Imagine placing solid spheres (possibly of different radii) around a point light source is such a way that:

a. The light source is not inside any sphere.
b. The spheres do not intersect.
C. The spheres will block all light coming from the source.

What is the minimum number of spheres needed to achieve this?

I think it is [hide] 5 [/hide] but I'm not quite sure...

This is an interesting problem. How did you find your answer?
[/hide]

What kind of configuration are you thinking of?
I'm fairly sure you can do it with eight.

I got [hide]Let the spheres be centered at the vertices of a tetrahedron, and let the "holes" in the four "faces" be covered by one sphere each. for a total of 8 spheres[/hide] as well.

That was my solution as well.
I'm not sure how to prove it's optimality.

The only thing I can think of is attacking it as a "degrees of freedom" type of proof.
(i.e. adding spheres is taking away degrees of freedom)

[quote="pascal_1623"]i got 5 by putting the centers of the spheres at the vertices of a triangle. Then I placed one sphere on top and one on bottom, but I'm not sure if that works...[/quote]
I think there will still be holes left.... try it with playground balls to see if it works.

[quote="pascal_1623"]i got 5 by putting the centers of the spheres at the vertices of a triangle. Then I placed one sphere on top and one on bottom, but I'm not sure if that works...[/quote]

Hmm but that way another two triangles are formed and there are two empty spots left.

1. we can't directly enclose the point, so we can only encircle it with spheres.
2. tetrahedron is the least-sides closed polyhedron in 3-dimension, so at least we should put 4 inter-tangent spheres centered at the vertices of a tetrahedron in which the point is.
3. but spheres can't intersect, they can only be tangent to each other; for each "hole", we mut at least put 1 sphere to block the rays.
Hence, at least we must use 8 spheres.

Maybe like this? (Although it doesn't seem like a maths proof...)


1)

If 2x-5, x+1, and 3x-8 are all integers and x+1 is the median of these integers, which of the following could be a value for x?

5,7,9,10,11 (I'm looking for a way not involved in substituting)

2) If x^2 + y^2 = 73 and xy=24, what is the value of (x+y)^2?

3) If 30 percent of 40 percent of a positive number is equal to 20 percent of [i]w[/i] percent of the same number, what is the value of [i]w[/i]?

4) If 2x<y<0, which is the greatest?

-2x, -(2x+y), 2x, 0, -y

5) Set X has x members and set Y has y members. Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k>0). Which of the following represents the number of members in set Z?

Guys, I'm sorry if I'm disrupting the forum by asking for "simple" problems you all might be thinking, but I really need to practice for the SAT and some math problems that I've forgotten these years. But I do hope that if there "is" someone like myself practicing the SAT, these would come in handy.

for number two, just note that $ (x\plus{}y)^2 \equal{} x^2\plus{}2xy\plus{}y^2$

[hide="#1"]say $ 2x\minus{}5<x\plus{}1<3x\minus{}8$

subtract x from all sides.

x-5<1<2x-8
since x-5<1, x<6.

since 1<2x-8, $ x>\frac{9}{2}$

so $ \frac{9}{2}<x<6$

hm 5 is in that interval.

answer : 5[/hide]

[hide="#2"]

$ (x\plus{}y)^2\equal{}x^2\plus{}2xy\plus{}y^2$
$ x^2\plus{}y^2\equal{}73, 2xy\equal{}48$

$ (x\plus{}y)^2\equal{}73\plus{}48\equal{}121$

[/hide]

[hide="#3"]

$ x*\frac{3}{10}*\frac{4}{10}\equal{}x*\frac{2}{10}*\frac{w}{100} \implies 1200\equal{}20w \implies w\equal{}60$

answer : 60

[/hide]

[hide="#4"]
we can simply assume that $ y\equal{}\minus{}1, x\equal{}\minus{}1$
$ \minus{}2x\equal{}2$
$ \minus{}(2x\plus{}y)\equal{}3$
$ 2x\equal{}\minus{}2$
$ 0\equal{}0$
$ \minus{}y\equal{}1$

answer : $ \minus{}(2x\plus{}y)$
[/hide]

#5

the problem is not complete.

what were the choices for #5?
oops nvm didnt see the k part :D

Why is this in middle school Classroom math?

That said,

[hide="#1"]
$ x+1=M$. Also, subtract all values by $ M$. $ 0$ is the median of $ 0$, $ M-6$, and $ 2x-9$. Assume that $ M-6<0<2M-9$. $ \boxed{5}$ works in this situation.
[/hide]

[hide="#2"]
$ (x+y)^2=x^2+y^2+2xy=73+48=\boxed{121}$.
[/hide]

[hide="#3"]
$ 0.3\cdot 0.4 \cdot n=0.2 \cot W \cdot n$.
$ 0.12=0.2W$.
$ W=0.6=60%
$.
[/hide]

[hide="#4"]
x, y are both negative.
The numbers that are positive (obviously the greatest is positive) are $ -2x$, $ -(2x+y)$, and $ -y$.
This, Also, $ -2x>-y$.
Finally, because both numbers are negative, their sum is less than each of them, and the negative of their sum is greater than each of their negatives. Thus, $ -(2x+y)$ is the greatest.
[/hide]

[hide="#5"]
x+y-k?
[/hide]


How do we construct the heptadecagon(17-gon)? And how to prove the construction? :pilot:

[url=http://cdyjapan.hp.infoseek.co.jp/famous/const008.html]Richmond Method 1893[/url]

[youtube]E2QMzRcjKkM[/youtube]

I want to know how to prove these construction  :)


superior algebra
superior algebra unsolved

Let $R$ be a commutative unitary ring such that all of its ideals are prime. Prove that $R$ is a field.
[hide]Estimated life-time in the unsolved forum: half an hour. Btw can you solve it using different approaches? :D [/hide]

I'd say 1 minute! Take a nonzero $x$ and look at the ideal generated by $x^{2}$.

That's still 13 minutes ;) [he counted the time on the forum, not how long someone needs]


The number of $ x$-intercepts on the graph of $ y \equal{} \sin(1/x)$ in the interval $ (0.0001,0.001)$ is closest to

$ \textbf{(A)}\ 2900 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3100 \qquad \textbf{(D)}\ 3200 \qquad \textbf{(E)}\ 3300$

[hide]$ z : \equal{} 1/x$

This is the number of $ z$ intercepts of $ y \equal{} \sin z$ in the z-interval $ (1\; 000 \; ,10 \; 000)$

We can approximate the number of intersections as $ \frac{10 \; 000 \; \minus{} \; 1 \; 000}{\pi} \equal{} \frac{9\; 000}{\pi} \approx 2 \; 900$[/hide]

Why do they always make these AMC 22-25s more annoying than hard?
Anyways, we have $ sin(1/x)\equal{}0 \implies 1/x\equal{} k \pi \implies1000 \le k \pi \le 10000$, hence the answer's 2900, but on the AMC you'd have to divide by pi. How fun.

^Calculators were allowed in 2003, so it wouldn't have been too bad.

Also, you have 9000/(3.14), and 3.1*2900 = 8990, so the answer will be less than 2900.

[b]Solution:[/b] Clearly a root must be in the form $x = \frac{1}{k\pi}$, where $k$ is an integer. We have that: $$0.0001 \le \frac{1}{k\pi} \le 0.001 \Rightarrow$$ $$1000 \le k\pi \le 10000 \Rightarrow k \approx \frac{9000}{\pi} < \frac{9000}{3} = 3000 \Rightarrow \boxed{A}$$


The distance between the front and the rear axle of a car is $ 1,9m$ and the distance between the left and the right wheels is $ 1,5m$. The car has got rear wheel drive, but one of the rear tires is damaged - it has got a coefficient of friction $ 0,2$, while the other tires have got a coefficient of friction $ 0,8$. With what maximum acceleration can the car start to move rectilinearly without sliding, if all wheels are loaded equally?

I don't know the solution to this problem, there is a hint that it should be solved numerically.

The only unbalanced force acting on the faulty tire is the frictional force, with maximum value [b]Nu[/b], [b]N[/b] is normal force and [b]u[/b]=.2. Hence maximum acceleration of the car is [b]gu[/b]

golduck can u xplain ur question and hint a bit

As I said I don't know the solution or even the answer. The car must move rectilinearly so it shouldn't rotate, so maybe it is possible to have a bigger acceleration with a bigger force on the good rear wheel and turning the front wheels to avoid rotation.


Die without me,
Never thank me.
Walk right through me,
Never feel me.
Always watching,
Never speaking.
Always lurking,
Never seen.

Who or what is "me" referred to in the problem ?

[hide]

Air.  You die without air but never thank it.  You walk right through air without feeling it.  Air is always watching, never speaking, always lurking, and never seen.

[/hide]

[hide]Definitely air.[/hide]

@E^(pi*i)=-1:  Sometimes, you do see air (though your definitely not supposed to, but go to LA)...  :roll:


[quote]1) A science experiment calls for 0.05g of potassium to be added to a solution every 5 mins. The only potassium available is a bottle of 10-cg tablets.... 100 cg = 1g.... How should these tablets be used for the experiment?

A. Add one quarter tablet every 5 mins.

B. Add one half tablet every 5 mins.

C. Add one tablet every 5 mins.

D. Add two tablets every 5 mins.

E. Add two and one half tablets every 5 mins.[/quote]

[quote]2) In a group of 400 grade 12 students and 600 grade 11 students, the average score on a test was 70 for the grade 12 students and 50 for the grade 11 students. The average score for the entire group of 1000 students is.

A. greater than 50 but less than 55

B. greater than 65 but less than 70

C. exactly 60

D. greater than 60 but less than 65

E. greater than 55 but less than 60[/quote]

[quote]3) A fertilizer is advertised as a 5:3:2 mix, indicating that it contains chemicals A and B in a ratio of 5 to 3, and chemicals B and C in a ratio of 3 to 2. How many kilograms of chemical C are contained in a 30kg bag of fertilizer?

A. 2

B. 3

C. 6

D. 12

E. 15
[/quote]

[quote]4) Last Tuesday, 90 percent of the customers at a fast food restaurant ordered french fries and 60 percent ordered a soft drink. The percentage of customers who ordered BOTH french fries and soft drink must have been between

A. 0 percent and 10 percent

B. 10 percent and 40 percent

C. 50 percent and 60 percent

D. 60 percent and 70 percent

E. 70 percent and 90 percent[/quote]

[quote]5) Weathering is the gradual wearing down of land due to exposure to air and water. In a wet climate, limestone weathers at a rate of approximately 0.05 cm every 100 years. At this rate, about how long would it take to completely wear away a layer of limestone 30 cm thick?

A. 600 years

B. 6000 years

C. 60 000 years

D. 600 000 years

E. 6 000 000 years[/quote]

[quote]6) In which of the following statements is the use of percent LEAST appropriate?

A. The new warehouse has 30.5 percent less storage space than the old one.

B. Only 85.5 percent of the runners finished the marathon.

C. The meat is 97 percent fat free.

D. There were 150 percent fewer protesters at the rally this year.

E. Her sales are up 200 percent over last year.[/quote]

[quote]7) For which of the quantities below is it more reasonable to estimate the quantity by gathering data on a sample rather than on the whole population?

I. The proportion of Canadian citizens who favor the passage of a law
II. The proportion of grade 10 students at Banting High School who have taken a computer programming course.
III. The proportion of drivers in a certain province who wear seat belts.

A. I only

B. II only

C. III only

D. Both I and II

E. Both I and III
[/quote]

[quote]8) A teacher discovered that the two students receiving the highest scores on a recent test had cheated. The teacher decided to remove these two scores and recompute the range and mean of the test scores. How would the recomputed values of the range and mean compare with the original values?

A. Both would be greater.

B. Both would be less.

C. The mean would be greater, and the range would be less.

D. The mean would be less, and the range would be greater.

E. It is impossible to predict how the range and the mean would change.[/quote]

[quote]9) The population of a country is 21.2 million. If 86% of the people live in urban areas, which of the following is the best approximation (in millions of people) of the total urban population?

A. 20 divided by 0.80

B. 20 divided by 0.90

C. 20 * 0.90

D. 20 * 0.80

E. 30 divided by 0.80
[/quote]

[quote]10)
The masses of two substances were found in a science lab. The masses determined were rounded to the nearest hundredth of a gram and reported as follows:

SUBSTANCE 1 0.58g
SUBSTANCE 2 0.42g

What is the maximum possible difference (in grams) between the exact masses of these two substances?

A. 0.15

B. 0.16

C. 0.17

D. 0.18

E. 0.19
[/quote]

[quote]11) A pair of contact lenses costs 300.00. The warranty with the lenses states that if the lenses are damaged or lost within 12 months, they will be replaced at a cost proportional to the part of the 12 months they were used. If the lenses are scratched after 8 months, how much should it cost to replace them?

A. 67.00

B. 75.00

C. 100.00

D. 200.00

E. 225.00[/quote]

[quote]12) In an automobile factory, 42.5 percent of the cars produced had power windows. What is an equivalent way of describing the proportion of cars with power windows?

