calculus integration geometry perimeter search

What is a fast way to do this problem?

If x, y, and z are positive integers such that x^2+y^2+z^2=174, what is the greatest possible value of x+y+z?

This one too..

How many distinct isosceles triangles having sides of integral length and perimeter 113 are possible?

92-93 State Sprint Round

Guess that they should all be equal to each other.  This is (very often) the case for symmetric problems like this.  For this problem, it gives us that x:^2: = y:^2: = z:^2: = 58.

Now that you've guessed that, you can in fact prove it.  My particular method for doing this is a little inelegant and not exactly intuitive, but it should do:
we'll show that any other combination of numbers adding up to 3:sqrt:58 will give us a larger total sum of squares.  This will show that if we have the sum of the 3 numbers any larger than 3:sqrt:58, the sums of their squares will be strictly greater than 174.
For laziness, I'm going to let :sqrt:58 = s.
To prove this, call z = a + s and y = b + s and x = -a - b + s, where a and b might be positive, negative, or zero.  Then
x:^2: + y:^2: + z:^2: = (a:^2: + 2as + s:^2:) + (b:^2: + 2bs + s:^2:) + ((a + b):^2: - 2as - 2bs + s:^2:) = a:^2: + b:^2: + (a + b):^2: + 3s:^2: = a:^2: + b:^2: + (a + b):^2: + 174.  This is definitely larger than 174, unless a = b = (a + b) = 0.  And we're done.

As for the second one, do a search for it on this website -- that problem ahs been discussed before here.

JBL: :sqrt:58 isn't a positive integer

Answer below..

[hide]

The easiest way to do this one is probably just consider all possible cases: there won't be too many.

Starting just from x being as big as possible, we get (13,2,1), when x=12 we get no solutions, x=11 we get (11,7,2), x=10 we get (10,7,5), x=9 there are no solutions , and x=8 or lower there are no solutions. So theres only actually 3 solutions overall: the biggest sum of x,y,z is 22.

[/hide]

No faster way on this one. That could be done in just a couple of minutes..

Well, on the first problem, I made a list and got only 3 solutions, (13,2,1), (11,7,2), and (10,7,5), and out of those, the latter gives the greatest sum of 22. I tried cubing x+y+z but that didn't work, and neither did the AM-GM Inequality.

And for the second one, there are 28 triangles.

x - leg
y - base

2x + y = 113
2x > y
2x + y > 2y
113 > 2y
56.5 > y
y is an odd whole number since all the sides need to have integral lengths. Counting the odd numbers between 1 and 55 inclusively, we get 28.

Ah, yes, the wonderful joys of reading . . .
If the problem were different, I would have done it right . . .

For the first question, the upper bound can quickly be estimated by using Cauchy:

[hide](x2 + y2 + z2)(12 + 12 + 12) :ge: (x + y + z)2

174 * 3 :ge: (x + y + z)2

x + y + z :le: 22.[/hide]

Then one simply needs to find a triplet that works.

I didn't get 22, TripleM (Stephen) did.

Which leads me to the somewhat random question:  what does TripleM mean, anyhow?

The intials of his name...wait a minute, if his first name is "Stephen" how could there be three more initials? :?

it was like a nickname: math master "last name" or someat like that. but that's off topic.


algebra
polynomial
induction
algebra unsolved

Let $ S_{k}(n) \equal{} \sum_{r\equal{}0}^{n} r^{k}$, where $ k$ is a positive integer, so that

$ S_{1}(n)\equal{}\frac{1}{2}n(n\plus{}1)$ and $ S_{2}(n)\equal{} \frac{1}{6}n(n\plus{}1)(2n\plus{}1)$.

(i) By considering $ \sum_{r\equal{}0}^{n} [(r\plus{}1)^{k}\minus{}r^{k}]$, show that

$ kS_{k\minus{}1}(n)$

$ \equal{}(n\plus{}1)^{k} \minus{}(n\plus{}1)\minus{}\binom{k}{2}S_{k\minus{}2}(n) \minus{}\binom{k}{3}S_{k\minus{}3}(n) \minus{} ... \minus{}\binom{k}{k\minus{}1}S_{1}(n).  (*)$

Obtain simplified expressions for $ S_{3}(n)$ and $ S_{4}(n).$

(ii) Explain, using (*), why $ S_{k}(n)$ is a polynomial of degree $ k\plus{}1$ in $ n$.  Show that in this polynomial the constant term is zero and the sum of the coefficients is $ 1$.

(ii) induction. for example, to show that the sum of the coefficients is 1, we have, by induction, $ kS_{k \minus{} 1}(1) \equal{} 2^k \minus{} 2 \minus{} \left[\binom{k}{2} \plus{} \binom{k}{3} \plus{} \cdots \plus{} \binom{k}{k \minus{} 1}\right] \equal{} 2^k \minus{} 2 \minus{} (2^k \minus{} 2 \minus{} k) \equal{} k$, so $ S_{k \minus{} 1}(1) \equal{} 1$.


Could someone prove heroes formula $ \sqrt{(s)(s \minus{} a)(s \minus{} b)(s \minus{} c)}$  where $ a,b,c$ are the sides of a triangle and $ s$ is half of the perimiter.

I've tried but just cant get it.  Thanks

$ S\equal{}\frac{1}{2}bc\sin A\equal{}\frac{1}{2}bc\sqrt{1\minus{}\cos ^ 2 A}\ (\because 0<A<\pi)$.

$ \equal{}\frac{1}{4}\sqrt{(2bc)^2\minus{}(b^2\plus{}c^2\minus{}a^2)^2}.$


inequalities
trigonometry
geometry
logarithms
function
geometric transformation
reflection

Your inequality is equivalent to
    prod(1-cosA)>= prod cosA
   <=> 8[prod(sinA/2)]^2>=prod cosA
   <=>  \sum tgA >= \sum cotg(A/2).
And this can be proved by separating it into three similar inequalities
tgA + tgB >= 2cotg(C/2).

[quote="darij grinberg"]For every acute-angled triangle with angles A, B, C, we have the inequalities

  $(\sec{A} - 1)(\sec{B} - 1)(\sec{C} - 1) \geq 1$.
[/quote]

A more algebraic flavor proof:

Let $x=\tan{A},y=\tan{B},z=\tan{C}$ then $x,y,z>0$ since $\Delta{ABC}$ is acute. It is well-known that $x+y+z=xyz$. The inequality is equivalent to:

(*) $(\sqrt{1+x^2}-1)(\sqrt{1+y^2}-1)(\sqrt{1+z^2}-1) \geq 1$.

$yz(1+x^2) = yz + x(xyz) = yz + x(x+y+z) = (x+y)(x+z)\geq (x+\sqrt{yz})^2$ by Cauchy-Schwarz. Hence

(1) $\displaystyle \sqrt{1+x^2}-1 \geq \frac{x}{\sqrt{yz}}$

Similarly, we have

(2) $\displaystyle \sqrt{1+y^2}-1 \geq \frac{y}{\sqrt{zx}}$

(3) $\displaystyle \sqrt{1+z^2}-1 \geq \frac{z}{\sqrt{xy}}$

Multiplying (1), (2) and (3) together yields (*).

Q.E.D.

(*) itself could make a nice inequality problem:

[quote]
Let $x,y,z>0$ such that $x+y+z=xyz$. Prove that

$(\sqrt{1+x^2}-1)(\sqrt{1+y^2}-1)(\sqrt{1+z^2}-1) \geq 1$.[/quote]

I think I have forgotten to say something:

[quote="darij grinberg"]For every acute-angled triangle with angles A, B, C, we have the inequalities

$\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)\geq 1$[/quote]

Actually, this inequality has a nice geometric equivalent:

Let S be the area of an acute triangle ABC. Now consider the incircle and the A-excircle of triangle ABC. These two circles have the lines CA and AB as their two external common tangents. The line BC is one of their two internal common tangents. Let the other internal common tangent be denoted by $t_a$. Similarly define the lines $t_b$ and $t_c$. Let T be the area of the triangle formed by the three lines $t_a$, $t_b$ and $t_c$. Then, $T \geq S$.

  Darij

[quote](sec A - 1)(sec B - 1)(sec C - 1) \geq 1; [/quote]

Or, you can change the angles with $\pi/2-A,\pi/2-B,\pi/2-C$, and get to prove that $\prod(\frac 1{\sin A}-1)\geq 1$, with $A+B+C=\pi/2$. Now let's apply $\ln$ to this and get to prove $\sum \ln(\frac 1{\sin A}-1)\geq 0$. But $\frac 1{sinx}-1$ is a decreasing function on $[0,\pi/2]$, ln is a convex function so we get that $\ln(\frac 1{\sin x}-1)$ is a convex function on $[0,\pi/2]$. Now let's apply Jensen for this function and so we get a minimizaton for that sum of the form $3ln\frac 13 (\sum \frac 1{\sin A}-3)$. But from Cauchy and from $\sum \sin A\leq \frac 32$ (again from Jensen) we easily obtain the desired result.

[quote="ikap"]But $\frac 1{sinx}-1$ is a decreasing function on $[0,\pi/2]$, ln is a convex function so we get that $\ln(\frac 1{\sin x}-1)$ is a convex function on $[0,\pi/2]$.[/quote]

Are you sure it is?

PS. I'll try your trigonometry 1 problem; I am really not the best inequality solver, hence I can't ensure to post a solution.

  Darij

Sorry, I wanted to write $\ln$ is a concave function. However it doesn't suffice to make the other one convex. But by calculating f'', and proving it's positive we are done. It s quite simple to show that.

[quote="darij grinberg"][quote="ikap"]But by calculating f'', and proving it's positive we are done. It s quite simple to show that.[/quote]

Have you done it?

  Darij[/quote]
I think it has done.for $\sum \sin A \leq \frac{3}{2}$ then $\sum \frac 1{\sin A} \geq 6$
we are done :)
here is my proof.let $a,b,c$be the three side of the we have to prove that
$(a^2+b^2-c^2)(c^2+a^2-b^2)(b^2+c^2-a^2) \leq (a^2-(b-c)^2)(a^2-(b-c)^2)(b^2-(c-a)^2)(c^2-(a-b)^2)$
for obtuse triangle,it is obvious.
if not,we have $(a^2+b^2-c^2)(c^2+a^2-b^2) \leq (a^2-(b-c)^2)^2$ :)

[quote="ikap"][quote="zhaobin"] for obtuse triangle,it is obvious.
[/quote]

Yes, but in the initial inequality it doesn't work because one of the factors is negative.[/quote]
yes,thank you.and I have done than the problem.hehe

by the way, if I am here, I want to say that my proof doesn't work, since I have remade the computations for $f\"$ and the function is not convex on the entire interval  :D

Ikap, you needn't perform any calculations in order to see that the function $f\left(x\right)=\ln\left(\frac{1}{\sin x}-1\right)$ is not convex on the whole interval $\left(0;\;\frac{\pi}{2}\right)$: have you ever seen a function which is convex on an open interval and tends to $+\infty$ at one endpoint of the interval and to $-\infty$ at the other one? :)

Something different: I have been asked to explain why the geometric equivalent I gave in post #8:

[quote="darij grinberg"]Let S be the area of an acute triangle ABC. Now consider the incircle and the A-excircle of triangle ABC. These two circles have the lines CA and AB as their two external common tangents. The line BC is one of their two internal common tangents. Let the other internal common tangent be denoted by $t_a$. Similarly define the lines $t_b$ and $t_c$. Let T be the area of the triangle formed by the three lines $t_a$, $t_b$ and $t_c$. Then, $T \geq S$.[/quote]

is really equivalent to the inequality $\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)\geq 1$. Well, this is because of the formula $\frac{T}{S}=\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)$. So it remains to prove this formula.

In fact, let $X=t_b\cap t_c$, $Y=t_c\cap t_a$ and $Z=t_a\cap t_b$. Then, the triangle XYZ is the triangle formed by the three lines $t_a$, $t_b$ and $t_c$, and thus, the area of this triangle is T.

It is well-known that the two internal common tangents of two given circles are symmetric to each other with respect to the line joining the centers of the two circles. Consider the incircle and the A-excircle of triangle ABC; their two internal common tangents are the lines BC and $t_a$. Thus, the lines BC and $t_a$ are symmetric to each other with respect to the line joining the centers of the incircle and the A-excircle of triangle ABC. But the line joining the centers of the incircle and the A-excircle of triangle ABC is simply the angle bisector of the angle CAB; hence, the lines BC and $t_a$ are symmetric to each other with respect to the angle bisector of the angle CAB. In other words, the line $t_a$ is the reflection of the line BC in the angle bisector of the angle CAB.

On the other hand, let A', B', C' be the feet of the altitudes of triangle ABC; in other words, the triangle A'B'C' is the orthic triangle of the triangle ABC. It is well-known that the line B'C' is antiparallel to the side BC of triangle ABC. Now, an antiparallel to a side of a triangle is the reflection of a parallel to this side in the angle bisector of the opposite angle; hence, the line B'C', being antiparallel to the side BC of triangle ABC, is the reflection of a line g parallel to this side BC in the angle bisector of the angle CAB.

So the lines B'C' and $t_a$ are the reflections of the lines g and BC in the angle bisector of the angle CAB. But the lines g and BC are parallel to each other; thus, their reflections B'C' and $t_a$ must also be parallel to each other. Since the line $t_a$ coincides with the line YZ, we thus have B'C' || YZ. Similarly, C'A' || ZX and A'B' || XY. Hence, the triangles A'B'C' and XYZ are homothetic. Thus, the angles of triangle XYZ are equal to the respective angles of triangle A'B'C'.

Now, it is well-known that, since the triangle ABC is acute, the angles of the orthic triangle A'B'C' of triangle ABC are 180° - 2A, 180° - 2B, 180° - 2C. (This is Lemma 1 in post #2 of http://www.mathlinks.ro/Forum/viewtopic.php?t=23696 .) Thus, the angles of triangle XYZ, being equal to the respective angles of triangle A'B'C', must also be 180° - 2A, 180° - 2B, 180° - 2C.

