geometry circumcircle inradius ratio incenter Euler trigonometry

In a triangle $ABC$, points $M$ and $N$ are on sides $AB$ and $AC$, respectively, such that $MB = BC = CN$. Let $R$ and $r$ denote the circumradius and the inradius of the triangle $ABC$, respectively. Express the ratio $MN/BC$ in terms of $R$ and $r$.

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=20406 , we have $\frac{MN}{OI}=\frac{a}{R}$, where O is the circumcenter and I is the incenter of triangle ABC. In other words, $\frac{MN}{a}=\frac{OI}{R}$. Since $OI=\sqrt{R^2-2Rr}$, this becomes $\frac{MN}{a}=\frac{\sqrt{R^2-2Rr}}{R}$. Thus we have

$\frac{MN}{BC}=\frac{MN}{a}=\frac{\sqrt{R^2-2Rr}}{R}=\sqrt{\frac{R^2-2Rr}{R^2}}=\sqrt{1-2\frac{r}{R}}$.

  darij

[quote="billzhao"]It's amazing that this configuration was throughly discussed 1.5 years ago in [i]Crux with Mayhem[/i] (October 2003 issue, solution to problem 2776). Guess the problems just keep getting recycled, don't they.

Anyways, on the contest, I came up with a brute-force solution, expressing everything in terms of $a,b,c$. It worked, but definitely not as nicely.[/quote]
billzhao, the most amazing is this:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30539]www.mathlinks.ro/Forum/viewtopic.php?t=30539[/url]
This appears in Moldova MO [b]2005[/b]!
Anyway, here is my solution.
Let the perendicular from I and O to AB be F',F respectively. E',E are defined similarly. Now by computing out the length of FF',EE',AM,AN, the answer is trivial by Euler Theorem.
and, at last, how well did the Canadians perform in APMO?

[quote="billzhao"]Anyways, on the contest, I came up with a brute-force solution, expressing everything in terms of $a,b,c$. It worked, but definitely not as nicely.[/quote]
 :D  That's also what I did in the contest... Though it is not nice at all, it took me not much time to solve this problem.

[quote="mecrazywong"]billzhao, the most amazing is this:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30539]www.mathlinks.ro/Forum/viewtopic.php?t=30539[/url]
This appears in Moldova MO [b]2005[/b]!
[/quote]
Unbelievable!

[hide="Solution"]
We use barycentric coordinates. Set $A$ at $(1,0,0)$, $B$ at $(0,1,0)$, and $C$ at $(0,0,1)$.

Then, $AN=b-a$, $AM=c-a$, so $M$ is at $\left(\frac{a}{c},1-\frac{a}{c},0\right)$, and $N$ is at $\left(\frac{a}{b},0,1-\frac{a}{b}\right)$

Now, the displacement $MN$ is $\left(\frac{a}{b}-\frac{a}{c},\frac{a}{c}-1,1-\frac{a}{b}\right)$

So, by the distance formula:

$MN^2=a^2\left(1-\frac{a}{c}\right)\left(1-\frac{a}{b}\right)+b^2\left(\frac{a}{b}-\frac{a}{c}\right)\left(\frac{a}{b}-1\right)+c^2\left(\frac{a}{c}-\frac{a}{b}\right)\left(\frac{a}{c}-1\right)$

$\frac{MN^2}{BC^2}= \left(1-\frac{a}{c}\right)\left(1-\frac{a}{b}\right)+\left(1-\frac{b}{a}\right)\left(1-\frac{b}{c}\right)+\left(1-\frac{c}{a}\right)\left(1-\frac{c}{b}\right)$

Expanding, we have that this is equal to:

$\frac{1}{abc} \cdot [3abc-(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)+a^3+b^3+c^3]$

The stuff in the brackets, however, is $abc-(a+b-c)(a-b+c)(-a+b+c)$. Letting $(a+b-c)(a-b+c)(-a+b+c)=x$ and $a+b+c=p$ for ease of notation, we have that:

$\frac{MN^2}{BC^2}=1-\frac{x}{abc}$.

Now, recall that $4r^2=\frac{x}{p}$ and $abc=R\sqrt{px}$, so the expression is equal to:

$1-\frac{x}{R\sqrt{px}}=1-\frac{1}{R}\cdot \sqrt{\frac{x}{p}}=1-\frac{2r}{R}$,

so $\frac{MN}{BC}=\boxed{\sqrt{1-\frac{2r}{R}}}$[/hide]

Let A' , B' , C' are excenters of triangle ABC
Easy to see that angle AC'N = B'A'C' - 2*CC'A' = O'C'I , were O' is circumcenter of A'B'C'
AC'BIN is cyclic
X is Miquel point of lines BA , CN , MN , BC
Not hard to prove that X is midpoint of B'C' and triangle XMN ~ XBC , so XN/XC = MN/BC
C'O' intersect IA at point V
VC'I ~ AC'N and C'O'/O'V =C'X/XA , so O'C'I~XC'N
XN/CX = XN/C'X = IO'/C'O' = sqrt(R*R-2*R*r)/R . done

Denote $BC=a$, $AC=b$ and $AB=c$. Let the incenter of the triangle be $I$ and let the point at which the incircle is tangent to $AB$ be $D$. Now note that

\[a=2R\cdot \sin{A} = 2R \cdot \sin{\frac{A}{2}} \cdot \cos{\frac{A}{2}}\]

and by trigonometric ratios in triangle $\triangle{AID}$,

\[AD=r\cdot \frac{\cos{\frac{A}{2}}}{\sin{\frac{A}{2}}}.\]

Combining these ratios with the fact that $AD=\frac{b+c-a}{2}$ yields that

\[\frac{b+c-a}{a}=\frac{2AD}{a}=\frac{r}{R\cdot \sin^2{\frac{A}{2}}}.\]

By cosine law and the fact that $1-\cos{A}=2\sin^2{\frac{A}{2}}$, it follows that

\[\frac{MN^2}{a^2}=\frac{(b-a)^2 + (c-a)^2 -2(c-a)(b-a) \cos{A}}{a^2}\]
\[\frac{MN^2}{a^2}=\frac{(b^2 + c^2 - 2bc\cos{A})-a(1-\cos{A})(b+c-a)}{a^2}\]
\[\frac{MN^2}{a^2}=1-\frac{(1-\cos{A})(b+c-a)}{a}\]
\[\frac{MN^2}{a^2}=1-\frac{2\sin^2{\frac{A}{2}}(b+c-a)}{a}\]
\[\frac{MN^2}{a^2}=1-\frac{2r}{R}.\]

Therefore

\[\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}.\]

Let the midpoint of arc $BAC$ of $\odot ABC$ be $R$.
Then rotation about $O$ with rotation angle $\angle BAC$ carries $M \to N$ and $B \to C$.

Thus
$$\frac{MN^2}{BC^2}=\frac{RM^2}{RB^2}$$

Applying cosine formula and extended sine formula,
$$\frac{RM^2}{RB^2}=\frac{(2R\cos{\frac{A}{2}})^2+a^2-2\cdot2R\cos{\frac{A}{2}}\cdot a \cdot \cos({B+\frac{A}{2}}}){(2R\cos{\frac{A}{2}})^2}$$
$$= 1+\frac{(\sin A)^2}{(\cos{\frac{A}{2}})^2} - 4\sin{\frac{A}{2}}\sin({B+\frac{A}{2}})$$
$$=1-4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$$

Now the remaining part is easy.

I knew that in the case when angle $A= 30^0$, $MN^2$ becomes equal to $R^2-2Rr$. Therefore the ratio becomes equal to $\sqrt{1-\frac{2r}{R}}$.
First express $\sqrt{1-\frac{2r}{R}}$ in terms of $a,b$ and $c$. Then apply cosine rule on triangles $AMN$ and $ABC$ to get the value of $MN^2$ in terms of $a,b,c$. Now just expand both the sides and they become equal.
Its really long but does the work.

We will use barycentric coordinates with reference triangle $\triangle ABC$. By the problem conditions, $M=\left(\frac{a}{c},\frac{c-a}{c},0\right)$ and $N=\left(\frac{a}{b},0,\frac{b-a}{b}\right)$. Then, $\overrightarrow{MN}=\left\langle \frac{a(b-c)}{bc},\frac{c-a}{c},\frac{a-b}{b}\right\rangle$, and thus
$$|MN|^2=-a^2\left(\frac{(c-a)(a-b)}{bc}\right)-b^2\left(\frac{a(b-c)(a-b)}{b^2c}\right)-c^2\left(\frac{a(b-c)(c-a)}{bc^2}\right)$$
$$=\frac{a^2(a-b)(a-c)-ab(a-b)(b-c)+ac(a-b)(b-c)}{bc}.$$
From here, it is clear that
$$\frac{|MN|^2}{|BC|^2}=\frac{a(a-b)(a-c)-b(a-b)(b-c)+c(a-c)(b-c)}{abc}$$
$$=\frac{a^3+b^3+c^3-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2+3abc}{abc}$$
$$=\frac{abc-(a+b-c)(a-b+c)(-a+b+c)}{abc}.$$
Because $$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{rs}{2},$$ it follows that $$r=\frac{1}{2}\sqrt{\frac{(a+b-c)(a-b+c)(-a+b+c)}{a+b+c}}.$$ Similarly, because $$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{abc}{4R},$$ we can rewrite $R$ as $$\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$ As a result, $$\frac{r}{R}=\frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc},$$ and $\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}$.

We use barycentrics on $\triangle ABC.$ Notice $M=(a:c-a:0)$ and $N=(a:0:b-a)$ so $\overrightarrow{MN}=\left(\frac{a(b-c)}{bc},\frac{c-a}{c},\frac{a-b}{b}\right).$ Hence, \begin{align*}|MN|^2&=-a^2\cdot\frac{c-a}{c}\cdot\frac{a-b}{b}-b^2\cdot\frac{a(b-c)}{bc}\cdot\frac{a-b}{b}-c^2\cdot\frac{a(b-c)}{bc}\cdot\frac{c-a}{c}\\&=-\frac{a((-a+b+c)(a-b+c)(a+b-c)-abc)}{bc}\\&=\frac{a}{bc}\cdot(abc-8(s-a)(s-b)(s-c))\end{align*} and $MN/BC=\sqrt{1-\frac{8(s-a)(s-b)(s-c)}{abc}}.$ Notice $$\sqrt{s(s-a)(s-b)(s-c)}=[ABC]=\frac{abc}{4R}=sr$$ so \begin{align*}\frac{2r}{R}&=\frac{2\sqrt{(s-a)(s-b)(s-c)/s}}{\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}}\\&=\frac{8(s-a)(s-b)(s-c)}{abc}.\end{align*} Therefore, $MN/MC=\sqrt{1-2r/R}.$


analytic geometry
graphing lines
slope
geometry
circumcircle
ratio
3D geometry

http://mcis.jsu.edu/mathcontest/geometryans06.pdf

I need help on these numbers for elegant or good solutions. If any of you could, could any of you post good solutions?  :D

7
9
14
15
17
18
19
24
25
26
29
31
32
33
34
35
37
38
43
47

Thanks! I am going over these problems day by day.  :D

I got 30 of them right and didn't do 17 and missed 3.  :(

7)

If you draw a line from the origin to the Point (-3,8)..you have a right-angle at that point (Point-tangent)..u can use pythagorean theorem to find ur answer..

I'll just tell you how to solve them cuz I'm too lazy to figure them out :P
[hide="7"]Find the slope of the line between the origin and (-3,8). since the tangent line is perpendicular to it, you can determine the equation of it (you know the slope and the point (-3,8) so that is sufficient) now just determine the x-intercept of that line[/hide]

[hide="9"]ok so angle EBC is 30 (use iscoceles properties) that means angle FBC=90-30=60 so angle EBF is 60+30=90. so triangle EBF is right. not only that, since EB=EF, EBF is a 45-45-90 so you can determine EB. Now draw altitude CZ of EBC and you have a 30-60-90. use that to determine the square's side length and finish the problem off.[/hide]

[hide="14"]refer to the diagram to determine the side length of diagonal. the rest is easy enough.[/hide]

[hide="15"]there's 14/1+1=15 stacks in each row and there's 14(12)/12+1=13 rows. since each stack holds 3 lemons, the answer is 13(15)(3)=585[/hide]

[hide="17"]$a^{2}+b^{2}+c^{2}=722$ using pythagorean theorem this results in $2c^{2}=722\rightarrow c=19$

so $a^{2}+b^{2}=361$ and $a+b=42-19=23\rightarrow a^{2}+2ab+b^{2}=529$ subtracting the two yields $2ab=168\rightarrow ab=84\rightarrow ab/2=42$[/hide]

[hide="18"]I'll put a pic tomorrow but just exterior angle theorem the crap out of it or just set some values for the angles. [/hide]

19-will put pick later, but it resembles number 14, draw some radii to tangents as shown in there

[hide="24"]the triangle with the same point has the circumcenter as the intersection of the two radii. find the equations of two perpendicular bisectors of the sides and find their intersection.[/hide]

[hide="25"]refer to the diagram. you can pythag/30-60-90 it[/hide]

I'll finish the rest tomorrow, I need to go to sleep now :)
[/hide]

26) The triangles $ABE$ and $CDE$ are similar, so, $\frac{15}{30}= \frac{x}{16}$, where $x$ is $CD$. Then, by simple Pythagoran theorem, $BC = 2\sqrt{65}$.

