number theory
least common multiple
greatest common divisor
geometry
perimeter
logarithms
ARML

New York City A

[u]Round 1[/u]
7th Polya Mathematics Competition

1. What are the last three digits of $1999^{1999}$?
[hide] you should pass back 999.
$999^{2}$ ends in 001, hence $999^{3}$ ends in 999.
In general, any odd power will end in 999. [/hide]

2. How many solutions are there to 3x+37y=TNYWR in which both x and y are nonnegative integers?
[hide] before you get anything from first person, you should look at the problem and make a note that 3 and 37 are relatively prime. I dont see anything else you can do in advance.[/hide] [hide] you should pass back 10.
3x+37y=999; RHS is 0(mod3) -> 37y has to be 0(mod3) -> y is 0(mod3)
the solutions are (333,0), (296,3), (259,6), ... , (0,27)
so there are 10 solutions in total.[/hide]

3.  A number consists of TNYWR repetitions of the digit 1. Find the complete prime factorization of the number. Example and hint: If TNYWR is 5, then 11111 = 41*271
Further hint: if you got the correct number passed to you, the prime factorization will consist of four distinct primes.
[hide] if i were the third person, I would try prime factorization starting from TNYWR = 5 [/hide] [hide] the answer is 11*41*271*9091.
1111111111=11111 times 100001, and 11111 = 41*271 from hint and 100001 is divisible by 11 -> 11*91
so 11*41*91*271[/hide]

[u]Round 2[/u]
7th Polya Mathematics Competition

1. How many positive integers are divisors of 5040?
[hide] you should pass back 60.
$5040=2^{4}*3^{2}*5*7$, so the number of divisors is (4+1)(2+1)(1+1)(1+1)=60. [/hide]

2. Let L be the least common multiple of TNYWR and 2000, and let D be the greatest common divisor of TNYWR and 2000. Compute $\frac {L}{D}$.
[hide] I would have done prime factorization of 2000 before getting anything from the first person - not that it is helpful on this particular problem, but it tends to be easier to find LCM and GCD when the numbers are prime factorized.[/hide] [hide] you should pass back 300
L = 6000, D = 20, $\frac {6000}{20}=300$.[/hide]

3. Let t = TNYWR. The equation $x^{3} - 31x^{2}  + tx - 900 = 0$ has three distinct integer solutions. Find them.
[hide] If I were the third person, I would have done factor 900, since integer solutions have to be factors of 900. Also, since leading coefficient is zero, make a note of that -900 is the negation of the product of the solutions, where -31 is the negation of the sum of the solutions. You could have guessed without getting anything from the second person! [/hide] [hide] the answer is 6, 10, 15 after a little trial and error. [/hide]

[u]Round 3[/u]
Polya Mathematics Competition

1. Solve for x; $2^{16}=16^{x}$.
[hide]you should pass back 4
$2^{16}=(2^{4})^{4}=16^{4}$
OR
$2^{16}=2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2
=(2*2*2*2)*(2*2*2*2)*(2*2*2*2)*(2*2*2*2)=16*16*16*16=16^{4}$
[/hide]

2. Regular hexagon ABCDEF has area TNYWR. What's the area of triangle ACE?
[hide] before you get anything from the first person, you should be able to recognize that triangle ACE is a half of the regular hexagon. It is obvious once you draw it; when O stands for the center of the hexagon (where AD and BE meet), ABC = AOC and so on. [/hide] [hide] you should pass back 2
TNYWR *$ \frac {1}{2} = 4 *\frac {1}{2} = 2$ [/hide]

3. A "flower" pattern is made by drawing a regular hexagon inscribed in a circle of radius TNYWR. Then six circles are drawn with each of the edges of the hexagon as their diameters. The flower consists of everything inside the original circle or inside any of the six smaller circles. What is the area of the flower?
[hide]without knowing radius, you should have constructed diagram and, if time allows, express area with variable "r" - get ready to plug in. [/hide][hide]the answer is $6\sqrt{3} + 3\pi$
It is easy to see that its 6 semicircles and 6 equilateral triangles[/hide]

[u]Round 4[/u]
Polya Mathematics Competition

1. An equilateral triangle has area 2. A regular hexagon with the same perimeter has what area?
[hide] you should pass back 3. To save time, instead of calculating a side of triangle, note that in order for hexagon to have same perimeter as triangle, the sides are $\frac {1}{2}$ of triangle's, so when you divide the hexagon into 6 equal triangles, each of them are $\frac {1}{4}$ of the triangle - so hexagon in total has $\frac {6}{4}$ area of triangle's.[/hide]

2. A convex polygon has exactly TNYWR obtuse angles, the remaining angles being acute or right. What is the maximum possible number of sides the polygon can have?
[hide] the sum of all angles will be 180(n-2) [/hide] [hide]you should pass back 6.
you want to maximize the obtuse angle since the sum is fixed, so that the remaining angles can have less share. If n was 6, the sum is 720, and three obtuse angle's sum can be up to $540^{-}$ degrees; three angles whose sum is $180^{+}$ -> no problem. If n was 7, the sum is 900, and three obtuse angle's sum can be up to $540^{-}$; four angles whose sum is $360^{+}$ -> at least one of them has to be an obtuse angle(no good). It should be intuitively obvious that as n gets larger, you need more obtuse angles. [/hide]

3.  T=TNYWR
Let f(x) = x(x+T) What is the positive value of n for which 2f(n)=f(n+6)
[hide] with some algebra,
$n^{2}+(T-12)n-36-6T=0$[/hide] [hide] the answer is 12; T=6 gives you $n^{2}-6n-72=(n-12)(n+6)=0$, positive value of n is 12.[/hide]

[u]Round 5[/u]
University of Iowa Math Tournament

1. What is the side length of a square that has the same area as a circle with radius  $\frac {8}{\sqrt{\pi}}$?
[hide] you should pass back 8 - I dont think any explanation is necessary[/hide]

2. Let T be TNYWR.  Calculate the product of  $\log_3{T}$ and $\log_2{9}$.
[hide]  $\log_a{b}$ times $\log_b{a}$ is one so i bet that T is going to do something with base 2! [/hide] [hide] you should pass back 3. ${\log_3{8}}$*$\log_2{9}$ = $\frac{log8}{log3}$*$\frac{log9}{log2}$ = $\frac{log8}{log2}$*$\frac{log9}{log3}$ = $3 * 2 = 6$ [/hide]

3.  In a room of K people, everyone shakes everyone else’s hand once.  In all, there are TNYWR handshakes.  What is K?
[hide] TNYWR = $_K{C}_2$[/hide] [hide] the answer is 4. [/hide]

[hide="Finally"]
I'm learning TeX as I'm writing this haha.
I'm learning it from what people have posted :)
[/hide]

[hide="Hidden_by_moderator"]Bit of a mistake...For Round 2, Question 3, the solutions are -6,-10 and -15, as opposed to 6,10,15...[/hide]

[hide="Hidden_by_moderator"]wait a sec...6,10 and 15 are correct, that's just not the equation given to us during the relay...[/hide]

Ravi (or stan): can you hide what stan wrote?

yes, originally in my problem bank,  it was copied and pasted from online and the minus sign was expressed as question marks. I said plus without verifying them, just not to waste too much time.

[quote="Geehoon"]Ravi (or stan): can you hide what stan wrote?[/quote]Done.

Those seem like pretty decent relays -- maybe a little easier than ARML relays.

easier than ARML?

people were complaining that Polya is too hard.
I guess we need more practice.

by the way, feel free to say any comments on solutions please.
(any suggestion to cut down times as well).

i think the difficulty level is only elevated for the person 3.  the other two have relatively easy questions.  also, i think polya is supposed to be 5-10, not 3-6, so the time constraints may have increased the difficulty for person 3.

Yes, I did think there was a fairly heavy burden on the third person.  I haven't done ARML in a few years, so it might in fact be a good approximation.  In any event, what matters most is the practice, not the particulars.


Well i am sort of stuck with this problem ,i dont know but maybe i am thinking in a bad way on it .
Well here it is :
[color=blue]Problem[/color]
Let $n>10000$ then prove there exist numbers like $a$ and $b$ such that :

$n < a^2+b^2<n+3\sqrt[4]{n}$

Ps.i was trying to study the structure of the numbers with form $a^2+b^2$ but it turns out to be complicated. :|
all i just find out is :
$a^2+b^2$ has even number of $4k+3$ prime factors.and each prime in form of $4k+1$ is showable in form of $a^2+b^2$

It is well-known pproblem from Russian MO 2001 or 2002

It was even posted before:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Russia+2001&t=292


Show that if $ x,y>1$, then $ \frac{x^2}{y\minus{}1}\plus{}\frac{y^2}{x\minus{}1} \ge 8$

Very simple!  You can use $ (x\minus{}2)^2\equal{}x^2\minus{}4x\plus{}4\ge0$ to get $ x^2\ge4(x\minus{}1)$.  Then the result follows by AM-GM.

Cauchy kills it .

[hide]$ LHS\ge \frac{(x\plus{}y)^2}{x\plus{}y\minus{}2}$ , so it remains to prove $ (x\plus{}y)^2\plus{}16\ge 8(x\plus{}y)$ (AM-GM)[/hide]

Yes, it is very simple.

My solution: We have to show $ \frac{x^2}{y\minus{}1}\plus{}4(y\minus{}1)\plus{} \frac{y^2}{x\minus{}1} \plus{}4(x\minus{}1) \ge 4x\plus{}4y$, which is true by AM-GM on each pair of terms.


Point P is 9 units away from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?

[hide="answer"] 12 [/hide]

we find the shortest possible length and the longest possible length, and then find the ones inbetween. say the center is $ O$ the longest possible one is obviously the diameter, $ 30$. the shortest, is perpendicular to $ OP$. let the two points where the chord intersects the circle be $ A$ and $ B$. we have right triangles AOP and BOP. using pythagorean theorem, we get $ AP\equal{}12$ so $ AB\equal{}24$.

that means we can form any chords passing though $ P$ to have a length between $ 24$ and $ 30$. there are $ 30\minus{}24\plus{}1\equal{}\boxed{7}$ chords between.

[quote="abacadaea"]we find the shortest possible length and the longest possible length, and then find the ones inbetween. say the center is $ O$ the longest possible one is obviously the diameter, $ 30$. the shortest, is perpendicular to $ OP$. let the two points where the chord intersects the circle be $ A$ and $ B$. we have right triangles AOP and BOP. using pythagorean theorem, we get $ AP \equal{} 12$ so $ AB \equal{} 24$.

that means we can form any chords passing though $ P$ to have a length between $ 24$ and $ 30$. there are $ 30 \minus{} 24 \plus{} 1 \equal{} \boxed{7}$ chords between.[/quote]d

Draw a good picture; there is more than one chord you can draw with length 25.

ill try paint because i dont know asy....

whoops i misread the question. 24 and 30 have only one chord. 25-29 have 2 different chords because you can draw it the other way. kinda like reflecting it. $ 5*2\plus{}2\equal{}\boxed{12}$

[hide="SOLUTION"]We can see that the shortest chord is $ 24$- (Perpendicular to radius), and the longest chord is $ 30$- Diameter.  Therefore, there are 5 distinct lengths in between: $ 25, 26, 27, 28, 29$.  We can double each of these for symmetry, counting $ 10$ lengths.  After adding the two extremes mentioned in the beginning, we get $ \boxed{12}$ as our answer.[/hide]

Thanks everyone.

Now I know that I was on the right track, but I mixed up my pythagorean triples; I thought the longest side was 30 and the shortest was 25 (I added 5 to each member of the (15,20,25) pythagorean triples set)
:oops: :oops:


[quote="laughinghead505"]does order count?[/quote]
No, it doesn't. :D

[hide="for those who know combinatorics"]Since red has to be one then there are four choices. Two have to be given out. $\binom{4}{2}=6$.[/hide]
[hide="for those who do not"]Since red has to be one then there are four choices. Two have to be given out. There are four choices for the first and three for the second. $4 \cdot 3=12$. BUT, this only counts when order matters, but it clearly doesn't. If the choices are a and b then there are two ways to arrange them $ab$ and $ba$. We only want to count half of them. $\frac1212=6$[/hide]

[quote="Arvind_sn"]How many ways can Billy Bob three of five M&M's to Killy Kob if he has to give her the red one as one of those three?[/quote]
[hide]that leaves 4 and you got to give away 2 so thats like 6[/hide]

[quote="funcia"][hide]3 times 2 times 1[/hide] that works too doesn't it?[/quote]

basically $3!$


calculus integration function derivative trigonometry logarithms algebra

[quote="borislav_mirchev"]1.
$ \int \frac {x \plus{} e^{x}(sinx \plus{} cosx) \plus{} sinxcosx}{(x^{2} \plus{} 2e^{x}sinx \minus{} cos^{2}x)^{2}} \, dx$[/quote]

Multiplying and dividing by two I get $ \frac {1}{2}\int \frac {2x \plus{} 2e^{x}(sinx \plus{} cosx) \plus{} 2sinxcosx}{(x^{2} \plus{} 2e^{x}sinx \minus{} cos^{2}x)^{2}} \, dx$.
The function I have now to integrate it's the derivative of $ \minus{} \frac {1}{x^{2} \plus{} 2e^{x}sinx \minus{} cos^{2}x)} \, dx$. Right?