Note that the sidelines YZ, ZX, XY of triangle XYZ are the lines $t_a$, $t_b$, $t_c$ and thus touch the incircle of triangle ABC; hence, the incircle of triangle ABC is simultaneously the incircle of triangle XYZ, and consequently, the inradius of triangle XYZ is the inradius of triangle ABC, i. e. the inradius of triangle XYZ is r.

Now, we will prove a lemma:

[b]Lemma 1.[/b] If A, B, C are the angles, r is the inradius and S is the area of a triangle ABC, then

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$.

[i]Proof of Lemma 1.[/i] If a, b, c are the sidelengths and $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, then, it is well-known that $\tan\frac{A}{2}=\frac{r}{s-a}$. Thus,

$\cot\frac{A}{2}=\frac{1}{\tan\frac{A}{2}}=\frac{1}{\left(\frac{r}{s-a}\right)}=\frac{s-a}{r}$.

Similarly, $\cot\frac{B}{2}=\frac{s-b}{r}$ and $\cot\frac{C}{2}=\frac{s-c}{r}$. Thus,

$r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}=r^2\cdot\frac{s-a}{r}\cdot\frac{s-b}{r}\cdot\frac{s-c}{r}$
$=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{r}=\frac{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{sr}$.

Now, by the Heron formula, $s\left(s-a\right)\left(s-b\right)\left(s-c\right)=S^2$, while by another well-known formula for the area of a triangle, sr = S. Thus,

$r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}=\frac{S^2}{S}=S$,

and Lemma 1 is proven.

Applying Lemma 1 to the triangle ABC with the angles A, B, C, the inradius r and the area S, we get

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$.

On the other hand, applying Lemma 1 to the triangle XYZ with the angles 180° - 2A, 180° - 2B, 180° - 2C, the inradius r and the area T, we get

$T=r^2\cdot\cot\frac{180^{\circ}-2A}{2}\cdot\cot\frac{180^{\circ}-2B}{2}\cdot\cot\frac{180^{\circ}-2C}{2}$.

Thus,

$\frac{T}{S}=\frac{r^2\cdot\cot\frac{180^{\circ}-2A}{2}\cdot\cot\frac{180^{\circ}-2B}{2}\cdot\cot\frac{180^{\circ}-2C}{2}}{r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}}$
$=\frac{\cot\frac{180^{\circ}-2A}{2}}{\cot\frac{A}{2}}\cdot\frac{\cot\frac{180^{\circ}-2B}{2}}{\cot\frac{B}{2}}\cdot\frac{\cot\frac{180^{\circ}-2C}{2}}{\cot\frac{C}{2}}$.

But

$\frac{\cot\frac{180^{\circ}-2A}{2}}{\cot\frac{A}{2}}=\frac{\cot\left(90^{\circ}-A\right)}{\cot\frac{A}{2}}=\frac{\tan A}{\cot\frac{A}{2}}=\tan A\cdot\frac{1}{\cot\frac{A}{2}}=\tan A\cdot\tan\frac{A}{2}$
$=\frac{\sin A}{\cos A}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos A}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\frac{2\sin^2\frac{A}{2}}{\cos A}=\frac{1-\cos A}{\cos A}=\frac{1}{\cos A}-1$
$=\sec A-1$,

and similarly, $\frac{\cot\frac{180^{\circ}-2B}{2}}{\cot\frac{B}{2}}=\sec B-1$ and $\frac{\cot\frac{180^{\circ}-2C}{2}}{\cot\frac{C}{2}}=\sec C-1$, so that

$\frac{T}{S}=\frac{\cot\frac{180^{\circ}-2A}{2}}{\cot\frac{A}{2}}\cdot\frac{\cot\frac{180^{\circ}-2B}{2}}{\cot\frac{B}{2}}\cdot\frac{\cot\frac{180^{\circ}-2C}{2}}{\cot\frac{C}{2}}=\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)$,

and our formula is proven.

  Darij

Use $tgA+tgB+tgC\ge cotg{\frac{A}{2}}+cotg{\frac{B}{2}}+cotg{\frac{C}{2}}$ what is because $tgx+tgy\ge 2cotg\frac{x+y}{2}$ for all $x,y\in [0;\frac{\pi}{2}]$

[quote=darij grinberg]For every acute-angled triangle with angles $A$, $B$, $C$, prove the inequality

$\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)\geq 1$.

[i]Equivalent formulation:[/i] For every acute-angled triangle with angles $A$, $B$, $C$, prove the inequality

$\left(1-\cos A\right)\left(1-\cos B\right)\left(1-\cos C\right)\geq\cos A\cos B\cos C$.

[hide="Remark"][i]Remark.[/i] The equivalence between these two statements of the problem can be seen as follows: For any angle $x$, we have $\sec x-1=\frac{1}{\cos x}-1=\frac{1-\cos x}{\cos x}$. Thus,

$\left(\sec A-1\right)\left(\sec B-1\right)\left(\sec C-1\right)\geq 1$

is equivalent to

$\frac{1-\cos A}{\cos A}\cdot\frac{1-\cos B}{\cos B}\cdot\frac{1-\cos C}{\cos C}\geq 1$,

and since $\cos A$, $\cos B$ and $\cos C$ are all positive (because $A$, $B$ and $C$ are acute angles), this is equivalent to

$\left(1-\cos A\right)\left(1-\cos B\right)\left(1-\cos C\right)\geq\cos A\cos B\cos C$.[/hide]

  Darij[/quote]
$cosAcos(B-C)\le cosA\Rightarrow cosBcosC(1+cosA)\le sinBsinC(1-cosA)$

$\Rightarrow \frac{ sin^2AcosBcosC}{sinBsinC}\le(1-cosA)^2\Rightarrow \prod \frac{ sin^2AcosBcosC}{sinBsinC}\le \prod (1-cosA)^2$

$\Rightarrow (1-cosA)(1-cosB)(1-cosC) \ge cosA cosB cos C$
[url=http://www.artofproblemsolving.com/community/c6h518862p2922063]here[/url]


geometry
AMC
AMC 12
AMC 12 B
3D geometry
AIME
ratio

Since we already have a Number Theory one, I decided to create this.
If the question needs a diagram, try to either create a java file or provide a link to the picture.

To start (2007 AMC12B #14):
Point P is inside equilateral triangle ABC. Points Q, R, and S are the feet of the perpendiculars from P to AB, BC, and CA, respectively. Given that PQ=1, PR=2, and PS=3, what is AB?

(A)4 (B)3sqrt3 (C)6 (D)4sqrt3 (E)9

i have to stop writing solutions that are slow and long.
but anyways:
make AB=x. then the area of the triangle is equal to (sqrt(3)x^2)/4. split the triangle into three parts, so x/2+2x/2+3x/2=(sqrt(3)x^2)/4. 12x=sqrt(3)*x^2, and x=4*sqrt(3). Answer choice D.
i need to learn latex.
don't really have a new problem, so...

Here's a problem:

In triangle ABC let $ AC\equal{}x\plus{}1$, $ AB\equal{}x\minus{}1$, and $ BC\equal{}x$. There is a point $ D$ on $ BC$ such that $ BD\equal{}a$ and $ DC\equal{}b$ and $ AD$ is perpendicular to $ BC$. Show that $ b\minus{}a\equal{}4$.

- Source: A book called Geometría(Cuadernos de Olimpiadas de Matemáticas)

$ a\plus{}b\equal{}BC\equal{}x$

also

$ AD^2\equal{}(x\minus{}1)^2\minus{}a^2\equal{}(x\plus{}1)^2\minus{}b^2$

$ b^2\minus{}a^2\equal{}(x\plus{}1)^2\minus{}(x\minus{}1)^2\equal{}4x$

$ (b\minus{}a)(b\plus{}a)\equal{}4x$

$ b\minus{}a\equal{}\frac{4x}{b\plus{}a}\equal{}\frac{4x}{x}\equal{}4$

There is a quadrangle $ ABCD$, such that $ AD\equal{}CD$, $ \angle{BCD}\equal{}60^\circ$, $ \angle{BAC}\equal{}30^\circ$. Prove that $ BC\equal{}CD$.  :)

Well since no one seems to know the answer, here's a new easy problem:

NP:
Find the distance between the two lines given by the equations: 2x-5y=13 and 2x-5y=1

While I'm not sure because I haven't tried it, It's customary to put a solution up when you answer a problem.

Point on first line: $ (4,\minus{}1)$
Distance formula between point and line: $ \frac{Ax\plus{}By\plus{}C}{\sqrt{A^{2}\plus{}B^{2}}}$
$ \frac{8\plus{}5\minus{}1}{\sqrt{29}}\equal{}\frac{12}{\sqrt{29}}\equal{}\frac{12\sqrt{29}}{29}$

Next Question
Triangle $ ABC$ has $ <C\equal{}\frac{\pi}{3}$ and $ BC\equal{}4$ . Point $ D$ is the midpoint of $ BC$. What is the largest possible value of $ tan<BAD$ ?

$ \textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} \minus{} 3} \qquad \textbf{(E)}\ 1$

Since no one seems to know this, lets move on.

The 3 sides of a triangle are 5,6, and x. Inscribed inside of the triangle is a circle with radius 3. What is x?

There's probably a better way to do this; my solution got pretty ugly.

[hide="(Ugly) solution"]Define $ s\equal{}\frac{11\plus{}x}{2}$. Equating $ 3s$ and $ \sqrt{s(s\minus{}5)(s\minus{}6)(s\minus{}x)}$, we find (after a great deal of simplification) that $ x^4 \minus{} 122 x^2 \plus{} 24x \plus{} 385\equal{}0$.

From the rational root theorem, we find that $ \boxed{x\equal{}11}$ is a root.[/hide]

[b]Problem.[/b] Find the area of the region enclosed by $ |x\minus{}60|\plus{}|y|\equal{}|x/4|$.

There is a formula called heron's formula to get the area of a triangle with the side lengths. So we do

sqrt([11+x]/2)([x+1]/2)([x-1]/2)([11-x]/2)

Also halfperimeter*radius=area so now we get

[33+3x]/2=([11+x]/2)([x+1]/2)([x-1]/2)([11-x]/2)

And we do this thing called difference of squares on the otehr side to get

[33+3x]/2=sqrt(121-x^2)(x^2-1)/4

Then we multiply to get

66+6x=\sqrt{(121-x^2)(x^2-1)}

Then we square and get

36x^2+792x+4356=-x^4+122x^2-121
x^4-86x^2+4477=0
(x^2-121)(x^2+37)=0
x=11.

Is this right????????

Let's call the edge length WLOG 8. The surface area is obviously 96sqrt3.
So if you connect the centers, by visualization, you can see you make a cube. I'll find the side diagonal and use that to make the cube.So imagine making an isosceles triangle in the middle of the tetrahedron such that the even sides are equal to the heights of the faces and the base is 8. We can now use similar triangles. After splitting that triangle in two, we see that the ratio of the height to the length from the face to the center is 4sqrt2:4 or sqrt2. Since the center of a triangle is 1/3 up, the length from the face to the center is 8/3. Or times two is 16/3. Since that is the diagonal on the cube, the side length is 8sqrt2/3. Therefore, the surface area is 256/3. So this over the previous gives  8sqrt3/27. Unfortunately, this isn't possible since 8sqrt3 and 27 aren't both integer answers.
Could someone point out the flaw?

(also, considering the drop in interest and the forum this is in, let's keep it at an amc 12 level or lower.

Ihatepie, I'm getting the same answer as you. I'll $ \text{\LaTeX}$-ify the solution to make mistakes easier to spot.

First, let the side of the octahedron be $ n$. We see that the new solid is a cube, so we draw a line along a side diagonal of that cube. Since the ratio of the diagonal to $ n$ is $ \frac{2}{3}$, the diagonal is $ \frac{2n}{3}$. Dividing by $ \sqrt{2}$, we have
\[ \frac{2n}{3\sqrt{2}} = \frac{n\sqrt{2}}{3}.
\]
The surface area of the cube is then clearly $ 6\left( \frac{n\sqrt{2}}{3} \right)^2 = \frac{12n^2}{9} = \frac{4n^2}{3}$.

The surface area of the octahedron is $ 8\frac{n^2\sqrt{3}}{4} = 2n^2\sqrt{3}$.

The ratio is $ \frac{\frac{4n^2}{3}}{2n^2\sqrt{3}} = \boxed{\frac{2\sqrt{3}}{3}}$

I still can't find the mistake; maybe someone else can?

There's no error.

However, there might be an error in the statement. The ratio of the VOLUMES is rational.

[b]Solution[/b]
The volume of the cube is $ \frac{n^32\sqrt{2}}{27}$. The volume of the octahedron is $ \frac{n^3\sqrt{2}}{3}$. The ratio is $ \frac{2}{9}$, so the answer is $ 81\minus{}4 \equal{} \boxed{77}$.

[b]New Problem[/b]
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $ m$ times the area of the square. What is the ratio of the area of the other small right triangle to the area of the square? (2000 AMC 12 #21)

[hide="Solution"]
Let the triangle has an area $ m$ times that of the square of side length $ x$ have a height of $ z$ and let the smaller triangle have a base of $ y$. According to the given information, we have

$ \frac {z\cdot x}{2} \equal{} mx^2 \\
z \equal{} 2mx.$

From the two similar right triangles, we obtain $ \frac {z}{x} \equal{} \frac {x}{y} \implies z \equal{} \frac {x^2}{y}$. Substituting that into the previous equation, we have

$ \frac {x^2}{y} \equal{} 2mx \implies 2m \equal{} \frac {x}{y}.$

Our desired ratio is

$ \frac {\frac {1}{2}\cdot y\cdot x}{x^2} \equal{} \frac {1}{2} \cdot \frac {y}{x} \equal{} \boxed{\frac {1}{4m}}.$
[/hide]

Next Problem:

Triangle $ ABC$ and point $ P$ in the same plane are given. Point $ P$ is equidistant from $ A$ and $ B$, $ \angle APB \equal{} 2\angle ACB$, and $ \overline{AC}$ intersects $ \overline{BP}$ at point $ D$. If $ PB \equal{} 3$ and $ PD \equal{} 2$, then evaluate $ AD\cdot CD$.