A. 85 out of every 200 cars

B. 50 out of every 110 cars

C. 45 out of every 100 cars

D. 40 out of every 100 cars

E. 40 out of every 85 cars[/quote]

[quote]13) The students in Borden Junior High School sold $ \$13,000$ worth of candy bars last year. If their profit is $ \$1,000$ on the first $ \$10,000$ in sales and 20% on all sales over $ \$10,000$, which of the following represents their total profit selling candy bars?
[color=darkblue]
Moderator's Note: Try to learn some $ \text{\LaTeX}$ before using it and experimenting in examples.[/color]

A. [code]1000[/code] + (0.20 x [code]13 000[/code])

B. [code]1000[/code] + (0.20 x [code]10 000[/code])

C. [code]1000[/code] + 0.20 ([code]10 000[/code]-[code]1000[/code])

D. [code]1000[/code] + 0.20 ([code]13 000[/code]-[code]1000[/code])

E. [code]1000[/code] + 0.20 ([code]13 000[/code]-[code]10 000[/code])[/quote]

[quote]14) Thank you in advance to whoever answers this for me. Appreciate it so much.

If the temperature at 8:00 a.m. was 10 degrees C. and at 4:00 p.m. the same day it was -6 degrees C., what was the average decrease per hour during this period?

A. 0.5 degrees C

B. 1.5 degrees C

C. 2 degrees C

D. 4 degrees C

E. 8 degrees C
[/quote]

[quote]15) Another math question for whoever wants to answer it for me. Thank you.

 The speed of light is 299 793 km/s. If it takes 8 mins. for light to reach the earth from the sun, which of the following is the best estimate of the distance (in kilometres) between the earth and the sun?

A. 300 000 x 8 x 60

B. 300 000 x 8

C. [u]300 000[/u]
                      x 60
             8

D. [u]300 000[/u]
             8

E. [u]300 000[/u]
                     x 8
            60[/quote]

[quote]16) In A.D. 1990, an animal bone estimated to be 3000 yrs old was found. In about what year did this animal die?[/quote]

[quote]17) Last Tuesday, 90 percent of the customers at a fast food restaurant ordered french fries and 60 percent ordered a soft drink. The percentage of customers who ordered BOTH french fries and soft drink must have been between

A. 0 percent and 10 percent

B. 10 percent and 40 percent

C. 50 percent and 60 percent

D. 60 percent and 70 percent

E. 70 percent and 90 percent[/quote]

[quote]18) Consider the following relationship:

Relative Humidity| actual moisture content
(percent)= ________________________times 100
saturation moisture cntent

If actual moisture content remains constant and saturation moisture content increases, what happens to relative humidity?

A. It increases.

B. It decreases.

C. It remains unchanged.

D. It could increase or decrease depending on the exact amount of the increase in saturation moisture.

E. It increases until it reaches 100 and then decreases.
[/quote]

1)
[quote]10-cg =0.1g

but we need 0.05g, which is one half of 0.1g.

answer : B. Add one half tablet every 5 mins.[/quote]

2)
[quote] $ \frac {400*70 \plus{} 600*50}{1000} \equal{} \frac {28000 \plus{} 30000}{1000} \equal{} 58$

answer : E. greater than 55 but less than 60[/quote]

3)
[quote]A:B:C=5:3:2
A:B:C=5x:3x:2x

30=5x+3x+2x=10x
x=3
C=2x=6

answer : C(6)[/quote]

5)
[quote]0.05 cm every 100 years

X100

5 cm every 10000 years

X6

30 cm every 60000 years

answer : C (60000 years)[/quote]

6)
[quote]lol

D. There were 150 percent fewer protesters at the rally this year.

the total is 100% and 150% decreased? xD

D is the most inappropriate use of %.[/quote]

[quote]86% is closer to 90% than 80%.

and 21.2 million is about 20 million.

so the total urban population would be 20 million * 0.9

Answer : C
[/quote]
[quote]i have no idea what the 'following' says

so ill just solve it

problem: The students in Borden Junior High School sold 13000 dollar worth of candy bars last year. If their profit is 1000 dollars on the first 10 000 dollars in sales and 20% on all sales over 10000,which of the following represents their total profit selling candy bars?

think thats it

over 10000=3000 dollars
$ 3000*\frac {2}{10} \equal{} 600$

1000+600=1600

answer : 1600 dollars

[/quote]

17)
[quote]if their overlapping part was smallest = 90+60-100=50
if their overlapping part was largest = 60

answer : C[/quote]

[quote="dgemor"]Consider the following relationship:

Relative Humidity|                   actual moisture content
                        (percent)=  ________________________times 100
                                         saturation moisture cntent

If actual moisture content remains constant and saturation moisture content increases, what happens to relative humidity?

A. It increases.

B. It decreases.

C. It remains unchanged.

D. It could increase or decrease depending on the exact amount of the increase in saturation moisture.

E. It increases until it reaches 100 and then decreases.[/quote]

could you latex the formula?
i am kinda guessing here by looking at your quote
hard solution:
so consider the constant a which stands for actual moisture content and the variable s that stands for the saturation moisture content and the variable r that depends on a and s
we have
$ r % = \frac {a}{s} \cdot 100
$
we want to know what happens to r when s increases
suppose s is doubled
the fraction $ \frac {a}{s}$ is halved so r is halved
if s is  multiplied by l, with l>1 (this signifies s increasing) then r is divided by l (this signifies r decreasing) so as s increases, r decreases and the answer is b

5) suppose you have a few towers
if the tallest two are knocked down, you can see the average height of the towers will decrease
(this is kinda hard to prove but it is common sense so i will leave it at that)

so the mean will decrease

8)
suppose the least grade is a and the highest 3 are b, c, and d
we know a<b<c<d
the original range is d-a, the new one is b-a
we know d>b so d-a>b-a and the original range is higher so the new range is lower

easier:
so the numerator of a fraction stays the same and the denominator increases
since you are dividing a constant into more parts, the quotient will be less and
as s m c increases, r h decreases.
and the answer is $ \boxed{b}$

side note: you should read the forum guidelines

14)
i my self would think e
the number of tenth graders at a school is kinda sorta low-ish but the populations of provinces and nations is definitely higher

I think it is. II is better doing whole population because there are not everywhere same conditions, III could do a sample, because it's a province, and I'm not sure about I.

Would this happen to be homework? And please don't post 16 problems all at once, especially until you can fix your codes.  :wink:

I'm trying to compile all of the smaller topics into one huge topic.

fishythefish, if you could help me fix his coding then it will be greatly appreciated  :)

I'd love to.

There's just one problem....Some of it is pretty straightforward: he just put a typed problem into a "code" box. The problem is that with some of the others, it's hard to tell what he meant. I see here that some of the problems have been fixed, but I've tried to do them in the original threads, and the problems are all but unintelligible.


Ross Mathematics Program
geometry
radical axis
geometry proposed

Let ABC be triangle and take the circles (O1),(O2),(O3) passes
through vertex (B,C),(C,A),(A,B) such that its centers are external.
If X,Y,Z are the another cut points of [(O2),(O3)],[(O1),(O3)],
[(O1),(O2)] respectively.Prove that:AY.BZ.CX=AZ.BX.CY

A cute one, which looks scary at first :):

Let $T$ be the radical center of the three circles. Then $\frac{AY}{BX}=\frac{TA}{TB},\ \frac{BZ}{CY}=\frac{TB}{TC},\ \frac{CX}{AZ}=\frac{TC}{TA}$. We just multiply these three expressions and we're done.

P.S. This works in all cases except when the three circles actually intersect at $T$ ($X=Y=Z=T$), but that case is, of course, trivial.

[quote="juancarlos"]Let ABC be triangle and take the circles (O1),(O2),(O3) passes
through vertex (B,C),(C,A),(A,B) such that its centers are external.[/quote]

This condition (that the centers are external) is not necessary!

[quote="juancarlos"]If X,Y,Z are the another cut points of [(O2),(O3)],[(O1),(O3)],
[(O1),(O2)] respectively.Prove that:AY.BZ.CX=AZ.BX.CY[/quote]

Indeed, in Ross Honsberger's nice book [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], this result is mentioned as the [i]Haruki theorem[/i], and proven by exactly the same argument as Grobber used.

  Darij

This is only a degenerate case of Haruki's theorem.
Thanks.
Juan Carlos

[quote="juancarlos"]This is only a degenerate case of Haruki's theorem.[/quote]

Why this? Maybe you know a more general version of Haruki's theorem, but the result which was called Haruki theorem in Ross Honsberger's book I cited above is exactly equivalent to your problem.

  Darij

Dear Darij:
Are you consider that the perform of "my problem" is equal
to original perform of Haruki's theorem?.
Please see the anexed file GIF.
Thanks

[url=http://mathworld.wolfram.com/HarukisTheorem.html]Here[/url]'s a link according to which your problem is exactly Haruki's theorem. Could you please state the more general version you know?

Please see:
http://poncelet.math.nthu.edu.tw/disk3/cabrijava/haruki.html
And the another degenerate case in the anexed file GIF

But Juan Carlos, the applet you linked to does show that your problem is equivalent to the Haruki theorem!

Just denote the six pairwise points of intersection of the three circles by A, X, B, Y, C, Z and you get exactly the configuration of your problem...

Or do I misunderstand something here?

  Darij

No, You  don't misundeerstood nothing.
Only, I refered to the well-know original perform as  three circles with
common region and the problem the three circles have  not a common
region, further the  proof don't use the Ceva's theorem as the original case.
For these facts I named as degenerate case.
According to the cut points of the circles all are cases of Haruki's theorem.
Sorry for the confusion.
Thanks.
Juan Carlos

Thanks, now it's clear.

The circles needn't have a common region. And [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42807]the proof (at least, the one given in Honsberger's book)[/url] doesn't need Ceva's theorem, although the Haruki theorem looks very similar to Ceva. The three circles always have a radical center T, which lies on the lines AX, BY, CZ, so that you can just use Grobber's argument.

  Darij


algebra
function
domain
real analysis
real analysis unsolved

Prove that the system $\left\{ \begin{array}{l}\frac{dx}{dt}=y^{2}-xy, \\ \frac{dy}{dt}=x^{4}-x^{3}y \end{array}\right.$ has no non-constant periodis solutions.
(O. Stanzhitskyy)

Called $S$ the system, it is an autonomous differential equation system

$F$:$R^{2}\rightarrow R^{2}$
$F(x,y)=(y(y-x);x^{3}(x-y))$ is $C^{1}$ then with Cauchy-Lipschitz
for any $(a,b) \in R^{2}$ there exist a unique solution of $S$ passing
trough $(a,b)$, and the interval of the definition is an open.

Equilibrum point:
$(0,0)$ and all point on the line $y=x$

Cartesian equation of the orbit:
$S$ gives $x'x^{3}+y'y=0$ integrate

$x^{4}+2y^{2}=k \geq 0$

All orbit are closed curves
There is an $t'$ real such that $x(t')=a,y(t')=b$
Let the smallest real $t''>t'$ such that $x(t'')=a,y(t'')=b$ then $T=t''-t'>0$
is the period of the solution $x(t),y(t)$.
Indeed $x(t-T),y(t-T)$ is solution of the Cauchy problem at $(a,b)$
with Cauchy-Lipschitz theorem $x(t-T)=x(t), y(t-T)=y(t)$ and
the interval of definition of the solution is $R$.

With a software you draw $x^{4}+2y^{2}=k \geq 0$ there is a symetri to the
origin.

Another view:

Bendison's criterion for periodic solution:

Consider the autonomous system $S$

$x'=f(x,y)$
$y'=g(x,y)$ in a domain $D \subset R^{2}$

Theorem Suppose $D$ is simply connected, $f,g$ is continuously
differentiable in $D$.
$S$ can only have periodic solution if $f'_{x}+g'_{y}$ changes
sign in $D$ or if $f'_{x}+g'_{y}=0$ in $D$

Back to our problem

$f(x,y)=y(y-x)$
$g(x,y)=x^{3}(x-y)$

then $f'_{x}+g'_{y}=-y-x^{3}$ change sign
so we have periodic solution.


Let ABCD be a convex quadrilateral, and let K, L, M, N be the midpoints of its sides AB, BC, CD, DA, respectively. Let the lines NL and KM intersect at a point T. Prove that

$\frac83\left|DNTM\right|<\left|ABCD\right|<8\left|DNTM\right|$,

where $\left|P_1P_2...P_n\right|$ denotes the area of an arbitrary polygon $P_1P_2...P_n$.

Don't get worried because of this one being an IMO training problem - it's quite easy knowing some basic facts about quadrilaterals...

  darij

$NK, LM$ are the midlines of the triangles $\triangle ABD, \triangle BCD$. Hence, $NK = BD/2 = LM$ and $NK \parallel BD \parallel LM$. This is sufficient to conclude that $KLMN$ is a parellogram and $T$ the intersection of its diagonals. Denote

$\triangle = |ABCD|$
$\triangle_T = |NKT| = |KLT| = |LMT| = |MNT|$
$\triangle_A = |AKN|, \triangle_B = |BLK|, \triangle_C = |CLM|, \triangle_D = |DNM|$

Since $2 \cdot NK = BD$, $4\triangle_A = |AKN| < \triangle$ and similarly, $4\triangle_B < \triangle$, $4\triangle_C < \triangle$, $4\triangle_D < \triangle$. Let $X$ be the intersection of the lines through the midpoints $N, K$ parallel to the sides $AB, DA$, respectively. Since $AKXN$ is a parallelogram with the diagonal $NK$, the triangles $\triangle XNK \cong \triangle AKN$ are congruent, and $|XNK| = |AKN| = \triangle_A$. The triangles $\triangle XMN \cong \triangle BLK$ are also congruent, because $XN \parallel BK$, $MN \parallel KL$ and $MN = KL$. Hence, $|XMN| = |BLK| = \triangle_B$. In addition, $XM \parallel BL$ and $XM = BL$. The triangles $\triangle XLM \cong \triangle CML$ are also congruent, because the side $LM$ is common and the sides $XM, BL$ are parallel and equal. Hence, $|XLM| = |CML| = \triangle_C$. The triangles $\triangle XKL \cong \triangle DNM$ are also congruent,  because $XK \parallel DN$, $KL \parallel MN$ and $KL = MN$. Hence, $|XKL| = |DNM| = \triangle_D$. Consequently, the parallelogram area $|KLMN|$ is equal to the area $|ABCD| - |KLMN|$, i.e.,

$8 \triangle_T = 2|KLMN| = |ABCD| = \triangle$

Since $|DNTM| = \triangle_D + \triangle_T$, the desired inequality can be written as

$\frac 8 3 (\triangle_D + \triangle_T) < 8 \triangle_T < 8(\triangle_D + \triangle_T)$

The right inequality is now trivial and the left inequality melts down to

$8 (\triangle_D + \triangle_T) < 24 \triangle_T$

$\triangle_D < 2 \triangle_T$

which is certainly true, because we earlier showed that $4 \triangle_D < \triangle = 8 \triangle_T$.