Please, correct your tex notation. You are right.

Number 8
My answer is 0
$ I \equal{} \int_{\frac {1}{2}}^2 \frac {lnx}{1 \plus{} x^{2}} \,dx \equal{} \int_{ \minus{} ln2}^{ln2}\frac {ue^u}{1 \plus{} e^{2u}}du$
$ f( \minus{} x) \equal{} \minus{} f(x) \equal{} > f(x)$ is odd,where $ f(x) \equal{} \frac {xe^x}{1 \plus{} e^{2x}}$
My substitution is $ x \equal{} e^u$

10)

$ \int_{\frac {3}{2}}^{2} \left(\frac {x \minus{} 1}{3 \minus{} x}\right)^{\frac {1}{2}} \,dx$

Substitute $ x \equal{} 2 \plus{} \sin t$ to get:

$ \int_{ \minus{} \frac {\pi}{6}}^{0} \left(\frac {1 \plus{} \sin t}{1 \minus{} \sin t}\right)^{\frac {1}{2}} \cos t \,dt$

$ \int_{ \minus{} \frac {\pi}{6}}^{0} \left(\frac {\cos{\frac {t}2} \plus{} \sin{\frac {t}2}}{\cos{\frac {t}2} \minus{} \sin{\frac {t}2}}\right) \cos t \,dt$

$ \int_{ \minus{} \frac {\pi}{6}}^{0} (\cos{\frac {t}2} \plus{} \sin{\frac {t}2})^2 \,dt$

$ \int_{ \minus{} \frac {\pi}{6}}^{0} 1 \plus{} \sin t \,dt \equal{} t \minus{} \cos t \bigg{|}_{ \minus{} \frac {\pi}6}^0 \equal{} (0 \minus{} 1) \minus{} ( \minus{} \frac {\pi}6 \minus{} \cos{ \minus{} \frac {\pi}6}) \equal{} \frac {\pi}6 \minus{} 1 \plus{} \frac {\sqrt {3}}{2}$

The weird looking answer worries me. I could be wrong  :oops:

I too believe 9th one diverges

4)

$ \int \frac {x^{5n \minus{} 1} \plus{} x^{n \minus{} 1}}{1 \plus{} x^{6n}} \,dx$

Sub $ x^n \equal{} t$, to get:

$ \int \frac {x^{5n \minus{} 1} \plus{} x^{n \minus{} 1}}{1 \plus{} x^{6n}} \,dx \equal{} \int \frac {1 \plus{} t^4}{1 \plus{} t^6} \,dt$

Now we break it into partial fractions:

$ \equal{}\int \frac{1}{3} \left(\frac{t^2 \plus{} 1}{t^4\minus{}t^2\plus{}1} \plus{} \frac{2}{t^2\plus{}1}\right) \,dt$

$ \equal{}\frac13(2\tan^{\minus{}1} t \plus{} I)$

$ I \equal{} \int \frac{t^2 \plus{} 1}{t^4\minus{}t^2\plus{}1} \, dt$

$ I \equal{} \int \dfrac{\frac1{t^2} \plus{} 1}{t^2\minus{}1\plus{}\frac1{t^2}} \, dt$

Now we sub $ u \equal{} t \minus{} \frac1{t}$,

$ I \equal{} \int \dfrac{\frac1{t^2} \plus{} 1}{t^2\minus{}1\plus{}\frac1{t^2}} \, dt$

$ I \equal{} \int \dfrac{du}{u^2 \plus{} 1} \, du \equal{} \tan^{\minus{}1} u$

So...
$ \int \frac {1 \plus{} t^4}{1 \plus{} t^6} \,dt \equal{} \frac13\left(\tan^{\minus{}1} t \plus{} \tan^{\minus{}1} \left(t \minus{} \frac1{t}\right)\right)$

Finally subbing back the $ t \equal{} x^n$
$ \int \frac {x^{5n \minus{} 1} \plus{} x^{n \minus{} 1}}{1 \plus{} x^{6n}} \,dx \equal{} \frac13\tan^{\minus{}1} x^n \plus{} \frac13\tan^{\minus{}1} (x^n \minus{} x^{\minus{}n}) \plus{} C$

[quote="isomorphism"]10)

The weird looking answer worries me. I could be wrong  :oops:

[/quote]

Don't worry, it is correct. I got it too (and checked on Mathematica). (Comparing to the results I get in my homeworks... this isn't weird at all  :lol: )

for the second

$ I=\int\(x \frac{1-sin2x}{4}dx$ and then integrate by parts

Excellent solutions and ideas... what about other unsolved? Are they difficult?

have maybe the [9] this form $ \int_{0}^{\pi} \frac {x\plus{}sinx}{1 \plus{} cosx} \,dx$  ?

My solution to #3:
$ \displaystyle\int\frac{\sqrt{1\plus{}\ln{x}}}{x\ln{x}}dx
\\
\\
x \equal{} e^{u\minus{}1} \to dx \equal{} e^{u\minus{}1}du
\\
\\
\displaystyle\int\frac{\sqrt{1\plus{}u\minus{}1}}{e^{u\minus{}1}\ln{e^{u\minus{}1}}}e^{u\minus{}1}du
\\
\\
\displaystyle\int\frac{\sqrt{u}du}{u\minus{}1}
\\
\\
u \equal{} t^2 \to du \equal{} 2tdt
\\
\\
2\displaystyle\int\frac{t^2dt}{t^2\minus{}1} \equal{} 2\int\frac{t^2\minus{}1\plus{}1}{t^2\minus{}1}dt \equal{} 2t\plus{}\int\frac{2dt}{t^2\minus{}1}
\\
\\
\frac{2}{t^2\minus{}1} \equal{} \frac{1}{t\minus{}1} \minus{} \frac{1}{t\plus{}1}
\\
\\
2t \plus{} \displaystyle\int\left(\frac{1}{t\minus{}1}\minus{}\frac{1}{t\plus{}1}\right)dt \equal{} 2t \plus{} \ln{\left|\frac{t\minus{}1}{t\plus{}1}\right|} \plus{} C$

finally we resubstitute to x
$ u=t^2 \to t=\sqrt{u}
\\
\\
\displaystyle 2\sqrt{u} + \ln{\left|\frac{\sqrt{u}-1}{\sqrt{u}+1}\right|} + C
\\
\\
x=e^{u+1} \to u = \ln{x} + 1
\\
\\
\displaystyle 2\sqrt{\ln{x} + 1} + \ln{\left|\frac{\sqrt{\ln{x} + 1}-1}{\sqrt{\ln{x} + 1}+1}\right|} + C$
Simplifying we get the (arguably) nicer answer of:
$ \displaystyle 2\sqrt{\ln{x} + 1} + \ln{\left|\frac{\ln{x}}{\ln{x}+2\sqrt{\ln{x}+1}+2}\right|} + C$

My Solution for number 6:

$ \displaystyle\int\arcsin{\sqrt{\frac{x}{x\plus{}a}}}dx
\\
\\
\frac{x}{x\plus{}a} \equal{} 1 \minus{} \frac{a}{x\plus{}a} \equal{} \sin^2{u}
\\
\\
\frac{a}{x\plus{}a} \equal{} \cos^2{u} \to  x \equal{} a(\sec^2{u}\minus{}1)
\\
\\
x \equal{} a\tan^2{u} \to dx \equal{} 2a\tan{u}\sec^2{u}du
\\
\\
2a\displaystyle\int u\tan{u}\sec^2{u}du
\\
\\
u \equal{} \arctan{t} \to t \equal{} \tan{u} \to du \equal{} \frac{dt}{\sec^2{u}}
\\
\\
2a\displaystyle\int t\arctan{t}dt
\\
\\
w \equal{} \arctan{t} \to dw \equal{} \frac{dt}{t^2\plus{}1}
\\
\\
dv \equal{} tdt \to v \equal{} \frac{t^2}{2}
\\
\\
2a\left(\frac{1}{2}t^2\arctan{t} \minus{} \frac{1}{2}\displaystyle\int\frac{t^2dt}{t^2\plus{}1}\right)
\\
\\
a\left(t^2\arctan{t} \minus{} \displaystyle\int\left(1\minus{}\frac{1}{t^2\plus{}1}\right)dt\right)
\\
\\
at^2\arctan{t}\minus{}at\minus{}a\arctan{t} \plus{} C
\\
\\
au\tan^2{u}\minus{}a\tan{u}\minus{}au\plus{} C
\\
\\
x \equal{} a\tan^2{u} \to u \equal{} \arctan{\sqrt{\frac{x}{a}}}
\\
\\
x\arctan{\sqrt{\frac{x}{a}}}\minus{}a\sqrt{\frac{x}{a}}\minus{}a\arctan{\sqrt{\frac{x}{a}}}\plus{} C
\\
\\
(x\minus{}a)\arctan{\sqrt{\frac{x}{a}}}\minus{}a\sqrt{\frac{x}{a}} \plus{} C$

Integral 9 was taken from site - it is what were written. It may be wrong. I'm sorry for late response. Did you (mathlinkers) like these integrals?

for number [8]

donate the function $ f(x) \equal{} \int_{\frac {1}{x}}^{x} \frac {lnt}{1 \plus{} t^{2}} dt$

find $ f'(x)$ and then the function then $ I \equal{} f(2)$

I will post my solution later because i have not  time

( I find that I=0 .Am i right?)

[quote="giannis"]for number [8]

donate the function $ f(x) \equal{} \int_{\frac {1}{x}}^{x} \frac {lnt}{1 \plus{} t^{2}} dt$

find $ f'(x)$ and then the function then $ I \equal{} f(2)$

I will post my solution later because i have not  time[/quote]

Have you seen post 4 in this thread ? :P

[color=red]Solution Number 8[/color]

Let  $ f(x) \equal{} \int_{\frac {1}{x}}^{x} \frac {lnt}{1 \plus{} t^{2}} dt$ .It's clear that $ I\equal{}f(2)$

$ f'(x)\equal{}\frac{lnx}{x^2\plus{}1}\plus{}\frac{lnx}{1\plus{}\frac{1}{x^2}}\frac{\minus{}1}{x^2}\equal{}0$

so $ f$ is constant .We have $ f(0)\equal{}0$ so $ f(x)\equal{}0$

Then we take $ I\equal{}f(2)\equal{}0$...


How many ways to prove the enequality:
$\sin\frac{A}{2}\cdot\sin\frac{B}{2}\cdot\sin\frac{C}{2}\leq\frac{1}{8}$ where $A,B,C$ are angles of triangle $ABC$

The second:
$a^{2}=b^{2}+c^{2}-2bc\cos A\geq 2bc-2bc\cos A=2bc(1-\cos A)=4bc\sin^{2}\frac{A}{2}$
$\Rightarrow \sin\frac{A}{2}\leq\frac{a}{2\sqrt{bc}}$

Done :)

[hide]
From the area formulas
\[[ABC]=2R^{2}\sin A\sin B\sin C=r^{2}\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}\]
we have
\[\sin^{2}\frac{A}{2}\sin^{2}\frac{B}{2}\sin^{2}\frac{C}{2}=\frac{r^{2}}{16R^{2}}\]
\[\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{r}{4R}\leq \frac{1}{8}\]
 :D
[/hide]


During a 24-hour period, 1440 cars pass through a toll booth.
What is the mean number of cars that pass through per minute?

In 24 hours, there are $ 24 \times 60 \equal{} 1440$ minutes. So our answer is $ 1440 \div 1440 \equal{} \boxed{1}$.


