geometry
3D geometry
sphere
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analytic geometry
rotation
quadratics

I am studying for the National Mu Alpha Theta Mathematics Competition in mid/late July and would really like some help.  I am taking a topic test called "Area and Volume" because I am very good at geometry, but there is one type of problem that I just can't figure out.  They deal with objects inscribed in spheres or circumscribed about spheres.  Here are some example problems for you to help me with:

1.  A sphere with radius 10 has a cone inscribed within it of maximum volume.  What is the volume of this cone?

2.  A square pyramid circumscribed about a sphere with radius 10 has minimum surface area.  What is the height of this pyramid?

3.  A cylinder of maximum volume is inscribed within a sphere of radius 10.  What is the surface area of this cylinder?

4.  A cone of maximum surface area is inscribed within a sphere of radius 10.  What is the surface area of this cone?

5.  A cylinder of maximum surface area is inscribed within a sphere of radius 10.  What is the surface area of this cylinder?

6.  A cone of minimum volume is circumscribed about a sphere of radius 10.  What is the volume of this cone?

I'm sure you can see what kind of problems I'm talking about now.  I have no idea how to do any of these, so I really want a very good, clear, detailed solution to all 6 of them.  I need diagrams included in all the solutions (I am an extremely visual learner).

I have to get going really soon so I can't write anything detailed.
First thought for the first question is try to maximize the area of a triangle within a circle or radius 10 (think of it as a sideways projection). Then create the circular base based on the coordinates you get. I am thinking this analytically, I have no idea how you can go about with compass constructions for maximizing such things.
Also this is just a very hurried though. If you try this, play around with slight variations with a calculator and see if the volume goes down =)

No, you don't want to maximize the area of the triangle. That won't give you the same thing as maximizing the volume.

Put an $ (h,r)$ coordinate system so that the vertex of the cone is at the center of the coordinate system. Writing out the equation of the circle in that system (which you then revolve about an axis to get a sphere) you get that the cone is inscribed in the sphere if $ r^2 \plus{} h^2 \minus{} 20h \equal{} 0.$ So we can phrase the problem as:

Maximize $ \frac {\pi}3r^2h$ subject to the constraint $ r^2 \plus{} h^2 \minus{} 20h \equal{} 0.$

And note that the $ \frac {\pi}3$ isn't all that important - we'd get the same answer if we said to maximize $ r^2h.$

From there, use whatever techniques you know (personally, I'll go for Lagrange multipliers) - several things work.

[hide="One way to express the answer."]I get that the maximal cone occupies $ \frac8{27}$ of the volume of the sphere.[/hide]

I'm not exactly sure what you mean by an $ (h,r)$ coordinate system, and I don't know how you got $ r^2 \plus{} h^2 \minus{} 20h \equal{} 0$.  Can you please not skip any steps, I really want to understand these problems thoroughly.

Alright I am back =) I don't really understand what he means either :( Hopefully he gets back to us.
Anyways the method I meant, but realized I worded quite horribly.

[b]1. A sphere with radius R has a cone inscribed within it of maximum volume. What is the volume of this cone? [/b] (Note I changed $ 10$ to $ R$)
Draw a circle with some arbitrary triangle inside it. I was just playing with Geogebra for awhile and got a good picture, but the code would not pop up for displaying... :(
Label the height of the triangle $ h$, and half of the bottom edge $ r$ for the radius of the cone. We're looking at the whole thing sideways. Now label the distance from the center of the circle to the base of the cone (where the $ r$s are) by $ \alpha$. Also draw a line from the centre to some vertex of the triangle (not along $ h$). Now we crank away:

We see that $ h \equal{} R \plus{} \alpha$, and by Pythagoreans $ r^2 \plus{} \alpha^2 \equal{} R^2$. So $ h \equal{} R \plus{} \sqrt {R^2 \minus{} r^2}$. We're going to be maximizing the cone, so we need its volume formula, and integrate the sphere's restrictions. Easily done by subbing in:

So $ V_{cone} \equal{} V \equal{} \frac {1}{3}\pi r^2 h \equal{} \frac {1}{3}\pi r^2(R \plus{} \sqrt {R^2 \minus{} r^2})$  Note: Since you're learning this, keep in mind you need to find relationships between everything in order to put everything in terms of constants and one of the variables what you're maximizing depends on; in this case we're making the volume depend on the radius.
So now we have the volume in terms of a variable, which gives a graph. We find the critical points (the term I was taught, where $ f'(x) \equal{} 0$):

$ \frac {dV}{dr} \equal{} \frac {2r\pi}{3}(R \plus{} \sqrt {R^2 \minus{} r^2} ) \plus{} (\frac { \minus{} 2r}{2\sqrt {R^2 \minus{} r^2}})\frac {r^2\pi}{3} \equal{} ...$  simplify for practice I guess :P $ ... \equal{} (\frac {\pi r}{3\sqrt {R^2 \minus{} r^2}})(2R\sqrt {R^2 \minus{} r^2} \plus{} 2R^2 \minus{} 3r^2)$

We set this equal to $ 0$, and rearrange to get:

$ 2R\sqrt {R^2 \minus{} r^2} \equal{} 3r^2 \minus{} 2R^2$ from which we deduce, after algebra, $ r \equal{} \frac {\sqrt {8}}{3}R$ (Keep in mind the restriction $ r > 0$)

So what now? Since we have $ r$, plug it into the previous equations with $ r$, namely $ h \equal{} R \plus{} \sqrt {R^2 \minus{} r^2}$ and we get

$ V \equal{} \frac {1}{3}\pi r^2 h \equal{} ...$practice simplifying $ ... \equal{} \frac {32\pi}{81}R^3$  Whew =) Pretty easy looking.

So now plug in $ R \equal{} 10$ for your question and you get that the maximum volume is $ \approx 1241.12$.
Also, using Professor Kent's idea, the volume would be $ \frac {8}{27}\times \frac {4\pi}{3}R^3$ which is equivalent.

Did you want solutions to the rest of the problems? They're all pretty much the same if this made complete sense.

Hi5Man:

I'll follow Fermat's Little Turtle and write the radius of the sphere as $ R$ instead of $ 10$ - I should have done that in the first place, as the specificity doesn't actually make things any easier.

We'll get the sphere by rotating a circle in the $ x,y$-plane, and I'd prefer to put the origin of the coordinate plane as a point on the sphere - that will be the vertex of the cone. Accordingly, let the center of the circle be $ (R,0)$ and the radius be $ R.$ That gives an equation $ (x - R)^2 + y^2 = R^2.$ Expand that out and rearrange it to $ x^2 - 2Rx + y^2 = 0.$

We're going to rotate this circle around the $ x$-axis to make the sphere, and we're going to rotate a triangle around the $ x$-axis to make the cone. One vertex of this triangle - the one that will become the vertex of the cone - will be at $ (0,0).$ The other two vertices of the triangle will be at $ x = h,y = 0$ and $ x = h,y = r.$ Only that last vertex must be on the circle in order for the cone to be inscribed in the sphere, so substituting $ h$ for $ x$ and $ r$ for $ y,$ we get $ h^2 - 2Rh + r^2 = 0.$

This is essentially the same equation that Fermat's Little Turtle got except that I didn't bother solving the quadratic equation for $ h$ as he did. I think the picture in my mind (that I'm not showing you) is rotated 90 degrees from the picture in Fermat's Little Turtle's mind (that he didn't show you), but that's completely unimportant.

From here, we're going to use calculus to find a critical point. We'll need some evidence that that critical point is actually a maximum, but we can see this intuitively. Consider the cone to depend on $ h.$ We must have $ 0\le h\le 2R.$ But if $ h = 0,$ we get $ r = 0$ and the "cone" is a single point with no volume - that's not a maximum. And at the other extreme, if $ h = 2R,$ then $ r = 0,$ the "cone" is a line segment, the volume is zero, and that's no maximum either. For $ 0 < h < 2R,$ it's clear that the volume is positive, so there must be a maximum somewhere. If we should happen to find more than one non-trivial critical point, then we'll have to do something to sort things out, but we suspect that there will actually be only one non-trivial critical point. If that happens, we'll know it's the maximum.

Fermat's Little Turtle proceeded by first-semester calculus methods. I'd prefer to use third-semester calculus methods - to be specific, Lagrange multipliers. We also note (as I did in my first post) that maximizing the volume is the same as maximizing $ r^2h.$

Maximize $ r^2h$ subject to $ r^2 + h^2 - 2Rh = 0.$
\begin{align*}2rh & = 2\lambda r \\
r^2 & = \lambda(2h - 2Rh)\end{align*}
From the first equation, either $ r = 0,$ which we reject as trivial (not a maximum, as already discussed), or we divide by $ 2r$ to get that $ h = \lambda.$ Substitute that into the second equation:
\[ r^2 = 2h(h - R) = 2h^2 - 2Rh
\]
But we also have $ r^2 = - h^2 + 2Rh$ from the constraint equation, so
\begin{align*}2h^2 - 2Rh & = - h^2 + 2Rh \\
3h^2 & = 4Rh\end{align*}
Again rejecting $ h = 0$ as trivial and not a maximum, we get that $ h = \frac {4R}3.$ Substituting this into the constraint equation gets us that
\[ r^2 = - \left(\frac {4R}3\right)^2 + 2R\left(\frac {4R}3\right) = \frac {8R^2}9
\]
Multiply these together to get that $ r^2h = \frac {32R^3}{27}$ or $ V = \frac {32\pi R^3}{81}.$

Of course, this is the same answer that Fermat's Little Turtle gave. My only point in writing it all out is that the actual algebraic nitty-gritty may be a little easier and a little less prone to computational mistakes than the square-root-laden single-variable calculus approach. Lagrange multipliers can be that way, once you're used to it.

Thank you both very much!!!!  The two different methods were very easy to follow, although I'm going to have to practice using Lagrange multipliers for Mr. Merryfield's method.  I don't need solutions to all of them, they were just the general kinds of questions that I needed help on.  But I would like help on numbers 5 and 6.  Number 5 is the same kind of thing but with a cylinder instead of a cone and number 6 is a kind of problem where something is circumscribed about a sphere instead of inscribed in it.  Then if anyone feels up to it I would like help on number 2 also, but that seems like a much harder problem and I would be very surprised if anything other than a cylinder or cone inscribed/circumscribed inside/about a sphere showed up on the test at the national competition.  But I would really like help with 5 and 6 before anyone attempts to explain number 2 to me.  Again, thank you two so very much, I've been working on these "sphere optimization problems" all year long.  I hope this isn't too much to ask for.

#5 is easier than #6. In both cases, I'll use $ R$ instead of $ 10.$ As before, being specific about the radius of the sphere doesn't make things any easier than just using the letter.

In both cases slice through the picture with a plane that includes the axis of the cone or cylinder to get a two dimensional picture.

In the case of #5, we see a rectangle inside a circle of radius $ R.$ Half of the width of the rectangle is $ r$ and half of the height is $ \frac{h}2.$ The Pythagorean theorem gives us that $ r^2+\left(\frac h2\right)^2=R^2.$

So the problem become this:

Maximize $ 2\pi rh+2\pi r^2$ subject to $ r^2+\left(\frac h2\right)^2=R^2.$

========================================

Now for the more difficult #6. I'm going to draw my flat picture so that the center of the $ xy$ coordinate system is the center of the base of the cone, and the axis of the cone is the $ y$ axis. We indicate the cone with the line that connects $ (r,0)$ to $ (0,h).$ (This line will be rotated around the $ y$ axis to make the cone.) We indicate the curve with the circle of radius $ R$ centered at $ (0,R).$ The circle has to be tangent to the line.