This thread is for the Team Chess Tournament Round 1: Team Gamma (White) vs. Team Zeta (Black). Play begins on December 7th. Any questions may be asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]here[/url]. All information/FAQ is being updated [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]here[/url].

[hide="Team Gamma"]
   1. nayel
   2. BoesFX
   3. mz94
   4. shaggy75
[/hide]

[hide="Team Zeta"]
   1. ProtestanT
   2. chessplayingballer
   3. snee
   4. toadoncart
[/hide]

Each team will have 3 days for each move.

Good luck!

Move 1 (White): [b]d4[/b]

[img]http://img101.imageshack.us/img101/967/diagramikphpqy6.jpg[/img]

sorry i didn't see the pm.
how do i make a move?

edit-
Move 2 (Black): [b]d5[/b]
[img]http://img338.imageshack.us/img338/6397/camf87mnrh0.jpg[/img]

[quote="mz94"]ok, like, do we win by forfeit? idk how much time they have but its been at least a week or 2[/quote]

Team Zeta now has a warning for exceeding the time limit.

You can win by any normal chess methods, or if the game is taking too long, the TDs will decide.

Move 2 (White): [b]c4[/b]

[img]http://img179.imageshack.us/img179/1199/diagramik2phpwb7.jpg[/img]

Move 1 (Black):[b]Nf6[/b]
[img]http://img88.imageshack.us/img88/92/cailcjkbic4.jpg[/img]

Move 3 (White): [b]cxd5[/b]

[img]http://img165.imageshack.us/img165/8464/diagramik3phpjf1.jpg[/img]

Merry Christmas everyone! :)

Move 3 (Black): [b]Nxd5[/b]
[img]http://img266.imageshack.us/img266/7982/move3blacknxd5hb8.jpg[/img]

Move 4 (White): [b]e4[/b]

[img]http://img91.imageshack.us/img91/9466/diagramik4phpfe4.jpg[/img]

Nb6

The message is too small. Please make the message longer before submitting.

Move 5 (White): [b]Nc3[/b]

[img]http://img46.imageshack.us/img46/3584/diagramik5phphm4.jpg[/img]

Is the other team gonna move any time soon? :|

*tumbleweed*

Can our team can be considered to win this game? :roll:

[quote="ProtestanT"]From: 	ProtestanT
To: 	nayel
Posted: 	Thu Jan 31, 2008 7:32 pm
Subject: 	Re: Chess tournament

You can consider that we resigned if my team doesn't make a move in the next week.[/quote]

I guess that pretty much wraps things up.

I really don't see if this tournament thing is getting to somewhere...

[b]Team Gamma wins by forfeit.[/b]

The problem with the tournament is that the players aren't really active. I guess the majority of the wins will be from late forfeits rather than actual skill.


function
algebra
domain
calculus
derivative
complex analysis

Hi all, I have been working on this question for hours and I am stuck:

Let {f_n} be a sequence of holo. functions on domain X in C. assume the sequence is uniformly bounded on comp. subsets of X. Furthermore, {z_k} is a sequence in X converging to z_o in X and for all z_k, the limit of f_n(z_k) exists. Prove {f_n} converges uniformly on comp. subsets of X.

Okay here is what I have done so far: Using Montel's, we get {f_n} is normal, and hence we have a convergent subsequence. Let F be the limiting function of this subsequence.
We get lim f_n(z_k) = F(z_k) for all k, and using that I showed limit f_n(z_o) = F(z_o).
I then showed that this convergence is uniform on {z_k} + {z_o}.
I have also managed to show that limit (f_n)'(z_o) = f'(z_o) although my proof may be wrong, I think it is ok.

Now I am stuck. I know that it suffices to show that the derivatives of all orders at z_o converge, but this may not be the best way.

Please do not solve this problem for me as it is a homework problem, but if you could offer a hint as to what my next step should be that would be greatly appreciated.

Thanks in advance

You can finish it the way you go. Just notice that the $ n$-th derivative of $ F$ can be expressed as the limit of certain linear combinations of $ F(z_k)$ and, moreover, the difference between the derivative and the linear combination can be estimated by sumething depending only on the upper bound for $ F$ in a disk around $ z_0$.

Alternatively, just assume that $ f_n$ do not converge to $ F$. Then take a subsequence whose members are all far from $ F$ and choose a convergence subsequence of that subsequence.


function
real analysis
real analysis unsolved

[url=http://bildr.no/view/510916][img]http://bildr.no/thumb/510916.jpeg[/img][/url]
[url=http://bildr.no/view/510917][img]http://bildr.no/thumb/510917.jpeg[/img][/url]

I can not understand the paragraph denoted by red line.

Thanks

[quote="mavropnevma"]Just make yourself a graphic sketch. After getting $ x_2 < a < x_0 < x_1 \leq b$ with $ f(x_2) < f(x_0) < f(a) \leq f(x_1)$, any value $ v \in (f(x_0), f(a))$ will be taken at least three times: at least once each in the intervals $ (x_2,a)$, $ (a,x_0)$ and $ (x_0, x_1)$, in contradiction with the hypothesis.

[hide="For Your Eyes Only"]You should try a little harder before giving up and asking the question on the forum. Do it only when you have exhausted all your resources...[/hide][/quote]

Yes， you are right, I feel a little boring and tiring. Thanks, I should study harder and harder.

This question , I feel it is strange.
$ f(x_0)\equal{} \min_{x\in [a,b]} f(x)$, so it is obvioulsy $ f(x_0) \leq f(a)$.
But the author wants to prove $ f(x_0)\geq f(a)$.


Prove that if $ p$ and $ 8p^{2} \plus{} 1$ are both primes, then so is $ 8p^{2} \plus{} 2p\plus{}1$.

[hide]When $ p\equal{} 1 or \minus{}1 (mod 3)$,

$ 8p^{2}\plus{}1 \equal{} 0 (mod 3)$

Therefore for both $ p$ and $ 8p^{2}\plus{}1$ to be prime,
$ p\equal{}3$
when
$ 8p^{2}\plus{}2p\plus{}1 \equal{} 79$, and the statement is true. [/hide]


A line is tangent at $ A$ to a circle with center $ O$. Point $ N$ is on the line and $ P$ is on the circle, so
that $ PN$ is perpendicular to $ AN$. If $ AN \equal{} 20$ and $ PN \equal{} 8$, determine the radius of the circle.

[hide="Solution"]I was just wondering if I got the right solution. My answer is $ 29\equal{} r.$ [/hide]

If you extend NP to intersect the circle again at P', then we know NP'*PN=AN^2, so NP'=50, PP'=42 thus $ r \equal{} \sqrt {20^2 \plus{} 21^2} \equal{} 29$, so yes you were right.

Yes, I agree, although I didn't do it as neatly as serialk11r.

I bisected $ P$ at $ M$, so that triangles $ ANP$ and $ OMA$ are similar and then used $ \tan NPA \equal{} \frac{5}{2} \equal{} \tan OAM$ and $ AP\equal{} 4\sqrt{29}$, by Pythagoras, so $ r^{2} \equal{} 25\times 29 \plus{} 4\times 29$ and $ r\equal{}29$.

I did the cheap way  :P

You assume that AN is the y-axis (or at least sort of).
This means that the perpendicular PN will make a right angle, and the tangent location will make another right angle.
make a trapezoid, use pythagorean, and you're done

I made a few assumptions, but the problem seems to allow you to do so because it's pretty general.


vector
quadratics
algebra
quadratic formula

Show that the triangle with vertices $ a$, $ b$, $ c$ in the complex plane is equilateral iff
\[ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca
\][hide="one direction"]
If $ \triangle abc$ is equilateral, we have:
\[ (c \minus{} a)e^{\pi i/3} \equal{} b \minus{} a \\
(a \minus{} b)e^{\pi i/3} \equal{} c \minus{} b \\
(b \minus{} c)e^{\pi i/3} \equal{} a \minus{} c
\]
Squaring and adding together, we get:
$ \left((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2\right)\left(e^{2\pi i/3} \minus{} 1\right) \equal{} 0$
$ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 0$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca$
[/hide]

[hide="other direction"]  It is easy to see that the condition implies $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\equal{}0$.  Let $ x\equal{}a\minus{}b,y\equal{}b\minus{}c,z\equal{}c\minus{}a$.  Then $ x,y,z$ are the vectors that form the triangle with vertices $ a,b,c$ in the complex plane.  We have from the condition that $ x^2\plus{}y^2\plus{}z^2\equal{}0$.  We also have $ x\plus{}y\plus{}z\equal{}a\minus{}b\plus{}b\minus{}c\plus{}c\minus{}a\equal{}0$.  Thus, we find that $ xy\plus{}yz\plus{}xz\equal{}0$ from squaring $ x\plus{}y\plus{}z\equal{}0$ and subtracting $ x^2\plus{}y^2\plus{}z^2\equal{}0$.

Let $ xyz\equal{}c^3$.  Then the numbers $ x,y,z$ satisfy the equation $ r^3\minus{}c^3\equal{}0$.  Thus, $ x,y,z$ are some permutation of $ c,ce^{\frac{2i\pi}{3}},ce^{\frac{4i\pi}{3}}$.  That is, $ x,y,z$ are the vectors that form an equilateral triangle.[/hide]

Let $ \omega^2\plus{}\omega\plus{}1\equal{}0$

Then $ (a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}ca\minus{}ab)\equal{}(a\plus{}b\omega\plus{}c\omega^2)(a\plus{}b\omega^2\plus{}c\omega)\equal{}0$

How did I come up with this factorization?

[quote="Altheman"]Let $ \omega^2 \plus{} \omega \plus{} 1 \equal{} 0$

Then $ (a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab) \equal{} (a \plus{} b\omega \plus{} c\omega^2)(a \plus{} b\omega^2 \plus{} c\omega) \equal{} 0$

How did I come up with this factorization?[/quote]How can you make that work in the reverse direction, though? As far as I can tell knowing that $ a\plus{}b\omega\plus{}c\omega^2\equal{}0$ doesn't help.

[quote="Altheman"]$ (a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab) \equal{} (a \plus{} b\omega \plus{} c\omega^2)(a \plus{} b\omega^2 \plus{} c\omega) \equal{} 0$

How did I come up with this factorization?[/quote]
Hmm, you could just use the quadratic formula and solve for $ a$...

[quote="worthawholebean"]]How can you make that work in the reverse direction, though? As far as I can tell knowing that $ a \plus{} b\omega \plus{} c\omega^2 \equal{} 0$ doesn't help.[/quote]

Actually it does. For equilateral triangle a,b,c either $ a \plus{} b\omega \plus{} c\omega^2 \equal{} 0$ or $ a \plus{} c\omega \plus{} b\omega^2 \equal{} 0$ and it is iff

In this problem, we can WLOG $ c\equal{}0$ (do you see why?) Then the condition is $ a\equal{}\minus{}\omega b$ or $ a\equal{}\minus{}\omega^2b$

[quote="Altheman"]In this problem, we can WLOG $ c \equal{} 0$ (do you see why?) Then the condition is $ a \equal{} \minus{} \omega b$ or $ a \equal{} \minus{} \omega^2b$[/quote]Ah, got it. Thanks.


MATHCOUNTS AMC AIME FTW geometry calculus integration

This is...[size=150][b]AIME15's Tournament![/b][/size]

There will be $ 5$ rounds, in increasing difficulty.  In the earlier rounds, the difficulty will range from MathCounts School Round level to MOEMS/AMC 8 level.  Toward the later rounds, the difficulty will range from medium-hard AMC 8 level to MathCounts State/National level.  Each round will have $ 15$ problems, posted here in the forum.  A full solution will later be provided to each problem.  Calculators will be allowed, but please do not use any internet aids.  You cannot communicate with anyone else about this test.  You are on your honor.  You will be allotted $ 15$ minutes per round.  The problems in each round are in no particular order.  Scoring will be $ 4$ points for a correct answer and $ 1$ point for no answer.

There will be no limit in sign-ups, and you can sign up through PM or posting.   Reminder PM's will go out at the beginning of each round.  Signups will end on December 22, 2008.

Feel free to ask questions.  Please confirm in-thread.

