calculus
derivative
Ring Theory
trigonometry
analytic geometry
trig identities
Law of Cosines

I have a problem that I have no idea how to solve. It is supposed to be solved using implicit differentiation and related rates. Let me know if you could help me figure it out, or just give me the steps to solving it. Thanks.

[i]A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m? [/i]

If anyone can figure out even the formula that we would use to find the rate, that would be much appreciated. I can't even figure out where to begin. Thanks.

Brett

I got an answer of 7/(pi) m/sec, but I don't think that is right. I set up an equation where d is the distance from the runner to the friend. I let the runner's speed be c'(rate around the circle). In this particular problem, the friend's distance is equal to the diameter so I used the equation for circumference of a circle c=(pi)d. Then d/dt[c=(pi)d]=[c'=(pi)d']. Plugging in the knowns. 7 m/s=(pi)d'. Ergo, d'=7/(pi) m/sec.

For some reason this doesn't seem right to me, so please let me know if my answer is correct or not.

Thanks.

Brett

No.
You said d was the distance between the friend and the runner, but later used d as the diameter.
You said c' was the rate of the runner, but later used c as the circumference, a constant.

I would use the law of cosines because you have two sides constant. You also know sin(theta)=y/r. The only part I haven't gotten up to is the exact location when they're 200m apart, in terms of x and y.

I know, I was completely off. I am currently using the law of cosines and polar coordinates, with the circle at the origin. Let me know if you finish. Thanks.

Brett

Here's what I got:

I set the center of the circular track he's running on as (0,0).  The friend is standing 200 m away, at (0, 200).  The position of the runner is given by the equation $x^2 + y^2 = 100^2$.  His speed is 7 m, and it's the same as the linear velocity.  The angular velocity is then: $\frac{d\theta}{dt} = \frac{v}{r} = \frac{7}{100}$.

The distance between him and his friend (0, 200) is:

$D = \sqrt{(x - 0)^2 + (y - 200)^2} = \sqrt{x^2 + y^2 - 400y + 200^2}$

Substituting:

$D = 200 = \sqrt{100^2 + 200^2 - 400y} = \sqrt{50000 - 400(r\sin{\theta})} = \sqrt{50000 - 40000\sin{\theta}}$

Solving for $\sin{\theta}$:

$200^2 = 50000 - 40000\sin{\theta} \rightarrow \sin{\theta} = \frac{-10000}{-40000} = \frac{1}{4}$

$\cos{\theta} = \frac{\sqrt{15}}{4}$

Now going back to the distance between him and his friend:

$D = \sqrt{x^2 + y^2 - 400y + 200^2} = \sqrt{(r\cos{\theta})^2 + (r\sin{\theta})^2 - 400(r\sin{\theta}) + 200^2}$

$D = \sqrt{(100\cos{\theta})^2 + (100\sin{\theta})^2 - 400(100\sin{\theta}) + 200^2}$

$\frac{d}{dt}(D) = \frac{-40000\cos{\theta}\frac{d\theta}{dt}}{2\sqrt{(100\cos{\theta})^2 + (100\sin{\theta})^2 - 400(100\sin{\theta}) + 200^2}} = \frac{-40000}{2 \cdot 200} \cos{\theta} \frac{d\theta}{dt}$

$\frac{d}{dt}(D) = -100(\frac{\sqrt{15}}{4})(\frac{7}{100}) = \frac{-7\sqrt{15}}{4}$

So at the moment when he is 200 m away from his friend, the distance between them is decreasing at a rate of approximately 6.778 meters per second.  I did all of those calculations in a rush, so I hope there are no mistakes.   ;)


[quote="sharkman"]Opposite vertices of a rectangular box whose edges are parallel to the coordinate axes are given below.  List the coordinates of the other six vertices of the box.

(0, 0, 0); (4, 2, 2)[/quote]
Um, I'm not sure if this is hard enough to be in intermediate...
[hide] (0,2,0) (0,2, 2) (0,0,2) (4,2,0) (4,0,0) (4,0,2)[/hide]


Can anyone show me a solution for #1? On the contest, i used the cotangents of the 1/2<A, 1/2<B, 1/2<C, 1/2<D. Although i thought that I had solved the problem, but I discovered a major calculation error which destroys my whole argument. Now, I don't know how to do this anymore.


What did you mean?
If you consider that $ i\equal{}\sqrt{\minus{}1}$, than this sum diverges.
If you consider the sum $ \sum_{i\equal{}1}^{\infty} \frac{1}{i^2}$, then this is the question, that asked a lot of times before. See for example [url]http://www.mathlinks.ro/viewtopic.php?t=200685[/url]

There is a bunch of different methods to sum that series. Just pick one.


Figure $ 1$ is $ 3\times2^0\equal{}3$ endpoints.
Figure $ 2$ is $ 3\times2^1\equal{}6$ endpoints.
Figure $ 3$ is $ 3\times2^2\equal{}12$ endpoints.
$ \vdots$
Figure $ 5$ is $ 3\times2^4\equal{}\boxed{48}$ endpoints.


analytic geometry
geometry
trigonometry
inequalities
circumcircle
geometry proposed

Triangle $ABC$ is isosceles ($AB=AC$). From $A$, we draw a line $\ell$ parallel to $BC$. $P,Q$ are on perpendicular bisectors of $AB,AC$ such that $PQ\perp BC$. $M,N$ are points on $\ell$ such that angles $\angle APM$ and $\angle AQN$ are $\frac\pi2$. Prove that
\[\frac{1}{AM}+\frac1{AN}\leq\frac2{AB}\]

I have a solution using cordinates geometry!
If no one find a more nice solution, i'll post!

I have a nice and short metric solution, so here it goes:

We have that $AN=\frac{AQ}{\cos{QAN}}$ and $AM=\frac{AP}{\cos{PAM}}$.

But $AQ=\frac{x}{\cos{QAC}}$ and $AP=\frac{x}{\cos{PAB}}$, where $x=\frac{AB}{2}$.

So $AM=\frac{x}{\cos{PAM}\cdot \cos{PAB}}$ and $AN=\frac{x}{\cos{QAN}\cdot\cos{QAC}}$, so the inequality rewrites as:

$\cos{QAC}\cdot \cos{QAN}+\cos{PAM}\cdot \cos{PAB}\leq 1$

$\iff \cos{B}+\cos{(QAP-\frac{A}{2})}+\cos{BAM}+\cos{(QAP-\frac{A}{2})}\leq 2$.

But $\angle{B}=180-\angle{BAM}$, therefore $\cos{B}=-\cos{BAM}$, thus the inequality rewrites as $\cos{(QAP-\frac{A}{2})}\leq 1$, which is true.

[quote="Virgil Nicula"][color=darkred]Let $ABC$ be an $A$- isosceles triangle and let $d$ be a line for which $\{\begin{array}{c}A\in d\\\ d\parallel BC\end{array}$. Given are two points $P$, $Q$ such that $\{\begin{array}{c}PA=PB\\\ QA=QC\end{array}$.
Denote the points $\{M,N\}\subset d$ for which $\{\begin{array}{c}PM\perp PA\\\ QN\perp QA\end{array}$. Prove that $\frac{1}{AM}+\frac1{AN}\leq\frac2{AB}$.

[size=134][b]I think isn't necessary the condition $PQ\perp BC$. Maybe I was mistaken . The all is perhaps ![/b][/size][/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $\{\begin{array}{c}m(\widehat{MAP})=x\\\ m(\widehat{NAQ})=y\end{array}$. Suppose w.l.o.g. $0\le y\le x\le B$. Thus,

$\{\begin{array}{c}AB=2\cdot AP\cdot\cos \widehat{BAP}=-2\cdot AP\cdot \cos (B+x)=-2\cdot AM\cdot\cos x\cdot\cos (B+x)\\\ AC=2\cdot AQ\cdot\cos \widehat{CAQ}=2\cdot AQ\cdot\cos (B-y)=2\cdot AN\cdot\cos y\cdot\cos (B-y)\end{array}\|$ $\implies$

$\frac{1}{AM}+\frac{1}{AN}=$ $\frac{2}{AB}\cdot[\cos y\cos (B-y)-\cos x\cos (B+x)]=$ $\frac{1}{AB}\cdot [\cos (B-2y)-\cos (B+2x)]=$ $\frac{2}{AB}\cdot \sin (B+x-y)\sin (x+y)$ $\implies$

$\frac{1}{AM}+\frac{1}{AN}\le\frac{2}{AB}$, with equality iff $x=90-\frac{B}{2}$ and $y=\frac{B}{2}$. [size=134][b]Exists at least a mistake, please ?![/b][/size]

[b]Remark.[/b] Denote the circumcenter $O$ of $\triangle ABC$. Thus, $PQ\perp BC\Longleftrightarrow$ $OP=OQ\Longleftrightarrow$ $AP\cos x=AQ\cos y$.
But $\frac{AP}{AO}=-\frac{\sin B}{\cos (B+x)}$ and $\frac{AQ}{AO}=\frac{\sin B}{\cos (B-y)}$. Therefore,
$\cos x\cos (B-y)+\cos y\cos (B+x)=0$ $\Longleftrightarrow$ $\cos (B+x-y)+\cos (x+y)\cos B=0$.[/color]

I think are you sure $x+y\le\pi$ when you get $\sin (B+x-y)\sin (x+y) \le 1$ you must have $\sin(B+x-y)\ge 0$ and $\sin(x+y)\ge 0$ or $0\le B+x-y,x+y\le\pi$

Yes, $0\le y\le x\le B\Longrightarrow$ $\{\begin{array}{c}0<B+x-y\le 2B<180\\\ 0\le x+y\le 2B<180\end{array}$.


我们讲别的，可是不在这里讲，或者是讲了你没看见。

我们的目标不是数学竞赛
这只不过是一个过程！！！！！！！！！！

[quote="Enya"]反正我算槛外人，也不好乱讲。
不过你们除了数学竞赛不能讲点别的吗？:mad:
顺便讲一句大家不爱听的话，
所谓的数学大国，也只是IMO拿金牌。真正的，没有很大的建树啊。
丘成桐，也是在国外培养起来的。。。
估计大家都不爱听了，因为国人向来以数学好而骄傲。

末了，强烈推荐一本书，
I want to be a mathematician     by Paul R. Halmos
超级好看啊，最近才发现的～～[/quote]

这本书哪里有?

:blush: 不懂你们是怎么想的，就我看，只有弄过竞赛才知道多苦。。。。

[quote="Enya"]反正我算槛外人，也不好乱讲。
不过你们除了数学竞赛不能讲点别的吗？:mad:
顺便讲一句大家不爱听的话，
所谓的数学大国，也只是IMO拿金牌。真正的，没有很大的建树啊。
丘成桐，也是在国外培养起来的。。。
估计大家都不爱听了，因为国人向来以数学好而骄傲。

末了，强烈推荐一本书，
I want to be a mathematician     by Paul R. Halmos
超级好看啊，最近才发现的～～[/quote]

Could you share your expenrience about it with us?


A deck contains $n$ cards labeled $1,2,3,..,n$. Starting with an arbitrary ordering of the cards, repeat the following operation: if the top card is labeled $k$, reverse the order of the top $k$ cards. Prove that eventually the top card will be labeled $1$.

It's been discussed before, I think. Anyway, it's pretty straightforward: let $n_0$ be the largest card to ever appear on the initial position. It's clear that after it appears on the first position and we perform the move, it will never appear there again. Now let $n_1<n_0$ be the largest card ever to appear on the first position after we've performed the move with $n_0$, and so on. This must eventually stop, of course, and since it can only stop when the top card is $1$, we're done.


conics
ellipse
analytic geometry
graphing lines
slope
symmetry
trigonometry

If we have an ellipse with axies of 8 and 10, and a circle is drawn with its center as a focus of the ellipse, and is internally tangent, what is the radius of the circle?

This is from the AoPS books, and i wondered if there were any cool solutions, the one offered in the solution manual seems impossible for somebody to know just reading the book, i have one way that works pretty well, but i want to see other solutions.

Also the picture might not be accurate, it is hard to draw in paint

[hide]In some ON-system this ellipse will have the equation $\frac{x^2}{25}+\frac{y^2}{16} = 1$. Differentiating this implicitly we see that in the point $(x_0,y_0)$ the "slope" of the ellipse is $-\frac{16x_0}{25y_0}$. For the positive focus $(c,0)$ it is true that $c^2 = 25 - 16 = 9$, so we can assume the circle's centre is $(3, 0)$.

Let the sought radius be $r$. We have that the circle is given by $(x-3)^2 + y^2 = r^2$. Differentiating this implicitly gives that the slope of the circle in the point $(x_0,y_0)$ is $\frac{3-x_0}{y_0}$. Comparing this to the old value we see that $x_0 = 25/3$. Now, this point is not on the ellipse, and that's what beats me. This seems to imply that the question is impossible to solve. If $x_0$ was on the ellipse you could simply solve for $y_0$ and then from that you would get the radius of the circle and be done. I have no idea why it doesn't work. Doesn't internally tangent mean that the circle and ellipse have the same slope where they meet?[/hide]

[hide]Let the ellipse be , and let the circle be  (make sure to take  at the end to get the real radius; I just left it unsquared cause it's easier to deal with). Then simply plug one variable in terms of the other (probably  since it's easier), and you're going to get a 2nd degree equation in  in the form ; Even though there is four possible interesections, due to the symmetry of the figure, you're going to have two pairs of interesections, each pair having the same  coordinates (and opposite ). So you only want one pair, so determinant is .[/hide]

edit: I just did it this way and got  and ...maybe Kalle is right and it's unsolvable?

Here's a thought:

The slope of the tangent at the point of intersection is $-\frac{16x}{25y}$.  Remember that this will be perpendicular to the line going through the radius of the circle and that point.  This means the slope of the line going through the radius is $\frac{25}{16}$.  You have a point $\left(3,0\right)$, so the equation for that line is:

\[ y=\frac{25}{16}\left(x-3\right) \]

Plug this into the equation for the ellipse, and solve for $x$, coming up with $x\approx 4.3$, which is $1.3$ units from the focus.  $y$ at this point is $\approx 2.04$.  The pythagorean theorem will now give the radius:

\[ r=\sqrt{1.3^{2}+2.04^{2}}\approx 2.418 \]

I may have made some mistakes, but I believe this is correct.