[asy]defaultpen(linewidth(.8pt));
dotfactor=4;

pair A = origin;
pair B = (2,0);
pair C = (3,1);
pair P = (1,2.25);
pair D = intersectionpoint(P--B,C--A);

dot(A);dot(B);dot(C);dot(P);dot(D);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N);

draw(A--B--P--cycle);
draw(A--C--B--cycle);[/asy]

I'm probably wrong with this, but wouldn't it be possible to circumscribe a circle around it?
P would be the center, A, B, and C would lie on the edge. So AD*CD=5.

If I'm not right (which I bet I'm not) could you provide a hint?

I'm don't think $ ABCP$ is cyclic, if that's what you're saying. It would be nice though, one step power of a point.

$ ABCP$ isn't cyclic.$ <APB\equal{}2<ACB$, so it can't be cyclic.

Ihatepie is certainly correct.  I did it like this (I think it's the same method he did it?):
Drawing the circumcircle of $ \triangle{ABC}$, we see that $ P$ is the center of this circle.  This is seen from $ PA \equal{} PB$ (radii of this circle), and arc $ AB \equal{} 2a \equal{} < APB \equal{} \frac { < ACB}{2}$.  We extend $ PB$ to hit the circumference of circle $ P$.  From this, the answer is obviously $ (1)(5) \equal{} \boxed{5}$.

New Problem: $ ABCD$ is a quadrilateral with right angles at $ A$ and $ C$.  Points $ E$ and $ F$ are on $ AC$, and $ DE$ and $ BF$ are perpendicular to $ AC$.  In $ AE \equal{} 3$, $ DE \equal{} 5$, and $ CE \equal{} 7$, then find $ BF$. (AHSME $ 1990$).
To make this problem more interesting, I will add that there is a VERY pretty solution to this problem with minimal algebra.

[hide]
If we label $ \angle BAC$ as $ \theta$, we notice that $ \angle CAD$ is $ 90\minus{}\theta$ and $ \angle ADE$ is also $ \theta.$ Thus, $ \triangle BAE \sim \triangle ADF.$ Similarly, $ \triangle CBE \sim \triangle DCF.$ If we label $ BE$ as $ x$ and $ EF$ as $ m$ and use the first similarity, we get \[ \frac{x}{3} \equal{} \frac{3\plus{}m}{5}.\] Now, using the second similarity, we get \[ \frac{x}{7} \equal{} \frac {7\minus{}m}{5}.\] If we isolate $ m$ and set them equal, then we get $ \frac {5x}{3} \minus{} 3 \equal{} 7 \minus{} \frac {5x}{7} \Rightarrow 5x(\frac{1}{3}\plus{}\frac{1}{7}) \equal{} 10 \Rightarrow x \equal{} \frac{21}{5} \equal{} \fbox{4.2.}$
Seems like a fairly textbook solution to me, so i doubt its your "VERY pretty solution" but it works :D .
[/hide]

1999 AIME number 4:
The two squares shown share the same center $ O$ and have sides of length 1. The length of $ AB$ is $ 43/99$ and the area of octagon $ ABCDEFGH$ is $ m/n$ where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

[img]http://www.artofproblemsolving.com/Wiki/images/8/8e/AIME_1999_Problem_4.png[/img]

[hide="solution"] $ \triangle AOB$ has a base of $ \frac {43}{99}$ and a height of $ \frac {1}{2}$ There are $ 8$ of these congruent triangles, so the total area is $ 8\left(\frac {1}{2}\right)\left(\frac {1}{2}\right)\left(\frac {43}{99}\right) \equal{} \frac {86}{99}$. The sum of the numerator and denominator is thus $ \boxed{185}$.[/hide]

1991 AIME #2
Rectangle $ ABCD$ has sides  $ \overline{AB}$ of length $ 4$ and $ \overline{CB}$ of length $ 3$. Divide $ \overline{AB}$ into $ 168$ congruent segments with points $ A \equal{} P_0, P_1, \ldots, P_{168} \equal{} B$, and divide $ \overline{CB}$ into $ 168$ congruent segments with points $ C \equal{} Q_0, Q_1, \ldots, Q_{168} \equal{} B$. For $ 1\le k\le167$, draw the segments $ \overline{P_kQ_k}$. Repeat this construction on the sides $ \overline{AD}$ and $ \overline{CD}$, and then draw the diagonal $ \overline{AC}$. Find the sum of the lengths of the $ 335$ parallel segments drawn.

[hide]
The length of each $ P_kQ_k$ can be thought of as a diagonal for another rectangle. The length and width of each of these rectangles is $ \frac{3k}{168}$ by $ \frac{4k}{168}$, where $ k$ ranges from 0 to 168. By the pythagorean theorem, the length of each of these diagonals is $ k\sqrt{3^2 \plus{} 4^2}/168$, which makes the sum of $ P_kQ_K$ equal to  $ (1\plus{}2\plus{}3\plus{} \cdots \plus{}168)(5)/168$ = $ (168*169)(5)/(2)168$. Since this is also repeated for $ AD$ and $ CD$, we can multiply that by two, but we must subtract 5 because AC would have been counted twice. Therefore, the sum of all the segments is $ 169*5\minus{}5 \equal{} \fbox{840}.$
[/hide]

AIME 1994 number 2:
A circle with diameter $ PQ$ of length 10 is internally tangent at $ P$ to a circle of radius 20. Square $ ABCD$ is constructed with $ A$ and $ B$ on the larger circle, $ CD$ tangent at $ Q$ to the smaller circle, and the smaller circle outside $ ABCD$. The length of $ AB$ can be written in the form $ m\plus{}\sqrt{n}$, where $ m$ and $ n$ are integers. Find $ m\plus{}n.$

[hide="Totally beast solution"]
(sarcasm, by the way)

Make the center of the large circle the origin. The line tangent at $ Q$ is given by $ x \equal{} \minus{} 10$.

The large circle is defined by $ y^2 \plus{} x^2 \equal{} 400$.

Let $ A \equal{} (x,\sqrt {100 \minus{} x^2})$. We can find $ B \equal{} (x, \minus{} \sqrt {400 \minus{} x^2})$ as a reflection of $ A$ over the $ x$-axis. Therefore, the distance from $ A$ to $ B$ is $ 2\sqrt {400 \minus{} x^2}$.

If we take $ y \equal{} \sqrt {400 \minus{} x^2}$ we find that it intersects the tangent at $ Q$ at a point $ ( \minus{} 10,\sqrt {400 \minus{} x^2})$. The distance from this point to $ A$ is simply $ x \plus{} 10$.

Now we just have to solve $ x \plus{} 10 \equal{} 2\sqrt {400 \minus{} x^2}$

$ x^2 \plus{} 20x \plus{} 100 \equal{} 1600 \minus{} 4x^2$

$ 5x^2 \plus{} 20x \minus{} 1500 \equal{} 0$

$ x^2 \plus{} 4x \minus{} 300 \equal{} 0$

$ x \equal{} \minus{} 2 \plus{} 4\sqrt {19}$.

Plugging in gives a distance of $ 8 \plus{} 4\sqrt {19} \equal{} 8 \plus{} \sqrt {304}$, giving an answer of $ \boxed{312}$.
[/hide]

[b]Problem.[/b] (AMC 12 2002 #25) Let $ f(x) \equal{} x^2 \plus{} 6x \plus{} 1$. Let $ R$ denote the set of points $ (x,y)$ such that $ f(x) \plus{} f(y)\leq 0$ and $ f(x) \minus{} f(y)\leq 0$. Find the integer closest to the area of $ R$.

[quote="red_pnguin"]...for those of you who don't enjoy coordinate bashing as much as grn_trtle [/quote]
 :o

[hide="alternate solution to aime 1994 number 2"]

...for those of you who don't enjoy coordinate bashing as much as grn_trtle

Since the circle with diameter $ PQ$ is internally tangent to the larger circle, we know they form a homothety, thus making $ P$, $ Q$, and $ O$, the center of the larger circle, colinear. If we label the point that $ PQ$ intersections $ AB$ as $ M$ and the point $ PQ$ intersects circle $ O$ as $ N$ and label the distance $ MN$ as $ m$, then $ QM \equal{} 30\minus{}m$. With power of a point on $ M$, and noticing that $ QN$ bisects $ AB$, \[ \frac{1}{4}AB^2 \equal{} (40\minus{}m)(m).\] Since $ Q$ is the point of tangency of $ CD$ and $ P, Q, M, N$ are collinear, $ QM \perp CD$ so $ QM \parallel AB$. This means $ QM \equal{} AD \equal{} AB$, so $ AB \equal{} 30\minus{}m$. If we substitute that into the other equation, we get \[ (30\minus{}m)^2 \equal{} 4m(40\minus{}m).\] Rearranging, we arrive with \[ m^2 \minus{} 44m \plus{} 180 \equal{} 0.\] Solving for $ m$, we arrive at \[ m \equal{} 22\minus{}\sqrt{304}.\] Since $ AB \equal{} 30\minus{}m$, $ AB \equal{} 30\minus{}22\plus{}\sqrt{304} \equal{} 8\plus{}\sqrt{304}.$ $ m\plus{}n$ would be $ 8\plus{}304 \equal{} \fbox{312.}$

[/hide]

[hide="Yet another solution"]If the side of the square is $ b$, then

$ \sqrt{R^2\minus{}{b^2\over 4}}\plus{}R\minus{}2r\equal{}b$

which is seen if we connect the center of the big circle to a square vertex belonging to the circumference of the big circle.

From the above equation $ b$ is found directly.[/hide]

Thats the most beautiful solution I've ever seen...
but i still don't understand how you got the formula.
Could you (or anyone else) explain how that formula is derived?
Thanks

edit: ah i see it now. Perhaps I was alone there, but for those of you who are wondering, QO is $ R\minus{}2r$ and O to the intersection of QO and AB is $ \sqrt{R^2\minus{}\frac{b^2}{4}}$ by pythagoras.


I found this question when I was browsing through books today at Borders...

X+X^-1 = (1+ :rt5: )/2

Solve for

X^(2000) + X^(-2000)

Hint
[hide]cos(:pi:/5)=(1+ :rt5: )/2[/hide]

You may use the complex number or Moivre 's theorem.

[quote="Myth"][quote="jace"]Hint
cos(:pi:/5)=(1+ :rt5: )/2[/quote]
???????[/quote]

I think saying something like
"Don't you mean $2\cos \frac{\pi}{5}$" is a bit more helpful than "???????" :)

Ok, I will ask TripleM everytime I want to write comment.  :mad:

I will answer to this problem without using Complex number.

By $\left(x+\frac{1}{x}-\frac{1}{2}\right)^2=\left(\frac{\sqrt{5}}{2}\right)^2$, Clearing the denominator of both sides,we have $x^4-x^3+x^2-x+1=0$,since $x=-1$ is not root of this equation,
therefore we obtain $\frac{1-(-x)^5}{1-(-x)}=0$,yielding $x^5=-1$,then $x^{2000}=(x^5)^{400}=1$.The desired answer is $2$.


Pardon my ignorance.

USAMOs older than '02 would also be nice.

Thanks in advance.

They should be in the contests after vv fixes bug.

You can either go to the AMC website (search "AMC Math" on Google, I forgot the exact address, it should be the first one that shows up) and buy the tests all the way to 1982 I think. Or you can go to the contest section of AoPS, they should have almost all the problems.


How can I make an entire paragraph indented?  Sort of like using itemize without the bullets.

You could try using \leftskip, for example \setlength\leftskip{3em} which you change back to \setlength\leftskip{0em} after the paragraph.

But often it is best to use a package which takes into account all the possible consequences. In this case use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=hanging]hanging package[/url] (available in MiKTeX) where you write for example \hangpara{3em}{0} at the start of the hanging paragraph.


AMC
AIME
AMC 10
geometry
3D geometry
pyramid
octahedron

This is the "official" thread for discussing Mock AIME 4. Results have been computed and posted. It looks like participation was a little better than it was last time. Some people who contacted me to take it early did not send answers.

Participants:
darktreb
Optimosis E
captcha000
mathfanatic
joml88 E
beta E
theone853
Rep123max E
RC-7th
pi^2
towersfreak2006
ginasboys E
nosepuppy E
shake9991
keta E
tetrahedr0n E
NightFlarer E
jeffrey E
white_horse_king88 E
(E - Requested contest early.)

Winners:
1st Place-7 darktreb
2nd Place-6 Optimosis
T-3rd-4 captcha000
T-3rd-4 mathfanatic
T-3rd-4 joml88
T-3rd-4 beta

General Statistics for 14 Answer-Submitting Participants:
Mean: 3.286
Mode: 4.000
Median: 3.000
Range: 1-7
Standard Deviation: 1.666
Easiest Problem: #2
Hardest Problem(s): #8, 9, 13 (two people were off by factors of 2 and 6 respectively), 14, 15.