[b][u]SIGNUPS[/b][/u]
1. 5849206328x
2. ernie
3. dragon96
4. xpmath
5. nikeballa96
6. isabella2296
7. Mewto55555
8. BOGTRO
9. Discrete_Math
10. high_flyer10
11. athunder
12. Bugi
13. 007math

Reminder PM's will go out at the start, but I'd like at least 10-12 people before starting.  Also, you can sign up in-thread now.

Only $ 2$ contestants will be eliminated in the $ 1$st round.  If there is a huge tie with everybody getting a perfect score, I will post a tiebreaker question.  This is plausible, because the questions in round $ 1$ are amazingly easy.

We have $ 10$ contestants now, so we'll start once signups are over.  The test, along with its solutions, are already made.

2 questions.

1) How many volunteers are there? (I'm one of them yay!!!)
2) What about incorrect answers?

2 volunteers.

Incorrect answer=0 points.

[quote="mightykim"]i'll try. but what happens if i don't do it..?[/quote]

If you don't do it you will be punished.  Punished severely.  By that, I mean
[hide]
you will make me very sad.
[/hide]

So you better do it :)

Well, you just get a $ 0$ for that round, but the first couple rounds are really easy, so there's no point not to do it.

[b][u]SIGNUPS[/b][/u]
1. 5849206328x
2. ernie
3. dragon96
4. xpmath
5. nikeballa96
6. isabella2296
7. Mewto55555
8. BOGTRO
9. Discrete_Math
10. high_flyer10
11. athunder
12. Bugi
13. 007math
14. mightykim
15. dafjbomb
16. andyliao
17. akhil0422

Eh, actually $ 3$ will be eliminated this round.

Good luck!  Round $ 1$ is now posted.  You have until December 26, 11:59 PM PST to send me your answers.

[quote="AIME15"]Yes.  Just answers.  Also, [b]send all of the answers at the same time, do not just send 1-8 one day, and 9-15 the next.[/b][/quote]

*cough* aime15 *cough*

This is because of the time limit.  You're allowed 15 minutes, so if you submit some today, and some tomorrow, you are clearly violating the rules.

I never said that either...

heh, time to find aime shortlist and longlist problems that pythag hasn't done yet...heh

[quote="Bugi"]Wait. Why does it matter who was that person?

And how are we going to do USAMO/IMO in 45 seconds? Even 42/42 at IMO can't do it.[/quote]

USAMO/IMO #1. Usually, IMO #1 has a < 1 minute solution....

Like, this year, there was a 5 sec solution to IMO #1 using complex numbers.

But we don't know complex numbers :wink:

um.. this is in the math counts forum, usamo and imo arent mathcounts lv, even if you can solve them in a second

This Master's Round will test your knowledge of MATHCOUNTS combinatorics.  It is worth $ 60$ points, though I will probably receive complaints "It's not that hard", etc.  It would have been harder if pythag011 had sent me the questions, but oh well.  If you can solve all of these with ease, then you should be able to solve almost any MATHCOUNTS counting or probability problem.  [b]Solutions are needed.[/b]

No deadline yet.  Good luck.  You have $ 1$ hour.

[hide="Masters Round"]
1. The following shows a $ 5 \times 4$ array of points, each of which is $ 1$ unit away from its nearest neighbors.  Determine the number of non-degenerate triangles (i.e., triangles with positive area) whose vertices are points in the given array.
[asy]size(75);
dot((0,0));
dot((0,1));
dot((0,2));
dot((0,3));
dot((0,4));
dot((1,0));
dot((1,1));
dot((1,2));
dot((1,3));
dot((1,4));
dot((2,0));
dot((2,1));
dot((2,2));
dot((2,3));
dot((2,4));
dot((3,0));
dot((3,1));
dot((3,2));
dot((3,3));
dot((3,4));[/asy]

2. Compute the number of distinct paths not passing through point $ (2,2,2)$ which travel from points $ (0,0,0)$ to point $ (4,4,4)$ in $ 12$ steps, changing a coordinate by $ 1$ at each step.

3. All the numbers on my gym lock are from $ 1$ to $ 40$.  However, I have forgotten my combination!  Two of the numbers of the three-term sequence are $ 17$ and $ 24$, but I have forgotten the third, and I don't know which is which.  Assuming it takes ten seconds per try, how long will it take me to test every possibility?

4. How many ways are there of arranging the six numbers on a die, assuming rotations are the same?

Note:

[asy]size(20);
dot((0,0));
dot((0,1));
dot((0,2));
dot((1,0));
dot((1,1));
dot((1,2));[/asy]

is considered the same as

[asy]size(20);
dot((0,0));
dot((1,0));
dot((2,0));
dot((0,1));
dot((1,1));
dot((2,1));[/asy]

for the purposes of this problem.

5. Claudia has cans of paint in eight different colors.  She wants to paint the four unit squares of a $ 2 \times 2$ board in such a way that neighboring unit squares are painted in different colors.  Determine the number of distinct coloring schemes Claudia can make.  Two coloring schemes are considered the same if one can be obtained from the other by rotation.

6. Given a sphere, a circle is called a great circle if it is the intersection of the sphere with a plane passing through its center.  Five distinct great circles dissect the sphere into $ n$ regions.  Let $ m$ and $ M$ be the minimum and maximum values of $ n$, respectively.  Find $ m$ and $ M$.

7. In how many ways can one use identical dominoes to tile a $ 2 \times 10$ board?

8. Two numbers, $ x$ and $ y$, are randomly chosen between $ 0$ and $ 1$.  What is the probability that $ x\plus{}y$ will be less than $ 1$?

9. Two school administrators are the only people who have dial-in access to the Fileserver, which can handle only one call at a time.  Each has a $ 15$-minute project to do, and hopes to use the fileserver between $ 4$ pm and $ 6$ pm today.  Neither will call after $ 5: 45$ pm, and neither will call more than once.  What is the probability that the BOTH complete their tasks?

10. Consider the square with vertices $ (0,0)$, $ (1,0)$, $ (0,1)$, and $ (1,1)$.  Choose a random point in this square and draw the line segment from it to $ (0,0)$.  Choose a second random point in this square and draw the line segment from it to $ (1,0)$.  What is the probability that the two line segments intersect?
[/hide]

Wait, this is utterly trivial.

Grr... you ruined my plan of including 5 IMO Shortlist problems...

I will post master questions HERE.

[hide]
1. We look at recursive sequences modulo some numbers. Let's look at the Fibonacci sequence mod ,2 , 3,5 and 7. Note that eventually, it just repeats the same thing over again, like 1,1,0,1,1,0.... However, it may be of the sort 5,3,3,3 where 5 is the "tail" (not part of the repeated sequence) and 3 is the body (the repeated part). For the primes 2, 3, 5, and 7, list the tail and the body of the Fibonacci sequence mod those primes. This problem will be graded by answer only.

2. Prove that the fibonacci sequence repeats mod any prime p.

3. Prove that, when the fibonacci sequence is taken mod p, the length of the body is $ \le \frac{p^2\plus{}p}{2}$

4. Let k be a fixed integer greater than 1. Let m = $ 4k^2\minus{}5.$ Show that there exist positive itnegers a and b such that the sequence defined by

\[ \begin{matrix*} x_0 \equal{} a,\quad x_1 \equal{} b,\quad x_{n \plus{} 2} \equal{} x_{n \plus{} 1} \plus{} x_n\quad\text{for}\quad n \equal{} 0,1,2,\dots, \end{matrix*}\]

has all its terms relatively prime to m.

5. Find all n such that $ n^2$ divides $ 2^n\plus{}1.$

6. Now, in case  you haven't noticed, since I don't like enumerative combo, I like nt better, so the problems are nt problems. Heres a nice one that I created: We have 3 positive itnegers a,b,m. We have $ a^m\minus{}1$ divides $ b^m\minus{}1.$ Prove that the largest odd divisor of a-1 divides b-1.

7. Also prove that a-1 divides $ b^2\minus{}1$ in problem 6.

8. Find all tuples of primes (p,q,r) such that p divides $ q^r\plus{}1,$ q divides $ r^p\plus{}1,$ and r divides $ p^q\plus{}1.$

[/hide]

Only 8 problems... 8 very hard problems.

Do the one that I posted.  The other ones are not mathcounts level :wink:

No, not yet.  There's still Pythag011's Extra Credit problems, which will be worth also 60 points.  Of course, the problems are pretty tough...

So I need 6 pythag's to be first...
Dafjbomb is the winner (I give up).

In third place...we have:
[hide="A very close third place..."]
:winner_third: xpmath!  Very nice job!
[/hide]
In second...
[hide="A not-so close second place...:P"]
:winner_second: Bugi!  Again, very nice!
[/hide]
And for first place...
[hide="And first place..."]
[hide="Drumroll, please..."]
[hide="I never got the drumroll..."]
:winner_first: dafjbomb! Congratulations!
[/hide]
[/hide]
[/hide]

Congratulations to everyone who participated; everyone did a pretty fine job.

[hide="Sign ups"][b][color=green]1. 5849206328x
2. ernie
3. dragon96
4. xpmath
5. nikeballa96
6. isabella2296
7. Mewto55555
8. BOGTRO
9. Discrete_Math
10. high_flyer10
11. athunder
12. Bugi
13. 007math
14. 5849206328x
15. andyliao
16. dafjbomb
17. ReseesPieces
18. abacadaea
19. mightykim[/b][/color][/hide]
[hide="Round 1"][color=green][b]1. nikeballa96: 60
1. 5849206328x: 60
1. andyliao: 60
1. dafjbomb: 60
1. Discrete_Math: 60
6. 007math: 56
6. xpmath: 56
6. isabella2296: 56
9. ReseesPieces: 53
10. high_flyer10: 52
10. akhil0422: 52
10. athunder: 52
10. dragon96: 52
14. Mewto55555: 48
14. ernie: 48
16. Bugi: 45[/b][/color]
[hide="Eliminated"][color=red][b]17. BOGTRO: 44
18. mightykim: 43
19. abacadaea: x[/b][/color] (requested to have score removed)
[/hide]
[/hide]
[hide="Round 2"][b][color=green]1. dafjbomb: 60+30+12=102
2. 5849206328x: 60+30+7=97
3. Discrete_Math: 56+30+7=93
4. xpmath: 52+28+8=88
5. andyliao: 52+30+5=87
6. Bugi: 60+22.5+3=85.5
7. Mewto55555: 60+24+0=84
8. high_flyer10: 56+26+0=82
9. ReseesPieces: 52+26.5+2=80.5
10. 007math: 52+28+0=80
11. isabella2296: 48+28+0=76
11. ernie: 46+24+6=76[/color][/b]
[hide="Eliminated"][color=red][b]13. nikeballa96: 39+30+0=69
14. akhil0422: 38+26+0=64
15. FantasyLover: 56+0+2=58
16. BOGTRO: 52+0+1=53
17. athunder: 0+26+0=26
17. dragon96: 0+26+0=26[/hide][/color][/b][/hide]
[hide="Round 3"][b][color=green]1. dafjbomb: 51+60=111
2. 5849206328x: 48.5+60=108.5
3. Discrete_Math: 46.5+57=103.5
4. Bugi: 42.75+60=102.75
5. Mewto55555: 42+60=102
6. xpmath: 44+52=96
7. andyliao: 43.5+52=95.5
8. ernie: 38+56=94[/color][/b]
[hide="Eliminated"][b][color=red]9. ReseesPieces: 40.5+49=89.5
10. isabella2296: 38+49=87
11. 007math: 40+45=85
12. high_flyer10: 41+0=41[/b][/hide]
[/color]
[/hide]
[hide="Round 4"][b][color=green]1. dafjbomb: 55.5+60+34=149.5
2. Bugi: 51.375+52+17=120.375
3. xpmath: 48+50+22=120
4. Discrete_Math: 51.75+42+x=93.75 [/color][/b]
[hide="Eliminated"][b][color=red]5. andyliao: 47.75+50+0=97.75
6. Mewto55555: 51+42+0=93
7. 5849206328x: 54.25+0+0=54.25
8. ernie: 47+0+0=47 [/b][/color]
[/hide][/hide]
[hide="Round 5"][b][color=green]1. dafjbomb: 74.75+56=130.75
2. Bugi: 60.1875+56=116.1875[/color]
[/b][hide="Eliminated"][b][color=red]3. xpmath: 60+56=116
4. Discrete_Math: 46.875+x+24=70.875+x[/color][/b]
[/hide]
[/hide]
[hide="Round 6"][b][color=green]1. dafjbomb: 65.375+45+0=110.375[/color]
[/b][hide="Eliminated"][b][color=red]2. Bugi: 58.09375+16+0=74.09375[/b][/color][/hide][/hide]


Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w \equal{} C(O,R)$ with the diameter $ [AD]$ ,

i.e. $ O\in AD$ . Prove that exists the relation $ R\cdot\left(AB^2 \plus{} BC^2 \plus{} CD^2\right) \plus{} AB\cdot BC\cdot CD \equal{} 4R^3$ .

[u][b]Remark.[/b][/u] If $ B\equiv C$ , i.e. $ BC \equal{} 0$ , then obtain the [b]Pythagoras' theorem[/b] $ AB^2 \plus{} BD^2 \equal{} AD^2$ .

[hide]Since this is a classic problem, there must be a nicer approach than this but....