[quote="Kalle"]Your solution is wrong. You can't just ignore $\frac{x_0}{y_0}$ and assume it is one, as you have.[/quote]
Yeah, I see what you mean.  I'll attempt it again in a bit if nobody else does. :)

well, here is the way that i did it

[hide]let T be the point of tangency, and the circle have a center at focus F_1, and the other focus F_2

by definition of a ellipse, we have $F_1T+F_2T=10$, and $F_1F_2=6$ and $F_1T=r$

look at $F_1T$, this is the radii of our circle, and since it is internally tangent, that segment is at it's minimum value, suppose that there was a value for that segment that was smaller than this, that would make the circle go outside the ellipse, contradiction...so it must be minimized

so use the law of cosines on $F_1F_2T$, we get

$36= r^2+(10-r)^2-2(r)(10-r)\cos\angle F_1TF_2$

$r=5-\sqrt{25-\frac{32}{1+\cos\angle F_1TF_2}}$
we use the negative one because we want to minimize $r$
to minimize this it terms of ${{\angle F_1TF_2}}$, it is pretty clear that when $\angle = 0$, then we have the minimum of $\boxed{2}$[/hide]

another way to do it

[hide]if we have the ellipse and the circle and they are tangent, when we solve for the x coordinate, then we should have only one solution, so when we solve for them, we get a quadratic, and set the discriminant to 0 and we get $r=2$[/hide]

Altheman:  Could you expand on that second method?

[hide]using the orientation of my picture, we have

$\frac{x^2}{25}+\frac{y^2}{16}=1$ for the ellipse and
$(x-3)^2+y^2=r^2$

we know that they are tangent, that means that they intersect 2 times, both with the same x cooridinate, and each with opposite y values

so when we solve that equation, we get a quadratic with $r$ in the coefficients, since we want one solution, we set the discriminant equal to 0

$16(x^2)+25(r^2-(x-3)^2)=400$
$16x^2+25r^2-25x^2+150x-225-400=0$
$-9x^2+150x+25r^2-625=0$
$150^2-4(-9)(25r^2-625)=0$

this way, we get that r=0, this is odd, but when i checked it over, then i found that there was another solution that had y being imaginary?! so i suppose that this does not work that well, i am suprised though, because it works for other similar situations like when you want tangent lines without using calculus,

i think you could do this for solving for y though, and making it so there is only one pair of y solutions, but my other method is way better than this...[/hide]

[quote="Kalle"]I think this last method is precisely what Elemennop suggested, except he/she accidentally wrote determinant instead of discrimant. This is what is so weird. My method is very standard also, and is the way I dealt with several problems when we were doing ellipses. I see no reason for it not working.

Does your first "working" solution ensure such a solution actually exists? I think there is something wrong with the problem. Using Maple, the system $x^2/100+y^2/64 = 1, (x-3)^2 + y^2 = 4$ has no real solutions.[/quote]
I agree; this problem is a bit strange.  On a second attempt, I continuously got $x=\frac{25}{3}$, just as you and others got earlier.  This obviously doesn't lie on the ellipse or the circle.  Can anyone explain where that $\frac{25}{3}$ came from?

the reason that you got no solutions is because the equation for the ellipse is

$\frac{x^2}{25}+\frac{y^2}{16}=1$, remember the major axis is $2a$

Ah right. The solution you get is $(5,0)$. That is why my solution does not work. My solution doesn't work for $y=0$.

recall optics properties of ellipses and circles... you have that a ray leaving from the center of the circle (which is one of the focuses of the ellipses) towards tangency point has to reflect back to the center of the circle _and_ to the other focus of the ellipses...
so it comes out that these three points are collinear, and you have two possible solutions...
ain't it correct and elegant? :)

otherwise.. the least distance from the focus to the ellipses is $a-c$ (with our standard notation, where $F_1F_2 = 2c$, and for any point $P$ on the ellipse we have $F_1P + F_2P = 2a$), and so $a-c$ is the least $d$ such that we have a solution for $(x-c)^2 + y^2 = d^2$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$, and since it's the least, the relative solution is double, so it's a tangent circle..

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be the equation of the ellipse $\epsilon$,

where $0<b<a$, $0<c$, $b^2+c^2=a^2$.I note the focus $F(c,0)$ and

a mobile point $L(au,bv)\in \epsilon$, $0<u$, $0<v$, $u^2+v^2=1$. I observe that

$LF^2=(au-c)^2+(bv)^2=a^2u^2+b^2(1-u^2)-2acu+c^2=(cu-a)^2$.

I shall ascertain the position of the point $L\in \epsilon$ for which the distance $LF$ is minimum,

i.e. $0\le u\le 1$ for wich $|cu-a|=a-cu$ is minimum. I observe that $u_{min}=1$,

 i.e. $L_{min}\equiv A(a,0)$ and the length $r$ of the appreciated radius is equal to $a-c$.

In the our case, $r=2$.

I tried to do this again, in a manner similar to Altheman's.  I came up with:

$r=\frac{64}{20-12\cos{\theta}}$

and then minimized it using the fact that r will be smallest when $\theta=180^{\circ}$.

I don't know if anyone cares at this point, but the reason the suggested method of setting the discriminant as zero does not work is because in the solution there is as suggested only one intersection between the circle and the ellipse, but for the two polynomials there is one real and one non-real solution!

The usual method of setting the discrimant to zero is used for a line and a more complex curve. The thing is that a line cannot have one real intersection and one non-real intersection with a curve, and hence the method works. As I said, I don't know if someone will care to read this, but it's always good to reflect about why methods don't work so you don't make stupid generalizations.

Yeah, I got $x=\frac{25}{3}$ when I tried with calculus.  So then I did this method.
Essentially, we are trying to find the smallest possible distance from a point on the ellipse to $(3, 0)$.  Note that for some point on the ellipse $(x, y)$, the distance is \[ \sqrt{(x-3)^2+y^2}=\sqrt{x^2+y^2-6x+9} \]  Now we can manipulate the equation of the ellipse. \[ \frac{x^2}{25}+\frac{y^2}{16}=1 \implies y^2=16-\frac{16}{25}x^2 \] So, plugging this in for $y^2$, we have that the distance from $(x, y)$ to $(3, 0)$ is \[ \sqrt{\frac{9}{25}x^2-6x+9}=\sqrt{(\frac{3}{5}x-5)^2}=|\frac{3}{5}x-5| \] This number, with $x \in [-5, 5]$, is minimized when $x=5$, thus yielding $\boxed{2}$.  So the circle DOES touch the ellipse at 1 point.
We could have just as easily maximized distance from $(-3, 0)$.  We would have ended up with $|\frac{3}{5}x+5|$, which is maximized at $x=5$ (or minimized at $x=-5$, a reflection of the previous picture), giving $8$, implying that the distance from the other focus is $10-8=2$.

The reason calculus didn't work is that the point of tangency is one where the derivative is undefined.  I originally thought that there were supposed to be 2 points of tangency: I didn't know that the problem was in figuring that out.

[quote="Elemennop"]Yep, I was thinking it was intersect on top and at bottom, not at the rightmost point of the ellipse...

So I guess next time I do this same method and end up with a void answer, it implies that it only intersects at one point ie. the vertex?[/quote]
Well, you should check your answer, for example by making sure that there is one and only one intersection.

edit: Depends on what method you meant. When having a circle and an ellipse as in this case, I'm not so sure setting the discriminant to zero is ever a good method, since it is possible for such equations to have one real and one non-real answer. Since we got $x = \frac{25}{3}$ in both the calculus method and the one by setting the discriminant to zero, it is possible that there is some connection between the two - i.e. both methods fail at the same time. However, unless you figure that connection out, I wouldn't dare using the non-calculus method.

Or if you wanted to, you could find the second derivative.  If, in that area, the circle's slope changes at a rate faster than that of the ellipse, then there is one intersection point: otherwise, there are 2.

[hide="Categorically, r=2 !"] Here is the my second solution (similarly with the [b]K81O7's solution[/b]). In the first quadrant, more exactly $3\le x\le 5$, the equation of the ellipse is $y=f(x)=\frac 45\sqrt{25-x^2}$ and the equation of the wanted circle is $y=g(x)=\sqrt{r^2-(x-3)^2}$. We observe that: $\left(\forall\right)x\in [3,5]$, $g(x)\le f(x)$ $\Longleftrightarrow$ $\sqrt{r^2-(x-3)^2}\le \frac 45\sqrt{25-x^2}$ $\Longleftrightarrow$ $25r^2\le 25(x-3)^2+16(25-x^2)$ $\Longleftrightarrow$ $25r^2\le (3x-25)^2$ $\Longleftrightarrow r\le \frac 15\left| 3x-25\right|$. Therefore, $\left(\forall\right) x\in [3,5],\ r\le \frac 15\left|3x-25\right|$ $\Longleftrightarrow$ $r\le \frac 15\min_{3\le x\le 5} \left|3x-5\right|$ $\Longleftrightarrow$ $r\le2$. We come to the conclusion $r_0=2$.
[u]Remark.[/u] Generally, for any ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the above property is truly, i.e. $r_0=a-\sqrt{a^2-b^2}$. \[ \boxed {\Join}\boxed {\bullet} \boxed {\ \bigstar\ }\ \boxed {\boxed {\mathrm {\ Categorically,\ r=2\ !\ }}}\ \boxed {\ \bigstar\ }\boxed {\bullet}\boxed {\Join}  \] \[ \boxed { \mathrm {\ S\ T\ O\ P\ }}  \] ......................:spider:  :gathering:  :winner_first:  :gathering:  :spider: \[ \boxed {\ Happy\ New\ Year\ 2\ 0\ 0\ 6\ !\ }  \]  [/hide]

[quote="Elemennop"]Well the non-calculus solution ($D=0$) works all the time for other graphs intersections (say, two ellipses or a circle + parabola)...it should just as well work here. The whole point of setting it equal to zero was to make two of the solutions imaginary, or non-existant and thus get tangency.[/quote]
Well, it will work if there is some sort of symmetry in the intersections, like if there are two intersections with the same x-value but different y-values (it's not possible to have a non-real solution in this case). However, if the circle only is tangent to the parabola in one point, I don't see any reason for why setting the discriminant to zero must work.

but it does not because even if they are tangent in the real plane, that does not mean they will not intersect on the imaginary/complex plane...like if you find the other solution for r, you get an imaginary number for for the other x-value


ten cards 1-10 are arranged in a stack face down so that the first card is removed; the second card is put at the bottom of the stack; the third card is recoved; the fourth card is put at the bottom of the stack; and so on, until only one card remains. The removed cards, in order, are 1-9. The remaining card is 10. In the original stack, wat was the sum of the cards adjacent to card 10?

[hide]  I think the order is 1,6,2,10,3,7,4,9,5,8.  So the sum of the numbers adjacent to 10 is $2+3=\boxed5$.[/hide]

honestly, is noone elso gonna answer this? ineed help on this one :|  :|  :|

meenamath is right. You can figure this out first by numbering the cards 1-10 (based on order) and figure out which ones are discarded and kept each round. Given that order, you then assign those numbers a different number (the one in the problem) and list them in order. You then get that list. I hope that's relatively clear.

Sorry if i was vague before.  Here's a solution.  Hopefully its better than my first one:
[hide]Consider 10 slots for the 10 cards.  The cards 1-9 are removed in that order.  So the first card to be removed is 1, which makes it the 1st card in the stack.  The second card to be removed is 2, so its the 3rd card in the stack.  Keep going like that numbering every other empty slot, and going back up to the top when neccesary.  Eventually, you will get 1,6,2,10,3,7,4,9,5,8. So the sum of the numbers adjacent to 10 is $2+3=\boxed{5}$.
[/hide]


From: [url=http://www.xxhh.net/]http://www.xxhh.net/[/url]

某造假币的浙江私人老板不小心造出了一张15元的,于是想去某偏僻山区花掉这张假币.到了河南,遇见一卖冰糖葫芦的,把15元的人民币递过去,要了一串1块钱糖葫芦.找完钱后,浙江老板晕了,找回的竟然是两张7块的人民币!

[quote="siuhochung"]:rotfl:[/quote]
呵呵. 总在数学世界里泡着终究是枯燥的.......所以我以后会定期发一些这样的东西, 让大家轻松一下.  :)


inequalities
extremal principle
triangle inequality
geometry open
geometry

Hi,

How could I show that the distance between 2 points inside an equilateral triangle is maximal when the two points are confused with the vertex of this triangle ?

Please, just hints  :wink: (using extremal principle with triangle inequality ?)

Draw a circle with one vertex as centre and the the other two vertices on the circumference. Then it is clear that the distance between any two points in (or on) the triangle cannot exceed the radius of the circle. :)

Lemma: If a point P is inside or on the edges of $ \Delta ABC$, then $ AP \le max\{AB,AC\}$
Draw a circle with centre $ A$ and radius $ max\{AB,AC\}$ and you will see why

[hide="Sorry i should hide the proof"]Let the 2 pts be $ P,Q$, the triangle be $ \Delta ABC$
$ Q$ must be in either one of $ \Delta ABP,\Delta APC,\Delta PBC$
By Lemma , $ PQ \le max\{AP,BP,CP\}$
Again by lemma, $ AP \le max\{AB,AC\},BP \le max\{AB,BC\},CP \le max\{BC,AC\}$
So $ PQ \le max\{AB,BC,CA\}$   :D[/hide]


I have not heard from Shobhit or Marcus in a while.  We are still registered to go to Princeton, but the deadline for dues is quickly approaching.  Also, I probably need to give the Princeton people a list of members ASAP.  I am not sure if everyone we want to go will go (as I found out with the Duke contest).  I am getting second hand information from Sean, but I would rather hear from one of you.  I have heard about using a van instead of flying.  If that is the case, I might be able to get us a good deal at Hertz.  We got a very favorable deal for the Duke trip this year.  Also, I know that Sean's dad is willing to donate a considerable amount.  We can use this topic in the forum to stay in better contact, as deadlines are fast approaching.