#43:

If $n$ is the 'ratio' between the circumferences, we find the equation:

$\frac{(\frac{3n}{2\pi})^{2}\pi-(\frac{2n}{2\pi})^{2}\pi}{6}= 150$

and then:

$\frac{\frac{9n^{2}}{4\pi^{2}}\pi-\frac{n^{2}}{\pi^{2}}\pi}{6}= 150$

thus:

$\frac{9n^{2}}{4\pi}-\frac{n^{2}}{\pi}= 900$
and
$n^{2}(\frac{9}{4\pi}-\frac{1}{\pi}) = 900$

next:

$n = \sqrt{\frac{900}{\frac{9}{4\pi}-\frac{1}{\pi}}}$
$= \frac{30}{\sqrt{\frac{5}{4\pi}}}$
$= \frac{6\sqrt{\frac{5}{4\pi}}}{\frac{1}{4\pi}}$
$= 24\pi\sqrt{\frac{5}{4\pi}}$
$= 24\pi\frac{\sqrt{5}}{\sqrt{4\pi}}$
$= 24\pi\frac{\sqrt{5}\sqrt{4\pi}}{4\pi}$
$= 12\sqrt{5}\sqrt{\pi}$
$= 12\sqrt{5\pi}$

Finally, since this value equals $n$, the radius of the larger circle would be:

$\frac{3n}{2\pi}$
$= \frac{36\sqrt{5\pi}}{2\pi}$
$= \frac{18\sqrt{5\pi}}{\pi}$

And thus the answer is D.

7
[hide="9"]From some angle chasing, $angle EBF=90$. We can use 45-45-90 and 30-60-90 triangles. Since $EF=16$, $\implies EB=8\sqrt{2}$, $\implies BC=8\sqrt{6}$, $\implies P=32\sqrt{6}$.[/hide]
[hide="14"]The diagonal of the square equals $2(1+\sqrt{2})$, so the sidelength is $\sqrt{2}+2$. Thus, the wanted area is $(\sqrt{2}+2)^{2}-2\pi=6+4\sqrt{2}-2\pi$.[/hide]
[hide="15"]There are at most 15 stacks and 13 rows, so the largest yield is $15\times13\times3=585$.[/hide]
17
18
[hide="19"]Let r be the radius of the smaller circle, $\sqrt{2}/2=1/2+r(1+\sqrt{2})$. Solve for $r=\frac{3-2\sqrt{2}}{2}$, and the area is $(\frac{17-12\sqrt{2}}{4})\pi$.
[/hide]
[hide="24"]From Distance Formula, $AB=2\sqrt{5}$. Shoelace it, the area is  $15$, and it follows that the altitude to AB is $3\sqrt{5}$.[/hide]
[hide="25"]Tangents make right angles. Hypotenuse is 12, leg is 6, the other leg (the tangent) is $6\sqrt{3}$.[/hide]
[hide="26"]$\triangle ABE ~\triangle CDE$, so $15/30=CD/16\implies CD=8$. From Pythag, $CB=\sqrt{8^{2}+14^{2}}=2\sqrt{65}$.[/hide]
[hide="29"]Ratio of first circle to second circle is same as ratio of second circle to third circle, $8: 12=12: r\implies r=18$. Area is $324\pi$.[/hide]
[hide="31"](i don't think i explained this well) Draw radii to where the other edge meets the circle. The resulting triangle is comprised of two 2-1-rt3 (30-60-90) triangles, $(18-9-9\sqrt{3})$. The area of the triangles is $81\sqrt{3}$. The sector has angle 120 degrees, so the sector + the shaded half of circle - the triangle is our desired area. or $(1/2+1/3)\frac{18^{2}}\pi-81\sqrt{2}=270\pi-81\sqrt{3}$.[/hide]
[hide="32"]Similar triangles, there's a 12-4-$8\sqrt{2}$ triangle is similar to the 30-10-$20\sqrt{2}$ triangle. The cone therefore has a volume of $\frac{10^{2}\times20\sqrt{2}\times\pi}{3}=\frac{2000\sqrt{2}\pi}{3}$.[/hide]
33
[hide="34"]Use Distance, the smallest distance between any two of the given points is $2\sqrt{6}$, which must be the side of the square. Then the radius of the sphere must be $3\sqrt{2}$, so the volume is $72\sqrt{2}\pi$.[/hide]
[hide="35"]The length of the altitude to the hypotenuse is $10+10\sqrt{2}$, so the legs are each $5\sqrt{2}+10$, thus the perimeter is $30+20\sqrt{2}$.[/hide]
[hide="37"]Equation, $(3x)^{2}+x^{2}=10^{2}\implies x=\sqrt{10}$, and the area is $3x\times x=30$.[/hide]
[hide="38"]$\angle NOT=360-108-90-60$, $\triangle NAO$ is isosceles, thus $\angle ANO=\angle AON=39$, $\angle AOT=60$. $\angle TON=\angle AON+\angle AOT=39+60=99$.[/hide]
43,47

47 and 33 left.

 I went over most of the others last night.  :lol:

You all did really well explaining, or maybe I'm just an idiot.

[hide="47"]ok let the center be O. now draw AO, OB and OC. so what you should have is 2 30-60-90 triangles. use the properties to determine the radius. The shaded region is ABOC-sector BOC. since sector BOC has degree measure of 120 degrees, it should be easy to find it.[/hide]

#33, think about how to derive those lengths.

And British Flag, and then notice that it is a square.

I have a few questions.

I don't understand 19.

Also, I don't get 31

31. I don't see how people explained it.

I got the rest corrected and gone over for a long time.  :lol:

could someone elaborate?

19 if u draw a horizontal line thru O' and O..and then draw a vertical line thru O you see that 1 = 1/2 + (the horizontal distance between the vertical line and the centre of O' which can be found by pythagorean theorem) + r (the radius of the smaller circle)...

so the horizontal distance =

$(r+\frac{1}{2})^{2}-(\frac{1}{2}-r)^{2}= d^{2}$

i hope that didnt confuse u even more and 31..well if u take the centre of the circle and connect it to the the point where the other line and the circle meet...u see the distance between there is 18..and then the horizontal distance is 9 (given)..since its a right angle triangle is a $1: \sqrt{3}: 2$ triangle..so you can follow the other solutions...

Anywhere to get the problems without answers?

I don't know. I just practiced it by showing a solution if I could to see if I knew how to do the problem. I posted these to see if there were elegant solutions.

I could only find ones with answers..


A girl entered a shoe store, wanting to buy a pair of shoes That'll  be 2400 the shopkeeper declared. Unfortunately, the girl didn't have that much, but she decided to make a deal. She offered to buy the pair of shoes but only if the shopkeeper loaned her as much money as she currently had. The shopkeeper agreed, and the girl bought the shoes and left.
Then the girl went straight to the Book Store, and found a book she wanted for 2400. She still didn't have that much, so she made the same offer to the shopkeeper. The shopkeeper agreed to loan her the same amount of money as she currently had, and she bought the book.
She tried the same thing at both the Pharmacy and the Back-To-School Shop, where she bought a 2400 item at each shop. After completing her purchase at the Back-To-School shop, she had no money left.

How much money did the girl start with?

Cute problem, but what is it doing in the Advanced Fields section? I'm moving it to Basics. Hope, it won't stay unsolved there for long   :) .

Good job LuCy4EvA, I checked it and it indeed works.

[quote="math92"]i don't like reading long problems[/quote]

We all have to do them sometime.

after reading it 3 times i got
[hide="da answer"]$2250 dollars is how much she started with[/hide]


I right clicked and saved as the USAMTS include file. When I try to build it (LaTeX->PDF) I get this error from Adobe Reader: There was an error opening this document. The file cannot be found and another one from TexnicCenter: Cannot execute the command.

I would like to have this work before Round 1 Solutions are due. Ha.

You are more likely to get an answer if you submit similar questions to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX and Asymptote forum[/url].

My first suggestions: make sure that 1) you are trying to build the solutions file and NOT the include file, and 2) that the \include{...} line has, as its argument (the part between the curly braces) the exact name of the include file as you saved it (except for the extension, which should be .tex) and that the two files are in the same folder.


What's the correction applied to Newton's Law of Universal Gravitation that was recently in the news???

That sounds interesting. I haven't heard anything about it yet but I would like to too!

Does anybody else have any idea???

I think I saw somewhere

\[ F = \dfrac{G_{\infty} mM }{r^2} [ 1 + ( 1 + \dfrac{r}{\lambda} ) \alpha e^{ - \dfrac{r}{\lambda}} ] \]

where $\alpha = -0.007$ and $\lambda = 200 \ m$ approx.

Have any idea where it comes from???


In Albatross's school, all numerical grades are expressed as percentages and are rounded to the nearest integer (as an example, 72.24% rounds to a 72%, a 99.50% rounds to a 100%, etc.) and any grade that [i]rounds[/i] to a 90 or higher is considered an A. Also, all of his tests are out of 106 points and have an integer as the score. (No extra credit either.)

In the first marking period, Albatross's grade is made up of only nine tests. If he averages an 88% (rounded to the nearest integer!) on the first seven tests, what is the lowest [b]score[/b] he can get on the 8th test and still receive an A for the marking period?

[hide="My answer"]
92... I'm about 86.79% sure it's right.
[/hide]

[size=200]READ CAREFULLY[/size]

[hide]The lowest grade that rounds to an A is 89.5. This means the lowest total he'll have is 805.5, but since you can only have integral scores, it's 806. The highest grade that will round to 88 is the real number just less than 88.5. Multiply that by 7 to get 725.5. That means the highest integral total of his first 7 tests is 725. Assume 106 for the last test as you want as low as possible for the 8th. $806 - 106 - 725 = \boxed{81}$.[/hide]

[quote="Treething"]But that's not an A.[/quote]

[hide] $88+88+88+88+89+89+89+81+106=806$ according to my calculator. $\frac{806}{9}=89.55555...$ which rounds to $90$.  :? [/hide]

[quote="Ignite168"][quote="binonunquineist"][quote="Treething"]But that's not an A.[/quote]

Actually it is.[/quote]
Is your middle name Ying?

Atheist Temple: 88 88 88 88 89 89 89 is 83% Treething fails at averaging.[/quote]

omg... $\frac{88 + 88 + 88 + 88 + 89 + 89 + 89}{106 * 7}$ is 83%...

[quote="lotrgreengrapes7926"]I got [hide] $\boxed{81}$.  :? With the grades 88, 88, 88, 88, 89, 89, 89, 81, 106. [/hide][/quote]
that's nine grades.

i got [hide]103.  to get the last grade you do (88.49*7 (because that is the highest grade he can have to get an 88) + x)/8.

easier understanding way:

(88.49*7 + x)/8

we have to find the lowest value of x to make the equation thing as low as possible but not below 89.5. that's 96.57.
102/106 = 96.22%
103/106 = 97.16

103 is the lowest score he can get on the test to get an A.[/hide]

[quote="Treething"][/quote][quote="Ignite168"][quote="binonunquineist"][quote="Treething"]But that's not an A.[/quote]

Actually it is.[/quote]
Is your middle name Ying?

Atheist Temple: 88 88 88 88 89 89 89 is 83% Treething fails at averaging.[/quote][quote="Treething"]

omg... $\frac{88 + 88 + 88 + 88 + 89 + 89 + 89}{106 * 7}$ is 83%...[/quote]

The problem is not clear, you should've specified the percenetages are out of 106, rather than 100. I assumed the highest grade was 106/100 and when you meant no extra credit, you meant nothing else would be added other than test scores.

OK guys.

[hide="..."]
93, 93, 94, 94, 94, 94, 94 are the first seven scores.
[b]92[/b] is the 8th score.
106 is the 9th score.
[/hide]

[quote="Ignite168"][/quote][quote="Treething"][/quote][quote="Ignite168"][quote="binonunquineist"][quote="Treething"]But that's not an A.[/quote]

Actually it is.[/quote]
Is your middle name Ying?

Atheist Temple: 88 88 88 88 89 89 89 is 83% Treething fails at averaging.[/quote][quote="Treething"]

omg... $\frac{88 + 88 + 88 + 88 + 89 + 89 + 89}{106 * 7}$ is 83%...[/quote][quote="Ignite168"]

The problem is not clear, you should've specified the percenetages are out of 106, rather than 100. I assumed the highest grade was 106/100 and when you meant no extra credit, you meant nothing else would be added other than test scores.[/quote]

[quote="Sir Treething"]
...Also, all of his tests are out of 106 points...
[/quote]

[quote="Sir Treething"]
[hide="..."]
93, 93, 94, 94, 94, 94, 94 are the first seven scores.
[b]92[/b] is the 8th score.
106 is the 9th score.
[/hide]
[/quote]

[hide]
The lowest grade that rounds to an A is 89.5. $\frac{89.5}{100} = \frac{x}{106}$. Multiply by 106 and 9 and then divide by 100 to get 853.83 but since you can only have integral scores, the lowest possible total for an A is 854. The highest grade that will round to 88 is the real number just less than 88.5. Multiply that by 106 and 7 and then divide by 100 to get 656.67. That means the highest integral total of his first 7 tests is 656. Assume 106 for the last test as you want as low as possible for the 8th. $854 - 106 - 656 = \boxed{92}$. Keep in mind that is out of 106 though and its 100% equivalent would be 87.[/hide]

I WIN!

[quote="Treething"] lowest [b]score[/b]

[size=200]READ CAREFULLY[/size][/quote]

I even [b]bolded[/b] it.

For Christ's sake. (Sorry solafidefarms.) I'll explain it in more mathematically rigorous terms.

score $\neq$ percentage


Well they are all arithmetic sequences. The formula is

$ \frac{\text{last term}\minus{}\text{first term}}{\text{common difference}}\plus{}1$

So for the first, $ \frac{70\minus{}1}{3}\plus{}1\equal{}\frac{69}{3}\plus{}1\equal{}24$

What about the last one? It doesn't really have a common difference.

