geometry
logarithms
trigonometry
function
algebra
floor function
power of a point

Here are some problems! Just for fun, do as many as you can, and PM me the answers. Next week, the answers will be posted, along with the people who got the most questions correct. I plan to have future rounds, so tell me whether the questions should be easier or harder.

(1)Simplify: [tex]\frac{\sqrt{5} + 2\sqrt{7} + 3}{\sqrt{35} + 3\sqrt{5} + 3\sqrt{7} + 7}[/tex]

(2)Suppose [tex]AB[/tex] and [tex]CD[/tex] are two perpendicular chords of the same circle that intergect at the point [tex]E[/tex], [tex]AE [/tex]= 12,[tex] DE [/tex]= 4, and [tex]CE[/tex] = 6. What is the area of the circle?

(3)Evaluate: [tex]\frac{1}{1*2*3*4} + \frac{1}{2*3*4*5} + ... + \frac{1}{28*29*30*31}.[/tex]

(4)Simplify completely [tex]\frac{x(x^2 + 4)}{x - 2} + \frac{4(x + 2)}{2 - x}.[/tex]

(5)A circle is tangent to the y- axis at (0,2) and the larger of its x - intercepts is 8. Calculate the area of the circle.

(6)Evaluate: [tex]\frac{1}{2\sqrt{1} + \sqrt{2}} + \frac{1}{3\sqrt{2} + 2\sqrt{3}} + \frac{1}{4\sqrt{3} + 3\sqrt{4}} + ..... + \frac{1}{100\sqrt{99} + 99\sqrt{100}}[/tex]

(7)Diagonals AC and BD of quadralateral ABCD meet at E. It AE = 2, BE = 5, CE = 10, DE = 4, and BC = 15/2, find AB.

(8)Let a and b be the two positive solutions to the following equation. [tex]\log_{3x} 3 + \log_{27} 3x = -\frac{4}{3}[/tex]. Find a + b.

(9)Given that [tex]\log_{10} 2 \approx 0.3010 ... [/tex] and [tex]\log_{10} 3 \approx 0.47712 ...[/tex], what is the number of digits in the decimal expansion of [tex]12^{10}[/tex]? (Show your work)

(10)Simplify completely: [tex]\frac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}} - \sqrt{3 - 2\sqrt{2}}.[/tex]

Could someone please put this on the bottom of the sticky list until next week? Thank you - I just don't want it getting washed down the list.

Yay! I love Challenge Problems. Thanks white_horse_king88.

[quote="lonelytraveler8"]Some of these are really tough :?[/quote]

At least you can answer them.  The only ones I have a hope of trying are 4 and 9.

Actually, check out the Logs thread in this forum. Use those equations and solve number 8. Look for usage of those equations.

I thought 4, 7 and 8 were pretty simple.  I did number 9 but only because I happen to know [tex]2^{10}[/tex] and I actually multiplied it out...it took less than 2 minutes.  But I got help on it to solve it the legitament way.  however beyond 4, 7, and 8 I'm stuck and getting help on them so anything I do after this doesn't count for me...except now that I've seen a really really elegant solution for number 6 I have a dim idea for number 3.

Yes. There is a very elegant solution to number 6. You just have to recognize the pattern in the sequence.

[quote="white_horse_king88"]Now that you mention it, the only problem I would like to see work for is problem 9, because I don't want somebody plugging in values into their TI-89, and counting digits.[/quote]

Who would ever do [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14894&postdays=0&postorder=asc&&start=15]that[/url]?  ;)

(9th post from the top)

I have to write some better questions next time :)

I like your new icon white_horse_king_88...do you play starcraft much?  I used to but don't have it anymore :cry:

I managed to do 3-8. I even tried simplifying 1 & 10, but they don't look simple enough. However, I have no idea how to do 2 & 9. :?

Edit:
Only #1 left.  :)

[mod edit]  Solutions for challenge sets should not be posted until after the deadline.

Um, liyi, this was a challenge set, could you not post the answers until this is over - like next week.

10's actually been posted before, by Rep123Max in a square root challenge

Oops. Sorry for the overlap. I didn't know Rep posted it before.

I don't believe anyone else who hasn't sent me their answers will, so everyone who wants to post solutions may do so. I will post solutions probably on Monday.

WHOOHOO! MATH! OH GOODY! I'm gonna ace this. *shouting to Dad* I need help Dad!
 :blush: seriously these are really HARD!!!

I was wondering where and when will the solutions be posted?

Answers:

 $1)\\\displaystyle\frac{\sqrt5+2\sqrt7+3}{\sqrt{7}(\sqrt{5}+\sqrt{7}) + 3(\sqrt{5}+\qrt{7})} \\ = \displaystyle\frac{(\sqrt{5}+\sqrt{7}) + (\sqrt{7}+ 3)}{(\sqrt{5}+\sqrt{7})(\sqrt7+3)} \\ = \frac{1}{3+\sqrt{7}} + \frac{1}{(\sqrt{5}+\sqrt{7})} \\ = \displaystyle\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}}{2} \\ = \frac{3-\sqrt{5}}{2}$

2) Notice that (AE)(BE) = (DE)(CE) by power of a point. So, BE = 2. Let the radius be BO, where O is the origin of the circle. Any line perpendicular to a chord from the origin by definition intersects the chord at its midpoint. Thus, let the midpoint of AB be F. Let the midpoint of CD be G. FO = DG - DE = 1, and FB = 7.  Thus, you have $\sqrt{(7^2+1^2)} = \sqrt{50} = 5\sqrt{2}$, and the area is 50 :pi: .

Here are the first two - everyone else can post their solution too, you know! I will post more later today -

Does anyone want to do a math chat on these problems are solutions?

A math chat sounds like a good idea.   You should probably host it since these are your questions.  If there is one I ll try to attend.

How about this friday! I think these problems may take around an hour and a half, so 6:00 - 7:30 Eastern? Maybe?

I was wondering, now that solutions are posted, is you could tell everyone's score

devious_ : 7 correct
RC-7th: 9 correct
lonelytraveler8: 3 correct

Only 3 people sent answers - people can still PM me answers before Friday, when I will host the math chat on this set.

I'm guessing 1,2,and 10 don't count anymore :lol:

To whitehorseking: which one's did I do? I think I did 4,8,and 9

Yeah - sorry - didn't think of math chat until I posted the first 2 solutions. However, you send me 8,9, and 10, which means you can still send 4 for credit. :)

Reminder of the math chat tonight. It is from 6 - 7:30 on the chat room "math chat".

$4)\frac{x(x^2+4)}{x-2} + \frac{4(x+2)}{2-x} \\\\ = \frac{x(x^2+4)}{x-2} - \frac{4(x+2)}{x-2} \\\\ = \frac{x^3 - 8}{x-2} \\\\ = \frac{(x-2)(x^2+2x+4)}{x-2} \\\\ = x^2 + 2x +4$

5) Let the x-intercept be point B, the center of the circle O, the point of tangency to the y-axis A, and let the perpendicular from O to the x-axis intersect the x-axis at point C. Let the radius be r, and CB be x.

$x^2 = r^2 - 2^2 \\ x+r=8 \\ r = 8-x \\ x^2 = (8-x)^2-4 \\ x = 15/4 \\ r = 17/4 \\ \pi * r^2 = \frac{289}{16} \pi$

6) Let the kth term in the sequence be $a_k$.

$a_k = \displaystyle\frac{1}{(k+1)\sqrt{k} + k\sqrt{k+1}} \\\\ = \displaystyle\frac{1}{(k+1)\sqrt{k} + k\sqrt{k+1}} \\\\ = \displaystyle\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})} \\\\ = \displaystyle\frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k(k+1)}} \\\\ = \displaystyle\frac{1}{\sqrt{k}} - \displaystyle\frac{1}{\sqrt{k+1}} (k = 1,2, \ldots99)$

That becomes: $\displaystyle(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}) + \displaystyle(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}) + \ldots + \displaystyle(\frac{1}{\sqrt{98}} - \frac{1}{\sqrt{99}}) + \displaystyle(\frac{1}{\sqrt{99}} - \frac{1}{\sqrt{100}}) \\\\ = 1 - \frac{1}{10} = \frac{9}{10}$

7) Draw the quadralateral ABCD with the diagonals meeting at E. With the given, we can conclude from the power of a point that BE/AE = CE/DE = 5/2. Thus, $\Delta{AED}\sim\Delta{BEC}$. So, BE/AE = BC/AD, or 5/2 = (15/2)/AD, and thus AD = 3. We can also conclude that $\Delta{AEB}\sim\Delta{DEC}$, and so AE/DE = AB/DC or 1/2 = AB/DC and DC = 2(AB). From $m\angle{BAC}\cong m\angle{BDC}$, ABCD is cyclic. Using Ptolemy's Theorem, (AB)(DC)+(AD)(BC) = (AC)(BD), and thus $ AB = \frac{1}{2}\sqrt{171}.$ I discussed another solution to this problem using the law for cosines.

We know that $c^2 = a^2+b^2-2ab\cos{\theta}$, where $\theta$ is the angle opposite side c. Thus, you can solve, since c=15/2, a=4, and b = 10. Once you find $\theta$, you want AD, which since you know $\theta$ and a and b, you can solve for AD, which is c in the equation. Simple and clean.

$8) \log_{3x} 3 + \log_{27} 3x = -\frac{4}{3} \\\\ \log_{3x} 3 + \displaystyle\frac{1}{\log_{3x} 27} = -\frac{4}{3}$

Let $\log_{3x} 3 = x$.

You have $x + \frac{1}{3x} = -\frac{4}{3} \\\\ 3x^2 + 4x + 1 = 0 \\\\ (3x+1)(x+1) = 0 \\\\ x = -\frac{1}{3}, -1 \\\\ \log_{3x} 3 = -1, and -\frac{1}{3}$, so solve and get x = 1/9 and 1/81.
So, $a+b = \frac{10}{81}$.

$9)\log_{10} 12^{10}$ has the base value of [tex]\log 12[/tex] times 10 + 1 digits. This is because $\log_a b^c = c \log_a b$. $\log 12 = \log 2 + \log 2 + \log 3$, which is 1.07912, and that times 10 is 10.7912 and +1 = 11.7912. The greatest integer function of that is 11, so that value has 11 digits.

10)You have one monstrous equation, which can be seen as two parts, the first and the second. You find that if you square the first half, you get 2, and the second half is [tex](\sqrt{2}-1)^2[/tex], so you have $\sqrt{2} - (\sqrt{2} -1) = 1 $.

I moved this thread - many of these are too tough for what we intend to be Getting Started.

In general, if you're not sure of what level to post a challenge problem in, choose the harder of the two forums you're considering.

Also, people should only respond to a challenge if it is of their difficulty level (i.e. olympiad students should leave getting started and intermediate challenges to other students).  Questions that are asked for information, as opposed to challenges, are open to anyone who can answer them clearly.

Thanks!


modular arithmetic
number theory proposed
number theory

Let $ W\equal{}{0000001,0000002,...,9999999}$ where each of the elements are taken as $ 7$ digit numbers! Write the numbers in $ W$ in any order or so to form a composite number $ C$. Then prove that $ 239$ always divides $ C$.

Verify that $ 10^7\equiv 1\pmod{239}$. Therefore $ C\equiv 1\plus{}2\plus{}\dots\plus{}9999999\pmod{239}\equiv\frac{10^7(10^7\minus{}1)}{2}\pmod{239}\equiv 0\pmod{239}$.

That is OK but I had also solved it in the exact way! Can you perhaps give me an alternate proof?


conics
parabola
algebra
system of equations

Write the equations of the parabolas with the following conditions:
a. with vertex (5,-3) and passing through (3,9)
b.  passing through the points (-1,-6), (3,-12), and (2,-4.5)

i know how to do a. but for b i don't know how to make the parabola go through ALL of the points.

a. You can set the equation of the parabola as $ y\equal{}k(x\minus{}5)^{2}\minus{}3\ (k>0)$.

b. You can set the equation of the parabola as $ y\equal{}px^{2}\plus{}qx\plus{}r\ (p\neq 0)$.

Then create three seperate equations by substituting your ordered pairs into the equation kunny provided, and solve for p, q and r...

[hide]for b.)

For equation 1, using $ (\minus{}1,\minus{}6)$
  $ \minus{}6 \equal{} a(\minus{}1)^{2}\plus{}b(\minus{}1)\plus{}c$

For equation 2, using $ (3,\minus{}12)$
  $ \minus{}12\equal{} a(3)^{2}\plus{}b(3)\plus{}c$

For equation 3, using  $ (2,\minus{}4.5)$
  $ \minus{}4.5\equal{} a(2)^{2}\plus{}b(2)\plus{}c$

Then look for the values of $ a$, $ b$ and $ c$.[/hide]

[hide="Using the System of Equations"]
$ 9a\plus{}3b\plus{}c\minus{}2\left( 4a\plus{}2b\plus{}c\right) \equal{}\minus{}12\minus{}2\left(\minus{}4.5\right)$
$ a\minus{}b\minus{}c\equal{}\minus{}3$

Solve for $ a$
$ a\minus{}b\minus{}c\equal{}\minus{}3\implies a \equal{} b\plus{}c\minus{}3$
Substitute
$ b\plus{}c\minus{}3\minus{}b\plus{}c\equal{}\minus{}6$
$ \boxed{c\equal{}\frac{\minus{}3}{2}}$
Substitute
$ a\minus{}b\equal{}\frac{\minus{}9}{2}$
$ 2a\plus{}b\equal{}\minus{}2$
Add the equations
$ \boxed{a\equal{}\frac{\minus{}13}{6}}$

Substitute those two
$ \frac{\minus{}13}{6}\minus{}b\minus{}\frac{3}{2}\equal{}\minus{}6$
$ \boxed{b\equal{}\frac{7}{3}}$

Final parabola
$ \boxed{\boxed{y\equal{}\frac{\minus{}13x}{6}\plus{}\frac{7x}{3}\minus{}\frac{3}{2}}}$.
[/hide]


geometry
incenter
ratio
projective geometry
geometry proposed

Let $ M$, $ I$ be the centroid and the incenter of triangle $ ABC$, $ A_1$ and $ B_1$ be the touching points of the incircle with sides $ BC$ and $ AC$, $ G$ be the common point of lines $ AA_1$ and $ BB_1$. Prove that angle $ CGI$ is right if and only if $ GM \parallel AB$.