(And yes, I'm telling you to draw your own pictures.)

Let $ (x,y)$ be the point at which the circle is tangent to the line. The following must be true of the point:

1. It must be on the line. This gives us $ hx+ry=hr.$

2. It must be on the circle. This gives us $ x^2+(y-R)^2=R^2.$

3. The direction from $ (x,y)$ to the center of the circle $ (0,R)$ must be perpendicular to the line. The slope of that direction is $ \frac{y-R}x$ and the slope of the line is $ -\frac hr.$ Hence $ \left(\frac{y-R}x\right)\cdot\left(-\rac hr\right)=-1,$ which can be rearranged to $ hy=rx+hR.$

That gives us three equations in four variables. We'd like to eliminate that down to one equation in two variables by eliminating $ x$ and $ y.$

I'll let you chew on that for a while and I'll be back later.

[quote="Fermat's Little Turtle"]

[b]1. A sphere with radius R has a cone inscribed within it of maximum volume. What is the volume of this cone? [/b] (Note I changed $ 10$ to $ R$)
[/quote]

Just use Pythagorean theorem, denote $ r,\ h$ be the radius of the base, the hight respectively, we have $ V \equal{} \frac {\pi}{3}r^2h \equal{} \frac {\pi}{3}\{R^2 \minus{} (h \minus{} R)^2\}h \equal{} \frac {\pi}{3}h^2(2R \minus{} h)\ (0 < h < 2R)$, differentiating this with respect to $ h$ gives $ \frac {dV}{dh} \equal{} \frac {4}{3}\pi (4R \minus{} 3h)$, yielding $ V_{max} \equal{} V_{h \equal{} \frac {4}{3}R} \equal{} \frac {32}{81}\pi R^3.$

Alternative Solution:

Use AM-GM, $ V \equal{} \frac {4}{3}\pi \frac {h^2}{4}(2R \minus{} h)\leq \frac {4}{3}\pi \left(\frac {\frac {h}{2} \plus{} \frac {h}{2} \plus{} 2R \minus{} h}{3}\right)^3 \equal{} \frac {32}{81}\pi R^3$,

the equality holds when $ \frac {h}{2} \equal{} 2R \minus{} h\Longleftrightarrow h \equal{} \frac {4}{3}R$, by $ r^2\equal{}h(2R\minus{}h)$, $ r\equal{}\frac{2\sqrt{2}}{3}R$, which satisfies $ 0 < h < 2R.$

Whoops - I've got a better way on #6.

The radius of a circle inscribed in a triangle is $ \frac As$ where $ A$ is the area and $ s$ is the semiperimeter.

We have an isosceles triangle with base $ 2r$ and altitude $ h.$ Each of its other two sides have length $ \sqrt {r^2 \plus{} h^2}$ so that its semiperimeter is $ s \equal{} r \plus{} \sqrt {r^2 \plus{} h^2}.$ Its area is $ rh.$

So the problem can now be stated as:

Minimize $ \frac13r^2h$ subject to $ \frac {rh}{r \plus{} \sqrt {r^2 \plus{} h^2}} \equal{} R.$

You can try algebraically rearranging the constraint equation. I think I'd prefer to see it without square roots or denominators - say, $ r^2h^2 \minus{} 2r^2hR \minus{} R^2h^2 \equal{} 0.$ Or $ h(r^2\minus{}R^2)\equal{}2r^2R.$

[quote="Hi5Man"]
6.  A cone of minimum volume is circumscribed about a sphere of radius [b]R[/b]. What is the volume of this cone?.[/quote]

Express the area of a triangle in two ways or remark the similarity of two triangles, we have $ \sqrt {r^2 \plus{} h^2}R \equal{} (h \minus{} R)r$, squaring both sides and rearranging gives $ r^2 \equal{} \frac {hR^2}{h \minus{} 2R}$, since $ 0 < h < 2R$, we have

$ V \equal{} \frac {1}{3}\pi r^2h \equal{} \frac {\pi}{3}R^2\left(h \plus{} 2R \plus{} \frac {4R^2}{h \minus{} 2R}\right)$

$ \equal{} \frac {\pi}{3}R^2\left(h \minus{} 2R \plus{} \frac {4R^2}{h \minus{} 2R} \plus{} 4R\right)$

$ \geq \frac {\pi}{3}R^2\left(2\sqrt {(h \minus{} 2R)\cdot \frac {4R^2}{h \minus{} 2R}} \plus{} 4R\right) \equal{} \frac {8}{3}\pi R^3$,

the equality holds when $ h \minus{} 2R \equal{} \frac {4R^2}{h \minus{} 2R}\ (0 < h < 2R)\Longleftrightarrow h \equal{} 4R, r\equal{}\sqrt{2}R.$ :lol:

[quote="Kent Merryfield"]#5 is easier than #6. In both cases, I'll use $ R$ instead of $ 10.$ As before, being specific about the radius of the sphere doesn't make things any easier than just using the letter.

In both cases slice through the picture with a plane that includes the axis of the cone or cylinder to get a two dimensional picture.

In the case of #5, we see a rectangle inside a circle of radius $ R.$ Half of the width of the rectangle is $ r$ and half of the height is $ \frac {h}2.$ The Pythagorean theorem gives us that $ r^2 \plus{} \left(\frac h2\right)^2 \equal{} R^2.$

So the problem become this:

Maximize $ 2\pi rh \plus{} 2\pi r^2$ subject to $ r^2 \plus{} \left(\frac h2\right)^2 \equal{} R^2.$.[/quote]

I continue this, we can set $ r \equal{} R\cos \theta ,\ h \equal{} 2R\sin \theta \ \left(0 < \theta < \frac {\pi}{2}\right)$, we have

$ S \equal{} \boxed{ 2\pi r(h \plus{} r)} \equal{} \pi \cdot 2R\cos \theta (2R\sin \theta \plus{} R\cos \theta)$

$ \equal{} 2\pi R^2(\sin 2\theta \plus{} \cos ^ 2 \theta)$

$ \equal{} \pi R^2(2\sin 2\theta \plus{} 2\cos ^ 2 \theta)$

$ \equal{} \pi R^2(\cos 2\theta \plus{} 2\sin 2\theta \plus{} 1)\leq (1 \plus{} \sqrt {5})\pi R^2.$

The equality holds when $ (1,\ 2)\parallel{}(\cos 2\theta ,\ \sin 2\theta)\ (0 < \theta < 2\pi)\Longleftrightarrow (\cos 2\theta ,\ \sin 2\theta) \equal{} \frac {1}{\sqrt {5}}(1,\ 2)$

$ \Longleftrightarrow \cos \theta \equal{} \sqrt {\frac {5 \plus{} \sqrt {5}}{10}},\ \sin \theta \equal{} \sqrt {\frac {5 \minus{} \sqrt {5}}{10}}$, yielding $ r \equal{} \sqrt {\frac {5 \plus{} \sqrt {5}}{10}}R,\ h \equal{} \sqrt {\frac {10 \minus{} 2\sqrt {5}}{5}}R$, which satisfies $ 0 < h < 2R.$

[color=red]Last edited[/color]

kunny, shouldn't that be $ S\equal{}2\pi r(h\plus{}r),$ not $ 2\pi r(h\plus{}2r)?$

With that change, I'm getting rather different answers, such as $ r\equal{}\sqrt{\frac{5\plus{}\sqrt{5}}{10}}\,R$ with $ h\equal{}(\minus{}1\plus{}\sqrt{5})r.$

Repeating kunny's #11 using Lagrange multipliers:

Maximize $ r^2h$ subject to $ (r^2-R^2)h-2r^2R=0.$
\begin{align*}2rh&=\lambda(2rh-4rR)=2\lambda r(h-2R)\\ r^2&=\lambda(r^2-R^2)\end{align*}
Solving each of these for $ \lambda$ and equating, we get:
\begin{align*}\frac{r^2}{r^2-R^2}&=\frac{h}{h-2R}\\ r^2h-2r^2R&=r^h-R^2h\\ 2r^2R&=R^2h\\ h&=\frac{2r^2}R\end{align*}
Rewrite this as $ r^2=\frac{hR}2$ and put that into kunny's version of the constraint equation, namely $ r^2=\frac{hR^2}{h-2R}.$

We can solve this to get that $ h=4R,$ and it then follows that $ r=\sqrt{2}\,R.$

[quote="Kent Merryfield"]kunny, shouldn't that be $ S \equal{} 2\pi r(h \plus{} r),$ not $ 2\pi r(h \plus{} 2r)?$

With that change, I'm getting rather different answers, such as $ r \equal{} \sqrt {\frac {5 \plus{} \sqrt {5}}{10}}\,R$ with $ h \equal{} ( \minus{} 1 \plus{} \sqrt {5})r.$[/quote]

You are right, but the value of $ h$ seems to be wrong.

Anyway, I have just edited in #12.

My expression for $ h$ is correct and agrees with your value. Note that I used $ r$ there, not $ R.$ The equivalence is this:
\[ \sqrt{\frac{5\plus{}\sqrt{5}}{10}}\,R\cdot\left(\minus{}1\plus{}\sqrt{5}\right)\equal{}\sqrt{\frac{10\minus{}2\sqrt{5}}5}\,R\]
To see that cheaply, just put those numbers into a calculator.

Whoa, you guys are a lot of help!!  Does anyone feel like helping me out with number 2?  That's the last one.  If not I understand.  I'm going to print this out and study it, again thank you for all your help   :)

Why don't you try to do it and present a solution here?  I'm sure someone would be willing to point out any errors you might make.

The most useful flat (2-dimensional) picture for #2 is the same as the picture for #6.


Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]  \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)

Let BD / DC = a, CE / EA = b, and AF / FB = c.

First, by Ceva's theorem, abc = 1.

Let [APF] = s_1, [BPF] = s_2, [BPD] = s_3, [CPD] = s_4, [CPE] = s_5, and [APE] = s_6, and [ABC] = S.

1/(a + 1) = DC / (BD + DC) = DC / BC = (s_4 + s_5 + s_6) / S.
1/(b + 1) = (s_2 + s_1 + s_6) / S.
1/(c + 1) = (s_2 + s_3 + s_4) / S.

Then \sum 1/(a+1) = (s_1 + s_2 + s_3 + s_4 + s_5 + s_6 + s_2 + s_4 + s_6) / S = 3/2, since s_1 + s_3 + s_5 = s_2 + s_4 + s_6 = S/2.

Then multiplying (a+1)(b+1)(c+1) to both sides,

2(b+1)(c+1) + 2(a+1)(c+1) + 2(a+1)(b+1) = 3(a + 1)(b + 1)(c + 1) reduces to

ab + bc + ca - a - b - c = 3 - 3abc = 0, since abc = 1.

But then ab + bc + ca - a - b - c = ab + bc + ca - a - b - c + abc - 1 = 0 since abc - 1 = 0.