1.5, 5.5, 9.5, 14.5... This isn't an arithmetic sequence because 14.5 - 9.5 = 5 while 9.5-5.5 = 4

Oh whoa, I'm not sure what the pattern is in the second one; it increases 4. then 4, then 5, perhaps you meant the 4th term to be 13.5? But then 118 isn't divisible by 4...

Hmm...not an arithmetic sequence?

1.5 + 4 = 5.5...

5.5 + 5 = 10.5...

+ 6...+ 7....+ 8...etc etc

Maybe quadratic thingy?

yep that could be a quadratic. In a quadratic, the difference of the difference between successive terms is the same (im pretty sure?)

anyways, i dont know how to do a quadratic, but for your first one instead of just memorizing a formula, you could do the question logically. Since the common difference is three, we can add a constant number to each term to make it divisible by three. That wouldnt change the number of terms at all. Say we add 2 to each number to make it divisible by 3. Now the set is: 3,6,9.....69,72. Dividing each number by 3 wouldnt change the number of numbers in the set either, so lets do that. Now the set is: 1,2,3.....24.Thats a lot easier to work with. The number of terms is 24. Now you have a method that you can use if you ever forget the formula.

[quote="Cliu0301"]1. How many numbers are in the set 1, 4, 7, 10......, 67, 70?

2. How many numbers are in the set 1.5, 5.5, 9.5, 14.5..........119.5?

 :( Cliu fails.[/quote]

1. This is an arithmetic sequence in which each term after the first term is three more than the previous term. The total increase in the whole sequence is an increase of 69. Since each term increases by three. We must increase 23 times to get from 1 to 70. This means that there are 23 terms after the first term. Or 24 terms.

2. This is not a sequence at all. It is increasing by four, and then suddenly increases by five, so I think there's an error. But in general in order to find the number of terms in a sequence like this;

1.5, 5.5, 9.5, 13.5, ... 401.5

We notice that this isn't the easiest thing to count, but if we multiply all the terms by two, then we get the sequence;

3, 11, 19, 27, ... 803

We see that the sequence is increasing by eight, so if the first term was divisible by eight, then so would all the terms after it. If we add five to all the terms, then we will make all the terms divisible by eight. If we do this we get;

8, 16, 24, 32, ... 808

Now if we divide all the terms by eight we get;

1, 2, 3, 4, ... 101

So there are 101 terms in the sequence. Do you get it? If you are lazy, then it just requires a bit of manipulation.  :D

Oh sorry...

I'm bad at adding  :blush:

I meant to say 1.5, 5.5, 10.5, 16.5 *blushes at mistake*


What is the amount of natural numbers that leave a remainder of 41 when divided into 1997?

The answer is infinity.

There exists an infinite values of N that has the form

N = 1997x + 41 , where x is greater or equal than 0.

[hide="Answer"]$kn+41=1997 \Rightarrow kn=1956$. Both $k$ and $n$ must be positive integers, so we factor: $1956=2^2\cdot 3\cdot 163$, but we can only use divisors greater than $41$, or else they are not capable of leaving a remainder of $41$. Therefore, the only numbers that work are $163$, $3\cdot 163$, $2\cdot 163$, $2\cdot 3\cdot 163$, $2^2\cdot 163$, and $2^2\cdot 3\cdot 163$, for a total of $6$ numbers.[/hide]

[b]EDIT:[/b] allrighty, I think you misread the problem.

[quote] EDIt: allrighty, I think you misread the problem. [/quote]

OBJECTION (as they say in the court rooms):

The answer is infinity. I think you misread the problem. Here are the first seven integers:

1997+41, 1994*2+41, 1997*3+41, 1997*4+41, 1997*5+41, 1994*6+41, 1997*7+41...

Which all are congruent to 41 mod 1997.[/quote]

A "number divided into 1997" means $\frac{1997}{n}$, not $\frac{n}{1997}$. Case closed.  :?

[hide] If a number leaves a remainder of 41 when divided into 1997, then  $ \frac{1997-41}{n}$ or $\frac{1956}{n}$  Meaning $n$ is a positive factor of 1956.  The prime factorization of 1956 is: $ 2^2 \cdot 3 \cdot 163$.  However, we know that only numbers larger than 41 can leave a remainder if 41.  We also know that 1956 has $ 3 \cdot 2 \cdot 2 = 12$ factors.  The ones that dont work are combinations of the following product $ 1 \cdot 2 \cdot 2 \cdot 3 $ which is the same thing as saying all the factors of 12 dont work. There are $ 3 \cdot 2 = 6 $ factors of 12.  Meaning there are $12-6=6$ values for $n$ [/hide]


calculus
derivative
logarithms
LaTeX
trigonometry
calculus computations

$ g(x,y) \equal{} \ln(ye^{xy})$

what is $ g_x$ and $ g_y$

And what have you tried?

Also, use \ln in Latex to get $ \ln$ (similarly \sin, \cos, etc.)

Well the one with respect to x we just need to treat y as a constant and take the derivative of it. So the derivative of ln(x) is just 1/x right? But we use the chain rule here as well to take the derivative whatever is inside the ln(....), right?

Yes, just treat that $ y$ as a $ 5$, or whatever makes you think of it as a constant.

$ \frac{d}{dx} \ln(f(x)) \equal{} \frac{1}{f(x)} \left[\frac{d}{dx}f(x)\right]$.

I would start with algebra here: $ g(x,y)\equal{}\ln y\plus{}xy$. That's very easy to work with.

Okay so what I got with respect to x is:

$ \frac {y^2e^{xy}}{ye^{xy}}$

Is this correct??

Which is $ y$, which can also easily be seen from Jmerry's help.

And the answer with respect to y is:

$ \frac {e^{xy} \plus{} xye^{xy}}{ye^{xy}}$

Is this correct??

Yes, this is true. But you should simplify it by canceling the exponential...


Kyle has 112 unit cubes. If no unit cube may be cut into smaller pieces, what is the minimum number of additional unit cubes he needs in order to build a solid cube using all the unit cubes?

The closest cube is $ 125$, so he needs $ 125\minus{}112\equal{}\boxed{13}$ more unit cubes.


geometry
rectangle
percent
percent decrease

Kenny increased the length of a rectangle by $ 33\frac13\%$. By what percent must he decrease the width so that the area of the rectangle remains the same?

$ xy \equal{} \frac{4x}{3} \times y \times z$

$ xy \equal{} \frac{4z}{3} \times xy$

$ 1 \equal{} \frac{4z}{3}$

$ z \equal{} \frac{3}{4}$

The percent decrease must then be $ \boxed{25\%}$.


topology
geometry
vector
abstract algebra
advanced fields
advanced fields theorems

[quote="mlok"]Try to read Hatcher's [url=http://www.math.cornell.edu/~hatcher/AT/ATpage.html]Algebraic topology[/url]. There is no K-theory in that book. But then again, there are no royal roads in mathematics.[/quote]

thanks. maybe i'll try to get through the book when i'm at that level.

i've heard that k-theory combines algebraic topology (like what that text covers), algebraic geometry, geometric (or general?) topology, and analytic number theory.

Cohomology is a powerful mathematical technology for classifying differential forms. In the 1960s, work by Sir Michael Atiyah, Isadore Singer, Alexandre Grothendieck, and Friedrich Hirzebruch generalized coholomogy from differential forms to vector bundles, a subject that is now known as K-theory.
    Witten has argued that K-theory is relevant to string theory for classifying D-brane charges. D-brane objects in string theory carry a type of charge called Ramond-Ramond charge. Ramond-Ramond fields are differential forms, and their charges should be classifed by ordinary cohomology. But gauge fields propagate on D-branes, and gauge fields give rise to vector bundles. This suggests that D-brane charge classification requires a generalization of cohomology to vector bundles -- hence K-theory.  :wink:

One ponit of start of $K$ theory is the study of the vector bundles. They are in simple words, families of vector spaces (called fibers) each one supported by each point of a space $X$, the simpliest case is the is the tangent bundle of a manifold (a surface for example), which is the collection of all tangent spaces of that manifold. You can define operations over this bundles such as direct sum o tensor product doing this  over each fiber.

The main point is that the vector bundles over a space are a semigroup with the direct sum (as the natural numbers) and you can completed to a group named $K_{0}(X)$. The $K$ theory is the study of such $K$ groups, (you can define $K_{1}(X)$ as $K_{0}$ of the suspension of $X$ an so on) and the mesure topological properties of $X$.

I can recomend you Wegge-Olsen book which is an algebraic aproach to $K$ theory, specially to read the introduction whic is a very friendly reading.

[quote="drunner2007"]Does anybody here know any K-Theory?
It seems very interesting, and I want to learn more about it.[/quote]
There are algebraic K-theory and topological K-theory. I don't know much about the first one, only know the former. The following is the motivation of topological K-theory:
K-theory was introduced by Grothendieck. One can see in his formulation of Riemann-Roch theorem that for each projective algebraic variety, G set a group from the category of coherent algebraic sheaves. Atiyah considered a topological analog defined for any compact space X, a group K(X) constructed from the category of vector bundles on X. It's topological K-theory.
Have nice with this subject!


calculus
integration
trigonometry
algebra
polynomial
real analysis
real analysis unsolved

Let $ n$ a positive integer and $ s$ a real number. Prove that
\[ \int_0^1 dt \prod_{i\equal{}0}^{n}\cos^2(s\pi \plus{} 2^i \pi t)\equal{}\frac{1}{2^{n\plus{}1}}.\]

I have a question : is your question correct? I guess there are some typos here

well i mean if s is zero then

$ s \equal{} 0 \to$
$ 2^{n \plus{} 1}\int_0^1 dt \prod_{i \equal{} 0}^{n}\cos^2(s\pi \plus{} 2^i \pi t) \equal{} \int_0^1 \prod_{i \equal{} 0}^{n}(1 \plus{} \cos(2^{i \plus{} 1}\pi t)dt \equal{}$
$ \int_0^\pi \prod_{i \equal{} 0}^{n}(1 \plus{} \cos 2^{i \plus{} 1}t)\frac {dt}{\pi} \equal{} 1 \plus{} P(\sin 2^i\pi) \equal{} 1$

where P is some polynomial of the n variables $ \sin 2^it$, with P(0)=0

so i imagine if we carried the s around it would work too

$ \int_0^\pi \prod_{i \equal{} 0}^{n}(1 \plus{} \cos (2^{i \plus{} 1}t\plus{}2s))\frac {dt}{\pi}\equal{}1\plus{}P(\sin [2^i\pi\plus{}s])\minus{}P(\sin s)\equal{}1$


I tried using \setcounter{section}{1} in a document, and no error messages nor warnings were reported in the transcript, but the resetting of the variable didn't work.

Minimal example:

[code]
\documentclass[11pt]{article}
\usepackage{amsmath}
\begin{document}
\section{Table of Contents}
blah blah blah
\setcounter{section}{1}
\section{Introduction} %this still displays a 2(?)
Hello, world!
\end{document}
[/code]

The current value of the counter is increased for each new section, so if you want a 1 you'll need \setcounter\setcounter{section}{0} to set it to 0 first and then the next sectionn will be numbered 1.

Thanks for the help! Is there are way to remove the number beside a particular section?

Use \section* to not display the section number.


For integer $ k\geq 1$ let $ a_k$ be a maximum integer $ \leq \sqrt{k}$ and $ b_k \equal{} \frac {k}{a_k}$. For integer $ n\geq 1$ let $ S_n$ be the sum of $ b_k$ for any $ k$ satisfying  $ a_n\leq n$. Find all $ n$ such that $ \frac {S_n}{n}$ is integer.

By any chance did you mean $ a_k\le n$ instead of $ a_n\le n$?

Sorry, I made a typo. You are right.

[quote="kunny"]For integer $ k\geq 1$ let $ a_k$ be a maximum integer $ \leq \sqrt {k}$ and $ b_k \equal{} \frac {k}{a_k}$. For integer $ n\geq 1$ let $ S_n$ be the sum of $ b_k$ for any $ k$ satisfying  $ a_n\leq n$. Find all $ n$ such that $ \frac {S_n}{n}$ is integer.[/quote]

Could you correct this "for any $ k$ satisfying  $ a_n\leq n$"  into "for any $ k$ satisfying  $ a_k\leq n$"?

I can't edit it.

Thanks in advance.

Japanese Communities Moderator

kunny


I am angry because I am not able to solve this equation. Can someone help me?
Thank you for your time and support.

1/3 (c + 2) = c + 4

So, $ \frac{1}{3}(c\plus{}2)\equal{}c\plus{}4$?
First, we multiply each side by 3 to get
$ c\plus{}2\equal{}3c\plus{}12$. Then we subtract c and 12 from each side to get
$ \minus{}10\equal{}2c$. Dividing by 2, we find that
$ c\equal{}\minus{}5$

[hide]or do it the harder way:

open it up: $ \frac{1}{3}c\plus{}\frac{2}{3}\equal{}c\plus{}4$
subtract $ \frac{1}{3}c$: $ \frac{2}{3}\equal{}\frac{2}{3}c\plus{}4$
subtract 4:$ \minus{}\frac{10}{3}\equal{}\frac{2}{3}c$
multiply by $ \frac{3}{2}$: $ c\equal{}\minus{}5$[/hide]


[hide]Obviously $ n\equal{}4$ or $ n\equal{}8$ because the last two digits must be divisible by 4.
We also have that $ 12\minus{}6\minus{}(m\plus{}n)\equal{}11k$, for ANY integer k.

