Compute the number of positive integers less than 1,000,001 that are either a perfect square or a perfect cube, but not both.

Solution (in spoiler):

[hide]We use PIE. The number of perfect squares is 1,000. The number of perfect cubes is 100. The number of perfect <insert term of 6th power here> is 10. The answer is 1,000+100-10 = 1090.[/hide]

[color=cyan]Close but not quite.  Re-read the question.[/color]

Aww, i really shouldn't be posting, but:

Fierytycoon:

[hide]The problem with your solution is that in the question, a perfect 6th cannot be in your answer. For example, 64 is not one of the answres. So 64 needs to be subtracted from both teh 1000 perfect squares AND the 100 perfect cubes.

Now, do u see wat u did wrong?[/hide]

Oh, silly me.

[hide]So I should have subtracted 10 twice, resulting in 1080.[/hide]


Consider $F(x) = x^{x^{3}}$.

Obviously $F(1) = 1$ and $F(x)$ grows with $x$ without bound.

Find at least one solution for $x^{x^{3}}= 3$ for $x > 1$.

Do show the method of solving this.

${f(x)=x^{x^{3}}}$

${x^{x^{3}}=3}$

${x=\sqrt[3]{3 u}}$

${f\left(\sqrt[3]{3 u}\right)=3}$
${3^{u}u^{u}=3}$

${u=1}$

The solution is the 3 cubic roots of 3

it have another solution. but to fint them is need to use the Lambert W-Function

http://mathworld.wolfram.com/LambertW-Function.html


http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=17
Click one of the available years, and then there is a pdf option. :)
This is the short list though.

[quote="arkhammedos"]PLEASE FRIEND where can i find  all imo shirt and long list but in forma
pdf
thnks[/quote]

PLEASE don't use chatspeak, FRIEND.


[hide]$\log_{10} \sin x + \log_{10} \cos x = -1 \implies \log_{10}\sin x\cos x=-1$

$\implies \sin x \cdot \cos x = \frac{1}{10} \clubsuit$

$\log_{10}(\sin x + \cos x) = \textstyle \frac{1}{2} (\log_{10} n - 1)$

$\implies \log_{10}(\sqrt{n}) - \log_{10}(\sin x + \cos x)=\frac{1}{2}$

$\log_{10}(\frac{\sqrt{n}}{(\sin x + \cos x)}) = \frac{1}{2}$

$\implies \frac{\sqrt{n}}{(\sin x + \cos x)}=\sqrt{10}$

From result $\clubsuit$

$\implies \frac{n}{\sin^2 x + 2\sin x \cos x + cos^2 x} =10$

$\implies n = 12$

is my solution correct? ;)[/hide]

[hide]

Combining logs, we get $\log_{10} \sin x\cos x = -1$, so $\sin x\cos x = \frac{1}{10}$.

Multiply both sides of the second equation by 2 and use log properties to get $\log_{10} (\sin x + \cos x)^2 = \log_{10} n-1$, so combining the 1 with the RHS log, eliminating the logs,  and expanding we see that  $\sin^2 x + \cos^2 x + 2 \sin x\cos x = \frac{n}{10}$.

Using the Pythagorean trig identity and the earlier result, we see that $\frac{6}{5}=\frac{n}{10}$, so that $n=12$.

[/hide]

Whoops. I got really confused because I thought that it was $\log_{10} (n-1)$. But yeah, similar method with that problem, only n comes out to be fractional, so I knew I went wrong somewhere.

[quote="nsato"]Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \textstyle \frac{1}{2} (\log_{10} n - 1)$, find $n$.[/quote]

[hide]combine and simplify the first equation to get:

$\sin\cos=\frac{1}{10}$

multiply the second by 2 and get rid of the logs...

$1+2\sin\cos=\frac{1}{10}*n$
$10*(1+\frac{2}{10})=12=n$[/hide]

log sin x+ log cos x=-1
log( sin x .cos x)=-1
sin x. cos x=1/10
log( sin x+ cos x)=1/2{ log n - 1}
log( sin x+ cos x)=1/2{ log n - log 10}
log( sin x+ cos x)=log (√n/10)
sin x+ cos x=√(n/10)
(sin x+ cos x)²={√(n/10)}²
sin²x+2sin x cos x+cos²x=n/10
1+ 2/10=n/10
12/10= n/10
n=12


in a triangle, the altitude, angle bisector, and the median from angle C divide angle C into four congruent angles.  find angle C

let <c/4= s . let the altitude,bisector and median be CH,CD,and CM.
notice that CD should lie between CH and CM.
BM/MA = BC/AC . sin s/sin 3s
but BC/CA=sinA/sinB =cos s /cos 3s
hence cos s.sin s =cos 3s.sin 3s and therefore sin2s=sin6s
and therefore c=90.


do we solve this??
 is it  by applying the formula :
k=r'(t)xr''(t) over ||r'(t)||^3

Ah, quick question, a bit off topic, sorry, but I would like to know why I keep seeing problems posted from some sort of webpage. They look like a quiz and I would like to take it.

This is from an online homework/quiz service that many intro calculus classes (and physics and other subjects like that) use at large universities. You probably can't do them because the universities pay those companies and only students taking classes at universities that have paid are likely to be able to get access.


well, i'm up pretty early...and some of you might be waking up right now to get ready to fly to dallas, see you guys there!!!!!

i think i need to get an early start on packing... :P

I packed yesterday, and was unhappy; I wanted to pack on Thursday.


inequalities
geometry
circumcircle
inradius
function
trigonometry
geometric transformation

Let $ABC$ be a triangle, let $O$ be its circumcenter, and let $H$ be its orthocenter.

Let $P$ be a point on the segment $OH$.

Prove that

$6r\leq PA+PB+PC\leq 3R$,

where $r$ is the inradius and $R$ the circumradius of triangle $ABC$.

[b]Moderator edit:[/b] This is true only if the point $P$ lies inside the triangle $ABC$. (Of course, this is always fulfilled if triangle $ABC$ is acute-angled, since in this case the segment $OH$ completely lies inside the triangle $ABC$; but if triangle $ABC$ is obtuse-angled, then the condition about $P$ lying inside the triangle $ABC$ is really necessary.)

The second part can be reduced to proving $OA+OB+OC\le 3R$ (obvious :)) and $HA+HB+HC\le 3R$, because $XA+XB+XC$ is a convex function, so it reaches its maximum in the endpoints of the segment on which we consider it. This is because for any $P$ on [$OH$] we have $\|\overline{PA}\|=\|t\overline{OA}+(1-t)\overline{HA}\|\le t\|\overline{OA}\|+(1-t)\|\overline{HA}\|$ (with $t\in [0,1]$).

By letting $d_a,d_b,d_c$ be the distances from $O$ to the sides $BC,CA,AB$, we have $HA=2d_a$ and the like. We also have $3R=OA+OB+OC\ge 2(d_a+d_b+d_c)$ (the Erdos-Mordell inequality), and since the right part is $HA+HB+HC$, we are done.

I guess this leaves the first inequality.

[color=red][Moderator edit: I fear this proof is not enough, since it doesn't work for the case when the triangle ABC is obtuse-angled, and the point P lies on the segment OH and inside triangle ABC.][/color]

I just want to have a remark :
     For any point P lying inside ABC we have $PA + PB + PC \geq 6r$
     I 'll consider later the necessity of the condition "lying inside ABC".  :D

[quote="prowler"]Let $ABC$ be a triangle, let $O$ be its circumcenter, and let $H$ be its orthocenter.

Let $P$ be a point on the segment $OH$.

Prove that

$6r\leq PA+PB+PC\leq 3R$,[/quote]

Strange, nobody has posted a complete solution until now.

Well, let's first deal with the left inequality.

[quote="treegoner"]     For any point P lying inside ABC we have $PA + PB + PC \geq 6r$
     I 'll consider later the necessity of the condition "lying inside ABC".  :D[/quote]

Of course, this condition is unnecessary since the minimum of PA + PB + PC is taken inside (or on the boundary) of triangle ABC. Anyway, there is no need to distinguish the cases of P lying inside and outside triangle ABC. One can handle all positions of P at once. Here is one of the possible ways to do this:

We have to prove:

[color=blue][b]Theorem 1.[/b] Let r be the inradius of a triangle ABC. For any point P in the plane of the triangle ABC, the inequality $PA+PB+PC\geq 6r$ holds, with equality if and only if the triangle ABC is equilateral and the point P is its center.[/color]

[i]Proof.[/i] Theorem 1 will be a direct corollary of the following two theorems:

[color=blue][b]Theorem 2.[/b] If $\Delta$ is the area of triangle ABC, and $W=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)}$, then $PA+PB+PC\geq W$. Equality holds if and only if the point P is the Fermat point of triangle ABC, and all angles of triangle ABC are $\leq 120^{\circ}$.[/color]

[color=blue][b]Theorem 3.[/b] The inequality $W\geq 6r$ holds. This inequality becomes an equality if and only if the triangle ABC is equilateral.[/color]

[i]Proof of Theorem 2.[/i] Let BCX be the equilateral triangle erected outwardly on the side BC of triangle ABC. Then, by the Ptolemy inequality, applied to the quadrilateral PBXC, we have $PB\cdot CX+PC\cdot BX\geq PX\cdot BC$. But since the triangle BCX is equilateral, we have CX = BX = BC, and thus, we get $PB+PC\geq PX$. Hence, $PA+PB+PC\geq PA+PX$. But after the triangle inequality for the (possibly degenerate) triangle APX, we have $PA+PX\geq AX$. Thus, $PA+PB+PC\geq AX$.

Now, after the Cosine Law in triangle ABX, we find

$AX^2=AB^2+BX^2-2\cdot AB\cdot BX\cdot\cos\measuredangle ABX$.

Since the triangle BCX is equilateral, BX = BC = a and < CBX = 60°; the latter implies < ABX = < ABC + < CBX = B + 60°. Thus, we get

$AX^2=AB^2+BX^2-2\cdot AB\cdot BX\cdot\cos\measuredangle ABX=c^2+a^2-2\cdot c\cdot a\cdot\cos\left(B+60^{\circ}\right)$
$=c^2+a^2-2\cdot c\cdot a\cdot\left(\cos B\cos 60^{\circ}-\sin B\sin 60^{\circ}\right)$
$=c^2+a^2-2ca\cos B\cos 60^{\circ}+2ca\sin B\sin 60^{\circ}$.

But by the Cosine Law in triangle ABC, we have $2ca\cos B=c^2+a^2-b^2$, and by a well-known formula for the area of a triangle, the area $\Delta$ of triangle ABC can be written as $\Delta=\frac12 ca\sin B$, so that we have $2ca\sin B=4\Delta$. Thus,

$AX^2=c^2+a^2-2ca\cos B\cos 60^{\circ}+2ca\sin B\sin 60^{\circ}$
$=c^2+a^2-\left(c^2+a^2-b^2\right)\cos 60^{\circ}+4\Delta\sin 60^{\circ}$
$=c^2+a^2-\left(c^2+a^2-b^2\right)\cdot\frac12+4\Delta\cdot\frac{\sqrt3}{2}=\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)$,

so that

$AX=\sqrt{AX^2}=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)}=W$.

And therefore, the inequality $PA+PB+PC\geq AX$ becomes $PA+PB+PC\geq W$. The consideration of the equality case is left to the reader. Theorem 2 is proven.

[i]Proof of Theorem 3.[/i] First, recall the inequality $\Delta\geq 3\sqrt3r^2$. This inequality is probably well-known, but I couldn't find it here, so here is a proof:

Let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. We have to show the inequality $\Delta\geq 3\sqrt3r^2$. By a well-known formula for the area of a triangle, $\Delta=rs$; thus, this inequality becomes $rs\geq 3\sqrt3r^2$, what simplifies to $s\geq 3\sqrt3r$. But by another well-known formula, $r=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$; hence, the inequality that we have to prove takes the form

$s\geq 3\sqrt3\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$;

upon squaring, this becomes

$s^2\geq 27\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$.

Multiplying this with $\frac{s}{27}$, we get

$\frac{s^3}{27}\geq\left(s-a\right)\left(s-b\right)\left(s-c\right)$.

Finally, since

(s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s,

we have $\frac{s^3}{27}=\left(\frac{s}{3}\right)^3=\left(\frac{\left(s-a\right)+\left(s-b\right)+\left(s-c\right)}{3}\right)^3$, and the inequality in question becomes

$\left(\frac{\left(s-a\right)+\left(s-b\right)+\left(s-c\right)}{3}\right)^3\geq\left(s-a\right)\left(s-b\right)\left(s-c\right)$.

But this trivially follows from AM-GM, applied to the positive numbers s - a, s - b, s - c. Thus, the inequality $\Delta\geq 3\sqrt3r^2$ is proven.