Answers:
[hide]1. 061
2. 180
3. 103
4. 667
5. 075
6. 002
7. 434
8. 919
9. 035
10. 595
11. 346
12. 240
13. 504
14. 171
15. 039[/hide]Solutions may be found at http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4Solutions.pdf

Score Distribution:
15
14
13
12
11
10
9
8
7 *
6 *
5
4 ****
3 ***
2 ***
1 **
0

Solvers by Problem:
1 *******
2 ***********
3 *****
4 *******
5 *
6 *****
7 *****
8
9
10 *
11 **
12 **
13
14
15

[i]Moderator note: I don't know why you can't, but I added a poll for you. -mathfanatic[/i]

really hard... since I've never taken AIME (except the 1989 one... as practice; this is my first year doing AMC10 or 12) i would not know how it compares... except my feeling is that this one is much tougher than the actual AIME will be

The contest is now online. Peruse it with the following links:

Contest: http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4.pdf
Answers Only: http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4Answers.pdf
Solutions: http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4Solutions.pdf

Judging by that distribution and the work I've done with the students on this list, this AIME was way harder than the average AIME.  Those of you who didn't do as well as you'd like on the AIME shouldn't take your results on this as an indication that you don't have hope to make USAMO.

It was today??? Aaaah! I thought it was on Sunday!

well you cn always takae it tomorrow and check your answers

Heh. I signed up to take it, but I never got around to actually doing it. By the looks of the contest, I doubt I would have done very well.

I like how my answers for #1 and #3 were both off by 1, and I had half the answer on #13 because I am fabulously gifted when it comes to noticing the difference between a negative and positive number.

Ah well this is what makes these practice AIME's great. They're practice! Mr. Ruscyk is right about not getting discouraged, this was a well written contest with very few "gimmes", and if you made many silly mistakes like me, hopefully you can figure out how to get your act together by the time the real contest rolls around.

wow...and you still managed to get a 7...

I have a fair amount of AIME experience. These practice tests are great for the younger kids. I hope it just motivates everyone to work harder.

Everyone should give Mildorf big kudos for doing all this great work. And of course, he would have gotten a 15   ;)

Oh yeah, would someone else please try to verify this for me.

I did #7 with an incorrect method at first, as I made one of my very frequent reading errors. I thought the problem simply wanted the sum of the areas of the 6 pyramids and the cube. The answer I got was 216 + 216*sqrt(2).

Then I realized that shape doesnt' remotely resemble an octahedron, and proceeded to do the problem the way it was supposed to be done. Through that method I got 252 + 180*sqrt(2)

This is interesting because with both the incorrect and correct method, you still get the correct answer of 434 = 252 + 180 + 2 = 216 + 216 + 2

Interesting coincidence? Or did I make a computational mistake the first way?

Cube: 216
Each Pyramid: 1/3 * 36 * 3 :sqrt:2 = 36 :sqrt:2
=> Volume = 216 + 216 :sqrt:2

Wow, that's an amazing coincidence!  :lol:

Hmph, this was even tougher than the first three.

Hmm, I just noticed this now. I got 1,2, 4, 6, and 9, with awesomely stupid errors on 3 (said that all of the factors with 2^k, where k is even, give 2 factors, ugh) and 7 (said that the height was 6$\sqrt{2}$ rather than 3$\sqrt{2}$. Once again, thanks Mildorf. I greatly appreciate the problems.

On the area enclosed by the graph question, wouldnt you want to maximize the value of x^2+y^2 in order to get the major of the ellipse.  and minimize x^2+y^2 to get the minor.

from the center, the max value of x^2+y^2 for center

Sorry for reviving this; by coincidence I got the right answer for number 7... But could someone explain to me (preferbly with a picture) how an octahedron is formed by placing the pyramids like so?

i'm just curious, why was the 1994 AIME extremely hard (according to the poll)? (only thing i know special about that year is the dream team)


[quote="pvthuan"]Let $a,b,c\geq0$ such that $a+b+c=1$, prove that
\[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq\sqrt{4+6(ab+bc+ca)}. \]
[/quote]

Squaring both sides of the inequality we get that

\[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq\sqrt{4+6(ab+bc+ca)}\]
<=>
\[\sum{\sqrt{(a+b)(a+c)}}\geq 1+3(ab+bc+ac)\]
<=>
\[\sum{\sqrt{(a+b)(a+c)}}\geq \frac{(a+b+c)^{2}+3(ab+bc+ac)}{(a+b+c)}\]

Now let's make the following substitutions
\[a+b=x^{2},a+c=y^{2},b+c=z^{2}.(x,y,z\geq 0)\]
=>
\[\sum{\sqrt{(a+b)(a+c)}}\geq \frac{(a+b+c)^{2}+3(ab+bc+ac)}{(a+b+c)}\]
<=>
\[(xy+xz+yz)(x^{2}+y^{2}+z^{2})+x^{4}+y^{4}+z^{4}\geq 4(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\]

From AM-GM we have that \[\sum{xy(x^{2}+y^{2})}\geq \sum{2x^{2}y^{2}}\]
,therefore it is sufficent to prove that
\[x^{4}+y^{4}+z^{4}+xyz(x+y+z)\geq 2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\]
<=>
\[xyz(x+y+z)\geq (x+y+z)(-x+y+z)(x-y+z)(x+y-z)\]
<=>
\[xyz\geq (-x+y+z)(x-y+z)(x+y-z)\]

But latter is the well-known Shur's inequality. :wink:

[quote=pvthuan]Let $a,b,c\geq0$ such that $a+b+c=1$, prove that
\[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq\sqrt{4+6(ab+bc+ca)}.\][/quote]
http://www.artofproblemsolving.com/community/c6h550960p5725736
If $a,b,c$ are positive real numbers such that $a+b+c=ab+bc+ca.$ Then$$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq \sqrt {6+4(a+b+c)}\geq 3\sqrt 2.$$

http://www.artofproblemsolving.com/community/c6h1084690p4787825
Let $a,b,c\geq 0$ and $a+b+c=1$. Then$$ \sqrt{6}\geq\sqrt {a + b}  + \sqrt {b + c}  + \sqrt {c + a}\geq 2$$
http://www.artofproblemsolving.com/community/c6h580320p3426170
Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$
\[\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2.\]

[quote=oldbeginner][quote=pvthuan]Let $a,b,c\geq0$ such that $a+b+c=1$, prove that
\[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq\sqrt{4+6(ab+bc+ca)}.\][/quote]
The following inequality is also true.

Let $a,b,c\geq0$ such that $a+b+c=1$, prove that
\[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq\sqrt{6}\][/quote]

$\sum \sqrt{a+b} \ge \sqrt{6} \Leftrightarrow \sum \sqrt{(a+b)(a+c)} \le 2 $
By AM-GM : $\sum \sqrt{(a+b)(a+c)}\le \sum \dfrac{2a+b+c}{2}=2$

[quote=suli][quote=sqing]Yeah..
Let $a,b,c\geq 0$ and $a+b+c=1$. Then$$ \sqrt {a + b}  + \sqrt {b + c}  + \sqrt {c + a}\geq 2$$.[/quote]

*facepalm*
$$\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a} \ge (a + b) + (b + c) + (c + a) = 2$$
because $\sqrt{x} \ge x$ whenever $x \le 1$.[/quote]
http://www.artofproblemsolving.com/community/c6h1084690p4787825

[quote=luofangxiang]The following inequality is also true
[img]http://s1.sinaimg.cn/middle/0018zOAxgy70H5rLt9C20&690[/img][/quote]

The inequality is equivalent to that
\[\sum{\sqrt{a+b}}\ge{\sqrt{4(a+b+c)+2\sqrt{3(ab+bc+ca)}}}\]
Square on both sides of the inequality,we have
\[\sum{(\sqrt{(a+b)(a+c)}-a)}\ge{\sqrt{3(ab+bc+ca)}}\]
Using inequality $(x+y+z)^2\ge{3(xy+yz+zx)}$,we need to prove that
\[\sum{[(\sqrt{(a+b)(a+c)}-a)(\sqrt{(b+c)(b+a)}-b)]}\ge{ab+bc+ca}\]
This is an identity!

[quote=luofangxiang]The following inequality is also true
[img]http://s1.sinaimg.cn/middle/0018zOAxgy70H5rLt9C20&690[/img][/quote]
After squaring of the both sides we need to prove that $2\sum_{cyc}\sqrt{(a+c)(b+c)}\geq2(a+b+c)+2\sqrt{3(ab+ac+bc)}$ or
$2(a+b+c)-2\sqrt{3(ab+ac+bc)}\geq\sum_{cyc}\left(\sqrt{a+c}-\sqrt{b+c}\right)^2$ or
$\sum_{cyc}(a-b)^2\left(\frac{1}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{1}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}\right)\geq0$, which is obvious:
$a+c+b+c+2\sqrt{c^2+ab+ac+bc}\geq a+b+c+\sqrt{3(ab+ac+bc)}$.
Done!

See this:
http://artofproblemsolving.com/community/c6h1449955p8309976
Greetings from Lorian Saceanu!


[quote="fallinlovewithmaths"]for all positive real numbers a,b,c such that a+b+c=3. Prove that

(a^2 + 3b^2)/ab^2(4-ab) + (b^2 + 3c^2)/bc^2(4-bc) + (c^2 + 3a^2)/ca^2(4-ac)[/quote]
I think it may be that
$ a\plus{}b\plus{}c\equal{}3$ then $ \sum_{cyc} \frac{a^2\plus{}3b^2}{ab^2(4\minus{}ab)} \geq 4$

[quote="chien than"][quote="fallinlovewithmaths"]for all positive real numbers a,b,c such that a+b+c=3. Prove that

(a^2 + 3b^2)/ab^2(4-ab) + (b^2 + 3c^2)/bc^2(4-bc) + (c^2 + 3a^2)/ca^2(4-ac)[/quote]
I think it may be that
$ a \plus{} b \plus{} c \equal{} 3$ then $ \sum_{cyc} \frac {a^2 \plus{} 3b^2}{ab^2(4 \minus{} ab)} \geq 4$[/quote]
By AM-GM inequality:
$ \sum \frac{a^2\plus{}3b^2}{ab^2(4\minus{}ab)} \ge \sum \frac{4\sqrt[4]{a^2b^6}}{ab^2(4\minus{}ab)}$
$ \equal{}\sum \frac{4}{\sqrt{ab}(2\minus{}\sqrt{ab})(2\plus{}\sqrt{ab})} \ge \sum \frac{4}{2\plus{}\sqrt{ab}}$ (because $ \sqrt{ab}(2\minus{}\sqrt{ab}) \le \frac{(2\minus{}\sqrt{ab}\plus{}\sqrt{ab})^2}{4}\equal{}1$)
SO we need to prove:
$ \sum \frac{1}{2\plus{}\sqrt{ab}} \ge 1$
By Cauchy-Schwarz inequality, we have:
$ \sum \frac{1}{2\plus{}\sqrt{ab}} \ge \frac{9}{6\plus{}\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}} \ge \frac{9}{6\plus{}a\plus{}b\plus{}c}\equal{}1$

sorry.. i was careless. chien than was right. we should prove that the inequality is more than 4. Thank you NguyenDungTN

[quote="ppt"]By Cauchy-Schwarz inequality, we have:
$ \sum \frac {1}{2 \plus{} \sqrt {ab}} \ge \frac {9}{6 \plus{} \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}}$
I'm sorry.
Can you explain more detail, please?[/quote]
By Cauchy-Schwarz:
$ (\frac{1}{2\plus{}\sqrt{ab}}\plus{}\frac{1}{2\plus{}\sqrt{bc}}\plus{}\frac{1}{2\plus{}\sqrt{ca}})((2\plus{}\sqrt{ab})\plus{}(2\plus{}\sqrt{bc})\plus{}(2\plus{}\sqrt{ca})) \ge 9$ :)


Let positive reals $x,y,z$ and $m,n$ positive integers with $n\geq m$.
Prove that

$\dfrac{x^n}{(z+y)^m}+\dfrac{y^n}{(z+x)^m}+\dfrac{z^n}{(x+y)^m}\geq\dfrac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$ :)

I think you should change the title. I know you have been active in this forum for some time, so you SHOULD know the rule of naming a topic. It's quite disgusting to me to see topics to be named like "nice", "beautiful" which makes searching much harder.

Anyway back to your inequality, WLOG assume $x \ge y \ge z$ and use chebyshev on the sequences $\{ x^{n-m} \}$ and $\{ \dfrac{x^m}{(y+z)^m} \}$. It remains to use power mean and nesbitt to deal with $\{ \dfrac{x^m}{(y+z)^m} \}$

Sorry Soarer and all the community. Big wrong. I will change the title. Your solution is correct and cool. But this is for me who know the solution. You could write it with a bit more details but it doesnot mind. If someone wants an explanation we will give it.


Suppose the rectangle on a dot paper is a pool table. The dots on the corner are holes. A ball starts at one corner of the table, is pushed to an edge, bounces off that edge to another, and so on, as shown in the diagram. When the ball finally reaches a corner it drops off the table. For an [i]m x n[/i] table, how can you predict the number of the paths possible?

[quote="cardwacko9999"]my answer...

infinite

since there are infinite angle measures between 0 and 90, there are infinite paths.[/quote]

  The word "edge" indicates the ball is shot of at a 45 degree angle.  As illustrated in the diagram.

 O and heres a hint.

[hide]  Whats special about 45 degree, right-angled triangles? [/hide]

I guess I have to put additional information. This is an original problem that I am trying to investigate, so Im sorry for the errors.

Suppose that the ball is shot at the corner(s) of the small dotted squares (please see drawing) and can only bounce at a 90 degrees angle, for an [i]m [/i]x [i]n[/i] table, how can you predict the number of the paths possible?


Hi there,

I am trying to prove the following:

Triangle ABC is acute, the bases of the altitudes at B and C are E and F respectively.
Let A' be the midpoint of BC. Prove that Triangle A'EF is isoceles.

I drew an example of this on sketchpad, it is attached. By inspection you can see that |A'F| = |A'E| so angle EFA' = angle FEA'

I am not sure where to start, I was thinking of drawing the parallel to AC, AB, BC that pass through points B, C, and A respectively, creating an outer triangle that would be similar to triangle ABC and would create the perpendicular bisectors of this new bigger triangle.  I couldnt see any relation between this and triangle A'EF

Any thoughts?

its ok, I figured it out, just draw circle centered at A' with BA'

It is very simple.Let     $M$  the midpoint of    $BC$ (you have it A').