P.S.
Marcus, I still have your wedding gift to send you and your Larson book.  Tell me how to get them to you, or I can bring them at USC.

well, the BellSouth CEO never said anything...yet..., so I'm assuming we'll just have to drive. Mr. Neal said that he and his wife will be willing to rent vans and drive us to Princeton, and this will save us about 2000 dollars. The trip should cost about 1200 dollars then, but we'll get about 1500 to be safe. John Berman is flying btw (he has tickets already). 1500 dollars is definitely enough though, we'll charge each person on the trip 100 dollars, making 900 total, and me and sean can pay an extra 600.

And btw, everyone will be able to go except for Junda I think--exams or something

Er, are you asking Mr R?  Cause I don't think anyone else here would know.  And don't you see Mr R everyday in person anyway? (just saying...)

Shobhit said something about GSSM having exams.  But I'm not sure if that only applies for USC.  Hm.

[quote="mysmartmouth"]I have confirmation from everyone except Jim, Junda, and the second calculus person.  Apparently Junda isn't coming, so we NEED TO FIND TWO PEOPLE:

1 Algebra
1 Calculus

Also, is Jim coming?  I couldn't send him the email, since I don't have his email address.[/quote]

OOH MAYBE I'LL COME (no kidding... maybe algebra) (wait, I'd probably like fail, but oh well)

You don't actually have to tell us your roster until you arrive on the day of the contest. However, you do need to be sure of the number of people you have (if you register and pay for 10, and then want to take less than 10, you can only get the registration fee refunded if you make the change before Nov 22).

ok here's the deal:

I already bought plane tickets for 8 people (the 7 SC people and Mr. R) to leave Charlotte on December 15th at 5:45 PM, and we will arrive at Newark at 7:32 PM. We will leave Newark on December 17th at 8:00 AM and arrive at Charlotte at 9:57 AM.

The cost for the trip was 187 dollars per person (for a total of about 1500 dollars). My parents would be willing to drive half of the people and if you can drive us there too Mr. R that would be good.

The NC people will be going separately. John Berman already bought tickets two weeks ago and is going to NYC (he'll be getting a drive to our hotel) and Yasha and Arnav are going together around the same time from NC to Newark, so we should meet up with them.

On another note, I talked to Mr. Lightsey (the bellsouth CEO) and I have to fax him a letter with all the costs tommorrow and he will tell me how much he is willing to help out on Friday. Depending on how much he's willing to pay, we'll charge the individual SC people going to princeton.

Harry, Asif, I, and Sean are definitely going, but I have no clue who the other three are, so just email them or something.

Who's JB getting a ride with?  If I come I plan on taking a bus from Boston to NYC and then a bus from NYC to Princeton (the NJTransit).

His parents I believe.  Can you check up on bus prices from Newark to the hotel?  That's something I thought about, but then disregarded when I thought we were driving.  But since we're flying again, I need to tell the kids to prepare for the extra out-of-pocket costs.

Local buses in Princeton will be $\$$1.25 a pop. Probably the one from Newark (or NY) to Princeton will be more, but I've never taken it so I don't know. The train from Newark to Princeton Junction is $\$$13.50, and the train station is adjacent to the airport, so this is convenient. (There is a local bus stop in Princeton Junction; most likely you can take a bus to your hotel from there. If not, getting your train ticket to Princeton campus is only $\$$1 more, and then you can take a bus from there.)

http://www.njtransit.com

When I searched the NJ Transit site for buses from Newark Airport to Princeton I couldn't find anything.  I think the best bet is the train (same one Nathan mentioned).

It's $\$$13.25 one way except I think you can get by for $\$$9, Sean. ( [url=http://www.njtransit.com/sf_tr_schedules.jsp?action=entry&resptype=U&MinB=0&MinA=0&Fare=Y&Line=&LineDesc=&LineLookup=&LineExt=&LineExt2=&STime=12%3A00+PM&RTime=1%3A00+PM&Oloc=Newark+Airport&OSID=37953+++&Dloc=Princeton+Junction&DSID=125+++++&back=sf_tr_schedules.shtml&ori=37953+++%3ANewark+Airport&des=125+++++%3APrinceton+Junction&DOW=W&mm=11&dd=15&yyyy=2006&date=11%2F15%2F2006]link to train site[/url])

[hide="P.S."]
JK sean.[/hide]

Ordered To-Do List:

[list]

* Thursday - Contact Junda & Jim w/ email.

* Thursday - Get permission forms out (mail to whom?)

* Friday - Reserve hotel rooms (Saiem?).

* Friday - FUNDING UPDATE!  Send out email ASAP.

* Create day-by-day itinerary.  This will point out any lapses in reasoning.

[/list]


slove this equation:

$ x\plus{}2005 \equal{} \frac{1}{2}\sqrt {3 \plus{} \frac {1}{2}\sqrt {3 \plus{} \frac {1}{2}\sqrt {3 \plus{} \frac {1}{2}\sqrt {x \plus{} 2008}}}}$

Let x+2005=k.Now the equation becomes,
 $ k = \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + k}}}}$
Which is equalent to the equation
 $ k = \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + ......}}}}$
Which is equalent to the equation,
 $ k = \frac {1}{2}\sqrt {3 + k}$,which is an easy quadratic to solve.
on solving,we get k=1 or $ {\frac { - 3}{4}\}}$
 on checking k=1 is the only solution.

Hence $ x = - 2004$ is the only solution.  :!:

My solution:
Let $ u=x+2005$ we have $ 2u=\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {u + 3}}}}$

Let $ v=\frac {1}{2}\sqrt {3 + \frac {1}{2}\sqrt {u + 3}}$ we have $ \begin{cases}2u=\sqrt{3+\frac{1}{2}\sqrt{3+v} \\ 2v=\sqrt{3+\frac{1}{2}\sqrt{3+u}\end {cases}}}$

$ \Longrightarrow u=v \Longrightarrow u=1\Longrightarrow x=-2004$


Hello everyone.  I get the impression that PDF documents can incorporate interactive 3D graphics, you know, use the mouse to rotate a 3D figure embedded in a PDF document.  That would rock.  Is there a way to save a graphics image in a PDF file to allow this?

Thanks,
Shaw

Have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=movie15]movie15 package[/url].

Thanks Steve.  I'll check it out.  I'm currently using the subfig package to embed several plots in one figure.


Let T : R^3 --> R^3 be a linear transformation. Show that there is a line L such that T(L) = L.

http://www.mathlinks.ro/Forum/viewtopic.php?t=146823 - problem 4


geometry
inradius
perimeter
MATHCOUNTS
number theory
relatively prime
prime numbers

I posted this on OK forum also.

I found this just today, also.

There's 2 methods to find the triples really easily.

The methods are [u][b]Method of Pythagoreans[/b][/u] and [u][b]Method of Plato[/b].[/u]

[u]Method of Pythagorean:[/u]
let $ m$ be any odd number  greater than 1.

$ m, \ \frac {m^2 \plus{} 1}{2}, \ \frac {m^2 \minus{} 1}{2}$

[u]Method of Plato: [/u]
let $ n$ be any integer greater than 1.

$ 2n, \ n^2 \plus{} 1, \ n^2 \minus{} 1$

$ \star$ I don't know how those work out, but this is pretty cool.
If you know how it works, can you guys explain how those work?

[quote="DonkeyKong"]I posted this on OK forum also.

I found this just today, also.

There's 2 methods to find the triples really easily.

The methods are [u][b]Method of Pythagoreans[/b][/u] and [u][b]Method of Plato[/b].[/u]

[u]Method of Pythagorean:[/u]
let $ m$ be any odd number  greater than 1.

$ m, \ \frac {m^2 \plus{} 1}{2}, \ \frac {m^2 \minus{} 1}{2}$

[u]Method of Plato: [/u]
let $ n$ be any integer greater than 1.

$ n, \ n^2 \plus{} 1, \ n^2 \minus{} 1$

$ \star$ I don't know how those work out, but this is pretty cool.
If you know how it works, can you guys explain how those work?[/quote]

I'll assume the triples are meant to be pythagorean triples and not just random numbers

I don't think plato's works, try $ n\equal{}2$.

and as for the other one, I believe that, in order, it's
$ m$, $ \frac{m^{2}\minus{}1}{2}$, and $ \frac{m^{2}\plus{}1}{2}$.

Anyways, to show it's a triple, just show that the sum of the squares of two terms is the square of the other number

$ m^{2}\plus{}\left(\frac{m^{2}\minus{}1}{2}\right)\equal{}m^{2}\plus{}\frac{m^{4}\minus{}2m^{2}\plus{}1}{4}\equal{}\frac{4m^{2}}{4}\plus{}\frac{m^{4}\minus{}2m^{2}\plus{}1}{4}\equal{}\frac{m^{4}\plus{}2m^{2}\plus{}1}{4}\equal{}\left(\frac{m^{2}\plus{}1}{2}\right)^{2}$

The only reason it doesn't work for even numbers is that it won't generate integers for every number

Oh.. wait, My mistake.. it's suppose to be $ 2n, \ n^2 \plus{} 1, \ n^2 \minus{} 1$.

Sorry.. good catch alanchou. Thanks.

And I also kind of get where you came up with 1st one, but how about the 2nd one??

Do you know?

[b]Babylonian Method:[/b]  $ m^2 \minus{} n^2, 2mn, m^2 \plus{} n^2$

The Platonic method is the special case $ n \equal{} 1$.  The Pythagorean method is the Platonic method divided by $ 2$ when $ m$ is odd.

Proof:  $ (m^2 \minus{} n^2)^2 \plus{} (2mn)^2 \equal{} m^4 \minus{} 2m^2 n^2 \plus{} n^4 \plus{} 4m^2 n^4 \equal{} m^4 \plus{} 2m^2 n^2 \plus{} n^4 \equal{} (m^2 \plus{} n^2)^2$

This method generates [b]all primitive[/b] triples when $ \gcd(m, n) \equal{} 1$ (that is, all triples $ (a, b, c)$ where $ \gcd(a, b, c) \equal{} 1$).  The other two methods only generate some triples.

What do you mean?  The second method finds a lot more than just $ 3\minus{}4\minus{}5$ triangles.

[quote="mz94"]DK's 2nd one is to find any 3,4,5 triangles[/quote]Not really... It gives you various sets of pythagorean triples.

t0rajir0u, when would you use the Babylonian method? If you have to go through the trouble to find two relatively prime numbers, then plug them in a rather sizable equation, you might as well remember all of the primitive triples... there are only so many! :wink: :P

Yes, but by the time you get into the largest ones, it won't be exactly easier to apply the babylonian method. It's only faster than just using the plugging into the pythagorean for the low values, where you can easily memorize.  :lol:

[quote="darkmatter47"]Yes, but by the time you get into the largest ones, it won't be exactly easier to apply the babylonian method. [/quote]

Can you find me all primitive triples who have $ 120$ as a side without the "Babylonian" method?  ;)

The use of the method I gave is twofold.  First, it allows you to solve problems that require you to consider an arbitrary or unknown primitive triple, or problems that I just gave above that require computation of large primitive triples with certain properties.  Second, it allows you to prove things about primitive triples that would be difficult without it.

Another concrete application:  find all right triangles of inradius $ 24$.

[hide="Discussion"] It turns out that, for a right triangle with side lengths $ a, b, c$, the inradius is $ r \equal{} \frac{a \plus{} b \minus{} c}{2}$.  The area of a right triangle is $ A \equal{} \frac{ab}{2} \equal{} rs$ where $ s$ is half the perimeter $ \frac{a \plus{} b \plus{} c}{2}$.  Letting $ a \equal{} 2mn, b \equal{} m^2 \minus{} n^2, c \equal{} m^2 \plus{} n^2$, we can calculate that

$ r \equal{} n(m\minus{}n)$

Remember, this is for [i]primitive[/i] triples.  What you need to do is find the primitive triples with inradius $ 1, 2, 3, 4, 6, 8, 12, 24$.  Do you want to try this problem without the "Babylonian" method? [/hide]

I see. Nice tip... could come in handy (perhaps on target mathcounts)  :)

Target MathCounts, except perhaps at nationals and tougher state levels, wouldn't have problems like this.

The Babylonian method generates ALL primitive pythagorean triples. To get even more triples, multiply each term by k.


number theory
least common multiple
MATHCOUNTS
inequalities
geometry
AMC
AIME

[quote="diophantient"][quote="baudhack"]Without using brute force, find the smallest value for N in which 7 + 30N = composite number is satisfied.[/quote]
http://web.mit.edu/hmmt/www/datafiles/problems/2007/pguts07.pdf
Number 12.[/quote]

Okay...explain more, please.

[quote="buzzer11"][quote="baudhack"]Without using brute force, find the smallest value for N in which 7 + 30N = composite number is satisfied.[/quote]

[hide]
7 and 30 are relatively prime. Thus their lcm is 7*30.

Now we look at the expression $ 7+30N$. Since we dont know what N is, first look at 7. For the expression to be composite, it must be divisible by 7, since 7 is prime. this means 30N is divisible by 7. Thus, the smallest integer N would be $ 7$. (I'm assuming you mean nonnegative integer)[/hide][/quote]

Thanks. The problem is that N=6, because 187 is divisible by 11.

The assumption being made is that x | (7+30N) => x | 7, x | 30N
But that doesn't work in addition
1+3 = 4, 2 doesn't divide 1 or 3

baudback, you asked for a solution without brute force. If I remember correctly, this is from 2004 State Target (problem 7? yes I need to get out more...), and many MathCounts problems require you to just plug and chug. Fortunately, this one isn't too bad, you only have to check 1-6, which shouldn't take more than 30 seconds. Also, it's pretty trivial to find this solution path, even after searching for a non-trial-and-error solution.

Many problems in advanced number theory involve similar case-checking, for example a problem from 104 Number Theory Problems from the Training of the USA IMO Team, asked to find all positive integer n such that n!+5 was a perfect cube. I found a solution using mod 8, requiring to check n=1,2,...,7, the book gave a similar solution using mod 7, requiring to check n=1,2,...,6, and also a more technical solution, requiring to check n=1,2,...,9. Such casework, actually, is one of the reasons why I don't like number theory as much as say, geometry.