Thus we can have $ 6\minus{}(m\plus{}n)\equal{}\minus{}11\rightarrow17\equal{}m\plus{}n\rightarrow m\equal{}9,n\equal{}8$.

Thus our answer is 17[/hide]

It says -largest- 6 digit number. I don't know how to solve this, but just pointing it out... largest number is 595628, which gives 17...

i still ont get the second part on the 11k part.

cud u help me out sum more

Another way to do this is to, since the last digit is either 4 or 8, plug in one and work backwords to fit "divisible by 11". Try eight cuz its cooler.

5n5628.

Subtract n off, where n is $ 11 x k x 10^n$, starting with n = 1,and k is an integer and a variable.

First subtract 88, getting 5n5540.
Now subtract 440 getting 5n5100.
Now subtract 1100, getting 5n400. Now its fairly obvious that n =9, since n must equal 5 + 4.

um, where did the "divisible by 9" come from?

the last digit could also be 0, but that wouldn't really help in this problem

[hide]If 5m562n is divisible by 44, then it must be divisible by 4 & 11.  Since 5 + 5 + 2 = 12, 6 + m + n = 1, 12, or 23.  Since we are looking for the largest amount,  we want to aim for 6 + m + n = 23.  So, m +n then must equal 17 if the divisibility check for 4 works.  Since m + n = 17, m & n must be 9 & 8.
Since n must equal a 4 or 8, n = 8.  So, m must equal 9.

The final number is

5[u]9[/u]562[u]8[/u]. So 9 + 8 = [b]17[/b][/hide]

@ alanchou EDITED : I was thinking the answer was nine whe I wrote that.

[quote="verizonwires"]
12= 6+m+n.  so 6=m+n?
[/quote]
the divisibility rule states that $ 12\equiv 6\plus{}m\plus{}n \pmod{11}$ not $ 12\equal{}6\plus{}m\plus{}n$

[quote="ZhangPeijin"][quote="verizonwires"]Given that M and N are digits, what is the sum of the values which yield the greatest 6 digit number 5m562n that is divisible by 44.

I keep gettin 6, because it has to be divisible by 11 so 12= 6+m+n.  so 6=m+n?

but the answer is 17

no STUPID ANSWSERS!!!!!!!!!!!!!!!!!!!!!![/quote]
Since the number needs to be divisible by 4 and 11, the alternating digits must add up to a multiple of 11, and the last two digits must be a multiple of four. Since the second to last digit is 2, n must be either 4 or 8.
Therefore, either
$ 5 \minus{} m \plus{} 5 \minus{} 6 \plus{} 2 \minus{} 4 \equal{} 11x$
Or
$ 5 \minus{} m \plus{} 5 \minus{} 6 \plus{} 2 \minus{} 8 \equal{} 11x$
Simplifying...
$ 2 \minus{} m \equal{} 11x$
$ \minus{} 2 \minus{} m \equal{} 11$
Since we want the greatest number, we will use the second equation.
$ \minus{} 2 \minus{} m \equal{} 11$
m can only be one digit, therefore m is 9.
$ \minus{} 2 \minus{} 9 \equal{} \minus{} 11$
Using the values of m and n we get 595628.
Since the question wants the sum of the values of m and n, $ 9 \plus{} 8 \equal{} 17$[/quote]

How did u get from 11x to 11 on the second equation


geometry
rhombus
inequalities
similar triangles
Pythagorean Theorem

Let $ ABCD$ be a rhombus with $ AC\equal{}16$ and $ BD\equal{}30$. Let $ N$ be a point on $ \overline{AB}$, and let $ P$ and $ Q$ be the feet of the perpendiculars from $ N$ to $ \overline{AC}$ and $ \overline{BD}$, respectively. Which of the following is closest to the minimum possible value of $ PQ$?
[asy]unitsize(2.5cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

pair D=(0,0), C=dir(0), A=dir(aSin(240/289)), B=shift(A)*C;
pair Np=waypoint(B--A,0.6), P=foot(Np,A,C), Q=foot(Np,B,D);

draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(Np--Q);
draw(Np--P);

label("$D$",D,SW);
label("$C$",C,SE);
label("$B$",B,NE);
label("$A$",A,NW);
label("$N$",Np,N);
label("$P$",P,SW);
label("$Q$",Q,SSE);

draw(rightanglemark(Np,P,C,2));
draw(rightanglemark(Np,Q,D,2));[/asy]$ \textbf{(A)}\ 6.5 \qquad
\textbf{(B)}\ 6.75 \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 7.25 \qquad
\textbf{(E)}\ 7.5$

[hide]Let $ O\equal{}AC \cap BD$. ON=PQ so PQ is minimum when ON is. This is when ON is the altitude in AOB. As AB=17 $ ON\equal{}\frac {8 \cdot 15}{17}\equal{}\frac{120}{17}$[/hide]

haha TheZutonForce, "Please allow a few moments for the image to load."

Extra.

Find a synthetic proof that area PNQ is maximum when P is [hide]the midpoint of AB[/hide]. :D

[quote]Find a synthetic proof when the area [PNQ] is maximum.[/quote][b][u]Proof[/u].[/b] Denote $NP=x\ ,\ NQ=y$ . Observe that $\frac x{15}+\frac y8=1$ ([u]constant[/u]). Thus, the area $[PNQ]=xy$ is
max. iff $\frac x{15}\cdot \frac y8$ is max. iff $\frac x{15}=\frac y8=\frac 12$ . In conclusion, the area $[PNQ]$ is max. iff $N$ is the midpoint of $[AB]$ .

I was rather surprised no one used Cauchy Schwarz. I felt the need to add this solution.
[hide="Solution"]Let $a$ and $b$ be the sides of the altitudes from $N$ to $AC$ and $BD$, respectively. Using similar triangles, we can find that
\[\frac{a}{8-b} = \frac {15}{8} \implies 8a + 15b = 120.\]
Now by Cauchy Schwarz inequality, we have
\[(a^2 + b^2)(8^2 + 15^2) \ge (8a +15b)^2 = 120^2 \implies PQ \ge \frac{120}{17} \approx \boxed{7}.\][/hide]

Let $O = AC \cap BD$. Then $PQ = ON$. The minimum value of $ON$ occurs when $ON \perp AB$, so when $ON$ is the altitude to the hypotenuse of right triangle $\triangle ABO$. Since $AO = 8$ and $BO = 15$, $AB = 17$ and therefore $PQ = ON =\dfrac {8\cdot15}{17} = \dfrac {120}{17}$.

Hello @matrule, can u ples explain why the minimum value of $ON$ occurs when $ON\perp AB$?

Let's create a new point, $M$, on $AB$. Put it anywhere so that $OM$ is NOT perpendicular to $AB$. We have a right triangle, $\triangle ONM$. (Remember, $ON$ is perpendicular to $AB$.) From Pythagorean Theorem, $ON^2+MN^2=OM^2$. From here we can see that $ON < OM$. So, the shortest value of $ON$ is when it is perpendicular to $AB$

Heh . . . calculus would be "simple" here . . . (set up your coordinate system with $AC\cap BD=(0,0),$ and let $P=(a,0)$ so that we can find $Q$ and then the distance, so we may take the derivative and set it to $0.$)
Have fun with overkill!

[quote=Derive_Foiler]Heh . . . calculus would be "simple" here . . . (set up your coordinate system with $AC\cap BD=(0,0),$ and let $P=(0,a)$ so that we can find $Q$ and then the distance, so we may take the derivative and set it to $0.$)
Have fun with overkill![/quote]

Could someone please explain how to solve this problem using this method?

[url=https://www.desmos.com/calculator/odyf9bb9do]Here you go.[/url]

I wrote up the solution into Desmos. It's semi-interactive.

[quote=application][quote=AlanLiangTheAsian][quote="TZF"][hide="Pretty Picture Solution"][i]Please allow a few moments for the image to load[/i]
[img]http://img.photobucket.com/albums/v194/mukmasterxps/solution2.png[/img]
There. It's as simple as rotating your test-booklet.

Let the intersection of the diagonals be O, and assign the values $ BQ \equal{} 15x$ and $ NQ \equal{} 8x$.
We want the minimum value of:
$ PQ \equal{} \sqrt {OP^2 \plus{} OQ^2} \equal{} \sqrt {(15 \minus{} 15x)^2 \plus{} (8x)^2} \equal{} \sqrt {289 x^2 \minus{} 450x \plus{} 225}$

The minimum occurs when $ x \equal{} \frac {225}{289}$, which yields $ PQ \equal{} \frac {120}{17}$

The answer is $ \boxed{\textbf{(C)} \; 7}$[/hide][/quote]

how would one quickly find that answer 120/17 without plugging all these fractions in? calculation trick?[/quote][/quote]

[b]Solution:[/b] We exploit similar triangles. Notice that $AN=x, NB = 17-x$ so $NP=\frac{15}{17}x, NQ=8-\frac{8}{17}x$. When we find $PQ=\sqrt{NP^2 + NQ^2}$, we get the expression $\sqrt{x^2 - \frac{128}{17}x + 64} \text{ } (\star)$. It is not hard to see that the minimum occurs when $x=\frac{64}{17}$, which gives the approximate value of $\boxed{7}$ according to $(\star)$.


Hi Everybody,
Just 1 day left for Euclid, and guess what: cemc's webiste is not working. I still have to do the years 2003, 2004, and 2005. If any you have the solutions to these years in .pdf format then please email them to unexplained_ebe@yahoo.ca  or gangunzola@hotmail.com or even faizan_f1@hotmail.com If you know of another website with the solutions please tell me about it. Your help will be greatly appreciated. Thanks.   :D  :D

I.B. Nature

Hello again. The website's working again.  :blush:


inequalities
logarithms
induction
quadratics
algebra
quadratic formula

My friend asked me this question and I had a solution. I wanted some other solutions so I'll post the problem and see what you guys have.
I'll post my solution later if requested.

Question
Prove for all $ n$ natural numbers:
$ \frac {\log_{10} \left(1 \plus{} n\sqrt {2^{n \minus{} 1}}\right)}{n}\leq \log_{10} 2$

do we need to solve this or proof this one??

This is just a start,

\begin{eqnarray*}
\frac{\log_{10}\left(1+n\sqrt{2^{n-1}}\right)}{n}&\leq&\log_{10}2\\
\log_{10}\left(1+n\sqrt{2^{n-1}}\right)&\leq& n\cdot \log_{10}2\\
\log_{10}\left(1+n\sqrt{2^{n-1}}\right)&\leq&\log_{10}2^n\\
1+n\sqrt{2^{n-1}}&\leq&2^n\\
2^n&\ge& 1+n\sqrt{2^{n-1}}
\end{eqnarray*}

Which I believe can be proven with induction.

Sorry for not stating. You're required to prove.
I tried induction, but it didn't work out, so I did this instead.

Solution
Step 1
Cross-multiply and then remove logs. The equation becomes
$ 2^n\geq \ 1\plus{}n\sqrt{2^{n\minus{}1}}$

Test $ n\equal{}1$, we see that equality holds.
Test $ n\equal{}2$and $ n\equal{}3$ and we see that the inequality holds. Assuming that this pattern will hold true for all natural numbers greater than $ 1$
We now wish to prove
$ 2^{n\plus{}1}>1\plus{}(n\plus{}1)(\sqrt{2^n})$
Let $ p^2\equal{}2^n$
So we have
$ 2p^2>(n\plus{}1)p\plus{}1$

Carry over terms to prove
$ 2p^2\minus{}(n\plus{}1)p\minus{}1<0$

Because the coefficient of $ p^2$ is positive, the curve has a max. turning point and because of our assumption it means that the entire curve is below the x-axis, i.e. it has no real roots.

For this equation to have no real roots the discriminant $ \sqrt{b^2\minus{}4ac}<0$
Inserting values:
$ \sqrt{(n\plus{}1)^2\minus{}4(2)(\minus{}1)}<0$
$ \sqrt{n^2\plus{}2n\plus{}9}<0$

To determine the values for n under which our assumption holds, we use the quadratic formula
$ n\equal{}\frac{\minus{}2\pm\sqrt{2^2\minus{}4(1)(9)}}{2}$

This value of $ n$ is complex, however the question asks us to prove this statement of natural numbers.
Because our assumption holds true only when n is complex, we must reject our assumption due to question constrains.
Therefore, the statement $ 2p^2\minus{}(n\plus{}1)p\minus{}1>0$ must hold true for all natural numbers greater than $ 1$.
Equality holds at $ n\equal{}1$.

Hence, our original statement
$ \frac{\log_{10} \left(1 \plus{} n\sqrt{2^{n \minus{} 1}}\right)}{n}\leq \log_{10} 2$
is true for all n natural numbers.

End of solution

Critique away. I make another call for solutions from all (especially that induction proof).
Thanks in advance.


If the length and width of a rectangle are each increased by $ 10 \%$, then the perimeter of the rectangle is increased by

\[ \textbf{(A)}\ 1 \% \qquad
\textbf{(B)}\ 10 \% \qquad
\textbf{(C)}\ 20 \% \qquad
\textbf{(D)}\ 21 \% \qquad
\textbf{(E)}\ 40 \%
\]

[hide]Should be increased by the same factor: $ \boxed{10%}$[/hide]

The answer is obvious if you know that the scale factors of the side lengths and perimeters in similar figures are the same.

[hide="Solution"]
Let us assume the initial length and breadth to be $100l$ and $100b$ respectively.