[i]Author: A.Zaslavsky[/i]

Let $ CG$ cut $ A_1B_1$ at $ D$ and $ AB$ at $ C_1.$ $ G$ is Gergonne point $ \Longrightarrow$ $ C_1$ is tangency point of the incircle $ (I)$ with $ AB.$ Let $ (I)$ cut $ CG$ again at $ J.$ From quadrilateral $ CA_1GB_1,$ the cross ratio $ (C, G, D, C_1) = \frac {\overline{CC_1}}{\overline{CD}} \cdot \frac {\overline{GD}}{\overline{GC_1}} = - 1$ is harmonic. $ A_1B_1$ is polar of $ C$ WRT $ (I)$ $ \Longrightarrow$ the cross ratio $ (C, D, J, C_1) = \frac {\overline{CC_1}}{\overline{CJ}} \cdot \frac {\overline{DJ}}{\overline{DC_1}} = - 1$ is also harmonic.

$ \color{red}\Longrightarrow:$ Assume $ GM \parallel AB.$ From the harmonic cross ratio $ (C, G, D, C_1) = - 1$ and $ \frac {\overline{GC}}{\overline{GC_1}} = - 2$ $ \Longrightarrow$ $ D$ is midpoint of $ CC_1$ and $ \frac {\overline{DC_1}}{\overline{GD}} = - 2.$ On the other hand, from $ \frac {\overline{CJ}}{\overline{DJ}} = - \frac {\overline{CC_1}}{\overline{DC_1}} = - 2$ $ \Longrightarrow$ $ D$ is also midpoint of $ GJ.$ It follows that $ G$ is midpoint of $ JC_1,$ which means that the angle $ \angle CGI = \angle JGI$ is right.

$ \color{red}\Longleftarrow:$ Assume $ \angle CGI$ is right $ \Longrightarrow$ $ G$ is midpoint of $ JC_1.$ From the harmonic cross ratio $ (C, D, J, C_1) = - 1,$ we then have $ \frac {\overline{CC_1}}{\overline{DC_1}} = \frac {\overline{GJ}}{\overline{GD}} = - \frac {\overline{GC_1}}{\overline{GD}}.$ Comparing this with the harmonic cross ratio $ (C, G, D, C_1) = - 1,$ we get ${ \overline{DC_1}} = \overline{CD},$ which means that $ D$ is midpoint of $ CC_1.$ From the harmonic cross ratio $ (C, G, D, C_1) = - 1$ then follows that $ \frac {\overline{GC}}{\overline{GC_1}} = - 2$ or $ GM \parallel AB.$

Since $ A_1,B_1$ lie on the circle with diameter $\overline{ IC},$ the statement is equivalent to prove that Gergonne point $ G$ lies on the circle $ (J)$ with diameter $ IC$ $ \Longleftrightarrow$ $ MG \parallel AB.$ Let $ C_1$ denote the tangency point of $ (I)$ with $ AB.$ Assume that $ G$ lies on  $ (J).$ $ AB$ is the polar of intersection $ P \equiv IC \cap A_1B_1$ WRT $ (J),$ thus $ JP \perp AB$ $ \Longrightarrow$ $ JP \parallel IC_1.$ Since $ J$ is midpoint of $ IC,$ then $ P$ is midpoint of $ CC_1.$ From the harmonic division $ (C,G,P,C_1) \equal{} \minus{} 1,$ we get $ CG \equal{} 2GC_1$ $ \Longrightarrow$ $ MG \parallel AB.$

The opposite statement is immediatly true, this is: If $ MG \parallel AB,$ from the harmonic division $ (C,G,P,C_1) \equal{} \minus{} 1,$ $ P$ is midpoint of $ CC_1$ $ \Longrightarrow$ $ JP \perp AB.$ Since $ P$ is the pole of $ AB$ WRT $ (J),$ it follows that $ G \in (J).$

[color=darkblue][b]Caution ![/b] [u]I"ll use another notations[/u].[/color]

[quote][color=darkred]Let $ ABC$ be a triangle with the [u]centroid[/u] $ \mathrm G$ and the [u]incircle[/u] $ C(\mathrm I,r)$ which touches given triangle in $ D\in (BC)$ , $ E\in (CA)$ ,

$ F\in (AB)$ . Denote the [u][b]Gergogne[/b]'s point[/u] $ \Gamma\in AD\cap BE\cap CF$ . Prove that $ \boxed {\ \mathrm G\Gamma\parallel BC\ \Longleftrightarrow\ \mathrm I\Gamma\perp AD\ }$ .[/color][/quote]

[color=darkblue][b][u]Proof (metric)[/u].[/b] Denote $ \left\|\begin{array}{c} 2p \equal{} a \plus{} b \plus{} c \\
\ p \minus{} a \equal{} x > 0 \\
\ p \minus{} b \equal{} y > 0 \\
\ p \minus{} c \equal{} z > 0\end{array}\right\|$ . From $ IA^2 \minus{} ID^2 \equal{} IA^2 \minus{} IE^2 \equal{} AE^2 \equal{} (p \minus{} a)^2$ obtain $ \boxed {\ IA^2 \minus{} ID^2 \equal{} x^2\ }\ \ (1)$ .

[u][b]Aubel[/b]'s relation[/u] : $ \frac {\Gamma A}{\Gamma D} \equal{} \frac {FA}{FB} \plus{} \frac {EA}{EC} \equal{} \frac {p \minus{} a}{p \minus{} b} \plus{} \frac {p \minus{} a}{p \minus{} c} \equal{}$ $ \frac xy \plus{} \frac xz \equal{} \frac {x(y \plus{} z)}{yz}$ , i.e. $ \boxed {\ \frac {\Gamma A}{x(y \plus{} z)} \equal{} \frac {\Gamma D}{yz} \equal{} \frac {AD}{x(y \plus{} z) \plus{} yz}\ }\ \ (2)$ .

[u][b]Stewart[/b]'s relation[/u] : $ AD^2\cdot (y \plus{} z) \plus{} yz(y \plus{} z) \equal{} z(x \plus{} y)^2 \plus{} y(x \plus{} z)^2$ , i.e $ \boxed {\ AD^2(y \plus{} z) \equal{} x^2(y \plus{} z) \plus{} 4xyz\ }\ \ (3)$ .

Therefore, $ \Gamma A^2 \minus{} \Gamma D^2\ \stackrel{(2)}{ \equal{} }\ \frac {x^2(y \plus{} z)^2 \minus{} y^2z^2}{[x(y \plus{} z) \plus{} yz]^2}\cdot AD^2\ \stackrel{(3)}{\Longleftrightarrow}$ $ \boxed {\ \Gamma A^2 \minus{} \Gamma D^2 \equal{} \frac {x(y \plus{} z) \minus{} yz}{x(y \plus{} z) \plus{} yz}\cdot\frac {x^2(y \plus{} z) \plus{} 4xyz}{y \plus{} z}\ }\ \ (4)$ .

$ \blacktriangleright\ \mathrm G\Gamma\parallel BC$ $ \Longleftrightarrow$ $ \Gamma A \equal{} 2\cdot\Gamma D\ \stackrel{(2)}{\Longleftrightarrow}\ x(y \plus{} z) \equal{} 2yz$ , i.e. $ \boxed {\ \mathrm G\Gamma\parallel BC\ \Longleftrightarrow\ x(y \plus{} z) \equal{} 2yz\ }\ \ (5)$ .

$ \blacktriangleright$ $ \Gamma I\perp AD$ $ \Longleftrightarrow$ $ \Gamma A^2 \minus{} \Gamma D^2 \equal{} IA^2 \minus{} ID^2\ \stackrel{(1)\wedge (4)}{ \equal{} }\ \frac {x(y \plus{} z) \minus{} yz}{x(y \plus{} z) \plus{} yz}\cdot \frac {x^2(y \plus{} z) \plus{} 4xyz}{y \plus{} z} \equal{} x^2$ $ \Longleftrightarrow$

$ \frac {x(y \plus{} z) \minus{} yz}{x(y \plus{} z) \plus{} yz} \equal{} \frac {x^2(y \plus{} z)}{x^2(y \plus{} z) \plus{} 4xyz}$ $ \Longleftrightarrow$ $ \frac {x(y \plus{} z)}{yz} \equal{} \frac {2x^2(y \plus{} z) \plus{} 4xyz}{4xyz}$ , i.e. $ \boxed {\ \Gamma I\perp AD \ \Longleftrightarrow\ x(y \plus{} z) \equal{} 2yz\ }\ \ (6)$ .

From $ (5)\wedge (6)$ obtain the chain of equivalencies $ \boxed {\ \mathrm G\Gamma\parallel BC\ \Longleftrightarrow\ a(p \minus{} a) \equal{} 2(p \minus{} b)(p \minus{} c)\ \Longleftrightarrow\ \mathrm I\Gamma\perp AD\ }$ .[/color]

[color=darkblue][b]Remark.[/b] Exists a triangle $ ABC$ for which $ \mathrm G\Gamma\parallel BC$ , i.e. $ a(p \minus{} a) \equal{} 2(p \minus{} b)(p \minus{} c)$ ?! The answer is [b]YES[/b]. Indeed, show easily

that $ a(p \minus{} a) \equal{} 2(p \minus{} b)(p \minus{} c)\Longleftrightarrow f(\underline a)\equiv 2\underline a^2 \minus{} (b \plus{} c)\underline a \minus{} (b \minus{} c)^2 \equal{} 0$ . Observe that $ a > 0$ and $ \frac {b \plus{} c}{4}\le a$ . Suppose

w.l.o.g. $ b < c$ . Thus $ \begin{array}{c} f(b) \equal{} c(b \minus{} c) < 0 \\
\ f(c) \equal{} b(c \minus{} b) > 0 \\
\ f(c \minus{} b) \equal{} \minus{} 2b(c \minus{} b) < 0\end{array}\implies\max \{c \minus{} b,b\} < a < c$ and $ A < 90^{\circ}$ , i.e. $ \boxed {\ b < a < c < a \plus{} b\ }$  .

If $ b\equal{}c$ , then $ f(a)\equal{}0\Longleftrightarrow 2a^2\minus{}2ab\equal{}0$ , i.e. $ a\equal{}b\equal{}c$ and in this case $ G\equiv I\equiv\Gamma$ .[/color]


induction
geometry
algebra
binomial theorem
difference of squares
special factorizations

Is there any simple reason why:
[code]
1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2
[/code]
?

Just a hint would be fine, too.

Note that
[tex]\displaystyle\sum_{k=0}^n (1+k)^4=\sum_{k=0}^n (k^4)+(1+n)^4[/tex]
Binomial Theorem
[tex]\displaystyle\sum_{k=0}^n (1+k)^4=\sum_{k=0}^n (1+4k^3+6k^2+4k+k^4)[/tex]

=[tex]\displaystyle\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)+\sum_{k=0}^n (k^4)[/tex]So
[tex]\displaystyle\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)+\sum_{k=0}^n (k^4)[/tex]=[tex]\displaystyle\sum_{k=0}^n (k^4)+(1+n)^4[/tex]
[tex]\displaystyle(1+n)^4=\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)[/tex]
[tex]\displaystyle\sum_{k=0}^n (1)=1+n[/tex]
[tex]\displaystyle6\sum_{k=0}^n (k^2)=n(n+1)(2n+1)[/tex]
[tex]\displaystyle4\sum_{k=0}^n (k)=2n(n+1)[/tex]
sub them in and solve for
[tex]\displaystyle\sum_{k=0}^n (k^3)[/tex]
we have [tex]\displaystyle\sum_{k=0}^n (k^3)[/tex]
=[tex]\displaystyle\frac{n^2(n+1)^2}{4}[/tex]
= [tex]\displaystyle\frac{(n(n+1))^2}{2^2}[/tex]

I think the question meant to ask for some clever (combinatorial?) or intuitive reason as to why it should be the case rather than a proof. I don't know of a good reason, but if someone does know of one, I would be delighted to learn.

ComplexZeta was right - I was looking for an intuitive reason. I've tried reasoning geometrically but haven't come up with anything. Nonetheless, I did enjoy beta's proof, and sometimes, it seems, intuitive understands can be developed from proofs.

I've never seen a clear reason why that should be the case.  I've always just assumed it was coincidental.  You might try expanding everything out with the binomial theorem for small values and seeing what happens.

A geometric approach seems unlikely to me, because of the high levels of the degrees -- fourth-degree things don't make for obvious geometric solutions.

try http://merganser.math.gvsu.edu/calculus/summation/cubes.html
they seem to have a geometric representation of both sums.

induction works, but it doesn't really give an intuitive understanding of why it is true.

[quote="Alison"]This can also be seen as a classic example of telescoping series.

There's the nifty fact that n^3 = (1 + 2 + ... + n)^2 - (1 + 2 + ... n-1)^2, which I'll leave you to check (you can brute force it through with the formula for the triangular numbers, although if you factor the difference of squares it becomes almost a proof without words).

Then you get

1 ^ 3 + 2 ^ 3 + 3 ^ 3 + ... n ^3 =

1^2 + ((1 + 2) ^ 2 - 1 ^ 2)  + ((1 + 2 + 3) ^2 - (1 + 2)^2) + ... + ((1 + 2  + 3 + ... n)^2 - (1 + 2 + 3 + ... n - 1) ^ 2)

And all the squares cancel out except for the (1 + 2 + 3 + ... n - 1) ^ 2 at the end -- hoorah![/quote]

WOW

[quote="polymorphic"]There is a nice geometric proof.  Think of a big square with (n+1)x(n+1) small squares.  Now look at the (n-1)x(n-1) square in the interior.  Replace it with nxn squares of an even smaller size.   Now keep doing that until you get a 2x2 square.  Express the area of the original (n+1)x(n+1) square in two ways.[/quote]

Could you clarify?  I don't quite see what you mean by the replacing squares part.