But brute force can be good. It builds character, and is always a handy tool if you don't know how to solve any hard problem on any hard test. Even at MOP they teach you fancy brute force techiniques (muirhead for inequalities, mass points for geometry, etc.). Look at kyyuanmathcount, he totally 0wned #12 on this year's AIME with coordinates. It's not something he should be completely proud of (he has many other achievements for that  :D ), but hey, he solved the problem and his index went up by 10.

OK, I'm done now.

[quote="13375P34K43V312"]But brute force can be good. It builds character[/quote]

nice rant Carl Lian...

anyways, as i did this problem, i found that n=6 by plug and chug. I don't see another way to do it...   :huh:

[quote="13375P34K43V312"]brute force ... is always a handy tool if you don't know how to solve any hard problem on any hard test.[/quote]


Besides essentials, what are some good items to bring?  My ideas:

[u]Stuff[/u]
XBOX 360/Halo 2
Money
Cards
Swimming Trunks

What else?

I personally wouldn't advise the bringing of expensive electronic equipment (other than your calculator), but I suppose if you can't live without it for one night ...

A pillow.  It took until my 3rd or 4th ARML before I remembered this, but they don't give you a pillow at the Penn St. site.  Balled-up jeans are much less comfy to sleep on.

It's two nights for us this year since we (SC) are going up a day early ;)

[hide="random off topic rambling"]About the pillows...I actually didn't bring a pillow to running camp (which lasts a week) a few years back...actually I don't think I brought a pillow to running camp two years in a row even though the second year I remembered perfectly.  Balled up towels weren't too bad...except that I only had 2 and I obviously needed to shower...wait, why am I telling you guys this?  Let me get back on topic...[/hide]

You don't [i]need[/i] anything else.  I'll be bringing a frisbee though, so SC's covered there.  Someone else from SC should bring a basketball...I can go out and buy one if no one has one.

Actually...I'm going to start a post in the SC section.

My packing list

Frisbee
Clothes (shorts & T-shirts ... gotta look my best =)
SOAP
SHAMPOO
WASH RAG
TOWEL

Forgetting these last three items = BAD, especially if you are staying extra nights, and ESPECIALLY ESPECIALLY bad if you are rooming with me (bad for me mostly)

Conditioner (No comments from the peanut gallery)

Cards
Maybe some poker chips, though my friend will have some
MP3 player

Even if I had finals this year, I've given up on bringing textbooks ... I just can't work on bus rides.

CALCULATOR
PENCILS/PENS

And, of course
[hide]Your A game =)[/hide]
JB

I told our guys to pack light and don't take anything too valuable. We're only going to be in Las Vegas for about 24 hours (shortly after noon on Friday to shortly after noon on Saturday). They did give us pillows in San Jose; I don't know about Las Vegas becuase we haven't been there before but I assume we'll get them.

Cell phones: banned in the team/power round rooms, of course, and switched off during the individual/relay rounds, but there often is some benefit to being able to call your parents and say, "The bus just passed through Victorville and we'll probably make it to Duarte by _____ ."

[quote="Kent Merryfield"]Cell phones: banned in the team/power round rooms, of course, and switched off during the individual/relay rounds, but there often is some benefit to being able to call your parents and say, "The bus just passed through Victorville and we'll probably make it to Duarte by _____ ."[/quote]

This is important.  Two years ago I was the official cell phone confiscator for the NYC A team :)

[quote="JBL"]A pillow. It took until my 3rd or 4th ARML before I remembered this, but they don't give you a pillow at the Penn St. site. Balled-up jeans are much less comfy to sleep on.[/quote]
[quote="joml88"][hide]About the pillows...I actually didn't bring a pillow to running camp (which lasts a week) a few years back...actually I don't think I brought a pillow to running camp two years in a row even though the second year I remembered perfectly. Balled up towels weren't too bad...except that I only had 2 and I obviously needed to shower...wait, why am I telling you guys this? Let me get back on topic...[/hide]
[/quote]
Um, am i the only one who does not sleep with a pillow?

[quote="quantum leap"]Um, am i the only one who does not sleep with a pillow?[/quote]

Based on a sample size of 3, yes.

I would advise against bringing the video games.  Where would you even play them?  Card games are more fun anyway, IMHO.

One thing I would strongly advise adding to your list is an extra change of clothes if you plan on playing frisbee.  My team (MN) would always have a game Friday afternoon, and it was nice to have a change of non-sweaty clothes to wear for that evening.

So here's what I would usually pack:
2 changes of clothes (1 for post-frisbee, 1 for Saturday)
My binder full of problems from practice
Math supplies
At least one frisbee
At least one deck of cards
SET deck
Apples to Apples
Towel, soap, toothbrush, and all that

That's really all you need (and some things you don't need).  As people have said before, pack light.

Sly Si, you've left unanswered the all-important question of whether there are pillows at the Iowa site!  :)

I found bringing some snack foods nice.
Last year, we pigged out on ice cream, tortilla chips and salsa, and Goldfish crackers.
Card games are great on the bus.
I also brought an alarm clock and a light blanket.

[quote="perfectnumber628"]don't forget the most important thing: RUBIK'S CUBES!!!!!

(i will be bringing 5)[/quote]
I'm not worried about a shortage of cubes ever being a problem for the Southern California team - not as long as [url=http://cubefreak.hp.infoseek.co.jp/]you-know-who[/url] is a member.

I just checked my packing list for ARML (not going this year, :( ) and it contains...
...toilet paper.

A couple of years ago, there was no toilet paper in the suite I was assigned to, so now I always bring a roll.  (Good thing it's on my list, or I would forget every time.)

I suspect that students don't need to worry about this, because they have communal bathrooms, right?

[quote="perfectnumber628"]don't forget the most important thing: RUBIK'S CUBES!!!!!

(i will be bringing 5)[/quote]

Hmm... maybe that's why your avatar's a Rubik's cube...
I stink at those, by the way.

Bring electronic board games or just regular board games for the nights.

Eh, too bad Macky is going to Las Vegas... I wish he was going to Penn State so I could meet him.

I remember having a pillow at the Iowa site last year.

Wow, at first I thought that you created that account especially for that post...but your join date is like October. What a coincidence.

Anyways...lessee...Iowa site. I remember it being REALLY uncomftorble. Bring an alarm clock, or you won't be able to tell time (unless you can tell by the position of the sun...at night). Also, bring bottled water for brushing your teeth. I think they have somewhat hard water there.

Yes, you're given pillows and blankets at the Iowa site.  I found them to be reasonably comfortable, though they're a far cry from hotel-quality.  After all, you're only there one night (two if you're lucky).  IMHO, bringing your own bedding would be a waste of space.  And I thought the water was fine, though bottled water for drinking might not be a bad idea.  Also, really good drinking water can be found in the student union; we always used to go there for breaks during frisbee games.

And filletwho makes a great point about brining a clock of some kind.  Our alarm clock generally consisted of the coaches banging on our doors when it was time to get up.  There was at least one year when I woke up early with no idea what time it was and didn't want to go back to sleep because I thought it was almost time to get up.  To make matters worse, I heard what I thought was door-banging down the hall, but it turned out to be Chicago people (I think) going in and out of the bathroom.  Needless to say, I got less sleep that year than I should have.

Another piece of advice for the Iowa site, especially for first-timers: Burge hall is big.  Burge is confusing.  You can get yourself lost pretty easily.  Staying with other people is highly recommended until you find your way around (which may take two or three visits to ARML).  Also, in the room numbers, the first digit is the corridor number, not the floor number.  Second digit is the floor.

My list:

Multiple decks of cards
AoPS volumes 1 and 2
Pillow (for the bus)
Set, of course.   SET!
My rubik's cube [and, most likely, my idiot's cube (one from which I have removed all of the stickers.)]
A few books.
And, finally, a towel, because you should always know where your towel is.  Not really.

[quote]Wow, at first I thought that you created that account especially for that post...but your join date is like October. What a coincidence. [/quote]

Yeah, I created the account and then got busy with other things and never got around to using it.

I brought my laptop last year; didn't have any problem with access in the room, and there's usually time for it somewhere in my two nights there. I talked with a few people this year about possibly bringing my NES, but it'd probably be more trouble than it's worth.

[quote="Elemennop"]I'm thinking of bringing a gamecube and SSBM, because lately I've been unable to get through a single day without 3-4 hours of SSBM...but I dunno who's bringing a TV.[/quote]
I don't think that something like that is absolutely necessary, if you ask me.

I'm going to have to agree with that.  A gaming console is not necessary, and of course, is liable to be stolen.

I have seen, however, people bring a console with them in previous years of the competition.  It doesn't mean that it's impossible, it's just somewhat risky.

I would also strongly recommend against bringing a game system.  There are plenty of other things to do, and you're not there for very long.

I dunno, I might not...but seriously, I have withdrawal if I don't play it for long periods of time...It's really bad.

Wow, there's this kid in my chemistry class that brings SSBM to class now that ap's are over. he's supposedly the best in my school.

I'm a SSB purist. I think it's a lot funner to play than SSBM (the controls are too fluid and wierd).

Bah, SSB is horrid. It's way too slow...there's absolutely no speed or reaction time involved when playing SSB. SSBM lightning melee is the only thing I play that's fast enough to keep me entertained.

Yeah I've only played SSBM a few times but I don't like it as much as SSB. (Maybe because I've gotten too used to SSB...)


This may sound trivial, but i have no proof of it.

Given any triangle. Prove that the inscribed circle is the largest possible circle that can be contained in the triangle. Similarly, proof that the circumscribed circle is the smallest circle containing the triangle.

[quote="mao_zai"]This may sound trivial, but i have no proof of it.

Given any triangle. Prove that the inscribed circle is the largest possible circle that can be contained in the triangle.[/quote]

This statement is correct.

[quote]Similarly, proof that the circumscribed circle is the smallest circle containing the triangle.[/quote]

This statement is incorrect.  Why?  Indeed, what is the correct relationship between the smallest circle and the triangle that it contains?

[quote]This statement is incorrect. Why? Indeed, what is the correct relationship between the smallest circle and the triangle that it contains?[/quote]

oops. :blush:  you're right. i think the statement is true for non-obtuse triangle (all angles $\le 90^{o}$).

For obtuse triangles, the smallest circle should have the diameter equal to the length of the longest side of the triangle.


Let $ p_a,p_b,p_c$ be the lengths of the altitudes from the centroid of a triangle $ ABC$ with inradius $ r$. Prove that $ r\equal{}\left(\frac{\frac{1}{p_a}\plus{}\frac{1}{p_b}\plus{}\frac{1}{p_c}}{3}\right)^{\minus{}1}$.

the distances from the centroid to the sides of ABC are equal to a third of the corresponding altitudes then susbtitute them into the well known relation $ \frac {1}{r} \equal{} \frac {1}{h_a} \plus{} \frac {1}{h_b} \plus{} \frac {1}{h_c}$ and we are done.

Let the triangle $ ABC$ have area $ A$, semiperimeter $ s$ and centroid $ G$. Let any triangle $ XYZ$ have area $ |XYZ|$.
Then $ r\equal{}\frac{A}{s}\equal{}\frac{1}{\frac{1}{2}\left(\frac{a}{A}\plus{}\frac{b}{A}\plus{}\frac{c}{A}\right)}$
$ \equal{}\frac{1}{\frac{a}{2|BCG|}\plus{}\frac{b}{2|ACG|}\plus{}\frac{c}{2|ABG|}}$ since the lines from the vertices to the centroid divide the triangle into triangles of equal area
$ \equal{}\frac{1}{\frac{1}{p_a}\plus{}\frac{1}{p_b}\plus{}\frac{1}{p_c}}$.


algebra
polynomial
probability
quadratics
analytic geometry
graphing lines
slope

I imagine this question has been posed many times before, but I came up with it yesterday. What is the probability that a degree n polynomial, where n is even, has at least one real root given that all its coefficients are real? It then dawned on me that the problem isn't that easy even in the case n=2. Would anyone care to try it, either the general case or even the quadratic case?

I have a paper on this subject, anyone interested contact me.

I think you have to define how you are are picking your ramdom coefficients for this problem to be meaningful. In other words, you to have to define some probability measure. Because in the case of n=2, for example, the polynomial will have a real root if b^2 :ge: 4ac. But that doesn't have a meaningful probability until you put a measure on the space of choices for a, b, and c.

What if you were to select the coefficients with normal distribution with mean 0 and standard deviation 1? Previously I meant uniform distribution, but working with uniform distribution seems to cause some problems. I imagine that the problem is solvable as stated with normal distribution.

This topic has been well studied.  For example, see the book "Random Polynomials" by Albert Bharucha-Reid.  (This book is unfortunately out of print, but you can find it in university libraries.)  I believe that both the uniform and normal distributions are covered.  A related problem that has been studied even more is the eigenvalues of a random matrix.

Does anyone know how the solution from (-1. 1) in each variable is related to the solution from (-infinity, infinity) in each variable, if applicable?

Never mind, I've arrived at a solution.

Well, in order for there to be no real roots, the value b - 4ac must be less than 0. The case where it is 0 is negligible compared to the cases that we are evaluating.  Consider the dimensions of space to be a,b,c. I will proceed by determining the portion of space relative to the entire space that meets the criterion.
Solve for c. Then it becomes c = b/4a. If you choose any arbitrary b, for example 1, you'll notice that at that b, the line c(a) divides the plane into 2 equal halves, because the slope is constant at that b and the y-intercept is 0. If we extend this over all b, it follows that because at all cross-sections, the probabilty is 1/2, then over space, the probability must also be 1/2.

Keep in mind that we want not only a and b but also c to be within [-1,1] in your simplified version of the problem. But that doesn't really answer the question. Is it true that when we get the same probability switching from [-1,1] to [-infinity,infinity] for each term? Perhaps to transform from [-infinity, infinity] to [-1,1] we casn divide by the largest of (|a|,|b|,|c|) so that one of them is on the boundary and the others are inside (or possibly on the boundary, but we can ignore that case) and then work out probabilities. But does that change anything? Maybe not, since in the expression b^2-4ac, every term is quadratic. Any thoughts?

Simon, my solution was for over each variable from -infinity to infinity.