Initially, the perimeter will be

$\\ P_{1} = 2(100l + 100b) \newline
\Rightarrow  P_{1} = 200(l + b) ...........(1)$

Now since length and breadth is increased by 10%, we get the new perimeter to be

$\\ P_{2} = 2(110l + 110b) \newline
\Rightarrow  P_{2} = 220(l + b) ...........(2)$

Here 110 because, 10% of 100 is 10, adding that to 100 yields 110.

Now simply divide 2 by 1 and multiply it by 100 to see how much more it is than the initial stage.

$ \\ \frac{P_{1}}{P_{2}} \times 100 = \frac{220 (l+b)}{200(l+b)} \times 100  \newline \newline
\Rightarrow \frac{P_{1}}{P_{2}} \times 100 = \boxed {110 \%} $

which is 10% more than the initial perimeter.

[/hide]

If the length and with of a rectangle increase by the same amount, [hide="then..."]the perimeter of the rectangle would also increase by the same percentage, which, in this case is 10%. The same idea applies to all polygons.[/hide]

the recangle is increased by[hide="=>"]b]10%[/hide]


Is i^i where (i=sqrt(-1)) really defined.If so what is its value?


probability
videos
conditional probability

Jon, John, and Jonathan each want to buy the same video game. Unfortunately, there's only three copies of it left in the store. The trio pools their money and sends Jonathan to the store to buy all three. The chance that Jonathan stabbs them both in the back and runs away with all three video games is one fourth. The chance that he gives two of the games to John is one quarter, and the chance that he gives two of the games to Jon is one half. If Jon receives two games, he has a one half chance of giving one to John and a one half chance of keeping both, and vice versa. Given that John received a game, what's the probability that Jonathan gave the two games to Jon?

well...if you assume that if Jonathan gives the two games to John, then John will keep them both
(it is not mentioned in the problem what John will do if he gets the games)

Then the probability of Jonathan giving the two games to Jon would be 100% cause otherwise it would not be possible to John to receive only one game.

Notice the "and vice versa" in the problem.

Making a probability tree helps make it easier to pick out all the numbers from that paragraph so they can be worked with. Also, when the problem says "Given that John received a game", I assume it means that he could have 1 game or 2 games.

[hide="Solution, as I see it"]
The probability that event $ A$ happened, given event $ B$ happened, is $ \frac{P(A \cap B)}{P(B)}$, where $ P(A \cap B)$ is the probability that event A and B both happened, and $ P(B)$ is the probability of B happening at all.

Let event $ A$ be "Jonathan gave two games to Jon", and let event $ B$ be "John received a game".

$ P(A \cap B) \equal{} (\frac{1}{2})(\frac{1}{2}) \equal{} \frac{1}{4}$

$ P(B) \equal{} (\frac{1}{4}) \plus{} (\frac{1}{2})(\frac{1}{2}) \equal{} \frac{1}{2}$

$ \frac{P(A \cap B)}{P(B)} \equal{} \frac{\frac{1}{4}}{\frac{1}{2}}$

$ \Longrightarrow \frac{1}{2}$[/hide]

Next time feel free to name the characters A, B, and C.  :wink:

[hide="Solution."]
The cases are limited enough so that we need not use too much conditional probability here. For sake of clarity, I will say $ (a,b,c)$ to mean that Jon got $ a$ games, John got $ b$, and Jonathan got $ c$. The cases and their chances are as follows:

$ (0,0,3)\equal{}1/4$
$ (0,2,1)\implies(1,1,1)\equal{}1/8$
$ (0,2,1)\implies(0,2,1)\equal{}1/8$
$ (2,0,1)\implies(1,1,1)\equal{}1/4$
$ (2,0,1)\implies(2,0,1)\equal{}1/4$

The probability that John got a game is thus $ \frac{1}{8}\plus{}\frac{1}{8}\plus{}\frac{1}{4}\equal{}\frac{1}{2}$. Of these cases, the ones in which Jonathan gave two games to Jon is $ \frac{1}{4}$. Thus:

$ \frac{\frac{1}{4}}{\frac{1}{2}}\equal{}\boxed{\frac{1}{2}}$.
[/hide]


superior algebra
superior algebra unsolved

this is part of bigger proof. I am trying to use this result

"a prime number p can be written as the sum of two squares iff p is 1(mod4)"

to find all primes in gaussian integers.

[url=http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares]Fermat's Theorem on Sums of Two Squares[/url]


linear algebra
matrix
linear algebra unsolved

Suppose $A$ is a $n \times n$ matrix such that the each row and each column contains the number $1,2,3,...,n$.
Let $B$ be the inverse of $A$.
Prove that each row and each column of $B$ has the same sum, and determine that sum.


Yup, "addendum"'s a word, as in "The emoticons were addenda (plural of addendum)"

this could be done by just looking at it... its clearly a geometric series, so 6 = 1/(1-x) and x is 5/6

for a geometric series starting at 0, the ratio is always
(Answer-1)/answer

[quote="MysticTerminator"]Just as a small addendum (is that a word?) we note that |x|=5/6 is indeed less than 1, so the infinite series converges.

For example, if the RHS were 1/3 or someat, we could blindly go ahead and solve 1/(1-x)=1/3 => 1-x=3 =>x=-2, but then the infinite series wouldn't converge, so there would be NO solutions.[/quote]

But mystic, the LHS starts off with a one. How can it equal something less than 1? The positve terms on the lLHS are larger than the negative terms on the LHS

yeah, that was my point. the point was that it is impossible. see, if you gave that to most peeps, they would still (if they didn't notice what you just said) sum the infinite series and solve to get x=-2, even though that's wrong since |x|>1.

darkquantum wrote:Solve for x: 1+x+x^2+x^3+...=6.

(note, yes, the left side is infinite)

[hide]x + x:^2: + x:^3: + ... = 5

x(1 + x + x:^2: + ...) = 5

1 + x + x:^2: + ... = 5/x

Hence, 6 = 5/x and x = 5/6[/hide]


conics
parabola
Vieta
function
quadratics
analytic geometry
calculus

Can someone tell me if this is the correct derivation of the formula to find the $ x$ value of the vertex of a parabola?

The two roots of the equation $ ax^2+bx+c=0$ are:

$ \displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The $ x$ value of the vertex of a parabola is given by the average of the roots:

$ (\displaystyle\frac{-b+\sqrt{b^2-4ac}}{2a})+(\displaystyle\frac{-b-\sqrt{b^2-4ac}}{2a})=\displaystyle\frac{-2b}{2a}=\displaystyle\frac{-b}{a}*\displaystyle\frac{1}{2}=\displaystyle\frac{-b}{2a}$

So the equation for the $ x$ value of a parabola is given by the formula: $ \displaystyle\frac{-b}{2a}$

[quote="JZambrano201"]Can someone tell me if this is the correct derivation of the formula to find the $ x$ value of the vertex of a parabola?

The two roots of the equation $ ax^2 + bx + c = 0$ are:

$ \displaystyle\frac { - b\pm\sqrt {b^2 - 4ac}}{2a}$

The $ x$ value of the vertex of a parabola is given by the average of the roots:

$ (\displaystyle\frac { - b + \sqrt {b^2 - 4ac}}{2a}) + (\displaystyle\frac { - b - \sqrt {b^2 - 4ac}}{2a}) = \displaystyle\frac { - 2b}{2a} = \displaystyle\frac { - b}{a}*\displaystyle\frac {1}{2} = \displaystyle\frac { - b}{2a}$

So the equation for the $ x$ value of a parabola is given by the formula: $ \displaystyle\frac { - b}{2a}$[/quote]

that is correct. :)
however, you could just straight up say the sum of the roots is -b/a by vieta's, so the average is -b/2a

This might not be relevant, but I'll post the proof of the quadratic formula, someone might need it in this thread... :ninja:

$ ax^2 \plus{} bx \plus{} c \equal{} 0$ (Quadratic Equation)

$ ax^2 \plus{} bx \equal{} \minus{} c$ (Subtracting $ c$)

$ 4a(ax^2 \plus{} bx) \equal{} 4a( \minus{} c)$ (Starting to complete the square, by multiplying by $ 4a$)

$ 4a^2x^2 \plus{} 4abx \equal{} \minus{} 4ac$ (Distributing and simplifying)

$ 4a^2x^2 \plus{} 4abx \plus{} b^2 \equal{} b^2 \minus{} 4ac$ (Completing the square, by adding $ b^2$)

$ (2ax \plus{} b)^2 \equal{} b^2 \minus{} 4ac$ (Factoring)

$ 2ax \plus{} b \equal{} \pm\sqrt {b^2 \minus{} 4ac}$ (Taking the square root)

$ 2ax \equal{} \minus{} b\pm\sqrt {b^2 \minus{} 4ac}$ (Subtracting $ b$)

$ x \equal{}\frac { \minus{} b\pm\sqrt {b^2 \minus{} 4ac}}{2a}$ (Finishing by dividing by $ 2a$)

[quote="JZambrano201"]Can someone tell me if this is the correct derivation of the formula to find the $ x$ value of the vertex of a parabola?

The two roots of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ are:

$ \displaystyle\frac { \minus{} b\pm\sqrt {b^2 \minus{} 4ac}}{2a}$

The $ x$ value of the vertex of a parabola is given by the average of the roots:
[/quote]One complication is that the two roots might not be real.  You need to say something about that case.

My comments: the quadratic formula is simply the summary of the experience of what happens when you try to solve an arbitrary quadratic [i]equation[/i] by the technique of completing the square (as fishythefish showed).

The parabolic curve is a quadratic [i]function[/i] rather than a quadratic [i]equation[/i]. However, the technique of completing the square is still precisely what you need.

So my suggestion: start over. Don't quote the quadratic formula, since you are not solving an equation. And don't derive the quadratic formula, for the same reason. But do use the technique of completing the square. Apply that to the quadratic function, and see if you can write $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ as $ a\cdot(x\minus{}\text{something})^2\plus{}(\text{something else}).$

Do that correctly, and you'll see that you also make Ravi B's objection go away.

@ravi b:

If you have had a solid grounding in imaginary numbers (which I'm surprised you don't know about @ this level), then you know that the quadratic formula still works. However, there will be answers that are in the form $ ai$ or $ a\plus{}bi$.

Here is an example:

$ x^2\plus{}2x\plus{}5\equal{}0$

$ x\equal{}\frac{\minus{}2\pm\sqrt{2^2\minus{}4(1)(5)}}{2(1)}$

$ x\equal{}\frac{\minus{}2\pm\sqrt{4\minus{}20}}{2}$

$ x\equal{}\frac{\minus{}2\pm\sqrt{\minus{}16}}{2}$

$ x\equal{}\frac{\minus{}2\pm4i}{2}$

$ x\equal{}\minus{}1\pm2i$

$ x\equal{}${$ \minus{}1\plus{}2i, \minus{}1\minus{}2i$}

As you can see (with a grounding in imaginaries), the quadratic formula holds. Also, complex solutions come in conjugate pairs. This means that for every solution $ a\plus{}bi$, there is a solution $ a\minus{}bi$.

Yes, I agree that the quadratic formula works even when the roots are not real.  But the original problem asks about the vertex of the parabola, not about the roots.  How do you know that the x-coordinate of the vertex will be the average of the roots, especially when the roots are not real?  JZambrano's argument needs more work.

Oh. well we know that every parabola has a vertex, so we don't need to worry about an imaginary one. Just use $ \minus{} \frac{b}{2a}$

But the original problem was to justify the formula $ \minus{}\frac{b}{2a}$.

Im not sure if this is correct or stated correctly:

Given a quadratic equation:

$ ax^2\plus{}bx\plus{}c\equal{}0$

That has two imaginary solutions:

$ \displaystyle\frac{\minus{}b\plus{}zi}{2a}$ and $ \displaystyle\frac{\minus{}b\minus{}zi}{2a}$

Where $ z$=$ \sqrt{b^2\minus{}4ac}$ when $ b^2\minus{}4ac<0$ thus being imaginary.

The average of the solutions is:

$ \displaystyle\frac{\minus{}b\plus{}zi}{2a}\plus{}\displaystyle\frac{\minus{}b\minus{}zi}{2a}\equal{}\displaystyle\frac{\minus{}b}{a}*\displaystyle\frac{1}{2}\equal{}\displaystyle\frac{\minus{}b}{2a}$

Therefore, even if the roots of a parabola are imaginary, the vertex will always be real.

Essentially, the vertex of a quadratic function is the maximum or minimum depending on the sign of $ a$.

How can you find a maximum or minimum using the given hints about completing the square and that $ x^2 \ge 0$

JZambrano, I agree that the average of the roots is always $ \minus{}\frac{b}{2a}$.  But how do you go from there to figuring out the vertex, especially when the roots are not real?

I agree with Kent and SimonM that completing the square is a good way to proceed.

[quote="1=2"][quote="Ravi B"]JZambrano, I agree that the average of the roots is always $ \minus{} \frac {b}{2a}$.  But how do you go from there to figuring out the vertex, especially when the roots are not real?

I agree with Kent and SimonM that completing the square is a good way to proceed.[/quote]

Or you could find the derivative and set that equal to 0 and solve, but completing the square is the best non-calculus way.[/quote]

"Set that equal to 0"

The only way you could do that is WLOG varying $ c$ since that only shifts the graph.

If you want the $ (x,y)$ coordinates of the vertex, then it's easy.

First write $ ax^2 \plus{} bx \plus{} c \equal{} 0$ as $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$

Then $ (x,y)$=

$ ( \minus{} \frac {b}{2a},f(x))$

or $ ( \minus{} \frac {b}{2a},f( \minus{} \frac {b}{2a}))$

This is because  $ \minus{} \frac {b}{2a}$ is the axis of symmetry. (Which is why it is also equal to the average of the roots.)