[quote="ronfler"]Does the AMC do handscoring? I answered 12 questions on the amc 12b and got them all right (i checked online), but my teacher said today I got a 73 or something. I did erase one question, but I thought I did a pretty good job, so I'm not sure what the deal is.[/quote]

You could've accidently shifted down a row and bubbled eveything incorrectly.  That would be the worst. :(

[quote="ronfler"]Does the AMC do handscoring? I answered 12 questions on the amc 12b and got them all right (i checked online), but my teacher said today I got a 73 or something. I did erase one question, but I thought I did a pretty good job, so I'm not sure what the deal is.[/quote]They will, on request, rescore your answer form.  There is a $\$5$ fee.  Your teacher should have a copy of the rescoring form on page 12 of his/her AMC Teachers' Manual, or you can download one from [url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/TM06hs.html[/url].


function
limit
induction
algebra proposed
algebra

Find all continuous functions $f: \mathbb R \to \mathbb R$ in each of the following cases (each equation is supposed to be true for all reals $x$):
1) $f(f(x)) \equal{} 3f(x) \minus{} 2x$.
2) $f(f(x)) \equal{} f(x) \minus{} x$.
3) $f(f(f(x))) \equal{} 2f(f(x)) \plus{} f(x) \minus{} 2$.
4) $f(f(f(x))) \equal{} 4f(x) \plus{} 1$.

[quote="Allnames"]A.Find all continuous  functions satisfying:
1)
\[ f: \mathbb R \to \mathbb R
\]

\[ f(f(x)) \equal{} 3f(x) \minus{} 2f(x)
\]
[/quote]

Maybe it is $ 3f(x)\minus{}2x$ instead of $ 3f(x)\minus{}2f(x)$ ??

I have just edited, pco!. The first seems to be solved by the same method you used in the link above. But I can't pass the step when $ f(x)$ is a increasing function ( I had another proof, but long and very ugly).
Especially. I want you to help me solve the fourth (  I have not solved it, till now)

[quote="Allnames"]A.Find all continuous  functions satisfying:
4)
\[ f: \mathbb R \to \mathbb R
\]

\[ f(f(f(x))) = 4f(x) + 1
\]
[/quote]

1) some infos on $ f(\mathbb R)$
We obviously have $ f(f(x))=4x+1$ $ \forall x\in f(\mathbb R)$
If $ a>-\frac 13\in f(\mathbb R)$, then $ 4a+1>a$ and $ 4a+1\in f(\mathbb R)$ and so $ [a,+\infty)\subseteq f(\mathbb R)$
For same reason : If $ b<-\frac 13\in f(\mathbb R)$, then $ (-\infty,b]\subseteq f(\mathbb R)$

So :
Either $ f(\mathbb R)=\mathbb R$
Either $ f(\mathbb R)=\{-\frac 13\}$
Either $ f(\mathbb R)=[a,+\infty)$ or $ (a,+\infty)$ for some $ a\geq -\frac 13$
Either $ f(\mathbb R)=(-\infty,a]$ or $ (-\infty,a)$ for some $ a\leq -\frac 13$

2) $ f(x)$ is injective, so monotonous, for $ x\in f(\mathbb R)$
This is an immediate consequence of $ f(f(x))=4x+1$ $ \forall x\in f(\mathbb R)$

As first consequence, if $ f(\mathbb R)=[a,+\infty)$ or $ (a,+\infty)$ for some $ a\geq -\frac 13$, $ f(x)$ is increasing over $ (a,+\infty)$.
Else, $ lim_{x\to +\infty}f(x)=L\geq a$ and $ f(f(x))=4x+1$ would be impossible when $ x\to +\infty$

As second consequence, $ f(\mathbb R)=(-\infty,a]$ or $ (-\infty,a)$ for some $ a\leq -\frac 13$, $ f(x)$ is increasing over $ (-\infty,a)$.
Else, $ lim_{x\to -\infty}f(x)=L\leq a$ and $ f(f(x))=4x+1$ would be impossible when $ x\to -\infty$

3) Solutions
3.1) $ f(\mathbb R)=\{-\frac 13\}$
Then, obviously, $ f(x)=-\frac 13$ $ \forall x$
And this indeed is a solution.

3.2) $ f(\mathbb R)=\mathbb R$
Then $ f(x)$ is a continuous monotonous bijection and $ f(f(x))=4x+1$ $ \forall x\in\mathbb R$
This is a classical equation with infinitely many continuous solutions (including trivial $ 2x+\frac 13$ and $ -2x-1$)
Solutions may be built piece by piece (I'm not quite sure for decreasing case).

3.3) $ f(\mathbb R)=[a,+\infty)$ or $ (a,+\infty)$ for some $ a\geq -\frac 13$
Choose any $ b\in (a,4a+1)$ and buid (classical piece by piece method) an increasing continuous fonction from $ [a,+\infty)$ $ \to[b,+infty)$
Choose any values for $ f(x)$ when $ x\leq a$ in respect with rules :
$ f(x)$ is continuous
$ f(a)=b$
$ [a,b)$ (or $ (a,b)$) $ \subseteq f((-\infty,a])$

Example :
$ f(\mathbb R)=[0,\+\infty)$ with :
$ f(x)=2x+\frac 13$ $ \forall x\geq 0$
$ f(x)=\frac{1+\sin(x)}{3}$ $ \forall x\leq 0$

3.4) $ f(\mathbb R)=(-\infty,a]$ or $ (-\infty,a)$ for some $ a\leq -\frac 13$
Same construction as in 3.3 above.

[quote="Allnames"]A.Find all continuous  functions satisfying:
1)
\[ f: \mathbb R \to \mathbb R
\]

\[ f(f(x)) \equal{} 3f(x) \minus{} 2x
\]
[/quote]
1) $ f(x)$ is an increasing bijection
$ f(f(x))\equal{}3f(x)\minus{}2x$ immediately implies that $ f(x)$ is injective, so monotonous.
Since $ f(x)$ is continuous monotonous, $ f(f(x))$ is continuous increasing, so is $ f(f(x))\plus{}2x$ and so is $ f(x)$
If $ \exists L$ such that $ \lim_{x\to \plus{}\infty}f(x)\equal{}L_1$, then $ f(f(x))\equal{}3f(x)\minus{}2x$ would be false when $ x\to \plus{}\infty$
If $ \exists L$ such that $ \lim_{x\to \minus{}\infty}f(x)\equal{}L_2$, then $ f(f(x))\equal{}3f(x)\minus{}2x$ would be false when $ x\to \minus{}\infty$
So $ f(\mathbb R)\equal{}\mathbb R$ and $ f(x)$ is surjective.
Q.E.D.

2) If $ f(a)\equal{}b$, then $ f(2a\minus{}b)\equal{}2a\minus{}b$
If $ a\equal{}b$, this is immediate.
If $ f(a)\equal{}b\neq a$, let the sequence $ u_n\equal{}2a\minus{}b\plus{}(b\minus{}a)2^n$ $ \forall n\in \mathbb Z$
It's easy to see that $ u_{n\plus{}2}\equal{}3u_{n\plus{}1}\minus{}2u_n$ and, thru induction,  that $ f(u_n)\equal{}u_{n\plus{}1}$ $ \forall n\in\mathbb Z$ (caution : for $ n<0$, we need first to show that $ f(x)$ is a bijection and we did it in step 1).

Then $ f(2a\minus{}b\plus{}(b\minus{}a)2^n)\equal{}2a\minus{}b\plus{}(b\minus{}a)2^{n\plus{}1}$ and, setting $ n\to \minus{}\infty$, we get the result.

3) if $ f(a)\equal{}a$ and $ f(b)\equal{}b$ for some $ a<b$, then $ f(x)\equal{}x$ $ \forall x\in [a,b]$
Let $ c\in(a,b)$ and consider the line $ g(x)\equal{}2x\minus{}c$
Since $ f(a)\equal{}a$, $ f(b)\equal{}b$ and $ f(x)$ and $ g(x)$ are continuous, and since $ c\in(a,b)$ $ \exists x_0$ such that $ f(x_0)\equal{}g(x_0)$
So $ f(x_0)\equal{}2x_0\minus{}c$ and then, using step 2 above, $ f(2x_0\minus{}(2x_0\minus{}c))\equal{}2x_0\minus{}(2x_0\minus{}c)$ and so $ f(c)\equal{}c$
Q.E.D.

4) Solutions
Let $ A\equal{}\{x$ such that $ f(x)\equal{}x\}$
Using step 2, we know that $ A\neq\emptyset$
Using step 3, we know that :
Either $ A\equal{}\mathbb R$
Either $ A\equal{}(\minus{}\infty,b]$ for some real $ b$
Either $ A\equal{}[a,\plus{}\infty)$ for some real $ a$
Either $ A\equal{}[a,b]$ for some real $ a\leq b$

4.1) $ A\equal{}\mathbb R$
So $ f(x)\equal{}x$ $ \forall x$ and this indeed is a solution (always check back)

4.2) $ A\equal{}(\minus{}\infty,b]$ for some real $ b$
So $ f(x)\equal{}x$ $ \forall x\leq b$ and $ f(x)\neq x$ $ \forall x>b$
Let then $ u>b$ and $ v\equal{}f(u)$
Using now step 2, we have $ f(2u\minus{}v)\equal{}2u\minus{}v$ and so $ 2u\minus{}v\leq b$
But, using the sequence $ u_n\equal{}2u\minus{}v\plus{}(v\minus{}u)2^n$ as defined in step 2, we have :
$ u_n\equal{}2u\minus{}v\plus{}(v\minus{}u)2^n$
$ u_{n\plus{}1}\equal{}2u\minus{}v\plus{}(v\minus{}u)2^{n\plus{}1}$
and so $ f(u_n)\equal{}2u_n\minus{}(2u\minus{}v)$
So, if $ 2u\minus{}v<b$, setting $ n\to \minus{}\infty$, we get some $ u_n$ as near of $ 2u\minus{}v$ as we want with $ f(u_n)\neq u_n$
And this is impossible since $ f(x)\equal{}x$ $ \forall x\leq b$
So $ 2u\minus{}v\geq b$
Hence $ 2u\minus{}v\equal{}b$
And so $ f(x)\equal{}2x\minus{}b$ $ \forall x\geq b$
and this indeed is a solution (checking back)

4.3) $ A\equal{}[a,\plus{}\infty)$ for some real $ a$
Using the same method than in 4.2 above, we get :
$ f(x)\equal{}x$ $ \forall x\geq a$
$ f(x)\equal{}2x\minus{}a$ $ \forall x\leq a$

4.4) $ A\equal{}[a,b]$ for some real $ a\leq b$
Using the same method than in 4.2 above, we get :
$ f(x)\equal{}2x\minus{}a$ $ \forall x\leq a$
$ f(x)\equal{}x$ $ \forall x\in\mathbb[a,b]$
$ f(x)\equal{}2x\minus{}b$ $ \forall x\geq b$
(if $ a\equal{}b$, we have the solution $ f(x)\equal{}2x\minus{}a$ $ \forall x$)

[quote="Allnames"]A.Find all continuous  functions satisfying:
2)
\[ f: \mathbb R \to \mathbb R
\]

\[ f(f(x)) \equal{} f(x) \minus{} x
\]
[/quote]

$ f(x)$ is injective and continuous, so monotonous.
So $ f(f(x))$ is increasing, so $ f(f(x))\plus{}x$ is increasing, so $ f(x)$ is increasing.

$ f(f(0))\equal{}f(0)$ and so $ f(0)$ is a fixed point of $ f(x)$
$ f(u)\equal{}u$ $ \implies$ $ u\equal{}0$ and the only fixed point of $ f(x)$ is $ 0$. So $ f(0)\equal{}0$

Let then $ x>0$, so $ f(f(x))\equal{}f(x)\minus{}x<f(x)$
But, since $ f(x)$ is increasing, $ f(f(x))<f(x)$ $ \implies$ $ f(x)<x$ and so $ f(f(x))\equal{}f(x)\minus{}x<0$, but this is in contradiction with $ f(x)$ increasing and $ f(0)\equal{}0$

So no solution.


Oh, that makes sense. Thank you saltandpeppapig! ;)


geometry
similar triangles
geometry unsolved

[quote="CDP100"]ABCD is a square and J is arbitrarily point from AB. If <BFC = <DFG, prove that BG = BG.[/quote]
I think you mean $ BJ \equal{} BG,$ but it is true only when $ \angle BFC \equal{} \angle DFG \equal{} 90^{o}.$ That is, only when the point $ F,$ is the orthogonal projection of $ B,$ on the line segment $ CJ.$

So, from the similar triangles $ \bigtriangleup BFG,$ $ \bigtriangleup CFD,$ we have $ \frac {BG}{CD} \equal{} \frac {BF}{CF}$ $ ,(1)$

From the similar right triangles $ \bigtriangleup BCG,$ $ \bigtriangleup FCB,$ we have $ \frac {BF}{CF} \equal{} \frac {BJ}{BC}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ BG \equal{} BJ$ $ ($ because of $ BC \equal{} CD$ $ )$ and the proof is completed.

Kostas Vittas.