For higher n, I hypothesise the solution may involve the fundamental theorem of algebra (for example, for n=4, the only case where there will be no solution is when it factors into (a1x + b1x + c1) (a2x + b2x + c2) = 0, where both b1 - 4a1c1 < 0 and b2 - 4a2c2 < 0.

I agree that you can factor any quartic like that, but where does that get us? It's not as if we can take the probability of it working for 2, say P, and say that probability for 4 is 1-(1-P)^2.


trigonometry
inequalities
geometry
rectangle
LaTeX
function
trig identities

Two polygons are given on the plane. Assume that the distance between any two vertices of the same polygon is at most 1, and that the distance between any two vertices of different polygons is at least $ 1/\sqrt{2}$. Prove that these two polygons have no common interior points.

By the way, can two sides of a polygon intersect?

All right, I'll give it a shot.

It's easy to see that if the polygons share common interior points, at least one vertex of the first polygon, $ P_1$, must be in the interior of the second polygon $ P_2$. Call this vertex $ v_0$. For the sake of contradiction, let's say $ v_0$ exists and satisfies the conditions.

Now define vertices $ v_1, v_2,$ and $ v_3$ to be the vertices of the triangle that  enclose $ v_0$. By assumption, this triangle has side lengths equal to or less than 1. Furthermore, the distances between $ v_0$ to the vertices of the triangle are atleast $ \sqrt {2}/2$.

It is trivial to see that there must be an angle formed (with $ v_0$ as the middle point) between $ v_0$ and two vertices of the enclosing triangle with an angle of 120 degrees or more. Let the measurement of this angle be x. WLOG, let the vertices of the enclosing trying be $ v_1$ and $ v_2$. And apply the law of cosines on the triangle formed by $ v_1$, $ v_2$, and $ v_0$.

Let side $ v_1v_2$ be C. Let side $ v_0v_1$ be A, and $ v_0v_2$ B.

$ C^2 \leq 1$, and $ A^2, B^2 \geq 1/2$, and $ \minus{} 2ABcosx^\circ \geq 1/2$.

The last inequality is true because AB is atleast 1/2, and $ \minus{} 2cosx^\circ$ is atleast one since x is in the range (120,180).

Therefore the left side of the expression $ A^2 \plus{} B^2 \minus{} 2ABcosx^\circ \equal{} C^2$
must be at least 3/2, while the right side can be at most 1. This contradiction completes the proof.

But your extension question makes me think I may have misinterpreted the problem.  :|

[quote="holmes.roams"]It's easy to see that if the polygons share common interior points, at least one vertex of the first polygon, $ P_1$, must be in the interior of the second polygon $ P_2$.[/quote] Consider two rectangles, one with vertices (0, 0), (1, 0), (0, 3) and (1, 3), the other with vertices (-1, 1), (2, 1), (-1, 2), (2, 2).

Ah, of course.  :oops:

Seems as though I'll have to rethink this one drastically.  :o

If two polygons have a common point, then at least two of their lines will intersect. Let the lines be $ AB$ and $ CD$ intersecting at angle \theta. We have $ AB, CD < \equal{} 1$ Let the intersection point be $ O$.  The distance of separation of two vertices of different polygons will be:

$ d \equal{} \sqrt { min(AO,BO)^2 \plus{} min(CO,DO)^2 \minus{} 2min(AO,BO)min(CO,DO)cos(\theta)} \\   <\equal{} \sqrt {(\frac {1}{2})^2 \plus{} (\frac {1}{2})^2 \minus{} 2 \frac {1}{2}\frac {1}{2} cos( \frac {\pi}{2}} ) \\   \equal{} \frac {1}{\sqrt {2}}$

But we require $ d > \frac {1}{\sqrt {2}}$ , a contradiction. Hence, no lines of different polygons intersect.

[quote="holmes.roams"]That's not necessarily true, one can be completely inside the other.[/quote]I think it is impossible

[quote="holmes.roams"]That's not necessarily true, one can be completely inside the other.[/quote]
Ah! You're right. I completely overlooked it.

For case 2: One completely inside the other and no lines intersect.

We can enclose the outside polygon with a circle $ C$ of diameter $ 1$, since the distance between any two of its vertices is at most $ 1$. Let $ A$ be a vertex of the inside polygon. Draw a chord through $ A$ such that $ A$ is the midpoint of the chord, dividing circle $ C$ into two segments. Let its length be $ 2d$. Each of the two segments of $ C$ will have at least one vertex of the outside polygon so as to enclose $ A$. Let $ B$ be the vertex in the smaller part. The maximum possible distance between $ B$ and $ A$ is clearly $ d$. And since $ 2d$ is a chord of $ C$, $ d < \equal{} \frac {1}{2} < \frac {1}{\sqrt {2}}$ contradicting our requirement.

[quote="JBL"]LaTeX suggestions to make your writing more readable: \min, \cos (and all other trig functions, and may others) are built in: $ \min, \cos$.  \cdot makes a nice multiplication dot $ \cdot$, and \times makes the multiplication cross $ \times$.  \leq makes $ \leq$.[/quote]

Thanks. I was wondering about $ \leq$. I am new to latex.

[quote="greentreeroad"]By the way, can two sides of a polygon intersect?[/quote]
In the case when two sides of the same polygon intersect, I think the other polygon cannot even occupy the convex hull of self-intersecting polygon while satisfying the conditions. So, yes. But I am not sure.


Alright guys,
So how bout we bring some stuff to do when we need physical exercise (I know...exercise?) or fun mid breaks (though problem solving is generally my fun mind break, so I guess fun mind break of a fun mind break).

Physical Exercise  :juggle::
I'm guessing Basketball and Ultimate Frisbee are cool with most of the people. These games probably won't require much skill, considering this is math camp not the NBA or The Ultimate Frisbee International Summer League (U-FISL). Maybe we should have some sort of league. You know, have like 3 to 8 teams drafted on the first day and round robin afterwords.

Mind Break:
Chess is fun.
We could also play scrabble, monopoly, or something along those lines.
Definitely bring cards!!!
I can bring chips, but someone else bring some in case. (and everyone bring your bet...jk :P)

Any other ideas?

Yeah, I got a poker set complete with legit chips and two decks of cards, as well as several other decks of cards. I also got Settlers of Catan (w/ expansion). As for sports, someone should bring a soccer ball, because the only one I have is a size 4.

I can bring chips but probably nothing else, unless I get a good frisbee. (since I'm flying)

Psh. Who's bringing musical instruments? XD


If a<b<c
y=|x-a|+|x-b|+|x-c|
Find the minimal value of y.

Spoiler:

[hide]The min has to occur at either x=a,b,c.

At values very close to a, the equation becomes y = |x-a|-2x+b+c, we know that the minimum here is x=a, which gives y=-2a+b+c.

At values very close to b, the equation becomes y = |x-b|+c-a, we know that the minimum here is x=b, which gives y=c-a

At very close values to c, the equation becomes y = |x-c|+2x-a-b, same as before, x=c, y=2c-a-b.

I claim that c-a < 2c-a-b. It simplifies to b<c which is true.

I claim that c-a < b+c-2a. It simplifies to a<b which is true.

Therefore c-a is the minimum value of y.[/hide]

Another approach:

[hide]y = |x - a| + |x - b| + |x - c| :ge: (x - a) + 0 + (c - x) = c - a.

Choose x = b to achieve equality.[/hide]

Yet another approach.  [hide] Mentally sum the three absolute value graphs.  The rest is left for the reader. [/hide]

I don't know if this helps anyone else, but it helps me to think that [hide]if you're not at b, you can move in that direction and decrease |x-a| and |x-b| while only increasing |x-c|, and |x-a|, |x-b|, |x-c| all change at the same rate.[/hide]


algebra polynomial trigonometry inequalities function quadratics domain

[hide]
The are two obvious roots: $x=16$or$x=81$..
To prove there are no other solutions, lets assume $97-x=a^{4}$ and $x=b^{4}$ where a and b are positive.
so
$a+b=5$or$a=5-b$
and
$a^{4}+b^{4}=97$
Substitude the first equation to the second and simplify it, you can get:
$b^{4}-10b^{3}+75b^{2}-250b+264=0$
you know two roots for b are $16^{1/4}=2$ and $81^{1/4}=3$.
so you divide $(b^{4}-10b^{3}+75b^{2}-250b+264)$ by $(b-2)*(b-3)$ and you get $b^{2}-5b+44$.
The remaining two roots are imaginary. Therefore, b=2 and b=3 are the only two real solutions, which gives $\boxed{x=16}$ or $\boxed{x=81}$.[/hide]

[quote="Farenhajt"]Find all real values $x$ such that

$\sqrt[4]{97-x}+\sqrt[4]{x}=5$[/quote]

[hide]
Put $\sqrt[4]{x}= a$. Then all solutions will be roots of the polynomial $0=a^{4}-10a^{3}+75a^{2}-250a+264$.

Using RRT the polynomial becomes $0=(a-2)(a-3)(a^{2}-5a+44)$, thus our only real solutions are $x=2^{4}, \ 3^{4}=\boxed{16, \ 81}$.[/hide]

Polynomials?

[hide="Pah"] Let $a = \frac{97}{2}-x$.  We have

$f(x) = \sqrt[4]{ \frac{97}{2}+a }+\sqrt[4]{ \frac{97}{2}-a }$

It's pretty obvious that $f$ has a maximum at $a = 0$ and is concave.  There are therefore only two solutions, which are easy enough to calculate; $x = 16, 81$. [/hide]

ya haha.. I transformed the original equation to a 4 degree polynomial to prove 16 and 81 are the only two roots..

Thank you for the nice problems, Virgil Nicula! :)
[quote="Virgil Nicula"]$10.\ \ \left(\forall\right)x\in R\ ,\ \left|\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right|<1\ .$[/quote]
[hide]Note that \[\left|\sqrt{x^{2}+x+1}-\sqrt{x^{2}-x+1}\right| = \left|\frac{(x^{2}+x+1)-(x^{2}-x+1)}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}}\right| = \frac{|2x|}{\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}}\] Therefore, the given inequality is equivalent to \[\begin{aligned}& |2x| < \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}}\\ & \Longleftrightarrow\ 4x^{2}< (x^{2}+x+1)+2\sqrt{(x^{2}+x+1)(x^{2}-x+1)}+(x^{2}-x+1) = 2x^{2}+2+2\sqrt{(x^{2}+1)^{2}-x^{2}}\\ & \Longleftrightarrow\ x^{2}-1 < \sqrt{(x^{2}+1)^{2}-x^{2}}\end{aligned}\] Clearly, if $|x|<1$, the last inequality is true because $LHS < 0 \leq RHS$. So suppose we have $|x| \geq 1$. Then $x^{2}-1 \geq 0$, so we can square both sides to get the equivalent $x^{4}-2x^{2}+1 = (x^{2}-1)^{2}< (x^{2}+1)^{2}-x^{2}= x^{4}+x^{2}+1\ \Longleftrightarrow\ 3x^{2}> 0$, which is obvious when $|x| \geq 1$.[/hide]

[quote="Farenhajt"]Find all real values $x$ such that

$\sqrt[4]{97-x}+\sqrt[4]{x}=5$[/quote]

[hide="No guessing, no quartics, no functions"]Put $u=\sqrt[4]{97-x}, v=\sqrt[4]{x}$. Then

$u+v=5,u^{4}+v^{4}=97$

Transform $u^{4}+v^{4}$:

$u^{4}+v^{4}=(u^{2}+v^{2})^{2}-2u^{2}v^{2}=((u+v)^{2}-2uv)^{2}-2u^{2}v^{2}$

That yields

\begin{eqnarray*}(25-2uv)^{2}-2u^{2}v^{2}=97\iff 625-100uv+2u^{2}v^{2}=97\iff u^{2}v^{2}-50uv+264=0\end{eqnarray*}

Solving the quadratic gives $uv=44$ or $uv=6$. Hence we get two systems:

$u+v=5, uv=44$ or $u+v=5, uv=6$

The equation $t^{2}-5t+44=0$ has no real solutions, and $t^{2}-5t+6=0$ has the solutions $2,3$. Hence $x=v^{4}\implies x\in\{16,81\}$.[/hide]

[quote="Virgil Nicula"]$2.\ \ \{\begin{array}{ccccc}\sqrt{3x+y}&+& \sqrt{x-y}& = & 6 \\\ 3x &+& \sqrt{x-y}& = & 17\end{array}\ .$
[/quote]

[hide="Solution"]$u=\sqrt{3x+y},v=\sqrt{x-y}$

Then

$3x+y=u^{2}\land x-y=v^{2}\implies x=\frac{u^{2}+v^{2}}{4}$.