Maybe it's clearer if I start at the center and work outwards.  Start with a 2x2 square.  Now 2 - 1 = 1.  So put 1 square on each of the four sides of the original square.  Add in the corners to make a big 3x3 square.  (Each of the nine squares is the size of the original 2x2 square.)  Now 3 - 1 = 2.  So Put 2 squares on each side side of the big 3x3 square.  Add in the corners to make a big 4x4 square.  Continue until you reach n+1.  So in the end, the figure should consist of squares that get smaller as you approach the center.  Sorry i don't have an illustration.

I see what you mean now.  After dividing by 4, each time you basically add n^2 + 2*n*(n-1)*n/2 = n^3.  That's cool, although I'm not sure if the problem got any simpler/intuitive.

fyi the link that i posted above has the geometrical representation

I wonder if this identity can be proven combinatorially... I'm going to take a look over the next couple of days.

(EDIT:) I found such a proof online:

http://www.math.hmc.edu/~orrison/research/papers/two_quick.pdf


Just wondering how many people in AoPS\Mathlink likes to play sudoku .  :)

theres a topic for this in Math and Strategy Games...

I love sudoku.  I bought this book that has 200 puzzles yesterday.

I do. Interesting avatar, math92. I agree, this should be in the Math and Strategy Games forum.

I bought a book for $\$ 5$ with some obvious methods (i.e. scanning the rows and columns). Wasn't much help, but hey, it was only $\$ 5$...

Sorry for the LaTeX errors--I forgot about not using dollar signs.

Heres a tricky question: Whats the least number of numbers you need at the start to be sure of a unique solution? (To a 9*9 grid) At least 9 are needed,  to define what the symbols actually are, but thats i far as ive gotten. Can anyone come up with a way of counting the number of filled in grids there are?

And on a completely different note, theres a book in Easons called Shitedoku, where each of s,h,i,t,e,d,o,k,u, appear in each row, column and box. An interesting twist!

[quote="seamusoboyle"]Heres a tricky question: Whats the least number of numbers you need at the start to be sure of a unique solution? (To a 9*9 grid)[/quote]

This is an open problem(as long as this problem wasn't solved within last 4 months :D ). I used to know the upper bound, but I forgot. :(

Well can anyone count the number of grids?


Νέο τοπικ για θέματα Γεωμετριας!

ΑΣΚΗΣΗ 1

Να δείξετε ότι σε κάθε τρίγωνο ΑΒΓ , με ημιπερίμετρο $p$, ακτίνα περιγεγραμμένου κύκλου $R$, ακτίνα εγγεγραμμένου κύκλου $r$ και εμβαδό $S$ ισχύουν οι σχέσεις:

$a^2(p-a)+ b^2(p-b)+ c^2(p-c)=4S(R+r)$ και

$\frac{a(b+c)}{p-a}+\frac{b(c+a)}{p-b}+\frac{c(a+b)}{p-c}=\frac{4pR}{r}$ !!

 :)

Είναι
    α^2(ρ-α) + b^2(ρ –b) + c^2(ρ –c) = ρ(a +b+c) + 2ρ(ab + bc + ca) – (a +b+c)^3 +
                                                          = 3(a +b+ c)(ab +bc +ca) – 3abc  (1) .
Είναι γνωστό ότι
                                 α +b +c = 2ρ
                               ab + bc + ca =ρ^2 + r^2 + 4Rr
                                abc = 4ρRr
αντικαθιστώντας στην (1) και μετά από πράξεις παίρνουμε το ζητούμενο .
Είναι
a(b +c)/(ρ –α) + b(c +a)/(ρ –b) + c(a +b)/(ρ –c) =
= [2ρ^2(ab+bc+ca)+2abc(a+b+c)–ρ(a+b+c)(ab+bc+ca)-3ρabc]/(ρ–α))(ρ-b)(ρ-c)  (2)
Επίσης είναι γνωστό ότι
                                           (ρ–α))(ρ-b)(ρ-c) = ρr^2
αντικαθιστώντας στην (2) και μετά από πράξεις παίρνουμε τη ζητούμενη .

$(R,r,s)\Longleftrightarrow (a,b,c)\Longleftrightarrow (r_a,r_b,r_c)\Longleftrightarrow (s-a,s-b,s-c)\Longleftrightarrow$$(\tan \frac A2,\tan \frac B2,\tan \frac C2)\Longleftrightarrow\ a.s.o.$
$\blacksquare\ 1^{\circ}.\ a+b+c=2s;\ ab+bc+ca=s^2+r^2+4Rr;\ abc=4Rsr.$

$\blacksquare\ 2^{\circ}.\ r_a+r_b+r_c=4R+r$ $;\ r_ar_b+r_br_c+r_cr_a=s^2;\ r_ar_br_c=rs^2;\ ar_a=s(r_a-r).$

$\blacksquare\ 3^{\circ}.\ a.s.o.$ \[ \underline {\overline {\left|\ 3\sqrt 3r\le s\le \frac{3\sqrt 3}{2}R\ \right| }}  \]$\sum \frac{a(b+c)}{s-a}=\frac{1}{2S}\sum a(2s-a)r_a=\frac{1}{2r}\sum (2s-a)(r_a-r)\ a.s.o.$

Βάζω λύση σε αυτό το πρόβλημα . Απλα αλλάζω τα γράμματα gia na mporo na grafo latex
[quote="Anto"]Δίνονται δύο σταθεροί κύκλοι, που τέμνονται στα σημεία Α, D.
Σταθερή γωνία περιστρέφεται γύρω απο το Α. (Το σημείο Α είναι κορυφή αυτής.) και τέμνει τους δύο κύκλους στα σημεία B, C. Να βρεθεί ο γεωμετρικός τόπος του μέσου του τμήματος BC.[/quote]

Λύση :
Έστω $K_{1}, K_{2}$ τα κέντρα των κύκλων που περιέχουν τα B,C αντίστοιχα και έστω ότι οι ευθείες $K_{2}C$ και $K_{1}B$ τέμνονται στο $N$ και έστω Μ το μέσο του BC.
Εύκολα δείχνουμε ότι η γωνία $K_{1}NK_{2}$ είναι σταθερή (γιατί?).
Αν τώρα Κ είναι το μέσο του $K_{1}K_{2}$ τότε έχουμε
$|\vec{KM}|=\frac{1}{2}|\vec{K_{1}B}+\vec{K_{2}C}|=\frac{1}{2}\sqrt{R_{1}^{2}+R_{2}^{2}+2R_{1}R_{2}cos\ K_{1}NK_{2}}=ct$
άρα ο γεωμετρικός τόπος του Μ είναι κύκλος κέντρου K kai aktinas c

Επειδή δεν είναι αναλυτική η λύση αν έχετε πρόβλημα (που δεν πιστεύω ) κάπου , πείτε μου.

[quote="Anto"]Δίνονται δύο σταθεροί κύκλοι, που τέμνονται στα σημεία Α, Β.
Σταθερή γωνία περιστρέφεται γύρω απο το Α. (Το σημείο Α είναι κορυφή αυτής.) και τέμνει τους δύο κύκλους στα σημεία Κ, Λ. Να βρεθεί ο γεωμετρικός τόπος του μέσου του τμήματος ΚΛ.[/quote]

Στέργιε ο Αρετάκης στην έβαλε αυτή την άσκηση???Λοιπόν έκανα μια πολύ ωραία συνθετική απόδειξη.Θα τη γράψω άμεσα..

[i]Πρόκειται για μια αρκετά όμορφη άσκηση από Βιετναμέζικη Ολυμπιάδα-αν δεν κάνω λάθος![/i]

[quote="Anto"]Δίνονται δύο σταθεροί κύκλοι $C_{1},C_{2}$  που τέμνονται στα σημεία $A,B$.
Σταθερή γωνία $\omega$ περιστρέφεται γύρω απο το $A$. (Το σημείο $A$ είναι κορυφή αυτής.) και τέμνει τους δύο κύκλους στα σημεία $K,L$.Να βρεθεί ο γεωμετρικός τόπος του μέσου του τμήματος $KL$.[/quote]

[u][b]ΛΥΣΗ[/b][/u]
Έστω $M$ το μέσο του τμήματος $KL$.Προεκτείνω την $AM$ κατά τμήμα $MS=AM$.Έστω $S_{1}=SK$ $\bigcap$ $C_{1}$ και
$S_{2}=SL$ $\bigcap$ $C_{2}$.Αλλά $\widehat{AKS_{1}}$ $=$ $\omega$ $=$ $\widehat{ALS_{2}}$, πράγμα που σημαίνει ότι τα σημεία $S_{1}$ και $S_{2}$ είναι σταθερά.Επίσης
 $\widehat{S_{1}SS_{2}}$ $=$ $\omega$,άρα το σημείο $S$ <<βλέπει>> το σταθερό τμήμα $S_{1}S_{2}$ υπό σταθερή γωνία $\omega$ και συνεπώς κινείται σε ένα σταθερό κύκλο $C'$.
Εφαρμόζουμε την ομοιοθεσία $H(A, \frac{1}{2})$.Μέσω αυτού του γεωμετρικού μετασχηματισμού, κάθε θέση του  $S$ <<πάει>> στην αντίστοιχη θέση του $M$.Έτσι ο γεωμετρικός τόπος του $M$ είναι ο κύκλος $C$ πού προκύπτει από την ομοιοθεσία $H$.

[i]Πως σας φαίνεται?[/i]

ΑΣΚΗΣΗ 2

Ν.Δ.Ο $ab+bc+ca=t^{2}+r^{2}+4Rr$
οπου $a,b,c,t,R,r$ειναι οι πλευρες τριγωνου,η ημιπεριμετρος,η ακτινα του περιγεγραμενου,και η ακτινα του εγγεγραμενου κυκλου,αντιστοιχα....

 :D

[color=red]ΑΣΚΗΣΗ 2 [/color]
[hide]
$tr = \sqrt{t(t-a)(t-b)(t-c)}\Rightarrow $

$t^{2}r^{2}= t(t-a)(t-b)(t-c) \Rightarrow $

$tr^{2}= (t-a)(t-b)(t-c) \Rightarrow $

$tr^{2}= t^{3}-(a+b+c)t^{2}+(ab+bc+ca)t-abc \Rightarrow $

$tr^{2}= t^{3}-2t^{3}+(ab+bc+ca)t-abc \Rightarrow $

$tr^{2}=-t^{3}+(ab+bc+ca)t-abc \Rightarrow $

$(ab+bc+ca)t = tr^{2}+t^{3}+abc \Rightarrow $

$ab+bc+ca = r^{2}+t^{2}+\frac{abc}{t}$

Αλλά $\frac{abc}{4R}= tr \Rightarrow  \frac{abc}{t}= 4Rr$[/hide]

Ρε παιδιά καλύτερα να συνεχίσουμε εδώ
http://www.mathlinks.ro/Forum/viewtopic.php?t=113622&start=20

μην ανοίγουμε πολλά μέτωπα .(Και στην σελίδα παραπάνω η τελευταία άσκηση μένει άλυτη εδώ και μια βδομάδα .Περιμένω τις λύσεις σας .)


geometry
geometric transformation
rotation

x,y and z are positive reals such that x=a+b,y=b+c,z=a+c.
describe a method to construct an equilateral triangle ABC in the interior of which there is a point p such that PA=x,PB=y,PC=z.

happy solving!

Actually, for an equilateral triangle ABC and ANY point P inside, such that $PA=x,PB=y,PC=z$, there exist positive numbers $a,b,c$, such that $x=a+b,y=b+c,z=c+a$.(Pompeiu`s Theorem). Actually, the existence of that numbers $a,b,c$ means that $x,y,z$ are sidelenghts of a triangle.
You just need to construct that triangle with sides $x,y,z$ and draw its incircle.
To construct the triangle wth sidelenghts $x,y,z$, consider a $60^\circ$ rotation from the angle $\angle{BAC}$. The rest is easy.


function
limit
trigonometry
calculus
calculus computations

$ \displaystyle\lim_{m\to\infty} (\displaystyle\lim_{n\to\infty}( \cos{}^2{}^m (n! \pi x ))) = ?$

If $ x$ is rational, then $ n!\pi x$ is a multiple of $ \pi$ for sufficiently large $ n$, so $ (\cos{n!\pi x})^{2k}\equal{}1$ for all such $ n$, so $ \lim_{n\rightarrow\infty}(\lim_{k\rightarrow\infty}(\cos{n!\pi x})^{2k})\equal{}1$.  If $ x$ is irrational, then $ n!\pi x$ is not a multiple of $ \pi$ for any $ n$, so $ |\cos{n!\pi x}|<1$, so $ \lim_{k\rightarrow\infty}(\cos{n!\pi x})^{2k}\equal{}0$.

This is the same as exercise VII3.6 from Serge Lang's Undergraduate Analysis.
The solution which is given there, goes like this:
Suppose $ x\equal{}p/q$, then for large $ n$, $ n\geq 2|q|$, hence $ n!\pi x\equal{}2\pi k$ for some $ k\in\mathbb{Z}$, thus f(x)=1 (where f(x) is of course defined as that limit).
This is essentially the same argument as JRav's, but I don't see it yet.

Why is it true that "Suppose $ x\equal{}p/q$, then for large $ n$, $ n\geq 2|q|$"?

$ q$ is just some integer? There are always naturals, i.e. $ n$, such that $ n \geq |q|$, or even $ n \geq 2|q|$.

That is essentially the Archimedean property.