PS. In general case of $ \angle BFC \equal{} \angle DFG \neq 90^{o},$ the result as the problem states, is not true.


calculus
derivative
function
real analysis
real analysis unsolved

If $ x_1$ and $ x_2$ are two consecutive roots of a differentiable function f(x), is it always true that we can not have $ f'(x_1) f'(x_2) > 0$?

Which is why I have asked to prove that $ f'(x_1) f'(x_2) >0$ cannot hold

I have a simplistic proof that goes:

If either $ f'(x_1) \equal{} 0$ or $ f'(x_2) \equal{} 0$, we are done. Suppose both are non-zero

WLOG assume f(x)<0 in the interval $ (x_1,x_2)$ It cannot change signs as otherwise there will be a root between $ x_1$ and $ x_2$ contradicting the assumption that $ x_1$ and $ x_2$ are consecutive roots.

Then we must have $ f'(x_1)>0$ as by definition $ f'(x) \equal{} \frac{f(x\plus{}h)\minus{}f(x)}{h}$ and $ f(x) \equal{} 0$ at $ x \equal{} x_1$

In a similar way $ f(x_2)<0$ can be proved and the result follows.

I just want to know whether there is a smarter way to prove this

[quote="hsbhatt"]I swear there was a post just before my previous post. It seems to have disappeared now  :oops:[/quote]

You're right, I  began to answer and then I notice the " ...can NOT have ..". Sorry!

same as your proof but i like the sound of contradiction more:

assume $ f'(x_1)f'(x_2)>0$

then case 1: $ f'(x_1)>0$ and $ f'(x_2)>0$

gives us $ y_1\in (x_1, x_2)$ with $ f(y_1)>f(x_1)\equal{}0$
and $ y_2\in (x_1, x_2)$ with $ f(y_2)<f(x_2)\equal{}0$

since f is continuous there exists a zero in $ (y_1,y_2) \subset (x_1,x_2)$ contradiction

case 2: $ f'(x_1)<0$ and $ f'(x_2)<0$
similar argument.


I've just seen (at imo-official.org) that the 51st IMO (Kazakhstan 2010) now has an official date (July 2nd-15th), website (though with no information at all available yet), and logo.

You can check them out here: http://www.imo-official.org/year_info.aspx?year=2010

Official site IMO 2010:

http://www.imo2010org.kz

Good Luck!!!

/ Let's Go  :D /


[quote="cooly291"]I am having problems identifying which of the following (hyperbola, parabola, ellipse, or circle) equations are which and putting them in standard form:

[img]http://img.photobucket.com/albums/v486/cooly291/math12.jpg[/img][/quote]

Is this for your homework/summer assignment?
[hide=" :P "]Here's the first one:
$\frac{x^{2}}{7^{2}}+\frac{y^{2}}{(\frac{7}{\sqrt{13}})^{2}}=1$[/hide]

[quote="mathnerd314"][hide="I hope I can give a minor hint"]
Think about the standard forms of the equations and try to get these equations into those forms.
[/hide][/quote]
Don't look at the following if you only need a hint.
[hide="1"]ellipse[/hide]
[hide="2"]parabola[/hide]
[hide="3"]hyperbola[/hide]

[quote="cooly291"]I am having problems identifying which of the following (hyperbola, parabola, ellipse, or circle) equations are which and putting them in standard form:

[img]http://img.photobucket.com/albums/v486/cooly291/math12.jpg[/img][/quote]

#1, move y to other side and move the constant to RHS. Now it should be more obvious. If not, divide by 13 or whatever the number was.

#2, there is no $y^{2}$ term but theres an $x^{2}$, so what is it?

#3, you get something like $x^{2}-y^{2}$ or $y^{2}-x^{2}$, what type of conic section is written that way?

[quote="cooly291"]I am having problems identifying which of the following (hyperbola, parabola, ellipse, or circle) equations are which and putting them in standard form:

[img]http://img.photobucket.com/albums/v486/cooly291/math12.jpg[/img][/quote]
[quote="mathnerd314"][quote="Inspired By Nature"]Don't look at the following if you only need a hint.
[hide="1"]ellipse[/hide]
[hide="2"]parabola[/hide]
[hide="3"]hyperbola[/hide][/quote]

Consider providing solutions if you're just going to post the answers. Thanks.[/quote]
sure. :)
In the first one,
$x^{2}+13y^{2}=49$
-One can tell this looks like the equation for a circle, but the y term is changed thus making the circle longer horizontally. So it's not a circle, its an ellipse.
-The second one is very obvious as shoefactory said.
-In the last one, notice the negative sign between the first two terms. That's our tip-off that tells us it's a hyperbola. Another method (longer) is as follows:
we multiply the given equation by 4 and expand,
$(y-5)^{2}-4(x+1)^{2}=16$
$-4x^{2}+0xy+y^{2}-8x-10y+5=0$
Compare this to the general form of a conic:
$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$
In our case, thankfully B=0
I don't know if you know this, but if AC is negative, then the conic is a hyperbola. Looking at our equation above,
$(-4)(1)=-4$. It is negative, thus the conic is a hyperbola.
I hope that helps. :)

it's very simple
1) you just I solate the variables on one side of the equation, this will most likely look like the euation of a circle. but since the equation has to be equal to 1 in a conics section it becomes a Ellipse.
2) since theres is only one varible to a second power, once can assume once in standard form this will be equivalent to a hyperbola.
3) the Third one, once you make the equation equal to 1 you see the negative sign between the expression, therefore hinting that it must absolutely be a hyperbola, no questions asked.

[quote="kaca123"]
2) since theres is only one varible to a second power, once can assume once in standard form this will be equivalent to a hyperbola.[/quote]
are you sure? It's actually a parabola.


function limit calculus derivative integration inequalities

Let $f$ be continous function on $(a,b)$ such that for every $x \in (a,b)$ then
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}=g(x)\]

and $g(x)$ is finite. Show that if $g(x) \geq 0$ for all $x \in (a,b)$ then $f$ is increasing. And If $g(x)=0$ for all $x\in (a,b)$ then $f$ is constant.

[quote="Beginner"]Let $f$ be continous function on $(a,b)$ such that for every $x \in (a,b)$ then
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}=g(x) \]
and $g(x)$ is finite. Show that if $g(x) \geq 0$ for all $x \in (a,b)$ then $f$ is increasing. And If $g(x)=0$ for all $x\in (a,b)$ then $f$ is constant.[/quote]

Realize that $\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}= g(x)$ using the limit composite rule. This is the derivative. So if $f'(x)$ exists and $f'(x) \geq 0$ then the function is non-increasing (or decreasing in some texts). And if $f'(x) = 0$ then the function is constant. This is a well known result, are you asking for a proof. If you are, have you tried used the Mean Value Theorem?

[b]Theorem:[/b] Let $f$ be differentiable on $(a,b)$ if $f' \geq 0$ then $f$ is a non-increasing function.

[b]Proof:[/b] Let $a<x_{1}<x_{2}<b$ be two real numbers. Then $f$ is continous on $[x_{1},x_{2}]$ and differentiable on $(x_{1},x_{2})$ because by hypothesis it is differentialble on the full interval $(a,b)$. Then by MVT we have:
$\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}= f'(c) \geq 0$ because $f'(c)\geq 0$.
That means,
$f(x_{2})-f(x_{1}) = f'(c)(x_{2}-x_{1}) \geq 0$.
Thus,
$f(x_{2}) \geq f(x_{1})$

Obviously the existence of the limit of this special difference quotient does not imply the existence of the derivative. For example, take $f(x) = |x|$ in 0. By changing the value of $f$ in 0 you realize the condition does not even imply continuity.

Sylow Theory, I think Kalle Was Right. We can not assert the existense of derivative here. Tha't make the problem seems not trivial.

This isn't too bad if you can use Fatou's Lemma. Then if $a',b',h$ are chosen so $a<a'-h<a'<b'<b'+h<b$,

\[\int_{a'}^{b'}g(x)dx \leq \lim_{h \rightarrow 0}\frac{1}{2h}\int_{a'}^{b'}f(x+h)-f(x-h)dx = \lim_{h \rightarrow 0}\frac{1}{2h}\int_{b'-h}^{b'+h}f(x)dx-\frac{1}{2h}\int_{a'-h}^{a'+h}f(x)dx\ = f(b')-f(a') \]
where we use Fatou, then change of variables, then the continuity of f. Of course then $g\geq 0$ implies monotonicity, and reversing all the inequalities gives the other direction if $g=0$.

Not quite true. For Fatou you need to know some lower bound on $\frac{f(x+h)-f(x-h)}h$ and you have none ;) The problem is much simpler than you think :P. Here is a small [hide="hint"]Without loss of generality, $\int_{a}^{b}f=0$ (just subtract an appropriate constant from $f$). Let $F$ be the antiderivative of $f$ such that $F(a)=F(b)=0$. If $F$ attains its maximum at some intermediate point and that maximum is positive...[/hide]

fedja, what is your proof? If $ F$ doesn't attain positive maximum at intermediate point, then $ f(a) \le 0 \le f(b)$. but I can't figure out what happen if $ F$ have one.

BTW here is a different approach.
[url=http://www.dpmms.cam.ac.uk/~wtg10/meanvalue.html]an article on mean value theorem[/url] gives two kinds of arguments to extend a local behavior of a function on an interval to the corresponding global behavior: the first one is to consider the first point in the interval where something goes wrong and the second one is to use Heine-Borel theorem.

Let's try the first one. (the second one doesn't work). Suppose that $ g(x)$ is "positive" for all $ x$ and that there exists two points $ c$ and $ e$ with $ c<e$ such that $ f(c)>f(e)$ and let $ d$ be the first point in the interval $ [c,e]$ where something goes wrong. A naive approach would be to define $ d = \inf\{ x \in [c,e] : f(c)>f(x)\}$, but this definition doesn't guarantee that $ c<d$. Instead define $ d = \inf\{ x \in [c,e] : \frac{f(c)+f(e)}{2}> f(x) \}$. Then continuity of $ f$ at $ c,d,e$ ensures that $ c<d<e$ and $ f(d)=\frac{f(c)+f(e)}{2}$. Look at the left side and the right side of $ d$. We have $ f(x) \ge \frac{f(c)+f(e)}{2}$ on $ [c,d]$ and $ f(x) < \frac{f(c)+f(e)}{2}$ on some decreasing sequence of points on $ (d,e)$ converging to $ d$. Therefore $ g(d) \le 0$. Contradiction.

(It was enough to work only the case that $ g$ is positive for all $ x$: If $ g$ is nonnegative for all $ x$, just replace $ f$ with $ f(x)+\epsilon x$ where $ \epsilon >0$ so that $ g \ge \epsilon$.)

Here goes. Let's start with the simple trick of adding $ \varepsilon x$ to $ f$. This will allow us to assume that the limit is strictly positive rather than just non-negative. After that, suppose that $ F$ attains its maximum at some point $ y\in(a,b)$. Since $ g(y)>0$, we have $ f(y+h)>f(y-h)$ for all $ h\in (0,H)$ if $ H$ is small enough. But then $ F(y-H)+F(y+H)-2F(y)=\int_{0}^{H}[f(y+h)-f(y-h)]\,dh>0$, so either $ F(y-H)>F(y)$, or $ F(y+H)>F(y)$ - a contradiction.


trigonometry
calculus
derivative
function
algebra solved
algebra

Solve the following system in $\mathbb {R}^3$:
$\newline x+\sin x=y
\newline y+\sin y=z
\newline z+\sin z=x$.

It was proposed for the $10$'th grade, so most of the students wouldn't have done this, but $(x+\sin x)'=1+\cos x\ge 0$, so $x+\sin x$ is, in fact, strictly increasing (the derivative is $0$, but only in isolated points). This means that $x>y\Rightarrow y>z\Rightarrow z>x\Rightarrow$ contradiction, so $x=y=z$ and we get $\sin x=0$.

Fo the 10-th grade, you can show it is a non-decreasing function like this: $\Rightarrow x-y\geq \sin \frac {y-x}2 \cos {y+x}2$, and $\cos {x+y}2<\leq 1$ and  $\sin \alpha \leq \alpha$, for $\alpha \in [0,1]$.


On the planet MathMar, the time system is based on a 5 hour clock. There 25 minutes in an hour. It is now 3:20 on MathMar. What is the measure of the smaller angle formed by the two hands? Express your answer as a decimal to the nearest tenth.