Hence we get

$\begin{cases}u+v=6\\ \frac{3}{4}(u^{2}+v^{2})+v=17\end{cases}$

Plugging $u=6-v$ into the second equation we get

$3v^{2}-16v+20=0\iff v={10\over 3}\lor v=2$

$v={10\over 3}\implies u=6-v={8\over 3}\implies x=\frac{u^{2}+v^{2}}{4}={41\over 9}\implies y=x-v^{2}=-{59\over 9}$

$v=2\implies u=6-v=4\implies x=\frac{u^{2}+v^{2}}{4}=5\implies y=x-v^{2}=1$

The solutions are $({41\over 9},-{59\over 9}),(5,1)$[/hide]

[quote="Virgil Nicula"]
$8.\ \ 7^{x}+539=294x\ .$
[/quote]

[hide="Solution"]$7^{x}+7^{2}\cdot 11=7^{2}\cdot 6x\iff 7^{x-2}+11=6x$

Put $y=x-2$:

$7^{y}+11=6y+12\iff 7^{y}=1+6y\iff (1+6)^{y}=1+6y$

By generalized Bernoulli's inequality, we have $(1+6)^{y}\neq 1+6y$ for $y\neq 0$ and $y\neq 1$, hence the only solutions are $y\in\{0,1\}\iff x\in\{2,3\}$[/hide]

[quote="Virgil Nicula"]$6.\ \ \sqrt [3]{1-x}+\sqrt [3]{1+x}=\frac{x^{2}+2}{\sqrt{x^{2}+1}}\ .$
[/quote]

[hide="Solution"]Rewrite it as

$\sqrt[3]{1-x}+\sqrt[3]{1+x}=\sqrt{x^{2}+1}+\frac{1}{\sqrt{x^{2}+1}}$

Preliminary observations: If $a+b=c+d=S>0$, then $a^{3}+b^{3}+3ab(a+b)=c^{3}+d^{3}+3cd(c+d)$, which gives $3S(ab-cd)=c^{3}+d^{3}-a^{3}-b^{3}$. Applying that to the equation, where $a=\sqrt[3]{1-x}, b=\sqrt[3]{1+x}, c=\sqrt{x^{2}+1}, d=\frac{1}{\sqrt{x^{2}+1}}$, we get

$3S(\sqrt[3]{1-x^{2}}-1)=(x^{2}+1)^{3/2}+\frac{1}{(x^{2}+1)^{3/2}}-(1-x)-(1+x)=((x^{2}+1)^{3/4}-\frac{1}{(x^{2}+1)^{3/4}})^{2}$

But the LHS is non-positive, and the RHS is non-negative, hence both must be zero, which is possible only for $x=0$.[/hide]

[quote="Virgil Nicula"]$1.\ \ \sqrt [5]{(x-2)(x-32)}-\sqrt [4]{(x-1)(x-33)}=1\ .$
[/quote]

[hide="Solution"]Put $u=\sqrt[5]{(x-2)(x-32)}=\sqrt[5]{x^{2}-34x+64}$ and $v=\sqrt[4]{(x-1)(x-33)}=\sqrt[4]{x^{2}-34x+33}$. Then we get

$\begin{cases}u-v=1\\ u^{5}-v^{4}=31\end{cases}$

Plug $u=v+1$ into the second equation to obtain

$v^{5}+4v^{4}+10v^{3}+10v^{2}+5v-30=0$

By RRT or observation, we find that $v=1$ is the root of the equation. Therefore it's equivalent to

$(v-1)(v^{4}+5v^{3}+15v^{2}+25v+30)=0$

The second polynomial can be written thus

$v^{4}+5v^{3}+15v^{2}+25v+30=(v^{2}+{5\over 2}v)^{2}+{35\over 4}(v+{10\over 7})^{2}+{85\over 7}$ and it's obviously always positive.

Hence the only possibility is $v=1$, which gives $x^{2}-34x+33=1\iff x^{2}-34x+32=0$, which has the solutions $x_{1,2}=17\pm\sqrt{257}$[/hide]

[quote="Virgil Nicula"]$7.\ \ 4^{x}=3^{x}+\tan\frac{\pi}{12}\ .$
[/quote]

[hide="Solution"]$\tan{\pi\over 12}=\frac{\sin{\pi\over 6}}{1+\cos{\pi\over 6}}=\frac{{1\over 2}}{1+{\sqrt{3}\over 2}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

Hence we have

$4^{x}-3^{x}=2-\sqrt{3}$

or

$2^{2x}-\sqrt{3}^{2x}=2-\sqrt{3}$

If $x\leqslant 0$, then $2^{2x}\leqslant \sqrt{3}^{2x}\iff 2^{2x}-\sqrt{3}^{2x}\leqslant 0$, which means that the equality doesn't have a solution there.

IF $x>0$, then the function $f(x)=4^{x}-3^{x}$ is strictly increasing, hence the equation $f(x)=2-\sqrt{3}$ can have at most one solution. It's obvious that this solution is $x={1\over 2}$.

Appendix: Proof that $f(x)=4^{x}-3^{x}$ is strictly increasing for positive $x$. Let $x>y>0$. Then

$4>3\implies 4^{y}>3^{y}\qquad (*)$

$x-y>0\implies 4^{x-y}>3^{x-y}\implies 4^{x-y}-1>3^{x-y}-1\qquad(**)$

Multiplying $(*)$ and $(**)$ we get

$4^{x}-4^{y}>3^{x}-3^{y}\iff 4^{x}-3^{x}>4^{y}-3^{y}$. QED[/hide]

[quote="Virgil Nicula"]$9.\ \ \{\begin{array}{ccc}x^{5}+y^{5}& = & 33\\\ x^{3}-y^{3}& = & 7\end{array}\ .$[/quote]

[hide="Very inelegant"]By observation we find that $(2,1)$ is a solution. Assume that there's another solution $(u,v)$ where $u<2$. Then because of $u^{5}+v^{5}=2^{5}+1^{5}$ we must have $v>1$. But then $u^{3}-v^{3}<2^{3}-1^{3}=7$ - impossible. The similar argument for $u>2$. Therefore $(2,1)$ is the unique solution.[/hide]

[quote="minsoens"][quote="Virgil Nicula"]$3.\ \ \{\begin{array}{ccc}\sqrt{x+y}+\sqrt [3]{x-y}& = & 6\\\ \sqrt [6]{(x+y)^{3}(x-y)^{2}}& = & 8\end{array}\ .$[/quote]$a=\sqrt{x+y}$, $b=\sqrt[3]{x-y}$, then: $a+b=6$, $\boxed{ab=8}$ ...[/quote]
Isn't correctly ! Correctly is $\boxed{\ a+b=6\ \wedge\ \boxed{a|b|=8}\ }\ .$ You lost one solution : $a=3+\sqrt{17}$ , $b=3-\sqrt{17}$ ... \[\boxed{\ n\in N^{*}\Longrightarrow \sqrt [2n]{x^{2n}}=|x|\ }\ .\]

[quote="Virgil Nicula"][quote="Virgil Nicula"]$4.\ \ \sqrt [2x+1]{\sqrt [11-4x]{(-2x)^{3x}}}=\sqrt [3x-1]{7x+2}\ .$[/quote]
[color=darkblue][b]Lemma.[/b] $\{p,q\}\subset N^{*}\ ,\ (p,q)=1\ ,\ \{px,qx\}\subset Z\Longrightarrow x\in Z\ .$
[b]Proof.[/b] Meaning conditions : $\{\ 2x+1\ ,\ 11-4x\ ,\ 3x-1\ \}\subset N^{*}$ $\Longrightarrow$ $x\in \left[\ \frac{2}{3}\ ,\ \frac{5}{2}\ \right]$ and $\{\ 2x\ ,\ 3x\ \}\subset N^{*}\ .$
From the above lemma obtain $x\in \{\ 1\ ,\ 2\ \}\ .$ Following of the verification remain only $\boxed{\ x=2\ }\ .$[/color][/quote]

Why $2x+y, 11-4x, 3x-1$ must be positive integers?

Because are [b]"index of radical"[/b] ! What means $\sqrt [p]x$ for example, if $p=-5$, $p=\frac{2}{3}$, $p=\sqrt 3$ or $p=-\pi$  ?
The functions $f(x)=\sqrt [x] 2$ and $g(x)=2^{\frac{1}{x}}$ are differently because $D_{f}=N^{*}$ and $D_{g}=R^{*}$.
[b]Convention.[/b] $\sqrt [1] x= x$ for any $x\in R\ .$

Oh, I've [b]never[/b] heard of that convention, sorry. What [b]I[/b] was thought (and I think, the majority of people here) is that $\sqrt[y]{x}$ is the same as $x^{1/y}$ and $y$ can be [u]any non-zero real number[/u].

No conventions, but definitions for $f$ and $g$ ! The graph of the function  $f$ is $G_{f}=\left\{\left(n,\sqrt [n]2\right)\ |\ n\in N^{*}\right\}$- a finite set of points !

Whatever - definition or convention, I was never taught that there's ANY distinction between $\sqrt[y]{x}$ and $x^{1/y}$, except in writing.

[color=darkblue][b]Theoretically, a equality in which appears only one variable becomes immediately a functional equality,
i.e. $f(x)=g(x)$, where the both functions have the same definition domain ![/b] Here are some examples $f\ne g$ :
$\begin{array}{cccc}f(x)=\frac{x(x-1)}{x-1}& g(x)=x & \blacktriangleright & \frac{x(x-1)}{x-1}=x\ ,\ x\ne 1\\\ f(x)=\sqrt{x^{2}}& g(x)=x & \blacktriangleright & \sqrt{x^{2}}=|x|\\\ f(x)=\sqrt{x(x-1)}& g(x)=\sqrt x\cdot\sqrt{x-1}& \blacktriangleright & \sqrt{x(x-1}=\sqrt{|x|}\cdot\sqrt{|x-1|}\\\ f(x)=\sqrt [x]2 & g(x)=2^{\frac{1}{x}}& \blacktriangleright & \sqrt [x]2=2^{\frac{1}{x}}\ ,\ x\in N^{*}\\\ f(x)=\ln x(x-1) & g(x)=\ln x+\ln (x-1) & \blacktriangleright & \ln x(x-1)=\ln |x|+\ln |x-1|\ ,\ x\not\in [0,1] \\\ f(x)=x & g(x)=e^{\ln x}& \blacktriangleright & x=e^{\ln x}\ ,\ x>0\\\ f(x)=\sqrt x & g(x)=\sqrt [4]{x^{2}}& \blacktriangleright & \sqrt [4]{x^{2}}=\sqrt{|x|}\end{array}$

$u(x)=\ln x\ ,\ v(x)=\ln |x|$ $\Longrightarrow$ $u'(x)=\frac{1}{x}\ ,\ x>0$ and $v'(x)=\frac{1}{x}\ ,\ x\in R^{*}$ but $u'\ne v'$ !
 a.s.o.[/color]

Virgil, that's all very good and true, but do you get what I'm saying? In my country there's [u]NO DOMAIN DIFFERENCE[/u] between $\sqrt[x]{a}$ and $a^{1/x}$ - perhaps there [u]IS[/u] in Romania, just as you include $0$ in the set of natural numbers, unlike most of other countries?

O.K. In the my opinion, from Bourbaki's reading, it is wrong !
In the your opinion, the equation $\sqrt [x]{2}=5$ has the solution $x=\log_{5}2$ ! Think I well ?

That's right. I was taught that there's nothing wrong if you write $\sqrt[\pi]{2}$, it's just the same as $2^{1/\pi}$

[color=darkblue]Another examples (continuation) :
$11.\blacktriangleright\sqrt [3x-1]{\sqrt [8-3x]{(-x)^{x}}}=\sqrt [5x]{2x}$ (equation).
$12.\blacktriangleright\max\left\{a^{x},a^{-x}\right\}\ge a^{\max\left\{x,\frac{1}{x}\right\}\ .}$ where $0<a\ne 1$ (inequation).
$13.\blacktriangleright\log_{a}\left(1+\log_{a}x\right)\ge a(1-x)$, where $0<a\ne 1$ (inequation).
$14.\blacktriangleright\log_{(x^{2}+3x+4)}\ |x+2|=\log_{(x^{2}+5x+9)}\ |x+3|$ ([u]strong equation ![/u]).
$15.\blacktriangleright$ Given is the function $f_{m}: R\rightarrow R\ ,\ f_{m}(x)=\left|(m-1)x^{2}-2(m+1)x+m+1\right|$, where $m\in R$. Prove that :
$\left(\forall\right) x\in [-1,1]\ ,\ f_{m}(x)\le 4$ $\Longleftrightarrow$ $f_{m}(-1)\le 4$ (Virgil Nicula - [b][u]O.J.M.1992, Romania[/u][/b] ; [u]very strong ![/u]).[/color]

[quote="Tim_Lou"]well...
$ln(x(x-1))$ does not equal $ln|x|+ln|x-1|$ if x>0, x-1<0...[/quote]

Of course, but that's an "inherent" difference of domains. What we're talking about here is a mere DEFINITION - do we agree or disagree that $\sqrt[x]{a}$ and $a^{1/x}$ are one and the same thing. There's no NATURAL or INHERENT reason to claim either yes or no - it's a sheer formality.

[hide="#15"]
$\left (\forall x\in [-1,1]\right ) f_{m}(x)\le 4\Longrightarrow f_{m}(-1)\le 4$ is trivially true

The other part:

$f_{m}(-1)\le 4\iff \left |2m+1\right |\le 2\iff-{3\over 2}\le m\le{1\over 2}$

[b]case #1[/b]
$\left | m \right |\le{1\over 2}$
It's easy to verify:

$\left |x^{2}+2x-1\right |\le 2$ for all $x\in [-1, 1]$
$\left | x^{2}-2x+1\right |\le 4$ for all $x\in [-1, 1]$

hence

\begin{eqnarray*}f_{m}(x) &=&|m(x^{2}-2x+1)-(x^{2}+2x-1)| \le \\&\le & \left | m\right |\left | x^{2}-2x+1\right |+\left |x^{2}+2x-1\right | \le{1\over 2}\cdot 4+2=4\end{eqnarray*}
for all $x\in [-1,1]$

[b]case #2[/b]

$-{3\over 2}\le m\le-{1\over 2}$

$f_{m}(x)\le\left |m-1\right |x^{2}+2\left |m+1\right |\left |x\right |+\left |m+1\right |\le\left |m-1\right |+3\left |m+1\right |\le{5\over 2}+{3\over 2}=4$
for all $x\in [-1,1]$
[/hide]

[quote="Vidoje"]
$ (\forall) x\in [-1,1] f_{m}(x)\le 4\Longrightarrow f_{m}(-1)\le 4$ is trivially true.
The other part: $f_{m}(-1)\le 4\iff  |2m+1 |\le 2\iff-{3\over 2}\le m\le{1\over 2}$
[b]case #1[/b] $ | m  |\le{1\over 2}$ It's easy to verify: $\begin{array}{c} |x^{2}+2x-1 |\le 2 \forall) x\in [-1, 1]\\  | x^{2}-2x+1 |\le 4 (\forall) x\in [-1, 1]\end{array}$ $\Longrightarrow$
$f_{m}(x)=|m(x^{2}-2x+1)-(x^{2}+2x-1)| \le$ $ | m | | x^{2}-2x+1 |+ |x^{2}+2x-1 | \le{1\over 2}\cdot 4+2=4$ for all $x\in [-1,1]$
[b]case #2[/b] $-{3\over 2}\le m\le-{1\over 2}$ $\Longrightarrow$ $f_{m}(x)\le |m-1 |x^{2}+2 |m+1 | |x |+ |m+1 |\le |m-1 |+3 |m+1 |\le{5\over 2}+{3\over 2}=4$ for all $x\in [-1,1]$
[/quote]

[color=darkblue][b]Splendidly ![/b] The your proof is more nice and more "short" than the author's solution.[/color]

[hide="#14"]$(\forall x\in\mathbb{R})x^{2}+3x+4>0\land x^{2}+5x+9>0$, therefore the equation is defined for $x\neq-2,-3$ and is equivalent to

$\log_{|x+2|}((x+2)^{2}-x)=\log_{|x+3|}((x+3)^{2}-x)$

(For the equivalence to hold, we must have $|x+2|\neq 1,|x+3|\neq 1$, but we easily check that $x=-1$ and $x=-4$ aren't the solutions, hence no solution is hereby lost.)