Then (a - 1)(b - 1)(c - 1) = 0, from which the result follows.

(One of a, b, c equals 1, which implies one of the ratios AF/FB, BD/DC, CE/EA equals 1 and hence P lies on a median.)

[hide]

Make an affine transformation such that the image of $ABC$, $A'B'C'$, is equilateral.
Let $D', E', F', P'$ be the respective images of $D, E, F, P$.
Because affine transformations preserve ratios of segments and ratios of areas,
$\frac{[PAF]}{[ABC]} + \frac{[PBD]}{[ABC]} + \frac{[PCE]}{[ABC]} = \frac{1}{2}$ if and only if
$\frac{[P'A'F']}{[A'B'C']} + \frac{[P'B'D']}{[A'B'C']} + \frac{[P'C'E']}{[A'B'C']} = \frac{1}{2}$
Also $P$ lies on one of the medians of $ABC$ if and only if $P'$ lies on the respective median of $A'B'C'$.
So it suffices to prove the given statement after the affine transformation.
Therefore, for notational convenience, we will assume $ABC$ is equilateral, and prove the original statement.
Let $a=AB=BC=CA$.

First assertion: Suppose $P$ lies on one of the medians of $ABC$.
Without loss of generality, it lies on the median containing $A$.
So $BD=DC$ and by Ceva's Theorem, $\frac{CE}{EA} \cdot \frac{AF}{FB} \cdot \frac{BD}{DC} = 1$
$\frac{CE}{a-CE} = \frac{FB}{a-FB}$, so $CE = FB$.
So $E$ is the image of $F$ reflected through $AD$. It is not difficult to see, since the figure is symmetric about $AD$, that $\frac{[PAF]}{[ABC]} + \frac{[PBD]}{[ABC]} + \frac{[PCE]}{[ABC]} = \frac{1}{2}$.

(I can't prove the second assertion  :( )

[/hide]

[color=darkred][b][u]PP[/u].[/b] Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$
respectively. Prove that $[PAF]+[PBD]+[PCE]=\frac S2\iff P$ lies on at least one of the medians of triangle $ABC$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] Denote  the area $[XYZ]$ of the triangle $\triangle XYZ$ . I will suppose w.l.o.g. that $[ABC]=2$ and I note:

$\bullet\ M,N,P$ - the midpoints of the sides $[BC],[CA],[AB]$ respectively;

$\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$ $[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$

$\bullet\ x=y_2-z_1\ ,\ y=z_2-x_1\ ,\ z=x_2-y_1\ .$ Thus, $\boxed{x_1+y_1+z_1=x_2+y_2+z_2=1\ ;\ x+y+z=0}\ .$

From the [u][b]Ceva[/b]'s theorem[/u] results the relation $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1$ , i.e. $\frac{(x_1+y_1)+y_2}{(x_2+z_2)+z_1}\cdot \frac{(y_1+z_1)+z_2}{(x_2+y_2)+x_1}\cdot \frac{(z_1+x_1)+x_2}{(y_2+z_2)+y_1}=1\Longleftrightarrow$

$\frac{1+(y_2-z_1)}{1-(y_2-z_1)}\cdot \frac{1+(z_2-x_1)}{1-(z_2-x_1)}\cdot \frac{1+(x_2-y_1)}{1-(x_2-y_1)}=1\Longleftrightarrow$ $(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longleftrightarrow $ $xyz=0\Longleftrightarrow$

$y_2=z_1\ \vee\ z_2=x_1\ \vee\ x_2=y_1\Longleftrightarrow $ $EF\parallel BC\ \vee\ FD\parallel CA\ \vee\ DE\parallel AB\Longleftrightarrow$ $P\in [AM]\cup [BN]\cup [CP]\ .$[/color]

it is the same my solution, I prove that:abc=xyz,a+b+c=x+y+z and ab+bc+ca=xy+yz+zx :arrow:

:P ,due to tishio semiya,japan and was published as crux pro. bet 95-00.
  :D ,my sol. is similar.

Let a, b, and c denote the respective areas of PAB, PBC, and PCA. We then have [PAF] = ac/(b+c) and analogous, so our task is to determine for what a, b, and c we have

ac/(b+c) + ba/(a+c) + bc/(a+b) = (a+b+c)/2.

It would suffice to show that this equation implies either a = b, b = c, or c = a. Indeed, suppose for sake of contradiction that none of these equalities were true for some (a', b', c') satisfying the equation. Then expand out the equation and consider it as a cubic in a. By inspection, the cubic evaluates at zero when a = b, a = c, and a = -b - c. Since this equation must also hold at a = a', the cubic has four roots and must be identically zero. However, plugging in a = 0 gives a contradiction, so the result is proven.

[quote="va2010"]Sorry to revive this, but the polynomial given $(r-s)(s-t)(t-r)(r+s+t)$ was 2014 USAMO #3. Is there something similar between these two problems?[/quote]
Not really. Both of these problems utilize the shoelace determinant, but to different ends, USAMO 2014 #3 uses it to construct a function which gives a shoelace determinant of 0 (making the points collinear) iff each of the values are distinct, which clearly can be seen by the determinant. This problem WANTS us to show that atleast two of $x,y,z$ are equal, because this would imply that the areas of these two triangles are equal, which clearly occurs with the median. So, to answer your original question, no, but kind of.

[hide=Solution]Let $[BPC]=x$, $[APC] = y$, $[APB] = z$.

Note that $\frac{[AFP]}{[APB]}=\frac{y}{x+y}\Rightarrow [AFP]=\frac{yz}{x+y}$. Similarly, $[PBD]=\frac{xz}{y+z}$ and $[PCE]=\frac{xy}{x+z}$.

Our equation is then $\sum \frac{yz}{x+y} = \frac{x+y+z}{2}$. Clearing denominators and factoring yields $(x+y+z)(x-y)(y-z)(z-x)=0$. It follows that at least one of $x,y$ must be equal to each other so $P$ will lie on the median of at least one median. These steps can be shown to be reversible.[/hide]

[quote=Wolstenholme]This question almost begs the use of barycentric coordinates.  Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ P = (r : s : t) $ for some real $ r, s, t > 0 $.  Then $ D = (0 : s : t), E = (r : 0 : t), F = (r : s : 0)  $.

Now $ [PAF] = \frac{1}{(r + s + t)(r + s)}\begin{pmatrix}1 & 0 & 0 \\ r & s & 0 \\ r & s & t \end{pmatrix}[ABC] = \frac{st}{(r + s + t)(r + s)}[ABC] $ so the problem condition is equivalent to $ \frac{r + s + t}{2} - \frac{st}{r + s} - \frac{tr}{s + t} - \frac{rs}{t + r} = 0 $ which upon expanding becomes $ \frac{(r - s)(s - t)(t - r)(r + s + t)}{2(r + s)(s + t)(t + r)} = 0 $ and so $ r, s, t $ are not all distinct which immediately implies the desired result.[/quote]

What is formula for surfice of triangle in bary

[quote=ratiborpaljevic33]What is formula for surfice of triangle in bary[/quote]

See http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/bary.pdf

Say $P$ has barycentrics $\left(\dfrac{1}{a}:\dfrac{1}{b}:\dfrac{1}{c}\right).$ Then we can assign masses of $a,b,c$ to $A,B,C.$ Notice that the condition $P$ lies on at least one median of $\triangle ABC$ is the same as the condition that at least two of $a,b,c$ are equal. Without loss of generality, say $[ABC]=1.$

Note that by mass points, $BD=BC\frac{c}{b+c}$ and the height of the altitude from $P$ to $BD$ is $h_a\frac{a}{a+b+c},$ where $h_a$ is the height of the $A$-altitude. Thus $[PBD]=\frac{BC\cdot h_a}{2}\cdot\frac{c}{b+c}\cdot\frac{a}{a+b+c}=\frac{ca}{(b+c)(a+b+c)}.$ Analogously,
\[[PCE]=\frac{ab}{(c+a)(a+b+c)}\]
\[[PAF]=\frac{bc}{(a+b)(a+b+c)}.\]
Now consider the equation
\[\frac{ca}{(b+c)(a+b+c)}+\frac{ab}{(c+a)(a+b+c)}+\frac{bc}{(a+b)(a+b+c)}=\frac{1}{2}.\]
We can multiply by $(a+b+c)(b+c)(c+a)(a+b)$ and instead consider the polynomial
\[ca(c+a)(a+b)+ab(a+b)(b+c)+bc(b+c)(c+a)-\frac{1}{2}(a+b+c)(a+b)(b+c)(c+a).\]
Note that no solutions will be lost from this.

We claim that $a-b$ and $a+b+c$ are roots of this polynomial. To verify the former, let $a=b$ and note that
\[ca(c+a)(2a)+a^2(2a)(c+a)+ca(c+a)^2-\frac{1}{2}(2a+c)(2a)(c+a)^2=\]
\[(c+a)(2ca^2+2a^3+c^2a+ca^2-a(c+a)(2a+c))=0.\]
By symmetry, $b-c$ and $c-a$ are roots. This verifies the if direction.

To verify the latter, let $a+b+c=0$ and note that
\[ca(-b)(-c)+ab(-c)(-a)+bc(-a)(-b)=\]
\[abc(a+b+c)=0.\]

Note there cannot be any other roots as our polynomial is quartic, and we have found four roots of degree one.

In the context of the problem, $a+b+c=0$ is an extraneous root. This verifies the only if direction.

Let $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $P=(x,y,z)$. Then, $D=(0:y:z)$, $E=(x:0:z)$ and $F=(x:y:0)$.
Thus, we need to show that $$\frac{yz}{x+y}+\frac{xz}{y+z}+\frac{xz}{x+z}=\frac{1}{2}$$
is true iff two of $x,y,z$ are equal to each other with the constraint that $x+y+z=1$.
Now, \begin{align*}
\frac{yz}{x+y}+\frac{xz}{y+z}+\frac{xy}{x+z}&=\frac{1}{2}\Longleftrightarrow\\
\frac{y(1-x-y)}{x+y}+\frac{(1-y-z)z}{y+z}+\frac{x(1-x-z)}{x+z}&=\frac{1}{2}\Longleftrightarrow\\
\left(\frac{1}{x+y}-1\right)y+\left(\frac{1}{y+z}-1\right)z+\left(\frac{1}{x+z}-1\right)x&=\frac{1}{2}\Longleftrightarrow\\
\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z}&=\frac{3}{2}\Longleftrightarrow\\
\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z}&=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z}\Longleftrightarrow\\
\frac{x-z}{x+z}+\frac{y-x}{x+y}+\frac{z-y}{y+z}&=0\Longleftrightarrow\\
\frac{x-z}{x+z}+\frac{(y-x)(y+z)+(z-y)(x+y)}{(x+y)(y+z)}&=0\Longleftrightarrow\\
\frac{x-z}{x+z}+\frac{2yz-2xy}{(x+y)(y+z)}&=0\Longleftrightarrow\\
(x-z)\left(\frac{1}{x+z}-\frac{2y}{(x+y)(y+z)}\right)&=0\Longleftrightarrow\\
(x-z)\left(\frac{(x+y)(y+z)-2y(x+z)}{(x+y)(y+z)(x+z)}\right)&=0\Longleftrightarrow\\
(x-z)\left(\frac{y^2+xz-xy-yz)}{(x+y)(y+z)(x+z)}\right)&=0\Longleftrightarrow\\
\frac{(x-z)(x-y)(z-y)}{(x+y)(y+z)(x+z)}&=0.
\end{align*}
We are done.