Now, by the Weitzenböck inequality (see http://www.mathlinks.ro/Forum/viewtopic.php?t=14388 or http://www.mathlinks.ro/Forum/viewtopic.php?t=8059 ), we have $a^2+b^2+c^2\geq 4\sqrt3\Delta$. Thus,

$a^2+b^2+c^2+4\sqrt3\Delta\geq 4\sqrt3\Delta+4\sqrt3\Delta=8\sqrt3\Delta$.

Since $\Delta\geq 3\sqrt3r^2$, we thus have $a^2+b^2+c^2+4\sqrt3\Delta\geq 8\sqrt3\cdot 3\sqrt3r^2 =72r^2$. Hence,

$W=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3\Delta\right)}\geq\sqrt{\frac12\cdot 72r^2}=\sqrt{36r^2}=6r$,

and the inequality $W\geq 6r$ is proven. In order to obtain the case of equality, we note that all inequalities used in the above proof become equalities if and only if the triangle ABC is equilateral, so that the same holds for the inequality $W\geq 6r$. Thus, the proof of Theorem 3 is complete.

As Theorem 2 and Theorem 3 are established now, we thus have shown Theorem 1.

Now to the right inequality:

[color=blue][b]Theorem 4.[/b] Let O be the circumcenter, H the orthocenter and R the circumradius of a triangle ABC. Let P be an arbitrary point lying on the segment OH and inside the triangle ABC. Then, the inequality $PA+PB+PC\leq 3R$ holds, with equality if and only if the point P is the circumcenter of triangle ABC.[/color]

[i]Proof of Theorem 4.[/i] There are two cases to be considered:

[i]Case 1.[/i] The triangle ABC is acute-angled or right-angled.
[i]Case 2.[/i] The triangle ABC is obtuse-angled.

First we consider [i]Case 1[/i]. This case was handled by Grobber in his post #2; here is just a slightly different proof:

We first prove that $HA+HB+HC\leq 3R$. In fact, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #3 Lemma 1, we have HA = 2R cos A, HB = 2R cos B and HC = 2R cos C (since the triangle ABC is acute-angled or right-angled, so that all of its angles are acute or right), and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21188]a well-known inequality for triangles[/url] states that $\cos A+\cos B+\cos C\leq\frac32$; thus,

$HA+HB+HC=2R\cos A+2R\cos B+2R\cos C$
$=2R\cdot\left(\cos A+\cos B+\cos C\right)\leq 2R\cdot\frac32=3R$,

proving the inequality $HA+HB+HC\leq 3R$. Also, trivially, $OA+OB+OC\leq 3R$ since OA = OB = OC = R (in fact, O is the center and R is the radius of the circumcircle of triangle ABC).

Since the term PA + PB + PC, as a function of the point P, is convex (see Grobber's post #2 for an explanation of this), and a convex function on a segment attains its maximum at an endpoint of the segment, we can notice that for any point P on the segment OH, we have either $PA+PB+PC\leq OA+OB+OC$ or $PA+PB+PC\leq HA+HB+HC$. But on the other hand, we know that $OA+OB+OC\leq 3R$ and $HA+HB+HC\leq 3R$. Hence, $PA+PB+PC\leq 3R$. This proves Theorem 4 in the Case 1 (determination of the case of equality is left to the reader).

Now let's look at [i]Case 2[/i]. Actually, it seems that the proposer of the problem didn't mean this case to be considered at all, but still Theorem 1 is true in this case, so why not prove it :) .

Remember that in Case 2, the triangle ABC is obtuse-angled, so both its orthocenter H and its circumcenter O lie outside the triangle ABC. We WLOG assume that the obtuse angle of triangle ABC is the angle A; then, the orthocenter H of triangle ABC lies in the halfplane with respect to the line BC which contains the point A, while the circumcenter O of triangle ABC lies in the halfplane with respect to the line BC which doesn't contain the point A.

Since the point P lies on the segment OH and inside the triangle ABC, the segment OH must cross the interior of triangle ABC; thus, since both points O and H lie outside the triangle ABC, the segment OH must intersect the boundary of triangle ABC in two points. Let Q be the one of these two points which is nearer to the point H. Then, the part of the segment OH lying inside the triangle ABC must be contained in the segment OQ. Hence, the point P, lying on the segment OH and inside the triangle ABC, must lie on the segment OQ.

The point Q lies on the boundary of triangle ABC, i. e. on one of its sides BC, CA, AB. Clearly, the point Q cannot lie on the side BC (else, the point P, lying on the segment OQ, would lie in the same halfplane with respect to the line BC as the point O, but this halfplane doesn't contain the point A, and thus, an interior point of triangle ABC cannot lie in this halfplane). Hence, the point Q must lie on one of the sides CA and AB of triangle ABC. We can WLOG assume that the point Q lies on the side CA.

We have to prove the inequality $PA+PB+PC\leq 3R$.

Since the term PA + PB + PC, as a function of the point P, is convex, and a convex function on a segment attains its maximum at an endpoint of the segment, we can notice that, since the point P lies on the segment OQ, we have either $PA+PB+PC\leq OA+OB+OC$ or $PA+PB+PC\leq QA+QB+QC$. Hence, in order to prove the inequality $PA+PB+PC\leq 3R$, it will be enough to show that $OA+OB+OC\leq 3R$ and $QA+QB+QC\leq 3R$. Well, proving $OA+OB+OC\leq 3R$ is as trivial as it was in Case 1, so we only need to prove $QA+QB+QC\leq 3R$.

Since the point Q lies on the side CA of triangle ABC, we have QA + QC = CA, so that QA + QB + QC = (QA + QC) + QB = CA + QB; thus, the inequality which we must prove, $QA+QB+QC\leq 3R$, simplifies to $CA+QB\leq 3R$.

Now, note that the term QB, as a function of the point Q, is convex, and a convex function on a segment attains its maximum at an endpoint of the segment; thus, since the point Q lies on the segment OH, we have either $QB\leq OB$ or $QB\leq HB$. Thus, in order to establish the inequality $CA+QB\leq 3R$, it is enough to show that $CA+OB\leq 3R$ and $CA+HB\leq 3R$.

Proving $CA+OB\leq 3R$ is very easy: Since O is the center and R is the radius of the circumcircle of triangle ABC, we have OB = R, so that the inequality $CA+OB\leq 3R$ becomes $CA+R\leq 3R$, what simplifies to $CA\leq 2R$, what is trivial, since the segment CA is a chord in the circumcircle of triangle ABC and thus not longer than its diameter 2R.

Remains to show that $CA+HB\leq 3R$. Using the formulas CA = 2R sin B (Extended Sine Law) and HB = 2R cos B (from http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #3 Lemma 1, since the angle B is acute), this inequality becomes $2R\sin B+2R\cos B\leq 3R$, what, upon division by 2R, becomes $\sin B+\cos B\leq\frac32$, but this is obvious, since $\sin B+\cos B=\sqrt2\cdot\sin\left(B+45^{\circ}\right)\leq\sqrt2<\frac32$.

So the proof of Theorem 4 is complete.

Now, wasn't this all an eaaaasy exercise? ;)

  Darij

Indeed, Darij, it is not a difficult inequality. Here is my solution to the left inequality (the right inequality, mine is similar to yours )
     Let $A'B'C'$ be the antimedian triangle of $ABC$. By applying Ptolemei 's inequality to $PBCA'$ we obtain
        $PB.CA' + PC. BA' \geq PA'.BC \geq BC(AA' - PA) \geq a(2m_a - PA)$
     Since $CA'=c, BA' = b$, we obtain that
           $PB.c + PC.b + a.PA \geq 2am_a$
     Similarly, we obtain $PA.b + PB.a + c.PC \geq 2cm_c$, $PC.a + PA.c + b.PB \geq 2bm_b$. Therefore, by summing up, we have $(PA+PB+PC)(a+b+c) \geq 2(am_a+bm_b+ cm_c) \geq 12S$. Hence $PA+PB+PC \geq 6r$. ;)

[quote="zygmunt"][quote="treegoner"]
     Let $A'B'C'$ be the antimedian triangle of $ABC$. [/quote]

What is antimedian triangle. What is it looks like??[/quote]

It's actually called antimedia[b]l[/b] triangle. The [b]antimedial triangle[/b] of a triangle ABC is defined as the triangle enclosed by the parallels to the lines BC, CA, AB through the points A, B, C, respectively. Equivalently, it is the image of the triangle ABC under the homothety whose center is the centroid of triangle ABC and whose factor is -2.

  Darij

Grobber's proof is also my proof. Actually this proof is a well-known one. I saw it in a magazine some time ago. ;)

How do we prove that $\sum a m_a \geq 6S$?

I searched the forum but I failed to find it (i only found inequalities with $\frac{a}{m_a}$, $\frac{m_a}{a}$ etc.)

[hmm... that was fast. sometimes the easiest things are the hardest (for me)]

[quote="perfect_radio"]How do we prove that $\sum a m_a \geq 6S$?[/quote]

Let $h_a$ be the altitude of triangle ABC from the vertex A. Then, $m_a\geq h_a$, as in a right-angled triangle, the hypotenuse is longer than any catet. But by a well-known formula for the area S of the triangle ABC, we have $2S=ah_a$; thus, $\sum am_a\geq\sum ah_a=\sum 2S=6S$.

  Darij

[quote="prowler"]Prove that

$6r\leq PA+PB+PC\leq 3R$[/quote]

The left part of the inequality is just Erdös-Mordell.

I'm sure the second inequality has been posted a few times on this forum, maybe I even posted it myself a long time ago, but I can't seem to find it right now.

But Erdös-Mordell states that $2(PX+PY+PZ)\leq PA+PB+PC$ where $X, Y, Z$ are the feet of the perpendicular from $P$ to the sides.
How do I prove that $3r\leq PX+PY+PZ$?

Just observe that:
a) $\frac{3}{2}(\sum{PA})\geq\sum{PA}$+$\sum{PX}\geq{\sum{h_a}\geq{9r}}$.

For b): The relation is true for $P=O$ and $P=H$ , therefore it is true $\forall{P}\in[HO]$  :lol:

[b]Happy Easter Holidays![/b]

[quote="Sailor"]For b): The relation is true for $P=O$ and $P=H$ , therefore it is true $\forall{P}\in[HO]$  :lol: [/quote]

Let $S$ be the set of points on the segment $OH$. We have that the function \[ f: S \to \mathbb{R}: P \mapsto PA + PB + PC  \] is convex :) and that implies the result.


[hide](x+1)(x+1)=x^2+2x+1.
$(x^{2}+2x+1)(x+1)=x^{3}+3x^{2}+3x+1$
Just kind of foil it out except go x^2 times the other 2 and then 2x times the other two and then 1 times the others to get the final answer... [/hide]I hope this was clear enough, although it was probably nowhere near...

Instead of using binomial expansion (which is probably better in this case), we can just multiply it out:
[hide] \[(x+y)^{3}=(x+y)(x+y)(x+y)=((x+y)(x+y))(x+y)=(x^{2}+xy+yx+y^{2})(x+y)=(x^{2}+2xy+y^{2})(x+y)=x^{3}+2x^{2}y+xy^{2}+x^{2}y+2xy^{2}+y^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\] [/hide]

[hide]When expanding a term like this, it may be quickest to multiply it out, but with larger expansions, use pascals triangle.

So since the 3rd row in the triangle [well the 4th row, but row #3], is 1-3-3-1, those are the cooeficients.  so we have $x^{3}+3x^{2}+3x+1$[/hide]

[quote="kstan013"][quote="10000th User"][quote="markat619"]x^3 + 3x + 3y + y^3 by Pascal's triangle.[/quote]This is incorrect. Please check The QuattoMaster 6000's post[/quote]There's only one problem with it: there are y's instead of 1's![/quote]

Um, no...  It still isn't correct if it has ys instead of ones.

[quote="MathStudent2"]How do I expand this:  (x + 1)^3

Thanks in advance![/quote]
[hide]
i know that there have been many answers posted already, but im just going to give a short and simple answer to this question:

$(x+1)^{3}$ $=$

$(x+1)(x+1)(x+1)$ $=$

take out $(x+1)(x+1)$ first. solve it: $x^{2}+1x+1x+1$ which equals

$x^{2}+2x+1$ . then do $(x^{2}+2x+1)(x+1)= x^{3}+x^{2}+2x^{2}+2x+1x+1$[/hide]

your answer: [hide]$x^{3}+3x^{2}+3x+1$[/hide]

You solve it by just multiplying everything, just like expanding (x+1)²

[hide](x+1)² = x²+2x+1

(x²+2x+1)(x+1) = x³+3x²+3x+1

So, (x+1)(x+1)(x+1) = x³+3x²+3x+1[/hide]

:)


calculus
integration
calculus computations

This is less than $\sum_{n=1}^{\infty}\frac1{n^{2}}+\sum_{m=1}^{\infty}\frac1{m^{2}},$ so it converges.