Let m<BOA=2a, m<BOC=2b, m<COD=2c. This means a+b+c=90
Then AB=2Rsina, BC=2Rsinb, CD=2Rsinc
R(AB^2+BC^2+CD^2)+(AB)(BC)(CD)=
4R^3((sina)^2+(sinb)^2+(sinc)^2)+8R^3(sina)(sinb)(sinc)=
4R^3((sina)^2+(sinb)^2+(sinc)^2+2(sina)(sinb)(sinc))=
4R^3((sina)^2+(sinb)^2+sinc(sinc+2(sinb)(sinc)))=
4R^3((sina)^2+(sinb)^2+cos(a+b)(cosacosb-sinasinb+2sinbsinc))=
4R^3((sina)^2+(sinb)^2+(cosacosb)^2-(sinasinb)^2)=
4R^3((sina)^2+(sinb)^2+1-(sina)^2-(sinb)^2-(sinc)^2+(sinasinb)^2-(sinasinb)^2... (since (cosa)^2+(sina)^2=1)
so we get that R(AB^2+BC^2+CD^2)+(AB)(BC)(CD)=4R^3
[/hide]


calculus integration function trigonometry

Does anyone know of any integrals other than integral(e^(x^2)) that must be evaluated numerically?

Thanks.

There are lots of them -- basically anything that puts together trancendental functions with anything else.  $\sin(\cos(x)), e^{\sin(x)}, \sin(e^{\cos(\log(x))})$ for example.


1. Find the exact value of $\sum_{n=1}^{\infty}\frac1{n(n+1)}$

(OK, #1 is quite famous and quite well known.)

2. Find the exact value of $\sum_{n=1}^{\infty}\frac1{n(n+1)(n+2)}$

3. Find a general formula, valid for all $k\ge 1$, for the exact value of $\sum_{n=1}^{\infty}\frac1{n(n+1)(n+2)\cdots(n+k)}$

and prove that your answer is correct.

(The real challenge is #3. There are proofs with varying levels of length and complexity, including one that is quite short and elegant.)

Nice result :
According to my computations the result for #3 is $\frac 1 {k \cdot k!}$, which is indeed the result for the two previous sums.
The following is not supposed to be elegant but it works (I hope) :
Let $\dsp S_k = \sum_{n \geq 1} \frac 1 {n(n+1) \cdots (n+k)}$

Now, for each $n$, we have by classical decomposition into simple fractions:
$\dsp \frac 1 {n(n+1) \cdots (n+k)} = \frac 1 {k!} \sum_{i=1}^k \frac {(-1)^{i-1} C_k^i} {n+i}$ (1)

Therefore, using kind of dominos, each term of the form $\frac 1 p$ appear exactly $k$ times in the sum with respective total coefficient $\dsp \frac 1 {k!} \sum_{i=1}^p (-1)^{i-1}C_{k-1}^{i-1}$, with the usual convention $C_n^k = 0$ for $k>n$.

Using the well-known result $\dsp \sum_{i=0}^k (-1)^iC_{k-1}^i = 0$, it follows that considering partial sums, each of the terms of the form $\frac 1 p $ simplifies and disappears when $p\geq k+1$, except, since it is a partial sum, for a finite and bounded number (more or less $\frac {k(k+1)} 2$) of them. Thus, taking the limit all these remaining terms tend to $0$, and then :
$\dsp S_k =\frac 1 {k!} \sum_{p=1}^k  \frac 1 p (\sum_{i=1}^p (-1)^{i-1}C_{k-1}^{i-1})$.

Now, an easy induction leads to another known result that is : $\sum_{i=1}^p (-1)^{i-1}C_{k-1}^{i-1} = (-1)^{p-1}C_{k-1}^{p-1}$.
Then :
$\dsp S_k = \frac 1 {k!}  \sum_{p=1}^k  \frac 1 p (-1)^{p-1}C_{k-1}^{p-1} = \frac 1 {k!} \frac 1 k \sum_{p=1}^k (-1)^{p-1}C_k^p = \frac 1 {k!} \frac 1 k(-(-1)^{-1}C_k^0 )= \frac 1 {k \cdot k!}$, as claimed.

Pierre.

You are right, since $\frac 1 {k!}$ is a constant, I've omited it in my computation since it appeared in (1).
I've edited it, so the proof may be ok now.

Pierre.

Yes, the answer is $\frac1{k\cdot k!}.$

As for the quickest and most elegant path - I thought I saw kunny post the key idea, but it has now disappeared. Would you please re-post that method, kunny?

Pierre must be kidding  ;)
My solution:
\[S_k=\frac{1}{k}\sum_{n=1}^\infty\frac{(n+k)-n}{n(n+1)\dots(n+k)}=\frac{1}{k}\left(S_{k-1}-(S_{k-1}-\frac{1}{k!})\right)=\frac{1}{k\cdot k!}.\]
 :D  :D  :D

Thank you ,Kent Merryfield.

My idea is exactly same as Myth's idea.

kunny

But Myth, with that kind of solution, I would not have been able to write two pages... :D
Just remember that my goal is to write the most complicated and long solution to the most easy problem. Thus, I'm working on it. :D  :D

Pierre.

[quote="pbornsztein"]But Myth, with that kind of solution, I would not have been able to write two pages... :D
Just remember that my goal is to write the most complicated and long solution to the most easy problem. Thus, I'm working on it. :D  :D [/quote]
Accepted
 :D  :D  :D

This post is overkill - the same idea, expressed in slightly different notation - but it may help others reading this thread understand what is going on. The first step is the partial fractions identity $\frac1{n(n+k)}=\frac1k\left(\frac1n-\frac1{n+k}\right)$.

So, $\sum_{n=1}^{\infty}\frac1{n(n+1)\cdots(n+k-1)(n+k)}=$

$\frac1k\sum_{n=1}^{\infty}\frac1{n(n+1)\cdots(n+k-1)} -\frac1{(n+1)\cdots(n+k)}=$

$\frac1k\left(\frac1{1\cdot2\cdots k}-\frac1{2\cdot3\cdots(k+1)}+\frac1{2\cdot3\cdots(k+1)}- \frac1{3\cdot4\cdots(k+2)}+\cdots\right)$

The sum "telescopes", with every term cancelling out except the first $\frac1{k!}.$


geometry
trigonometry
circumcircle
trig identities
Law of Cosines

Given triangle ABC with BC=a,CA=b,AB=c .
1)Suppose that three length side of triangle ABC satisfy :
$ \left\{\begin{array}{l}a^2 (b \plus{} c \minus{} a) \equal{} b^3 \plus{} c^3 \minus{} a^3 \\
\frac {{1 \plus{} \cos C}}{{\sin C}} \equal{} \frac {{2a \plus{} b}}{{\sqrt {4a^2 \minus{} b^2 } }} \\
\end{array} \right.$
Prove that a=b=c (or triangle ABC be equilateral triangle ).

2)Compute $ P_{G/(ABC)}$ with G be centroid of triangle ABC and (ABC) be circumcenter of triangle ABC.

[hide="1"]We add $ a^3$ to the first equation to get
$ a^2(b \plus{} c) \equal{} b^3 \plus{} c^3 \equal{} (b \plus{} c)(b^2 \minus{} bc \plus{} c^2)$,
and dividing by $ (b \plus{} c)$ we get
$ a^2 \equal{} b^2 \minus{} bc \plus{} c^2$.
By the Law of Cosines,
$ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc \cos A$, and equating the two expressions for $ a^2$ we get
$ b^2 \minus{} bc \plus{} c^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc \cos A$.
Solving for $ \cos A$ we find that $ \cos A \equal{} \dfrac{1}{2}$, so $ A \equal{} 60^\circ$.
Looking at the second equation we notice that
$ \dfrac{1 \plus{} \cos C}{\sin C} \equal{} \dfrac{(\sqrt {1 \plus{} \cos C})^2}{\sqrt {1 \minus{} \cos^2 C}} \equal{} \dfrac{(\sqrt {1 \plus{} \cos C})^2}{\sqrt {(1 \minus{} \cos C)(1 \plus{} \cos C)}} \equal{} \dfrac{\sqrt {1 \plus{} \cos C}}{\sqrt {1 \minus{} \cos C}}$
and
$ \dfrac{2a \plus{} b}{\sqrt {4a^2 \minus{} b^2}} \equal{} \dfrac{2a \plus{} b}{\sqrt {(2a \plus{} b)(2a \minus{} b)}} \equal{} \dfrac{\sqrt {2a \plus{} b}}{\sqrt {2a \minus{} b}}$
Equating the two expressions we have
$ \dfrac{\sqrt {1 \plus{} \cos C}}{\sqrt {1 \minus{} \cos C}} \equal{} \dfrac{\sqrt {2a \plus{} b}}{\sqrt {2a \minus{} b}}$

$ \dfrac{1 \plus{} \cos C}{1 \minus{} \cos C} \equal{} \dfrac{2a \plus{} b}{2a \minus{} b}$
Subtracting $ 1$ from both sides we get
$ \dfrac{2 \cos C}{1 \minus{} \cos C} \equal{} \dfrac{2b}{2a \minus{} b}$

$ \dfrac{\cos C}{1 \minus{} \cos C} \equal{} \dfrac{b}{2a \minus{} b}$

$ 2a \cos C \minus{} b\cos C \equal{} b \minus{} b\cos C$

$ 0 \equal{} b \minus{} 2a \cos C$

$ 0 \equal{} b^2 \minus{} 2ab \cos C$.
By Law of Cosines $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos C$. Since $ 0 \equal{} b^2 \minus{} 2ab \cos C$,
$ c^2 \equal{} a^2$, so $ c \equal{} a$, which implies $ C \equal{} A \equal{} 60^\circ$.
$ B \equal{} 180^\circ \minus{} A \minus{} C \equal{} 180^\circ \minus{} 60^\circ \minus{} 60^\circ \equal{} 60^\circ$, so since $ A \equal{} B \equal{} C \equal{} 60^\circ$, $ a \equal{} b \equal{} c$.[/hide]


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$, then $ \sum(\sqrt{a}\plus{}\sqrt[3]{a}\plus{}\sqrt[6]{a})\leq9$

$ \sqrt {a} \le \frac {1 \plus{} a}{2}.$

$ \sqrt [3]{a} \le \frac {1 \plus{} 1 \plus{} a}{3}.$

$ \sqrt[6]{a} \le \frac {1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} a}{6}.$

So $ \sum \sqrt {a} \plus{} \sqrt [3]{a} \plus{} \sqrt [6]{a} \le \frac {a \plus{} b \plus{} c}{3} \plus{} \frac {a \plus{} b \plus{} c}{2} \plus{} \frac {a \plus{} b \plus{} c}{6} \plus{} \frac {3}{2} \plus{} \frac {6}{3} \plus{} \frac {15}{6} \equal{} 9$


linear algebra
matrix
linear algebra unsolved

Let $ \textbf{X}$ is $ m\times n$ matrix, $ \textbf{u} \in \mathbb{R}^{m}, \textbf{v} \in \mathbb{R}^{n}$. Find upper bound for $ min_{\textbf{u},\textbf{v}} \parallel{} \textbf{X} \minus{} \textbf{u}\textbf{v}^{T}\parallel{}^{2}$ where $ \parallel{}.\parallel{}$ is Euclidean norm.

[quote="alekk"]this is exactly a least square linear regression. You should be a able to find the exact value of this infimum.[/quote]

not quite exactly, because there are two parameters. i found solution by computing SVD of X, minimal value is $ \sum_{i>1}\sigma_{i}^{2}$, where $ \sigma_{i}$ is i-th singular value of X.

sorry, I have read your post too fast  :blush:
This problem  is well known, and the approach with the singular value decomposition is the right one (eg: see the article in Wikipedia about the SVD) You are basically trying to compute the distance between the matrix $ X$ and $ \mathcal S \equal{} \{M : rank(M) \leq 1 \}$. One could think of other sets $ \mathcal S$ to continue the problem  :)


[quote="Yimin Ge"]Owing to the cruel circumstances which forced the Austrian team to leave in the early morning of tuesday, I did not have the chance of saying Goodbye to everyone, so belated: Goodbye to everyone, this was a wonderful IMO.

Nevertheless, there are some things that annoy me. We did not get an IMO-report, last year, we were given an about 150 page long booklet, containing several reports, statistics and other facts on the IMO. I would have loved to get one of this years IMO but unfortunately I did not.

At the banquet, we were told that maybe the reports were still in print. Since we did not get them, I want to ask: did anyone get a report of this years IMO - and if yes, is there any chance to get one? The report of last year was together with the IMO-Poster (of which I know that there does not exist one of this year) my most important souvenir of last years IMO (also because I did neither get an honourable mention nor a medal last year).[/quote]

Yimin, you are pretty unfair to us! Requesting the current report but not offering the IMO 2005 report to us, e.g. by scanning it.

Hi orl,
apologies, but I don't really understand what you mean. I was referring to this report (see the attachment).
As I said, it has about 150 pages and scanning this many pages is a work I have no time to do, at least not at the moment (but we have holidays now so maybe soon).
I did not want to be impertinent or importunate, I just wanted to clarify if I missed something (in my opinion not quite insignificant) due to our early departure, the request of getting it was not meant that seriously after all (I think I really should use more smilies :D ).
Yimin


trigonometry
AMC
AIME
USA(J)MO
USAMO
AMC 10
number theory

Hi everybody,

I just finished the AoPS V1 and am about to start on the Intro to Counting and Intro to Number Theory books. I was also considering taking an AMC problem series class but am not sure whether to take the 10 or 12. I'm a 9th grader in Trig/Analytic Geo (basically Algebra 2 with Trig and Analytic Geo) and am fairly confident I can score 100 on the AMC 10 and 80 on the AMC 12. My goal is to qualify for AIME. So:
1) Which course should I take?
2) Which test should I take?