[quote=Wolstenholme]This question almost begs the use of barycentric coordinates.  Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ P = (r : s : t) $ for some real $ r, s, t > 0 $.  Then $ D = (0 : s : t), E = (r : 0 : t), F = (r : s : 0)  $.

Now $ [PAF] = \frac{1}{(r + s + t)(r + s)}\begin{pmatrix}1 & 0 & 0 \\ r & s & 0 \\ r & s & t \end{pmatrix}[ABC] = \frac{st}{(r + s + t)(r + s)}[ABC] $ so the problem condition is equivalent to $ \frac{r + s + t}{2} - \frac{st}{r + s} - \frac{tr}{s + t} - \frac{rs}{t + r} = 0 $ which upon expanding becomes $ \frac{(r - s)(s - t)(t - r)(r + s + t)}{2(r + s)(s + t)(t + r)} = 0 $ and so $ r, s, t $ are not all distinct which immediately implies the desired result.[/quote]
This is pretty nice. My approach is mostly similar to this.

Use barycentrics on $\triangle ABC.$ Let $P=(p,q,r)$ so $D=(0:q:r),E=(p:0:r),$ and $F=(p:q:0).$ Then, \begin{align*}\frac{[AFP]}{[ABC]}&=\frac{1}{p+q}\begin{vmatrix}1&0&0\\p&q&0\\p&q&r\end{vmatrix}=\frac{\begin{vmatrix}q&0\\q&r\end{vmatrix}}{p+q}=\frac{qr}{p+q}.\end{align*} Hence, the area condition is true if and only if $$0=\frac{qr}{p+q}+\frac{pr}{q+r}+\frac{pq}{p+r}-\frac{p+q+r}{2}=\frac{(p-q)(p-r)(r-q)(p+q+r)}{(p+q)(q+r)(r+p)}.$$ This is equivalent to $p,q,$ and $r$ being non-distict. $\square$

ah

Let $P=(x,y,z)$ in barycenric coordinates. We then have $$D=(0,y,z),E=(x,0,z),F=(x,y,0).$$ Being careful that all areas are of the same sign, we have $$\frac{[AFP]+[BDP]+[CEP]}{[ABC]}=\sum_{cyc} \frac{1}{(x+y)(x+y+z)}det{(1,0,0;x,y,0;x,y,z)}$$ $$=\sum_{cyc} \frac{yz}{(x+y)(x+y+z)}.$$ We want to show that $$\frac{yz}{x+y}+\frac{zx}{y+z}+\frac{xy}{z+x}=\frac{1}{2}(x+y+z)$$ if and only if either $x=y$, $y=z,$ or $z=x.$ After expanding, this becomes $$2(yz^3+zx^3+xy^3+sym(2,1,1)+cyc(2,2,0))=sym(2,1,1)+sym(3,1,0)+sym(2,2,0)+sym(2,1,1)$$ Note that $sym(2,2,0)=2cyc(2,2,0)$, so this cancels to just $$2(yz^3+zx^3+xy^3)=sym(3,1,0)$$ $$yz^3+zx^3+xy^3-(y^3z+z^3x+x^3y)=0.$$ Finally, this factors as $$(x-y)(y-z)(z-x)(x+y+z)=0,$$ and since $x+y+z=1$ this is true if and only if one two of $x,y,z$ are equal, which is equivalent to $P$ lying on a median.

writing this up for OTIS (ive solved this at least 3 times before but never wrote it up)
-----
Apply barycentric coordinates on $\triangle ABC$. Set $P=(l:m:n)$, so by parameterization we have $D=(0:m:n), E=(l:0:n), F=(l:m:0)$. Taking determinants and simplifying reduces the problem to $$\frac{(l-m)(l-n)(n-m)(l+m+n)}{(m+n)(l+m)(l+n)}=0$$ which immediately implies the desired.

[hide=Solution]
Use barycentric coordinates with respect to triangle $ABC$. Let $P = (x, y, z)$. Note that every point on $AP$ can be expressed $(t : y : z)$ for some $t\in \mathbb{R}$, so $$D = AP\cap BC = (0 : y : z) = \left(0, \frac{y}{y+z}, \frac{z}{y+z}\right).$$ Similarly, $$E = BP\cap AC = (x:0:z) = \left(\frac{x}{x+z}, 0, \frac{z}{x+z}\right)$$ and $$F = CP\cap AB = (x:y:0) = \left(\frac{x}{x+y}, \frac{y}{x+y}, 0\right).$$ Thus, we have that $$\frac{[PAF]}{[ABC]} = \begin{vmatrix} x & y & z \\ 1 & 0 & 0 \\ \frac{x}{x+y} & \frac{y}{x+y} & 0 \end{vmatrix} = \frac{yz}{x+y},$$ where the area is signed, but since $\frac{yz}{y+z} > 0$, this still holds when the area is unsigned. Similar computations yields $$\frac{[PBD]}{[ABC]} = \frac{zx}{y+z}$$ and $$\frac{[PEC]}{[ABC]} = \frac{xy}{z+x},$$ where the area is unsigned (althought it can be signed). Therefore, $$[PAF] + [PBD] + [PEC] = \frac{1}{2}[ABC]$$ is equivalent to
\begin{align*}
    \frac{1}{2} &= \frac{[PAF]}{[ABC]} + \frac{[PBD]}{[ABC]} + \frac{[PEC]}{[ABC]} \\
                &= \frac{yz}{x+y} + \frac{zx}{y+z} + \frac{xy}{z+x} \\
                &= \frac{\sum_{\text{cyc}} yz(y+z)(z+x)}{(x+y)(y+z)(z+x)},
\end{align*}
which is equivalent to $$2\sum_{\text{cyc}} yz(y+z)(z+x) = (x+y)(y+z)(z+x)$$ since $x, y, z > 0$, so since $x+y+z = 1$,
\begin{align*}
    0 &= (x+y)(y+z)(z+x) - 2\sum_{\text{cyc}} yz(y+z)(z+x) \\
      &= (x+y)(y+z)(z+x)(x+y+z) - 2\sum_{\text{cyc}} yz(y+z)(z+x) \\
      &= -(x+y+z)(x-y)(y-z)(z-x) \\
      &= -(x-y)(y-z)(z-x),
\end{align*}
which is equivalent to $x = y$, $y = z$, or $z = x$ (not mutually exclusive). If $x =y$, then $P$ lies on the $C$-median. If $y = z$, then $P$ lies on the $A$-median. If $z = x$, then $P$ lies on the $B$-median. Thus, $x=y$, $y=z$, or $z=x$ is equivalent to $P$ lying on at least one of the medians of $\triangle ABC$, so the area condition is equivalent to $P$ lying on at least one of the medians of $\triangle ABC$, as desired. $\blacksquare$


geometry
circumcircle
parallelogram
function
geometric transformation
reflection
incenter

I've finally finished the problems! Here they are:

1st independent study
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=47957
Let $P$ be a point in the interior of $\triangle ABC$. The lengths of the sides of $\triangle ABC$ is $a,b,c$, and the distance from $P$ to the sides of $\triangle ABC$ is $p,q,r$. Show that the circumradius $R$ of $\triangle ABC$ satisfies \[ \displaystyle R\le \frac{a^2+b^2+c^2}{18\sqrt[3]{pqr}}. \] When does equality hold?

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47959
Find all primes $p$ such that the number of distinct positive factors of $p^2+2543$ is less than 16.

2nd independent study
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=47962
A club provides 30 snacks to 18 members, and each member orders 3 different snacks. It is known that every snack is ordered by at least one member, and that any two members order at most one same snack. Is it possible to find 12 snacks, such that the snacks ordered by any member is not completely in these 12 snacks?

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47964
It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.

3rd independent study
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=27188
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that for all positive integers $m$ and $n$, $(f(m))^2 + f(n)$ divides $(m^2+n)$.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=47966
Given a triangle $ABC$, $A_1$ divides the length of the path $CAB$ into two equal parts, and define $B_1$ and $C_1$ analogously. Let $l_A$, $l_B$, $l_C$ be the lines passing through $A_1$, $B_1$ and $C_1$ and being parallel to the bisectors of $\angle A$, $\angle B$, and $\angle C$. Show that $l_A$, $l_B$, $l_C$ are concurrent.

Final exam-1st day
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=24075
Let $\{a_n\}$ be an infinite sequence such that $a_0>0, a_1>0, a_0 \neq a_1$ and \[ a_n=|a_{n+1}-a_{n+2}| \] for all $n \ge 2$. Prove that the sequence cannot be bounded.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=24078
The circle $\Gamma$ and the line $l$ do not intersect. Let $AB$ be the diameter of $\Gamma$ perpendicular to $l$, with $B$ closer to $l$. An arbitrary point $C$ distinct from $A$ and $B$ is chosen on $\Gamma$. The line $AC$ intersects $l$ at $D$, and the line $DE$ is tangent to $\Gamma$ at $E$, with $B$ and $E$ on the same side of $AC$. Let $BE$ intersect $l$ at $F$, and let $AF$ intersect $\Gamma$ at $G \neq A$. Prove that the reflection of $G$ with respect to $AB$ lies on the line $CF$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=47968
The set $\{1,2,\dots\>,n\}$ is called $P$. The function $f: P \to \{1,2,\dots\>,m\}$ satisfies \[ f(A\cap B)=\min (f(A), f(B)). \] What is the relationship between the number of possible functions $f$ with the sum $\displaystyle \sum_{j=1}^m j^n$?

Final exam-2nd day
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=41115
Let $a_1,a_2,\dots\>,a_n$ be $n>1$ positive reals, let $g_n$ denote their geometric mean, and define $A_i$ to be the arithmetic mean of $a_1,a_2,\dots\>,a_i$ for $i\ge 1$. Let $G_n$ denote the geometric mean of $A_1, A_2,\dots\>,A_n$. Show that \[ \displaystyle n\sqrt[n]{\frac{G_n}{A_n}}+\frac{g_n}{G_n} \le n+1. \] When does equality hold?

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=47970
Given a triangle $ABC$, we construct a circle $\Gamma$ through $B,C$ with center $O$. $\Gamma$ intersects $AC, AB$ at points $D$, $E$, respectively($D$, $E$ are distinct from $B$ and $C$). Let the intersection of $BD$ and $CE$ be $F$. Extend $OF$ so that it intersects the circumcircle of $\triangle ABC$ at $P$. Show that the incenters of triangles $PBD$ and $PCE$ coincide.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=41072
Given a positive integer $n>1$, let $P_n$ denote the product of all positive integers $x<n$ such that $n|(x^2-1)$. Find the remainder when $P_n$ is divided by $n$ for all $n>1$.

Note that a lot of the problems came from the IMO Shortlist.

I would appreciated any solutions, ideas, or comments. I also hope to gain access to the National Olympiad problems as soon as possible.

Oh k2c901_1 you must have worked really hard this afternoon! And Taiwan really have a strict TST, so many tests and problems......