Side note: some very interesting tests - the integral test and the condensation test - have as a hypothesis that the terms of the series decrease. Since these terms are not monotone, this series lies outside the scope of those tests.

[quote="Nea"]Study convergence:
$\sum_{n = 1}^\infty{a_{n}}$ where $a_{n}= \left\{{\begin{array}{*{20}c}{\frac{1}{n}}&{\text{for }n = m^{2}}\\{}\\{\frac{1}{{n^{2}}}}&{\text{for }n \ne m^{2}}\\ \end{array}}\right.m \in N$[/quote]
convergence is simpl question. whats ebout: Find this sum? i know it's $>$ than $\frac{\pi^{2}}{3}$ and $<$ $lnn$. but I don't said news :wink:

Actually, $\sum_{n=1}^\infty a_{n}= \sum_{n=1}^\infty \frac1{n^{2}}+\sum_{n=1}^\infty \left( \frac1{n^{2}}-\frac1{n^{4}}\right)$.
For example, $\sum_{n=1}^{9}= \mathbf{\frac11}+\frac14+\frac19+\mathbf{\frac14}+\frac1{25}+\ldots+\frac1{64}+\mathbf{\frac19}$ and then write the bolded parts, like $\frac19$, as $\frac1{81}+\left( \frac19-\frac1{81}\right)$.
We discussed series of the form $\sum_{n=1}^\infty \frac1{n^{2k}}$ like a million times.

PS: I really hope I didn't mess it up.


Hi,
I have two problems that make me confuse to solve. Can you help me? Thanks a lot.

[b]Problem 1:[/b] A group of  16 people in which there are 2 couples of husband and wife. We want to choose 8 people to take photograph. How many ways can we choose if in each picture there is exactly 1 couple of husband and wife?

[b]Problem 2:[/b] How many different numbers consisting of three digits are there such that in each number the digits are increasing or decreasing.

[hide="Problem 1"]
Okay, so there are 2 ways to choose a couple, and then we have $ \binom{12}{6}$ ways to choose the rest of the people. Thus the answer is $ 2 \cdot \binom{12}{6} \equal{} 1848$.
[/hide]

[size=150]Dear [b]isabella2296[/b]

Thank you very much for your replies. At first, I solve this problem as yours. And then I wonder if I missed some cases. For example, the case there is one couple of husband and wife and a husband or a wife of the other couple. So I resolve it and I get two solutions that make me confuse.

Solution 1:
I solve the problem directly

Choose a couple of husband and wife + 6 people left (as yours): $ 2 \cdot \binom{12}{6}\equal{} 1848$
Choose a couple of husband and wife + a husband or a wife (of the other couple) + 5 people left: $ 2 \cdot 2\cdot \binom{12}{5}\equal{} 3168$
Totally, we get: $ 1848\plus{} 3168 \equal{} 5016$

Solution 2:
I solve the problem indirectly

Choose 8 people arbitrarily: $ \binom{16}{8}\equal{} 12870$
Choose one person of two couple + 7 people left: $ 4 \cdot \binom{12}{7}\equal{} 3168$
Choose 2 people (not a couple of husband and wife) of two couple + 6 people left: $ 2 \cdot 2 \cdot \binom{12}{6}\equal{} 3696$
Choose 8 people not anyone in two couple: $ \binom{12}{8}\equal{} 495$
Totally, we get $ 12870\minus{}(3168\plus{}3696\plus{}495)\equal{}5511$

I confuse with two results. Can anyone help me?[/size]

In Case 2, you forgot to subtract the case where both couples are picked.  :wink:

Dear [b]azjps[/b],
Thank you very much for your replies.

Now, I would like to edit the problem 2 like this: How many different [b]natural[/b] numbers consisting of three digits are there such that in each number, the digits are increasing or decreasing?

In fact, I have solved it, but like the problem 1, I have to consider many cases, so my solution is very long and I' m not sure it ' s right!

Can anyone give me a hand?


AMC 8/10/12 (whichever is applicable to you) is just a few months away. I'm sure there are some of you that will just go take the test and do the best you can, but won't prepare much in advance. But there may be others, like myself, that want to do very well this year. Perhaps it's your last year; perhaps you have other reasons.

Most of us have read/are reading AoPS, volumes 1 and/or 2. What other math books are you reading in preparation for AMC/AIME?

And, to be useful, I'll post a few links to past questions that I've found online.
[url=http://www.math.uncc.edu/~hbreiter/index.html#American%20Mathematics%20Competitions]AMC problems[/url]
[url=http://www.kalva.demon.co.uk/aime.html]AIME problems[/url]
[url=http://www.kalva.demon.co.uk/index.html]A bunch of problems, including Olympiad level[/url]

This thread will be spam free. Thank you.

Some books on the AOPS level or easier are being discussed here:
[url]http://www.artofproblemsolving.com/wforum/viewtopic.php?t=2083[/url]


trigonometry
geometry
3D geometry
sphere
abstract algebra
geometric transformation
rotation

Ok, here I'm starting a physics marathon. The basic idea is that only one problem is active at a time, and that the person who posts the correct solution gets to post a new problem. Hoping that this marathon will not be forsaken, I'll post the first problem:

A homogenous cylinder of mass $ M$ and base radius $ R$ rolls across a horizontal plane. At one moment, the cylinder reaches the end of the horizontal plane and starts rolling on an inclined plane -- the angle between the inclined plane and the horizontal being $ \alpha$. Find the maximal velocity the cylinder can have so that it doesn't jump up while rolling onto the inclined plane. Neglect all slipping.

Have fun!

is it rolling onto an upward inclien or a downward one? also, does jump up mean lose contact with the surface?

The inclined plane is downward and yes, 'jump up' means lose contact with the surface.

ok, phew. I was doing up an incline and my force diagram was not making sense.

Well, then you didn't need to ask me :)

Will you participate in the weekly contest?

yeah, ill do what I can. I'm frantically trying to learn E and M in time for the semifinal round, so I might not be able to participate every week.

EDIT: Gah, i hate to ask, but can you please post a solution. I'm having trouble finding a condition for which the cylinder doesnt jump

[color=blue]The kinetic energy of the cylinder is

$W=\frac{3}{4}Mv^2$

The law of conservation of energy is

$\frac{3}{4}Mv_0^2+MgR(1-\cos\alpha)=\frac{3}{4}Mv^2$

$\Longrightarrow v^2=v_0^2+\frac{4}{3}gR(1-\cos\alpha)$

The cylinder does not jump up if this condition is satisfied $\frac{Mv^2}{R}\leq Mg\cos\alpha$

$\Longrightarrow v_0\leq \sqrt{\frac{gR}{3}(7\cos\alpha-4)}$

Is that the correct answer [b]Djole[/b]?[/color]

Yes, it's correct.

You should post a problem now...

Well, [b]phucnv87[/b] has asked me to post another problem because he doesn't have any problems to post. This time, I will post the second problem myself, but I urge everybody to post their own problem when they can. I do not have an infinite number of problems :) Thus, I expect that the poster of the correct solution will prepare at least a simple problem for this marathon.

Find the electric capacity of a system consisting of a small conducting ball of radius $a$ and an infinite conducting plane. The distance between the center of the ball and the plane is $b\gg a$.

Just a side note ---- The first one was already posted by me in the Physics contest --- so just check before posting! --- and the sencond one I know so I'm giving others a chance! --- just to know, are you using Irodov, djole?

Heh... Yes, I am, but neither these two problems nor the Contest ones were directly from Irodov. It's quite probable that it has these problems, too...

I'm pretty sure my solution has a flaw in its logic, so I'm posting it just so that someone can tell me how to get around it. (the main thing is that I assume the charge is uniformly distributed about the sphere, but I know that the presence of the plane will cause most of it to concentrate on the side of the sphere facing the plane).

Ok, lets give the sphere an arbitrary charge $Q$. For the plane to be an equipotential surface, it has to act as if there is another "mirror" sphere $b$ meters from the back of the plane of charge $-Q$.  Now lets find the potential drop from the sphere to the plane due to these charges. The potential drop due to the positive sphere is just $\frac{kQ}{b-a} - \frac{kQ}{b}$ and the potential drop due to the negative sphere is ${\frac{-kQ}{2b-a} - \frac(-kQ}{b}$. Now add these and divide by $Q$ to get the answer.

The idea is ok, just fix the $\text{\TeX}$ in your message so I can read it :) (or, better yet, post the complete solution now that you know you're on the right track!)

[color=blue]My calculation gives the answer $\large C=2\pi\epsilon_0\epsilon a$
where $\epsilon$ is the insulator constant of the environment[/color]

Sorry, I was sidetracked for a few days.  I'll post tonight.

An accelerometer has two arms arranged in a "V" shape. The angles of inclination of the two arms are $ \frac{\pi}{4}$ with respect to the horizontal. Find the acceleration of the vehicle it is installed in if the height of liquid in the two arms is $ h_1$ and $ h_2$ respectively. Assume the radius of the tube to be negligible. Neglect liquid viscosity and surface tension.

Since I haven't solved this kind of hydrodynamics problem, i'll make a guess based on intuition.

[hide]

In the non-inertial frame of the accelerometer, the liquid is affected by the real acceleration g, and the fictitious acceleration a in opposite direction of motion.  The total acceleration is:
\[ \vec g \plus{} \vec a
\]
Thus, the shape of the liquid surface will be as if we "tilted" the V tube instead of accelerating it (the increase in magnitude of the field does not affect volume and shape, unless the liquid contracts due to pressure).

Suppose we set a x-y coordinate system, with (0,0) at the corner of the V.  Then, the initial meniscus are at positions
\[ ( \minus{} h, h) and (h, h)
\]
.

Where h is the initial height of the column ($ h \equal{} \frac {h_1 \plus{} h_2}{2}$)

After tilting, the menisci are at position:
\[ ( \minus{} h_1, h_1) and (h_2, h_2)
\]
And the slope of inclination is :
\[ \frac {h_2 \minus{} h_1}{h_2 \plus{} h_1}
\]
This slope is perpendicular to the slope formed by the new field created by acceleration and gravitation.  Thus:
\[ \frac {h_2 \minus{} h_1}{h_2 \plus{} h_1} \equal{} \minus{}\frac {a}{g}
\]
Solving for a:
\[ a \equal{} g \frac {h_1 \minus{} h_2}{h_2 \plus{} h_1}
\]
[\hide][/hide]

I'll assume my solution is at least partially correct, so allow me to post a new problem.

(source: Kleppner and Kolenkow Introduction to Mechanics)

I haven't been able to solve it yet, it is a proof type of problem so we know what we are looking for.

6.41:

A plank of length 2 L leans against a wall (in the picture provided, the angle between the horizontal surface and the plank is $ \frac {\pi}{2}$.  It starts to slip downward without friction.  Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.

Hints:[hide]
Only a single variable is needed to describe the system.  Note the motion of the center of mass.

Also, the problem may utilize what is called Chasles theorem.[/hide]

[quote="integration"]
$ a \equal{} g\left(\frac {h_{1} \minus{} h_{2}}{h_{2} \plus{} h_{1}}\right)$[/quote]

[hide="Correct"]Pressure at the bottom of the accelerometer due to both arms is equal.[/hide]

Tell me if the problem is too difficult...

Anyone?

I'm pretty busy these days, but when I happened to glance at this forum and see this problem, it looked so interesting that I had to try it.

[hide="And indeed, I found it to be pretty difficult."]
I initially tried to solve this problem purely by using conservation of energy and the fact that while the top of the plank is in contact with the wall, the motion of its center of mass is constrained to a circle of radius $ L$, but I didn't get enough conditions with that.  I didn't want to use forces because I thought getting mixed in second derivatives would complicate the problem, but as I see it, considering forces is the only way to obtain the solution.

With that in mind, we can describe the system with the single variable $ \theta$, the angle between the plank and the horizontal.  Let the normal forces from the vertical and horizontal surfaces be $ F_1$ and $ F_2$, respectively, so we want $ \theta$ when $ F_1 \equal{} 0$, from which we can find $ h$.

