each other - they're just some arbitrary rectangles.

This general BWM problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=22321 . See also http://www.mathlinks.ro/Forum/viewtopic.php?t=118200 .

Let $ \triangle A'B'C'$ and $ \triangle A''B''C''$ be triangles with sides parallel to the sides of $ \triangle ABC$ through diagonal intersections of the squares and through opposite sides of the squares, respectively. Triangles $ \triangle ABC \sim \triangle A'B'C'$ are centrally similar, having parallel sides. Since sides of the triangle $ \triangle A'B'C'$ increased in the ratio of sides of the triangle $ \triangle ABC$, the homothety center is their common symmedian point $ K$. Likewise, triangles $ \triangle A''B''C'' \sim \triangle ABC$ are centrally similar with the same homothety center $ K$.

Quadrilateral $AHA''E$ is cyclic, having right angles at the vertices $H, E$, with circumcenter $A'$ at the midpoint of $AA''$ $\Longrightarrow$ perpendicular bisector of the segment $ HE$ passes through $ A'$. Similarly, the perpendicular bisectors of the segments $ DG$ and $ FI$ pass through $ B', C'$. Let $M$ be midpoint of $HE$. From circumcircle of this quadrilateral  $\Longrightarrow$ $ \angle C'A'K = \angle C''A''K = \angle HA''A = \angle HEA$. From $ A'B' \perp AE$, $ A'M \perp HE$  $\Longrightarrow$ $\angle HEA = \angle MA'B'$. It follows that the perpendicular bisector $ A'M$ of $HE$ is A'-median of the triangle $ \triangle A'B'C'$ and similarly, perpendicular bisectors of the segments $ DG$ and $ FI$ are B'-, C'-medians of this triangle, concurring at its centroid $ G'$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22321]Here[/url]'s the thread that grobber was referring to.

Thank you for the refernce. I am glad that my solution is different from those offered previously.

...

Is it right if we construct rectangles ABDE,BCFG,CAIJ with AB/BD=BC/BG=CA/AI???

Certainly, and the homothety center $K$ is still the symmedian point. The proposition is true even if the rectangles $ABDE$, $BCFG$, $CAHI$ are not similar at all. The homothety center $K$ then is not the symmedian point of the triangles $\triangle ABC, \triangle A'B'C'$, the lines $A'AK, B'BK, C'CK$ are just some concurring cevians and the perpendicular bisectors of the segments $DG, FI, HE$ concur because they are the isogonals of $A'K, B'K, C'K$ in the triangle $\triangle A'B'C'$.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 78 and 92
Sincerely
Jean-Louis


Let a,b,c be positive real such that $abc = 1$.
Prove:

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}$

[quote="paladin8"][hide]
$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} = \frac{(bc)^2}{a(b+c)} + \frac{(ac)^2}{b(c+a)} + \frac{(ab)^2}{c(a+b)} \ge \frac{(ab+bc+ca)^2}{2(ab+bc+ca)} = \frac{ab+bc+ca}{2}$
by the Cauchy-Schwarz lemma.

But $ab+bc+ca \ge 3\sqrt[3]{(abc)^2} = 3$ by AM-GM, so $\frac{ab+bc+ca}{2} \ge \frac{3}{2}$.[/hide][/quote]

i got the first part where you changed the LHS. But how did you get that to be greater than the RHS?? and what do you mean by Cauchy-Schwarz Lemma??

btw, just fyi, i did it in a different way:

[hide]
substitute $x=\frac{1}{a}$, $y=\frac{1}{b}$, $z=\frac{1}{c}$

then
$\displaystyle \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} = $
$\displaystyle \sum_{cyc} \frac{x^3}{y+z} = \sum_{cyc} \frac{x^3yz}{y+z} = \sum_{cyc} \frac{x^2}{y+z} \geq \frac{3}{2} \cdot xy \cdot yz \cdot zx = \frac{3}{2}$

Up there I multiplied the whole inequality by xy, yz, and zx, so that i could get the exponent on the numerator smaller while making the RHS stay the same. If we repeat the process once again, we can get it down to:

$\sum_{cyc} \frac{x}{y+z}$, which is Nesbitts. there are million ways to prove that. [/hide]

EDIT: UGGGH $\LaTeX$!!!! I love it, but i always screw up when i do it too much1!

I think it is called something like Cauchy-Schwarz extended for or something.

[quote="probability1.01"]What a coincidence, I just did this exact same problem on the Inequalities class board but a few minutes ago!  As for Cauchy-Schwarz thing, well it's 1 step away from the actual inequality, so maybe call it a corollary to Cauchy-Schwarz?

xxreddevilzxx -- your solution doesn't seem right.  I don't see what you did on your first and second steps.  Maybe I am misunderstanding your proof.[/quote]

his proof is ok.