Try them both. If you need a site with some AMC tests, I can PM it to you.

Both the AMC 10 and AMC 12 have two different test dates, so you could take the AMC 10A and the AMC 12B (A is the first date and B is the second).

If you finished AoPS Vol. 1 and understood it all, qualifying for AIME should be a piece of cake.  Try some practice tests: if you want, PM me and I can send you some.

So what is everyone taking.. Im midway through volume2, want to make USAMO. Don't know what to take.

If you wanna make USAMO, you wanna do well on AIME and AMC.  So you should take those classes :P

I think ill take the aime class, no the amc though since AMC 10 is so easy.

Not sure if u know this, but u need 120 for the amc10 and 100 for the amc 12, not 100 and 80. If you still can't go above 120 for amc 10, u might wanna do a few extra problem sets to supplement ur course.

General opinion of AoPS users:  If you want to make AIME, take AMC 12.  If you wanna make USAMO, take AMC 10.  (doesnt apply to 11th and up grade)

Thanks guys,

I know that you need 120 for AMC 10 and 100 for the 12 which is why I think I need a class. Also, if someone knew a site with several AMC 10's and 12's that would be helpful. Do any of you have any suggestions on how to prepare for the AIME without a problem series class?

Thanks!

http://www.math.ksu.edu/main/handbook/ProblemSets

I found this in another thread titled "AMC 8" made by isabella2296.


modular arithmetic
number theory
prime numbers
Congruences

Show that \[(-1)^{\frac{p-1}{2}}{p-1 \choose{\frac{p-1}{2}}}\equiv 4^{p-1}\pmod{p^{3}}\] for all prime numbers $p$ with $p \ge 5$.

Let $ p \equal{} 2n \plus{} 1, X \equal{} \prod_{k \equal{} 1}^n k, Y \equal{} \prod_{k \equal{} n \plus{} 1}^{2n}k, (p \minus{} 1)! \equal{} XY$. Then $ 2^{2n}(p \minus{} 1)! \equal{} \prod_{x \minus{} even}x\prod_{x \minus{} odd}(p \plus{} x).$
Therefore $ 4^n \equal{} 1 \plus{} p\sum_{x \minus{} odd}\frac 1x \plus{} p^2\sum_{i < j \minus{} odd}\frac {1}{ij} \plus{} O(p^3).$
$ 2\sum_{i < j \minus{} odd}\frac {1}{ij} \equal{} Z^2 \minus{} T,T \equal{} \sum_{k \minus{} odd}\frac {1}{k^2}, Z \equal{} \sum_{k \minus{} odd}\frac 1k.$
$ 2T \equal{} \sum_{k \minus{} odd}\frac {1}{k^2} \plus{} \sum_{k \minus{} even}\frac {1}{(p \minus{} k)^2} \equal{} O(p)$, when $ p > 3$.
Therefore for p>3 $ 4^n \equal{} 1 \plus{} pZ \plus{} \frac {p^2}{2}Z^2 \plus{} O(p^3).$
$ 2^nY \equal{} \prod_{x \minus{} odd}(p \plus{} x),2^nX \equal{} \prod_{x \minus{} odd}(p \minus{} x)$, therefore $ C_{2n}^n \equal{} \prod_{x \minus{} odd}\frac {p \plus{} x}{p \minus{} x}.$
It gives $ {( \minus{} 1)^nC_{2n}^n \equal{} \prod_{x \minus{} odd}(1 \plus{} \frac px )(1 \minus{} \frac px )^{ \minus{} 1} \equal{} \prod_{x \minus{} odd}(1 \plus{} 2p\frac 1x \plus{} 2p^2\frac {1}{x^2}) \plus{} O(p^3).}$
Therefore $ {( \minus{} 1)^nC_{2n}^n \equal{} 1 \plus{} 2pZ \plus{} 2p^2T \plus{} 4p^2\sum_{i < j \minus{} odd}\frac {1}{ij} \plus{} O(p^3) \equal{} 1 \plus{} 2pZ \plus{} 2p^2Z^2 \plus{} O(p^3).}$
and $ {4^{2n} \equal{} (1 \plus{} pZ \plus{} \frac {p^2}{2}Z^2)^2 \plus{} O(p^3) \equal{} 1 \plus{} 2pZ \plus{} 2p^2Z^2 \plus{} O(p^3) \equal{} ( \minus{} 1)^nC_{2n}^n \plus{} O(p^3).}$


number theory
greatest common divisor
least common multiple

I thought some time about this problem but besides some trivial observations I have no idea how to solve it. Here it is:

Find all positive integers a_1,a_2,...,a_n with the property that their least common multiple, their greatest common divisor and their product are sides of a triangle.

I posted the very easy case of n=2 into the intermediate forum but I can't see how the solution idea can be applied for n>2.

I hope you are better at number theory than me.

Misha

Here's a hint: you need to have the gcd plus the lcm > product. Now compare various terms to $\sqrt{a_1a_2\cdots a_n}$.

Easy!!! :D It's really easy. I have a solution now but it doesn't use the geometric mean. Can you post your whole solution? Then I'll post mine.

Misha

It clearly works if all the $a_i$s are pairwise relatively prime. So suppose that they aren't. Let's write $g$ for the gcd, $l$ for the lcm, and $p$ for the product. Then $l\le\frac{p}{2}$ and $g\le\sqrt{p}$. So we have $l+g\le\frac{p}{2}+\sqrt{p}\le p$ for $p\ge 4$. Since $p\ge 4$ if two of them share common factors, this takes care of all cases.

I like your solution and here is mine:
g:=gcd
a_i=b_ig for all i
Then lcm<=gb_1b_2...b_n so g+gb_1b_2...b_n=>g^nb_1b_2...b_n is a necessary condition. You can rewrite it to: 1+1/(b_1b_2...b_n)=>g^(n-1). Because of n=>2 we get 1+1/(b_1b_2...b_n)=>g. We have to consider two cases now: g=1 and g=2,b_i=1 for all i. We get the solutions (2|2) and n coprime numbers.

Misha

I didn't think about this very hard, but what if we change the question somewhat (maybe reducing it again to two numbers first?):
Given a pair of positive integers a, b, when is it true that the set S={x s.t. x|a and x|b} union {ab, lcm(a, b)} has the property that the sum of all but one of its members is greater than that member?
In other words, what if we include all the other common divisors of a and b, not just their gcd?


Let  $ ABC$ be an [b]acute-angled[/b] triangle . Prove that $ \sum_{cyc}\frac{\sqrt{b^2\plus{}c^2\minus{}a^2}}{s\minus{}a}\le 6$ , where $ 2s\equal{}a\plus{}b\plus{}c$ .

[b]Proposition.[/b] [color=#800000]An arbitrary non-obtuse $\triangle ABC$ has sides $a,b,c$ opposite angles $A,B,C$ respectively, and semiperimeter $s$. Then\[\frac{\sqrt{bc\cdot\cos A}}{s-a}+\frac{\sqrt{ca\cdot\cos B}}{s-b}+\frac{\sqrt{ab\cdot\cos C}}{s-c}\le 3\sqrt{2}\].[/color]
[b]Proof.[/b] [color=#0000BF]$\sum{\frac{\sqrt{bc\cdot\cos A}}{s-a}}\le 3\sqrt{2}\iff \frac{\sqrt{2bc\cdot\cos A}}{s-a}\le 6\iff \frac{\sqrt{b^2+c^2-a^2}}{s-a}\le 6$, since the Cosine Law states that $\cos A= \frac{b^2+c^2-a^2}{2bc}$, along with the cyclic versions.

By Ravi Substitution, let $ \begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}\iff\begin{cases}2x=a-b+c\\2y=a+b-c\\2z=-a+b+c\end{cases} $ , where $x,y,z\in\mathbb{R}^{+}$.

By CS, $\sum{\frac{\sqrt{b^2+c^2-a^2}}{s-a}}=\sum{\frac{\sqrt{(y+z)^2+(z+x)^2-(x+y)^2}}{z}}$

$=\sum{\sqrt{\frac{2\cdot(z^2+yz+zx-xy)}{z^2}}}\le \sqrt{3\cdot \sum{\frac{2\cdot(z^2+yz+zx-xy)}{z^2}}}$

$= \sqrt{6\cdot\left(3+ \sum{\frac{yz+zx-xy}{z^2}}\right)}$

It suffices to show that $\sqrt{6\cdot\left(3+ \sum{\frac{yz+zx-xy}{z^2}}\right)}\le 6\iff \sum{\frac{yz+zx-xy}{z^2}}\le 3$

$\iff xyz\cdot\sum{xy(x+y)}\le \sum{x^3y^3}+3x^2y^2z^2$,

otherwise written as $pq^2+p^2q+qr^2+q^2r+rp^2+r^2p\le p^3+q^3+r^3+3pqr$, where $\begin{cases}p=xy\\q=yz\\ r=zx\end{cases}$

$\iff p(p-q)(p-r)+q(q-r)(q-p)+r(r-p)(r-q)\ge 0$, which is Schur's Inequality of degree 1. $\square$[/color]

[b]Note.[/b] [color=#008000]The reason the non-obtuse condition is necessary is that otherwise one of $A,B,C$ would be obtuse, resulting in one of $\cos A,\cos B, \cos C$ being negative. But the LS takes the root of all three, which would be impossible if one of the angles were obtuse.[/color]


AMC
USA(J)MO
USAMO
AIME
AMC 12
AoPS Books

I just bought the AoPS books this summer, and I want to qualify the USAMO next year. But I find the practice problems in V2 is much harder than the instruction examples... So after how long time working(can't figure it out)can I see the solution?

And how much time after you working on a Volumn 2 practice problem to look the solution?

well if you're really lost, reread the chapter carefully to make sure you didnt miss anything.

if you have all the resources, but just cant apply them, i would try my best (that can be hours on a single problem) and if i still cant get it, i would look at the solution and try to recreate the solution myself.

Since your goal is training, I wouldn't spend more than an hour on any one problem.  Then, read the solution, while thinking about how you might have come up with it yourself once you've seen it.  Then, mark the problem, and come back and try to solve it on your own in 2-4 weeks.

Don't feel bad if you can't do all the problems the first time around - that's the whole point of AoPS books.  If you could do the problems all the first time around in any book, chances are the book is too easy for you.

i believe this is good advice..im in sort of the similar boat with AoPS vol 1. there's not too many problems that i can not solve (maybe 1-3 per chapter) but sometimes it does get frustrating. after iv put at least 15 minutes of thought into the problem, made sure im not missing anything, i read the solution then come back after a few hours and try to solve the same problem. With each problem, i try to absorb new methods of problem solving.

time to train some more! :bruce:  :weightlift:

Yeah I was in the same situation.  Now I'm at a stage where AMC 12 problems #23-25 and AIME problems #12-15 are the glass ceiling.  But I have a feeling that by the end of this summer break, I'll be able to solve the harder AMC12 and AIME problems greater than half of the time...


analytic geometry
geometry
vector
calculus
derivative
function
graphing lines

My economics class taught the Lagrange Multipler Method for optimising utility subject to a budget constraint.
The cases usually involve two commodities. I'm confused about the method.

1. Wouldn't $ \lambda$ be a constant the value of which we do not know at first, instead of a variable?

2. Why would the value of $ \lambda$ matter? Can't we just set it to just any non-zero number, thereby eliminating one variable?

3. Are we transforming the question from one originally solvable by 3-dimensional coordinate geometry to one that requires 4-D instead?

4. I simply don't know why the method works. :help: Could you help me understand it more intuitively/ graphically? I googled it but most of the websites use vectors, which I understand quite poorly.  :wallbash_red:

Thanks!

The [url=http://en.wikipedia.org/wiki/Lagrange_multipliers]wikipedia[/url] page is quite good.  You're going to have to know how to do partial derivatives, but other than that, it doesn't require anything advanced.

"If  (x,y)  is a maximum for the original constrained problem, then there exists a λ such that  (x,y,λ)  is a stationary point for the Lagrange function (stationary points are those points where the partial derivatives of Λ are zero). "

I don't understand this  :wacko:.  :?:

Here's how I explain it.

Let's start with a "free" max-min problem: we're trying to maximize or minimize $ f(x,y)$ with no constraint. What we do is find the partial derivatives $ \frac{\partial f}{\partial x}$ and $ \frac{\partial f}{\partial y}$ and find where they are both zero at the same time. Or, let's state this in vector language: we find the [b]gradient[/b] $ \nabla f$ and find the points where $ \nabla f\equal{}0.$ That's a "critical point" or "stationary point." Why do we do that? If the gradient is nonzero at some point, then $ f$ has a [b]directional derivative[/b] in some direction that is nonzero - which means that if we travel in the direction, there's a nonzero slope, and we can't possibly have a max or min.

So a critical point (stationary point) is a point at which all possible directional derivatives are zero. That doesn't mean it necessarily is a max or a min, but it allows that as a possibility.