Find the digits $a;b;c;d;e$ such that:
$\overline{abcd}+\overline{deb}=2007$
and $\overline{abcd}-\overline{deb}=2007$

[quote="Rust"][quote="olorin"]:huh:
$\overline{deb}={(\overline{abcd}+\overline{deb})-(\overline{abcd}-\overline{deb})\over 2}=0$.
and $\overline{abcd}=2007$.
So $d=0=7$. Contradiction![/quote]
If $a=2,b=0=c,d=7$, then $\overline{deb}=700+10b\not =0$.
I think, that 2 exesizes.
1. $\overline{abcd}+\overline{deb}=2007.$
2. $\overline{abcd}-\overline{deb}=2007$
1. had not solution because $b+d=7 \ or \ 17$ and $b+d=9 \ or \ 10$.
2. $a=2,d=b-3$ had not solution.[/quote]

No this is an exercise


In [url=http://groups.google.com/groups?dq=&hl=de&lr=&ie=UTF-8&oe=UTF-8&threadm=slrnc4rcla.j1m.mautsch%40almahani.math.ethz.ch&prev=/groups%3Fdq%3D%26num%3D25%26hl%3Dde%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26group%3Dde.sci.mathematik%26start%3D25]this de.sci.mathematik thread[/url], Daniel Gutekunst asks about the proof of the following inequality:

(a2 + b2)(b2 + c2)(c2 + a2) / 8 >= (ab + bc + ca)3 / 27

for positive a, b, c.

Any idea?

  Darij

It took me a time to realize that this is equivalent to the right bound of the Mathlinks Contest 3rd Edition 1st Round problem 2 !!!

  Darij

We can use the inequality below:(not very difficult to prove)
9(a 2 +b 2 )(b 2 +c 2 )(c 2 +a 2 ) \geq 8(a 2 +b 2 +c 2 )((ab) 2 +(bc) 2 +(ac) 2 ).
Then by Cauchy and B.C.S inequality,we have
suma 2  \geq ab+ac+bc
sum(ab) 2  \geq (ab+ac+bc) 2 /3.
By all this,we have the inequality we need :)

We can prove by
    problem<=>  \sum 27a^4b^2+a^2b^2c^2-4a^3b^3-24a^3b^2c (\sum  sym)
       we can use shur  to prove that  .


Let ABC be a right-angled triangle with ACB = 90. The
inscribed circle O of AABC is tangent to BC, CA , AB at D, E,
F respectively. AD intersects circle Oat P, BPC = 90". Prove that
AE + AP = PD.

“ACB = 90”is unnecessary，this problem is established for arbitrary triangle .
[b]Just harmony quadrilateral[/b]： Δ$ PMN\equal{}$ Δ$ DMN$，Then $ PM\equal{}DM$，$ DF\equal{}2PF$,，hen $ DP\equal{}AP\plus{}AE$ .

[b]Xiong Bin[/b] is chinese [b]IMO[/b] Captain，The “ACB = 90” is tset your calculation and comprehensive ability .


I am having trouble calculating the period of a sine or cosine graph. I even put my calculator to zoomtrig for my graphing zoom options and still have trouble.

I know that the entire period would be 2pi but what if an equation is given such as y=1/3 sin4(pi)(x) ?[/list][/code]

For a graph y=sin(ax) or y=cos(ax) the period is 2(pi)/a

period of $ \sinh$ is $ 2\pi$ so it repeats itself every $ 360^o$

If you have $  a\sinh x$ replacing x by bx it stretches the graph horizontally by a fractor of $ \frac {1}{b}$.

so $ y = asinbx$ has a period $ {2\pi}{b}$. So period relies on the $ bx$ bit.

Yeah, right. thats the general formula for periods of sin(x) and cos(x)

I think bos is right.  But don't forget about putting $ 2\pi$ [b]radians[/b].  $ 2\pi$ degrees is something else entirely.

Side note: does anyone else get bos' post with a gigantic blank spot in the middle?

I got it , too. Please edit your post, bos1234. :)

it is edited.. there is one line between each sentence but for some reason that huge space is there:S


[quote="ambarish phanse"][quote="socrates"][quote="mohamadhosein"]let$a,b,c,d,e$real numbers prove:
$a^2+b^2+c^2+d^2+e^2 \ge a(b+c+d+e)$[/quote]

Why do you post this here ??  :D
Anyway, just see it is equivalent to

$(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2e)^2\geq 0$[/quote][/quote]

Hey! what are you doing??
Where is your problem??
 :?

[quote="socrates"][/quote][quote="ambarish phanse"][quote="socrates"][quote="mohamadhosein"]let$a,b,c,d,e$real numbers prove:
$a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\ge a(b+c+d+e)$[/quote]

Why do you post this here ??  :D
Anyway, just see it is equivalent to

$(a-2b)^{2}+(a-2c)^{2}+(a-2d)^{2}+(a-2e)^{2}\geq 0$[/quote][/quote][quote="socrates"]

Hey! what are you doing??
Where is your problem??
 :?[/quote]


limit
algebra
polynomial
real analysis
real analysis unsolved

If P is from R[x] and gr(P)=p>=1, P(x)>0, x is real then

\lim oo   { 1/ln(n)* [(sum P(k)^(1/n), k=1,n) -n]}=p.

Guess that Cesaro Stolz should be some help with this one!!

So please help me with this one!

cheers! :D

I'll write what you do for P(x)=x, and then you can use the same method (I'm too lazy to use higher deg polynomials). We don't have to worry abt the fact that P(x) isn'i >0 for all real x, because we don't need p(x)>0 for negative x anyway. Call the term of our sequence a_n. Then a_n = (n/lnn)*((1^1/2+2^1/n+..+n^1/n)/n-1). Then a_n <= (n/lnn)*(n^1/n-1). If we write n^1/n=e^lnn/n and we denote k_n=lnn/n we have a_n<=(e^k_n-1)/k_n. Because k_n->0 we get that (e^k_n-1)/k_n -> 1 (*).

We now need a_n >= something. We use AM-GM: (1^1/n+..+n^1/n)/n >
(n!)^(1/n^2). Try to prove that a_n > (n/lnn)*((n!)^(1/n^2)-1) -> 1 (**). It's not hard. After we have proved this we get a_n -> 1, which is the degree of P. It's really easy to extend this proof. Here I used the fact that P(n) > P(i) for any i<n. You don't know that for the general case, but it's true from a certain n0, because any polinomial that has only positive values must become strictly increasing from a certain point.

Indeed for P(x)=x the problem is quite simple.. the problem I posted had two points.

i) solve it for p(x)=x and ii) generalization....

I think the generalization is quite difficult... you should try to solve it.. cause I haven't managed to solve it!

cheers! :D


geometry
circumcircle
incenter
trigonometry
geometric transformation
reflection
Ring Theory

Let $ABC$ be a triangle and its bisector at $A$ cuts its circumcircle at $D.$ Let $I$ be the incenter of triangle $ABC,$ $M$ be the midpoint of $BC,$ $P$ is the symmetric to $I$ with respect to $M$ (Assuming $P$ is in the circumcircle). Extend $DP$ until it cuts the circumcircle again at $N.$ Prove that among segments $AN, BN, CN$, there is a segment that is the sum of the other two.

http://www.mathlinks.ro/Forum/viewtopic.php?t=15773 .

  darij

[quote="orl"][quote="shobber"]Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem.[/quote]

Congrats for making a big step for being on the Chinese team.  :D[/quote]
I got this information from a book. It provided a elementary trig proof, which is also hard to come up with.
The book says we can turn this problem into proving:
\[\tan{\angle{ADN}}=\frac{\sin{C}-\sin{B}}{1-\cos{B}-\cos{C}}\]
Most guys managed to get here, but failed to go deeper.

A very hard and nice problem! I have nearly spent 2 hours on it.

WLOG, assume $b\ge a\ge c$. We will start with 2 lemmas:
Lemma 1
$\angle ACN=\angle IOD$.
Proof:
Let D' be the reflection of D in BC, D'' the antipode of D.
By similar triangle, $BD^2=DM\cdot DD''\Rightarrow DI^2=2DM\cdot DO=DD'\cdot DO$. Thus, $\angle IOD=\angle D'ID=\angle PID+\angle IPD=180^\circ-\angle IDP=\angle ACN$.

Lemma 2
Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$
Proof: problem posted before by me :P

Now we apply a homothety to $\triangle AXY$ centered at A with the ratio $\frac{AC}{AY}$. Suppose the image is ACZ. Then consider the spiral smilarity of $\triangle ACZ$ centered at C through an angle $\angle ZCB$ (anticlockwisely) and ratio $\frac{BC}{AC}$. Suppose the image is $\triangle CBN'$. Clearly $N'\in(O)$. We have $\frac{BN'}{CN'}=\frac{a-c}{b-a}\Rightarrow bBN'+cCN'=a(BN'+CN')$. But the Ptolemy Theorem yields $bBN'+cCN'=aAN'$. Therefore BN'+CN'=AN'.

Note that $90^\circ=\angle(XY,IO)=\angle(ZC,IO)=\angle(BC,IO)+\angle BCZ=90^\circ-\angle IOD+\angle ACN'$. Thus, $\angle IOD=\angle ACN'$. By Lemma 1, N=N', so we also have AN=BN+CN.
For the case $b\ge c\ge a,a\ge b\ge c$, etc. The proof is essentially the same so I don't want to repeat, which is extremely tedious and boring.

[quote="mecrazywong"]

Lemma 2
Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$
Proof: problem posted before by me :P [/quote]

Do you have a link or idea for proof this lemma ?

I have a really nice solution . I made it yesterday
as all know we must prove $\sin AP=\sin BP + \sin CP$
we know that $\frac{XM}{XI}=\frac{1}{2}=\frac{TM}{TI}(=\sin \frac{A}{2}).\frac{\sin AP -(\frac{B-C}{2})}{\sin AP}\Leftrightarrow \sin AP = 2.\sin \frac{A}{2}.\sin AP -(\frac{B-C}{2})=\sin PB + \sin PC$

In early 90's, this problem seemed very hard, but it's not that hard nowaday.
I saw a very smart and simple solution, maybe the official solution, on a Chinese journal by using the area means.

Dear Mathlinkers,
I have also this reference

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323883

Sincerely
Jean-Louis

Let $a=x^2$, $b=y^2$, $c=z^2$, $d=-yz$, $i=-xy-yz-zx$, $m=\frac{y^2+z^2}2$, $p=y^2+z^2+xy+yz+zx$. Then,

\begin{align*}
n&=\frac{d-p}{d\overline p-1}\\
&=\frac{-(y+z)(x+y+z)}{-yz\left(\frac1{y^2}+\frac1{z^2}+\frac1{xy}+\frac1{yz}+\frac1{zx}\right)-1}\\
&=\frac{-xyz(y+z)(x+y+z)}{(y+z)(xy+yz+zx)}\\
&=\frac{xyz(x+y+z)}{xy+yz+zx}.
\end{align*}

Therefore,
$$n-a=\frac{xyz(x+y+z)-x^2(xy+yz+zx)}{xy+yz+zx}=\frac{x(yz-x^2)(y+z)}{xy+yz+zx},$$ so $AN=\frac{|yz-x^2||y+z|}{|xy+yz+zx|}.$ Therefore, $|yz-x^2|^2|y+z|^2=-\frac{(yz-x^2)^2(y+z)^2}{x^2y^2z^2}$, so $ANxyz|xy+yz+zx|=\pm i(yz-x^2)(y+z)$. Therefore, there exists a sign $\varepsilon_A$ such that $ixyz|xy+yz+zx|\varepsilon_AAN=y^2z+yz^2-x^2y-x^2z$, so taking the cyclic sum implies $\varepsilon_AAN+\varepsilon_BBN+\varepsilon_CCN$, so one of the three segments is the sum of the other two.


If Sammy won a billion dollars in the Super Lottery, and he planned to spend $ \$3,000$ a day, how many centuries would it take
to spend it all? Express the answer to the nearest whole number of
centuries. Assume Sammy can earn no interest on his money and
that 365 days = 1 year.

Using a calculator, $ 1,000,000,000\div3000\equal{}333333.3333\approx913$ years. It would take Sammy nine centuries to spend all of his money.

Does anyone know how to do this quickly without a calculator?  Because this is what I'm having trouble with  :blush:

$ \frac{\$1,000,000,000}{\left(\frac{\$3,000}{\text{day}}\right)\left(\frac{365\text{ days}}{1\text{ year}}\right)\left(\frac{100\text{ years}}{1\text{ century}}\right)}$.
We can see how a lot a numbers factor out. The $ \$3,000$ factors out $ 3$ zeros in the numerator, and the $ 100$ factors out another $ 2$ zeros. We are left with $ \frac{10,000}{\frac{1,095}{1\text{ century}}}\equal{}\frac{10,000}{1,095}\text{centuries}$. Approximating $ \frac{10,000}{1,095}\approx9$ so the final answer is $ \boxed{9}\text{ centuries}$

technically undefined because he cant live that long, but yes 9

Wouldn't the answer be $10?$ If the number you got when you did $\frac{10, 000}{1, 095}$ was greater than $9,$ (which it is) it would take $10$ centuries to completely exhaust the money.


Problem #1
$ A^{100}\equal{}I_n$ find $ A^{\minus{}1}$

Problem #2
Let A=$ \begin{Bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 1 \end{Bmatrix}$

Diagonalize A and give a formula to find $ A^{100}$

Problem 2 takes some computation but maybe not too much computation.

We see that the rows all have the same sum, $ 3.$ That means that $ 3$ is an eigenvalue with eigenvector $ (1,1,1)^T.$ Then there's the nearly-zero second column. If we just subtracted off $ 2$ times the identity, we'd zero that column, so $ 2$ is an eigenvalue with eigenvector $ (0,1,0)^T.$ Now, the trace of the matrix is $ 4$ and that's the sum of the eigenvalues, so the third eigenvalue is $ -1.$ Knowing that, we can find the third eigenvector, which is $ (3,1,-3)^T.$

Probably the most computation I need to do is to find the inverse of the matrix whose columns are the eigenvectors. I have some choice here in choosing what scalar multiple of each eigenvector to use; I didn't try to get to clever there but I suspect there's a choice that makes thinks slicker.