Also, past IMOs are generally easier than recent ones.  But 1995 is still pretty recent.  And xxreddevilzxx, would you please elaborate on your solution?  I don't quite see what you're talking about  :blush: .

[quote="mecrazywong"][quote="paladin8"][quote="mecrazywong"]This problem is from IMO 1995.[/quote]

Wow. It is. A bit easy for an IMO problem, no? ;)[/quote]
Well...maybe you don't know the cut-off for gold of IMO 1995 is 37 points ;)[/quote]

No, I didn't :P But yeah, I can see why, though.

Actually, I don't know if the end of his solution is correct.

It's not hard to finish though, we have

$\sum \frac {x^2}{y+z}\geq \frac {\sum x}{2}\geq \frac {3}{2}$

The worst part about this inequality, though, is that you can solve it without much difficulty by expanding and using $abc=1$ to homogenize$.

[quote="paladin8"]The Cauchy-Schwarz lemma, I don't know if it actually has a real name:

$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n} \ge \frac{(\sum a_i)^2}{\sum b_i}$[/quote]

Using this,

$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}=\frac{\frac{1}{a^2}}{a(b+c)}+\frac{\frac{1}{b^2}}{b(c+a)}+\frac{\frac{1}{c^2}}{c(a+b)}\geqq \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2(ab+bc+ca)}$

$=\frac{(ab+bc+ca)^2}{2(ab+bc+ca)}\geqq \frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$.


linear algebra
matrix
linear algebra unsolved

If $A$ and $B$ are $n$ x $n$ matrices such that the matrix $C=AB-BA$ commutes with $A$, then $C$ must be nilpotent.


function
logarithms
calculus
calculus computations

y(x) is a differentiable function defined for all real numbers that satisfies the equation
$ y'x^2 \plus{} y \equal{} 0$

find $ y( \minus{} ln3)$

$ \frac {\textbf dy}{\textbf dx} x^2 \equal{} \minus{} y$
$ \frac {\textbf dy}{y} \equal{} \minus{} \frac {\textbf dx}{x^2}$
$ \ln |y| \equal{} \textbf C \plus{} \frac {1}{x}$
$ y \equal{} \textbf{exp} \left\{\textbf C \plus{} \frac {1}{x} \right\}$
$ y( \minus{} \ln 3 ) \equal{} \textbf{exp} \left\{\textbf C \minus{} \frac {1}{\ln 3} \right\}$
where C is some constant.

$ \frac {dy}{dx} \plus{} \frac {y}{x^2} \equal{} 0$

Use an integrating factor on that

[hide]$ \frac {d}{dx}(e^{ \minus{} \frac {1}{x}}y) \equal{} 0$ So integrating: $ e^{ \minus{} \frac {1}{x}y} \equal{} c$ and solve for y $ y \equal{} c e^{\frac {1}{x}}$ etc.[/hide]

edit: Sorry, posted before I saw your reply @petko

but what is the actual value of y(-ln3)?
[hide="hint"]There's only one function that satisfies [b]both[/b] requirements[/hide]

$ y \equiv 0$ so $ y(\minus{} \ln 3) \equal{} 0$


How many 10-digits distinct  integers can  be written using 0,0,0,0 and 1,1,1,1,1 and 2

Are they supposed to be 'well' written? I mean : is $0$ allowed to be the leftmost digit?

Pierre.

Assuming it has to begin in either 1 or 2, isn't it just 9!/(4!5!) + 9!/(4!4!1!) = 126+630 = 756? If it is allowed to begin in 0, wouldn't it be 10!/(4!5!1!) = 1260? Or am I missing something?

0 not allowed to be the left most digit...

ok, then my first answer should work, eh?


If I'm understanding the question correctly, the $ 600,000$ will lose $ 15\%$ of $ 600,000$ the first year, $ 13.5\%$ of $ 600,000$ the second year, and so on. Therefore, they will lose $ 15\plus{}13.5\plus{}\cdots\plus{}3\plus{}1.5\equal{}10\cdot\frac{15\plus{}1.5}{2}\equal{}5\cdot16.5\equal{}82.5\%$ of what they have. They will be left with $ 17.5\%$, or $ \boxed{105,000}$

[quote="indialover"]Q:A piece of equipment cost a certain factory Rs600,000.If it depriciates in value 15% the first,13.5% the next year,12% the third year,and so on.What will be its value at the end of 10 years ,all percentages applying to the original cost?[/quote]
[hide="Solution"]
It is an arithmetic progression, so you need to calculate $ \displaystyle\sum_{k = 1}^{10} - 1.5k + 16.5$.

This is $ 15 + 13.5 + 12\cdots1.5$.  Therefore, it is $ \dfrac{10\cdot16.5}{2}$ or $ 165/2 = 82.5$.  So $ .825\cdot600,000$ is ${ 495000}.$  So $ 600,000 - 495,000 = \fbox{105,000}$[/hide]