[hide]

360 degrees on the clock, and 25 minutes, so $\frac{360}{25}=14.4$ degrees for each minute. The minute hand has passed through $14.4 \times 20 = 288$ degrees. The hour hand has passed through 3 hours and 20 twenty-fifths of an hour. 25 minutes in an hour, so it takes 5 minutes for the hour hand to go from one hour to the next. $3 \times 5 \times 14.4 + \frac{20}{25} \times 14.4 \times 5 =
273.6$. The difference is 288 - 273.6 = 14.4 degrees.
[/hide]

[hide="answer"]
360/25, too lazy to simplify =(
[/hide]

[quote="Monkey"][hide] I just made the equation:

72h+(72m/25)-(360m/25)= absolute value of degree measure. where h=the hour value and m=the minute value. Plugging in h=3 and m=20 gives [b]14.4 degrees [/b]as the absolute value.[/hide][/quote]

Basically what I did:

[hide]
Number of hours is $h$, minutes is $m$.
\[[\frac{360}{25}(m)] - [\frac{360}{5}(h) + \frac{360}{5} \times \frac{m}{25}]\],
which simplifies to:
\[[\frac{72}{5}(m)] - (72h + \frac{72m}{25})\], or
\[\frac{72m}{5} - (72h + \frac{72m}{25})\]
\[\frac{288m}{25} - 72h\]
[/hide]


abstract algebra
superior algebra
superior algebra unsolved

10.5.27. suppose $f : A \rightarrow M$, $g : B \rightarrow M$ are $R$-module homomorphisms

prove that $X = \{(a,b) | a \in A, b \in B, f(a) = g(b)\}$ is an $R$-submodule of the direct sum $A \oplus B$.

i'm having trouble showing the existence
how do I prove existence?

Existence of what? The zero element is in $X$, and $\{(0,0)\}$ is a submodule.


I have a few algebra problems, that I haven't been able to solve I hope you guys can help me.

I made the exercises in word and then took a screenprint of it cos it was a little too hard to just type it here.

[img]http://img.photobucket.com/albums/v37/Laetitia/hogeremachtswortel.jpg[/img]

What do you want us to do with these expressions? (If it's just simplyfing them... it doesn't probably belong in the olympiad forum, but I'm sure somebody, perhaps me, will be glad to help.)

How do you want them simplified? Fractional exponents or radical symbols? Negative exponents or fractional expressions?

Either way, converting to fractional exponents as a first step makes life much easier.

Not sure what you mean, since i'm not a native english speaker, if I would give you the solutions to what it should be according to the book, would that solve your question. Although the book has been wrong on some moments

I got the solution but I got no idea how to get there
[quote="Katerina"]
[img]http://img.photobucket.com/albums/v37/Laetitia/hogeremachtswortel.jpg[/img][/quote]

ok don't worry about the first one, that one I solved but the solution was wrong in the book, so I forgot I already had it

[img]http://img.photobucket.com/albums/v37/Laetitia/sol.jpg[/img]

[quote="leoxnlin"]I can't quite tell what is in the exponent of the second problem. Could you clarify that part?

Were those the answers from your book? Unfortunately, none of them seem quite right to me.[/quote]
She meant the following, maybe?:

1. $ \sqrt[3]{\sqrt[5]{3a^{10}}}$

2. $ \left(\frac12\sqrt[3]{4}\right)^{1\plus{}\frac12}$

3. $ \minus{}3a^5(\minus{}a)^5\plus{}4(\minus{}a^5)^2\plus{}\frac{(3a^5)^2}{4a\sqrt{a}}$

4. $ \frac{x^2}{\sqrt[3]{x^4}}$

5. $ (2q\sqrt[3]{q^2})^4$

If that is so, then:

1. $ \sqrt[3]{\sqrt[5]{3a^{10}}}\equal{}\left(\left(3a^{10}\right)^\frac15\right)^\frac13\equal{}\left(3a^{10}\right)^\frac1{15}\equal{}3^\frac1{15}a^\frac23$

2. $ \left(\frac12\sqrt[3]{4}\right)^{1\plus{}\frac12}\equal{}\left(2^{\minus{}1}2^\frac23\right)^\frac32\equal{}\left(2^{\minus{}\frac13}\right)^\frac32\equal{}2^{\minus{}\frac12}$

3. $ \minus{}3a^5(\minus{}a)^5\plus{}4(\minus{}a^5)^2\plus{}\frac{(3a^5)^2}{4a\sqrt{a}}\equal{}7a^{10}\plus{}\frac94a^\frac{17}2$

4. $ \frac{x^2}{\sqrt[3]{x^4}}\equal{}x^{2\minus{}\frac43}\equal{}x^\frac23$

5. $ (2q\sqrt[3]{q^2})^4\equal{}2^4q^\frac{20}3$

[quote="leoxnlin"]I can't quite tell what is in the exponent of the second problem. Could you clarify that part?

Were those the answers from your book? Unfortunately, none of them seem quite right to me.[/quote]

Yeah it's from a book I got to revise my math for uni next september
I've found some mistakes in the answers before, but these ones I couldn't get solved

the exercices are like 10000th User put them

I agree with 10000th User's answers. Thus the answers Katerina posted are
2) Wrong, unless you're going to tell me that it means $ \displaystyle \frac{1}{\sqrt{2}}$, which would make the answer Katerina posted a rather questionable notation...
3) At first I thought this one was wrong until I realized that the "8" in Katerina's post was probably meant as an exponent rather than the index of the radical.
4) Wrong, however you look at it.
5) Wrong, however you look at it.

I was looking at the solutions this afternoon and the one worked out for number 1, 4, 5 were the same as what I had worked out but with the solutions being wrong from the book I wasn't sure.

However the 2nd and the 3rd can be written as the solution in the book.

2. $ (\frac12\sqrt [3]{4})^{1 + \frac12} = (2^{ - 1}2^\frac23)^\frac32 = (2^{ - \frac13})^\frac32 = 2^{ - \frac12}
=\sqrt{2^{-1}}=\sqrt{\frac12\frac22}=\sqrt{\frac14\frac21}=\frac12\sqrt{2}$

I got to this one supposing that the guy who wrote the book didn't want any fractions or negative exponents under the root, that after two days looking for how he could have gotten to that solution  :P

3. $ - 3a^5( - a)^5 + 4( - a^5)^2 + \frac {(3a^5)^2}{4a\sqrt {a}} = 7a^{10} + \frac94a^\frac {17}2
= 7a^{10} + 2\frac14a^8\sqrt{a}$

I'm not sure whether it is a common way to do it but in the book fractions get simplied so $ \frac54 = 1\frac14$

Thank you all for your help


Where do you get your numbers in your usernames from?

6 and 28 are perfect numbers. See if you can derive my AIM sn :wink:

mines is my only elementary school Connect2School account number: mxia22. (now it's different)

[quote="33286"]It spells my name in base 36.[/quote]

According to 'chgbase'... your name is "POM"?
Or perhaps "5181702"?

The letter "e" on my keyboard is broken, and I need 3's instead?

[quote="Brut3Forc3"]Th[b]e[/b] l[b]e[/b]tt[b]e[/b]r "[b]e[/b]" on my k[b]e[/b]yboard is brok[b]e[/b]n, and I n[b]ee[/b]d 3's inst[b]e[/b]ad?[/quote]

???

I just needed a short Username.

from now on in this post, raised to the dynamo power means reverse the digits. Here we go:

(Key: $ x$ $ \equal{}$ my favorite number, $ y$ $ \equal{}$ my basketball jersey number, $ z$ $ \equal{}$ my second favorite number, $ a$ $ \equal{}$ $ \frac {1} {29}$ of my state score.)

$ ((xy^{z}\plus{}a)^{dynamo})(x^{z}))^{dynamo} .$

....

or just $ x^6$ :D

well, this sounds nerdy, but note: i am not a nerd.

I used to play runescape and i had an account named hyperhavoc. that one got hacked, and i couldnt get into it. then i made hyperhavoc2. and that happend over and over, till i got hyperhavoc6. but 6 was too random, and i got 5 back. so now my username is hyperhavoc5. lol  :rotfl:  :rotfl:  :rotfl:

[quote="hyperhavoc5"]well, this sounds nerdy, but note: i am not a nerd.

[/quote]

note to self: hyperhavoc5 is not a nerd

got it dude.

[quote="33286"][quote="Yongyi781"]
According to 'chgbase'... your name is "POM"?[/quote]

Well, my nickname. It also should not be written in all caps.

I got my state writing test scores back recently (the test itself was three month ago) and they wrote my name in all caps all the time for some reason. They appear to have done this to everyone.[/quote]
pom aka eric?

94 is my birth yr, quickster is something from spongebob...

142857 is my favorite number.

142857*1=142857
142857*2=285714
142857*3=428571
142857*4=571428
142857*5=714285
142857*6=857142
142857*7=999999

the digits recurs when multiply by 1-6.

it also fit when it is mutiplied by other number if the first few digit is added to the last 6 numbers.

eg. 142857*8=1142856
when the first 1 is added to the 6, it will become 142857 again.

eg2. 142857*9=1285713
when the first 1 is added to the 3, it will become 285714 again.

[quote="Klebian"]
pom aka eric?[/quote]
Yes. Also, just because I said it shouldn't be in all caps doesn't mean it should be in all lowercase.   :D

what numbers?   :P

I don't want this to be spam.

I know, I read the title.  I don't have any numbers in my username.  I dislike having to explain failed jokes.  It's awkward.

Anyways, I'll stop posting here, lest it deteriorate into spam.

well it's not really going into spam.
but isnt 82 a cool number?


geometry
perimeter
combinatorics proposed
combinatorics

[quote="mdk"]The sides of a regular triangle are bicolored. Do there exist on its perimeter monochromatic three vertices of a regular triangle?[/quote]

It seems not :

Color A and (BC) (Side BC without points B and C) with colour 1 and the reminder of the perimeter with colour 2.

Obviously it does not exist on ABC's perimeter any three monochromatic vertices of a regular triangle :
- Triangles having three vertices on the same side of original ABC  are not regular  :lol:
- Triangles having two vertices on one side of ABC need two have one vertex on A, B or C in order to be regular. Then :
-- triangle ABC is non monochromatic (A is colour 1 and B and C are colour 2)
-- triangles AXY (different from ABC) are non monochromatic (A is colour 1, and (AB) is colour 2)
-- triangles BXY (different from ABC) are non monochromatic (B is colour 2, and (BC) is colour 1)
-- triangles CXY (different from ABC) are non monochromatic (C is colour 2 and (BC) is colour 1)
- Triangles having one vertex on (AB), one vertex on (AC) and one vertex on (BC) are non monochromatic ((AB) is colour 2 and (BC) is colour 1)


I wondered if anyone had a neat solution for this?

how many numbers there exist with sum of the digits equal to n, in which no two consecutive digits are the same?

(so for 4 we have 121,31,13,4,1021,10201, etc... but not 400,1111,22,...)

Edit: I managed to reduce this problem to an elementary combinatorics problem, still I'm missing one point: If you got $n$ ones and $m$ zeros, how many permutations are there in which we have $x$ consecutive ones?

hey, Peter VDD is back !

your problem says me nothing.

No, it's a pretty stupid problem

too bad I didn't see the problem for it yesterday though I thought it couldn't be hard

just seemed to have some blackout :?

Anyway, yes here I'm back :lol: but only for a couple of days, I'm leaving for holidays on monday.


A point $P$ is joined to the vertices of triangle $ABC$ .The straight lines $AP=y$ , $BP=z$ , $CP=x$ are reflected at the angle bisectors passing through $A$ , $B$ , $C$ to $v$, $w$, $u$ respectively .Prove that $u,v,w$ pass through  one point $Q$ .

Theorem 1 at http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 . Theorem 2 at http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 . For a really challenging generalization, see http://www.mathlinks.ro/Forum/viewtopic.php?t=68858 .

  darij

The following problem is quite similar .
Three lines $x,y,z$ are incident with a point $P$ and are orthogonal to the sides $c,a,b$ of a triangle $ABC$ . Now $x,y,z$ are reflected at the midpoints of $c,a,b$ to $u,v,w$ respectively .Prove that $u,v,w$ pass also through one point $Q$ .

There exist also a solution for this with isogonal i think .


i am  pretty confused with the steps to take in this problem, i do not understand what to do with the -3/5 rght at the beginning. thanx for any help

-3/5 - 22/20 + 9/10=

this may be easy for some people but i am trying to learn. thanx again....

Do you mean you don't understand how to take the negative of a fraction?

For any a and b:

$ \frac{\minus{}a}{b}\equal{}\minus{}(\frac{a}{b})$

$ \frac{ a}{\minus{}b}\equal{}\minus{}(\frac{a}{b})$

$ \frac{\minus{}a}{\minus{}b}\equal{}\frac{a}{b}$

Sorry about the ugly LaTeX, I can't remember how to increase the height of the parentheses.

If you mean you don't know how to add/subtract, them, just take common denominators.

In other words, if you want to find $ \frac{a}{b}\plus{}\frac{c}{d}$, by taking common denominators, it is $ \frac{ad\minus{}bc}{bd}$.