Now we turn to the [b]constrained[/b] problem: maximize (or minimize) $ f(x,y)$ subject to the constraint $ g(x,y)\equal{}c.$ We're confining ourselves to a [b]level set[/b] of the function $ g(x,y).$ (In two variables, that level set is a curve; in three variables it would be a surface, and so on.) We want to do that same thing we did before, which is the find points where all directional derivatives are zero, only now the issue is finding where all [b]allowable[/b] directional derivatives are zero. What are the allowed directions? The allowed directions are those that are tangent to the level set $ g(x,y)\equal{}0.$ That level set is always perpendicular to the gradient $ \nabla g.$

But then the directional derivative of $ f$ in the direction $ \vec{u}$ is given by $ \nabla f\cdot\vec{u}.$ When is this directional derivative zero? When $ \vec{u}$ is perpendicular to $ \nabla f.$ Remember that the allowed directions are perpendicular to $ \nabla g.$ So what are we looking for? We're looking for a condition such that everything perpendicular to $ \nabla g$ is also perpendicular to $ \nabla f.$ Geometrically, that would mean that $ \nabla g$ and $ \nabla f$ point in the same direction, and the way to say that is to say that one is a scalar multiple of the other. That is, there exists $ \lambda$ such that $ \nabla f\equal{}\lambda \nabla g.$ Write out the components of that at there are your Lagrange multiplier equations.

:what?: I'm confused  :(.

If we are looking for points where the contour lines of a function and a particular level set of another function (the constraint) touch tangentially, why aren't we using simple slope as in 2D, instead of partial derivatives?

Am I supposed to visualise the problem in 2 dimensional or 3 dimensional space :?:  :o

If I use Figure 1 in the wiki site,
http://en.wikipedia.org/wiki/Lagrange_multipliers#Very_simple_example
as an example. Is it correct to say our objective is to walk along $ g(x,y)\equal{}c$, and find the point where $ f(x,y)$ does not change in value?

Thanks again  :P


Or, you must pick either $ 2$ from the top and $ 1$ from the bottom or $ 1$ from the top and $ 2$ from the bottom, so $ 2 \times 3 \times 3\equal{}\boxed{18}$.

[quote=Blue_Whale]mathemagician1729, why minus 2?[/quote]
$\binom{6}{3}$ is the number of ways to choose three of the points from the grid, and you need three points to make a triangle. However, the three points along the top of the grid are collinear, as are the three along the bottom, and three collinear points can't make a triangle, so we subtract those $2$ choices of three points out, to get mathemagician1729's answer of $\binom{6}{3}-2=18$.

LOL just troll and argue degenerate triangles. :P (I know this isn't possible but oh well)

[hide=Answer][quote=GameBot]How many distinct triangles can be drawn using three of the dots below as vertices?  (Congruent triangles and different positions are considered distinct.)
[asy]dot((0,0));
dot((1,0));
dot((2,0));
dot((0,1));
dot((1,1));
dot((2,1));[/asy][/quote]

For the first one there are $3$ ways for the first point of the triangle. Then $2$ ways. Lastly for the last point  we have 3 ways.
$3 \cdot 2 \cdot 3= \boxed{18}$ [/hide]
[b][color=#f00]EDIT: fixed latex[/color][/b]


logarithms
limit
algebra unsolved
algebra

Prove that the followingequation:
$ A.a^x\plus{}B.b^x\equal{}A\plus{}B$(where $ a>1;0<b<1;A;B\in R$) has at most two solutions.

First, rewrite:

$ f(x) = A(a^x - 1) + B(b^x - 1) = 0$

Now, you can see that as $ x \rightarrow \infty$, the LHS goes to $ sgn(A)$ and when $ x \rightarrow - \infty$ LHS goes to $ sgn(B)$. So, we have four basic examples:

(1)
$ sgn(A) = sgn(B)$
$ A \neq 0$
$ B \neq 0$

$ f'(x) = a^x A \ln(a) + b^x B \ln(b) = 0$

$ \left(\frac {a}{b}\right)^x = - \frac {B \ln(b)}{A \ln(a)}$

$ x\ln\left(\frac {a}{b}\right) = \ln( - \frac {B \ln(b)}{A \ln(a)})$

$ x = x_{extrema} = \frac {\ln( - \frac {B \ln(b)}{A \ln(a)})}{\ln\left(\frac {a}{b}\right)}$

This is real since $ sgn(A) = sgn(B)$, $ a > 0$ and $ 0 < b < 1$. Also, it is the only solution to $ f'(x) = 0$ so this means that since $ f(x)$ is continuous, there are exactly $ 2$ solutions to $ f(x) = 0$.

(2)
$ sgn(A) = - sgn(B)$
$ A \neq 0$
$ B \neq 0$

As in (1), we get the same value for $ x_{extrema}$. Since the condition that $ sgn(A) = sgn(B)$ was one of several essential ones for $ x_{extrema} \in \mathbb{R}$, here it gives $ x_{extrema} \in \mathbb{C}$, so $ f(x)$ has no extremes and therefore $ f(x) = 0$ has only one solution.

(3)
$ A = 0$
$ B \neq 0$

Now, $ f(x) = B(b^x - 1)$. Now, $ \lim_{x \rightarrow \infty}f(x) = - sgn(B)B$ and $ \lim_{x \rightarrow - \infty}f(x) = sgn(B)$. Also:

$ f'(x) = b^x B \ln(b) = 0$

Since $ f(x)$ has no extremes, and as we can see above it approaches values of different signs as $ x \rightarrow \pm \infty$, we can say that there is exactly one solution to $ f(x) = 0$.

(4)
$ A \neq 0$
$ B = 0$

Same as in (3) but reversed results in the manner of $ x$ approaching $ \pm \infty$.

That's it!!

Sure, I should say that if $ A = B = 0$ there is an infinitive number of solutions to $ f(x) = 0$, but I don't think this is the case that's worth studying.

Cheers!

Please lock this topic. It's running problem in MYM.
 :wink:

What is $ sgn(A)$? I don't know.Is $ sgn(A)\equal{}sign(A)$


Find all ordered pairs of positive integers such that $ 7(x\plus{}y)\equal{}3(x^2\minus{}xy\plus{}y^2)$.


What is the nature of the point $ x \equal{} 0$ with respect to the differential equation
\[ x^2y'' \minus{} xy' \plus{} (1 \minus{} \gamma x)y \equal{} 0?\]Find a series solution about $ x \equal{} 0$ for $ \gamma \neq 0$ and write down the form of a second, independent solution. Find two independent solutions of the equation for $ \gamma \equal{} 0$.

hello, two independent solutions in the case $ \gamma\equal{}0$ are $ x$ and $ x\ln(x)$.
Sonnhard.


Given that 6x +y = 15, the value of 3x can be written in terms of y as ay + b for some numbers a and b. What is the simplified value of a +b?

[hide="Solution"]$ 6x\plus{}y\equal{}15 \implies 6x\equal{}15\minus{}y \implies 3x\equal{}7.5\minus{}0.5y \implies a\plus{}b\equal{}7$

Answer: 7[/hide]


Please compute \[\mathop{\lim }_{n \to \infty }\prod_{i = 1}^{n}{(1-\frac{1}{{2^{i}}})}\] Thank you

See http://www.mathlinks.ro/Forum/viewtopic.php?t=90517 and tell me what the problem is really about ;)


Prove algebraically:

$ \sum_{k \equal{} 0}^{min\{m, n\}} \binom{x}{m \plus{} n \minus{} k} \binom{m \plus{} n \minus{} k}{m \minus{} k, n \minus{} k} \equal{} \binom{x}{m}\binom{x}{n}$

(by definition, $ \binom{a \plus{} b \plus{} c}{a, b} \equal{} \frac {(a \plus{} b \plus{} c)!}{a!b!c!}$)

I recently posted this question and two related questions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259730]here[/url], so as such I'll hide the answer.

[hide="Algebraically"] $ (1 \plus{} t)^x (1 \plus{} s)^x \equal{} (1 \plus{} t \plus{} s \plus{} ts)^x \equal{} \sum_{k\equal{}0}^{x} {x \choose k} (t \plus{} s \plus{} ts)^k$.  Now replace $ k$ with $ m \plus{} n \minus{} k$ (since we're only interested in the coefficient of $ t^m s^n$) and apply the multinomial identity.  The term we want is $ t^{m\minus{}k} s^{n\minus{}k} (ts)^k$. [/hide]

[hide="Combinatorially"] Choose a subset of $ m$ elements and a subset of $ n$ elements out of a set of $ x$ elements and suppose that their intersection has size $ k$.  Then their union has size $ m \plus{} n \minus{} k$, so equivalently we can choose their union and pick $ k$ elements in their union to be the intersection, $ m \minus{} k$ to belong to the first set only, and $ n \minus{} k$ to belong to the second set only. [/hide]

Thank you! I appreciate your help.

But it was not exactly what I was waiting for... The first proof is essentially the same as the second, hence I consider it combinatorial. I was thinking about something even more algebraic, like Stiffel or absortion... Can somebody board the problem this way?

Do you mean as a polynomial identity?  There is an analogue of Taylor expansion for polynomials of the form $ {x \choose k}$, and I considered that technique, but as it requires evaluating the finite differences of $ {x \choose m} {x \choose n}$ I think this might be more trouble than it's worth (although perhaps enlightening).

(I am sorry for haven't answered last week)

Yes, I was looking for some solution like this. Can you show it?


[quote="nutz_for2.718281828"]We have three random 3 digit palindromes. Their sum is 1362. Compute the sum of the sums of the digits of the three numbers.[/quote]
[hide]
aba
cdc
efe

a+c+e=2 or 12 or 22
b+d+f=6 or 5 or 4 or 16 or 15 or 14 or 26 or 25 or 24
a+c+e=13 or 12 or 11
a+c+e=12
b+d+f=15
12+12+15=24+15=$39$
[/hide]

[hide="more complete solution"]
You can express the sum of the three numbers $aba$, $cdc$ and $efe$ as $100a+10b+a+100c+10d+c+100e+10f+e$. Thus, you have $101(a+c+e)+10(b+d+f)=1362$. Since they're all integers, we know that $1362-101(a+c+e)$ must be divisible by 10. We find that $a+b+c=12$ works, so then $b+d+f=15$. Since you have 2 each of $a$, $b$, and $c$, the answer is $2(a+b+c)+b+d+f=12\cdot2+15=\fbox{39}$ [/hide]

You are both right! :)

P.S. I actually picked out three palindromes for this problem


[quote="007math"][quote="XxemosakuxX"]New Question: How many whole numbers between 99 and 999 contain only 1 zero * from my worksheet

( heres a simple question to give everyone a break :D )[/quote]

There are a total of $ 999 \minus{} 99 \equal{} 900$ numbers.  We have $ 9 \times 9 \times 9 \equal{} 729$ three-digit numbers that contain no zeros.  So the answer is $ 900 \minus{} 729 \equal{} \boxed{171}$[/quote]

I don't think so... I think its $ 81\plus{}81\equal{}162$... I think you did at least 1 zero, but the questions asks for exactly one zero.

you have to subtract the numbers 100,200,..., 900 so 171-9 or 162.

Ok let me look at this again

So the question asks for the number of numbers from 99-999 that contain EXACTLY one zero.

There are 729 numbers which contain NO zeros.

That means there are 171 numbers that have either one or two zeros.

The numbers 100,200,300...,900 have two zeros and there are nine of those.

171-9 = 162. Where did I go wrong?  :huh:

Well I'll post my solution:

Casework:

Case 1: xy0. There are 9 choices for x, and 9 for y, so 81 total
Case 2: x0y. Same as above, so 81 total.

81+81=162.

[quote="ThinkFlow"]BOGTRO, you missed 10, 20, etc. There are nine of these, so the final answer is $ \boxed{171}$.[/quote]

[quote="XxemosakuxX"][b]between 99 and 999 [/b][/quote]

cough

Sorry, I just realized that I misread the problem... :blush:

900-9x9x9 = 729.
900-729=171
171-9=[b]162[/b]     (because 100....900 don't count)
Or you can think that if its 100 there is 101,102,103,104,105,106,107,108,109 which is 9 numbers times 9 since there are 9 different hundreds digits. 9x9=81
and then there is 10,,20,30,40,50,60,70,80,90 and you multiply by 9 because you need a 3 digit number. 9x9=81
81+81=[b]162[/b]
Did I go wrong? :huh:  :huh:

No, your solution is correct, as are the other solutions. :)

or you cold do 101-110 is 10 numbers, 120-190 by 10's is 8 numbers so thats 18, 18 x 9 = 162 numbers?


Say that three circles intersect such that the points of intersection go like A, F, B, D, C, E.  Prove that AD, BE, and CF are concurrent.

I thought of Ceva's theorem didn't know...

Thanks

Sorry, I can't really draw this so I will try attach something.

The green and red intersect at A and D, where D is more centralized.  The green and yellow intersect at B and E, where E is more centralized.  The red and yellow intersect at C and F, where F is more centralized.

Now we prove that AD, BE, and CF are concurrent, I think that is much better.

[quote="mathgeniuse^ln(x)"]Wait, do radical axes of a triangle concur at a point?[/quote]

Radical axes are a property of circles, not triangles...

If two pairwise radical axes go through a point, then the third must, by transitivity of equality (just look back at the definition of a radical axis...)

Not sure how you would use Ceva's here.

[hide="Solution"]Let the intersection of $AD$ and $CF$ be $P.$ Next, draw chord $BE$ through $P,$ contained in the circle containing the points $A,$ $B,$ and $D.$ First assume that the chord is not contained inn the circle with points $C,$ $F,$ and $B$ on its circumference.