By a well known teorem for the median in a right triangle we have :

      $MF = BC/2$      and     $ME = BC/2$

  so     $MF = ME$

 [u]Babis[/u]


trigonometry
calculus
derivative
induction
integration
function
real analysis

Let $f : \left( 0, \infty \right) \to \mathbb R, \, f(x) = \frac{\sin x}{x}$.

Prove that $\left| f^{(n)}(x) \right| \leq \frac1{n+1}, \, \forall n \geq 1, \, \forall x > 0$.

In the contest, you get a hint, which turns the problem into one boring computation.
Let's see if somebody solves the problem without it.

It is by Gronwall.I saw it a short time ago and reduced it to sth that can be solved by considering derivative,seemingly elegant,at any rate.Post your hint and I will post mine if they are different.

It is easy to prove by induction that

$\frac{d^{n}}{dx^{n}}(\frac{\sin x}{x})=\frac{1}{x^{n+1}}\int_{0}^{x}y^{n}\sin (y+\frac{n+1}{2}\pi)dy$.

Using this equality, which is called Gronwall's identity, the result folows imediately.

and another one:

$\frac{d^{n}}{dx^{n}}(\frac{1-\cos x}{x})=\frac{1}{x^{n+1}}\int_{0}^{x}y^{n}\cos(y+\frac{n\pi}{2}))dy$.

This is also an identity of Gronwall.

That was exactly the hint. But I think it's very hard without.

You can try my approach,which I ,by virture of my poor ability to print,just propose to explain a little.Obviusly,by employing Leibniz formula or sth else,you may obtain the derivative as a fraction.Multiply the denominator to render it at the right side.'Left-Right' is the new function which you are going to investigate.Use derivative to attack it,and then most of the terms in this long expression cancel one another,thus causing the result manifest itself.
   If it is flawed,let me know.

You can use the identity
\[\frac{\sin x}{x}= \int_{0}^{1}\cos(xt) \, dt. \]
Now differentiate $n$ times.

There's the Fourier analysis approach:

$f(x)=\frac{\sin x}{x}=\frac12\int_{-1}^{1}e^{itx}\,dt$

Now differentiate under the integral sign to get

$f^{[n]}(x)=\frac12\int_{-1}^{1}(it)^{n}e^{itx}\,dt$

Then $|f^{[n]}(x)|\le\frac12\int_{-1}^{1}|t|^{n}\,dt=\frac1{n+1}.$

[Note that $f$ is defined at zero and $f(0)=1.$ This is the function with the removable singularity removed.]

[color=green][Ah - Ravi B and I were thinking the same thing.][/color]


Can someone explain the squareroot algorithm to me?  I used to know it (really!) but I have no idea now, its been so long ago.  My 4th grade TAG teacher taught it to me.  I looked at [url=http://www.mathpath.org/Algor/algor.square.root.htm]MathPath's Explanation[/url] but it is way too complicated.  Can someone explain it step by step to me in plain English?  No a (subcript) n-1.  That stuff makes my head hurt.  I learned this is 4th grade, so I know this can be explained in plain English.  Please help!

This is actually pretty easy:

Guess...divide...average.

Lets pick a number to find the square root of...How about 6?

We begin by guessing....What number might be the sqare root of 6 (you can pick any number...it really doesnt matter)... Lets pick 2.

Now we divide 6/2=3 and find the average (3+2)/2 giving us 2.5 That is a pretty rough estimate wouldnt you say?  We can continue the process for a better approximation by dividing and averaging...

divinde: 6/2.5=2.4 find the average (2.5+2.4)/2 giving us 2.45... and keep on going dividing and averaging...

I know another method...Its much more efficienct...but I cant remember it right now.

Wait...Here it is http://mathforum.org/library/drmath/view/52610.html
If you dont understand it just ask me.

Its an easier way to square root by hand.


logarithms
calculus
derivative
calculus computations

Two points $ P,\ Q$ moves on the curves $ y\equal{}x\plus{}2,\ y\equal{}\ln x$ respectively. Find the minimum value of $ \overline{PQ}$.

I got $ \frac{3\sqrt 2}{2}$.. is it correct?

hello, we have to minimize
$ \overline{PQ}\equal{}\sqrt{(x_1\minus{}x_2)^2\plus{}(x_1\plus{}2\minus{}\ln(x_2))^2}$.
Sonnhard.

[quote="spix"]I got $ \frac {3\sqrt 2}{2}$.. is it correct?[/quote]

Correct. How did you solve it?

Maybe our solution seems to be same.

I solved it in 30 seconds.

I used partial derivatives to get the minimum but something tells me there s an easier way...

Determine where the perpendicular to ln(x) is perpendicular to x+2,too.These is the points

You can use the tangent line at the point $ (1,\ 0)$ on $ y\equal{}\ln x$.

Would someone like to write out a solution for this one?


topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

let $ P: (E,e_0)\rightarrow(X,x_0)$ be a covering map and the
induced map on fundamental groups is surjective,also E is path connected.prove that p is a homeomorphisam.

With some additional assumptions, you could try using the covering classification theorem, which, roughly, states that coverings of $ X$ correspond (up to homeomorphism) to subgroups of $ \pi_{1}(X)$. The covering in question corresponds to the whole $ \pi_{1}(X)$ (since induced map i surjective), so it's homeomorphic to the trivial cover. However, the theorem assumes that $ X$, besides being path-connected, is also locally path-connected and semi-locally simply connected, conditions slightly stronger than only being path-connected.


To graph the inverse functions:
Since $csc^{-1}$ $x=sin^{-1}(\frac{1}{x})$ and $sec^{-1}$ $x=cos^{-1}(\frac{1}{x})$ why isn't $cot^{-1}$ $x=tan^{-1}(\frac{1}{x})$, but instead,
it is $cot^{-1}$ $x=\frac{\pi}{2}-tan^{-1}(x)$

Thanks for the help.

-------------
I have a few unanswered questions unrelated to the question above (some are not completely answered).  It would be great if someone could help me.  I appreciate your every help.
please look at this one especially -->  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43155
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42861

The principal values for $\tan^{-1} x$ and $\cot^{-1} x$ are different.
The range of $y=\tan^{-1} x$ is $y\in (-\frac{\pi}{2},\frac{\pi}{2})$.
The range of $y=\cot^{-1} x$ is $y\in (0, \pi)$.
So to obtain the correct graphs on the calculator, the formula is in this complicated form.
Note that $y=\tan^{-1} (\frac{1}{x})$ will give a different set of values that also satisfy $\cot y=x$, but this graph is discontinuous and gives no value for $x=0$.


[quote="sunoco"]I have a problem. Can someone explain to me how to figure out this problem?

Question-

What is the point slope form of an equation with the slope 3/5 that passes through the point(10,-2)?

Edit- Also have another one.

Question2-

If a slope of a line is 2 and its y-intercept is 5, what is the equation of the line?[/quote]

[hide="1"]
Point slope form is in $(y-y_{1})=m(x-x_{1})$ so your equation would be

$(y+2)=\frac{3}{5}(x-10)$
[/hide]

[hide="2"]
In the form of $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept, your equation is:

$y=2x+5$.
[/hide]

[quote="sunoco"]THanks.  :lol:

I have another one though.

Question- Rewrite the equation 2x+4y+2=3y+5 in general form.[/quote]

[hide]
Well, to start, get all the terms on one side of the equation.

$2x+4y+2=3y+5 \to$

$2x+4y-3y-5=0 \to$

$2x+y-5=0$
[/hide]


[quote="Silverfalcon"]If $a,b,$ and $c$ are numbers such that

$-3b+7c = -10$
$b-2c = 3$
$a+2b-5c = 13$

what is the value of $a$?[/quote]

[hide]if you add all the equations up you get
$a=6$[/hide]

[hide="Answer"]$-3b+7c=-10$
$b-2c=3\Rightarrow 3b-6c=9$
$(-3b+7c=-10)+(3b-6c=9)=(c=-1)$
$b-2(-1)=3\Rightarrow b=1$
$a+2(1)-5(-1)=13\Rightarrow \boxed{a=6}$[/hide]

[b]EDIT:[/b] Wow, I feel so stupid... I didn't realize it was quite that simple. :roll:

[hide]It seems like the sum of the last two equation equals the opposite of the third equation (for variables b and c). Adding cancels b and c, and a=6[/hide]


In how many ways can you choose $4$ positive integers $a,b,c,d$ where $a<b<c<d$ such that

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is an integer?

[quote="PTynan89"]Try changing it to $\frac{a+b}{ab}+\frac{c+d}{cd}$.

Neither of which is itself reducible.[/quote]What do you mean by reducible? $\frac16+\frac12 = \frac{2+6}{12}= \frac{8}{12}= \frac23$ which reduces...

I think I may have seen this before, and that you can only do it with 1, 2, 3, 6 (although I haven't taken the time to check this, so it may be wrong). That's one solution though.

Nice Problem. Solution
[hide]
me, persoanlly I don't know any number theory or anything so I did it the most easiest algebra way..
examining if $a\ge 4$, its immposible.
So lets examine cases for $a=1,2,3$.
$1/3+1/b+1/c+1/d=k$, the maximum $k=.95$.
$1/2+1/b+1/c+1/d=k$, because $k=1$ if $k$ is an integer.
$1/2+1/b+1/c+1/d=1\Rightarrow 1/b+1/c+1/d=1/2$
$1/b>1/6$, testing case $b=5$:
$1/2+1/5+1/6+1/8<1<1/2+1/5+1/6+1/7$. testing $b=4$:
$1/2+1/4+1/7+1/8<1<1/2+1/4+1/6+1/7$.after testing for
$b=3$ its impossible as well, testing for $1,2,3,6$ works using the logic above.[/hide]

Is this true?

There exist positive integers $a_{1}, a_{2}, \cdots, a_{n}$ with $1 < a_{1}< a_{2}< \cdots < a_{n}$ such that $\sum_{i=1}^{n}a_{i}^{-1}$ is an integer if and only if there is a perfect number that has $n+1$ divisors.

Let me say it like this: then there is a perfect number for any number of divisors :D

[quote="PTynan89"]Try changing it to $\frac{a+b}{ab}+\frac{c+d}{cd}$.

Neither of which is itself reducible.

EDIT: I meant reducible as in "can be reduced to an itneger."[/quote]

It does not follow that their sum is not an integer.

[quote="Franz Joseph"][quote="jimhu"]Nice Problem. Solution

me, persoanlly I don't know any number theory or anything so I did it the most easiest algebra way..
examining if $a\ge 4$, its immposible.
[/quote]

How did you get this part?[/quote]
$\frac{1}{a}>\frac{1}{b}>\frac{1}{c}>\frac{1}{d}$
$4\frac{1}{a}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$
 :wink:


A grammar book, a dictionary, a thesaurus and a novel are placed
on a bookshelf, side-by-side, with the spines facing out, and the text
on the spines right side up. In how many different orders can these
four books be placed on the bookshelf?

Since no restrictions were defined, the number of ways to order is simply $ 4\times3\times2\times1$, also known as $ 4$ factorial, which equals $ \fbox{24}$.


Let be $ ABC$ a triangle and let $ D\in (BC)\ , E\in (CA)\ , \ F \in (AB)$ such that $ \sphericalangle BAD\equal{}\sphericalangle DAC \ ,\ BE\perp CA\ ,\ AF\equal{}BF$ . If $ AD\cap BE\cap CF\neq \varnothing$ prove that $ \sin \ B\equal{}\mathrm{tg} \ A\cdot \cos \ C$

[hide]By Angle Bisector Theorem, $ {BD\over CD}\equal{}{c\over b}$. By construction, $ CE\equal{}a\cos\gamma, EA\equal{}c\cos\alpha$.

By Ceva's, $ {AF\over FB}\cdot{BD\over DC}\cdot{CE\over EA}\equal{}1$, hence $ {c\over b}\cdot{a\cos\gamma\over c\cos\alpha}\equal{}1\iff {a\cos\gamma\over b\cos\alpha}\equal{}1$.

By Sine Law, $ {a\over b}\equal{}{\sin\alpha\over\sin\beta}$, hence $ {\sin\alpha\cos\gamma\over\sin\beta\cos\alpha}\equal{}1$ and the result follows.[/hide]


modular arithmetic
quadratics
number theory
number theory unsolved

Solve this equation if x,y are positive integers
$x^{2}-y^{3}=7$

[b]Solution.[/b] Assume to the contrary that there exists $(x,y)\in\mathbb{N}^{2}$ such that $x^{2}=y^{3}+7$.  Notice that if $x^{2}=y^{3}+7$, then $x^{2}+1=y^{3}+8=(y+2)(y^{2}-2y+4)=(y+2)((y-1)^{2}+3)$.  Clearly, $y\equiv 1\pmod 2$, so $(y-1)^{2}+3\in 4\mathbb{Z}+3$.  Since $4\mathbb{Z}+1$  is closed under multiplication, $x^{2}+1$ must have a prime factor of the form $4k+3$.  Let $p$ be a prime such that $p|x^{2}+1$.  Then $x^{2}\equiv-1\pmod p$.  But by Fermat's Little Theorem, $x^{p-1}\equiv 1\pmod p$, thus $4|p-1\implies p\in 4\mathbb{Z}+1$.  Contradiction! $\Box$

[quote="ZetaX"]$\mod 4$[/quote]
Suppose that its $\mod 4$
Why?

[Edited]

I'm sorry, I intend to ask another question, but then I understand.