The -3/5 is quite similar to subtracting 3/5.

So if I were to answer your problem, it would be as follows:
-3/5-22/20+9/10=-12/20-22/20+18/20=-14/20=-7/10

and there you have it. :)

$ \frac{\minus{}a}{b}\equal{}\frac{a}{\minus{}b}$ Just remember that ab = negative if either one is negative, and if both are the same, then it's positive. Similarly, the same applies for division. In case you didn't know, fractions are the same thing as division.

thanx, i saw the reply that had it -14/20,and i kept working it out wondering why i was getting -16/20...hahahaha..everyone makes mistakes..

thanx

[quote="Temperal"]
Sorry about the ugly LaTeX, I can't remember how to increase the height of the parentheses.
[/quote]
You just use \left( and \right) so the parentheses will match the height.
$ \left(\frac{n}{k}\right)$

Correct!
[code]
$\left(\frac{1}{2}\right)$
[/code]
produces
$ \left(\frac{1}{2}\right)$


In how many ways can you write a non - negative integer n as a sum of t non-negative integers ?

if I have understood your problem correctly,your asking for this:
[b]how many positive solutions does the equation $ x_{1}+...+x_{t}=n$ have?[/b]
[hide]the answer is $ n-1 \choose t-1$[/hide]


Gauss
calculus
derivative
algebra
polynomial
combinatorial geometry
China

Find the greatest $n$ such that $(z+1)^n = z^n + 1$ has all its non-zero roots in the unitary circumference, e.g. $(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0$ implies $|\alpha| = 1.$

n equals 7. For $n=7$ it is not difficult to see that all roots have modulus 1 ( use well-known trick to factorize $ (x+y)^7-x^7-y^7$). To prove this one is best, use Gauss theorem: the root of the derivative of a polynomial is in the convex hull of the roots of the polynomial. Thus, if all roots of a polynomial have modulus at most 1, so does the derivative of the polynomial. Just compute the derivative and find it's roots.


In the plane is given a circle with radius $R$. Two other circles with sum of radiuses equal to $R$, touch the circle internally. Prove that the line passing through the touch-points include one of the common points of those circles.

Let $C(O, R)$, $C_{1}(O_{1}, r_{1})$ and $C_{2}(O_{2}, r_{2})$ be the three circles, and let $P_{1}, P_{2}$ be the tangency points of $C_{1}, C_{2}$ respectively. Then $O, O_{1}, P_{1}$ and $O, O_{2}, P_{2}$ are collinear. Let $H$ be an intersection between $C_{1}$ and $C_{2}$. We have $O_{1}O = HO_{2}$ and $O_{1}H = OO_{2}$. So $OO_{1}HO_{2}$ is a paralleogram.

$\measuredangle P_{1}HO_{1}= \frac{180^{\circ}-\measuredangle P_{1}O_{1}H}{2}= \frac{\measuredangle OO_{1}H}{2}= \measuredangle P_{2}HO_{2}$.

Hence $\measuredangle P_{1}HO_{1}+\measuredangle P_{2}HO_{2}= \measuredangle OO_{1}H = 180^{\circ}-\measuredangle O_{1}HO_{2}$.


function
integration
limit
real analysis
real analysis unsolved

[quote="eugene"]Is there an entire function $f(x)$ such that $\int_{\mathbb{R}} (f(x))^2dx < + \infty$ and $\lim_{x\to \pm \infty}f(x) \neq 0$.[/quote]Let $g(z) = e^{-z^2}$, and consider $f(z) = \sum_{n \ne 1}g(n^2(z-n^2)).$ The series converges locally uniformly, and $\lim_{n \to \infty}f(n^2) = 1$. Also, $\int_{-\infty}^\infty |g(n^2 x)|^2 dx = \frac{c}{n^2}$, and thus $f$ is square integrable.

Edit: replaced the typo $g(x/n^2)$ with the correct version $g(n^2 x)$.

Thanks.

An interesting aproach is based upon this very nice Carleman's theorem http://www.mathlinks.ro/Forum/viewtopic.php?t=86116#.
From this theorem it is enough to find a function $f$, just being continious(not neccessary entire) which is in $L^2(\mathbb{R})$, then take any $\epsilon(t) \in L^2(\mathbb{R})$  such that $lim_{\infty}\epsilon(t) =0$, for example $\epsilon(t) = \frac{1}{1+t^2}$. And the theorem i mentioned gives us an entrie functuion which satisfies all the properties, albeit without an explicit construction.

[quote="eugene"]Thanks.

An interesting aproach is based upon this very nice Carleman's theorem http://www.mathlinks.ro/Forum/viewtopic.php?t=86116#..[/quote]It's clear that the answer of the question in this post is a consequence of Carleman's result. I think that Carleman's result can be proved with the same technique (approximate an arbitrary continuous function by a series of scaled Gaussians).


For $ i\equal{}1,2,\cdots ,7$, $ a_i$ and $ b_i$ are nonnegative numbers such that $ a_i \plus{} b_i \le 2$. Prove that there exist distinct indices $ i,j$ such that $ |a_i \minus{} a_j|\plus{}|b_i \minus{} b_j| \le 1$

Hint: Proof by Contradiction. Arrive at a contadiction $ a_i \plus{} b_i \ge 2$. Use $ |x\minus{}y| \le |x| \plus{}/\minus{} |y|$


function
calculus
derivative
analytic geometry
graphing lines
slope
limit

Find the tangent line of the cubic function $ y \equal{} x^3$ at the point of tangency $ (2009, 2009^3)$ [color=blue]without using differentiation[/color].

I'm gonna hide my solution since I used differentiation, but so you could check if my answer's right...cause i don't know how to solve it any other way..

[hide]$ y' \equal{} 3x^2$ so slope $ \equal{} 3*2009^2$
$ y \minus{} 2009^3 \equal{} 3*2009^2(x \minus{} 2009)$
$ y \equal{} 2009^2x \minus{} 2*2009^3$.[/hide]

Can we use $ \displaystyle\lim_{h\rightarrow0}\frac {f(x \plus{} h) \minus{} f(x)}{h}$?
If so, then it would be:
$ \displaystyle\lim_{h\rightarrow0}\frac {(x \plus{} h)^3 \minus{} x^3}{h}$
$ \displaystyle\lim_{h\rightarrow0}\frac {3x^2h \plus{} 3xh^2 \plus{} h^2}{h}$
$ \displaystyle\lim_{h\rightarrow0}3x^2 \plus{} 3xh \plus{} h$
$ \equal{} 3x^2$

Then just do all the work in my previous answer to get $ y \equal{} 2009^2x\minus{}2*2009^3$.

[quote="modularmarc101"]Can we use $ \displaystyle\lim_{h\rightarrow0}\frac {f(x \plus{} h) \minus{} f(x)}{h}$?
If so, then it would be:
$ \displaystyle\lim_{h\rightarrow0}\frac {(x \plus{} h)^3 \minus{} x^3}{h}$
$ \displaystyle\lim_{h\rightarrow0}\frac {3x^2h \plus{} 3xh^2 \plus{} h^2}{h}$
$ \displaystyle\lim_{h\rightarrow0}3x^2 \plus{} 3xh \plus{} h$
$ \equal{} 3x^2$

Then just do all the work in my previous answer to get $ y \equal{} 2009^2x$.[/quote]

same thing as differentiating.

[quote="kunny"]As you realize, we don't require the solutions. :wink:

Anyway, regrettably you made a mistake in calcultion of the slope. :)[/quote]

Oh i see it should be $ y \equal{} 3*2009^2x\minus{}2*2009$.

This is a polynomial, and polynomials bring their own techniques.

The tangent line is $ y \equal{} mx \plus{} b$. The criterion for tangency is that $ x^3 \minus{} mx \minus{} b \equal{} 0$ must have a [i]double[/i] root at $ x \equal{} 2009.$

Let me just set up a synthetic division, and force the condition $ 2009^3 \minus{} 2009m \minus{} b \equal{} 0.$

$ \begin{array}{ccccc}2009 & 1 & 0 & \minus{} m & \minus{} b \\
& & 2009 & 2009^2 & 2009^3 \minus{} 2009m \\
\hline
& 1 & 2009 & 2009^2 \minus{} m & 0\end{array}$

Now, insist that the quotient also be divisible by $ x \minus{} 2009$:

$ \begin{array}{cccc}2009 & 1 & 2009 & 2009^2 \minus{} m \\
& & 2009 & 2\cdot2009^2 \\
\hline
&1 & 2\cdot2009 & 3\cdot2009^3 \minus{} m\end{array}$

We get the condition that $ 3\cdot2009^2 \minus{} m \equal{} 0,$ or $ m \equal{} 2009^3.$ Then we get the same equation of the line that you see above.

That's right, in short we can write the equality $ x^3\equiv (x \minus{} 2009)^2Q(x) \plus{} R(x)$, where $ R(x)$ is the remainder with degree less thatn or equal to 1.

Now all you have to do is just to do the division. In the same way, we can find the tangent line of the quartic function $ y\equal{}x^4$ at the point of tangency $ (2,\ 16)$ without using differentiation.

A historical excursion:

Here is a method that Rene Descartes used to find the tangent line to the curve $ y\equal{}x^3$ at the point $ (a,a^3).$

Find a circle of radius $ r$ centered at $ (b,0)$ (forcing the center to be on the $ x$ axis is arbitrary but we have a degree of freedom to exploit).

We want the equations $ y\equal{}x^3$ and $ (x\minus{}b)^2\plus{}y^2\equal{}r^2$ to have a double root at $ x\equal{}a.$

This is some pretty serious polynomial algebra, but more or less modern algebraic notation was the new and powerful tool of the day, and Descartes knew how to exploit it.

The need for the circle to pass through $ (a,a^3)$ fixes $ r^2\equal{}a^6\plus{}(a\minus{}b)^2.$

Then the equation we want to have a double root is

$ x^6\plus{}(x\minus{}b)^2\equal{}a^6\minus{}(a\minus{}b)^2.$

$ x^6\minus{}a^6\plus{}(x\minus{}b)^2\minus{}(a\minus{}b)^2\equal{}0$

$ (x\minus{}a)(x^5\plus{}ax^4\plus{}a^2x^3\plus{}a^3x^2\plus{}a^4x\plus{}a^5)\plus{}[(x\minus{}b)\minus{}(a\minus{}b)][(x\minus{}b)\plus{}(a\minus{}b)]\equal{}0$

$ (x\minus{}a)(x^5\plus{}ax^4\plus{}a^2x^3\plus{}a^3x^2\plus{}a^4x\plus{}a^5\plus{}x\plus{}a\minus{}2b)\equal{}0.$

We need to be able to factor another factor of $ (x\minus{}a)$ out of that second factor, which means that that second factor must be zero when $ x\equal{}a.$

$ 6a^5\plus{}a\plus{}a\minus{}2b\equal{}0,$ or $ b\equal{}a\plus{}3a^5.$

So the slope of the line from $ (a,a^3)$ to the center of the circle has slope $ \frac{\minus{}a^3}{3a^5}\equal{}\minus{}\frac{1}{3a^2}.$

That's a radius of the tangent circle, and it is a normal to the curve. The tangent line is perpendicular to this, so its slope is the negative reciprocal, so its slope is $ 3a^2.$ Given that, we can get other results involving the tangent line.

I'm really not sure why Descartes used the circle rather than just an arbitrary line - the circle makes things more complicated than they have to be. But procedures are often more complicated when first invented than they need to be.

At roughly the same time, Fermat was finding tangent lines to polynomial curves by taking their derivatives. Oh, he didn't call it the derivative, and he didn't have the ideology of limits, and he didn't have the inverse connection between what he was doing and areas - but you'd recognize what we now call the derivative in his symbols.