Put $a=|x+2|, b=|x+3|, t=\log_{|x+2|}((x+2)^{2}-x)=\log_{|x+3|}((x+3)^{2}-x)$. Then

\begin{eqnarray*}\log_{a}(a^{2}-x)=\log_{b}(b^{2}-x)=t &\iff& a^{t}=a^{2}-x\land b^{t}=b^{2}-x\\ &\implies& a^{t}-b^{t}=a^{2}-b^{2}\end{eqnarray*}

[b]1.[/b] $x<-3\implies a=-x-2\land b=-x-3\implies a-b=1$. Then

$a^{t}-b^{t}=a^{2}-b^{2}\iff (1+b)^{t}-b^{t}=(1+b)^{2}-b^{2}$. If $t\leqslant 0$, then LHS is non-positive while RHS is positive because $b>0$ - no solution. If $t>0$, then LHS is strictly increasing: $(1+b)^{t}-b^{t}=(1+b)^{t}\left(1-\left({b\over{b+1}}\right)^{t}\right)$ with both factors being strictly increasing for $b>0$. Therefore the only solution is $t=2$, but then $x=a^{2}-a^{t}=0\not <-3$

[b]2.[/b] $x>-2$. Similar to [b]1[/b] since $b-a=1$. Again we obtain $x=0$ which is a solution as $0>-2$

[b]3.[/b] $-3<x<-2$. Then $a=-x-2\land b=x+3\implies a+b=1$. Since $a>0$ and $b>0$, that implies $0<a<1, 0<b<1$. Then

$a^{t}-b^{t}=a^{2}-b^{2}\iff a^{t}-(1-a)^{t}=a^{2}-(1-a)^{2}=2a-1$

Now note that $(\forall x\in\mathbb{R})x^{2}+3x+4>1\land x^{2}+5x+9>1$, which implies $t<0$ because of $a,b<1$

If $a<{1\over 2}$, then $0<a<1-a<1\stackrel{t<0}{\implies}a^{t}>(1-a)^{t}\implies a^{t}-(1-a)^{t}>0$ while $2a-1<0$. Similarly, for $a>{1\over 2}$ we have $a^{t}-(1-a)^{t}<0$ while $2a-1>0$. Hence the solution exists only if $a={1\over 2}$, and then any real $t$ satisfies the equation. From $a={1\over 2}$ we get $b={1\over 2}\implies |x+2|=|x+3|={1\over 2}\implies x=-{5\over 2}$. As $-2<-{5\over 2}<-3$, this also is a solution.

Finally, we have $x\in\{-{5\over 2},0\}$[/hide]

[quote="Virgil Nicula"]$14.\blacktriangleright\log_{(x^{2}+3x+4)}\ |x+2|=\log_{(x^{2}+5x+9)}\ |x+3|\ .$[/quote]
[color=darkblue][b]Definition.[/b] $x\ .s.s.\ y\Longleftrightarrow xy>0\ \ \vee\ \ x=y=0\ .$ Here are some examples :

$|x|\ .s.s.\ x^{2}\mathrm{\ \ ;\ \ }$ $(|x|-|y|)\ .s.s.\ (x^{2}-y^{2})\mathrm{\ \ ;\ \ }$ $\arccos x\ .s.s.\ (1-x)$ $\mathrm{\ \ ;\ \ }$ $(\arccos x-\arccos y)\ .s.s.\ (y-x)$ a.s.o.
$(a^{x}-a^{y})\ .s.s.\ (a-1)(x-y)\mathrm{\ \ ;\ \ }$ $(a^{x}-b^{x})\ .s.s.\ [x(a-b)]\mathrm{\ \ ;\ \ }$
$(\log_{a}x-\log_{a}y)\ .s.s.\ [(a-1)(x-y)]\mathrm{\ \ ;\ \ }$ $\log_{a}x\ .s.s.\ (a-1)(x-1)\ ;$
$(\log_{a}x-\log_{b}x)\ .s.s.\ [-(a-1)(b-1)(a-b)(x-1)]\ .$

[b]Exercise.[/b] Solve the inequation : $E(x)\equiv (|x-2|-|x|)\cdot\log_{x+3}(2x^{2}+x)\cdot[(x^{2}-2)^{x+1}-x^{x+1}]\le 0\ .$
[b]Indication.[/b] On the existence domain  $D=(-3,\infty )-\{-2,-\sqrt 3,-\sqrt 2,1,\sqrt 3\}$ we have
$E(x)\ .s.s.\ ( |x-2|^{2}-|x|^{2})\cdot[(x+3)-1][(2x^{2}+x)-1]\cdot[(x^{2}-2)-x][(x+1)\le 0$ a.s.o.

[b]Lemma.[/b] Let $a,b,c$ be any real numbers. If $a>1$ and $b>1$ then $\boxed{\ \log_{a}(a+c)=\log_{b}(b+c)\Longleftrightarrow c=0\ \ \vee\ \ a=b\ }\ .$

[u]Proof.[/u] Suppose $a\ne b$. Thus, $\log_{a}(a+c)=\log_{b}(b+c)=d\Longleftrightarrow \{\begin{array}{c}a+c=a^{d}\\ b+c=b^{d}\end{array}\ .$
$\{\begin{array}{c}c=a^{d}-a=b^{d}-b\Longrightarrow c\ .s.s.\ (a-1)(d-1)\Longrightarrow c(d-1)\ge 0\\ \frac{a+c}{b+c}-\frac{a}{b}=(\frac{a}{b})^{d}-\frac{a}{b}\Longrightarrow c(b-a)\ .s.s.\ (a-b)(d-1)\Longrightarrow c(d-1)\le 0\end{array}$ $\Longrightarrow$ $c(d-1)=0$ $\Longrightarrow$ $c=0\ \ \vee\ \ d=1$ $\Longrightarrow$ $c=0$.

[b]Proof of the proposed problem.[/b] For $a=x^{2}+3x+4$, $b=x^{2}+5x+9$ and $c=x$ the equation $(14)$ becomes $\log_{a}(a+c)=\log_{b}(b+c)\ .$
From the above lemma obtain $c=0$ or $a=b$, i.e. $\boxed{\ x=0\ }$ or $x^{2}+3x+4=x^{2}+5x+9$, i.e. $\boxed{\ x=-\frac{5}{2}\ }\ .$[/color]


[quote="great math"]Wouldn't the denominator have sigma notations?[/quote]

apologies, you are right  :blush:

[quote="earldbest"]let $ a,b,c$ be real numbers satisfying $ a < 2(b \plus{} c)$,$ b < 2(c \plus{} a)$,$ c < 2(a \plus{} b)$. Prove that $ 3\le \frac {4\sum a^{3} \plus{} 15abc}{\sum a \sum ab} < 6$.[/quote]

$ \frac {4\sum a^{3} \plus{} 15abc}{\sum a \sum ab} \minus{} 3$
\[ \equal{} \frac {{\sum {\left( {2a \plus{} 2b \minus{} c} \right)\left( {a \minus{} b} \right)^2 } }}{{\sum a \sum {ab} }}\]\[ \ge 0\]

$ 6\minus{}\frac {4\sum a^{3} \plus{} 15abc}{\sum a \sum ab}$
\[ \equal{} \frac {{\prod {\left( {2a \plus{} 2b \minus{} c} \right) } }}{{\sum a \sum {ab} }}\]\[ > 0\]


Does the hormonic series with the fractions that has digit 7 in their denominator dropped converges?
 :blush:

[quote="rustam"]hormonic series[/quote]

 :rotfl:

Let us estimate the contribution of all n-digit numbers to the series. There are $8\cdot 9^{n-1}$  n-digit numbers whose decimal representation doesn't contain $7$. Each of them is greater than or equal to $10^{n-1}$, so each contributes at most $1/ 10^{n-1}$. We get an upper bound on the value of the series:
\[\sum_{n=1}^\infty \frac{8 \cdot 9^{n-1}}{10^{n-1}}= 8 \cdot \frac{1}{1-9/10}= 80.\]


John flips two coins and Sarah flips one coin. What is the probablity that John gets more heads than Sarah?

There are two possibilities for Sarah: head or tails.  If she gets tails, then John needs at least 1 head = 1 - prob(no heads) = 1 - (1/2)^2 = 3/4.  If she gets heads, then John needs two heads = (1/2)^2 = 1/4.  Putting that all together: 1/2 x 3/4 + 1/2 x 1/4 = 1/2.

if sarah has tails then he will have more heads in 3/4 of cases
if sarah has heads then he will have more heads in 1/4 of cases

so he will win 1/2 cases ...

we could also do it the other way around,by finding out the probability that john will have the same number of heads or less heads

then we sustract it from 1 and we have our answer


given triangle ABC with inner circle (I) .Circle (I) tangent to BC , AC ,AB at point X , Y,  Z. M is the arbitrary point inside (I) . XM , YM ,ZM intersects circle (I) at A1 , B1 , C1 . prove that AA1 , BB1 and CC1 are concurrent .

Dear HTA and Mathlinkers,
this interesting and nice problem (Steinbart' s theorem) has been posted several times on this site.For a synthetical proof see
http://perso.orange.fr/jl.ayme    vol. 3 "Les points de Steinbart et de Rabinowitz".
Sincerely
Jean-Louis


Let C be any real matrix (we don't know anything else about it, i.e. if it's normal, etc). Then U=Transpose(C)*C is a positive semidefinite matrix. Let e1 and e2 be the two largest eigenvalues of U. If I know that e1+e2 > 1, can I deduce anything about the eigenvalues of C?

The only ideas I have right now are that clearly Trace(U) > 1, and the rank(U)=rank(C).

Thanks,
Sev

[url=http://en.wikipedia.org/wiki/Singular_value_decomposition]Singular value decomposition[/url] can tell you something about $ C$. But I don't know about eigenvalues. If you are working over reals, $ C$ may have no eigenvalues at all.


Let $ ABCD$ be a cyclic quadrilateral. Show that the line connecting the midpoints of arc $ AB$ and arc $ CD$ and the line connecting the midpoints of arc $ BC$ and arc $ DA$ are perpendicular.

someone please?
or maybe a link is very OK.

thank you,.

It is very easy with angle chasing . Name M,N the midpoints of the arc AB,CD and P,Q the midpoints of the arcs BC,AD.

Name X the intersection of NM and PQ , we have

$ \angle MXQ = \frac {1}{2}(arcMQ + arc PM)$ ${ = \frac {1}{4}(arcAB + arcBC + arcCD + arcDA) = \frac {1}{4}(2\pi}) = \frac {\pi}{2}$

Which means that $ MN$ is perpendicular to $ PQ$

i'm sorry but what do you mean by $ arc$. I'm not used to use that. can u define it for me please?

Thanks.

Dear Mathlinkers,
these two lines are resp. parallel to the two bissector of the diagonal angle.

Sincerely
Jean-Louis


plz help me out for 3 questions plz solve works thx a lot

question 1:

x + 2y = 5
x - 7y = -1
2x + y =4

question 2:

2x - 3y =1
3x + 4y = -3
x - 10y = 5

question 3:

3x +y = -2
x - 2y = 7
5x - 3y = 6

There are two variables and three equations..... these three lines must meet in one point. Solve it like you would solve a normal equation, and see if there is a valid solution.


dn is the last non-zero digit of n! . Show that there is no k such that dn = dn+k for all sufficiently large n.

Note: problems 3, 11, 13, 17, 22 and 23 were used in the Olympiad and are not shown here.

1. it has been posted before.
2. obviously, you copied it from Kalva.
3. you copied it badly.
4. wrong forum.

[quote="ashegh"]Note: problems 3, 11, 13, 17, 22 and 23 were used in the Olympiad and are not shown here.[/quote]

lol you even copied this  :lol:

did you even read what you wrote? why don't you just give us  the link instead of making so much topics: http://www.kalva.demon.co.uk/short/sh83.html


A man is in the center of a circle. There is a lion that can move only around the circle. What is the maximum value of  L/M (L is the speed of the lion and M is the speed of the man) such that the man can escape?

[hide="pi"]

The man is in the center $O$.

The lion is at a point $A$. At this moment, the man starts moving to the antidiametric $A'$ of $A$.

The man has to run in a distance $OA'=r$, where $r$ is the radius of the circle.

The lion has to make a distance $\pi r$

So the ratio of their speeds must be less than $\pi$

(accurately, there is not a maximum; if it is exactly $\pi$ the lion will eat the man  :o )

[/hide]

Sorry. Your answer is wrong. The value 4 is possible and it is not the maximum. The man can move everywhere he wants inside the circle. He doesn't need to move only in a line. The problem is harder.

[quote="joao1"]The man can move everywhere he wants inside the circle. [/quote]But the lion will move to the point on the circle that's closest to the man.

The man runs on a curve (changing of the direction) or eventually (through aproximate) on a broken line, in ladder ! We will come to a differential equation.


conics
ellipse
geometry
geometric transformation
reflection
trapezoid
advanced fields

Prove that the product of the distances from the focal points of an ellipse to a variable tangent is constant.

I hope I translated good from Romanian:
focar-->focal point
elipsa-->ellipse
I hope a Romanian can help if I am not corect.

Let f be the reflection of F in the tangent d (which touches the ellipse at T) and let f' be the reflection of F'. Somewhere in the message for Problem 2 I mentioned that F, T, f' and F', T, f are collinear and Ff'=F'f=2a; it's also obvious that ff'=FF'.