You can find a PDF copy [url=http://www.aapt.org/physicsteam/2009/upload/F-ma-Exam-Solutions-2009.pdf]here[/url].

Two questions:
On number 14, I'm assuming that angular momentum is conserved up to the collision and rotational energy is conserved after.
On 16, I broke down the big spring of constant $ k$ into two halves each of constant $ 2k$. Is that the right thinking?

If it makes sense and it gets you the right answer, then it's the right thinking.  The system is symmetric so it looks like you need only consider one side.

But why? How do you consider the force that each of the $ 2k$ springs puts on the other? Won't the force on both sides of the spring (mass and the other spring) affect the distance compressed $ x$? I answered correctly on the exam but couldn't exactly get around this point. There's probably something really simple that I'm forgetting.

I dont understand the answer to the 14th question. How is the horizontal component of momentum not conserved?

EDIT: Oops I didnt consider the time : 0+

For some reason, the AAPT has pulled the exam off the site (even though the quarterfinals have already begun).

Is there any way to get these solutions?

They once posted it on the physics team homepage, but later somehow removed it. I think you can check [url]http://www.webassign.com[/url] to see if it's still there.

I found it, but how do we find out our score?

I can't remember if they tell you your score but they'll definitely notify you if you make it to the next round.  Ask your physics teacher if he has any information for you.


Hey. Can some people post various formulas for triangular geometry? Not simple ones like similarity, but definitions, formulas and theorems of things like centroids, incenters, circumcenter, etc.

:) yeah... i think this would help people.

[quote="cincodemayo5590"][b]Mildorf[/b]'s [url=http://www.tjhsst.edu/~tmildorf/math/Adv.htm]lectures[/url] probably have all you need.

(If this shouldn't be here -- and I don't know exactly what the rules are for referencing work like this are -- can a mod kindly remove it? Thanks. :) )[/quote]
thank you for the link  :D


linear algebra
matrix
linear algebra unsolved

Prove the following theorem:

If $ t_1, t_2, ..., t_n$ are distinct points in an interval $ [a,b]$, and if distinct points $ s_1, s_2, ..., s_n$ are outside that interval, then the matrix having elements $ a_{ij} \equal{} (t_i \minus{} s_j)^{\minus{}1}$ is invertible.

I don't feel like posting a proof now, but the result is a particular case of http://en.wikipedia.org/wiki/Cauchy_determinant , and you will find a sketch of a proof (with $ \plus{}$ instead of $ \minus{}$, but I hope this won't hinder you) at [url=http://www2.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi]Victor Prasolov's [i]Problems and theorems in linear algebra[/i], translated by Dimitry Leites[/url], chapter 1.1, paragraph 1.3 (a great problem collection).

  darij


linear algebra
matrix
function
trigonometry
Putnam
algebra proposed
algebra

[b]Let $a,b \in {\mathbb C} , b \ne 0,$ $D\subset {\mathbb C}$  and $f : D \to C$  a differentiable function. Find  all $2 \times 2$ complex matrices  $A$  with the spectrum in  $D$ , such that
\[ f(A)={a\; \; \; b\choose 0\; \; \; a}\; . \][/b]

[i] In case $a=1, b=1966 , f(x)=\sin{x}$ the problem was proposed at Putnam Competition-1996.[/i]

There is no such problem in Putnam 1966
http://www.kalva.demon.co.uk/putnam/putn66.html

He means 1996 (B4):
[url]http://www.kalva.demon.co.uk/putnam/putn96.html[/url]


Emma and Ben are playing cards. The loser paid 1 to the winner in the first game, 2 in the second game, 4 in the third game, and so on. (The prize was always doubled.) Emma started with 601, but lost all her money in 10 games. Which game did she win?

so the total given by emma was 1023 but she only lost 601 so she didn't lose 422

but when ben loses he has to give her money.  two trans take away from her loss.

she didn't lose 211

she didn't lose 128,64,16,2,1

game 1 she won 1
game 2 she won 2
game 3 she lost  4
game 4 she lost  8
game 5 she won 16
game 6 she lost  32
game 7 she won 64
game 8 she won 128
game 9 she lost  256
game10 she lost 512

total loss:601

I don't think that works. First of all, only ten games were played, so the highest stake was 512.

Now, we find that if she lost all her games, she would've lost 2^10 -1, or 1023

A correct answer that I got is:  (didn't check for multiple solutions)
game 1 : lost 1
game 2 : won 2
game 3 : won 4
game 4 : lost 8
game 5 : lost 16
game 6 : won 32
game 7 : lost 64
game 8 : won 128
game 9 : won 256
game10 : lost 512

nope. That has her left with 422

because she would lose 1023, she did NOT lose 422. That means that she won 211. 211 in binary is

11010011

so she won the games with stakes: 1, 2, 16, 64, and 128

yeah...cool!

[hide]ha you fell for it![/hide]


limit
logarithms
real analysis
real analysis solved

Determine $\displaystyle \lim_{n \to \infty} \frac {\sqrt[n] {n}-1}{\sqrt[2004] {n+1}-\sqrt[2004]{n}}$.
 The problem is old & easy, but I find it quite interesting.

I got 0...is it right?
if yes, I'll post my solution...if  not, I'll try more

sorry but my solution was wrong although the result is right, I'll do it again

Let $p$ be a positive real number such that  $0 < p < 1$.

$(n+1)^p  - n^p = p*u^{p-1}$ , where $n < u < n+1$ , so  $ \displaystyle 0 < \frac {1}{(n+1)^p - n^p} < \frac{1}{(p*n^{p-1}}$

L'Hospital rule ===>  $\displaystyle\frac{n^{1/n} - 1}{n^{p-1}}  \approx \frac{(1-ln(n))/n^2}{(p-1)*n^{p-2}}  \approx \frac{1-ln(n)}{(p-1)*n^p} \approx 0 $

Finally  $\displaystyle \lim_{n \to \infty} \frac {n^{1/n}-1}{(n+1)^p - n^p} = 0$

 :cool:

The same result with slightly different notation. As in Diogene's post, let $0<p<1$.

Factor $n^p$ out of the denominator and expand by the binomial series:

$(n+1)^p-n^p = n^p\left({(1+\frac1n)^p-1}\right) = n^p\left({\frac{p}n+O\left({\frac1{n^2}\right)}\right)}$. Note that this is essentially the same estimate Diogene used.

For the numerator, $n^{1/n}-1 = e^{(\ln n)/n}-1 = \frac{\ln n}n + O\left({\frac{\ln^2n}{n^2}}\right)$.

Thus, $\displaystyle \frac{n^{1/n}-1}{(n+1)^p-n^p} = \frac{\frac{\ln n}n+O\left({\frac{\ln^2n}{n^2}}\right)}{n^p\left({\frac{p}n+O\left({\frac1{n^2}\right)}\right)}} = \frac{\ln n + O\left({\frac{\ln^2n}n}\right)}{pn^p+O(n^{p-1})}$,

which clearly goes to zero because $\ln n$ grows much slower than $n^p$ for $p>0$.


For 1: You know $f(1)$. Using that equation, what other values of the function could you find out?

2: $x+1$ is a factor implies what about the roots of that polynomial? What does that tell you about $C$?

hint for 2:[hide]If x+1 is a factor then if x is -1, x + 1 is 0[/hide]

[quote="MaThS"]Need some help with these two problem

In a certain seqence, $f(1) = 3$ and $f(x+y) = f(x)f(y)$. Determine the value of $f(16)$, and give the closed form of the sequence.

If $x+1$ is a factor of $2x^2+5x+C$, determine the value of $|5C|$[/quote]

[hide="Answer"]
f(1) = 3
f(x+y) = f(x)f(y)

f(2) = f(1+1) = f(1)f(1) = (3)(3) = 9
f(4) = f(2+2) = f(2)f(2) = (9)(9) = 81
f(8) = f(4+4) = f(4)f(4) = (81)(81) = 6561
f(16) = f(8+8) = f(8)f(8) = (6561)(6561) = 43046721
[/hide]

#2
[hide]
I wasnt quite sure how to do the latex for synthetic division, so maybe if someone knows they could tell me :).
[code]
-1|  2   5   C
         -2  -3
       2  3  |0
So C = 3, |5C| = 15.
[/code]
[/hide][/code][/hide]


Find $ x$:

$ \sin{3x}\cos{x} \plus{} \cos{3x}\sin{x}\equal{}\frac{\sqrt{3}}{2}$

You can use the formula $ \boxed{\sin \ (\alpha \plus{}\beta)\equal{}\sin \alpha \cos \beta \plus{}\cos \alpha \sin \beta}$

Using what kunny suggested,

$ \sin{\left(x\plus{}3x\right)} \equal{} \frac{\sqrt{3}}{2}$

Notice how $ \sin{60} \equal{} \frac{\sqrt{3}}{2}$.

So basically, $ 4x\equal{} 60 \Rightarrow x\equal{}15^{\circ}$

So would it be $ \frac{\pi n}{2}  \plus{} 15^{\circ}$ where $ n$ is an integer?

Because $ \sin\theta$ is positive in both Quadrants I and II, it will be on a period of $ 2\pi$ for both $ \frac{\pi}{12}$ and $ \frac{\pi}{6}$


geometry
circumcircle
Euler
radical axis
geometry proposed

Let $ d$ be an arbitrary line passes through the circumcenter $ O$ of triangle $ ABC$. Let $ A'$ be the orthogonal projection of $ A$ onto $ d$, $ \Delta_1$ be the line passes through $ A'$ and perpendicular to $ BC$. Lines $ \Delta_2$, $ \Delta_3$ are defined similarly. Prove that $ \Delta_1$, $ \Delta_2$, $ \Delta_3$ are concurrent at a point lying on the Euler circle of triangle $ ABC$.

The problem=Theorem 694 in "COLLEGE GEOMETRY"(second edition) by Nathan Altshiller-Court; pp.288-289.

$ \text{Hint for other solution: }$
1).If $ d$ arbitrary $ (O\notin {d})$ then concurency point of lines $ \triangle_1, \triangle_2, \triangle_3$ is the ortopole of the line $ {d}$.
2). The orthopol $ \Omega$of line $ {d}$is the radical center of circumcircles of the pedal triangle of all point $ P\in {d}$. ($ {T. Lemoyne \ theorem})$
3). If  $ O\in {d}$, the circumpedal circle of $ O$ =euler circle of $ \triangle{ABC}\Rightarrow\Omega\in euler \ circle.$


analytic geometry
geometry
rectangle
IMO Shortlist
combinatorics

In the coordinate plane a rectangle with vertices $ (0, 0),$ $ (m, 0),$ $ (0, n),$ $ (m, n)$ is given where both $ m$ and $ n$ are odd integers. The rectangle is partitioned into triangles in such a way that

[i](i)[/i] each triangle in the partition has at least one side (to be called a “good” side) that lies on a line of the form $ x \equal{} j$ or $ y \equal{} k,$ where $ j$ and $ k$ are integers, and the altitude on this side has length 1;

[i](ii)[/i] each “bad” side (i.e., a side of any triangle in the partition that is not a “good” one) is a common side of two triangles in the partition.

Prove that there exist at least two triangles in the partition each of which has two good sides.