Linguistic note: the word "analysis" in mathematics originally meant [b]algebra[/b] - symbolic, manipulative algebra, the stuff we teach in high school courses called "algebra." What Descartes was doing was taking this "analysis" - this symbolic manipulation, this understanding of polynomials - and using it to attack problems in geometry. You can see directly how that leads to our word "analytic geometry." But the very same thing - the use of computational symbolic manipulations in geometric contexts - attached itself to the calculus of Newton, Leibniz, Jean and Jacques Bernoulli, Euler, [i]et al.[/i] and led to the fact that the word "analysis" appears in the title of this forum.


linear algebra
matrix
algebra
polynomial
invariant
number theory
greatest common divisor

It's a very simple exercise.

use elementary transformation to the below matrix:

$ \begin{bmatrix} \lambda \minus{} 13 & \minus{}16 & \minus{}16\\ 5& \lambda\plus{}7&6\\5&8& \lambda \plus{} 7\end{bmatrix}$

to get the diagonal form

my answer is :

$ \begin{bmatrix}1 & 0 & 0\\ 0&1&0\\0&0& \lambda^{3}\plus{} \lambda^{2}\minus{}21 \lambda \minus{}13 \end{bmatrix}$

[color=red]but it seems not right [/color]:huh:

The canonical diagonal form you are looking for is the [i]Smith canonical form[/i], and the polynomials appearing on the main diagonal are the invariant polynomials of the given matrix. An invariant polynomial $ i_r$ is defined as $ i_r \equal{} \frac {d_r}{d_{r \minus{} 1}}$, where $ d_k$ is the gcd of all k order minors of the matrix, and defining $ d_0 \equal{} 1$. In your matrix it is easily seen that $ d_1 \equal{} 1$. For $ d_2$, if $ M_{i,j}^{k,l}$ denotes the minor corresponding to rows i,j and columns k,l, then for your matrix we have

$ M_{1,2}^{1,2} \equal{} \lambda ^2 \minus{} 6 \lambda \minus{} 11, M_{1,2}^{1,3} \equal{} 6 \lambda \plus{} 2, M_{1,2}^{2,3} \equal{} 16 \lambda \plus{} 16, M_{1,3}^{1,2} \equal{} 8 \lambda \minus{} 24, M_{1,3}^{1,3} \equal{} \lambda ^2 \minus{} 6 \lambda \minus{} 11,$
$ M_{1,3}^{2,3} \equal{} \minus{} 16 \lambda \plus{} 16, M_{2,3}^{1,2} \equal{} \minus{} 5 \lambda \plus{} 5, M_{2,3}^{1,3} \equal{} 5 \lambda \plus{} 5, M_{2,3}^{2,3} \equal{} \lambda ^2 \plus{} 14 \lambda \plus{} 1$.

From this we see that $ d_2 \equal{} 1$. Denoting your matrix by $ A(\lambda)$ and the characteristic polynomial of a matrix P by $ \phi_P(\lambda)$ we have

$ d_3 \equal{} \det A(\lambda) \equal{} \det(\lambda I \minus{} B) \equal{} \phi_B(\lambda) \equal{} \lambda ^3 \plus{} \lambda^2 \minus{} 21 \lambda \minus{} 13$, where

$ B \equal{} \begin{bmatrix}13 & 16 & 16 \\
\minus{} 5 & \minus{} 7 & \minus{} 6 \\
\minus{} 5 & \minus{} 8 & \minus{} 7 \\
\end{bmatrix}$.

Therefore,

$ i_1 \equal{} \frac {d_1}{d_0} \equal{} \frac {1}{1} \equal{} 1, i_2 \equal{} \frac {d_2}{d_1} \equal{} \frac {1}{1} \equal{} 1, i_3 \equal{} \frac {d_3}{d_2} \equal{} \frac {\lambda ^3 \plus{} \lambda^2 \minus{} 21 \lambda \minus{} 13}{1} \equal{}$

$ \equal{} \lambda ^3 \plus{} \lambda^2 \minus{} 21 \lambda \minus{} 13$,

so your diagonal matrix is correct, provided I have made no mistakes.


How many 4-digit numbers have the last digit equal to the sum of the first two digits?

For $ 0\le n\le9$, the number of two digit integers $ \overline{ab}$ such that $ a\plus{}b\equal{}n$ is $ n$(prove this yourself). Thus, there are $ 1\plus{}2\plus{}3\plus{}\cdots\plus{}9\equal{}\frac{9\cdot10}2\equal{}45$ ways to construct the first two digits and the last two digits (the sum illustrates the possible $ \overline{ab}$ for every possible last digit $ 0\le n\le9$). Also, there are $ 10$ choices for the third digit, for a total of $ 45\cdot10\equal{}\boxed{450}$ four-digit numbers.

[quote="math154"]For $ 0\le n\le9$, the number of two digit integers $ \overline{ab}$ such that $ a\plus{}b\equal{}n$ is $ n$(prove this yourself). [/quote]  This is awesome, math154, This is true because we can choose any of n integers from 1...n for the first digit (excluding 0), and the 2nd digit is then set (since there is only one choice such that it can add to a sum of n).  This argument is pretty powerful because it turns this problem from a bashable ugly one, to an elegant question.

Here's another way with stars and bars: $ \overline{abcd}$ has the property that $a+b = d$,  with $1 \le a,d \le 9$ and $0 \le b \le 8$, as well as $1 \le a+b \le 9$. Define a new variable $x$, $0 \le x \le 8$ such that $a + b + x = 9$. Now that we've turned our inequality into an equation, we're almost set up to use stars and bars. Checking our restrictions on our variables we see that if we let $a' = a-1$, then our new equation $a' + b + x = 8$ will have all of $a', b, x$ between $0-8$. By stars and bars we need $2$ dividers to distribute the sum of $8$ to our three variables $=> \binom{10}2 = 45$. Since $c$ can be anything, we have $45 \cdot 10 = \boxed{450}$ ways.


limit logarithms function trigonometry calculus algebra polynomial

How do you prove that:

$\lim_{x\rightarrow x_{0}}u^{v}=e^{\lim_{x\rightarrow x_{0}}v(u-1)}$

This would not be true without additional conditions.

I'm not sure what you intend this to be. It is possible that the following might be useful:

$\ln u=(u-1)+o(u-1)$ as $u\to 1.$

[quote="sinaspidosnake13"]:wink:  Dude puhiki, $u$ and $v$ are functions which contain $x$[/quote]
You're not being very helpful, sina-two-different-screen-names.

As you originally stated this, it is false, and puuhikki was merely trying to point out that it is false.

Let's do this and get specific. Let $u=3-x,v=x,$ and take limits as $x\to 1.$

Then $\lim u^{v}=2$ but $e^{\lim v(u-1)=e.}$

They're not the same.

I do suspect that you may intend something with more conditions on it than what you stated, but you must make that clear.

Ok, First of all what do you mean you're not very helpful with two different screen names, Dude, i'm not trying to have fun here, I'm here to learn math from others and maybe teach sth to others, $sinaspidosnake13$ is for 2004(when i was highschool and did not know anything about working with mathlinks) and $sinajackson$ is for 2006(entering the university and trying to show up)
anyway, if you guys don't understand what i'm talking about, and maybe I'm wrong, let me tell you the real problem, it's one of the problems of University of Tehran's Mid-Term exams(1996) that says calculate the limit:
$lim_{x\rightarrow \frac{\pi}{2}}(\sin x+\cos x)^{\tan x}$

It has given two solutions, the second is that it says we can use $lim_{x\rightarrow x_{0}}u^{v}=e^{lim_{x\rightarrow x_{0}}v(u-1)}$
so $lim_{x\rightarrow \frac{\pi}{2}}(\sin x+\cos x)^{\tan x}=e^{lim_{x\rightarrow \frac{\pi}{2}}(\tan x)(\sin x+\cos x-1)}=e^{lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin x+\cos x-1}{\cot x}}=(Hopital)e^{lim_{x\rightarrow \frac{\pi}{2}}\frac{\cos x-\sin x}{1+\cot^{2}x}}=e^{\frac{0-1}{-1-0}}=e$

The point is that the other readers can't understant from the concept what do you mean. It might be clear for you, but for me $u^{v}$ is just a constant expression with respect to $x$. Remember that I haven't read that paper. It is a good skill to write the questions so that other understands that as well.

The correct way to write the question is something like that:

"I have some difficulties reading this and this paper. They are asking to prove that the limit something exists. They give a hint that we can use the following fact: What might they mean?"

That gives much more information than just "Prove u=v for every x" type posting. If you write more details and you do a mistake, it is much bigger possibility that the other understands what do you mean.

My advice to mathematicians (including myself) is that write mathematical text so that it is as easy as possible for others to read. I agree that it is sometimes hard to know what the other readers know and what they don't, but it is better to write a little more details than necessary than to find too narrow representation for your ideas.

The original problem clarifies what was really going on. We have a $1^{\infty}$ indeterminate case. The result you want is this:

** Suppose $\lim_{x\to x_{0}}u(x)=1$ **

Then $\lim_{x\to x_{0}}u(x)^{v(x)}=\exp\left(\lim_{x\tox_{0}}v(x)(u(x)-1)\right),$

provided that the latter limit exists.

Without the condition ** - **, this is false. By giving us this without that condition, you were giving us mathematical nonsense. I twice commented on the possibility of other conditions; you never responded to that. The phrase "provided the latter limit exists" is also important; without that we get in trouble.

The first step:

$\lim_{x\to x_{0}}u(x)^{v(x)}=\exp\left(\lim_{x\tox_{0}}v(x)\ln(u(x))\right)$

This follows from the definition of $u^{v}$ and from the fact that the exponential is a continuous function. From here on, I'll shorten the notation; I'll just say $u$ and $v$ without always being explicit about them being functions of $x,$ and I'll say "$\lim$" without being explicit about it being the limit as $x\to x_{0}.$ That part will be understood.

When I'm talking to a calculus class, I usually phrase the advice this way: when faced with and exponential-type indeterminate case ($0^{0},\infty^{0},$ or $1^{\infty}$), give it a name and take its logarithm. That is, if

$L=\lim u^{v}$ then $\ln L=\lim v\ln u.$

We use the Taylor polynomial fact I referred to earlier: for $u$ near $1,\ \ln u = (u-1)+o(u-1).$ (For the big-O, little-o notation see the sticky in the "Calculus Computations and Tutorials" section.)

Now it gets tricky. Suppose $\lim v(u-1)=a$ exists. Then $v\cdot o(u-1)=v(u-1)\cdot\frac{o(u-1)}{u-1}\to a\cdot 0 = 0.$

Therefore, $\lim v\ln u=\lim v[(u-1)+o(u-1)]=\lim v(u-1)$

And the result follows from that.

I might have organized it a little differently. Note that the solution you provided does use L'Hôpital; I'll just use that a different place.

$L=\lim_{x\to\frac{\pi}2}(\sin x+\cos x)^{\tan x}$

$\ln L = \lim_{x\to\frac{\pi}2}\tan x\ln(\sin x+\cos x)$

$\ln L = \lim_{x\to\frac{\pi}2}\sin x\cdot\frac{\ln(\sin x+\cos x)}{\cos x}$

$= 1\cdot\lim_{x\to\frac{\pi}2}\frac{\frac{\cos x-\sin x}{\sin x+\cos x}}{-\sin x}=\lim_{x\to\frac{\pi}2}\frac{1-\cot x}{\sin x+\cos x}=1.$

Since $\ln L=1,$ we have that $L=e.$

Ok, Thank you guys, and sorry for the bad arguement, It's just because sometimes i think the only person that is right is me.....
thank you for the advice puhiki :)  and thank you for the nice solution Kent Merryfield :)
Thank you very much.

Only in the following two conditions :

*) the irresolute case $\boxed{\ 1^{\infty}\ }\ ,$ i.e. $\lim_{x\to x_{0}}u(x)=1$ $\lim_{x\to x_{0}}v(x)=\infty$
**) $u(x)>0$ in a proximity of the point $x_{0}\ ,$ i.e. there is $\delta >0$ so that for any $x\in (x_{0}-\delta ,x_{0}+\delta )$ , $x\ne x_{0}$ we have $u(x)>0\ .$

we can use the relation $\lim_{x\to x_{0}}u(x)^{v(x)}=e^{\lim_{x\to x_{0}}v(x)\cdot [v(x)-1]}\ .$


[hide]Try to write 180 as a triangular number, or at least know where it is in relation to nearby triangular numbers:

n(n+1)/2=180

n^2+n=360

n^2+n-360=0

Use the quadratic formula:

The positive value (obviously the negative value wouldn't work) is about 18.48. So, the triangular number directly before 180 is 1+2+3+...+18, or 171. The 18th letter in the alphabet is R, so the 171th letter in the sequence is the last R in the sequence. After that comes 19 s's, of which the 180th is one. So, the answer is s.[/hide]

Quote:What is the 180th letter in the 'word' below, given that the obvious pattern continues:
ABBCCCDDDDEEEEEFFFFFF....
(The nth letter is repeated n times)[hide]

My solution is very close to Lynelle Ye's.

I sought the largest triangular number less than or equal to 180.

Since n(n+1)/2 is approximately n2/2 for large n,

I set n2/2 = 180 and found n = \sqrt{360} = 18.97

then concluded that n = 19.

The 19th letter is S.[/hide]

What is the 180th letter in the 'word' below, given that the obvious pattern continues:

ABBCCCDDDDEEEEEFFFFFF....