Sorry, I thought this had been answered earlier.  :blush:

And what jzambrano said earlier was also correct, I forgot to say that too...

Let's start at the beginning. By solving a quadratic equation $ ax^2\plus{}bx\plus{}c$ by completing the square, we obtained the roots. Because a parabola is symmetric, the average of these roots gave us the axis of symmetry. Because this is an x-value, we plugged it into the function $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ to obtain the y-value. This is the vertex, because the min (if $ a$ is positive) or max (if $ a$ is negative) must lie on the axis of symmetry.

I know that half of this isn't a "proof" kind of proof, but it is easy to obtain this if you just think for a second.

fishythefish, I think the discussion is now about what happens if there are no real roots.

Anyway, this is the answer which I would expect

[hide="Proof"]
Let the function be $ f(x)=ax^2+bx+c$

\begin{align*}
f(x) &= ax^2+bx+c \\
&= a(x^2+ \frac{b}{a}x + \frac{c}{a}) \\
&= a[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}]
\end{align*}

Since $ (x+\frac{b}{2a})^2 \ge 0$ with equality when $ x=-\frac{b}{2a}$ the the minimum (or maximum) must occur when $ x=-\frac{b}{2a}$
[/hide]

[quote]$ \displaystyle\frac { \minus{} 2b}{2a} \equal{} \displaystyle\frac { \minus{} b}{a}*\displaystyle\frac {1}{2} \equal{} \displaystyle\frac { \minus{} b}{2a}$[/quote]

just a question:

how can $ \frac { \minus{} 2b}{2a} \equal{} \frac { \minus{} b}{2a}$ if $ b < 0$ or $ b > 0$? :lol:

[quote="123456789"]I still remember in middle school, I was bored, because the teacher wouldn't tell us how to algebraically calculate the vertex (she said to guess and check by looking at the table function on the TI-83), so I did the following, where $ x$ is obviously the vertex:
\[ a{\left(x \plus{} 1\right)}^2 \plus{} b{\left(x \plus{} 1\right)} \plus{} c \equal{} a{\left(x \minus{} 1\right)}^2 \plus{} b{\left(x \minus{} 1\right)} \plus{} c
\]
[/quote]This is a great way of showing the symmetry of the parabola with the axis of symmetry passing through the vertex. In general,
\[ a{\left(v_x \plus{} k\right)}^2 \plus{} b{\left(v_x \plus{} k\right)} \plus{} c \equal{} a{\left(v_x \minus{} k\right)}^2 \plus{} b{\left(v_x \minus{} k\right)} \plus{} c
\]

where $ v_x$ is the x-coordinate of the vertex and $ k$ is some constant.

First of all, thanks to everyone who contributed to this post. Now I noticed that the question of 'what happens when the roots are imaginary' was mentioned. So this is what Im thinking:

To find the roots of a parabola, we must solve for the quadratic equation $ ax^2\plus{}bx\plus{}c\equal{}0$.

A generalization for this method is the quadratic formula, $ \displaystyle\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$.

When the value of $ b^2\minus{}4ac<0$, the roots are imaginary.

To find the value of $ x$ for the vertex, we must find the midpoint the roots.

When adding the roots together, notice that the terms $ \sqrt{b^2\minus{}4ac}$ are cancelled.

Therefore, reguardless of whether the roots are imaginary or real, the $ x$ value of the vertex will be real.

that's what i said. if anyone bothered to average the conjugate pairs of root $ a\plus{}bi$ and $ a\minus{}bi$, it would have been obvious. And furthermore, because of the way we solved it algebraically, it doesn't make a difference whether real roots are there or not. And roots aside, we can see that every parabola, even by definition, has a vertex.

I'm quite sure we are all ready to accept that $ \minus{}b/2a$ is the ordinate of the vertex. I think Ravi B's objection is that you are trying to find the vertex of a graph on a real plane by averaging two points that do not exist on this graph.

Indeed.  The argument by completing the square, on the other hand, is perfectly legitimate regardless of whether the parabola has real roots.  When the roots aren't real, it's not at all obvious that the imaginary roots tell us anything about the (real) geometry of the parabola.

Edit:  The complex graph of a parabola (where we allow $ x, y \in \mathbb{C}$) does, in fact, have the same symmetry as the real graph of a parabola, but here the need for an algebraic method of proof becomes evident.  I doubt any of you have geometric intuition applicable to a surface in 4-space.  :)

But wasn't the original problem how to find the vertex when the roots were imaginary. I accept that imaginary roots are not on the graph, and therefore, [i]visually[/i] do not provide any evidence of the location of the vertex. However, averaging the roots is an algebraic method, so I don't see the basis of your objection.

I accept that you have proved the following two facts:

(1) The average of the roots of the quadratic is $ \minus{}\frac{b}{2a}$.

(2) If the roots of the quadratic are real, then the x-coordinate of the vertex is the average of the roots.

From those two facts, you have proved as a corollary the following fact:

(3) If the roots of the quadratic are real, then the x-coordinate of the vertex is $ \minus{}\frac{b}{2a}$.

Unfortunately, that's weaker than what you were asked to prove.  The original problem is to prove the following fact:

(4) The x-coordinate of the vertex is $ \minus{}\frac{b}{2a}$, whether the roots are real or imaginary.

One way to prove that fact is to first prove the following:

(5) The x-coordinate of the vertex is the average of the roots, whether the roots are real or imaginary.

But you haven't proved (4) or (5) yet.  I await your proof.  Please don't reprove (1), (2), or (3).  I accept that you have proved them.  Either show how they prove (4) or (5), or find an independent proof of (4) or (5).

1) There is a handful of typoes in JZambrano201's posts.

2) [quote="Phelpedo"]I'm quite sure we are all ready to accept that $ - b/2a$ is the [b]ordinate[/b] of the vertex. I think Ravi B's objection is that you are trying to find the vertex of a graph on a real plane by averaging two points that do not exist on this graph.[/quote]Correct term for that is the [b]abscissa[/b].

3) Average of the roots of a parabola that yields the x-coordinate of the vertex is a numerical fact. This result has no geometrical content.

4) [quote="JZambrano201"]Can someone tell me if this is the correct derivation of the formula to find the $ x$ value of the vertex of a parabola?

The two roots of the equation $ ax^2 + bx + c = 0$ are:

$ \displaystyle\frac { - b\pm\sqrt {b^2 - 4ac}}{2a}$

The $ x$ value of the vertex of a parabola is given by the average of the roots:

$ (\displaystyle\frac { - b + \sqrt {b^2 - 4ac}}{2a}) + (\displaystyle\frac { - b - \sqrt {b^2 - 4ac}}{2a}) = \displaystyle\frac { - 2b}{2a} = \displaystyle\frac { - b}{a}*\displaystyle\frac {1}{2} = \displaystyle\frac { - b}{2a}$

So the equation for the $ x$ value of a parabola is given by the formula: $ \displaystyle\frac { - b}{2a}$[/quote]If the original poster asked about how to find the vertex and whether his way of finding the x-coordinate of the vertex, the method works.

5) Ravi B asked about a different question: he wanted someone to [b]prove[/b] how to find such vertex. In this case, it was through the trivial inequality. However, that requires the definition of 'vertex' to be the local extrema of a parabola...

6) Average of the roots depends on finding the roots through the quadratic formula, which in essence depends on the technique 'completing the square'.

7) Move on. The question has been answered.

This is an interesting topic but I'm moving up to Intermediate Forum now. The talk is getting bit advanced than what it started off with.

Regards,

[quote="t0rajir0u"][quote="10000th User"]3) Average of the roots of a parabola that yields the x-coordinate of the vertex is a numerical fact. This result has no geometrical content.[/quote]

This is precisely our objection.  Ravi B (and probably a few other posters here) has implicitly defined the vertex of a parabola as its single local extrema, which is a geometric statement.  You'll find that this is the commonly accepted definition.  I'm not sure what definition you're working with.[/quote]The definition that is appealing to me is the point of intersection of the axis of the parabola with the parabola. This definition makes use of the symmetry property present in the parabola and also covers the generalization where the axis is not parallel to the Cartesian axes. Now, I'm not really sure what definition(s) JZambrano201 and fishythefish are working with  :ninja:

[quote="t0rajir0u"][quote="10000th User"]This definition makes use of the symmetry property present in the parabola and also covers the generalization where the axis is not parallel to the Cartesian axes.[/quote]

When the axis of symmetry of a parabola isn't parallel to the axes, there is no longer a quadratic equation in one variable whose roots we can discuss, so this entire discussion fails to apply.  I believe it was clear we were restricting ourselves to a discussion of parabolas of the form $ y \equal{} ax^2 \plus{} bx \plus{} c$.[/quote]If you read carefully what it is written, it says "also covers" as an option, not meant to be included in your discussion and I am fully aware the main discussion is about parabolas of the form $ y \equal{} ax^2 \plus{} bx \plus{} c$. Is there anything else to discuss in this topic? Just wondering  :maybe:  or perhaps Ravi B still wants to see fishythefish's proof :D

Thanks for creating such a good discussion and as someone pointed out, there were some typoes in my posts which I apologize for. As for the proof, tOrajirOu was correct in stating that this discussion is restricted to parabolas in the form $ y\equal{}ax^2\plus{}bx\plus{}c$. Now, I understand that I must prove this statement for both real and imaginary roots. However, my argument is that the roots of a quadratic equation come in conjugate pairs. Because of this fact, when given a quadratic equation with imaginary roots, $ a\minus{}bi$ and $ a\plus{}bi$, their sum will be real. Therefore, the coordinate of the vertex will be real which is what I think Ravi B was questioning. As for what tOrajirOu also said, the fact that the vertex is defined as the single local extrema gives geometric value to the argument.

Here is a geometric way to find the vertex:

(@RaviB: Yes, even imaginary ones)

Write your imaginary roots in the form $ a\plus{}bi$.

Using $ a$ as the x-value and $ b$ as the y-value, graph them.

Their midpoint is the x-value of the vertex.

P.S. I do not know the proof completely for this, so see this link:
http://my.nctm.org/eresources/view_article.asp?article_id=6168

[quote="JZambrano201"]Because of this fact, when given a quadratic equation with imaginary roots, $ a \minus{} bi$ and $ a \plus{} bi$, their sum will be real. Therefore, the coordinate of the vertex will be real which is what I think Ravi B was questioning. [/quote]

That's not what Ravi B is questioning.  Reread his post with the five points.

[quote="fishythefish"]Write your imaginary roots in the form $ a \plus{} bi$.

Using $ a$ as the x-value and $ b$ as the y-value, graph them.[/quote]

I already addressed this point.  If you want the complex graph of $ y \equal{} ax^2 \plus{} bx \plus{} c$, you have to be willing to think about a surface embedded in four dimensions, and you can't prove the desired symmetry without using the completing-the-square proof Ravi wants you to use.

Thanks for clarifying that tOrajirOu. Now I understand Ravi B's question and I can see why completing the square is necessary for this proof.

There is always the Calculus way of getting that.

I have 2 things to say, and both are directed towards t0rajir0u:

1) Did you actually read the link?

2) xy-plane aside, have you ever worked with a complex plane? (y axis=imaginary axis, x-axis=real axis)

[quote="fishythefish"]1) Did you actually read the link?[/quote]

The link is irrelevant.  Reread Ravi B's discussion.

What the link is telling you is that the complex roots of a parabola with no real roots can still give you information about its shape.  I did not actually say that they didn't.  What I (and Ravi B) said was that [b]you didn't prove that they did.[/b]

[quote="fishythefish"]2) xy-plane aside, have you ever worked with a complex plane? (y axis=imaginary axis, x-axis=real axis)[/quote]

Only for the past... four years or so.  Have you ever worked with a projective complex plane?

Please don't question the credentials of people who are trying to help you.

Sorry, you're right.

I just noticed that if you actually go to the third page on that link, it links the imaginary roots to the vertex.


calculus
derivative
function
algebra
polynomial
induction
calculus computations

Assume that $ f(x)=\{\begin{array}{cc}
e^{\displaystyle-\frac{1}{x^2}}&x\neq0\\
0&x=0
\end{array}
$
Find $ f^{(n)}(0)$, here $ f^{(1)}(x)$ means $ f'(x)$, etc. :oops:

The first derivative is $ \frac{\minus{}2}{x^{3}}e^{\minus{}\frac{1}{x^{2}}}$ when $ f\neq 0$.  This function is very flat around the origin so you're going to get something resembling 0 in a neighborhood of the origin.

[quote="JRav"]The first derivative is $ \frac { \minus{} 2}{x^{3}}e^{ \minus{} \frac {1}{x^{2}}}$ when $ f\neq 0$.  This function is very flat around the origin so you're going to get something resembling 0 in a neighborhood of the origin.[/quote]
Yes, but from which can we draw the conclusion that the first derivative is flat around the origin. [/img]

From your picture, you should be able to see that $ f$ is very flat around the origin.  The derivative is $ \frac{\minus{}2}{x^{3}}e^{\minus{}\frac{1}{x^{2}}}\equal{}\frac{\minus{}2}{x^{3}e^{\frac{1}{x^{2}}}}$.  As $ x\to 0$, the denominator tends to 0 very, very quickly ($ e^{\minus{}1/x^{2}}\to\infty$ much faster than $ x^{3}\to 0$).