We shall use the word "triangle" to refer to any triangle in the partition. A triangle with only one bad side is to said to be of  kind [b]I[/b], otherwise it is said to be of kind [b]II[/b].

Let $V$ be the set of the midpoints of all the bad sides, let $E$ be the set consisting of any line segment which joins two midpoints of bad sides for a triangle of kind [b]II[/b] , let $G$ be the graph with $V$ as vertex set and $E$ as edge set. It is easy to show that each edge of $G$ is parallel to one of the axes and the distance from it to this axis is a half-integer , i.e., a number of the form $k+1/2$, where $k$ is an integer. for $P \in V$, let $d(P)$ denote the degree of $P$. Obviously, $d(P)$ is not greater than $2$ for all $P \in V$.

we consider the following three cases.

$(1)$: there exists a point $P$ such that $d(P)=0$. In this case, both the triangle with $P$ on their common ban side are of kind [b]I[/b], and the proof ends.

$(2)$: there is a point $Q \in  V$ such that $d(Q)=1$, then there exists a chain with $Q$ as its one extreme. Let $R$ be the other extreme of the chain, then $d(R)=1$. Both the triangle with $Q$ on its bad side and the triangle with $R$ on its bad side belong to kind [b]I[/b], in this case, the proof ends too.

$(3)$: $d(P)=2$ for all $P \in V$. Then $G$ is a disjoint union of cycles. Now we divide the rectangle into $mn$ unit squares with integral lattice point as vertices. According to the construction of $G$ and the hypothesis that $d(P)=2$ for all $P \in V$, the central point of each unit square in the rectangle is passed by one and only one cycle. By coloring those unit square black and white alternately, we make the rectangle into an $m \times n$ chessboard. Each cycle passes through black and white squares alternately. Hence the number of the unit squares passed by a certain cycle of $G$ is an even integer. So is the total number of the unit squares in the rectangle .This contradicts the fact that $mn$ is an odd integer. This ends the proof.

thanks a lot!!!!!!! :blush:  :blush:  :wacko:  :oops:  :D  :D  :blush:

[quote=sholly]
 (3)Each cycle passes through black and white squares alternately
                                                [/quote]

I don't understand this line much, can someone clarify it?

Color the rectangle with the standard black and white rectangle coloring. Call the [b] offset [/b] of a triangle in the configuration the area of it which is black minus the area of it which is white. Call a triangle [b] excellent [/b] if it has two good sides, and [b] plain [/b] otherwise.

[b] Lemma. [/b] The offset of an excellent triangle is $\pm \frac12$, and the offset of a plain triangle is $0$.

[b] Proof. [/b] This is a simple computation for excellent triangles. As for plain triangles, just split the plain triangle in half with the altitude of length $1$, and use the same computations from before.

$\blacksquare$

Since the "overall offset" of the entire rectangle is $1$, there must clearly be at least $2$ excellent triangles, so we're done.

$\square$


there is no 5 by 5 square. there are 5 dots, but a sie length of 4

Such a pain to find all the squares with irrational side lengths


Let $ S$ be a set of $ 2008$ elements. Subsets $ B_1,B_2,...,B_k$ of $ S$ are chosen such that

for each element $ x\in S$, there is $ j$ such that $ x$ is not in $ B_j$, and for any other $ m$ such that $ x$ is also not in $ B_m$, either $ j \equal{} m$ or $ B_j\cap B_m$ is empty.

Find the maximum value of $ k$.

i think some condition is missed out
else we always set $ B_{i}\equal{} \phi$ and have as many $ B_{i}$ as we want

I mean different subsets $ B_i$ (I thought it was clear, but you have a good point anyway  :) )

Based on the title of the post, I assume it should say, [quote="jbmorgan"]either $ j \equal{} m$ or $ B_j\cap B_m$ is [b]not[/b] empty.[/quote]

Also, the symbol \emptyset ($ \emptyset$) exists in LaTeX.

Thanks for telling me the empty set symbol.

By the way, I really gave the topic a bad name. The problem statement is according to what I intended...

If the problem statement is as you intended, then gauravpatil's comment is correct: we can have [i]all[/i] the subsets, since for any $ x \in S$, there is a $ j$ such that $ B_j \equal{} \emptyset \not\ni x$ and for any other $ m$ such that $ x \not \in B_m$, $ B_j \cap B_m \equal{} \emptyset \cap B_m \equal{} \emptyset$.

I'm terribly sorry. It's all because of my pathetically poor English.

Let me formulate the problem again.

Let $ S$ be a set of $ 2008$ elements. Nonempty subsets $ B_1,B_2,...,B_k$ of $ S$ are chosen such that

for each chosen subset $ B_i$, there is $ x\in S$ such that, firstly, $ x$ is not in $ B_i$, and secondly, for any other $ m$ such that $ x$ is also not in $ B_m$, either $ j \equal{} m$ or $ B_j\cap B_m$ is empty.

Find the maximum value of $ k$.

I hope it is written correctly now.


algorithm
advanced fields
advanced fields unsolved

I have the following problem which I am searchiing the solution for:

I have a set of points which are connected with weighted connections. A connection between two points costs differently,depending on the direction you take from one point to another - it takes distance and streams into account . Points are also weighted. In my problem I have to find a path, in which I have to maximize the sum of point weights. But there is a trick. I have a limited distance I can cover. That means I will not be able to traverse all points. Is there already an algorithm that solves this problem or do I have to start working on my own solution?

greetz

That looks incredibly hard to do purely mathematically. I'm no specialist in this kind of thing, but I'd suggest looking at the A* pathfinding algorithm that's used in computer programs for, well, finding paths. This approach may be an efficient way to find a good path.

I already have A* implemented. Thats how I get the edge cost. I also have cost for points. Now i have above stated criterias I have to consider to form a path through my points to maximize the sum of points and minimize the sum of edges. Problem is, there is not enough time to pass through all points.  I hope its clear what I want because I can't state it any clearer.


[quote="rahul1089"]I'm reposting this question because I think I might have posted it on the wrong forum earlier.

[b]Show that an n*n matrix A is invertible exactly when 0 is NOT an eigenvalue of A[/b][/quote]
Notice that fact that $0$ is eigenvalue of $\mathbf{A}$ is equivalent to fact that there exists such vector $\vec{x}\ne \vec{0}$ that $\mathbf{A}\vec{x}= \vec{0}$. Thus fact that $0$ is not eigenvalue of $\mathbf{A}$ is equivalent to fact that $\mathbf{A}\vec{x}=\vec{0}$ implies $\vec{x}=\vec{0}$. But from [b]Kent Merryfield[/b]'s post [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=628327#628327[/url] we know that $\mathbf{A}\vec{x}=\vec{0}\implies\vec{x}=\vec{0}$ is equivalent fact that $\mathbf{A}$ has independent columns. And it's well known that $\mathbf{A}$ has independent columns iff it's invertible so we are done.


vector
function
integration
articles
geometry
analytic geometry
graphing lines

Let's say that the ancient greeks were the only ones in the galaxy to have invented the use of the angle in geometry. All the other aliens were using many difficult calculations to overpass the use of the angle, but are excellent and well versed in mainly algebra. Now the aliens have come to earth, and you need to explain to them what is an angle. How would you tell them?

Now the same for velocity. Lets say the aliens defined velocity in a complicated way. in the same circumstances as above how would you define 'earth' velocity to an alien?

[hide="Hint"]Careful! the questions are not as easy as they seem, but are tricky :D [/hide]

What is the problem with using the rate of change of displacement???

I'm guessing the prob is that those aliens don't know vectors but isn't vectors a part of algebra as well???

No. They dont know what vectors are, because they have no intuition, and can only think with mathematical formulas.

i would give an alien a ruler and a stop watch. I would tell him to run forward and time himself as he ran. I would then tell him to divide the distance he has ran by his time .

Also I'm kind of confused about ur question can u go into greater detail of what aliens know

Time, distance? those are measurable quantities and very 'seeable'. The aliens as I said before dont have any physical insight and can only think in mathematical formulas, they dont know what it means to 'measure' something.

The aliens know algebra very well.  :wink:  :wink:

Consider set $ S$ of all ordered pairs $ (x,y)$ such that $ x^2 + y^2 = 1$
'The angle of $ (x,y)$' is a function $ \theta: S\rightarrow\mathbb{R}$
given by $ \theta(x_0,y_0) = \int_{(0,0)}^{(x_0,y_0)} \sqrt {\left( \frac {dy^2}{1 - y^2} + \frac {dx^2}{1 - x^2}\right)}$

Crude, I know :dry: , but it should work
I've read an article about this, that done properly (and much more elegantly), it can help us to 'visualise' the Lorentz Transformations and also something about hyperbolic geometry. I'll try to look it up.

EDIT: Yup, I found it! :)
[hide]
This article, by Giorgio Goldoni, had appeared in the Indian Science journal [i]Resonance[/i]
We assume that the aliens do know some geometry but they avoid the use of 'angle' by using a quantity called 'slope' $ k=\frac{y}{x}$, and the law of addition of slopes, $ K=\frac{k+k'}{1-kk'}$...(1)
[quote]
One day, a mathematician named Angle, while studying a complicated power series, obtains a new function that he calls 'tangent'and writes '$ \tan$'. Angle easily shows that tangent is invertible on the interval $ \left( -\frac{\pi}{2},\frac{\pi}{2}\right)$. He then discovers an incredible connection between the tangent and the slope, namely, if you pass from the slope $ k$ to the new quantity $ \theta=\tan^{-1} k$ you can replace the complicated formula (1) with a simple additive one. In other words, if $ \theta=\tan^{-1}k$, $ \theta'=\tan^{-1}k'$ and $ \Theta=\tan^{-1}K$, we have
$ \Theta=\theta+\theta'$
[/quote]
[/hide]

Yes very good. you are getting there. But this what you write is still much more 'intuitive'.
Try to be more abstract, since the aliens cannot understand it that well. As I mentioned they are the best in algebra. :D

[hide="hint"]Group Theory[/hide]

I'm not sure really

Let $ \left\langle \mathcal{G}, \plus{} \right\rangle$ be an ordered, dense and complete group such that

$ \exists T\in\mathcal{G},T\neq 0$ with $ T \plus{} T \equal{} 0$

The members of $ \mathcal{G}$ seem to satisfy all properties of angles.

(I'm using the usual definitions of order relation and completeness/LUB property)

EDIT: As I've noticed, my previous definition actually defined the 'measure' of an angle rather than an angle itself.

Sorry for the double post, I couldn't edit the previous one

For the velocity, whats wrong with
[hide="this?"]
Let $ X$ be a metric space

Define 'motion' as the function $ \mathbf{f}: \mathbb{R}\rightarrow X$

The 'velocity' at $ t\in\mathbb{R}$ is $ \mathbf{v}\in X$ given by $ \mathbf{v}(t) \equal{} \mathbf{f}'(t)$
[/hide]
Which is essentially the rate of change of postion vector..

[quote="Thaakisfox"][hide="hint"]Group Theory[/hide][/quote]

The [url=http://en.wikipedia.org/wiki/Circle_group]circle group[/url]?

Yes t0rajir0u, very close.

One-parameter Lie groups.
now just, say it explicitly, what is the angle.

The velocity is similar.

Ok here is what I was thinking of:

The angle is the canonical parameter of the one-parameter compact Lie-group.