We have $ P=\begin{bmatrix}1&0&3\\1&1&1\\1&0&-3\end{bmatrix}$ and compute that $ P^{-1}=\frac16\begin{bmatrix}3&0&3\\-4&6&-2\\1&0&-1\end{bmatrix}.$

Then $ P^{-1}AP=D=\begin{bmatrix}3&0&0\\0&2&0\\0&0&-1\end{bmatrix}$ and $ A=PDP^{-1}.$

So $ A^{n}=PD^{n}P^{-1}=\frac16\begin{bmatrix}{3\cdot 3^n+3(-1)^n}&0&3\cdot 3^n-3(-1)^n\\3\cdot 3^n-4\cdot 2^n+(-1)^n&6\cdot 2^n&3\cdot 3^n-2\cdot 2^n-(-1)^n\\3\cdot 3^n-3(-1)^n&0&3\cdot 3^n+3(-1)^n\end{bmatrix}.$

As a check, put first $ n=0$ and then $ n=1$ into this expression. We do get, respectively, $ I$ and $ A,$ so the computation does seem to check. Put $ n=100$ to get the desired answer. Note that to a first approximation,
\[ A^{100}\approx \frac{3^{100}}2\begin{bmatrix}1&0&1\\1&0&1\\1&0&1\end{bmatrix}\]


Aki un problema:

Demostrar que para todo $m$ natural, existen una infinidad de enteros $(x, y)$ tales que se cumplan las siguientes condiciones:
1) $x, y$ son primos relativos
2) $x$ divide a $y^{2}+m$  y   $y$ divide a $x^{2}+m$

perdão, mas falarei em português!

(x,y) = (1,1) é solução.

agora, veja que se (x,y) é solução, com x<y, $(\frac{y^{2}+m}{x}, y)$ também é solução.
e que, como y>x, temos y²+m > x²!

[color=blue]Intenté hacer una solución similar, por inducción, tomando que siempre $(1;1)$ cumple.
Se comprende la demostración, pero creo que faltan ciertas justificaciones. ¿Podrías redactar una solución formal al enunciado?[/color]


quadratics
trigonometry
calculus
probability
Support
geometry
geometric transformation

I find that calculator programming is useful because of the fast solving and shortness of the programs for formulas (i.e. quadratic formula can be 2 lines). I also enjoy my attempts at making a game. So far I've made 3/4 a snake game (moving ok, hitting wall = die, hitting dot = grow, HITTING TAIL = NOT DIE).

Also when teachers allow programs on tests, I can program the formulas onto the calculator. I mean making a program that solves the problem, not a lame LETS PUT THE FORMULAS ONTO THE CALCULATOR cheats. At my school, the teachers say: if you have skills to do it, do it.

Another question is, do you think this is cheating?

yo i love writing programs to do multiple choice and double-check my answers and stuff

i used to program games in like 8th grade? it's introduced me to java, which is a lot cooler, though.

[quote="t0rajir0u"]Define "useful."  Many of my computer science-type friends were introduced to programming through their calculators and went on to learn several other programming languages on other platforms and to become interested in computer science in general.  On the other hand, writing a program to solve a math problem isn't the same thing as knowing how to solve a math problem.[/quote]

"Useful" as in a useful skill too know. It's a good method for brute force (was used for the old AMC). I actually did both java and calculator at around the same time, since the languages are similar. I agree knowing how to solve a problem is definitely better, but if time is short, one can always resort to programmed programs.

Calculator programming is not useful in the sense that it will never be really important that you be able to program on your calculator.  On the other hand, calculator programming is useful in the sense that learning to program at all is useful and the more time you spend doing it, the better you are likely to become at it.

Programming in TI-BASIC encourages bad practice and ugly spaghetti code. I suggest learning a better language first so you are not led astray.

Meh, calculator programming is a useful skill because it keeps you from getting bored in class.

I've made several useful programs (my biggest being one where you can enter the molecular formula for a compound and get the mass, as to not waste 20 hours looking at the periodic table and copying in masses). Personally, I do not feel that using these programs is cheating. Usually, if you have the ability to make a working program, you no longer need to be tested on that skill anyway. Especially if its point is to just save time on a trivial task. However, being unsure, my conscious usually makes me use my programs on homework, but not tests.

On a note of usefulness and learning programming, TI assembly can also be useful cause it helps you learn how computers work at the lowest level (though calculators are surely far simpler than computers). TI-Basic is a sort-of-okay introduction to programming but its best uses are mathematics (well, it is a calculator :P)

[quote="moogra"]At my school, the teachers say: if you have skills to do it, do it.[/quote]

Wow, this is rare.

In my opinion, if you know the formula, then you can program it into your calculator. If you already understand the concept, then why waste your time going through tedious work for something you already know? In general, it's a time saver, and a useful thing when certain problems require many tedious calculations.

The calculator is a tool to make life easier. not to do the work for you.

in this case, formulas you know, formulas that you have proved before, are okay.

distance formula and quadratic formula are the two major ones.

all the roots and stuff they just throw me off when I'm calculating it out by hand.
the calculator helps me. it doesn't tell me the answer after i put the question in.

[quote]In my opinion, if you know the formula, then you can program it into your calculator. If you already understand the concept, then why waste your time going through tedious work for something you already know? In general, it's a time saver, and a useful thing when certain problems require many tedious calculations.
[/quote][/quote]
You took the words right out of my mouth.  Having a slight touch of arithmetic dyslexia or dyscalculia where I have this tendency to commit minor errors in addition and multiplication most of the time, probably brought about by the terror   of corporal punishment (a common practice in a third world country where I live) and other childhood terrors - or maybe my brain and several billion others are not wired to do mental addition and multiplication well - I've found that I really was quite good in math with a great deal of help from a calculator.  It's like a component of my brain that's got something to do with the four basic operations was defective all along and the calculator was my key/cure for doing well in math.
At any rate, that part about the calculator being a time saver rings true with me.

Its a very interesting topic. I'm pretty good at TI programming, but it just gets old. I'm taking some programming classes next semester to learn all the good stuff. I can write in java, but not to the extent that I'd like.

That being said, there are pros and cons to having a calculator program. Of course theres quadratic solvers, TRIG solvers, composite vs prime, and I've even written one to do all sorts of calculus (was a boon on the AP test). Some of the commands on a TI let you do some great stuff with probability and counting too, as well as number theory. When I wrote all of these things I actually learned a lot about the math behind it and I'm better in those subjects for it.

But then there's the other side of the coin. My teachers also said "If you have the skills, use 'em", but the question is, [i]when should we use them[/i]? Well one thing's for sure, [b]your calculator is the last resort on a standardized test[/b]. You cannot create a program that simply solves any problem on its own. The best you can do is speed up computations that require significant time and boolean arguments. There any many methods for computing tedious arithmetic without the aid of a calculator, and they should be learned.

You won't always have one during a test.

Do anyone know how I can make my TI 83+ work like a TI-89 or TI-nspire by programming or something?

The TI-89 doesn't support BASIC programming so no, sorry. Unless you want to program an amazing ASM program?

Does anyone know how to write a program to perform rotation of axes on a conic of the form $ Ax^2\plus{}Bxy\plus{}Cy^2\plus{}Dx\plus{}Ey\plus{}F\equal{}0$? (It'd probably require writing the new coefficients A', B', C', D', E', and F' in terms of A, B, C, D, E, F, and the sine and cosine of the angle you're rotating it through.)

I'll write it for you and post it if you give me the formula. I'm too lazy to remember it.

**Hangman is done

[quote="moogra"]I'll write it for you and post it if you give me the formula. I'm too lazy to remember it.[/quote]

Never mind. I found how to factor it by plugging the values into the original equation $ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, so I should be able to write the program from there:

$ {x^\prime}^2(A\cos ^2\theta\ + \ B\sin \theta\cos \theta\ + \ C\sin ^2\theta)\ + \ x^\prime y^\prime\{B(\cos ^2\theta\ - \ \sin ^2\theta)\ - 2(A\ - \ C)(\sin \theta\cos \theta)\}\ + \ {y^\prime}^2(A\sin ^2\theta\ - \ B\sin \theta\cos \theta\ + \ C\cos ^2\theta)\ + \ x^\prime(D\cos \theta\ + \ E\sin \theta)\ + \ y^\prime( - D\sin \theta\ + \ E\cos \theta)\ + \ F\ = \ 0$, where $ \theta = \frac {arccot \frac {A - C}{B}}{2}$.

$ A^\prime = A\cos ^2\theta\ + \ B\sin \theta\cos \theta\ + \ C\sin ^2\theta$
$ B^\prime = B(\cos ^2\theta\ - \ \sin ^2\theta)\ - 2(A\ - \ C)(\sin \theta\cos \theta)$
$ C^\prime = A\sin ^2\theta\ - \ B\sin \theta\cos \theta\ + \ C\cos ^2\theta)$
$ D^\prime = D\cos \theta\ + \ E\sin \theta$
$ E^\prime = - D\sin \theta\ + \ E\cos \theta$
$ F^\prime = F$

$ B^\prime$ should always be $ 0$, but it's always a good idea to calculate it just to check that you found the right angle for $ \theta$.

Here's the finished program if anyone else wants it:

:ClrHome
:Degree
:Input "A=",A
:Input "B=",B
:Input "C=",C
:Input "D=",D
:Input "E=",E
:Input "F=",F
:If A=C
:45→θ
:If A≠C
:tan-1(B/(A-C))/2→θ
:{Acos(θ)²+Bsin(θ)cos(θ)+Csin(θ)²,B(cos(θ)²-sin(θ)²)-2(A-C)(sin(θ)cos(θ)),Asin(θ)²-Bsin(θ)cos(θ)+Ccos(θ)²,Dcos(θ)+Esin(θ),-Dsin(θ)+Ecos(θ),F}→L[size=67]1[/size]
:ClrHome
:Disp "A'=","B'=","C'=","D'=","E'=","F'="
:For(X,1,6)
:Output(X,4,L[size=67]1[/size](X))
:End

[quote="gh625"]Here's the finished program if anyone else wants it:

:ClrHome
:Degree
:Input "A=",A
:Input "B=",B
:Input "C=",C
:Input "D=",D
:Input "E=",E
:Input "F=",F
:If A=C
:45→θ
:If A≠C
:{Acos(θ)²+Bsin(θ)cos(θ)+Csin(θ)²,B(cos(θ)²-sin(θ)²)-2(A-C)(sin(θ)cos(θ)),Asin(θ)²-Bsin(θ)cos(θ)+Ccos(θ)²,Dcos(θ)+Esin(θ),-Dsin(θ)+Ecos(θ),F}→L[size=67]1[/size]
:ClrHome
:Disp "A'=","B'=","C'=","D'=","E'=","F'="
:For(X,1,6)
:Output(X,4,L[size=67]1[/size](X))
:End[/quote]

Your program never works. :(
If A=C (and B≠0), then the program will not write a L[size=67]1[/size], and the following lines will either have an error, or will use a wrong list.
If A≠C, then the program doesn't set a value for θ.

Also, if A=C, then having $ \theta\equal{}\frac{\cot^{\minus{}1}\frac{A\minus{}C}{B}}{2}$ will give $ \theta\equal{}45^{\circ}$, so you don't need that bit of code.

Here is a suggested program fix, supposing that long formula I did not bother to read is correct: (also, note that I make mistakes quite often, so I'm probably wrong too. :P )
[hide="fat program"]:ClrHome
:Degree
:Prompt A,B,C,D,E,F
:[i]tan-1[/i](B/(A-C))/2→θ
:{Acos(θ)²+Bsin(θ)cos(θ)+Csin(θ)²,B(cos(θ)²-sin(θ)²)-2(A-C)(sin(θ)cos(θ)),Asin(θ)²-Bsin(θ)cos(θ)+Ccos(θ)²,Dcos(θ)+Esin(θ),-Dsin(θ)+Ecos(θ),F}→L[size=67]1[/size]
:ClrHome
:Disp "A'=","B'=","C'=","D'=","E'=","F'="
:For(X,1,6)
:Output(X,4,L[size=67]1[/size](X))
:End

[i](Note: tan-1 means $ tan^{\minus{}1}$, in case you couldn't tell.)[/i][/hide]

tan-1(B/(A-C))/2→θ

if A-C=0 then this is undefined, so it is necessary to have an "If A=C" case.

I assume the original program meant to have the "tan-1(B/(A-C))/2→θ" line right after the "If A≠C" line

[quote="Brut3Forc3"][quote="gh625"]Here's the finished program if anyone else wants it:

:ClrHome
:Degree
:Input "A=",A
:Input "B=",B
:Input "C=",C
:Input "D=",D
:Input "E=",E
:Input "F=",F
:If A=C
:45→θ
:If A≠C
:{Acos(θ)²+Bsin(θ)cos(θ)+Csin(θ)²,B(cos(θ)²-sin(θ)²)-2(A-C)(sin(θ)cos(θ)),Asin(θ)²-Bsin(θ)cos(θ)+Ccos(θ)²,Dcos(θ)+Esin(θ),-Dsin(θ)+Ecos(θ),F}→L[size=67]1[/size]
:ClrHome
:Disp "A'=","B'=","C'=","D'=","E'=","F'="
:For(X,1,6)
:Output(X,4,L[size=67]1[/size](X))
:End[/quote]

Your program never works. :(
If A=C (and B≠0), then the program will not write a L[size=67]1[/size], and the following lines will either have an error, or will use a wrong list.
If A≠C, then the program doesn't set a value for θ.

Also, if A=C, then having $ \theta \equal{} \frac {\cot^{ \minus{} 1}\frac {A \minus{} C}{B}}{2}$ will give $ \theta \equal{} 45^{\circ}$, so you don't need that bit of code.