This is a reasonably common question (it's a simple example of a smooth, non-analytic function) and you could try some searches of this forum to find past discussions.  (Try using combinations of the important words in the question and the keywords "smooth" and "analytic" and "non-analytic.")  You presumably can compute the derivatives of the function at $ x \neq 0$ by hand; they all have a common form, which one might describe as "something of polynomial growth times something of exponential growth."  (It might help to make the substitution $ y \equal{} \frac {1}{x}$ to make this a bit clearer.  Here the "polynomial" part is going to $ \pm \infty$ as $ x \to 0$ while the "exponential" part is going to $ 0$.)  From what you know about rates of growth of various functions, who wins in a fight between exponential and polynomial?  So, it should be enough to prove (probably you'll want to invoke the idea of induction here) that the derivatives away from 0 really [i]do[/i] have such a form (note that the precise formulas are irrelevant when doing rough estimates of growth rates, and we only need a rough estimate here) and so make your conclusion (explicitly invoking the definition of a derivative, of course, since the formulas you have for the derivatives don't work at $ x \equal{} 0$).


inequalities
logarithms
inequalities unsolved


Finding $ 600 \plus{} 1 \mod{7}$ will determine the column.

This is $ 6$.  So we must find the $ 6$th number, which is in column $ \boxed{\text{B}}$.

I'm not sure.

It might be $ C$...

The answer says D...

This is how I did it, not sure if it's that great  :maybe: :
By observing the example, we can group the numbers into groups of 7 in order to find a pattern, that for every 7, the location of the number restarts at B. Also, the integers are written starting from 2. Therefore, we subtract 1 from 600 to get 599. Then, we divide by 7 and get a remainder of 4. Starting from column B and following the pattern, we can get to column D.


calculus
integration
logarithms
trigonometry
calculus computations

What's wrong with my solution?

$ \int \sqrt{\tanh x}dx$

for $ u\equal{}\tanh x\rightarrow du\equal{}\frac{dx}{\cosh^2x}$

$ \int \sqrt{\tanh x}dx\equal{}\int\sqrt{u}\cosh^2x dx\equal{}\int\frac{\sqrt{u}dx}{1\minus{}u}\equal{}\int \frac{du}{2(1\plus{}\sqrt{u})}\minus{}\int \frac{du}{2(1\minus{}\sqrt{u})}$

for $ v\equal{}(1\plus{}\sqrt{u})\rightarrow dv\equal{}\frac{du}{2\sqrt{u}}$ and for $ z\equal{}(1\minus{}\sqrt{u})\rightarrow dz\equal{}\minus{}\frac{du}{2\sqrt{u}}$

$ \int\frac{dv(v\minus{}1)}{v}\plus{}\int\frac{dz(1\minus{}z)}{z}\equal{}v\minus{}\ln v\plus{}\ln z\minus{}z$

$ \int \sqrt{\tanh x}dx\equal{}\ln (\frac{1\minus{}\sqrt{\tanh x}}{1\plus{}\sqrt{\tanh x}})\plus{}2\sqrt{\tanh x}$


Given the reals $ a, b, c$ satisfy $ ab \plus{} bc \plus{} ca \geq\ 11.$
Find Minimum of $ P \equal{} \sqrt []{a^2 \plus{} 3} \plus{} \frac {\sqrt []{7}}{5}\sqrt []{b^2 \plus{} 3} \plus{} \frac {\sqrt []{3}}{5}\sqrt []{c^2 \plus{} 3}$
When does equality occur ?
Dear my friends, I hope you'll like it  :)

When does equality hold???
I guest that, $ a\equal{}1,b\equal{}2,c\equal{}3$

[quote="Hong Quy"]When does equality hold???
I guest that, $ a \equal{} 1,b \equal{} 2,c \equal{} 3$[/quote]

Yes, you are right. Please try to solve it.
And a similar + harder is one of problems of MIC Round 3 (of course, I can't post it here now)  :)

Nice inequality, Nguoivn.  :lol: We can prove it using the inequalities:

$ 2\sqrt{a^2\plus{}3}\ge a\plus{}3$,

$ \sqrt {7(b^2\plus{}3)} \ge 2b\plus{}3$,

$ 2\sqrt{3(c^2\plus{}3} \ge 3c\plus{}3$,

$ (5a\plus{}4b\plus{}3c)^2\minus{}44(ab\plus{}bc\plus{}ca)\equal{}(2a\minus{}b)^2\plus{}\frac 5{3}(3b\minus{}2c)^2\plus{}\frac 7{3}(3a\minus{}c)^2 \ge 0$.

Nice proof, Vasc  :lol:
[quote="nguoivn"]
And a similar + harder is one of problems of MIC Round 3   :)[/quote]

[b]Problem 3 (Round 3)[/b]
Given the reals $ a, b, c$ satisfy $ ab \plus{} bc \plus{} ca \geq\ 11.$
Prove that: $ \sqrt [3]{a^2 \plus{} 3} \plus{} \frac {7}{5\sqrt [3]{14}} .\sqrt [3]{b^2 \plus{} 3} \plus{} \frac {\sqrt [3]{9}}{5}. \sqrt [3]{c^2 \plus{} 3} \geq\ \frac {23}{5\sqrt [3]{2}}$
Equality also occurs when $ (a, b, c) \equal{} (1, 2, 3)$  :)


if it is very easy?! Can you post a proof?  :P  :wink:

[quote="irantst"]hi
you can assume xyz=1  or x+y+z=1  so ........[/quote]
Sorry, but we can't assue this because the inequality is not homogeneous :!:

[quote="Zarif"]I`m sorry. I don`t know English. But your inequality very easy.[/quote]

Your proof is wrong.

Nice!
Your inequality is equivalent to,
\[ 5(xy \plus{} xz \plus{} yz) \plus{} 4xyz \le 1 \plus{} 2(x \plus{} y \plus{} z) \plus{} {x^2} \plus{} {y^2} \plus{} {z^2} \plus{} \sum {{x^2}(y \plus{} z)\plus{}xyz(x\plus{}y\plus{}z)}\]
Use these inequalities
\[ \sum {{x^2}(y \plus{} z)} \plus{} 2(x \plus{} y \plus{} z) \equal{} \sum {({x^2} \plus{} 1)(y \plus{} z) \ge \sum {2x(y \plus{} z) \equal{} 4\sum {xy} } }\]

\[ {x^2} \plus{} {y^2} \plus{} {z^2} \ge xy \plus{} xz \plus{} yz\]

\[ 1 \plus{} xyz(x \plus{} y \plus{} z) \equal{} 1 \plus{} {x^2}yz \plus{} x{y^2}z \plus{} xy{z^2} \ge 4.\root 4 \of {{x^4}{y^4}{z^4}} \equal{} 4xyz\]
The result follows.

sorry
I made a big mistake
we can't assume xyz=1

[quote="Luba1123"]Let x, y, z be positive real numbers. Prove that $ \frac {1}{{x \plus{} y \plus{} z \plus{} 1}} \minus{} \frac {1}{{(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)}}\le\frac {1}{{8}}$
and determine when there is equality.[/quote]
I think your question is very nice!!!!!!
How can you get this question?Or can we get some similar problem?

this problem is in the crux mathematicorum  :)

[quote="Luba1123"]Let x, y, z be positive real numbers. Prove that $ \frac {1}{{x \plus{} y \plus{} z \plus{} 1}} \minus{} \frac {1}{{(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)}}\le\frac {1}{{8}}$
and determine when there is equality.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67113


Ok this is a contest where I (Or any helpers) will post a problem and then the first person to answer correctly is given a point. there will be 10 rounds and then the winner chosen. If u want to enter then post to tell me and i will write it down. Also, a few helpers would be nice, since in the contest only me and helpers are spposed to post any problems so this won't become a marathon. :lol: I NEED HELPERS!!!  :(

Okay i need 1 helper and 3-10 people casue i wann start by wednesday

:) yay! okay now i need more people to participate!

Should I join? Or am I too old to think quickly?

(That means I join if possible. :D )

I'll try to start tomorrow. Hopefully a few more poeple will join. :D

[quote="isabella2296"][quote="alton66"]Oh and isaballa how did you put the challenge icon on your 10 round tournament? Nice tournament too its really long.[/code][/quote]

Thanks, but what do you mean by "challenge icon"?[/quote]

that star thing...

when you make the topic theres something called a message icon
click on the star thing

I'm starting now!

Rules:

1. If you enter a wrong answer but reply first, you will not be given any points.
2. If you give an incorrect answer, do not post your new answer unless someone else has had a try as well.
3. Feel free to give comments, even if you are not posting an answer.
4. Don't feel bad if you get one wrong.  :(
5. PLEASE HIDE YOUR ANSWERS!
___________________________________________________________________________________________________________

PROBLEM 1:

I am the youngest of 5 brothers, we have been given 100 dolars to share like this:

Each brother gets 5 dollars more than the brother younger than them. How much do I get? (The youngest brother)

HURRY!!!

[quote="alton66"]First of all, that's not very clear, and also. If you wanna help you have to drop out of the competition. Just let Isabella do it.

It's also unfair that you didn't tell everyone when you were posting it, that way it's easier to judge the speed if they all saw it at the same time.[/quote]

Fine.

I'm not even in this thing.

Mehh.

Oh, and BTW, the answer is 90.

[hide="How do you Find It?"]The measure of an angle is $ \frac{1}{2}$ the angle measure of the intercepted arc.  Thus, you have to calculate $ \frac{1}{2} \cdot 180$, to get an answer of $ 90$.  It's a theorem.[/hide]

All right... First problem posted tomorrow (I can't help the delay) 12:00 forum time!

First problem!!!

What is the sum of all the even numbers from 387-569.

Please pm me the answers.

Ernie, has won the first round. Next problem tomorrow at 7 forum time!

Wait... I somehow changed my forum time to EST.  What time will the problem be posted in EST?  I'm probably not gonna even be on AoPS tomorrow.  But I'm leading!!!  YAY!

[quote="isabella2296"]EST is roughly 3 hours after forum time.[/quote]

Yeah, I won't be here on weekdays that time.  Oh well.

dang, missed the first prob
o well
soo...
7 forum time is about 10 eastern time?

Approximately, yes. It's a few minutes behind.

kool
except, it's 8:38 forum time and 12:42 here
soo...
either my forum clock is one hour late, or it's a difference of 4 hours
or maybe it has something to do with daylight saving......

[quote="isabella2296"]Probably daylight saving...because here, it's 10:44 am forum time and 1:58 EST pm.[/quote]

Wait... so its 11 AM est?  Oh well, I still won't be here.


Let $ABCD$ be a trapezoid $(AB \parallel CD)$.
Let the points $E\in AD \cap BC$ and $I \in AC \cap BD$.
$M,N$ are the midpoints of $AB,CD$ respectively.
It is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52603]known[/url] that the points $E,M,I,N$ are collinear.

Prove that the division $(E,M,I,N)$ is harmonic (that is, $\frac{EM}{EN}= \frac{IM}{IN}$)

[color=brown]P.S. It can be applied also to a parallelogram $ABCD$[/color]

Easily By Ceva we find that
$\frac{ND}{NC}\cdot\frac{BC}{BE}\cdot\frac{AE}{AD}=1$

so E,M,I,N are collinear .Now the triangles $AIB$ and $DIN$ are similar so $\frac{IM}{IN}=\frac{AB}{DC}$
But the triangles $EAB$ and $EDC$ are also similar so
$\frac{EM}{EN}=\frac{AB}{CD}$


quadratics
algebra
polynomial
quadratic formula

6. Use the quadratic formula to find the roots of the polynomial
\[kx^2+(k+1)x-(k+2)\]

Please?
Analytically..

Well, a quadratic equation is under the form : ax² + bx + c, here you have :
a = k
b = k + 1
c = -(k+2)
So, easy to find the roots.

I'm asking for help because I don't know where to start... please explain further.

Imho, you should read that : http://www.themathpage.com/alg/quadratic-equations.htm

That'll help you to solve/understand those problems ;)

We are not here to solve things [i]for[/i] you, and please do not double post.

And, by the way, what's so hard about calculating \[\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] for $a = k$, $b = k + 1$ and $c = -(k + 2)$?

I am stuck at

-(k+1) +- SR[(k+1)(k+1)+4k(k²+2)]
---------------------------------------------
                      2(k+1)

I'm horrible at algebra... god; you guys have no heart. You don't have to judge me just because I can't plug things in and always know exactly what to do.

I'm here to learn.. and by seeing things analytically I can better understand the steps taken to reach the solution.

You'll get \[\frac{-k - 1 \pm \sqrt{5k^2 + 10k + 1}}{2k}\] and you can't go any futher. Very ugly expression.

If it's that ugly, I'm guessing I did something wrong thus I asked for help from someone smarter then me. I have to drive out to the school within the next hour and I simply cannot afford to get this problem wrong. I NEED to get my GPA up and cannot do that by getting more B's and C's in subjects.  I'll ask once more, please help me.

But how can we help you more than we did already? I solved the problem for you right?

*Sigh* :(

If the quadratic coefficient were $1$, not $k$, then there would be a nice factoring of the quadratic.  Are you sure you have copied down the problem correctly?

[quote="frt"][hide]$x=\frac{-(k+1)\pm\sqrt{(k+1)^2+4k(k+2)}}{2k}$

$=\frac{-k-1\pm\sqrt{k^2+2k+1+4k^2+8k}}{2k}$

$=\frac{-k-1\pm\sqrt{5k^2+10k+1}}{2k}$

I don't think we can go further from here...
In other words, this is the roots in the simplest form in terms of $k$.[/hide]

EDIT- I guess I was a little too late :P But I showed some steps at least... Hope this helps.[/quote]

Thank you so much, and a side note.. I was up to step 6; therefore this just proves how horrible I am at basic algebra.