Now we can write a bunch of equations.  Taking torques gives
\[ F_2 L\cos\theta \minus{} F_1 L\sin\theta \equal{} \frac{1}{3}mL^2 \ddot{\theta}.\]
By conservation of energy,
\[ gL(1 \minus{} \sin\theta) \equal{} \frac{1}{2}v^2 \plus{} \frac{1}{6}L^2 \dot{\theta}^2\]
where $ v$ is the speed of the center of mass.  Using Newton's second law in the $ x$ and $ y$ directions on the center of mass gives $ F_1 \equal{} m\ddot{x}$ and $ mg \minus{} F_2 \equal{} \minus{}m\ddot{y}$ where
\[ x \equal{} L\cos\theta \implies \ddot{x} \equal{} \minus{}L\cos\theta\dot{\theta}^2 \minus{} L\sin\theta\ddot{\theta}\]
and
\[ y \equal{} L\sin\theta \implies \ddot{y} \equal{} \minus{}L\sin\theta\dot{\theta}^2 \plus{} L\cos\theta\ddot{\theta}.\]
Finally, using either $ v\sin\theta \equal{} \dot{x}$ or $ v\cos\theta \equal{} \minus{}\dot{y}$, we get $ v \equal{} \minus{}L\dot{\theta}$.

The big picture is to eliminate all of the first and second derivatives to get an equation in terms of $ \theta$.  First, setting $ F_1 \equal{} 0$ and using the $ y$-direction force equation, the torque equation becomes
\[ (g \minus{} L\sin\theta\dot{\theta}^2 \plus{} L\cos\theta\ddot{\theta}) \cos\theta \equal{} \frac{1}{3}L\ddot{\theta}.\]
The fact that $ F_1 \equal{} 0$ also implies $ \ddot{x} \equal{} 0$, or
\[ \cos\theta\dot{\theta}^2 \equal{} \minus{}\sin\theta\ddot{\theta}.\]
Substituting $ v \equal{} \minus{}L\dot{\theta}$ into the conservation of energy equation,
\[ g(1 \minus{} \sin\theta) \equal{} \frac{2}{3}L\dot{\theta}^2.\]
With these last three equations, we can finally do a lot of substituting.  Though it should come down to $ \sin\theta \equal{} 2/3$, the desired ratio, I kept getting $ \sin\theta \equal{} 1/2$.  So I might have made a computational error along the way, but I'm too tired of substituting to try to correct it.  But you see the big picture, anyway.
[/hide]

Yeah, genericme, I tried to brute force it too by separating motion of center of mass and angular rotation by CM (Chasles Theorem)... turned out pretty complicated.

That's a really nice solution  :D  (and a nice problem too)

Electrostatics for a change!

Here's a new problem from Irodov.

Two infinite conducting plates (plate 1 and 2) are separated by a distance $ l$.  A charge $ q$ is a distance $ x$ away from plate 1.  Find the charge induced on each plate.

PS: Please, make the solution understandable.  I looked at the solution at the end of irodov, and although the answer is very simple and intuitive, I couldn't make a correct justification why it's correct.

Thank you.

Also, next time, can the person who solves a problem post a new one? (and let's get creative :D, we can post other stuff than mechanics)

Here are a few hints.

Use mirror charges. Yes ChargeS in plural, and use a lot... :wink:

People around have been idle for quite a while!

Let's have one from S. S. Krotov:

A bob is supended between two poles as shown. The distance betweenc the poles is $ a$ and the difference in the heights of suspension is $ b$. The lengths of the massless threads are $ l_1$ and $ l_2$ respectively. Find the time period $ T$ of small oscillations if $ a^2 \plus{} b^2 \equal{} l_1^2 \plus{} l_2^2$.

[hide="Hint:"]Do you see a simple pendulum?[/hide]

Umm.. that was interesting.

Solution

[hide]

Let's set a cartesian coodinate system, at one of the pole, with the z axis coinciding with the pole, and the xy axis parallel to the ground.

The motion of the mass is limited to the intersection between two spheres, one centered on the z axis at (0, 0, b), of radius $ {l_1}^2$, and the other centered on the y axis (the one that is parallel to the screen, in the horizontal direction), at (0, a, 0).

The equations of the spheres are:
\[ (z \minus{} b)^2 \plus{} x^2 \plus{} y^2 \equal{} {l_1}^2\]
and
\[ z^2 \plus{} x^2 \plus{} (y \minus{} a)^2 \equal{} {l_2}^2\]
Adding the 2 equations and using the fact $ {l_1}^2 \plus{} {l_2}^2 \equal{} a^2 \plus{} b^2$, we obtain:
\[ (z \minus{} b/2)^2 \plus{} x^2 \plus{} (y \minus{} a/2)^2 \equal{} \frac {a^2 \plus{} b^2}{4}\]
We see that the particle is moving on a circle of radius $ \frac {1}{2} \sqrt {a^2 \plus{} b^2}$, which is inclined relative to the y-axis. The angle of inclination $ \theta \equal{} \arctan {b/a}$

Thus, the period for small oscilations is:
\[ T \equal{} 2 \pi \sqrt {\frac {\sqrt(a^2 \plus{} b^2)}{2g \cos {\theta}}}\]

\[ T \equal{} 2 \pi \sqrt {\frac {(a^2 \plus{} b^2)}{2ga}}\]
[\hide][/hide]

The original problem I posted, with the ladder fall on 2 frictionless walls, has been solved by Genericme, quite nicely:
[quote]
6.41:

A plank of length 2 L leans against a wall (in the picture provided, the angle between the horizontal surface and the plank is \frac {\pi}{2}. It starts to slip downward without friction. Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.
[/quote]
However, there is an easier way to do it. I guess, that counts as a new problem

Hints:

1. The center of mass of the ladder moves in a circle of radius l/2, centered at the corner of the walls, that means the velocity vector can be described as $ v \equal{} R \omega$

2. an isoceles triangle is formed by the radius of that circle to the CM of the ladder, the half ladder segment, from the CM to the x-axis, and the x axis itself. (this simplifies the geometry of the problem, especially to compute $ \omega$ to calculate energy)

3. Use conservation of energy, and the only free parameter is $ \theta$, the angle between the base of the ladder and the x axis (which happens to be equal to the other angle because of isoceles triangles  :D )

4.  Use the fact that $ F_x \equal{} a_x \equal{} 0$ when the ladder leaves the surface.

I'm unstickying this since its been really inactive - use the USAPHO marathon


Find the 294th digit in the decimal expansion of $\frac{5}{17}$. (294th digit to the right of the decimal point)

That works.  But would you be able to calculate that during a test?  I guess it's possible to know that fractions over 17 repeat every 16 terms.  But I've been told there is a solution involving a geometric series...

I think prototypex24's solution is the easiest and most effecitive one available

[quote="prototypex24"][hide="Solution"]$\frac{5}{17}=0.2941176470588235294117$

The digits of the decimal repeat after 16 places.
$294= 6  (mod 16)$

$\therefore$ the $294th$ digit $=$ the $6th$ digit $=7$[/hide][/quote]

Yeah, but how would you do it without a calculator, if possible to do it without one.


so here's the question:

Tippy Van Winkle is awakened from a deep sleep by the cuckoo of a clock that sounds every half hour.  Before Tippy can look at the clock, his brother Bippy enters the room and offers to bet 10 dollars that the hands of the clock form an acute angle.  Assuming that the hands have not moved since the cuckoo sounded, how much should Tippy put up against Bippy's 10 dollars so that it is an even bet?

Can someone help me with this problem.  This problem seems a bit confusing to me.

I know the answer, but I don't know how you get to the answer.

Please try to do the problem before you look into the answer....  Anyways this is the [hide=" answer"]

Did You Give Up?

Well, this is the [hide=" answer"]
14 dollars
[/hide][/hide]

Thanks!!!

You just gave a link to this thread...

For the problem, you have to find half hours (in which the minute hand is at 0 or 6 o'clock) where the minute hand and the hour hand form and angle <90 degrees. For example, 3:30 is an acute angle (because the hour hand is somewhere between 3 and 4.)
From this information, I can't think of any method besides counting.

[quote="Nerd_of_the_Ages"][quote="pascal12"]You just gave a link to this thread...[/quote]
When you middle-click hide tags, it opens a tab with the same thread. Just left-click it and it will open.[/quote]

Ahhh, duh. Haha, for some reason, I was expecting it to be a link to a different webpage. How stupid of me.

This is probably one of the slowest ways to get the answers... I'm pretty sure that there's a way faster one, but I can't think of one.
Well, think about how many times the a clock hits thirty minutes saying there were only 30 minutes on a clock and still 24 hours in a day. Then, that would mean that there are 48 times in a day where the cuckoo clock rings. An easy way to start to find what the chances an acute angle would form when it's EXACTLY x:00 (let x be some number from 1 to 12 if you look at a clock for NOW... :D ). There are 24 times when it is exactly x:00. Because there are 360 degrees around a clock, in between each number, there's a 30 degree angle since $ \frac{360}{12}$ is 30. So all the angles are...
0- When it's 12:00
30- When it's 1:00
60- When it's 2:00
90- When it's 3:00
120- When it's 4:00
150- When it's 5:00
180- When it's 6:00
210- When it's 7:00
240- When it's 8:00
270- When it's 9:00
300- When it's 10:00
330- When it's 11:00
Okay so there are 12 angles (24 if you're counting the repeat) and only $ \frac{4}{12}$ are acute, so the chances of getting an acute angle are $ \frac{1}{4}$ when it's only at x:00. Now to find when it is when it's x:30.

Now you have to know how fast the minute hand and hour hand move per what ever. The hour hand moves $ \frac{1}{2}$ every minute because if it moves 30 degrees every hour, then $ \frac{30}{60}$ is $ \frac{1}{2}$. So now you have to find the angles of each time. So now solve for each of the angles... (I probably got most of these wrong or all of them but I think their close...
12:30- 153
1:30- 122
2:30-92
3:30-63
4:30-33
5:30-3
6:30-2
7:30-32
8:30-62
9:30-92
10:30-121
11:30-152
So now that you have all the angles, $ \frac{6}{12}$ which are acute, which simplifies into $ \frac{1}{2}$

Now you have each of the chances. So the first fraction was $ \frac{8}{48}$ and the second one is $ \frac{12}{48}$. Add these two together to find the probability of having the angle acute in which you get $ \frac{20}{48}$ which simplifies to $ \frac{5}{12}$ which is the probability of it being acute.

BUT WE HAVEN'T ANSWERED THE QUESTION YET HAVE WE?! I don't really get the rest, I only know the chance that the angle is acute... I'm sorry... :(

Oh... I get it... I should've done that before...
But is there an easier way to do the part I did above? I mean, it's somewhat slow...


geometry
rectangle
Gauss
Purple Comet
similar triangles

can anyone explain to me how to do questions 14 and 15 on the purple comet math meet spring 2009 test?

My team got #14 right BTW, i just did not understand how they did it.(we got #15 wrong...)

I'm not very good with geometry...

We ended up getting honorable mention...

#14 isn't as hard as it looks - don't get intimidated by the diagram.  :)

[hide="Hint for #14"]
Similar triangles are VERY important to this problem
[/hide]

hmmm... Thanks...!

I will reattempt the problem. :D

EDIT: Wow, wait a second, how did you get the 1/17+2/17... thingy?

I got to 1/2*10/3*40...

The height of the rectangle is 40 so the height of the smallest triangle is 40/17. The height of the next is 80/17 and so on up to the largest triangle. 40/17 + 80/17.... is the same as 40(1/17 + 2/17... + 1)

would someone like to help with 15? i thought i got it right during the competition, but apparently i messed up some computations...

[quote="FantasyLover"]Are you talking about Middle School or High School division?

EDIT:

[hide="Middle School Division #14"]Length of the base of each shaded triangle is $ 70\cdot \frac {1}{21} \equal{} \frac {10}{3}$.

Then, the sum of the shaded triangles will be $ \frac {1}{2}\cdot\frac {10}{3}\cdot40\cdot \left(\frac {1}{17} \plus{} \frac {2}{17} \plus{} \cdots \plus{} \frac {17}{17}\right)$
$ \equal{} \frac {1}{2}\cdot\frac {10}{3}\cdot40\cdot9 \equal{} 600$

Answer: 600[/hide]

Didn't use similar triangles.[/quote]

you DO realize that while you are not using similar triangles directly, your answer IS made possible by the appearance of similar triangles. so technically, you used similar triangles indirectly :P

also, #15 is a straight brute force i guess. thats what i did at least. (was about 300 off, until i realized even #s didnt have even #s on the back and pretty much redid the etire thingy)

[quote="AOPS23"]I thought my team got the right answer for number 15.....
What did your guys' team get?

Also, Can we discuss Gauss? Everyone took it on Wednesday, right?[/quote]
If you want to know if you got it right, you could just look [url=http://www.purplecomet.org/winners/2091results.htm]here[\url]. Find your username and see which ones you got right.

[quote="AOPS23"]I thought my team got the right answer for number 15.....
What did your guys' team get?