[quote="pco"][quote="bibobeo"]Can anyone explain for me why $y\equiv 1 \pmod 2$?  :blush:[/quote]

$x\equiv 0,1,2,3 \pmod 4 \implies x^{2}\equiv 0,1,0,1 \pmod 4$
$y\equiv 0,1,2,3 \pmod 4 \implies y^{3}+7\equiv 3,0,3,2 \pmod 4$

And, since $x^{2}=y^{3}+7$, we must have $x\equiv 0,2 \pmod 4$ and $y\equiv 1 \pmod 4$[/quote]
Working in $\mathbb{Z}_{8}$, we have that if $y$ is even, then $x^{2}\equiv 7\pmod 8$.  But the fact that $7$ is a quadratic non-residue modulo $8$ gives us a contradiction.  Hence, $y$ must be odd.

Wait, sorry for my stupidness, but I don't under stand what you say  :( , Boxedexe.
From "Since ..." to the end, I don't get it.
What's the contradiction? When $y= 4k+1$, we have $x^{2}+1= (4k+3)(4h+3)$ so it has a form of $4\mathbb{z}+1$.


USAMTS
trigonometry
calculus
derivative
geometry
3D geometry
sphere

This problem has been passively bugging me for months, due to my sometime inability to find a solution (not an answer; I know how to write a program to find that :P ).  Fortunately, last night, I found a truly marvelous solution which is unfortunately too long to post before I eat lunch today.

46. Find the exact value of the infinite series $\sum_{ k=0 }^{ \infty }\mbox{Cot}^{-1}( k^{2}+k+1 )$.

[hide]We are motivated by the cotangent subtraction formula, \[\cot (m-n) = \frac{ \cot{m}\cot{n}+1 }{ \cot{n}-\cot{m}}\] This makes is apparent that \[\mbox{Cot}^{-1}( k^{2}+k+1 ) = \mbox{Cot^{-1}}(k)-\mbox{Cot^{-1}}( k+1 )\] Now we have a telescoping series, so we can say $\sum_{k=0}^{n}\mbox{Cot^{-1}}( k^{2}+k+1 ) = \mbox{Cot^{-1}}0-\mbox{Cot^{-1}}( n+1 )$.  Since as $x$ goes to infinity, $\mbox{Cot^{-1}}x$ goes to zero, our answer is $\mbox{Cot^{-1}}0 = \frac{\pi}{2}$, Q.E.D.[/hide]

:10:

Great solution.

Question: Is it assumed that the problem is asking for the answer in radians?

When writing a program during the test, I put an exponent of $-1$  in the wrong place, so my result was off by alot.

BTW: USAMTS Year 18 Round 1 Problems are posted.

Hah! cool. A nifty combination of trig identities and telescoping.

[quote="captcha000"]lol nice solution. just wondering, because i dont remember at all: calculators allowed on GCTM or not?[/quote]

Yes they are.

[quote="Boy Soprano II"]Fortunately, last night, I found a truly marvelous solution which is unfortunately too long to post before I eat lunch today.
[/quote]

 :rotfl:

I have found a truly marvelous solution which is unfortunately *cut cut* too short for this post to contain:

$x^{n}+y^{n}= z^{n}$ has no solutions for integers $x,y,z, n$ where $n>0$. The proof is quite simple actually. It is:

[hide]too short for this post to contain[/hide]

 :rotfl:

By the way, what is your name, captcha000?

Also, I have a question: I don't understand how you went from:

$\cot(m-n) = \frac{\cot{m}\cot{n}+1}{\cot{n}-\cot{m}}$ to
$\cot^{-1}(k^{2}+k+1)=\cot^{-1}(k)-\cot^{-1}(k+1)$

Could you explain this?  :)

You simply set $m$ equal to $\mbox{Cot}^{-1}(k)$ and $n$ equal to $\mbox{Cot}^{-1}(k+1)$, and then you see that the cotangent of their difference has the same cotangent as $\mbox{Cot}^{-1}( k^{2}+k+1 )$.

By the way, my proof of Fermat's Last Theorem is shorter :  we verify that it doesn't work for $n=3$, and then we differentiate to reduce $n$ to the case of $n-1$.  (Note how my proof is a single sentence, whereas you waste several sentences on introduction.  Also, I don't understand that step you did in the middle.)  So exceedingly simple, I can't believe it took Wiles and that lot so long.  But then, he used the inferior elliptic curves method, which is where you use curves in three-space of the class $x^{n}+y^{n}= z^{n}$, where $z$ is an integer, and you simply prove that it doesn't pass through lattice points.  You just have to use a few partial derivatives, which is slightly more annoying.  It just goes to show what poor hands a problem gets into when it doesn't get enough publicity.  By the way, I know that looks more like a perverted sphere than an ellipse; it's called "elliptic" because that was a poor translation from Taniyama's original manuscript (the translator was working on [i]All Your Base are Belong to Us[/i] at the same time).

[quote="Boy Soprano II"][quote="Xantos C. Guin"]Question: Is it assumed that the problem is asking for the answer in radians?[/quote]
I think so, because radians are so more natural a measure than degrees, but they probably would have accepted either one, because you're seriously unlikely to be dealing with an angle of $\frac{\pi}{2}$ degrees or 90 radians.  Anyhow, I know that in "serious" contexts, all angles are assumed to be in radians, just as a log without a base is assumed to be a natural log.[/quote]

Regardless of how unlikely it is to be dealing with an angle of $90$ degrees, you must write it as $90^\circ$ not $90$. Otherwise it is assumed to be radians.

[quote="Karth"]
By the way, what is your name, captcha000?

Also, I have a question: I don't understand how you went from:

$\cot(m-n) = \frac{\cot{m}\cot{n}+1}{\cot{n}-\cot{m}}$ to
$\cot^{-1}(k^{2}+k+1)=\cot^{-1}(k)-\cot^{-1}(k+1)$

Could you explain this?  :)[/quote]

captcha000 is Howard Tong. He was on GA-ARML before he went to Philps-Exeter in NH.

[quote="Boy Soprano II"]
You simply set $m$ equal to $\Cot^{-1}(k)$ and $n$ equal to $\Cot^{-1}(k+1)$, and then you see that the cotangent of their difference has the same cotangent as $\Cot^{-1}(k^{2}+k+1)$.
[/quote]

For those of you who don't think that explanation was obvious and need something that is easier to understand:

Let $\cot^{-1}{(k)}=m$ and $\cot^{-1}{(k+1)}=n$.

$\cot{(m)}=k$ and $\cot{(n)}=k+1$.

$\cot{(m-n)}=\frac{(k)(k+1)+1}{(k+1)-(k)}=\frac{k^{2}+k+1}{1}=k^{2}+k+1$

$m-n=\cot^{-1}{(k^{2}+k+1})$

$\cot^{-1}{(k)}-\cot^{-1}{(k+1)}=\cot^{-1}{(k^{2}+k+1)}$

[quote="Boy Soprano II"]
By the way, my proof of Fermat's Last Theorem is shorter :  we verify that it doesn't work for $n=3$, and then we differentiate to reduce $n$ to the case of $n-1$.  (Note how my proof is a single sentence, whereas you waste several sentences on introduction.  Also, I don't understand that step you did in the middle.)  So exceedingly simple, I can't believe it took Wiles and that lot so long.  But then, he used the inferior elliptic curves method, which is where you use curves in three-space of the class $x^{n}+y^{n}= z^{n}$, where $z$ is an integer, and you simply prove that it doesn't pass through lattice points.  You just have to use a few partial derivatives, which is slightly more annoying.  It just goes to show what poor hands a problem gets into when it doesn't get enough publicity.  [b]By the way, I know that looks more like a perverted sphere than an ellipse; it's called "elliptic" because that was a poor translation from Taniyama's original manuscript (the translator was working on [i]All Your Base are Belong to Us[/i] at the same time).[/b][/quote]

ROFL  :rotfl:

Man, this is just too funny. By the way, proving the theorem, I think is a bit harder than just utilizing partial derivatives. I thought of this method when I was in GHP and asked one of the instructors, who said had tried, and got quite far into the proofs, but ended up in a dead end. Also, if you start of with $n=3$ and you decrement $n$, wouldn't that yield $n=2$, for which the theorem no longer holds? (The theorem states that it is true for $n > 2$).

[quote="Xantos C. Guin"]
captcha000 is Howard Tong. He was on GA-ARML before he went to Philps-Exeter in NH.
[/quote]

@captcha000:

You're Howard??  :what?: That's so cool! I've heard so much about you   :D

lol that's me. its been a while since ive been to competitions around the state, but you'll probably hear more of me this year. =)

[quote="Karth"]Man, this is just too funny. By the way, proving the theorem, I think is a bit harder than just utilizing partial derivatives. I thought of this method when I was in GHP and asked one of the instructors, who said had tried, and got quite far into the proofs, but ended up in a dead end. Also, if you start of with $n=3$ and you decrement n, wouldn't that yield $n=2$, for which the theorem no longer holds? (The theorem states that it is true for $n > 2$).[/quote]
Hmm.  Yes, I think it would be fairly difficult to prove a number theoretic result using derivatives of any kind.  According to my proof, you are merely showing that case $n=k+1$ has solutions only if case $n=k$ has solutions, and it has long been known that $n=3$ has no solutions (I believe Fermat himself actually published a proof to that effect).  But the proof is really preposterous for the following two reasons :

1) You are differentiating with respect to what, exactly?
2) Just because $f(x)$ is an integer does not mean $f'(x)$ is an integer, and vice-versa.

(But actually Dr. Shimuru (a good friend of my father's advisor's cousin's son) rectified these difficulties with a quite effortless 50-page proof, although it was essentially a rehash of Taniyama's original paper.  Sadly, Taniyama and Shimuru lived on one of the smaller offshore Japanese islands where there was only one translator available.  The reason why Wiles got the credit is because he (in a flash of profound genius) realized that the two English translations were of one and the same result (they were translated so badly, nobody else realized this), and spent about seven years piecing together what exactly the translator had translated.  For this superb effort, he was rewarded $\$$ 1 000 000, and [i]All Your Base Are Belong to Us[/i] went down in history.  Of course later, it was discovered that neither of the translations had used any verbs at all, which led to an allegation that the pieces of "research" were in fact childish pranks.  This prompted Wiles to publish an additional 150 pages giving detailed evidence that the translator was unaware of the fact that the English language possessed verbs, and had tried to use pronouns instead.  It just goes to show that a minor in applied linguistics really can further your mathematical career.)

Wow, you actually know Dr. Shimiura?

EDIT: Wait, are you moving back, Howard?

[quote="captcha000"]i suppose you could say so. in a sense, i never really moved away, but i wont get too technical.[/quote]

So you will be attending competitions next year? And perhaps be on the ARML team?

some competitions. im going to lakeside, which doesnt have a really good math program, but oh well. as for ARML, yeah i will (hopefully) be on the team lol

I think that's the school Wes mentioned that he was planning to attend. If so, their math team will pwn others.

really? could be the other lakeside, i think there are two in georgia. but if not, thatd be pretty cool lol. when do you guys start? we start on monday

Yeah, same with Fulton County. We also start on the 14th... grr. and I still have summer homework left to do  :(

All I've done about summer reading is created a very easy AIME style problem about summer reading.

Arrgh, you are wrong, Santhosh.  Lassiter will still rule, whatever events come into play!  But Wes won't even be competing in most math tournaments next year, will he?  Just ARML.

Actually, I'm about ready for school to start.  That may or may not have something to do with my hermit-like style of life during the summer.

Hmm, I've done about half of my summer reading.  I read [i]Moby Dick[/i] :  a great book if ever there was one.

By the way, everybody come to UGA if by some bizarre circumstance you weren't planning to already, because Vestavia might be coming, and we must deny them as many top ten places as possible.  (I'm still smarting from GACS.)

Yeah, these math competitions work bizarrely.  I think that it's where you go to school, not where you live, that matters.  Hence Jeremy (Canadian) being on our ARML team.  Also, there was a Korean (Gye Hyun Baek) at MOP, who was allowed to come because he went to Exeter this year.

[quote="Boy Soprano II"]Arrgh, you are wrong, Santhosh.  Lassiter will still rule, whatever events come into play!  But Wes won't even be competing in most math tournaments next year, will he?  Just ARML.[/quote]

Ah, but then Wes isn't completely positive if he's going to ARML or not. Most unfortunate, indeed. But then if he chooses to go, he will be exponentially better since he attends MIT.

yeah baek got a 7 at arml. he was our best hope for an individual winner, but he didn't get one. =(. i doubt lakeside will go to UGA, but i'll probably be going as an individual

[quote="Xantos C. Guin"]I never said that they would take 1st. I just said that Lakeside would do well if Wes and Howard went there.

Agreed with coming to UGA. What is the cost per individual? (Wheeler probably won't attend as usual.)

BTW: do you think we should also go to an Alabama tournament to try and beat Vestivia, or is that not worth the travel?[/quote]
Okay.  You are forgiven, and there is truth in what you said. :D

The UGA website says that it's $\$$ 25 per team of four, or $\$$ 10 per individual without a team, plus $\$$ 5 per attending adult more than one per school.

Well, going to an Alabama tournament would certainly be fun, but they said that they would probably come to UGA.  Plus, even if Vestavia didn't win on the team, GA attendance would probably not be high enough to prevent them from taking most of the top ten finishes.  Although maybe if we created a conspiracy...

captcha :  Whoa, where were the really good upperclassmen?  The ones on IMO team?  I think Mrs. Poss said something vague about them being at some sort of prestigious event or other.

We could get two school buses to be availible for a weekend, and get all ARML members to rendez-vous at Lassiter/GACS either really the previous night. Then stop at a motel near the competition site. 20 ARML members should be enough to limit Vestivia to 5 top 10 people. Altough, this is highly optimistic, and would cost alot.

EDIT: After realizing that average school bus has slightly more than 20 seats that can easily fit 2 people if needed, I think that only one would be needed.

Nope, guys, our senior grade is an absolute killing machine...You might be able to get 2-3 top ten places from Vestavia.

Agreed. That's why I said 5 top10's is optimistic.