The trapezoid Fff'F' is isosceles, so it's cyclic, so Ptolemy holds, so FF'22+Ff*F'f'=F'f*Ff' so Ff*F'f'=4a2-4c2=4b2, but Ff*F'f'=4*the product of the distances from F and F' to d), so this product is always equal to b2 (as could easily have been seen by taking the tangent to be parallel to FF').


linear algebra
matrix
linear algebra unsolved

What is the maximal dimension of a subspace $F$ of $M_n(\mathbb{R})$ such that
the only non invertible matrix of $F$ is $O$, the null matrix.

For odd $n$ it is clear, but I don't get it at all for even $n$. For $n=2$ I have found 2, for $n=4$ I have found 4. What would be the value in the general case? :?

Just to get the odd case out of the way (as harazi said, it's clear :)): the dimension is $1$ when $n$ is odd, since we can assume that one of the matrices in our space is the identity, and given any non-scalar matrix $A$ from $F$ there is a real $\alpha$ s.t. $\alpha I+A$ is non-invertible but non-zero.

I also found $2$ and $4$ when $n=2$ and $4$ respectively.

Hmmm, when looking at this article you will see why the problem is so hard...


Prove that there is no polynomial p(x) with integer coefficients such that p(1)=4, and p(3)=5.

Answer

Lets assume there is a polynomial $ p(x)$ with integer coefficients such that $ p(1) = 4$ and $ p(3) = 5$

$ \Rightarrow a-b\,|\,p(a)-p(b)$

$ \Rightarrow 3-1\,|\,5-4$

$ \Rightarrow 2\,|\,1$

This is a contradiction. Hence there is no polynomial $ p(x)$ with integer coefficients such that $ p(1)=4$ and $ p(3)=5$.


Let $ n$ and $ p$ be positive integers greater than $ 1$, with $ p$ being a prime. Show that if $ n$ divides $ p\minus{}1$ and $ p$ divides $ n^3\minus{}1$, then $ 4p\minus{}3$ is a perfect square.

claim: p does not divides n-1
proof:
assume that p|(n-1)
on one hand p|(n-1) ==> p <= n-1
on the other hand n|(p-1) ==> n <= p-1
contradiction

claim: p|(n^2+n+1)
proof:
p|(n^3-1)
p does not divides n-1
therefore p|(n^2+n+1)

claim: p=n^2+n+1
proof:
write n^2+n+1=kp
k is of the form bn+1 (to prove this let k=bn+r and then mod n)
since n|(p-1), therefore p is of the form an+1
if p $\neq$ n^2+n+1, then k>=n+1
it follows that n^2+n+1=kp=(bn+1)(an+1)>=(n+1)(n+1) contradiction

therefore p=n^2+n+1 ==> 4p-3=(2n+1)^2


what is the ptolemy's inequality?? plz link to me...

It's on AoPSWiki.
[url]http://www.artofproblemsolving.com/Wiki/index.php/Ptolemy%27s_Inequality[/url]


geometry
perimeter
trigonometry
function
geometric transformation
reflection
circumcircle

In $\triangle ABC$, $\angle A=30^\circ$ and $BC=4$. The area of $\triangle ABC$ is $14$. $P$ is on $AB$, $Q$ is on $BC$ and $R$ is on $CA$. What is the minimum possible length of the perimeter of $\triangle PQR$?

Solution:
[hide]
Given angle A=30 degrees, BC=4
Area= ${\frac{1}{2}(AB)(AC)sin(30)=14}$
Then $(AB)(AC)=56$
Let angle B=x, then by the law of sines we have $\frac{4}{sin(30)}=\frac{AC}{sin(x)}$
So $AC=8sin(x)$, and $AB=\frac{56}{AC}=\frac{7}{sin(x)}.$

Now, the three sides are $4, 8sin(x), \frac{7}{sin(x)}$. The perimeter function is $f(x)=4+8sin(x)+\frac{7}{sin(x)}$.

Now the function must be minimized. Since the range of $sin(x)$ is over $[0,1]$, the dominant term is clearly $\frac{7}{sin(x)}$. To minimize this term, let $sin(x)=1$. Then the minimum perimeter is $4+8+7=19$.

Answer: $19$
[/hide]

Is this right?

EDIT: dangit, found the wrong triangle. Oh well. It was fun.

[hide="the idea..."]once you find the triangle (LoC, LoS suffice)...it becomes a classic problem, and you basically reflect the triangle over itself a few times...[/hide]

first step[hide]Prove that $\Delta ABC$ is acute[/hide]
second step[hide]Prove that for an acute triangle the inscribed triangle with minimal perimeter is the ortocentric triangle (the triangle formed from the bases of the altitudes)[/hide]
last step[hide]Prove that $S_{ABC}=p*R$ where $2p$ is the perimeter of ortocentic triangle and $R$ is the circumradius[/hide]
I dont think this is intermidiate but maybe im wrong :o

[hide="Hint etc"]
The idea is rather simple. As Altheman mentioned, the key to this problem is reflection. You need to reflect $\triangle ABC$ over $AB$ and $AC$. Let $Q'$ and $Q''$ be the new locations of $Q$ after the reflections. What kind of triangle is $\triangle AQ'Q''$? ;)

This problem can be solved without trig :)
[/hide]


If $ a_2 \neq 0$ and $ r$ and $ s$ are the roots of $ a_0 \plus{} a_1x \plus{} a_2x^2 \equal{} 0$, then the equality $ a_0 \plus{} a_1x \plus{} a_2x^2 \equal{} a_0\left (1 \minus{} \frac {x}{r} \right ) \left (1 \minus{} \frac {x}{s} \right )$ holds:

$ \textbf{(A)}\ \text{for all values of }x, a_0\neq 0$
$ \textbf{(B)}\ \text{for all values of }x$
$ \textbf{(C)}\ \text{only when }x \equal{} 0$
$ \textbf{(D)}\ \text{only when }x \equal{} r \text{ or }x \equal{} s$
$ \textbf{(E)}\ \text{only when }x \equal{} r \text{ or }x \equal{} s, a_0 \neq 0$

[hide="Solution"]
By Vieta, we have:
$ r + s = -\frac{a_1}{a_2}\\
rs = \frac{a_0}{a_2}$

Simplifying the RHS of the equation, we have:

$ a_0\left (1-\frac{x}{r} \right ) \left (1-\frac{x}{s} \right ) = a_0\left (1 - \frac{x}{r} - \frac{x}{s} + \frac{1}{rs}x^2 \right )\\
= a_0\left (1 - \left(\frac{r + s}{rs}\right)x + \frac{1}{rs}x^2 \right )\\
= a_0\left (1 + \frac{a_1}{a_0}x + \frac{a_2}{a_0}x^2 \right)\\
= a_0 + a_1x + a_2x^2$

Thus, the answer is ${ (A) \text{for all values of }x, a_0 \not= 0}$.
[/hide]


[currently r15s11z55y89w21]

The first CaML of the year was surprisingly trivial compared to those of last year, right? Although in times like this careless errors hurt and burn very badly...

I don't think this one was particularly easier than any from last year. I guess maybe this contest was a little more straightforward; there's usually a problem that makes you think a little more than the others (but then that one is pretty easy too).

I only had like 5 minutes to do it so I totally misread #6 and thought there were 3 brothers instead of 5.  :mad:

lol i did it in 5 minutes

this was way too easy

It was trivial compared to the previous years. This one was the same level as the middle school CML (continental)

[quote="moogra"][quote="Ubemaya"]This was definitely by far the hardest contest I've taken this year. I think you guys should be more considerate of the less intelligent people here before you dismiss contests as trivial and make them feel bad.[/quote]

The above comment may have worked if it wasn't full of sarcasm.[/quote]

How was it sarcasm? The test is definitely harder than all the mathcounts school tests that he takes.

Actually, the mathcounts school tests might be harder. But mathcounts school tests are pretty hard for some people, so yeah.

[r15s11z55y89w21]

[quote="pythag011"]How was it sarcasm? The test is definitely harder than all the mathcounts school tests that he takes.[/quote]

Ouch.

Anyway, yes, consideration for others is a great thing, but this is THE nerdy chat area after all. :P

There are various competitions with varying levels of difficulty. Math League is definitely one of the easiest, but I find that a good thing. It's usually by the CAML that we convince people to join our school's math team. If they do (reasonably) well the first time, they'll stay. (There are people who show up at a round of Mandelbrot, and it's their first math competition. Most of them never come back...)

[quote="EggyLv.999"][quote]How was it sarcasm? The test is definitely harder than all the mathcounts school tests that he takes.[/quote]

i found it easier...[/quote]

Read the next line of my post :P

Yeah...in all honesty, contest's aren't exactly fun when you can submit the paper 3 minutes after starting...I really hope that the CAML's get harder or many people may lose interest...

[quote="zephyredx"][quote="Differ"]I doubt CaML will become any more interesting than it is now. It's a school math team competition. People who can't do math need to be able to score points on it.[/quote]

True. Kinda like the USAJMO.

The problems seemed a bit harder than last year's, but not much.[/quote]

darn your sig

also CAML was less trivial than last year

[quote]also CAML was less trivial than last year[/quote]

yeah...there are far fewer perfect scores this year than last

And there's more than are listed - I know of at least 1 perfect scorer that's not on the list because the coordinator hasn't sent in scores for awhile.

CAML is a pretty hard contest. They should make a CAJML.

ya totally!
they should also give more time.

The CaML questions are so strange and straightforward. It's unlike anything that I have seen on other competitions

Also, I think I'm wrong, but  wouldn't it be JJCaML? the idea is still really funny


induction
inequalities unsolved
inequalities

I remember to have solved this one using integrals. A very short solution. The official solution used induction, but it was quite long and difficult.


modular arithmetic
number theory
prime factorization
number theory unsolved

Problem) Find all prime number p (not 2,5) which satisfy

$ ord_{p} 10 \equal{} ord _{p^2} 10$.

I know the answer of the problem, but I don't know solution.

p=3,487, and one more number but I can't remember it because it is too large.(maybe bigger than 10^6)

Anyway it has three answer.

This is from "the college mathematics journal"

Could you give a full reference (year, issue, author, etc.) otherwise your reference does not help.

$ 3487 \equal{} 11\cdot 317$

Also, you did not mention what $ \mbox{ord}_p(n)$ means.  It usually means $ s \equal{} \mbox{ord}_p(n)$ is the smallest positive integer such that $ n^s\equiv 1 \pmod{p}$, but unfortunately it is often incorrectly used as the exponent for $ p$ in the prime factorization of $ n$.

$ p\equal{}3,487$ means $ p\equal{}3$ or $ p\equal{}487$. And it's obvious what $ \mathrm{ord}$ means here as the second one is senseless.

I doubt that there is an easy characterisation of those primes, either (in fact I think it's open).

[quote="dmath328"]Problem) Find all prime number p (not 2,5) which satisfy

$ ord_{p} 10 \equal{} ord _{p^2} 10$.
[/quote]
Alternative description of such primes is
\[ 10^{p\minus{}1}\equiv 1\pmod{p^2}\]
and finding them is probably as hard as finding [url=http://en.wikipedia.org/wiki/Wieferich_prime]Wieferich primes[/url].

The next such prime, following $ 3$ and $ 487$, is $ 56598313$. There are no other such primes below $ 2\cdot 10^{11}$ - see http://oeis.org/A045616


$\boxed{\ x>0\ ,\ y>0\mathrm{\ and\ }x+y+z=xyz\Longrightarrow\sum\sqrt{\frac{2x}{y+z}}\ge 3\ }\ .$

[b]Remark.[/b] The condition $x+y+z=xyz$ may be replaced with the condition $x+y+z=1\ !$

Indeed, for $a=\frac{1}{yz}>0$ a.s.o., obtain $a+b+c=1$ and $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{1}{xyz}=\frac{1}{x+y+z}\ .$

[size=117][color=red]Sorry, this inequality is wrong. See the following replies ![/color][/size]

I don't think it is true .just try a=0.5,b=0.5.c=0
then coeffient of the right pary should be replace by $2\sqrt{2}$ then it is true.

Can $c$ be $0$? Since $c=\frac{1}{xy}$, I don't think so.

[quote="Virgil Nicula"]$\boxed{\ x>0\ ,\ y>0\mathrm{\ and\ }x+y+z=xyz\Longrightarrow\sum\sqrt{\frac{2x}{y+z}}\ge 3\ }\ .$
[/quote]
I think it's wrong: $x=y=6,$ $z=\frac{12}{35}.$ :wink:

[color=darkblue][b]Hawk Tiger and Arqady[/b], [u]you are right ![/u] I started from a correct inequality over an [b][u]acute[/u][/b] triangle.
I forgot to put this restriction on the variables $x,y,z$ for which $a=y+z$ a.s.o. I think that now is correctly :

$\{\begin{array}{c}x+y+z=1\\\\ x>0\ ,\ y>0\ ,\ z>0\\\\ xy\le z\ ,\ yz\le x\ ,\ zx\le y\end{array}$ $\Longrightarrow$ $\boxed{\ \sqrt{\frac{2x}{y+z}}+\sqrt{\frac{2y}{z+x}}+\sqrt{\frac{2z}{x+y}}\ge 3\ }\ .$

[b]Remark.[/b] $a^{2}\le b^{2}+c^{2}\Longleftrightarrow$ $(y+z)^{2}\le (z+x)^{2}+(x+y)^{2}$ $\Longleftrightarrow$ $yz\le x(x+y+z)$ $\Longleftrightarrow$ $yz\le x$ a.s.o.[/color]


Consider the equation $ a^3 \minus{} b^2 \equal{} 2$. Does it have solution without use of extended integer rings?

This is an example of a Mordell equation. $ y^2 \plus{} k \equal{} x^3$, which for $ k\neq 0$ has only finitely many solutions (but there is no method to always find them all).

[url]http://www.math.okstate.edu/~wrightd/4713/nt_essay/node5.html[/url] states the only solutions are $ (a,b) \equal{} (3, \pm 5)$.

A better link is [url]http://math.uga.edu/~pete/4400MordellEquation.pdf[/url]. It states this particular case was considered by Bachet and Fermat, and it offers a proof (albeit using those dreaded extensions of $ \mathbb{Z}$). But it looks like a good material.