Here is a suggested program fix, supposing that long formula I did not bother to read is correct: (also, note that I make mistakes quite often, so I'm probably wrong too. :P )
[hide="fat program"]:ClrHome
:Degree
:Prompt A,B,C,D,E,F
:[i]tan-1[/i](B/(A-C))/2→θ
:{Acos(θ)²+Bsin(θ)cos(θ)+Csin(θ)²,B(cos(θ)²-sin(θ)²)-2(A-C)(sin(θ)cos(θ)),Asin(θ)²-Bsin(θ)cos(θ)+Ccos(θ)²,Dcos(θ)+Esin(θ),-Dsin(θ)+Ecos(θ),F}→L[size=67]1[/size]
:ClrHome
:Disp "A'=","B'=","C'=","D'=","E'=","F'="
:For(X,1,6)
:Output(X,4,L[size=67]1[/size](X))
:End

[i](Note: tan-1 means $ tan^{ \minus{} 1}$, in case you couldn't tell.)[/i][/hide][/quote]

I had the line in my original code, I'd just forgotten to type it in when I moved it to the computer.


ratio
limit
logarithms
algebra unsolved
algebra

For any integer $k> 1$, the equation:
$x^k=x^{k-1}+1$
has a single positive real root $x_0$, $1<x_0<2$.
(for $k$-even has exactly another one negative real root and for $k$-odd has no other real roots)
For $k=2$, $x_{0,2}$ is the golden ratio itself while for $k=3,4$ we can express it with radicals of order $k$.
My questions is if we can express $y_{0,k}=x_{0,k}^{k-1}$ using radicals for this particular equation which solves the recurrence:
$a_n=a_{n-1}+a_{n-k}$
$y_{0,k}$ is the positive real root of:
$y^{\frac{k}{k-1}}=y+1$.
So far I proved that:
$\lim_{k\to \infty} y_{0,k} \frac{\ln k}{k} =1$.

Does anyone agree with my proof of:
$\lim_{k\to \infty}y_{k}\frac{\ln k}{k}=1$
where $y_{k}$ is the positive real root of:
$y^{\frac{k}{k-1}}=y+1$.


inequalities AMC USA(J)MO USAMO symmetry

How do you prove the "bunching" inequality? I don't understand the proof in Kiran's packet.

hmm. Looking at it now, I understand the proof. However, I have no idea how someone woulda come UP with this.

Here's the big question: is this allowable on olympiads? It looks just a little too powerful, and I am pretty sure you have to define what symmetric summations are in proofs.

Would you then have to prove bunching as well?

What Simon says is pretty much correct.  Also, to be on the safe side, you should give a short definition of the symmetric sum notation you're using, even if it's the standard sigma sym.

[quote="ComplexZeta"]In his inequalities lecture at MOP 2002, Reid Barton said you probably can't just quote bunching on the USAMO. But it isn't that hard to prove one specific case of bunching (as opposed to the general theorem). You might have to use AM-GM a few times, but that should be all.

You should be able to use symmetric sums in the USAMO if you use the standard notation, [tex]\displaystyle \sum_{\text{sym}} a^2 b[/tex]. But if you were to use the notation Reid told us he preferred, [tex]\displaystyle\sum_6 a^2 b[/tex], then you must define it.[/quote]

What is the "6" for? is that supposed to be a sigma?

So "bunching" is more of a technique, like smoothing for instance, that you know should always work when majorization exists?

Just to clarify, Kiran gives an (unfinished) example of showing

 :Sigma: [sym] x^5y^2z >= :Sigma: [sym] x^3y^3z^2.

How would you go about showing this, systematically?

(EDIT:) Ok, I think I got the hang of it (sort of)... took a bit of guess-and-check and abuse of symmetry, though.

[quote="Nukular"]What is the "6" for? is that supposed to be a sigma?[/quote]

It refers to the fact that when you write it out the long way, there are 6 terms.

Nukular, would you mind quickly posting an explanation of the proof of bunching? I had a post about this in another section of the forum, and I don't think Kedlaya's proof is complete anyways.

By the way, why can't someone just quote  this thing on an olympiad? Bunching is just Muirhead's Theorem, which sounds like a pretty formal theorem to me.

yea, Muirhead is a very powerful tool used in proving homogeneous inequalities. I cannot imagine that an olympiad grader would not know about it.

well, its good to be able to put a <i>name</i> to it. Thanks.

Although, I would suppose it might be better form to proof a specific case of it, while keeping the idea in mind...


topology
function
real analysis
advanced fields
advanced fields unsolved

If $ f,g$ from $ S^n$ to $ S^n$ are $ C^{\infty}$ maps that are $ C^0$-homotopic, then they are $ C^{\infty}$-homotopic.

I can do this with Stone-Weierstrass theorem, approximating the continuous homotopy. However, I am wondering if there is a direct neat less analysis-y argument. I mean, I guess anything will amount to this, I'd just like it in more abstract terms

Smoothing a $ C^0$ homotopy is the way to go. You can't avoid smoothing of maps if you want to do differential topology.

Well, the book I am reading in, does not contain the Stone-Weierstrass theorem, and mentions this "theorem" as an easy fact in another proof. I guess stuff like this is considered very elementary perhaps? It took me quite some time to prove it, because I am not used to doing arguments of this type.

In the argument I also found it necessary to construct a smooth function that is constant on a some intervals, and I found this quite difficult.  :blush: How would you do this?

Convolution with a [url=http://en.wikipedia.org/wiki/Bump_function]bump function[/url] produces a smooth approximation without Stone-Weierstrass. You can also use bump functions to construct smooth functions that are constant on given intervals.


calculus
integration
LaTeX
calculus computations

Find the value of the sum.
n
[size=150]E[/size]   (2-5i) =
i=1

how would i do this?

$ \sum_{i\equal{}1}^n (2\minus{}5i)\equal{}\sum_{i\equal{}1}^n 2 \minus{} 5\sum_{i\equal{}1}^n i\equal{} 2n\minus{}5\frac{n(n\plus{}1)}{2}$

What does $ E$ mean? I don't think it means $ \sum$. Is it the complete elliptic integral of the second kind?

i got that one now i dont get

$ \sum_{i=1}^{100}  (5^i - 5^{i-1})$

[color=green]Moderator says: learn basic $ \text{\LaTeX}$, it is easy![/color]


geometry
perpendicular bisector
Pythagorean Theorem

[quote="Silverfalcon"]Not sure for what level.

In a circle of radius 2, what is the length of the chord that is perpendicular bisector of the radius.

[hide="My_Sol"]

Call a circle with center Z. Draw a radius. Bisect it and call that midpoint C. Connect to the circumference on point A and B. Because it is perpendicular, ZCA is a right triangle. Hypotenuse, which is another radius, has length of 2. ZC = 1. Use Pythagorean Theorem, and get CA = $\sqrt 3$. The whole chord is twice of it.

So, it's $2 \sqrt 3$.[/hide]

What do you think?[/quote]

[hide]
$2 \sqrt 3$
Yeah, i got the same thing as you.
[/hide]

[color=indigo][hide] The answer is 2:sqrt:3. I just used pythagorean theorem.[/color][/hide]

[hide]Same here... 2:sqrt:3 ... although there are other ways, the pythagorean theorem is definately the easiest way... also you could do the cord trick (ab=c2 , if a is one side of the diameter to the intersection, b is the length of the other, and c is the length from one side of the intersection to the other. in this case, ab is 3, and the sqrt of three is  :sqrt:3. Multiply this by 2 to get your answer.... actually both take the same amount of time if you understand what you are doing (and don't make any infamous stupid errors)[/hide]


probability
limit
linear algebra
matrix
analytic geometry
parameterization
probability and stats

Let $ n>1$ be an integer. Denote by $ M(k)$ the number of matrices in $ \mathcal{M}_n(\mathbb{Z})$ with entries belonging to $ [\minus{}k,k]$, and $ D(k)$ the number of matrices in $ \mathcal{M}_n(\mathbb{Z})$ with entries belonging to $ [\minus{}k,k]$ that are diagonalizable over $ \mathcal{M}_n(\mathbb{Q})$.
Show that $ \lim_{k\to \infty}\frac{D(k)}{M(k)}\equal{}0$.

I can reduce it to an irritatingly simple statement that I couldn't find a proof of for the last half an hour  :mad:

First, the statement: Let $ u,v,x\in\mathbb R$, $ x\ne 0$. Then the number of pairs of integers $ a,b\in[\minus{}k,k]$ such that $ (a\minus{}u)(b\minus{}v)\equal{}x$ is much less than $ k$ if $ k$ is large enough.

Now, the reduction. First of all, note that diagonalizability over rationals implies that all eigenvalues are integer. Since they are obviously bounded by $ kn$, it suffices to show that for every integer $ m\in[\minus{}kn,kn]$ the probability that it is an eigenvalue of our matrix is much less than $ 1/k$. Let $ v_1,\dots,v_n$ be the rows of the matrix $ A\minus{}mI$ where $ A$ is our random matrix and $ I$ is the identity matrix. Assume that the rows $ v_1,\dots,v_{r\minus{}1}$ are linearly independent and the row $ v_r$ is their linear combination. Then we can determine the values of some $ n\minus{}r\plus{}1$ elements in the $ r$-th row from the values of other $ r\minus{}1$ elements by linear formulae depending only on the rows $ v_1,\dots,v_{r\minus{}1}$. This immediately implies that the probability of such an event is at most $ Ck^{\minus{}(n\minus{}r\plus{}1)}\le Ck^{\minus{}2}$ unless $ n\equal{}r$. So, it would suffice to estimate the probability that $ A\minus{}mI$ is not invertible under the assumption that the first $ n\minus{}2$ rows are linearly independent. To estimate it, note that for any such fixed $ n\minus{}2$ rows, we can find a pair of coordinates, say $ x_{1,2}$ and two linear functionals $ L_{1,2}(x_3,\dots,x_n)$ such that all the linear functionals vanishing on the first $ n\minus{}2$ rows are linear combinations of $ x_1\minus{}L_1$ and $ x_2\minus{}L_2$. Hence, the invertibility of $ A\minus{}mI$ is equivalent to the invertibility of the $ 2\times 2$ matrix whose rows are the the values of the linear forms $ x_1\minus{}L_1,x_2\minus{}L_2$ on the last 2 rows. Taking the determinant of this $ 2\times 2$ matrix and fixing the values of all values except those in the first 2 positions, we see that it will suffice to show that for every $ u,v,w,t\in\mathbb R$ the number of quadruples of integers $ a,b,c,d\in[\minus{}(n\plus{}1)k,(n\plus{}1)k]$ such that $ (a\minus{}u)(b\minus{}v)\equal{}(c\minus{}w)(d\minus{}t)$ is much less than $ k^3$. The reduction of this problem to the statement is classical, so I'll omit it.

So, who can prove the statement and finish the story?

$ (a\minus{}u)(b\minus{}v) \equal{} x \Rightarrow a \equal{} u\plus{}\frac{x}{b\minus{}v}$
As long as $ |b\minus{}v|$ is large enough, and $ x$ not zero, $ u\plus{}\frac{x}{b\minus{}v}$ can't possibly be an integer.
Therefore there is a range for $ b$ to take value. Similarly we can find a range for $ a$. These two ranges are independent of $ k$. Therefore as k gets large, the number of solutions is much less than k.

You seem to misread the statement (or I put it not very clearly): I need an estimate for the number of solutions that depends only on $ k$ (so it should be uniform in $ u,v,x$) and is $ o(k)$. What you proved is that for any fixed $ u,v,x$, the number is $ o(k)$ as $ k\to\infty$ but this $ o()$ is not uniform in the 3 parameters involved.

Actually, if you think of it a bit, you can realize that $ u,v$ and, thereby $ x$ arising in this particular problem are rationals with the numerators and the denominators bounded by some huge power of $ k$, so the statement for this particular case would follow from the fact that the number of divisors grows slower than any power. So, formally speaking, the problem is solved. Still, I'm unhappy with the absence of the proof for such a trivial assertion :(  :wallbash:


For $x,y,z\ge 0$ which satisfy $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3$, prove that
\[\frac{x^{2}+y^{2}+z^{2}}{x+y+z}\leq \sqrt{x^{2}+y^{2}+z^{2}-xy-zy-zx+1}\]
[hide="Remark."]Come up with as many solutions as possible, this one lends itself to several with possibility for nice creative use...[/hide]

[hide="Not sure..."]OK, I spent an hour to try and solve this. Then, just to check the inequality, I chose $z=\frac{3}{2}$ and then solved to find a possible $x=\frac{27+\sqrt{537}}{12}$ and $y=\frac{27-\sqrt{537}}{12}$. Then I got $LHS=\frac{119}{36}$ and $RHS=\frac{ \sqrt{54}}{2}$.
$3.31 < 3.67$ ...[/hide]

I'm sorry... I had the very [b]worst[/b] possible typo... I flipped the sign :oops: (i'll edit it now...)

Otherwise it's fine, I double checked my work (in Sketchpad) just to be sure...

Well for now it won't let me edit...

[quote="me@home"]For $x,y,z\ge 0$ which satisfy $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3$, prove that
\[\sqrt{x^{2}+y^{2}+z^{2}-xy-zy-zx+1}\ge \frac{x^{2}+y^{2}+z^{2}}{x+y+z}\]
[hide="Remark."]Come up with as many solutions as possible, this one lends itself to several with possibility for nice creative use...[/hide][/quote]

Exactly the intended idea.
[hide="tell tale part of solution"]
Important identity: \[(x+y+z)(x^{2}+y^{2}+z^{2}-xy-zy-zx)=x^{3}+y^{3}+z^{3}-3xyz\] which can be easily verified (most negative terms can be seen to cancel).
Thus, \begin{eqnarray*}x^{2}+y^{2}+z^{2}-xy-zy-zx+1&=&\frac{x^{3}+y^{3}+z^{3}-3xyz}{x+y+z}+1 \\ &=&\frac{x^{3}+y^{3}+z^{3}-3xyz}{x+y+z}+\frac{3}{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}\\&=&\frac{x^{3}+y^{3}+z^{3}-3xyz}{x+y+z}+\frac{3}{\frac{x+y+z}{xyz}}\\&=&\frac{x^{3}+y^{3}+z^{3}-3xyz}{x+y+z}+\frac{3xyz}{x+y+z}\\ &=&\frac{x^{3}+y^{3}+z^{3}}{x+y+z}\end{eqnarray*}
When I made that remark in the first post, it was mostly referring to the part [b]after[/b] this first step in the inequality.
[/hide]

wow i - at inequalities...and i am going to bed...instead of looking it up, i will just ask:

how do you show that

$(\sum x^{3})\ge (\sum x)(\sum x^{2})$?