By power of a point, $(AP)(PD)=(CP)(PF),$ and $(AP)(PD)=(BP)(PE).$

Now, chord $PE$ would be contained in both specified circles if and only if $(BP)(PE)=(CP)(PF).$ However, this is clearly true from the first two equations, and hence the three chords are concurrent at $P.$
[/hide]


[quote="wly3298456"]if $f(2x-3)=x^2+4x-5$, find the smaller zero of $f(x-1)$[/quote]

let $f(2x-3)=x^2+4x-5=(x+5)(x-1)=g(x)$
consider
$g\left(\frac{1}{2}(x+3)\right)=\left(\frac{1}{2}(x+3)+5\right)\left(\frac{1}{2}(x+3)-1\right)$
$\quad =f\left(2\left(\frac{1}{2}(x+3)\right)-3\right)=f(x)$

$f(x)=\left(\frac{1}{2}(x+3)+5\right)\left(\frac{1}{2}(x+3)-1\right)$ so
$f(x-1)=\left(\frac{1}{2}(x+2)+5\right)\left(\frac{1}{2}(x+2)-1\right)$
solving for those roots, we get $x=-12,0$ and the smallest would be $\boxed{-12}$

[quote="wly3298456"]if f(2x-3)=x^2+4x-5, find the smaller zero of f(x-1)[/quote]

$f(2x-3)=x^2+4x-5$

$f(x-1)=?$

$\frac{2x-3}{2}+\frac{3}{2}=x$, thus

$f(x)=(\frac{x+3}{2})^2+2(x+3)-5$

By plugging in $x-1$ for $x$, we simplify to $x^2+12x=0$, from here we solve for $x$ and we get $\boxed{-12,0}$.

I have a question.  What does it mean to say find the smaller zero?

It means you have to find the smaller root of the quadratic equation.

I would think that it's more tedious to factor $f(2x-3)$, don't you think ;) Since everyone except me solved it like that

Well, you could always do it without factoring.

$f(2x-3)=x^2+4x-5$
$2x-3=x$
$x=\frac{x+3}{2}$
$f(2\frac{x+3}{2}-3)=\left(\frac{x+3}{2}\right)^2+2(x+3)-5$
$f(x)=\left(\frac{x+3}{2}\right)^2+2(x+3)-5$
$f(x-1)=\left(\frac{x+2}{2}\right)^2+2(x+2)-5$
$\left(\frac{x+2}{2}\right)^2+2(x+2)-5=0$
$x^2+12x=0$

You solve for x and you get $-12,0$.  Since $-12<0$, the answer is $\boxed{-12}$.

Now that I think about it, without factoring is easier, because factoring is just an extra step :)

Yes, I always find it easier to do it that way, factoring and then multiplying always increases the chances for careless errors :)

Well this was the first time I solved such a thing...it was fun, and I agree that factoring just makes it complicated.  ;)

[quote="boxedexe"]$2x-3=x$[/quote]
so the entire proof is based on $x=3$ ...
i would suggest to have a diffrent notation for the "second" x like $x'$ or $y$ or anything else but $x$

well I suppsoe that would be more convinient.


P is any point inside the interior of the triangle ABC. AP,BP,CP when extended meet sides AB,BC,CA at D,E,F.

if BD=2CD  and AE=3EC then

(1)   AP:PD=?

(2) BP:PE =?

(3) CP:PF=?

[hide="Fairly Easy with Vectors"]D is on BC and BD=2CD, so $ \vec{D}\equal{}\frac{1}{3}\vec{B}\plus{}\frac{2}{3}\vec{C}$.  Similarly, $ \vec{E}\equal{}\frac{1}{4}\vec{A}\plus{}\frac{3}{4}\vec{C}$, since AE=3EC.  Thus, P is on line AD and BE, so it is of the form $ t\vec{A}\plus{}(1\minus{}t)(\frac{1}{3}\vec{B}\plus{}\frac{2}{3}\vec{C})$ and also of the form $ u\vec{B}\plus{}(1\minus{}u)(\frac{1}{4}\vec{A}\plus{}\frac{3}{4}\vec{C})$.  Setting the coefficients of $ \vec{A}$, $ \vec{B}$, and $ \vec{C}$ equal, and solving the resulting equations, we obtain t=2/11, u=3/11, $ \vec{P}\equal{}\frac{2}{11}\vec{A}\plus{}\frac{3}{11}\vec{B}\plus{}\frac{6}{11}\vec{C}$. Thus
AP/PD=9/2, BP/PE=8/3, CP/PF=6/5.[/hide]

Using Ceva's theorem in $\triangle{ABC}$, we have:

$\dfrac{BD}{DC} \cdot \dfrac{CE}{EA} \cdot \dfrac{AF}{FB} =1 \Longrightarrow \dfrac{AF}{FB}=\dfrac{3}{2} \text{      }\left(\because \dfrac{BD}{DC}=2 \text{ and }\dfrac{CE}{EA}=\dfrac{1}{3} \right)$.

Now using Menelaus theorem in $\triangle{ADC}$ with $BE$ as transversal, we have:

$\dfrac{AP}{PD} \cdot \dfrac{DB}{BC} \cdot \dfrac{CE}{EA}=1\Longrightarrow \dfrac{AP}{PD}=\dfrac{9}{2}$

Evaluate $\dfrac{BP}{PE}=\dfrac{8}{3}$ and $\dfrac{CP}{PF}=\dfrac{5}{6}$ in a similar manner.


In k-people association work (2^n - 1) comission (every no empty set members of association creates comission). In every comission have to choose chairman. It demands condition. If comission C is sum two comission A and B ($ C \equal{} A \cup B$), chairman comission C will be chairman at least one of comission A, B. Mark number posibility of choice chairman. I know that result is k!, but I don't know how show it

This is a double post. This poster's previous version of this post was removed from public view because it is from an ongoing contest whose deadline has not been reached.


geometry
rhombus
analytic geometry
trigonometry

The diagonals of rhombus $ABCD$ are 10 and 4.  Rhombus CDEF is drawn in the same plane, with $\overline{DE} \perp \overline{DA}$.  Compute the area of rhombus $CDEF$.

[hide]put in coordinate plane (see diagram :D)
Draw in square $ADEH,$ and $E = (-7, 3)$
$EC=\sqrt{(0+7)^2+(-2-5)^2}=\sqrt{98}$
$G=(-3.5, 1.5)=\frac{(D+F)}{2}$
$F=(-2,3)$ and $FD=\sqrt{(0+3)^2+(-5+2)^2}=\sqrt{18}$
area of $CDEF=\frac{\sqrt{98}\sqrt{18}}{2}=21$[/hide]

aww, why cant i hide diagrams?

Nice use of coordinates to kill the problem.

Try to solve this problem without coordinates

[hide="Click here for one hint"]The area of a rhombus is $K=a^2 \sin \alpha$[/hide]
[hide="Click here for another"]The two rhombi have complementary acute angles.[/hide]

wow...thats an extremely nice solution...it works out really well, especially because of the 20-21-29 phytagorean triangle...


Suppose that m and n are relatively prime integers. Assume they are both greater than 1.

Prove that the set $(0n, 1n, 2n, 3n,..., (m-1)n)$ is a complete residue system modulo m.

[hide="Not very rigorous, I suppose, but it touches on the main point."]Assume that two elements are in the same residue class. Then their difference must be divisible by m. That's impossible because its factors are n and something between 0 and m, and n is relatively prime to m, and something less than m cannot be divisible by m. Therefore the assumption is incorrect, and all elements of the set are in different residue classes mod m, which is the same as saying that the set is a complete residue system, mod m. (fyi, My teacher said complete set of residue class representatives.)[/hide]


AMC
AIME
USA(J)MO
USAMO
geometry
probability
AMC 10

So I just went through the AMC process this year for my first time, with a dissapointing score on my AMC 10, AMC 12, and AIME.  In practice, I can almost always find the answers to all of the AMC 10 problems, but my skills (and knowledge, I'm only up to algebra 2) are only advanced enough to get about 16 problems on the 12. On the aime I can get anywhere from 1-6 depending on what type of problems there are.

My goal is to get to the USAMO within a year (sophomore,) get a decent USAMO score as a Junior, and maybe even get close to MOP as a senior.

While I'll spend a good deal of time on practice problems and I've got a good amount of money, what AOPS books would be best to help me do well and understand more math in general? Should I buy the intro books? The intermediate books? Would it be better to buy the specialized books? Or just the general AOPS volume 1/2?

Thanks in advance.

(I already have intro to c&p which I used for mc.)

P.S. Does anyone know if and when AOPS is releasing their intermediate number theory and geometry books?

I say use the AoPS books first, you can go a long way using those alone, and get the subject books if you feel you're weak in an area.

//and maybe even make MOP as a senior.//

(Hmm, I hear making MOP as a senior is a pretty *big(=* deal.)

AoPS books are amazing.
I think both the general & specialized books are great. It depends on what you're "lacking" in. If your weakness is geometry, really work the specialized books, they really really help. and if money isn't a problem, *all* of them do help :P [i have all of them -that are out so far-... well i share them with my brother but still XD]. the general ones got me started , then i moved to improve in each specific area.
ohhh, and woot seems to have helped a bunch of my friends this year . i dont know tho, havent tried it out yet.

What do the AOPS Volume 1/2 Books have in them? Are they just a generalized version of all the specific books, or do they cover entirely different topics?

They are a generalized version of the others.

However, the subject books teach more than Vol. 1 & 2; I find that Volume 1 and 2 are better for just doing problems and reviewing, and the subject books teach more.

(this is just my opinion, which is bound to be bad, as I learn slower than most AoPSers :()

Where do you get the AoPS books?
From the online bookstore??

Yes.  The online bookstore is the best place to get AoPS books.

Would the intro books help at all with AMC 10/12/AIME stuff?

And also, If I know about 90% of the intro books, would I get left behind if I buy the intermediate ones, or does it review a little?

[quote="Ihatepie"]Would the intro books help at all with AMC 10/12/AIME stuff? [/quote]
They should.
[quote="Ihatepie"]And also, If I know about 90% of the intro books, would I get left behind if I buy the intermediate ones, or does it review a little?[/quote]
The first chapter of Intermediate Couting & Probability is dedicated to a quick review.

Is the "Art of Problem Solving VOlume 2" good for preparing for the AMCs and AIME?
If not, which book would be good?

aops 2 is specifically designed for amc 12 and aime.

I found that after working on First Steps for Math Olympians, I was able to do every AMC 10 problem this year. (It might be that this was an easy year though based on the USAMO index...)

It doesn't really explain the theorems and stuff extremely well (compared with AoPS vol 2), but I was already acquainted with most of it before I started the book so I was able to understand the explanations.

I didn't get a 150 because I wasn't fast enough to complete the test in the 75 minutes.

Speed is something that's mainly dependent on doing large numbers of problems, especially under timed conditions. The difference between solving a problem in 2 or 4 minutes is often how quickly you see the right pattern, which is obviously improved by having seen a huge number of problems.

[quote="kohjhsd"][quote="Ihatepie"]And also, If I know about 90% of the intro books, would I get left behind if I buy the intermediate ones, or does it review a little?[/quote]
The first chapter of Intermediate Couting & Probability is dedicated to a quick review.[/quote]

In fact, the first 3-4 chapters of Intermediate Algebra is also review.


Sequence
algebra
Inequality
construction
geometric sequence
IMO
IMO 1973

Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that:

[i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$
[i]b.)[/i] $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$
[i]c.)[/i] $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$

LaTeX is very boring, so I´ll just send the cases $ n \equal{} 5$ and $ n \equal{} 6$.

$ b_1 \equal{} a_1 \plus{} q(a_5 \plus{} a_2) \plus{} q^2(a_4 \plus{} a_3)$
$ b_2 \equal{} a_2 \plus{} q(a_1 \plus{} a_3) \plus{} q^2(a_5 \plus{} a_4)$
$ b_3 \equal{} a_3 \plus{} q(a_2 \plus{} a_4) \plus{} q^2(a_1 \plus{} a_5)$
$ b_4 \equal{} a_4 \plus{} q(a_3 \plus{} a_5) \plus{} q^2(a_2 \plus{} a_1)$
$ b_5 \equal{} a_5 \plus{} q(a_4 \plus{} a_1) \plus{} q^2(a_3 \plus{} a_2)$

$ b_1 \equal{} a_1 \plus{} q(a_6 \plus{} a_2) \plus{} q^2(a_5 \plus{} a_3) \plus{} q^3a_4$
$ b_2 \equal{} a_2 \plus{} q(a_1 \plus{} a_3) \plus{} q^2(a_6 \plus{} a_4) \plus{} q^3a_5$
$ b_3 \equal{} a_3 \plus{} q(a_2 \plus{} a_4) \plus{} q^2(a_1 \plus{} a_5) \plus{} q^3a_6$
$ b_4 \equal{} a_4 \plus{} q(a_3 \plus{} a_5) \plus{} q^2(a_2 \plus{} a_6) \plus{} q^3a_1$
$ b_5 \equal{} a_5 \plus{} q(a_4 \plus{} a_6) \plus{} q^2(a_3 \plus{} a_1) \plus{} q^3a_2$
$ b_6 \equal{} a_6 \plus{} q(a_5 \plus{} a_1) \plus{} q^2(a_4 \plus{} a_2) \plus{} q^3a_3$

I hope there are no difficulties in generalizing. In $ b_i$, indexes wich are next to $ i$ are monitored by $ q$´s with low exponents. To see that all conditions are fulfilled, observe we have

$ \frac {1 \plus{} q}{1 \minus{} q} \equal{} 1 \plus{} 2\sum \limits^\infty_{i \equal{} 1} q^i$.