Also, Can we discuss Gauss? Everyone took it on Wednesday, right?[/quote]

...no. many are taking it today and tomorrow.

Is brute force the only way to solve #15?

Also see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277150]here[/url], where I copied the solution on a different thread. You can see it [url=http://www.purplecomet.org/home/resource/188/MiddleSchoolSolutions2009.pdf]here[/url] as well on page 8. BTW, the first link is only for problem 15, but the second is for all of them.

oh... The problem was being discussed in 2 threads!


Let $ a, b$ and $ c$ be positive real numbers.

(i) Prove that $ 4(a^{3}\plus{}b^{3}) \ge (a\plus{}b)^{3}$.

(ii) Prove that $ 9(a^{3}\plus{}b^{3}\plus{}c^{3})\ge (a\plus{}b\plus{}c)^{3}$.

(1) Divided by 8,　(2) Divided by 27 both sides, then Jensen will wait you.

(i)[hide]By the Trivial Inequality, $ (a\minus{}b)^2\ge0 \implies a^2\minus{}ab\plus{}b^2\ge ab$. Multiplying by $ (a\plus{}b)$, we have

$ (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\ge (a\plus{}b)(ab)$

$ \implies a^3\plus{}b^3 \ge a^2b\plus{}ab^2$

Multiplying both sides by $ 3$,

$ 3a^3\plus{}3b^3 \ge 3a^2b\plus{}3ab^2$

Adding $ a^3\plus{}b^3$ to both sides,

$ 4a^3\plus{}4b^3 \ge a^3\plus{}3a^2b\plus{}3ab^2\plus{}b^3$

$ \implies 4(a^3\plus{}b^3)\ge(a\plus{}b)^3$, with equality when $ a\equal{}b$. Q.E.D[/hide]

These are just particular cases of power means...with $ \alpha \geqslant \beta$, we have
\[ \left(\frac {\sum_{i \equal{} 1}^nx_i^{\alpha}}{n}\right)^{\frac {1}{\alpha}}\geqslant\left(\frac {\sum_{i \equal{} 1}^nx_i^{\beta}}{n}\right)^{\frac {1}{\beta}}\Longleftrightarrow n^{\alpha \minus{} \beta}\left({\sum_{i \equal{} 1}^nx_i^{\alpha}}\right)^{\beta}\geqslant\left({\sum_{i \equal{} 1}^nx_i^{\beta}}\right)^{\alpha}\mbox{.}\]
:P

[quote="AndrewTom"]Let $ a, b$ and $ c$ be positive real numbers.

(ii) Prove that $ 9(a^{3} \plus{} b^{3} \plus{} c^{3})\ge (a \plus{} b \plus{} c)^{3}$.[/quote]

Which is true from the Holder Inequality  :)

(2) After proving (1), we have $ 4(a^3 \plus{} b^3)\geq (a \plus{} b)^3,\ 4(b^3 \plus{} c^3)\geq (b \plus{} c)^3,\ 4(c^3 \plus{} a^3)\geq (c \plus{} a)^3$.

Adding these inequalities give $ 8(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b)^3 \plus{} (b \plus{} c)^3 \plus{} (c \plus{} a)^3$

$ \therefore 9(a^3 \plus{} b^3 \plus{} c^3)\geq a^3 \plus{} b^3 \plus{} c^3 \plus{} (a \plus{} b)^3 \plus{} (b \plus{} c)^3 \plus{} (c \plus{} a)^3$

$ \geq a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)^3\ Q.E.D.$ :lol:


Can you suggest me a good book on Quantum Mechanics?Is Quantum Mechanics by Cohen-Tannoudji good?

i suggest "Bratelli: Operator algebras and quantum statistical mechanics" (this is a 2 volume book) quite good
and "Teschl: Mathematical methods in quantum mechanics" this is also very good and mathematical without the usual twisting of standard qm books.

Hey , yeahnigga, are those books used at an intro level quantum class?

As far as I know, I would probably stick with the more common books like Griffiths Quantum mechanics.  As usual, Griffiths is extremely student friendly (you'll know that if you used his marvelous electrodynamics book).  Shankar Quantum mechanics is also awesome, especially because of its extreme clarity and because it introduces all the linear algebra to you in the 1st chapter.

I think Cohen-Tannoudji 2 volume books are good, only if you have A LOT of time to spend, and really want a book that has very long explanations for you (hence, that's what the 2 volumes mainly contain).  That's why it's mainly used for grad student references, because it has a lot of material. But if you have the time, then Cohen-Tannoudji is extremely clear and thorough (sometimes, too thorough, to the taste of a hasty undergrad/HS student).

Well no they are not intro level books, but i didnt see the poster, Aashay specify the level of the books he is looking for  :) so i though i will just share the books i like...

I want intro-level books only..But they should extensively cover all the concepts of Quantum Mechanics

The books I posted are intro level quantum books.  For griffiths, you need to know multivariate calc and some Differential equations.

For Shankar, ODE's, multivariate is expected, and the ability to learn the 1st chapter on algebra.

Cohen Tannoudji, you need once again ODE's and multivariate, although he really tries to explain everything in detail, so I wouldn't be too concerned about your background with this.


Find the first integer n>1 such that the average of 1, 4, 9, 16, . . ., n^2 is itself a perfect square.

 :P

Has been posted several times.  There was a topic about this only yesterday or the day before.


logarithms
trigonometry
complex analysis
complex analysis unsolved

Sketch an image under $ w \equal{} \text{Log} (z)$ of

i)  the line $ y\equal{}x$
ii) the line $ x\equal{}e$

I have not done many problems with mappings under the principal value mapping of the logarithm.  I know that

$ \text{Log} (z) : \equal{} \ln r \plus{} i \theta \equal{} \ln | z | \plus{} i \text{Arg} (z)$.

However, I do not see what these images would look like.  I need a few pointers here on what to do.

So basically, you want to write everything in polar. To do that $ x\equal{}r\cos \theta$, $ y\equal{}r\sin\theta$ and simplify

$ y\equal{}x$ in polar is $ \theta \equal{}\frac{\pi}{4}$ or $ \minus{}\frac{3\pi}{4}$ AND $ r>0$ (since we are in the principle Log, these are the angles...) or $ r\equal{}0$

So $ Log r$ is in $ \minus{}\infty$ to $ \infty$ since $ r>0$ Then you have particular lines.

So I'd say that the graph is two vertical lines $ x\equal{}\frac{\pi}{4}$ and $ y\equal{}\minus{}\frac{3\pi}{4}$ and $ x>0$

Maybe you'd include the point at infinity because technically thats what the origin maps to...

ii)  $ x\equal{}e$ in polar is $ r\cos\theta \equal{}e$
This is harder... Maybe you can try this one :D

For (a), it is pretty much like what Altheman has said, but it looks to me his answer is very strange - it should probably be the two horizontal lines $ \theta \equal{} \frac{\pi}{4}$ and $ \minus{}\frac{3\pi}{4}$.


hallo,
Find all positive intergers $x, y$ such that $x+y$, $2x+y$ and $x+2y$ are perfect squares.

well that trivially isn't a solution since $x+2y = 6n^2$ is not a perfect square.
anyway the problem is asking if whether $p^2 + q^2 = 3 r^2$ has any solutions. Trivially the answer is 'no', since consider the solution minimising $p+q+r$. Indeed, $p^2 + q^2$ is divisible by $3$ so $3$ divides both of them.  substituting back in $r$ is also divisible by 3, so the solution set $\frac{p}{3}, \frac{q}{3}, \frac{r}{3}$ provides a solution set with a smaller $p+q+r$, contradicting the minimality of $p+q+r$.

It give Pell's equation: $x+y=m^2,2x+y=n^2,x+2y=l^2\Longrightarrow 3m^2-n^2=l^2$, whith had solution:
$n+\sqrt 3 m=l\sqrt 3 (2+\sqrt 3 )^k$.
Therefore $x=n^2-m^2,y=2m^2-n^2,n=\frac{3l}{2}((2+\sqrt 3 )^k-(2-\sqrt 3 )^k)),m=\frac{l}{2}((2+\sqrt 3 )^k+(2-\sqrt 3 )^k)$.

There is no Pell equation with a nontrivial solution here ;) (note: $l=0$)


trigonometry
Pythagorean Theorem
geometry

Say I have a problem like this:

[img]http://i124.photobucket.com/albums/p15/afterlife007/triangle.jpg[/img]

Can you just use the Pythagorean theorem-- a^2 + b^2 = c^2 ?

Could someone please elaborate?  I would think it would be something like this:

2m^2+ (2m+3)^2 = 5m^2

If  its a right angled triangle, yes why not!

If you want to solve for m, it would get complicated, even with trig, to solve this without some knowledge of what the angles are.

However, if the triangle is right, as it appears to be (Note: I only say "appears"; looks can be deceiving), then you can use the Pythagorean Theorem.

$ (2m)^2\plus{}(2m\plus{}3)^2\equal{}(5m)^2$

$ 4m^2\plus{}4m^2\plus{}12m\plus{}9\equal{}25m^2$

$ 0\equal{}17m^2\minus{}12m\minus{}9$

$ m\equal{}\frac{\minus{}(\minus{}12)\pm\sqrt{(\minus{}12)^2\minus{}4(17)(\minus{}9)}}{2(17)}$

$ m\equal{}\frac{12\pm\sqrt{144\plus{}252}}{34}$

$ m\equal{}\frac{12\pm\sqrt{396}}{34}$

$ m\equal{}\frac{12\pm6\sqrt{11}}{34}$

$ m\equal{}\frac{6\pm3\sqrt{11}}{17}$

Hmmm...This answer is a little ugly, so unless I'm wrong, it would have been complicated even with trig.

LOL--Good answers guys! :rotfl:  :blush: Yeah, it works out---The factoring through me off. If you guys wanted the answer it is... m=1.16 which is used to solve all. It's not pretty! That question wasn't part of my homework, but I was checking it out-- :wink:

Except your calculations were off a little--Unless, I missed something. 756 should be under the square root sign. Yeah, I just checked my work it's 144+612 under the root! :)

fishythefish's mistake was in Step 4.  $ 4\cdot17\cdot9$ is 612, not 252.  I assume fishy accidently used 7 in the multiplication instead of 17.

Yeah.  :blush:

I just realized that after reading Inequality's post.

I guess it just shows how it important it is to check your work. :oops:


[quote="N.T.TUAN"]Let $ a,b,c\in [-\frac{3}{2},\infty)$ such that $ abc+ab+bc+ca+a+b+c\geq 0$. Prove that $ a+b+c\geq 0$.[/quote]
$ abc+ab+bc+ca+a+b+c\geq 0\Leftrightarrow(a+1)(b+1)(c+1)\geq1.$
Let $ a=\max\{a,b,c\}.$ Hence, $ a+1>0.$ Thus, $ (b+1)(c+1)>0.$
Id est, $ abc+ab+bc+ca+a+b+c\geq 0\Leftrightarrow a\geq\frac{-b-c-bc}{(b+1)(c+1)}.$
Hence, it remains to prove that $ \frac{-b-c-bc}{(b+1)(c+1)}+b+c\geq0,$ which equivalent to
$ f(b)\geq0,$ where $ f(b)=(1+c)b^{2}+(c+c^{2})b+c^{2}.$
If $ 1+c>0$ then it remains to prove that $ (c+c^{2})^{2}-4(1+c)c^{2}\leq0.$
But $ (c+c^{2})^{2}-4(1+c)c^{2}\leq0\Leftrightarrow c^{2}(1+c)(c-3)\leq0,$ which true.
If $ 1+c<0$ then $ 1+b<0.$ Hence, $ \{b,c\}\subset[-\frac{3}{2},-1).$
$ b_{max}=-\frac{c+c^{2}}{2(1+c)}=-\frac{c}{2}>0.$
Id est, it remains to check that $ f\left(-\frac{3}{2}\right)\geq0.$
But $ f\left(-\frac{3}{2}\right)=\frac{(3-c)(3+2c)}{4}\geq0.$ Done.
Equality holds when $ a=b=c=0,$ $ a=3,$ $ b=c=-\frac{3}{2}$
and cyclic permutations  for the last cortège.  :)
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154912


calculus
integration
trigonometry
real analysis
real analysis unsolved

Dear everyone.

please answer my shoddy question.

calculate $ \int_{0}^{\frac {\pi}{2}}{\frac {\sin{p\theta}}{\sin{\theta}}\cos^{p \minus{} 1}{\theta}d\theta }$ for p>0

thanks.