Or Jensen on $f(x)=x^{2}$
$f\left(\sum \frac{x}{x+y+z}\cdot x\right)\leq \sum \left(\frac{x}{x+y+z}\cdot f(x)\right)$ $\iff \left(\frac{x^{2}+y^{2}+z^{2}}{x+y+z}\right)^{2}\leq \frac{x^{3}+y^{3}+z^{3}}{x+y+z}$
:wink:

[quote="silouan"][quote="Altheman"]
$(\sum x^{3})\ge (\sum x)(\sum x^{2})$?[/quote]

Chebychev  :wink:  for the triples x,y,z   x^2,y^2,z^2[/quote]

Well that works assuming the LHS has a factor of $3$... but as pointed out earlier it's not even relevant.

will i did sove this one but i did not use a speacial way to solve it first we got 1/XY+1/YZ+1/XZ=3  SO Z+Y+X/XZY+3AND THE ONLE NUMBER THAT SATISFAI THIS equatino is x=1 and y=1 andz=1
[/list]

nexo, they never said that x,y,z were integers.


What is the remainder when $1234567890$ is divided by $99$?

Here is what I have so far but then I'm stuck; probably because I forgot how to do this  :lol:

[hide]
$1234567890 \equiv x \pmod{99}$

[/hide]

also i didn't use mod arithmetic and i got some random obscure number is

[hide]

$98$

[/hide]

right?

[quote="davidlizeng"][quote="bpms"]100=1...[/quote]

bpms, what do you mean by 100=1 ????[/quote]
He meant $100\equiv1\pmod{99}$.


inequalities
geometry
area of a triangle
Heron\u0027s formula

I found this two inequalities. I'm sure there are very easy for some of you.

In every triangle:

(1)  $ (a\minus{}b\plus{}c)^3 \plus{} (a\plus{}b\minus{}c)^3 \plus{} (\minus{}a\plus{}b\plus{}c)^3 \geq 3abc$

(2)  $ (a^2 \plus{} 4m_a^2)^2 \plus{} (b^2 \plus{} 4m_b^2)^2 \plus{} (c^2 \plus{} 4m_c^2)^2  \geq  192\cdot  [ABC]^2$

(1)
Setting $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$ the inequality can be rewritten:

$ 8(x^3\plus{}y^3\plus{}z^3) \ge 3(x\plus{}y)(y\plus{}z)(z\plus{}x) \Longleftrightarrow 8(x^3\plus{}y^3\plus{}z^3)  \ge 3\sum_{sym}x^2y\plus{}6xyz$

But $ 6(x^3\plus{}y^3\plus{}z^3) \ge  3\sum_{sym}x^2y$ from muirihead and $ 2(x^3\plus{}y^3\plus{}z^3) \ge 6xyz$ from am-gm.

[quote="Dimitris X"](1)
Setting $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$ the inequality can be rewritten:

$ 8(x^3 \plus{} y^3 \plus{} z^3) \ge 3(x \plus{} y)(y \plus{} z)(z \plus{} x) \Longleftrightarrow 8(x^3 \plus{} y^3 \plus{} z^3) \ge 3\sum_{sym}x^2y \plus{} 6xyz$

But $ 6(x^3 \plus{} y^3 \plus{} z^3) \ge 3\sum_{sym}x^2y$ from muirihead and $ 2(x^3 \plus{} y^3 \plus{} z^3) \ge 6xyz$ from am-gm.[/quote]
Furthermore, the equality case holds if and only if the triangle is equilateral.

And for (2), please explain your notations.

Notation: usual.

$ m_a$ median $ AM_a$, $ M_a$ midpoint of $ BC$, etc.

[ABC] = area of ABC

I think my notation is quite standard.

[quote="hpoincare"]I found this two inequalities. I'm sure there are very easy for some of you.

In every triangle:

(1)  $ (a \minus{} b \plus{} c)^3 \plus{} (a \plus{} b \minus{} c)^3 \plus{} ( \minus{} a \plus{} b \plus{} c)^3 \geq 3abc$

(2)  $ (a^2 \plus{} 4m_a^2)^2 \plus{} (b^2 \plus{} 4m_b^2)^2 \plus{} (c^2 \plus{} 4m_c^2)^2 \geq 192\cdot [ABC]^2$[/quote]

[hide="(2)"]Actually the 192 can be replaced by 256:
(for 256)
Using the formula for the median of a triangle:
\[ m_a\equal{}\sqrt{2b^2\plus{}2c^2\minus{}a^2}\]
and plugging this (and the similar corresponding formulas for $ b,c$) into the inequality gives:
\[ (2b^2\plus{}2c^2)^2\plus{}(2c^2\plus{}2a^2)^2\plus{}(2a^2\plus{}2b^2)^2\ge256[ABC]\]
Now plugging in Heron's formula:
\[ [ABC]\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}\frac14\sqrt{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)}\equal{}\frac14\sqrt{2a^2b^2\plus{}2b^2c^2\plus{}2c^2a^2\minus{}a^4\minus{}b^4\minus{}c^4}\]
and expanding all terms give the inequality congruent to:
\[ 8a^4\plus{}8b^4\plus{}8c^4\plus{}8a^2b^2\plus{}8b^2c^2\plus{}8c^2a^2\ge16\left(2a^2b^2\plus{}2b^2c^2\plus{}2c^2a^2\minus{}a^4\plus{}b^4\plus{}c^4\right)\]
combining like terms and dividing each side by 24 gives the original inequality congruent to $ a^4\plus{}b^4\plus{}c^4\ge a^2b^2\plus{}b^2c^2\plus{}c^2a^2$, which is true by muirhead.
[/hide]


This should be in other problem solving.

Um... Do you like in Number Theory Forum?

But that's like in advanced.... :(

He meant in Other Problem Solving Topics I think.


[quote="Inspired By Nature"]The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral.[/quote]

[hide]Let the sides be $a, a+r, a+2r$

If the atitudes also form an arithmetic sequence, then $a(a+r),a(a+2r),(a+r)(a+2r)$ forms an arithmetic sequence.

Expand: $a^{2}+ar,a^{2}+2ar,a^{2}+3ar+2r^{2}$

From the first 2 terms, we find the common difference to be $ar$ which forces the third term to be $a^{2}+3ar$ meaning $2r^{2}=0\Rightarrow r=0$[/hide]

The sides are then $a,a,a$ which is an equilateral triangle.

[quote="SplashD]If the atitudes also form an arithmetic sequence, then $a(a+r),a(a+2r),(a+r)(a+2r)$ forms an arithmetic sequence.[/quote]
Why's that?  :o

I remember it had something to do with the area...

$S=\frac{ah_{a}}{2}\iff h_{a}=\frac{2S}{a}=\frac{2S}{abc}bc$, and similarly $h_{b}=\frac{2S}{abc}ac, h_{c}=\frac{2S}{abc}ab$. Since $\frac{2S}{abc}$ is the common factor, the conclusion follows.


vector
linear algebra
matrix
linear algebra unsolved

Given a Lx1 vector a(t) and a Kx1 vector b(t), who can tell me whether the attached vector expression can be reduced to a more compact one, e.g., a factorized expression?

* a(t)^T and b(t)^T are the transposes of a(t) and b(t) respectively.

It might help to start with a correct expression. What does $ n$ have to do with anything in those sums?

a(t)=[a1(t),a2(t),...,aL(t)]^T; b(t)=[b1(t),b2(t),...,bK(t)]^T
Both a and b are vector variables. Stated another way, they change with t.
The sum operations implemented on corresponding variables and their products can be viewed as the operations on each element of these vectors and matrices. N is the number of variables for the sum operations. Note that N is a fixed integer.

For example,
sum(a(t),t=1:N) = [ sum(a1(t),t=1:N), sum(a2(t),t=1:N), ..., sum(aL(t),t=1:N)] ^ T,

sum(a(t)b(t)^T) = [ sum(a1(t)b1(t), t=1:N), ..., sum(a1(t)bK(t), t=1:N);
                            sum(a2(t)b1(t), t=1:N), ..., sum(a2(t)bK(t), t=1:N);
                            ... ... ...    ... ... ...    .... ...    ...  ...  ...  ...  ...  ...
                            sum(aL(t)b1(t), t=1:N), ..., sum(aL(t)bK(t), t=1:N) ]

Ah- I'm not used to seeing $ t$ as an integer index.

All of those sums are matrix multiplication. We have an $ L\times N$ matrix $ A$, a $ K\times N$ matrix $ B$, and an $ N\times 1$ matrix $ j\equal{}(1,1\cdots,1)^T$. The expression is $ AB^T(BB^T)^{\minus{}1}Bj\minus{}Aj$. We can factor out $ A$ on the left and $ j$ on the right, but $ B^T(BB^T)^{\minus{}1}B\minus{}I$ is a commutator and resists simplification- unless $ B$ is square, in which case it's just zero.

Thank jmerry for your reply. However, maybe, I have not explained the problem very clearly.
I have to note that the matrices A and B you mentioned above do not exist in the expression.
The vectors a(t) and b(t) in the term SUM[a(t)*b(t),t=1:N] are combined together, and they cannot be separated into two individual matrices, e.g., A and B. An illustrative example is provided as follows.
So, I am not sure about whether the expression can be factorized as you indicated.

Thank you all the same!

The $ a$ are the columns of $ A$. The $ b$ are the columns of $ B$.

Thank you very much!
This is a good solution to me indeed.

Do you mean that B^T(BB^T)^{-1}B-I is zero iff B is square?
I wonder if it is a sufficent and necessary condition.

Think about what happens when $ B$ is invertible; it's simply not defined when $ B$ is square and not invertible.


A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $ \frac{3}{4}$ of its former rate and arrives $ 3 \frac{1}{2}$ hours late.  Had the accident happened $ 90$ miles farther along the line, it would have arrived only $ 3$ hours late.  The length of the trip in miles was:

$ \textbf{(A)}\ 400 \qquad
\textbf{(B)}\ 465 \qquad
\textbf{(C)}\ 600 \qquad
\textbf{(D)}\ 640 \qquad
\textbf{(E)}\ 550$

[hide="Click for solution"]
Let $ x$ be the distance from the point of accident to the end of the trip and $ R$ the former rate of the train.  Then the normal time for the trip, in hours, is:

$ \frac{x}{R}\plus{}1\equal{}\frac{x\plus{}R}{R}$.

Considering the time for each trip, we have:

$ 1\plus{}\frac{1}{2}\plus{}\frac{x}{\frac{3}{4}R}\equal{}\frac{x\plus{}R}{R}\plus{}3\frac{1}{2}$

and

$ 1\plus{}\frac{90}{R}\plus{}\frac{1}{2}\plus{}\frac{x\minus{}90}{\frac{3}{4}R}\equal{}\frac{x\plus{}R}{R}\plus{}3$.

Thus, $ x\equal{}540, R\equal{}60 \implies 540\plus{}60\equal{}600$ or $ \boxed{\textbf{(C)}}$.
[/hide]


$ f(x)\plus{}g(y)\equal{}\log(1\plus{}x\plus{}y\plus{}yx)$ for positive $ x$ and $ y$.

we have $ f(x) \plus{} g(y) \equal{} Log(1 \plus{} x \plus{} y \plus{} xy) \equal{} Log((1 \plus{} x)(1 \plus{} y)) \equal{} Log(1 \plus{} x) \plus{} Log(1 \plus{} y)$
for$ y \equal{} c$ we find $ f(x) \equal{} Log(1 \plus{} x) \plus{} k$ and $ g(x) \equal{} Log(1 \plus{} x) \minus{} k$


Find all $ f: R \rightarrow R$ such that for all real x,y,z the following equation holds

$ f(x \plus{} y \plus{} z) \equal{} f(x) \cdot f(1 \minus{} y) \plus{} f(y) \cdot f(1 \minus{} z) \plus{} f(z) \cdot f(1 \minus{} x)$

[hide]

I got either $ f(x) \equal{} 0$ or $ f(x) \plus{} f(x \plus{} 2) \equal{} 1$ and have no idea what to do with this.

[/hide]

Hmm, looks like I missed $f(x)=\frac{1}{3}$.

Seems like there aren't many solutions here, so I'll share mine :)

Let $P(x, y, z)$ be the assertion. First notice the following three important equations: $$P(1-z, 1-y, 1-x) \implies f(x) = f(3-x)$$ $$P(x, 1-x, 0) \implies f(1) = [f(x)]^2 + f(1-x)$$ $$f(1) = [f(1-x)]^2 + f(x).$$ Now, taking $P(0, 0, 0)$ we get $f(0) = 0$ or $f(1) = \frac{1}{3}$.

If $f(0) = 0$, we have $$P(x, 0, 0) \implies f(x) = f(x)f(1)+f(0)f(1)+f(0)f(1)$$ so $f(x) = 0$ or $f(1) = 1$, yielding our first solution. Then, we use the second and third equations we derived, but using $f(1) = 1$, we have $$[f(x)]^2 + f(1-x) = 1$$ $$[f(1-x)]^2 + f(x) = 1$$ These yield solutions for $f(x)$ as a constant, but since we must have $f(0) = 0$, this case only gives $f(x) = 0$ as a solution.

Now if $f(1) = \frac{1}{3}$, we have $$P(1, 0, 0) \implies 9f(0)^2 + 3f(0)-2 = 0$$ which means $f(0) = -\frac{2}{3}$ or $f(0) = \frac{1}{3}$. We can do the same thing as our earlier case to get $f(x)$ must be a constant, and since $f(1) = \frac{1}{3}$ our only solution from this case is $f(x) = \frac{1}{3}$.

Therefore our only possible solutions are $f(x) = 0$ and $f(x) = \frac{1}{3}$, and it is easy to check that these work.


function
limit
real analysis
topology
integration
real analysis unsolved

Is $ f: \mathbb{R}\to \mathbb{R}$ is a measurable function, then by Lusin's Theorem it is almost everywhere continuous.