[quote="Boy Soprano II"]captcha :  Whoa, where were the really good upperclassmen?  The ones on IMO team?  I think Mrs. Poss said something vague about them being at some sort of prestigious event or other.[/quote]

Only one person at Exeter currently has been on IMO team (Sherry Gong), but she made IPO (physics) this year and decided to go to that instead. We were so hopeful about winning first, (I got 4th the year before, and Ryan was champion). Granted the test wasn't really to our advantage, but oh well.

[quote="Karth"]Ah.. we'll own Vestavia any day. If we can take on Phillips Exeter to an extent with their 15 USAMO qualifiers, we can easily take on Vestavia  :D [/quote]

One of the USAMO qualifiers didn't go to ARML, and not all of the rest were on the team (we had like 10 alternates).

If you had 10 alternates, why didn't you just take 5 more people and make a second team?

easy to say that for a state, where lots of people want to participate. but for a single school, pretty much everyone who wanted to go was already going.

Well, I think all of Georgia beats all of Alabama pretty easily.  Indeed, I think even if you took out all of the A team, we'd still be a good match.  (I cite ARML.)  The problem is that all of Alabama goes to the same school (at least, all of the math people, with maybe a few exceptions), so they sneakily make it [i]look[/i] like they're better than us.  Let me think about Santhosh's proposed invasion.  But right now, I don't think an overnight stay would be necessary, or even school buses.  But that would depend on a lot of things.

[quote="captcha000"]Only one person at Exeter currently has been on IMO team (Sherry Gong)[/quote]Well, Ryan was on the team this year, but then I suppose you could say that, strictly speaking, he isn't at Exeter any more.  And for some reason I had this incorrect idea that Yi Sun went to Exeter (I don't know why).  Only 25 people at Exeter wanted to go to ARML?  Surprising.  On the other hand, I have no idea how big Exeter is.

Invasion? My plan only calls for about 20 people. (33 ARML members minus 7 seniors minus people who are busy that day or not intested in participating). An invasion would be like taking the top 50 non-seniors from the GCTM tournament.

According to [url=http://www.exeter.edu/about_us/about_us.aspx]this[/url] there are about 1000 Exonians at Exeter.

Wow, this topic went from trig to conspiracy plans in 26.25 hours.

There are probably more than 1000, but not much more, at Exeter. Wow O_O I didn't even know Ryan made IMO, lol haven't talked to him since I last saw him at school. (the day after ARML, took off for MOP). Yi Sun graduated from Harker school in California. Yi HAN goes to Exeter, rising junior, was at MOP, and (I think) top placer on USAMO.

Speaking of Exeter, I just recieved a package whether I wanted to attend Exeter this school year?

The letter in the package read:
"Dear AMC Participant:

Are you dreaming of an education like no other?

Imagine a place where it's cool to be smart..."

"Your success on the American Math Competition Contests leads us to believe that you and your parents might enjoy learning about one of America's oldest and most respected schools, Philips Exeter Acadamy."

Just wondering, did anyone else get a similar letter?

haha i started getting those in elementary school, but didn't really pay too much attention to them until 8th grade lol

I too got a Exeter packet.\

I must say that I probably amnae going to Vestavia this year. However, you might want to add to your sales pitch that for a little additional money (I think $5 a person) you can take the Alabama ARML TST, where the top 5-10 get real money :).

Ok, I'll do that.

BTW: One of the Math team coaches at Wheeler actually gave me information about UGA. I would never have expected to recieve any information until after the competition. I will most likely be attending.


I'm writing a large document in LaTeX and one of my section titles is too long, thus causing a bad box. The obvious solution is to put a line break \\ in, but... this makes the table of contents look silly as it breaks the line there too, even though there is ample space there for the whole title to fit a single line.

I think you can use the optional argument of the section command.  For example,
\section[This is a very long title]{This is a\\very long title}.

Either
1) if your document looks fine to you then don't worry about the bad box
or
2) you can tell \section to use a shortened form for the table of contents:
\section[this is used in the table of contents]{this is used for the section heading in the document\\ and can be very long and go on to two lines}
or
3) use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titlesec]titlesec[/url], [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tocloft]tocloft[/url] and/or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titletoc]titletoc[/url] packages to customise your sections and table of contents.

Option 2 is the obvious one to use first.


limit
logarithms
calculus
calculus computations

Hi guys,

Could you help me find this limit?

$ \lim_{x\rightarrow\infty}(e^x\plus{}x)^{1/x}$

Thanks in advance,

[quote="futurus"]Hi guys,

Could you help me find this limit?

$ \lim_{x\rightarrow\infty}(e^x \plus{} x)^{1/x}$

Thanks in advance,[/quote]

Let the limit be L(I am assuming it exists :P )

$ \lim_{x\rightarrow\infty}(e^x \plus{} x)^{1/x} \equal{} L$

$ \lim_{x\rightarrow\infty}\frac {\ln e^x \plus{} x}{x} \equal{} \ln L$

By L'Hospital,

$ \lim_{x\rightarrow\infty} \frac {e^x \plus{} 1}{e^x \plus{} x} \equal{} \ln L$

$ \lim_{x\rightarrow\infty} \frac {1 \plus{} e^{ \minus{} x}}{1 \plus{} xe^{ \minus{} x}} \equal{} \ln L$

$ \ln L \equal{} 1 \Rightarrow L \equal{} e$

P.S: Alternatively $ \lim_{x\rightarrow\infty}(e^x \plus{} x)^{1/x} \equal{} e\lim_{x\rightarrow\infty}(1 \plus{} xe^{ \minus{} x})^{1/x}$

$ L \equal{} \lim_{x \to \infty}(e^x \plus{} x)^{1/x} \equal{} \lim_{x \to \infty} (e^x)^{1/x}\left(1 \plus{} \frac{x}{e^x}\right)^{1/x} \equal{} e \cdot (1 \plus{} 0)^0 \equal{} e$


quadratics
number theory proposed
number theory

Determine all pairs $ (m,n)$ of natural numbers for which $ m^2\equal{}nk\plus{}2$ where $ k\equal{}\overline{n1}$.

EDIT. [color=#FF0000]It has been discovered the correct statement is with $ k\equal{}\overline{1n}$.[/color]

What does $ k\equal{} \overline{n1}$ mean?

That would be a number, but you overline it because it has unknown digits in it (n in here)
$ 10\equal{}\overline{10}$
You can't write $ n1$ you must write $ \overline{n1}$

Hmm, I wasn't sure about this, cause in the version orl sent to me, they write:

[b]Problem NT3.[/b]   Find all pairs of positive integers $ (m,n)$ such that $ m^2\equal{}nk \plus{}2$, where $ k$ is the number obtained from $ n$ by writing 1 on the [u]left[/u] side of its decimal representation.

Wouldn't that mean, that $ k\equal{} \overline{1n}$?  :huh:

Go to JBMO Shortlists in National Olympiads forum and see. I don't know

$ m^2 \equal{} n(10n \plus{} 1) \plus{} 2 \implies 40m^2 \minus{} 79 \equal{} (20n \plus{} 1)^2 \implies$
$ (40m \plus{} 20n \plus{} 1)(40m \minus{} 20n \minus{} 1) \equal{} 79$, possible factor pairs are (1,79) and (-1,-79), no solution.

Sorry, I've found out it's $ k\equal{}\overline{1n}$.

Your solution is wrong! test the numbers,  40 m^2 isn't a squre.

Having bust my brains out trying (and failing)  to do the thing as originally stated in JBMO with the mistake in it, I feel justified in reviving this thread to post a hopefully correct answer to the corrected problem, with $ k= \overline{1n} $, (rather than $ k= \overline{n1}$)

If  $n$ is a single digit, trying all the possibilities gives $(m,n) = (11,7)$ as a solution.

If $n$ is two digits, then $k=n+100$ so $nk=n^2+100n$
Since $100 = 0 mod(4)$, $nk$ must be the same as $ n^2 mod(4)$
which is either $0$ or $1 mod (4)$ since $n^2$ is a square.

So $nk + 2 = 2$ or $3 mod(4)$ , which cannot be a perfect square.
The same argument applies when n is more than 2 digits.
Hence $(m,n) = (11,7)$ is the only solution.

Merlin

(11,7) is apparently NOT a solution!
from the equation we get $(20n+1)^2-40m^2=-79$
let us consider the (OC's) equation
$x^2-40y^2=-79$
its solutions $x_1=9,x_2=411,x_{n+1}=38x_n-x_{n-1}$
hence it's 2-periodic modulo 20 with the period (9,11),so it cant be $20n+1$,yielding no solution to the original equation.

[quote="mavropnevma"]$(11,7)$ [b]is[/b] a solution, since $11^2 = 121 = 119 + 2 = 7\cdot 17 + 2$. The problem statement has been amended, from $k = \overline{n1}$ to $k = \overline{1n}$.[/quote]
oh gosh!the problem has been typed uncorrectly!
but the original problem can also be a divine one~

İf it is 1n, then the only solution is (m;n)=(11;7)


dayereye W va ghotr e AB az an mafroozand :
agar do dayereye C1.C2 bar yekdigar va W momas kharej bashand

(W,W1)=M
(W,W2)=N      va  (AN,W1)=X    (BM,W2)=Y    , (AN,BM)=Z

sabet konid X ,Y , Z hamkhat and.

horoofe rooye shekl fargh dare:
chon AB do sare ghotr and dar natije momas dar y movazie momas dar X e dar natije  X,Y,F(markaze tajanose C2,C3) ham khat and
chahar zelie FGCZ mohatie( zavie baazi) dar natije kafi ast sabet konim ke zavieye  FXC=180-CGZ  ke inam ba zavie bazi dar miaad.

khaheshan masaele hendesearo too marathone hendese bezarid Dge pas in marathon baraye chieW??? :?:


algebra
polynomial
calculus
derivative
function
domain
calculus computations

I practice some exercises of partial derivatives and in the one of these exercises I have to prove the result of the polynomial $ H(x,y)$ is always positive.

$ H(x,y)\equal{}x^{2}y^{4}\plus{}x^{4}y^{2}\minus{}3x^{2}y^{2}\plus{}1 \geq 0  \ \ \ x,y \in R$

Any hints on how to prove this ?

Direct from AM-GM:

$ \left( \frac{x^2y^4\plus{}x^4y^2\plus{}1}{3} \right) \ge \sqrt[3]{x^2y^4x^4y^21} \equal{} \sqrt[3]{x^6y^6} \equal{} x^2y^2.$

Or, if you feel like calculus...

We can look for extreme points of multivariable functions just as we do for single variable functions. The partial derivatives of $ H$ are $ H_x\equal{}2xy^4\plus{}4x^3y^2\minus{}6xy^2\equal{}2xy^2(y^2\plus{}2x^2\minus{}3)$ and $ H_y\equal{}4x^2y^3\plus{}2x^4y\minus{}6x^2y\equal{}2x^2y(2y^2\plus{}x^2\minus{}3)$
The critical points are where both of these derivatives are zero: the two axes, along with the four intersections $ (\pm 1,\pm 1)$ of those two ellipses. The critical values of $ H$ are $ 1$ at every point where $ x\equal{}0$ or $ y\equal{}0$ and $ 0$ at the points $ (\pm 1,\pm 1)$.
We're not done; we still have to account for the edges of the domain, in this case what happens as $ (x,y)$ goes off to $ \infty$ in any direction. Out there, note that we can rewrite $ H(x,y)\equal{}1\plus{}x^2y^2(x^2\plus{}y^2\minus{}3)$. If we're far enough out, $ x^2\plus{}y^2\minus{}3$ is positive, making the whole thing greater than $ 1$. Multiplying by $ x^2y^2$, it even goes to $ \infty$ unless we stay close to an axis. That can't be a minimum, so our global minimum value is $ 0$, achieved at $ (1,1)$ and its three reflections.


induction
complex numbers
advanced fields
advanced fields unsolved

Use mathematical induction to show that when n=2,3,...

Conj(z1*z2*z3*...*zn)=Conj (z1)*Conj (z2)*...*Conj (zn)

With this result. Let a0, a1, a2,..., an  denote real numbers, and let z be any complex number. Show that

Conj(a0+A1z+a2z^2+...+anZ^n)= a0 + a1 Conj(z)+ a2 Cong(z)^2+...+an Conj(z)^n

I understand the first property but I can only show it for Conj (z1*z2)= Conj ((ac-bd) +i (ad+cb)) = (ac-bd) - i(ad+cb) = ac-bd-adi-cbi= (a-bi) (c-di) = Conj(a+bi)*Conj (c+di)= Conj (z1)*Conj (z2). Letting z1= (a+bi) and z2=(c+di)

I'm not sure how to show it for an infinite amount of complex numbers. PLEASE HELP!!!

"Arbitrarily large finite number" and "infinite number" are two very different things.  All you have to do is show that the result for $ n$ numbers implies the result for $ n \plus{} 1$ numbers.

How do you express a product of $ n \plus{} 1$ numbers as a product of $ n$ numbers?

I'm not sure how to show that with complex numbers?!?

There is no need to think of them as complex numbers; the only place you use real and imaginary parts is in verifying that $ \overline{a\cdot b}\equal{}\overline{a}\cdot \overline{b}$. The generalization to more terms is completely abstract, and applies to any homomorphism.

Ok so...

conj (z1*z2*...*zn*zn+1)= conj (z1)*conj(z2)*...*conj(zn)*conj(zn+1)??

I don't see how that proves it?[/code]

OK- how do you multiply a bunch of numbers? All I see in the definition is the product of two numbers...

Complex numbers? (a+bi)(c+di)=(ac-bd)+i(ad+bc)...but it wants an large finite amount?

At this point, it doesn't matter that they're complex numbers and it doesn't matter what the result of the multiplication is - jmerry's asking what you are doing when you are multiplying a list of numbers.