For positive numbers $ a,$ $ b,$ $ c,$ $ d,$ $ e,$ $ f$ and $ g$ prove that:
\[ \frac {a\plus{}b\plus{}c\plus{}d} {a\plus{}b\plus{}c\plus{}d\plus{}f\plus{}g} \plus{}\frac {c\plus{}d\plus{}e\plus{}f} {c\plus{}d\plus{}e\plus{}f\plus{}b\plus{}g} > \frac {e\plus{}f\plus{}a\plus{}b} {e\plus{}f\plus{}a\plus{}b\plus{}d\plus{}g}\]

[hide]
We have
\[ \frac{a\plus{}b\plus{}c\plus{}d}{a\plus{}b\plus{}c\plus{}d\plus{}f\plus{}g}>\frac{a\plus{}b\plus{}c\plus{}d}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g}\]
\[ \frac{c\plus{}d\plus{}e\plus{}f}{c\plus{}d\plus{}e\plus{}f\plus{}b\plus{}g}>\frac{c\plus{}d\plus{}e\plus{}f}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g}\]
Adding the two, the left hand side is greater than
\[ \frac{e\plus{}f\plus{}a\plus{}b\plus{}2(c\plus{}d)}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g}\]
\[ >\frac{e\plus{}f\plus{}a\plus{}b\plus{}c}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g}>\frac{e\plus{}f\plus{}a\plus{}b}{e\plus{}f\plus{}a\plus{}b\plus{}d\plus{}g}\]
(where the last inequality follows easily from $ \frac{d\plus{}g}{e\plus{}f\plus{}a\plus{}b\plus{}d\plus{}g}>\frac{d\plus{}g}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g}$), proving the result.
[/hide]


[quote="chien than"]Let $x;y;z$ be positive real numbers satisfying $xy+yz+xz=3$.Prove that:
$\frac{x^{3}}{2z+y}+\frac{y^{3}}{2x+z}+\frac{z^{3}}{2y+x}\geq 1$[/quote]

[hide]$\frac{x^{3}}{2z+y}+\frac{y^{3}}{2x+z}+\frac{z^{3}}{2y+x}\ge \frac{(x^{2}+y^{2}+z^{2})^{2}}{3(xy+yz+xz)}\ge \frac{(xy+yz+xz)^{2}}{3(xy+yz+xz)}=1$ by cauchy[/hide]

[quote="chien than"]Let $x;y;z$ be positive real numbers satisfying $xy+yz+xz=3$.Prove that:
$\frac{x^{3}}{2z+y}+\frac{y^{3}}{2x+z}+\frac{z^{3}}{2y+x}\geq 1$[/quote]
Or by Holder:
$\frac{x^{3}}{2z+y}+\frac{y^{3}}{2x+z}+\frac{z^{3}}{2y+x}\geq\frac{(x+y+z)^{3}}{3\cdot\sum(2z+y)}=\frac{(x+y+z)^{2}}{3(xy+xz+yz)}\geq1.$ :wink:


Let a,b,c be none - gative numbers such that: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Find the maximum value of the expression: Q = (1 - a)(1 - b)(1 - c)
Good luck :)

TrannguyenquyND_VN and macduchuykhtn:you can post solution?
IT VERY EASY :D  :blush:


function
induction
analytic geometry
graphing lines
slope

Solve recurrence (find closed expression for [tex]a_n[/tex])
[tex]\displaystyle{a_n=\min_{1\leq i <n} \frac{a_i+a_{n-i}}A}
[/tex], [tex]a_1=1[/tex],[tex]A\geq2[/tex]

I have what I am fairly confident is the answer, although I have not yet attempted to prove it:

If [tex]2^m \leq n \leq 2^{m + 1}[/tex], then [tex]\displaystyle a_n = \frac{(2^{m+1}-n)A+(2n-2^{m+1})}{A^{m+1}}[/tex]

I'm also trying trying to eliminate the m and see if that will help at all.

Okay, I will now prove the formula I gave above by induction.  One can see easily that it is true for 1, 2, and whatever other cases you may want to check.  Now, assume it is true for n = 1, 2, ..., k.  First, I would like to demonstrate that the minimum occurs when you take the "middle-most" of the pairs (i, n - i).  Assume WLOG that we're working only on i < (k - 1)/2.  Note that in the formula above, changing n by 1 will only result in a change of [tex]\displaystyle \pm \frac{A-2}{A^{m+1}}[/tex] in the term.  Thus, we see that if we increase i, the change in the [tex]a_i[/tex] term is going to be (2 - A) divided by some power [tex]m_1[/tex] of A, while the change in the [tex]a_{n-i}[/tex] term will be (A - 2) divided by a power [tex]m_2[/tex] of A with [tex]m_2 \geq m_1[/tex] since [tex]n-i > i[/tex].  So, as a result we see that if the inductive assumption is true, we have that the minimum occurs when [tex]i = [\frac{n}{2}][/tex].  If n falls in the range [tex][2^m, 2^{m+1}][/tex], i falls in the range [tex][2^{m-1}, 2^m][/tex].  If (k + 1) is even, we will have [tex]\displaystyle a_{k+1} = 2\cdot a_{(k+1)/2}  = 2\cdot[/tex][tex]\frac{(2^m-(k + 1)/2)A+(k + 1 - 2^m)}{A^m}/A = \frac{(2^{m+1}-(k + 1))A+(2(k + 1)-2^{m+1})}{A^{m+1}}[/tex], as predicted.  In the odd case, we see that the two terms will also fall into a nice range, and adding up we still get the same thing, which I am too lazy to tex.  Q.E.D.

Nice proof, although a small mistake in answer. It should be  [tex]a_n = \frac{(2^{m+1}-n)A+(2n-2^{m+1})}{A^{m+1}}[/tex](your answer gives [tex]\frac{A+1}A[/tex] for [tex]a_1[/tex] instead of 1)
I used essentialy the same approach, but another proof of the fact that middle-most pair gives the minimum. Consider function [tex]f(x)=\frac{(A-1)2^{m+1}-(A-2)x}{A^{m+1}}, 2^m\leq x<2^{m+1}, x>1[/tex]. We want to prove by induction that  [tex]a_n=f(n)[/tex]. Notice that f(x) is continuous and convex. (On each of the intervals [tex][2^m,2^{m+1}][/tex] it is linear, and its slope increases when we move to the next interval). So [tex]a_i+a_{n-i}=f(i)+f(n-i)\geq2f(\frac{i+n-i}2)=2f(\frac n2)[/tex], and we have minimum at [tex]i=[\frac n2][/tex], that is [tex]i=n-i=\frac n2[/tex] for even n, and [tex]i=[\frac n2],n-i=[\frac n2]+1[/tex] for odd n.


geometry
geometric transformation
homothety
ratio
rotation
geometry proposed

Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.

Beautiful problem! My solution:

Let $E$ be the perpendicular from $P$ to $AC$, $F$ be the perpendicular from $P$ to $AB$. Then $AFPE$ is a rectangle. Let $T$ be the perpendicular from $M$ to $AB$, $S$ be the perpendicular from $M$ to $AC$. Then a homothety centered at $A$ takes $\triangle MTS$ to $\triangle PFE$, thus $\frac{AB}{AC}=\frac{MS}{MT}=\frac{PE}{PF}=\frac{AF}{AE}=\frac{AF-AB}{AE-AC}=\frac{BF}{CE}$.
Let $HR$ meet $AC$ at $V$ and $HQ$ meet $AB$ at $U$.
I claim that
$\frac{AV}{VR}= \frac{BF}{FP}$
First note that $\triangle RCV \sim \triangle PCE$ since they have two equal angles, thus $\frac{VR}{VC}=\frac{PE}{CE}$, or $VR= \frac{VC \cdot PE}{CE}$. Thus, we have
$\frac{AV}{VR}= \frac{BF}{FP} \iff \newline
1=\frac{BF}{AV} \cdot \frac{VR}{FP} \iff \newline
1= \frac{BF}{AV} \cdot \frac{VC \cdot PE}{CE \cdot PF} \iff \newline
1=\frac{BF}{CE} \cdot \frac{PE}{PF} \cdot \frac{VC}{AV} \iff \newline
1= (\frac{AB}{AC})^2 \cdot \frac{VC}{AV} \iff \newline
1=(\frac{AB}{AC})^2 \cdot \frac{VC}{VH} \cdot \frac{VH}{AV} \iff \newline
1=(\frac{AB}{AC})^2 \cdot (\frac{AC}{AB})^2$
Which is true. Thus our claim $\frac{AV}{VR}= \frac{BF}{FP}$ is true.
Thus, $\triangle BFP \sim \triangle AVR$ since they are both right triangles and have two adjacent sides proportional. Thus, $\angle ARV= \angle BPF$. Then,
$\angle ARP= \angle ARV+ \angle VRP= \angle BPF+ \angle CPE= \angle FPE - \angle BPC= 90-\angle BPC$
It follows that $AR \perp BP$, similarly $AQ \perp CP$, thus $A$ is the orthocenter of $\triangle PQR$.

Reverse problem also looks nice:

Given $\triangle ABC$ with orthocenter $H$, construct $\triangle HPQ$ such that:

a) $P\in AB, Q\in AC$
b) $\angle PHQ=90^\circ$
c) line $AH$ is a median of $\triangle HPQ$

I have another approach.

Consider the homothety $\iota$ with center $H$,ratio $AB/AC$ and rotation angle $90^{\circ}$. Hence $\iota(A)=C$ and let $\iota(P)=S$. Let $R'$ be the intersection of $PC$ with $SA$. We now show that $R \equiv R'$, which implies $\angle QAB= \angle SCA=\angle RCA$, hence $QA \perp PR$ and $RA \perp PQ$, hence $A$ is the orthocenter of $\triangle PQR$.

Let $T$ be the intersection of $AP$ and $SC$. Since $PA \perp SC$ and $\angle CAT = \angle HCA$, $HT \parallel AC$. Moreover, $\overline{AP}/\overline{CS}=\overline{AB}/\overline{AC}$. Therefore, we can reformulate the problem as follows: [i]Let $\triangle ABC$ be a triangle with $\angle BCA=90^{\circ}$. Consider points $P$ and $Q$ on the rays $CA$ and $CB$, such that $\overline{AQ}/\overline{BP}=\overline{BC}/\overline{CA}$. Let $S$ be the intersection of $AP$ and $BQ$. Then the foot $T$ of the perpendicular from $S$ onto $AB$ is the isometric conjugate of the foot $D$ from $C$ onto $AB$.[/i]

To prove this, it's enough to show that \[\frac{\overline{BT}}{\overline{TA}}=\frac{\overline{AC}^2}{\overline{BC}^2}=\frac{\overline{AD}}{\overline{DB}}\mbox{.}\] Let $M$ and $N$ be the feet of the perpendiculars from $P$ and $Q$ onto $AB$, respectively. We get $\overline{NA}=\overline{MB}$ and hence $\overline{BN}=\overline{AM}$ as well. Since $\triangle BTS \sim \triangle BNQ$ and $\triangle ATS \sim \triangle AMP$, we get \[\frac{\overline{BT}}{\overline{TA}}=\frac{\overline{BS}\cdot\overline{AP}}{\overline{AS}\cdot\overline{QB}}=\frac{[APB]}{[AQB]}=\frac{\overline{BP}\cdot\overline{AC}}{\overline{AQ}\cdot\overline{BC}}=\frac{\overline{AC}^2}{\overline{BC}^2}\mbox{,}\] which is what we had to prove. $\square$


Prove the following using induction $\binom{n}{k-1}+\binom{n}{k}= \binom{n+1}{k}$
$0 < k \le n$

What has me stumped is it has two letters.. i've only used induction with one letter.

It looks like you would do it on $k$. But it's much easier to prove combinatorially or using direct algebra.

I'm able to prove it directly through algebra or combinatorial arguments. It's just this text i'm reading says prove it using induction. How would i do it on k?

We can rewrite this as
$\left(\frac{n!}{(k-1)!(n-k+1)!}\right)+\left(\frac{n!}{(k!)(n-k)!}\right) = \left(\frac{(n+1)!}{(k!)(n+k-1)!}\right)$

To induct over two variables, you need to prove three things:

i. A base case

ii. That if it is true for the ordered pair $(n,k)$, it is true for the ordered pair $(n+1,k)$

iii. That if it is true for the ordered pair $(n,k)$, it is true for the ordered pair $(n,k+1)$

So if we wanted to prove it for $(13,42)$, we first use a base case, say $(1,1)$. By using (ii) repeatedly, we find that it is true for $(13,1)$, and then by using (iii) repeatedly, we find that it is true for $(13,42)$

It's actually not necessary to induct over both variables, since for any $n$ we can use $k = 1$:

${n \choose 0}+{n \choose 1}= 1+n ={n+1 \choose 1}$

As our base case.  Then we only need to induct over $k$ for a given $n$.


Who is expected to win state MC in VA this coming year?