The velocity is the canonical parameter of the Galilei Group. :D


benazar soale ghahangie khodamam fekr nakardam :D
farz konid $ p(n)$ bozorgtarin maghsoomelaihe farde $ n$ bashad sabet konid $ \frac{1}{2^{n}}\sum_{k \equal{} 1}^{2^{n}}\frac{p(k)}{k}>\frac{2}{3}$

let $ p(n)$ be the greatest odd divisor of $ n$...
prove that $ \frac{1}{2^{n}}\sum_{k\equal{}1}^{2^{n}}\frac{p(k)}{k}>\frac{2}{3}$

rahe hallesh ine:

darim too majmooe $ \{1,2,...,2^{n}\}$ bahs mikonim dge,khob bebinid too in majmooe tedade aadadi ke bar $ 2^{k}$ bakhshpaziran,($ 1\leq k\leq n\minus{}1$),barabare ba

$ \frac{2^{n}}{2^{k}}\equal{} 2^{n\minus{}k}$

khob hala har adadi ke bar $ 2^{k}$ bakhshpazir basharo mishe be soorate $ 2^{k}.r$ neveshtesh,hala vase ye seri az in adada $ r$ farde va vase ye serishoon $ r$ zoje,hala majmooei az in aadad ke $ r$eshoon fardaro begirid $ S_{k}$,khob hala ba kami deghat motevajeh mishim ke daghighan nesfe in adada $ r$eshoon farde va nesfe digashoon $ r$eshoon zoje,yeni dar vaghe $ |S_{k}| \equal{} 2^{n\minus{}k\minus{}1}$,dar vaghe $ S_{k}$ majmooei az adadas ke $ k$ bozorgtarin tavani az $ 2$e ke mishmorateshoon ya be ebarati too rabete $ \parallel X\parallel_{2}\equal{} k$ sedgh mikonan...

khob hala vase adade $ t\in S_{k}$ darim:

$ \frac{p(t)}{t}\equal{}\frac{1}{2^{k}}$

(dalilesham ineke mishe $ t$ ro be shekle $ 2^{k}.r$ neveshtesh ke $ r$ farde,pas bozorgtarin amele farde $ t$ barabare $ r$ has pas $ p(t) \equal{} r$ va baghie majera...)

khob pas age hame in adadro ba ham jam konim mishe:

$ \sum_{t\in S_{k}}\frac{p(t)}{t}\equal{}\frac{1}{2^{k}}\plus{}...\plus{}\frac{1}{2^{k}}\equal{} |S_{k}|.\frac{1}{2^{k}}\equal{} 2^{n\minus{}k\minus{}1}.\frac{1}{2^{k}}$

$ \equal{} 2^{n\minus{}2k\minus{}1}$

khob hala $ k$ hamoonjoori ke goftim mitoone az $ 1$ ta $ n\minus{}1$ taghir kone ke age jameshoon bezanim mishe be in shekl neveshteshoon:

$ \sum_{k \equal{} 1}^{n\minus{}1}\sum_{t\in S_{k}}\frac{p(t)}{t}\equal{}\sum_{k \equal{} 1}^{n\minus{}1}2^{n\minus{}2k\minus{}1}$

--------------------------------------------------
hamchenin age $ m$ ye adade fard bashe oon moghe $ p(m) \equal{} m$ pas $ \frac{p(m)}{m}\equal{} 1$ pas baraye aadade fard age in nesbataro ba ham jam konim mishe

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

(chon $ 2^{n\minus{}1}$ta adade fard darim...)

--------------------------------------------------
hamchenin baraye adade $ 2^{n}$ in nesbat barabare ba $ \frac{p(2^{n})}{2^{n}}\equal{}\frac{1}{2^{n}}$

--------------------------------------------------
pas hasele jami ke mikhastim barabare ba:

$ \frac{1}{2^{n}}\sum_{k \equal{} 1}^{2^{n}}\frac{p(k)}{k}\equal{}\frac{1}{2^{n}}(\sum_{k \equal{} 1}^{n\minus{}1}2^{n\minus{}2k\minus{}1}\plus{}2^{n\minus{}1}\plus{}\frac{1}{2^{n}})$

hala deghat konid ke dakhele paranteze balaro mishe intori sadetaresh kard:

$ \sum_{k \equal{} 1}^{n\minus{}1}2^{n\minus{}2k\minus{}1}\plus{}2^{n\minus{}1}\equal{}\sum_{k \equal{} 0}^{n\minus{}1}2^{n\minus{}2k\minus{}1}$

pas ebarate hokmo mishe intori neveshtesh:

$ \equal{}\frac{1}{2^{n}}(\sum_{k \equal{} 0}^{n\minus{}1}2^{n\minus{}2k\minus{}1}\plus{}\frac{1}{2^{n}})$

$ \equal{}\frac{1}{2^{n}}(2^{n\minus{}1}\sum_{k \equal{} 0}^{n\minus{}1}2^{\minus{}2k}\plus{}\frac{1}{2^{n}})$

$ \equal{}\frac{1}{2^{n}}(2^{n\minus{}1}\sum_{k \equal{} 0}^{n\minus{}1}\frac{1}{2^{2k}}\plus{}\frac{1}{2^{n}})$

amma midoonim:

$ \sum_{k \equal{} 0}^{n\minus{}1}\frac{1}{2^{2k}}\equal{}\frac{1\minus{}\frac{1}{2^{2n}}}{1\minus{}\frac{1}{2^{2}}}$

$ \equal{}\frac{1\minus{}\frac{1}{2^{2n}}}{\frac{3}{4}}$

$ \equal{} 4.\frac{1\minus{}\frac{1}{2^{2n}}}{3}$

pas ebarate hokm moadele ba:

$ \frac{1}{2^{n}}(2^{n\minus{}1}(4.\frac{1\minus{}\frac{1}{2^{2n}}}{3})\plus{}\frac{1}{2^{n}})$

$ \equal{}\frac{1}{2^{n}}(2^{n\plus{}1}.(\frac{1\minus{}\frac{1}{2^{2n}}}{3})\plus{}\frac{1}{2^{n}})$

$ \equal{}\frac{1}{2^{n}}(\frac{2^{n\plus{}1}}{3}\minus{}\frac{2}{3}(\frac{1}{2^{n}})\plus{}\frac{1}{2^{n}})$

ke age bazesh konim mishe:

$ \equal{}\frac{2}{3}\plus{}\frac{1}{2^{n}}(\frac{1}{2^{n}}\minus{}\frac{2}{3}(\frac{1}{2^{n}}))$

$ \equal{}\frac{2}{3}\plus{}\frac{1}{2^{n}}(\frac{1}{2^{n}}(1\minus{}\frac{2}{3}))$

$ \equal{}\frac{2}{3}\plus{}\frac{1}{2^{n}}(\frac{1}{2^{n}}(\frac{1}{3}))$

$ \equal{}\frac{2}{3}\plus{}\frac{1}{3.4^{n}}>\frac{2}{3}$

pas didim ke:

$ \frac{1}{2^{n}}\sum_{k \equal{} 1}^{2^{n}}\frac{p(k)}{k}\equal{}\frac{2}{3}\plus{}\frac{1}{3.4^{n}}>\frac{2}{3}$
yeni hamoon chizi ke mikhastim...


Να δειξεις οτι για καθε θετικο ακεραιο n υπαρχει πολυωνυμο n βαθμου το οποιο διαιρειτα απο τον n! και κανενας φυσικα συντελεστης του πολυων. δεν ειναι πολλαπλασιο του n!. :wink:

Γνωριζουμε οτι το γινομενο ν διαδοχικων αριθμων διαιρειται με το ν!(προφανες)
Αποδειξη: εχουμε  (ν+κ+1)(ν+κ+2)....(2ν+κ)/ν!=συνδυασμοι 2ν+κ ανα ν
=ακεραιος.
Επομενως για καθε θετικο ακεραιο ν υπαρχει το πολυονυμο (ν+κ+1)(ν+κ+2)....(2ν+κ)  με κ ακεραιο που ειναι ν-οστου βαθμου και διαιρειται με το ν!

ΡΑΛΛΗΣ :D

Ναι σωστη και ωραια η λυση σου Ραλλη. :D

σπουδαια ασκηση μου εδωσε πολυ τροφη για σκεψη.προσπαθησα να κατασκευασω το πολυωνυμο. και μου πηρε πολυ χρονο και δεν τα καταφερα τελικα.Ραλλη εξοχα,η λυση σου δειχνει εξοικοιωση,πολυ βασικο για η συνεχεια. :) και εγω βρηκα κσατι παρωμοιο .

Ναι καλη η ασκηση μου Κωστα,αλλα που ειναι και η δικη σου ασκηση που ειχες πει οτι θα προτινεις?Ανυπομονω να την δω!Ετοιμαζω και αλλη μια ασκηση(δικη μου)αλλα θελω να την ξανακοιταξω. :D


function
trigonometry
real analysis
real analysis unsolved

Give an example of a continuous real-valued function $f$ from $[0,1]$ to $[0,1]$ which takes on every value in $[0,1]$ an infinite number of times.

One possibility is to compose a [url=http://en.wikipedia.org/wiki/Space-filling_curve]space-filling curve[/url] with an orthogonal projection. For a more explicit example, take the Cantor staircase and replace each flat part of length $3^{-n}$ with a translated copy of $2^{-n}\sin(3^{n}2\pi x)$. Could work...

@mlok

i couldnt understand your second example about the cantor staircase. could you please elaborate on it? i'm interested. thanks

[quote="mlok"]I was just looking at the graph of [url=http://mathworld.wolfram.com/CantorFunction.html]Cantor staircase[/url] and imagined replacing the horizontal segment in the middle of the graph by a sine wave that reaches $1/2$ up and $1/2$ down, thus assuming all values between $0$ and $1$. Then look at two smaller horizontal segments, of length $1/9$ each, and replace each by a wave of amplitude $1/4$. One wave will have the values in $[0,1/2]$, the other in $[1/2,1]$. Together these two waves cover $[0,1]$ by their images. And so on...[/quote]

Ah, I see. That's really neat. The hint given in the book suggests a modification of the Cantor staircase, so I suspect that's pretty close to the idea that they were thinking of.

For extra credit, prove that there are no continuous functions of bounded variations with this property. Or disprove.  :maybe:


I have no idea BUT:

\[ 6 \equal{} \frac{\left [ \left(4\times 5\right) \minus{}2 \right]}{\left(3 \times 1 \right)}\]

Umm i don't know. But 7 = (5*3) - (4*2*1)
8 = (5*3) - 1 - 2 - 4
9 = 5(3!) - 24 +2 + 1
10 = 15 + 2 -4 -3

But really if u got 3 hours you can write out all the 100 numbers  :P .