Alright, let's go through an example of the induction step.  We know that $ \overline{ab} \equal{} \bar{a} \bar{b}$, correct?  Suppose I wanted to show that $ \overline{abc} \equal{} \bar{a} \bar{b} \bar{c}$.  What could I do?

Well, I know that $ \overline{bc} \equal{} \bar{b} \bar{c}$.  Suppose I let $ x \equal{} bc$.  Then $ \overline{abc} \equal{} \overline{ax} \equal{} \bar{a} \bar{x} \equal{} ...$?

it equals conj(a)*conj(b)*conj(c)...is there a way i could im u or something so i can talk about it?

No, you're almost already there!  Why does it equal $ \bar{a} \bar{b} \bar{c}$?  Because we already knew that $ \bar{x} \equal{} \bar{b} \bar{c}$.  In other words, the key here is [i]associativity.[/i]  When we're multiplying $ abc$, that's the same thing as doing $ bc$ first (and we know how conjugation works with respect to that) and then doing $ a$ times the result (and we know how conjugation works with respect to that), which tells us how conjugation works with respect to the entire product.  So all you have to do is replace $ 2$ numbers with $ n$ numbers and there's your induction step.

So like this?

Let n=zn*zn+1

Conj(z1*z2*...*n)=Conj(z1)*Conj(z2)*...*Conj(n)= Conj(z1)*Conj(z2)*...*Conj(zn)*Conj(zn+1)

That doesn't seem right...?

Please use a symbol other than $ n$. Duplicate notation is bad.

It might help to put some parentheses in an intermediate step. After all, the definition of the product of more than two numbers includes those parentheses.

So That's right Other than the few tweaks? that's math induction? Now hw should I start the next part?

Do I do the same thing?
Let b=anz^n * an+1z^n+1

Conj( a0+a1z+a2z^2+...b)= a0+ a1Conj(z) + a2Conj(z)^2+..+Conj(b)= a0+ a1Conj(z) + a2Conj(z)^2+...anConj(z)^n+ an+1 Conj (z)^n+1

You don't need the "let $ b$ equals" part unless you really want to.  Most people would be okay with just saying $ \overline{a_0 \plus{} a_1 z \plus{} ... \plus{} a_n z^n} \equal{} \overline{a_0} \plus{} \overline{a_1} \bar{z} \plus{} ... \plus{} \overline{a_n} \overline{z^n}$ and then applying the first result to conclude $ \overline{z^k} \equal{} \bar{z}^k$, although this is technically also an inductive argument.


I was reading the wikipedia article on the virial theorem
http://en.wikipedia.org/wiki/Virial_theorem
and I came to this part where it talks about why the time average of dG/dt might vanish:
[quote]"One often-cited reason applies to bound systems, i.e., systems that hang together forever. In that case, the virial G_bound is usually bounded between two extremes, G_min and G_max, and the average goes to zero in the limit of very long times τ"[/quote]
I don't understand why G is bounded => <dG/dt>=0 ?

I can't understand why [b]bounded system=>G is bounded[/b]
Given that however, the explanation seems sufficient.

<dG/dt> means time average over a long period.

<dG/dt> = {integrate dG/dt with respect to time from 0 to T]/T

when T --> infinity,
the numerator is finite because G(max) and G(min) exist and are finite quantities.

But the denominator is arbitrarily large.

So over a long period of time,
<dG/dt> = 0.


function
real analysis
real analysis unsolved

Find a function f, f'=0 (except possibly on a set of measure 0) and f is absolutly continuous, but f is not a constant function.

I was thinking the Cantor function, but it's not a.c. ...

If a function is absolutely continuous and singular, then it must be constant.  This is a well-known theorem.


if a+b+c=1, a,b,c>0, prove that

$ \frac{1\plus{}c(27ab\minus{}1)}{a\plus{}b} \leq 1\plus{}(a\plus{}b)^2 \plus{} 3c\plus{} \frac{c^3}{a\plus{}b}$

[hide]Since $ a\plus{}b\equal{}1\minus{}c$, that inequality is equivalent to
$ \dfrac{(1\minus{}c)\plus{}27abc}{1\minus{}c}\le 1\plus{}(1\minus{}c)^2\plus{}3c\plus{}\dfrac{c^3}{1\minus{}c}$.
Multiplying by $ (1\minus{}c)$, we get
$ (1\minus{}c)\plus{}27abc \le (1\minus{}c)\plus{}(1\minus{}c)^3\plus{}3c(1\minus{}c) \plus{} c^3$,
and combining like terms, we get
$ 27abc \le 1$.
We homogenize this equation to
$ 27abc \le (a\plus{}b\plus{}c)^3$,
which is equivalent to
$ \sqrt[3]{abc} \le \dfrac{a\plus{}b\plus{}c}{3}$, AM-GM.[/hide]

[quote="Mij"][hide]Since $ a \plus{} b \equal{} 1 \minus{} c$, that inequality is equivalent to
$ \dfrac{(1 \minus{} c) \plus{} 27abc}{1 \minus{} c}\le 1 \plus{} (1 \minus{} c)^2 \plus{} 3c \plus{} \dfrac{c^3}{1 \minus{} c}$.
Multiplying by $ (1 \minus{} c)$, we get
$ (1 \minus{} c) \plus{} 27abc \le (1 \minus{} c) \plus{} (1 \minus{} c)^3 \plus{} 3c(1 \minus{} c) \plus{} c^3$,
and combining like terms, we get
$ 27abc \le 1$.
We homogenize this equation to
$ 27abc \le (a \plus{} b \plus{} c)^3$,
which is equivalent to
$ \sqrt [3]{abc} \le \dfrac{a \plus{} b \plus{} c}{3}$, AM-GM.[/hide][/quote]

good job, actually is my solution too.


algebra
polynomial
linear algebra
matrix

Let $A$ be an n-square matrix for which $A^{k}=0$ for some $k>n$ . Show that $A^{n}=0$

Since powers are well defined we have $A^{n}=A^{k}A^{n-k}=0_{n}A^{n-k}=0_{n}$.

Let's see if I can do this without quoting Cayley-Hamilton. As elementary as possible:

Since $A^{k}=0,$ then $A$ is singular and has $0$ as an eigenvalue. Find an $x\ne0$ such that $Ax=0,$ and extend that to a basis. Make that basis the columns of a matrix $P_{1}.$ Then in block-structured form,

$P_{1}^{-1}AP_{1}=\begin{bmatrix}0&*\\0&A'\end{bmatrix}.$

Raise that to the $k$th power, and we conclude that $(A')^{k}=0.$ Continue this inductively. We ultimately show that for some $P$ we have $P^{-1}AP=B,$ where $B$ is strictly upper triangular - that is, all of the elements on or below the main diagonal are zero. Then by direct calculation, $B^{n}=0$ and hence $A^{n}=0.$

(Note that we nowhere used that $k>n.$ It's just that, for a reason related to what HilbertThm90 said, the question isn't very interesting otherwise.)


integration
limit
real analysis
real analysis unsolved

Let $ F(t) \equal{} \int_0^t \frac{dx}{1\minus{}x^\alpha}$, $ (\alpha \neq 0)$. Find $ lim _{t \rightarrow 0} F(t)$

Hi :) !!

i think that $ \lim_{t \to 0} F(t) \equal{} 0 \ if \ \alpha \equal{} 1 \ \ and \ \ \equal{} \frac {1}{1 \minus{} \alpha}\minus{} \frac{1}{1\minus{}\alpha} \equal{} 0 \ if \ not$

than $ \lim_{t \to 0}F(t)\equal{}0 \ \forall \alpha \neq 0$

limit always is equal to $ 0$
since:  let $ t<1/2$
\[ 0 \leq  \int_{0}^{t}\left|  \frac{1}{1\minus{}x^{a}} \right| dx \leq Const * t \to 0\]

[quote="Extremal"]limit always is equal to $ 0$
since:  let $ t < 1/2$
\[ 0 \leq \int_{0}^{t}\left| \frac {1}{1 \minus{} x^{a}} \right| dx \leq Const * t \to 0
\]
[/quote]

Why is it true : $ \int_{0}^{t}\left| \frac {1}{1 \minus{} x^{a}} \right| dx \leq Const * t$??? :maybe:  Could you explain?

since $ 0<t<1/2$ hence $ \left| \frac{1}{1\minus{}x^{a}} \right| \leq \frac{1}{\min_{x \in [0,1/2]}|1\minus{}x^{a}|}$
but $ \min_{x \in [0,1/2]}|1\minus{}x^{a}| >C>0$ since $ x^{a}$ can't be very close to $ 1$, if yes then $ x \approx 1$ but $ x<1/2$
thus we have
$ \frac{1}{\min_{x \in [0,1/2]}|1\minus{}x^{a}|} < \frac{1}{C}$


geometry
geometric transformation
reflection
rotation

how many ways are there to arrange 2 gold keys, 2 silver keys, and 2 bronze keys on a ring? (reflection, rotation not included)

the method I used was
opp opp opp 1
adj adj adj 1
opp mid mid 3
opp adj adj 3
adj mid mid 3

1+1+3+3+3=11

but I don't know how to get this mathematically.

[quote="13375P34K43V312"][hide]It would be $\frac{{6\choose2}{4\choose2}}{6}=15$ I think.[/hide][/quote]Don't forget, reflections are not counted either.

EDIT:
[hide]Your answer overcounts these four cases, which are reflections of each other but not rotations of each other.

AABCBC / AACBCB (3 different cases)
AABBCC / AACCBB (1 case)

$15-4=11$[/hide]


Consider the sequence \[ 1, \minus{} 2,3, \minus{} 4,5, \minus{} 6,\ldots,\] whose $ n$th term is $ ( \minus{} 1)^{n \plus{} 1}\cdot n$. What is the average of the first $ 200$ terms of the sequence?

$ \textbf{(A)}\minus{}\!1\qquad \textbf{(B)}\minus{}\!0.5\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 0.5\qquad \textbf{(E)}\ 1$

[hide]

Every two terms, the sum decreases by $ 1$. For example, on the first two terms, the sum is $ \minus{}1$, the first four terms $ \minus{}2$... So, the sum of the $ 200$ terms is $ \minus{}200/2\equal{}\minus{}100$. Hence, the average is $ \minus{}0.5\Longrightarrow\mathrm{ (B) \ }$

[/hide]

[hide="Click for solution"]
The $ 200$ terms can be grouped into $ 100$ odd-even pairs, each with a sum of $ \minus{}1$>  Thus the sum of the first $ 200$ terms is $ \minus{}1 \times 100\equal{}\minus{}100$, so the average of the first $ 200$ terms is $ \minus{}\frac{100}{200}\equal{}\minus{}0.5$, or $ \boxed{\textbf{(B)}}$.
[/hide]

[hide]

When the sequence has an even number of terms, two adjacent numbers in the series can be grouped together to attain the sum of $ \minus{}1$. Therefore, when there are $ n$ numbers in the sequence, the sum will be $ \minus{}\frac{n}{2}$, and the average will be:

$ \frac{\frac{\minus{}1}{2}}{n}\equal{}\minus{}\frac{1}{2}$

[/hide]

[hide]Because the even one is always one less than the odd one, so the summary of the first 200 terms is -100, and then -100/200 = -0.5 :starwars: [/hide]
 :surf:  :ddr:  :gathering:  :jump:  :yoda:

For all even $n$ the average of all in the interval $[a_{0}, a_{n}]$ is $-0.5$ by pairing


superior algebra
superior algebra unsolved

i am a first year student doing analysis and was wondering if someone could help prove

given 2 real number a and b, and that there exsist a number x such that ax = b
if b /=(cannot equal) 0, explain why b/0 is not defined.

because it would be a solution to a of a.0=b, while a.0=0 always?

I probably misunderstood your question, since we saw this in primary school...

For being really able to answer this question, you need a definition of the reals at first.


calculus
function
trigonometry
conics
parabola
geometry
geometric transformation

Okay, so my friend is in Calculus AB and she has this project to do... here's the instructions:

Make a physical model of a solid with a known cross-section on a base with a standard function. The following guidelines apply:
1. The base functions can be any non-linear function [b]except [/b]a square root, or absolute value. No trig functions. (she also said that you can't do a parabola)
2. The cross-section can be any shape [b]except [/b]a square
3. The materials must be no less than .5 cm thick and no greater than 1 cm thick. Your completed model must be between 12 and 24 cm long. There must be at least 24 laminations. The computations in your write up must reflect the width of the material you shoose to use.
4. The model must be solidly constructed. (Do not use glue sticks or tape to construct your model)
[b]NOTE: You will be required to compute the volume of the solid without a calculator. Pick your base functions carefully.[/b]

So yeah, my friend said she needs 2 base functions.. and they have to be complex, but I can't figure out ones that are complex where you can calculate the intersections to take the integral without a calculator..  :(  So if you have any ideas, please help.

[quote="ambierona"]but I can't figure out ones that are complex where you can calculate the intersections to take the integral without a calculator.[/quote] Why don't you take a cubic polynomial? Or $e^x$? Or $\frac 1{x^2}$? They seem to satisfy all the restrictions. Or, maybe, I'm missing something?  :?

She also said you can't have the 2 functions be mirror images of eachother (like $e^x$ and $-e^x$), and I was thinking $e^x$, and cubic polynomials, but I couldn't figure out another equation with easy intersection points that I would be able to calculate without a calculator.  :(  Like can you give an example if you thought of one?

Let me see if I understood you right: I'm still confused a bit about what exactly you are asking. You need two functions $f$ and $g$ whose graphs intersect at two easy to calculate points, correct? Then you may just take $f(x)=31x-7x^3$ and $g(x)=\frac{24}{x^2}$. The intersection points (with $x>0$) are $(1,24)$ and $(2,6)$. In general, you can choose the intersection points first and then choose the coefficients in your equations to make the graphs pass through the points chosen. It'll give you a linear system to solve. The example I proposed is, probably, even not the simpest possible one.