Probably Victoria Xia, because she's just really that beast....

hmmm unless she is ridiculously beast enough to get like top 4 or so nationals it's ideal for her to get 3rd (if you want her to do well at nats)

Michael Zhao/Max Schindler? C'mon math154, 1/3 means nothing.

last time i checked you were ridiculously beast enough to get like top 4 or so at nationals (after all, you did...)

also like ben kraft and ray both got 3rd?

and as the wiki article [b]used to say[/b] 3rd seed at nats is pretty awesome

As a 7th grader, she got perfect score at states, and really well at nats, so i would guess victoria xia would win again. She's also from longfellow, so... :wink:

dang. a perfect state? :o

you longfellow guys are too beast. can you give me some mathcounts tips? i wanna do better this year

no

i like many others practice mathcounts tests to solidify my usamo qual chances

[quote="alex960531"][quote="batteredbutnotdefeated"][quote="dysfunctionalequations"]...

I hope you (and Ubemaya) are joking.[/quote]

The scary part is they are NOT joking. :o

anyways, Victoria is going down this year because, well, longfellow needs to lose. LOSE.[/quote]

LOL, no way, i have no idea who you are, but we can pwn anyone in virginia...[/quote]

Not a chance bro.
1. Victoria Xia got lucky last year, and even if she's improved she'll be brought down.
2. There's a whole lot of new AoPSers going to VA state such as alien10, the archer, and srnth.
3. Lastly, the person who beats victoria is going to own at nats.

[quote="batteredbutnotdefeated"][/quote][quote="alex960531"][quote="batteredbutnotdefeated"][quote="dysfunctionalequations"]...

I hope you (and Ubemaya) are joking.[/quote]

The scary part is they are NOT joking. :o

anyways, Victoria is going down this year because, well, longfellow needs to lose. LOSE.[/quote]

LOL, no way, i have no idea who you are, but we can pwn anyone in virginia...[/quote][quote="batteredbutnotdefeated"]

Not a chance bro.
1. Victoria Xia got lucky last year, and even if she's improved she'll be brought down.
2. There's a whole lot of new AoPSers going to VA state such as alien10, the archer, and srnth.
3. Lastly, the person who beats victoria is going to own at nats.[/quote]

First of all normally the Virginia States Team is composed of the tops students in the Northern Virginia/Fairfax/George Washington or whatever chapter. And as seen in Countdown we totally owned everyone....

And new AoPSers dont really matter...

[quote="infinity4ever"]In addition, it's ridiculous to say that Victoria's good performance last year was due to luck.  Any girl who gets 5th at states as a 6th grader and 1st again as a 7th grader has very high potential for doing well at nationals this year.  Plus, she is one of the most consistently well-performing people I've ever met  :o which is clearly seen through her AMC and MPFG scores as well.[/quote]

true, very true, and she is probably going to get 1st place at states with the rest of Longfellow students

[quote="professordad"]I think we already know who's gonna win states  :D
But is there any competition for 2nd?

And what about 2nd place team?[/quote]

Not sure for second place individual, but definitely second place team will be Rachel Carson Middle School

[quote="AkshajKadaveru"]I will win![/quote]
Okay.

[quote="batteredbutnotdefeated"][quote="dysfunctionalequations"]...

I hope you (and Ubemaya) are joking.[/quote]

The scary part is they are NOT joking. :o

anyways, Victoria is going down this year because, well, longfellow needs to lose. LOSE.[/quote]

they lost


I already sent in my answers for the practice ARML.  If anyone wants a copy of the questions, I can pm them to you.  Just don't expect me to tell you any of the answers.

By the way, I have no life :roll: .

Didn't we all get this in the mail?

But I'd like the 1995 questions.  I didn't get that.

I don't have the 1995 problems.

The reason for my offer was in case someone hadn't gotten the 1998 problems yet.

What scores did you guys get on the team & individual if you got your score back already?

I got my scores back

Team: 8/10
Individual: 6/8

I didn't get one, so could you pm me one?

can someone pm me the team 1998 please
thank you


inequalities
logarithms
function
inequalities proposed

If $a,b,c$ a positive numbers such that $abc=1$, then

$a^2+b^2+c^2+9(ab+bc+ca) \geq 10(a+b+c)$.

we have to show that
$\sum a^2+\frac{9}{a}-10a\geq 0$ if$f(a)=a^2+\frac{9}{a}-10a$
we can show that $f(a)+f(b)\geq 2f(\sqrt{ab})$ if $ab\le 1$so the result follows
 :D
PS:(not that difficult)

Indeed, this method goes, and can be also used to show the stronger inequality

$a^2+b^2+c^2+10(ab+bc+ca) \geq 11(a+b+c)$.

:)  Vasc i got that the second inequality is true but how you can say that: 2nd is stronger than the 1st ...

Davron

Thanks to the coefficients ;)

Hence, we can say that arqady's one is even stronger!  :)

Davron, from the second inequality

$a^2+b^2+c^2+15\left(bc+ca+ab\right)\geq 16\left(a+b+c\right)$

and the inequality $a^2+b^2+c^2\geq a+b+c$ (which follows from abc = 1 using the Theorem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=203634#p203634]http://www.mathlinks.ro/Forum/viewtopic.php?t=32719 post #7[/url]), we derive that

$a^2+b^2+c^2+9\left(bc+ca+ab\right)$
$=\frac{6}{15}\left(a^2+b^2+c^2\right)+\frac{9}{15}\left(a^2+b^2+c^2+15\left(bc+ca+ab\right)\right)$
$\geq\frac{6}{15}\left(a+b+c\right)+\frac{9}{15}\cdot 16\left(a+b+c\right)$
$=10\left(a+b+c\right)$,

so the first inequality is proven. Thus, we can say that the second inequality is stronger than the first one.

By the way, I see that the inequality

$a^2+b^2+c^2+15\left(bc+ca+ab\right)\geq 16\left(a+b+c\right)$

was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=36874 (see post #2 for the equivalence) and solved by the same method as Albanian Eagle did it here (this method is actually the mixing variables method). Anybody has a better solution?

  Darij

[quote="arqady"]The Albanian Eagle's method gives
$a^2+b^2+c^2+15(ab+bc+ca) \geq 16(a+b+c)$ ;)[/quote]
[quote="perfect_radio"]Also see http://www.mathlinks.ro/Forum/viewtopic.php?t=36874 .[/quote] To prove the above inequality we need to show that
$17a^4+2a^3-13a^2+2a+1>0$ for any $a>0$.
There is a very nice solution to this inequality. Can you find it ?

Do you have some very nice regrouping of terms? I tried to do that some time ago, but I couldn't succeed :( , (some small interval gives me headaches) so I used brute force. I'll try more when I come back from school.

BTW, what's the best constant we can find in that inequality?

This gives me a solution for $\displaystyle a < \frac12$. But I can't handle the other case in the same way :(

Now I remain with the case $\displaystyle \frac12 < a < 0.58 \ldots$. Obviously, you see something that I don't see. My inequality skills are not so good...  :?  :(

The final hint is to use AM-GM inequality for the product from the right hand side of the inequality
$a^4>(2a^2+a-1)(1+3a-8a^2)$. ;)

[quote="Vasc"]To prove the above inequality we need to show that
$17a^4+2a^3-13a^2+2a+1>0$ for any $a>0$.
There is a very nice solution to this inequality. Can you find it ?[/quote]It's ugly :blush:  but  we can write directly $17a^4+2a^3-13a^2+2a+1 =(2a-1)^4+a(a+35)(a-\frac8{15})^2+\frac{15a^3+11a^2+10a}{225}>0$ for $a>0$.

[quote="Vasc"]Since the inequality is clearly true for $a \geq 1$, it remains to consider only $\frac 1{2} <a<1$.
The second hint is to write the inequality as
$a^4>(2a^2+a-1)(1+3a-8a^2)$.[/quote]

thats really beautiful

$(2a^2+a-1)(1+3a-8a^2)\leq \left(\frac{4a-6a^2}{2}\right)^2$

so it is equivalent to prove $a^2(2-3a)^2<a^4$ $\Longleftrightarrow (2a-1)(1-a)>0$ which is clear .

[quote="arqady"][quote="mathmanman"]
$2(a^2+b^2+c^2)+30(ab+bc+ca)-32(a+b+c) = 32(ab+bc+ca)-32(a+b+c)+(a-b)^2+(b-c)^2+(c-a)^2$[/quote]
I think it gives nothing. ;)[/quote]

Well, just compare $ab+bc+ca$ and $a+b+c$ knowing that $abc = 1$. :)

[quote="mathmanman"]
Well, just compare $ab+bc+ca$ and $a+b+c$ knowing that $abc = 1$. :)[/quote]
OK. $ab+bc+ca\geq a+b+c\Leftrightarrow ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}.$
Let $a=x^3,b=y^3,c=z^3.$
Hence, $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}\Leftrightarrow x^3y^3+x^3z^3+y^3z^3\geq x^4yz+y^4xz+z^4xy.$ This is false. ;)

\[a^{2}+b^{2}+c^{2}+9(ab+bc+ca)\geq10(a+b+c).\]
Here is a Vasc solution for the first post. Write the inequality as
\[\sum\left(a^{2}+\frac9a-10a+17\ln a\right)\geq0. \]
So, it suffices to show that the function $f(x)=x^{2}+\frac9x-10x+17\ln x$ is nonegative for $x>0$. Since
\[f'(x)=\frac{(x-1)(2x^{2}-8x+9)}{x^{2}}. \]
It follows that $f'(x)$ is negative for $0<x<1$ and positive for $x>1$. Therefore, $f(x)$ is increasing for $0<x\leq 1$, and increasing for $x\geq1$. This $f(x)\geq f(1)=0$.


请问一下哪边有下载LaTex?还是要到商店去买？

多谢！但上面只有Tex或ctex,没见到LaTex啊？


You would probably get a better answer in the Alcumus Forum.


[quote="GameBot"]A sweater is marked down $ 40\%$. The store announces a special one-day sale which reduces the sale price by $ 60\%$. What percent markdown on the original price does this represent?[/quote]

Let the price of the sweater be $ x$. When it's marked down 40%, its price is $ 0.6x$. When its marked down 60% again, its price is then $ 0.6x \cdot 0.4 \equal{} 0.24x$. Thus it had a total discount of $ \boxed{24\%}$.

Or, the total discount is actually $ 100\minus{}24\equal{}\boxed{76 \%}$.  :roll:


function
algebra
polynomial
calculus
calculus computations

$ f(x)\equal{}\frac{2x}{x\minus{}1}$

At what x values is this function not differentiable?

By differentiating you'll get that $ f'(x)$ won't be defined on $ x \equal{} 1,$ thus it's differentiable everywhere but not at $ x \equal{} 1.$

or else draw graph of f(x).
it tends to 2 as x-->infinity.
x=1 is asymptote.power being odd, f is of diff signs in both sides of f. it cuts x axis at x=0,and again goes to inf as x-->-inf. again all terms are polynomial, so only non-diff. at x=1


Solve the cubic equation :

$ 2x^3\minus{}6x^2\minus{}6x\plus{}29\equal{}0$

Wow, I got a really ugly solution...more than that, it's irrational, because it involves the square root of -29.

Using maple 12... you get solutions with cubic roots of $ \sqrt{233}$ and alot of other stuff ... I hope this wasn't the intended problem.

Are you sure you didn't mean $ \plus{}6x$ since that would give you the chance to complete the cube.


The probability that it will rain tomorrow is $ \frac {1}{2}$. The probability that it will rain the next day is $ \frac {1}{3}$. What is the probability that it will rain exactly 1 of the next 2 days?

P (rain exactly 1 of the next 2 days) = P (rain on tomorrow but not the day after tomorrow) + P (rain on tomorrow but not the day after tomorrow)
$ \equal{}\frac{1}{2}*\frac{2}{3}\plus{}\frac{1}{2}*\frac{1}{3}$
$ \equal{}\frac{1}{2}$

Yeah, Karnaugh maps or Venn Diagrams are good aids to solve these kind of problems.

lol. I kept getting $ \frac{2}{3}$, and could'nt figure out why i am wrong.
I just figured out the problem asked for exactly one day, not at least one day.

Im kinda embarrassed  :blush:


trigonometry
calculus
calculus computations

Could some prove that we have $ x\equal{}r(t\minus{}\sin t)$ and $ y\equal{}r(1\minus{}\cos t)$ for all points on a cycloid made by circle of radius $ r$?

Thanks. :)

It's all here... happy reading :roll:
http://jwilson.coe.uga.edu/EMAT6680Fa07/Schmidt/Assignment%2010/Gayle&Greg-10.htm


geometry
trigonometry
calculus
calculus computations

The cross section of a trough is an inverted isoceles triangle. Prove that the trough has maximum capacity when the vertex angle is pi/2 radians.