[hide]
$\displaystyle\frac{x(x+1)}{2}=180$

$x^2+x=360$

$x^2+x-360=0$

$x\approx18$

therefore the letter is S[/hide]

we are looking for the 180th letter in the word where the nth letter appears n times.
[hide]So:
we are really looking at traingular numbers here, where the nth triangular number represents the last place where the nth letter appears.
now we must set bounds upon the letters.... ie find which two triangulars numbers 180 is between.
we see that it is between 171 (18th triangular number ) and 190 (19 triangular number). the 171 letter is an 'r' and the letters 172 - 190 are s's. since 172<180<190, the 180th letter is an [b]S[/b].[/hide]

[hide]THe 180th letter is the 19th letter of the alphabet or s.  We know this because 19(19+1)/2=190, but that's where the s series ends.  It began 18 letters earlier on 172, since 190 is included to make it 19.[/hide]

The answer [tip=is]$S$[/tip]
15 year old thread wow!

We find the $n(n+1)/2$ closest to $180$ and that is $18(19)/2 = 171$. We simply find the $19$th letter in the alphabet which is the letter $S$.


geometry inequalities algebra polynomial AMC USA(J)MO USAMO

Hi,

I want to ask you which IMO problems do you like best or consider most beautiful? So write down your favourite one and say what you like about this one particularly. In due course I will make a poll for that...  :)

Which is my favourite problem? Well, as a triangle geometer, I love

[b]41st IMO 2000 Problem B3.[/b] Let ABC be a triangle. The feet of the altitudes from A, B, C are A', B', C', and the incircle ouches the sides BC, CA, AB at X, Y, Z. The lines B'C', C'A', A'B' are reflected in the lines YZ, ZX, XY, respectively. Show that the three new lines form a triangle with vertices on the incircle of triangle ABC.

I've always been to lazy to solve this one but I know I could solve it synthetically if I had a good reason (IMO, for instance)...

  Darij

There are many types of problems, these are the ones that I like most for each type of problem

Number Theory : IMO 2003/6
Geometry : IMO 1995/5
Functional Equation : IMO 1992/2
Sequence : IMO 1996/6
Inequality : IMO 2000/2
Invariance : IMO 2000/3

Most favorite combinatorics is difficult to choose because all the IMO combinatorics are equally good: IMO 1989/3, IMO 1989/6, IMO 1991/3, IMO 1992/3, IMO 1995/6, IMO 2001/3

Overall, the best of the best, the problem that I like most among all these types of problem is IMO 2003/6. This is the most beautiful IMO problem. Congratulations to Johan Yebbou Professeur Mathmatiques Spciales MP* for creating such a good problem, I hope that he can create more problems for IMO.

PS: best polynomial is USAMO 2002/3

I like following one:
Numbers k2+k+n are primes for k=0,1,...,[\sqrt (n/3)]. Prove numbers k2+k+n are primes for k=0,1,...,n-2.

I think you know Euler's example: n=41. It is hard to believe but k2+k+n is really prime if 0 \leq k \leq 39. Due to mentioned statement checking of this fact reduces to checking for k=0,1,2,3.

[quote="darij grinberg"]
[b]41st IMO 2000 Problem B3.[/b] Let ABC be a triangle. The feet of the altitudes from A, B, C are A', B', C', and the incircle ouches the sides BC, CA, AB at X, Y, Z. The lines B'C', C'A', A'B' are reflected in the lines YZ, ZX, XY, respectively. Show that the three new lines form a triangle with vertices on the incircle of triangle ABC.
I've always been to lazy to solve this one but I know I could solve it synthetically if I had a good reason (IMO, for instance)...
[/quote]

How would you solve it otherwise than solving synthetically?

Well, for instance, Christian Reiher (!) gave a rather messy trigonometric solution. I don't doubt he would have found a synthetic one if he had more time, but it is still not the easiest way as you see!

  Darij

well what would one look out for in a problem in order to determine whether it is beautiful or not?

personally i like geometry problems compared to other types because most problems have nice solutions that do not involved much calculation (of course there are many problems which have ugly trigonometric and coordinate geometry solns and those with messy side length chasing.)

i thought IMO 2003/3 was beautiful because the lemma used to prove the problem was easy and the solution of the problem required little knowledge, yet the result was so powerful in a sense that it turned out to be the hardest problem last year.  (if im not wrong, it was the question for which most contestants got 0, me being one of them)

i love most the famous one of engel(1988/6 i think): if $\frac{a^2+b^2}{ab+1}$ is a natural number, then it is a square. it is completely surprising and the proof is just beautiful(in most cases, in such problems you must determine all solutions and then check that it are squares).

btw what was the original proof for 2003/3? i solved it on my own by a quite beautiful argument, the main thing was to proof that if x,y,z are vector s.t. $|x+y|\geq \sqrt{3}|z|$ and symmetric, then the vectors x,y,z have equal length and the angle between them is always 120(or something like that, i'm not totally sure). my proof for that lemma was quite nice, but involved too much calculation in my opinion.

[quote="Peter Scholze"]i love most the famous one of engel(1988/6 i think): if $\frac{a^2+b^2}{ab+1}$ is a natural number, then it is a square.[/quote]

Yes, this one is cool... BTW, I never knew it was from Engel!

[quote="Peter Scholze"]btw what was the original proof for 2003/3?[/quote]

In fact, the official solutions are here:
  http://village.infoweb.ne.jp/~fvgm9250/web/imo2003/contents/prob.html
and here are some more solutions:
  http://www.mathlinks.ro/Forum/topic-15621.html
but I didn't read them through since I'm still trying to solve the problem for myself (this should take some time, I am not so good yet in geometric inequalities as in triangle geometry).

  Darij

the problem I like most is 2002\6
that was a very nice problem...
a micture of geometry with combinatorial ideas.
2004\3 is also a very nice combinatorics problem.
1997\3 is a very beautiful combinatorial inequality as the only way to solve it requires just logic.
(i mean it doesnt have a straightforeward solution with calculations etc..)
so in general these are the problems (or types of problems)i like the ones which need strong logical argument more than theqnique and theory...
 :roll:

2002 #3 is a very nice problem, as is 2003 #6.

What about the classic 95 #2 :)? Now a textbook foundation for every basic inequalities lecture.

I find 1986/3 (the pentagon problem) extremely beautiful.

I also like 1988/4 (disjoint intervals whose total length is 1988); it is nice that
the author reached "1988" while using only reasonably small numbers in the
problem statement.

1990/6 (convex 1990-gon with all angles equal) is also phantastic.

2000/3 (jumping fleas) is one of the most beautiful invariant problems I have seen.
2000/4 (magic card trick) is not difficult, but very original.

2003/3 (covex hexagon) is my favorite geometry problem.

I find IMO 2003 Q6 and IMO 2000 Q6 the most beautiful...

I also like IMO 1988 Q3 (the functional equation)

2003/6- classic NT with primes- my favourite.

1998/6-greatest function equation ever seen in IMO
2002/6-the last step of this problem is beautiful

My favorite is IMO-1960 no.-2
Because this is the first problem from IMO which I could solve.  :P

mine is 2011-6,the hardest geometry problem I've solved using pure euclidean method.
And @joxi,are you S.M. Joty?


for positive reals $ a,b,c$
$ \sum_{cyclic}{\frac{1}{a(1\plus{}b)}}\ge\frac{3}{1\plus{}abc}$

Multiplying by $ (1\plus{}abc)$ on both sides
we get that our inequality has the $ LHS \plus{} 3$ as
$ \sum{cyclic} \frac{1\plus{}a\plus{}ab\plus{}abc}{a(1\plus{}b)} \equal{} \sum{cyclic}\frac{(1\plus{}a)\plus{}ab(1\plus{}c)}{a(1\plus{}b)}\ge 6\equal{}3\plus{}3\equal{}RHS$.
$ QED$.


Two policemen are guarding a circular wall with length 1. The second policemen's speed is 2 times greater than the first one's. On the wall are some holes ( a hole is simply an arc of the circle). We call a system of holes stable if there is a way to place the policemen such that there is always a policemen covering a hole ( that is the policemen stands within the arc formed by the hole). Prove that any stable systems of holes must have total length greater than $\frac{1}{2}$ and
$\frac{1}{2}$ is the best constant.

the problem is equivalent to the following: Some intervals are selected in $[0,1)$, in such a way that for every $x$, we have that at least one of the numbers $\{x\},\{2x\}$ is in one of those intervals. Prove that the sum of the lenght of those intervals is at least $1/2$.

ok, let $A$ be the set of non selected intervals, for every non selected interval $I$ consider the intervals $I/2$ and $1/2+I/2$., and let $B$ be the set of this intervals. If the lenght of the selected intervals is less than $1/2$ then we have that the sum of lenghts of intervals in $A$ and $B$ is bigger than $1$ and therefore there is a point that belongs to an interval in $A$ and in $B$. Call it $x$, then $\{2x\}\in A$ and $\{x\}\in A$, contradiction!

Cool problem! :)


At the bottom of my LaTeX, when I output, I get a page number.  How do I make it go away?  Is there a code? Please help?

One method is to put
[code]\pagestyle{empty}[/code]
in the header.


Suppose M is the midpoint of the arc AB of a circle.
C is an arbitrary point, other than M, on the arc AMB and D is
the foot of the perpendicular from M upon the longer of the chords
AC and BC
Prove that D bisect the polygonal path ACB

lots of proofs.  here is one
suppose AC<BC
connect AM and BM
ACM and BCM are not congruent :D

...
but they are the ambiguous case.  hence AC+2CD=BC, or AC+CD=BC-CD=BD


If you use brute force and multiply out both sides and simplify, you see that this is equivalent to
$ (x_1y_2\minus{}x_2y_1)^2 \ge 0$.

Let's see what we can do with this inequality

Rewrite the original inequality and using Cauchy Schwartz, we obtain

\[ \begin{gathered} (2y_1^2\plus{}(2x_1\plus{}y_1)^2\plus{}x_1^2)(2y_2^2\plus{}(2x_2\plus{}y_2)^2\plus{}x_2^2) \ge (2y_1y_2\plus{}(2x_1\plus{}y_1)(2x_2\plus{}y_2)\plus{}x_1x_2)^2 \end{gathered}\]

Thus, the equality holds if and only if   $ \frac{x_1}{x_2}\equal{}\frac{y_1}{y_2}$


在三角形ABC内求一点P,使 向量AP^2+向量BP^2+向量CP^2　最小．

谢谢！！！！！！！！！！！！！！！！！ :D

重心G

事实上, 有如下命题:

平面内任意三角形ABC，G为三角形的重心，P为平面内任意一点，则有：
$ 3PG^2\equal{}PA^2\plus{}PB^2\plus{}PC^2\minus{}GA^2\minus{}GB^2\minus{}GC^2$

由此可见只有当P与G重合时$ PA^2\plus{}PB^2\plus{}PC^2$取最小值.

嗯，这个点就是重心G

不过楼上的公式是怎么来的？

[quote="paganinio"]嗯，这个点就是重心G

不过楼上的公式是怎么来的？[/quote]

我之前在不等式区贴过，找找看

[quote="paganinio"]嗯，这个点就是重心g

不过楼上的公式是怎么来的？[/quote]

用余弦定理或者斯特瓦尔特定理来证就可以了~


The point $ (a,b)$ is on the line $ y \equal{} \minus{}3x\plus{}9$. If $ a$ is increased by 2, what is the corresponding decrease in b so that the new point $ (a,b)$ is on the line also?

Since $ (a,b)$ is on the line, $ b \equal{} \minus{} 3a \plus{} 9$. Let the new pair be $ (a \plus{} 2,b \minus{} x)$ so that we want to find $ x$ and

$ b \minus{} x \equal{} \minus{} 3(a \plus{} 2) \plus{} 9 \equal{} ( \minus{} 3a \plus{} 9) \minus{} 6 \equal{} b \minus{} 6\implies x \equal{} \boxed6$.


Given the $7$-element set $A = \{ a ,b,c,d,e,f,g \}$, find a collection $T$ of $3$-element subsets of $A$ such that each pair of elements from $A$ occurs exactly once on one of the subsets of $T$.

There are 21 pairs, so we expect 7 triples.  From symmetry, we guess that there must be some cyclic way.

We finally figure out we should have spacings of 1, 2, and 3 on each triple, since there can only be that much spacing, and they occur 7 times each.

ABF, BCG, CDA, DEB, EFC, FGD, GAE.

ABC , ADE , AFG , BEG , BDF , CEF , CDG also works!

All possible answers are equivalent, up to a relabelling, since they must be the so-called Fano configuration, the finite projective plane of order $2$, also known as the symmetric BIBD $(7,3,1)$.

This is a RMO question & it carries 17 marks.
Therefore the answer should be written accordingly