That problem as written is:

Use the quadratic formula to fin dthe roots of the polynomial in Problem 4 [kx²+(k+1)x-(k+2) *where k is a constant].  Use the formula for S[size=75]k[/size] to calculate as it approachs infinity.

We still don't know what $s_k$ is and what limit we should compute, but anyways, I think this topic should be locked now...


number theory
prime numbers
arithmetic sequence
relatively prime
number theory unsolved

Do there exist a positive integer $n$ such that :
a) $n$ is not divisible by a square of a number greater than $1$;
b) the number $n+1,n+2,...,n+2003$ are divisible by a square of a number greater than $1$;
c) the number $n+2004$ is not the exact power of some positive integer ?

[quote="harazi"]I think it exists, but I have to use Dirichlet's theorem.  First, choose $ p_1,...,p_{2003}, q$ distinct prime numbers and by chinese theorem choose $n_0$ a solution of the congruences $ n=-1(mod p_1^2),..., n=-2003(mod p_{2003}^2), n=-2004+q(mod q^2)$. Then the arithmetic progression $ n_0+qp_1^2...p_{2003}^2$ also verifies these congruences, thus for each term $s$ of this progression the numbers $s+1,...,s+2003$ are not squarefree and $ s+2004$ is not a power since a prime of its decomposition has exponent 1.  Observe that if we choose $ p_i,q$ large enough, $n_0$ is relatively prime with each $q,p_1,...$ thus by Dirichlet theorem there is a prime $s$ in this progression. This prime is good (it verifies also the first condition).[/quote]
[i] I think it should be :

Then the arithmetic progression $ n_0+ n \cdot q^2 p_1^2...p_{2003}^2$ also verifies these congruences[/i]

  By the way , good solution :D


abstract algebra
superior algebra
superior algebra unsolved

Show that Z/d is not an injective Z/n-module if
there is a prime number p such that p divides both d and n/d.


MATHCOUNTS
geometry
Support
AMC
AMC 8
search
Harvard

Does anyone who is actually good at it have any advice? Especially for the sprint round. Or any advice on what not to do?

1.If you finish the test check it, don't sit there.
2.Skip the ones you don't know or think will take a while.
3.Also, read the formulas posted in the other topics in this forum.
4.Read the toolbox and Daiz bible.
5.Do the saab drills.

I'll think of more later.

[quote="xxm0rg07hxx"]I attached it here.[/quote]

Uhh...

Where is it?  It isn't in your post.

There should be a place that says download.
It has perfect squares, prime numbers, etc.

I just looked at it and it's three as a download.

[quote="JesusFreak197"]The download thing, maybe? Or did he attach that after you posted?[/quote]

Yeah, if you look in my quote box, it doesn't show the download available, so he must have attached it after I posted.

Actually, it never shows in the post box so just look at my post and it's there.

I've seen it before.
Is there some sort of strategy for doing well at the countdown round? I've heard that some people don't read the entire question.

The way to do best is to read the numbers and the calculations because they sometimes give extra information.

keep a level head.  If you do that, you feel more confident, and your opponent thinks that you just about have the answer.  It actually works!

Also, don't feel overconfident about your opponent, they may be better than you think.  If you get first in the written, watch the people before you in the countdown so you know what their game is.

One could make a slideshow of past countdown round questions that would switch slides every 45 seconds. That'd be the best way of studying for the countdown round (I'd imagine), without spending a lot of money on a projector or something.

From a 2nd grader's viewpoint (5 years ago at my brother's 7th grade nats):

It was a blast! The only bad part was that I couldn't solve ANY of the problems in the cooldown round or the countdown round? Like wtf I'm supposed to be good at math! The pin trading was fun. This kid gave me like 35 vermonts and they're rare, YAY!.

Now, from a competitor's viewpoint...  :?

I dunno now, but in 2001-2002 it was one of the main attractions. There was also a super fancy dinner party. 2001: DC. 2002: Chicago (fitting, Illinois winner w00t!)

do you know how long the teams get to stay there?

This year it's May 11-14, either that or 10-13 or 9-12.

I think it'd be kind of hard to make the state teams practice together before nationals, since people could live on opposite sides of the state. Do the teams practice together ever or does each member just practice individually?

I think that in most states, you drive far to get to practice as well as individually.
In DE the drive was only 10 min :lol:

It would suck to be in a state like CA then. What did you do (besides math) at nationals?

We got there and went to the dinner thingy and exchanged pins.

On Friday, we did the compeition in the morning and left for Joe Dumar's FieldHouse at night. We did LOTS AND LOTS OF FUN STUFF THERE. It was the funnest part of the trip.

On Saturday we went to the GM plant :sleeping: and then hung out for the rest of the day.

Sunday morning, we left

Each state had its own pins, and each person usually got about 57 of them.
Then you could trade your states pins for another states pins.

did you see the ESPN people? how do they pick which kids get to talk on TV anyway?

I think it was random.
I say them talking to some random guy from Connecticut and Puerto Rico, neither did really great.

No I didn;t have a chance to see the ESPN dudes but I knew that Mr. Rusczyk was doing the compeititon

[quote="janieluvsmusic"]It would suck to be in a state like CA then.[/quote]

Yeah, I can testify to that. I had to fly to San Jose to practice with the rest of the team.

The pins are just for fun. You trade with people from other states/territories and try to get all 57. :)


function
algebra
domain
real analysis
calculus
derivative
real analysis theorems

[size=150]What is the importance of the [i]Z-transform[/i] and it's inverse?[/size]

The z-transform belongs to the Fourier/Laplace family of transforms. In some mathematical circles, it is common to speak of the relationship between a sequence and its generating function (the power series with that sequence for coefficients) without calling it the z-transform, but other than some notational conventions, that's really the same thing.

The z-transform shares with the Laplace transform that it transforms functions on a one-sided, half-infinite domain (in this case, either $ \mathbb{N}$ or $ \mathbb{Z}^\plus{}.$) And you have the basic relationships that characterize all of the Fourier/Laplace transform family: pointwise multiplication on one side is convolution on the other side; differentiation on one side (if it makes sense) is multiplication by the variable on the other side; and so on.

Thanks a lot for your explanation.
Is there any webpage which illustrates the application of the z-transform?
I would like to learn more about the application of other generating functions too.  :)


I got textboxes to fill and checkboxes and radioboxes to select, and once I click on a button called "button1" I need to see the info in the list view. In case many checkboxes are selected, I don't know how to show all the info of the checkboxes in the same ligne, could someone help ?
(I'm okey with textboxes, and also with radioboxes since we need to show only one radiobox in the listview)
First, I fill the titles of columns in the listview
[code]
Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
ListView1.View = View.Details
ListView1.Columns.Add("Fisrt and last name", 90, HorizontalAlignment.Left)
ListView1.Columns.Add("Address", 90, HorizontalAlignment.Left)
ListView1.Columns.Add("city", 90, HorizontalAlignment.Left)
ListView1.Columns.Add("gender", 90, HorizontalAlignment.Left)
ListView1.Columns.Add("Hobbies", 90, HorizontalAlignment.Left)
[/code]

1/ Let's say that hobbies is composed of 3 checkboxes hb1 hb2 and hb3. and the user selected more than one. how can we display in the listview all the hobbies in the same box ?

2/ Another thing, for the First and last name, I've got two separated textboxes, so how can we show both of them in one box in the listview ?
Here is my attempt, but this works only if user selected 1 hobby
[code]
Private Sub button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles save.Click
ListView1.Items.Add(First & last name.Text)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(address.Text)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(city.Text)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(gender.Text)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(Hobbies.Text)
[/code]

3/ Another question : We got 2 textboxes 2 radioboxes 2 checkboxes, we would like to display in one msgbox the name of those which are not filled or checked. How can we do that ?

The simplest solution would be to use a bunch of ifs:

[code]        Dim displayText As String = ""
        If CheckBox1.Checked Then
            displayText += "CheckBox1 was checked "
        End If
        If CheckBox1.Checked Then
            displayText += "CheckBox2 was checked "
        End If
        If RadioButton1.Checked Then
            displayText += "RadioButton1 was checked "
        End If
        If RadioButton2.Checked Then
            displayText += "RadioButton2 was checked "
        End If
        ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(displayText)
        MsgBox(displayText)
[/code]
(Modify the code to fit your needs.)


If $a*b = a^b + b^a$, for all positive integer values of $a$ and $b$, then what is the value of $4*3$?


vector
Functional Analysis
real analysis
real analysis unsolved

Let $ A$ be a family of proper closed subspaces of the Hilbert space $ H\equal{}l^2$ totally ordered with respect to inclusion (that is
, if $ L_1,L_2 \in A$, then either $ L_1\subset L_2$ or $ L_2\subset L_1$). Prove that there exists a vector $ x \in H$ not contaied in any of the subspaces $ L$ belonging to $ A$.

[i]B. Szokefalvi Nagy[/i]

Suppose (for a contradiction) that $ \bigcup_{L\in A}L \equal{} H$. Let $ \{e_n\}_{n \equal{} 1}^\infty$ be an orthonormal basis for $ H$. Inductively, construct an increasing sequence $ (L_n)\subset A$ such that $ e_n\in L_n$ for all $ n$. For each $ n$ choose a unit vector $ x_n\in L_{n \plus{} 1}\ominus L_n$ and let $ x \equal{} \sum_{k \equal{} 1}^\infty2^{ \minus{} k}x_k$.

By supposition, $ x\in L$ for some $ L\in A$. If $ L\subset L_n$ then $ \sum_{k \equal{} n}^\infty2^{ \minus{} k}x_k\in L_n\cap L_n^{\perp} \equal{} \{0\}$, which is not possible. Therefore $ L_n\subset L$. But if that holds for all $ n$ then $ L\supset\bigcup L_n$, which is dense in H (because it contains all the vectors in an orthonormal basis). Since $ L$ is closed, that means that $ L \equal{} H$, contradicting the fact that $ L$ is a proper subspace.

Or: because the space is separable, we can find a countable cofinal set of $ L$'s, so we may as well assume the family of $ L$'s was countable to begin with. Now simply invoke the Baire Category Theorem.


[quote="durt"]Both $x$ amd $y$ are positive numbers less than $2$. Every positive number less than $2$ is equally likely to be the value of $x$; and every positive number less than $2$ is equally likely to be the value of $y$. What is the probability that $x$ and $y$ differ by less than $1$?[/quote]
[hide]
I got $\frac{3}{4}$
Draw one of those little squares for probability and the area that they differ by more than 1 has area 1. Since the area is 4 the probability is 3/4[/hide]


Find $\sum_{k=0}^n\arctan-\frac{2k+1}{k^2+k-1}$.

Masoud Zargar


I am new to LaTeX and this forum. I don't know how to get images or LaTeX into the forum. I have seen TeXer and I know how to get what I want, but I can't put it in my posts in the forum. Can anybody help me with putting LaTeX or images into the forum?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 1 & 2[/url].


Two standard 6-sided dice are tossed. What is the probability
that the sum of the numbers shown on the dice is a prime
number? Express your answer as a common fraction.

To solve this, let's first find the number of ways to get a prime sum. Let's consider 6 cases.

[b]Number on first die is 1[/b] On the second die, we can get a prime sum by rolling a 1, making a sum of 2, a 2, making a sum of 3, a 4, making a sum of 5, and a 6, making a sum of 7. So when the number on the first die is 1, we have $ 4$ ways to get a prime um.

[b]Number on first die is 2[/b] We can get a prime sum by rolling a 1, making a sum of 3, a 3, making a sum of 5, and a 5, making a sum of 7. We get $ 3$ ways.

[b]Number on first die is 3[/b] We get a prime sum by rolling a 2, getting a 5, and a 4, getting a 7. We have $ 2$ ways.

[b]Number on first die is 4[/b] To get a prime sum, we can roll a 1, making a sum of 5, and a 3, making a sum of 7. We get $ 2$ ways again.

[b]Number on first die is 5[/b] In order to get a prime sum, we roll a 2, getting a 7, and a 6, getting an 11. We have $ 2$ ways once more.

[b]Number on first die is 6[/b] To get a prime sum, we can roll a 1, getting a 7, and a 5, getting 11. So there are $ 2$ ways.

Now we add the number of ways together: $ 4\plus{}3\plus{}2\plus{}2\plus{}2\plus{}2\equal{}15$

We must now find the number of ways possible to roll any two numbers. There are $ 6$ ways to get a number from the first die, and another $ 6$ for the second. We multiply to get $ 36$ ways to get a number. (Notice that since we repeated sums in our cases, counting, for example, 2 and 5, and 5 and 2, as two separate cases, we don't divide by $ 2$ like we normally would in combinations.)

So our probability of rolling a prime sum is $ \frac{15}{36}$, which we can simplify $ \implies \boxed{\frac{5}{12}}$

Hmm. I think it's quite complicated.
[hide="Try this"]The minimum sum is 2 and the maximum is 12. The prime numbers in this range are $ 2, 3, 5, 7 and 11$.

1 way to get a sum of 2 : 1 & 1
2 ways to get a sum of 3 : 2 & 1, 1 & 2
4 ways to get a sum of 5: 1 & 4, 4 & 1, 2 & 3, 3 & 2
6 ways to get a sum of 7: 6 & 1, 1 & 6, 2 & 5, 5 & 2, 3 & 4, 4 & 3
2 ways to get a sum of 11: 5 & 6, 6 & 5
Adding up of the possibilities, we have $ \frac {15}{36}$ or $ \frac {5}{12}$ [/hide]

That's pretty much what I did. I just explained in a LOT of detail, but you could do it in your head fairly quickly.