What does this exactly mean? Does it mean that I can take limits of rational sequences $ x_{n}$ such as $ \lim_{x_{n}\to x}f(x_{n})=f(x)$, with $ x$ irrational?

Here's one possible statement of Lusin's Theorem:  Fix $ \epsilon>0$ and let $ f: [a,b]\to\mathbb{C}$ be Lebesgue measurable.  Then there is a compact set $ E\subset [a,b]$ such that $ \mu(E^{c})<\epsilon$ and $ f\big|_{E}$ (i.e., the restriction of $ f$ to $ E$) is continuous.

What it doesn't mean, in particular, is that $ f$ is continuous at (Lebesgue) a.e. every point of $ \mathbb{R}$, which can be seen by the example $ \chi_{\mathbb{Q}}$.

Another version: "a measurable function is approximately continuous almost everywhere". This is a classical example of how precise mathematical statements are.  :)
A function $ f$ is approximately continuous at $ x_{0}$ if there is a measurable set $ E\ni x_{0}$ with Lebesgue density $ 1$ at $ x_{0}$ such that $ f_{|E}$ is continuous at $ x_{0}$. A more elegant definition involves the density topology on $ \mathbb R$, in which we declare a set open iff it is measurable and has density $ 1$ at every point. Being approximately continuous means being continuous from the density topology to standard topology.

So if I have a function $ f: \mathbb{R}\to \mathbb{R}$ such that $ f(x)=cx$, for some $ c\in \mathbb{R}$ and for all $ x \in \mathbb{Q}$, how can I prove that it's continuous, using Lusin's theorem? I'm quite a novice in measure theory, so I don't know where to start from.

That's not enough information. What else do you know about the function?

are you requiring that $ f$ be measurable and additive, perhaps?

Take the convolution of $ \exp(if)$ with a $ C^\infty$ function having compact support.  Note that the convolution is continuous, so, since $ f$ is linear on the rationals at least, $ \exp(if(x)) = \exp(if(1)x)$ everywhere.

This means that for each $ x$, there is an integer $ k(x)$ such that $ f(x) = f(1)x+2\pi k(x)$.  If $ k(t)\not=0$, then by additivity, $ f(t/2k(t)) = f(1)t/2k(t)+\pi$, a contradiction.  So $ f$ is linear.

Additivity too, sorry. Ok, so I understand that this problem requires pretty advanced measure theory tools, which I don't possess. For instance, I didn't understand why $ \exp(if)=\exp(if(1)x)$ everywhere and why from this last result you say that there exist an integer $ k(x)$ etc.

If a function is measurable and additive, is it true, false or undecidable the fact that the function is bounded on an interval $ [a,b]$? Because from that, it will also follow the continuity of $ f(x)$.

Here's a nice argument:
Lemma: For any measurable set $ X\subset \mathbb{R}$ with positive measure, the set $ X-X=\{x-y: x,y\in X\}$ contains an open interval around zero.
There's a proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113431]here[/url]

Now, suppose $ f$ is a measurable additive function on $ [-a,a]$. The sets $ S_{n}=\{x\in [0,a]:-n\le f(x)\le n\}$ are measurable, and their union is $ [0,a]$. Since this is a countable union, some $ S_{n}$ must have positive measure. Now $ -2n\le f(x)\le 2n$ on $ S_{n}-S_{n}$; by the lemma, this contains some open interval $ (-\epsilon,\epsilon)$, and $ f$ is bounded on that interval.

[quote="Stewie"]Additivity too, sorry. Ok, so I understand that this problem requires pretty advanced measure theory tools, which I don't possess. For instance, I didn't understand why $ \exp(if)=\exp(if(1)x)$ everywhere and why from this last result you say that there exist an integer $ k(x)$ etc. [/quote]

there were some mistakes in my previous post; I'll edit them now.  Here are some more details.  It's fairly straightforward to show that the convolution of a measurable, locally integrable function with an infinitely differentiable compactly supported function inherits many of the good properties of the $ C^\infty$ function.  So let's take the fact that the convolution is continuous for granted.

Note also that any additive function is linear on the rationals.

Let $ \eta$ be a $ C^\infty$ function having compact support.  By definition, for each $ x$,
\begin{eqnarray*}(\exp(if) * \eta)(x) &=& \int_{\mathbb{R}}\exp(if(x-y)) \eta(y)dy \\ &=&\exp(if(x))\int_{\mathbb{R}}\exp(-if(y))\eta(y)dy \\ &=&L\exp(if(x)) \end{eqnarray*}
(where we put $ L=\int_{\mathbb{R}}\exp(-if(y))\eta(y)dy$)

Since $ f$ is linear on the rationals, for each $ q\in\mathbb{Q}$, $ (\exp(if) * \eta)(q)=L\exp(if(1)q)$.  Since the convolution is continuous, this means that for each $ x\in\mathbb{R}$, $ (\exp(if) * \eta)(x)=L\exp(if(1)x)=L\exp(if(x))$

This means that for each $ x\in\mathbb{R}$, $ \exp(if(x)) = \exp(if(1)x$, which is the same as saying, for each $ x\in\mathbb{R}$, there is an integer $ k(x)$ such that $ f(x) = f(1)x+2\pi k(x)$.  I handled the case where some $ k(x)\not=0$ in my first post, and then we're done :)

Thank you very much, I have a much better insight on the problem, now. One last question: did you use Lusin's theorem?


Prove that the integer next to  (3 + √ 5)ⁿ is divisible by 2 ⁿ for all n=0,1,2,3……………

It's wrong for $n=1,2$...
Note that this integer you speak of is (for large $n\geq3$)) equal to $a_n: =(3+\sqrt 5)^n + (3-\sqrt 5)^n$.
Now that is the general term for the sequence defined by $a_0=2,a_1=6,a_{n+2}=6a_{n+1}-4a_n$, and the result follows.


The numbers from $ 1$ to $ 8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

$ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 24$

[hide]
The sum of the numbers from $1$ to $8$ is $\frac{8*9}{2}=36$.

Half this sum must be on the top face, and half on the bottom face.  So each face must have a sum of $\frac{36}{2}=18$. [b]C.[/b]
[/hide]

This is a simple problem
We know that the sum for each combination is same, so we just divide the sum between 1-8 by 2
Answer is [b](C.) 18[/b]


geometry
circles
circumcircle
bisector
IMO Shortlist

Two circles $C_{1}$ and $C_{2}$ are (externally or internally) tangent at a point $P$. The straight line $D$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$.

Consider an inversion of pole $P$. It turns the circles $C_1$ and $C_2$ into two parallel lines $c_1$ and $c_2$, and the line $ABC$ into a circle $(C)$, which passes through $P$ and is tangent to $c_1$ in $A'$ (the image of $A$) and which cuts $c_2$ in $B',C'$ (the images of $B,C$ respectively.

In these conditions, it's really easy to show that $PA'$ is the internal or external bisector of $\angle B'PC'$ because it's obvious that $A'$ is the midpoint of the arc $B'C'$.

Hey Grobber, you are really an inversion freak! :D
A simple homothety is sufficient to solve it ;)

By the way, this problem was proposed by BELGIUM! :)

[quote="Arne"]A simple homothety is sufficient to solve it ;)[/quote]

I solved it with the "Raakomtrekshoek", although I got no idea how to translate that in English... can u help me a hand Ernie?

Besides, I think this should not be a too advanced problem... it was asked as a BMO round 1 question!

Peter means the alternating segment theorem ;)

Well, by far the simplest solution is the one using a homothety anyway :)

Just to give the solution everybody has been talking of:

Let $A$ be on $C_2$ and let $\displaystyle \{A'\}=t(A)\cap C_1$, where $t(A)$ is the tangent at $A$ to $C_2$.
The homothety of center $P$ taking $C_2$ to $C_1$ maps $A$ into $A'$ and $t(A)$ into $t(A')$ (the tangent at $A'$ to $C_1$).

$t(A)$ and $t(A')$ are parallel and intersect $C_1$ at $B,C$ and $A'$ respectively. This shows that $BA'=A'C$ and $A'$ is the midpoint of the arc $\t{BC}$. So $PA'$ is one of the angle bisectors of $\angle BPC$. Lines $PA'$ and $PA$ are the same and we are done.

I suggest the proof of the converse i.e.:
Let $PBC$ be a triangle and let $(C)$ be a circle tangent to $BC$ at $A$. If $PA$ is the bisector of $\angle BAC$, then $(C)$ is tangent to the circumcircle of $PBC$ in $P$.

[quote="orl"]Two circles $C_{1}$ and $C_{2}$ are (externally or internally) tangent at a point $P$. The straight line $d$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$.[/quote][color=darkblue]Denote : the intersection $M$ between the line $BC$ and the common tangent in the point $P$ ; the reflection $D$ of the point $A$ w.r.t. the point $M$. Then the points $A$, $D$ are harmonical conjugate w.r.t. the pair of the points $B$, $C$ because $MA=MD$ and $MA^{2}=MP^{2}=MB\cdot MC$ (see the equivalence $A3$ of the harmonical division in the lemma from the my message http://www.mathlinks.ro/Forum/viewtopic.php?t=46146 ). But $MB=MA=MD\Longrightarrow$ the rays $[PA$ and $[PD$ are the bisectors of the angles between the lines $PB$ and $PC$.[/color]

[hide="could you just.."]
Draw PX which is tangent to the 2 circles and let CP intersect the other circle at D. Then $ APB\equal{}XPB \pm APX\equal{}BCP \pm PAC\equal{}APD$
[/hide]


For the AMC 12 problem series -- where's the homework and how are you supposed to submit it?  :help:

You aren't supposed to turn the homework in.  You're supposed to discuss the practice problems on the class message board.  You'll find the class message board in the Online School section of the Forum.  Since you enrolled a little late, you probably weren't set up to see this message board yet.  You are now, so you should be able to access the problems.


Alan copied a picture to 66 2/3 % of its original size and gave the copy to Beth. Beth then reduced her copy to 66 2/3 % of its size and gave the reduced copy to Carl. At what percent will Cart have to photocopy the picture to return it to its original size? :?:  :?:

Let the area of the original picture be $ x$. Alan's copy is then $ \frac{2}{3}x$, and Beth's copy is $ \frac{4}{9}x$. Now let the percentage Carl will have to photocopy the picture to return it to its original size be $ y$. We have $ \frac{4}{9}xy \equal{} x \implies \frac{4}{9}y \equal{} 1 \implies y \equal{} \frac{9}{4} \equal{} \boxed{225\%}$.


Poll your score! :-)

Note that these ranges are based on last year's "score range"s with some of the upper ranges (e.g. Winner = 24-42) being divided up since clearly this year's lower bound for winner scores should be higher.

I hope the Winner bound isn't too much higher :(. Going to college a year early = being counted as a senior a year early.

I wouldn't be so sure that the winning score cutoff will be so high.  10 of the top 12 from last year graduated including 6 IMO gold medalists (with a total of at least 10 gold medals).  It would be difficult to replace one group of students that concentrated.  I think the Winners and Honorable Mention lists will include many new names.

[quote="MithsApprentice"]Oh that's true. What would you guys estimate the cutoffs to be around?[/quote]

I try to stay out of the estimation game.  Whatever I might say as my guesses, I'm not really any better at guessing than the students.  But if I were forced to gamble, I would guess that the winner's cuttoff will be very close to the same as last year's -- same for the honorable mention.  I think that the range of scores from 5th place to 24th might be very tight -- possibly within a problem.  This is purely guesswork based on a very tight cluster of score-guesses from students.

[quote]5th place to 24th might be very tight -- possibly within a problem.[/quote]

I would bet a significant amount of money that something like 5th (around there) to 24th will be very tight. Getting 3 was probably within a lot of people's capabilities (1, 4, 2 or 5) so who knows?

I got 3... and I suddenly wished I didn't skip grades, cuz I'm a senior now. MOSP! I can see you! ::feeling ridiculous::

I don't know about 5th - 24th. Perhaps 7th or 8th - 24th?

It seems that a lot of last year's better MOPpers are claiming at least 4. In which case, I'd say 24 points as a cutoff for winner is low.

Almost everyone here is saying they basically solved at least 2 problems, but I understand this isn't exactly a completely random sample of the calibur of USAMO qualifiers. Nor is it a fair sample of even the users of AoPS who qualified for USAMO.

about 15 for me... :(  i should have gotten up to 20 but i messed up on #4. there is always next year ;)

[quote]Almost everyone here is saying they basically solved at least 2 problems, but I understand this isn't exactly a completely random sample of the calibur of USAMO qualifiers. Nor is it a fair sample of even the users of AoPS who qualified for USAMO.[/quote]

And keep in mind that like last year, many scores turn out being considerably lower than estimated. And I'm pretty unbiased =p

Yeah that's quite true. Assuming you got full credit on a problem, but then not proving a "trivial" part of the solution could cost you.

Am I right in assuming that it is better to write more and explain what might not need explanation than to write too little and miss proving a "trivial" part?

My solution for 1 was 14 pages long, and my solution for 4 was 11 ... Obviously I'm on the "write too much" and overcautious end.

JB

[quote="MithsApprentice"]Well, there's no evil grader that's going to deduct points because your solution is too easy to follow/understand.[/quote]

MithsApprentice is right.  All the evil graders prefer to deduct points for nearly every other reason, but not that one.

Well, aren't they graded by committee to thwart evil graders?

Anyhow, I'm sure I'm in the 4-9 range.


[quote="Samsa"]Find all real pairs $ (x;y)$ in this system: \[ \left\{\begin{array}{c}x-2=(x+2)(y-2)\\ x^{2}=4(y^{2}-4y+x+3)\end{array}\right.\][/quote]
[b]Solution.[/b] We have: $ x^{2}=4(y^{2}-4y+x+3)\implies\left(x-2\right)^{2}=4\left(y-2\right)^{2}\implies\left(x+2\right)^{2}=4$ $ \implies x\in\left\{0,\,-4\right\}\implies \left(x,\,y\right)=(0,\, 1);\,(-4,\,5)$