The capacity can be arbitrarily large unless you put some restriction on the size of the trough.

i believe he means that when the sides are constant (say s), then maximize the area, this is easily done, noting the sine area formula

$A=\frac{1}{2}*s^{2}\sin\theta$ where $\theta$ is the angle between the two sides of the trough, this is clearly maximized when $\theta=\frac{\pi}{2}$


help?
$\log_{9}p=\log_{12}q=\log_{16}(p+q)$
find $\frac{q}{p}$
I've been through this one so many times and I can't get it. I realize that 9, 12 and 16, the bases, are in geometric progression, but that hasn't helped me much.

[size=59][color=red]b-flat:  use \log instead of just log[/color][/size]

$p=9^{a}, q=12^{a}, p+q=16^{a}\implies 9^{a}+12^{a}=16^{a}$

Then ${q\over p}=\left({12\over 9}\right)^{a}=\left({4\over 3}\right)^{a}$

$9^{a}+12^{a}=16^{a}\iff\left({9\over 12}\right)^{a}+1=\left({16\over 12}\right)^{a}\iff\left({3\over 4}\right)^{a}+1=\left({4\over 3}\right)^{a}$

Put $x=\left({4\over 3}\right)^{a}$. Then you get ${1\over x}+1=x\iff x^{2}-x-1=0\implies x={1+\sqrt{5}\over 2}$ (the negative solution is extraneous). Therefore ${q\over p}={1+\sqrt{5}\over 2}$

hi,  I told this prob to me friend ans she returned me another solution as follow:

p=9^a, q=12^a => p.q = 12^(2a)

We aslo have: (q-p)=12^a => (q-p)^2 = 12^(2a)

Hence: (q-p)^(2a) = p.q <=> q^2 - 3p.q + p^2 = 0 <=> q = (3 +/- sqrt{5}).p/2 <=> q/p = (3 + sqrt{5})/2. (because q/p>1)

It would have been a good solution of she knew that q>p so q/p > 1: She accepted the result q/p = (3-sqrt{5}/2 !


let $ a$ , $ b$ , $ c$ $ \in \mathbb R$ , such that $ \mid{a}\mid \: < 1$ ,and,  $ \mid{b}\mid \: < 1$  ,and, $ \mid{c}\mid \: < 1$ , prove that :

$ ab\plus{}bc\plus{}ca \plus{}1> 0$

1) Presume that $ a\geq0 , b\geq0 , c\leq 0$ Then

$ ab \plus{} bc \plus{} ca \plus{} 1 \equal{} |a\parallel{}b| \minus{} |b\parallel{}c| \minus{} |c\parallel{}a| \plus{} 1\geq|a\parallel{}b| \minus{} |b| \minus{} |a| \plus{} 1 \equal{} (1 \minus{} |a|)(1 \minus{} |b|) > 0$

 2)Presume $ a\geq0,b\leq0, c\leq0$, analogous


In a triangle $ ABC$, with angles $ A,B,C$ sides $ a,b,c$ and $ a \ge b \ge c$. If $ \dfrac{a^3 \plus{} b^3 \plus{} c^3}{\sin^3 A \plus{} \sin ^3 B \plus{} \sin^3 C} \equal{} 8$, find the maximum length of side $ a$.

[hide]Using $ \frac {sin A}{a} \equal{} \frac {1}{2R}$, we get R=1, so a is a side of a triangle with circumradius 1. So the maximum possible side is the diameter, which is 2.[/hide]

[hide="Alternate Solution"]
We have $ \frac{a}{\sin{A}} \equal{} \frac{b}{\sin{B}} \equal{} \frac{c}{\sin{c}} \equal{} 2R$ by the Law of Sines. Cubing the equality chain, we obtain:

$ \frac{a^3}{\sin^3{A}} \equal{} \frac{b^3}{\sin^3{B}} \equal{} \frac{c^3}{\sin^3{c}} \equal{} \frac{a^3 \plus{} b^3 \plus{} c^3}{\sin^3{A} \plus{} \sin^3{B} \plus{} \sin^3{C}} \equal{} 8R^3 \equal{} 8\implies R \equal{} 1$.

The maximum length of $ a$ is a diameter, we have $ a \equal{} 2R \equal{} \boxed{2}$.
[/hide]


Anyone think they could post the answers? I never got to check mine with the key.

Thank you very much for the quick response. I think you got a 150 based on those anwers. Mine agree with that.

[quote="amirhtlusa"]although the official answers will be out on Monday,
here are mine:
[hide]
1.a
2.c
3.b
4.e
5.d
6.a
7.b
8.c
9.a
10e
11c
12b
13e
14c
15a
16b
17b
18e
19e
20c
21e
22d
23b
24d
25e[/hide][/quote]

Didn't you say you checked your answers with the solutions pamphlet?  Then those would be the official answers ;)

FYI the (official) answers on the AMC10 were:
[hide="AMC10 Answers"]
1.a
2.c
3.b
4.e
5.d
6.b
7.a
8.e
9.c
10.e
11.c
12.c
13.d
14.b
15.d
16.d (argh)
17.a
18.c
19.c
20.e
21.e
22.c
23.b
24.b
25.c
[/hide]

And the answers posted for the AMC12 are correct.

how do you 16???  that's the only one i got wrong of the ones i answered.


inequalities
LaTeX
inequalities proposed

this is my first exo at this forum :
a;b;et c des reels positifs tel que:
(1/a+b)+(1/b+c)+(1/c+a)=2
prove that:
rac(a+b+c+3)>=rac(a)+rac(b)+rac(c

Let $ a,\ b,\ c$ be postive real numbers such that $ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\equal{}2$.

Prove that $ \sqrt{a\plus{}b\plus{}c}\geq \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

Are you sure, Kunny?
If so, then $ a \equal{} b \equal{} c \equal{} 3/4$ will give $ 3/2 \ge 3\sqrt3 /2$.

edit: ah, you forgot +3 ^^, but then it's still false, so maybe add 9/2^^

i am waiting your answers ................ :P

[quote="nirabolico"]you forget the number 3 in the inequalty[/quote]

At first sight, we couldn't see the number 3. The evidence was given by FelixD's post.

First of all what you should say is that you have typed and give a word of thanks, shouldn't you? :wink:

sorry i don't mean it
but for the moment the important is find the solution
a vous de jouer :huh:

sorry i find a fmistake at this inequality
it is in the cas : a=b=c=3/4 :oops:

[quote="kunny"]Let $ a,\ b,\ c$ be postive real numbers such that $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \equal{} 2$.

Prove that $ \sqrt {a \plus{} b \plus{} c}\geq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$.[/quote]
I really do not understand what is happening in this topic:
$ \sqrt {a \plus{} b \plus{} c}\geq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$
$ \Longleftrightarrow a \plus{} b \plus{} c\geq (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2 \equal{} a \plus{} b \plus{} c \plus{} 2\sum \sqrt{ab}$
It is true :?:

The inequality $ \sqrt{a\plus{}b\plus{}c\plus{}3} \ge \sum \sqrt{a}$ (*) ist stated. Kunny made a typo, he wrote $ \sqrt{a\plus{}b\plus{}c} \ge \sum \sqrt{a}$ instead of (*). But, as I remarked, (*) is still false for $ a\equal{}b\equal{}c\equal{}3/4$. Even the inequality $ \sqrt{a\plus{}b\plus{}c\plus{}9/2} \ge \sum \sqrt{a}$ is false with the condition $ \sum 1/(a\plus{}b) \equal{}2$ (there would be the equality case $ a\equal{}b\equal{}c\equal{}3/4$).

I do kow that , see#7. You should see the former post when you post. :wink:

What do you want to tell me with your post, kunny?

[quote="FelixD"]Are you sure, Kunny?
If so, then $ a = b = c = 3/4$ will give $ 3/2 \ge 3\sqrt3 /2$.

edit: ah, you forgot +3 ^^, but then it's still false, so maybe add 9/2^^[/quote]

As you noticed, the original problem was ${ \sqrt{a+b+c}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}}$, wasn't it? :wink:

No, the original problem was $ \sqrt{a\plus{}b\plus{}c\plus{}3} \ge \sum \sqrt{a}$.   :wink:
But I just summarised what was happening in this topic^^ (because of math10's post)

No! The original problem had no the number 3.

Anyway, @nirabolico, you should edit the problem in Latex typed!

And $ (1/a \plus{} b)$ isn't  interpreted as  $ \frac {1}{a \plus{} b}$.

Don't post the problem in French.  :wink:

Japanese Communities Modeartor


[quote="Nguyen Ngoc Linh"]Given: a^2 + a = -1
find the value of A=a^1989 + 1:(a^1989)[/quote]

I don't understand!

Oh! $ A \equal{} a^{1989}  \plus{} \frac{1}{{a^{1989} }}$

[hide="Hint"]$ a^2 \plus{} a \equal{} \minus{} 1 \color{red} \Rightarrow \color{black} a^3 \equal{} \minus{} a^2 \minus{} a \equal{} 1$[/hide]

EDIT:Sorry. I made a typo. I revised it some time ago.

[quote="Kouichi Nakagawa"][hide="Hint"]$ a^2 \plus{} a \equal{} \minus{} 1 \iff a^3 \equal{} \minus{} a^2 \minus{} a \equal{} 1$[/hide][/quote]

No no, it's false!!!  :|  :|

no, that should be right, although to get to that uses roots of unity, or just the ability to recognize $ x^2\plus{}x\plus{}1$ as a factor of $ a^3\minus{}1$.

[quote="5849206328x"]no, that should be right, although to get to that uses roots of unity, or just the ability to recognize $ x^2 \plus{} x \plus{} 1$ as a factor of $ a^3 \minus{} 1$.[/quote]

But $ x\equal{}1$ isn't such that $ x^2 \plus{} x \equal{} \minus{}1$  :)  :)  :)

there are two other solutions to $ x^3\equal{}1$, both of which are complex. :wink:

[quote="5849206328x"]there are two other solutions to $ x^3 \equal{} 1$, both of which are complex. :wink:[/quote]

No no, I think $ x^2  \plus{} x \plus{} 1 \equal{} 0 \leftrightarrow \left( {x \minus{} 1} \right)\left( {x^2  \plus{} x \plus{} 1} \right) \equal{} 0$

if $ x \ne 1$.  :wink:

[quote="L_Euler"][quote="Kouichi Nakagawa"]I have originally input $ \Rightarrow$ and a place to input with $ \iff$ by mistake.
Mathlinker was not basic and might notice that it was a trifling mistake, but revised it some time ago.[/quote]

I don't understand (because I know a few of English word  :( ) , but I think my idea is right![/quote]

If Japanese goes in this site, the contribution is easy.

[hide="#3 Japanese translation"]
$ a^2 \plus{} a \equal{} \minus{} 1$ より $ a^3 \equal{} \minus{} a^2 \minus{} a \equal{} 1$
[/hide]

@Nguyen Ngoc Linh: What do you think about our solutions?

@: Are you from Vietnam?  :)  :)

First, I'm Vietnamese. ( u see : 'interested', it's wrong).
Your solutions are correct.
My solution: a^3=1 so a^1989=(a^3)^663=1^663=1.
therefore, A=1+1:1+2.
( And, a^3=1 doesn't mean that a=1)

[quote="Nguyen Ngoc Linh"]Oh, a mistake. :blush:
A=1+1:1=2 (not A=1+1:1+2)[/quote]
a:b is a/b ?
In the future, I hope you write a/b.

It might have been confused with ratios.  :wink:


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

8. What is the y-intercept of the line that is the perpendicular bisector of the segment joining (-3,-4) and (5,10)? Express your answer as a mixed number.

Don't the coordinates get flipped over and the x-coordinate becomes opposite value?
[hide="Answer"]$ 3\frac47$[/hide]

[hide]mid point=[(5-3)/2,(10-4)/2]=(1,3)
slope of original seg=(10+4)/(5+3)=7/4
slope of new line=-4/7
equation of new line y=-4x/7+b, passes (1,3), get b=3 4/7.[/hide]


From rosy bowers we issue forth,
From east to west, from south to north;
Unseen, unfelt, by night, by day,
Abroad we take our airy way;
We foster love and kindle strife,
The bitter and the sweet of life.
Piercing and sharp, we wound like steel -
Now, smooth as oil, those wounds we heal;
Not strings of pearl are valued more,
Or gems encased in golden ore;
Yet thousands of us every day,
Worthless and vile, are thrown away.
Ye wise, secure with bars of brass
The double doors through which we pass;
For, once escaped, back to our cell
No human art can us compel.

What are the "we" referred to in the puzzle?