[hide="illusiat"]You may use scrap paper, but all of your answers must be neatly recorded on this sheet. [b]You have 20 minutes.[/b]
Calculators are NOT allowed. [/hide]

There are a few simple tircks you can do. For exmaple, if you notice that your solution to one of the problems has 1-2, you can flip it to make 2-1. Don't nessacarily do the problems in order, do the easiest ones first. Fore xample 29 is easy to do (1*2*3*4+5). Anything you immediately see write down. 1,2,3,4,5 using addition, subtraction can make all of 1-15. If you can't get a number, skip it and come back to it if you have time. If you do, try common thigns such as multiplying by 1, or multiplying by 4/2 instead of 2. If you can't get rid of 5 (like to make 24), try doing 5-1=4, which is easier to work with (btw, 24=(5-1)(3*4/2)). Practice with a lot of numbers beforehand. ideally, you'll get 5 numbers that you already got. remember that you can use farction, they just need to come out to integers later (i.e., 1*3/(2-5/4)=4). Remember that 1^(anything) is 1, so use this to get rid of any numbers you can't get rid of.

That's basically all I can really tell you.

Here are some values I found. Using patterns, I'm sure you can find close to all, if not all of the numbers from $ 1$ to $ 100$ using the five digits and the given operations.

$ \frac {4\times3}{2} \plus{} \frac {5}{1} \equal{} 11$
$ \frac {5\times4}{2} \plus{} 3 \minus{} 1 \equal{} 12$
$ 1 \plus{} 4\times5 \minus{} 3! \minus{} 2 \equal{} 13$
$ 13 \plus{} 2 \plus{} 4 \minus{} 5 \equal{} 14$
$ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \equal{} 15$
$ 1 \plus{} 2\times3 \plus{} 4 \plus{} 5 \equal{} 16$
$ 4\times5 \minus{} \frac {3}{2 \minus{} 1} \equal{} 17$
$ 1 \plus{} 2 \plus{} 3! \plus{} 4 \plus{} 5 \equal{} 18$
$ 21 \minus{} 3 \minus{} 4 \plus{} 5 \equal{} 19$
$ 1 \plus{} 2 \plus{} 3\times4 \plus{} 5 \equal{} 20$
$ 1 \minus{} 2 \plus{} 3 \plus{} 4! \minus{} 5 \equal{} 21$
$ 1 \plus{} 2\times3! \plus{} 4 \plus{} 5 \equal{} 22$
$ 4! \plus{} 5 \minus{} 1 \minus{} 2 \minus{} 3 \equal{} 23$
$ 1(4! \plus{} 5 \minus{} 2 \minus{} 3) \equal{} 24$
$ 1 \minus{} 2 \minus{} 3 \plus{} 4! \plus{} 5 \equal{} 25$
$ 1 \plus{} 2 \plus{} 3 \plus{} 4\times5 \equal{} 26$
$ 1 \plus{} 2\times3 \plus{} 4\times5 \equal{} 27$
$ 31 \minus{} 2 \plus{} 4 \minus{} 5 \equal{} 28$
$ 1 \plus{} 2 \minus{} 3 \plus{} 4! \plus{} 5 \equal{} 29$
$ 31 \minus{} 2 \minus{} 4 \plus{} 5 \equal{} 30$
$ 1(34 \minus{} 5 \plus{} 2) \equal{} 31$
$ 1 \plus{} 2 \plus{} 3!\times4 \plus{} 5 \equal{} 32$
$ 2 \minus{} 1 \plus{} 3 \plus{} 4! \plus{} 5 \equal{} 33$
$ 1(2 \plus{} 3 \plus{} 4! \plus{} 5) \equal{} 34$
$ 1 \plus{} 2 \plus{} 3 \plus{} 4! \plus{} 5 \equal{} 35$
$ 2 \minus{} 1 \plus{} 3! \plus{} 4! \plus{} 5 \equal{} 36$
$ 31 \minus{} 4 \plus{} 2\times5 \equal{} 37$
$ 1 \plus{} 2 \plus{} 3! \plus{} 4! \plus{} 5 \equal{} 38$
$ 14\times3 \plus{} 2 \minus{} 5 \equal{} 39$
$ 1\times4(2 \plus{} 3 \plus{} 5) \equal{} 40$
$ 1 \plus{} 4(2 \plus{} 3 \plus{} 5) \equal{} 41$
$ 1 \plus{} 2\times3! \plus{} 4! \plus{} 5 \equal{} 42$
$ 1(23 \plus{} 4\times5) \equal{} 43$
$ 1 \plus{} 23 \plus{} 4\times5 \equal{} 44$
$ 13\times4 \minus{} 2 \minus{} 5 \equal{} 45$
$ 12\times5 \minus{} 3\times4 \equal{} 48$
$ (1 \plus{} 4)(2 \plus{} 3 \plus{} 5) \equal{} 50$
$ 14\times3 \plus{} 2\times5 \equal{} 52$
$ 12\times5 \minus{} 3 \minus{} 4 \equal{} 53$
$ 4! \plus{} 1\times2\times3\times5 \equal{} 54$
$ 2\times4! \minus{} 1 \plus{} 3 \plus{} 5 \equal{} 55$
$ 12\times3 \plus{} 4\times5 \equal{} 56$
$ 1 \plus{} 2\times4! \plus{} 3 \plus{} 5 \equal{} 57$
$ 3\times4\times5 \minus{} 1\times2 \equal{} 58$
$ \frac {3!\times4\times5}{2} \minus{} 1 \equal{} 59$
$ 1\times\frac {3!\times4\times5}{2} \equal{} 60$
$ \frac {3!\times4\times5}{2} \plus{} 1 \equal{} 61$
$ 1(2 \plus{} 3\times4\times5) \equal{} 62$
$ 1 \plus{} 2 \plus{} 3\times4\times5 \equal{} 63$
$ 31\times2 \plus{} 4 \plus{} 5 \equal{} 71$
$ 43\times2 \minus{} 1 \minus{} 5 \equal{} 80$
$ 31\times2 \plus{} 4\times5 \equal{} 82$
$ 23\times4 \minus{} 1 \minus{} 5 \equal{} 86$
$ 1\times23\times4 \minus{} 5 \equal{} 87$
$ 1 \plus{} 23\times4 \minus{} 5 \equal{} 88$
$ 1 \minus{} 2 \minus{} 3! \minus{} 4! \plus{} 5! \equal{} 89$
$ 2 \minus{} 1 \minus{} 3! \minus{} 4! \plus{} 5! \equal{} 91$
$ 1(2 \minus{} 3! \minus{} 4! \plus{} 5!) \equal{} 92$
$ 1 \plus{} 2 \minus{} 3! \minus{} 4! \plus{} 5! \equal{} 93$
$ 1 \plus{} 3(2 \plus{} 4! \plus{} 5) \equal{} 94$
$ 23\times4 \minus{} 1 \plus{} 5 \equal{} 95$
$ 1\times23\times4 \plus{} 5 \equal{} 96$
$ 1 \plus{} 23\times4 \plus{} 5 \equal{} 97$
$ 3! \minus{} 4! \plus{} 5! \minus{} 1 \minus{} 2 \equal{} 99$


inequalities
linear algebra
matrix
linear algebra unsolved

[b]Let[/b] $ A \in GL_{n} \mathbb{(R)}$,$ B \in M_{n} \mathbb{(R)}$
[b]Prove that :[/b]
  [b]max[/b] {$ det^{4}(A)$,$ det^{4}(B)$} $ \leq \det \left[\begin {array} {LL} A^{T}A & B^{T}B \\ -B^{T}B & A^{T}A \end {array} \right]$
$ A^{T}$ [b]denotes the transposed matrix A[/b]
 :P  :P  :P

[hide="Hint"]First reduce to the case $ A\equal{}E_n$, then to $ B^tB$ is diagonal. From here the proof should be easy.[/hide]


Find the period of 2sin (4pix + pi/2) + 3cos (5pix).

The period of $ 2\sin\left(4\pi x\plus{}\frac\pi2\right)$ is $ \frac{2\pi}{4\pi}\equal{}\frac12$, and the period of

$ 3\cos{5\pi x}$ is $ \frac{2\pi}{5\pi}\equal{}\frac25$.

Thus, in order to have $ \frac12m\equal{}\frac25n$, we must have $ 5m\equal{}4n\implies(m,n)\geq(4,5)$

$ \implies\boxed2$ is the period.


Frobenius matrix norm is defined by $\Vert A \Vert_F = ( \sum_{i=1}^{n} \sum_{j=1}^{m} \vert a_{ij} \vert^2)^{1/2}$, could anyone help to show that $\Vert A \Vert_F = ( \sigma_1^2 + \sigma_2^2 +...+ \sigma_r^2)^{1/2}$

(Hint : Show that if $B=UC$, where $U$ is orthogonal, then $\Vert B \Vert_F= \Vert C \Vert_F$)

Sorry missing the definition as following:

If $A \in R^{n \times m}$ have singular values $\sigma_1 \geq \sigma_2 \geq...\geq 0$ then $\Vert A \Vert_2=\sigma_1$

It's simpler than that. ${\sum_{i,j} |a_{ij}|^2=\sum_i\sum_j \overline{a_{ij}}a_{ij}=\sum_j\sum_i A^*_{ji}}a_{ij}=\text{Tr}(A^*A)$ by direct calculation. The $\sigma_i^2$ are the eigenvalues of $A^*A$, and their sum is the trace.


Where do you find the national participants for 2008?

Look on the Mathcounts web site.  You might have to use the site map to find them.  They were there last month. I do not know if they have removed them yet.

if you went to nats u could look at the yearbookish thing


Two players (A and B) play a game with 2n cards. On each card a positive integer is written. The cards are shuffled and laid out in a row, with the numbers facing upwards. A starts and the players take turns to remove one card from either end of the row, until B picks up the remaining card. Each girl's score is the sum of the numbers on her chosen cards at the end of the game. Prove that A can always obtain a score at least as big as B's score.

Believe me, I have a delicious and super short solution :)

Here's my solution.

[hide]Number the cards 1, 2, ..., 2n.  Call the cards 2, 4, ..., 2n the even cards, and the cards 1, 3, ..., 2n-1 the odd cards.  Assume, without loss of generality, that the sum of the even cards is at least the sum of the odd cards.  Then A's strategy is to choose all the even cards.  At every turn, A will have a choice of an even card and an odd card; A will always choose the even card.  Poor B will get only a choice of two odd cards.  So at the end, A will have all the evens, and B will have all the odds.[/hide]

Yes, it's a nice problem.  I thought it was going to be harder until you said there was a short solution.  By the way, if you ever play the game for real, here is a simpler way to state A's strategy: [hide]After the first move, take a card from the same end that B just took from. [/hide]


function
trigonometry
logarithms
probability and stats

Let $ X_1,X_2,...$ be independent  random variables such that
$ P(X_k\equal{}k)\equal{}P(X_k\equal{}\minus{}k)\equal{}\frac{1}{2k^2}$,
$ P(X_k\equal{}1)\equal{}P(X_k\equal{}\minus{}1)\equal{}\frac{1}{2}(1\minus{}\frac{1}{k^2})$ if $ k>1$.
Show that $ \frac{U_n}{\sqrt{n}}\minus{}>N(0,1)$ in distribution, where $ U_n\equal{}\sum_{i\equal{}1}^{n}X_i$.

We use a transform method. We trust that we have already done enough analysis to have that an appropriate type of convergence of characteristic functions (such as uniform convergence on compact subsets) will imply convergence in distribution of the underlying random variables.

The characteristic function of $ X_k$ can be written as $ \chi_k(t)\equal{}\cos t\plus{}\frac1{k^2}(\cos kt\minus{}\cos t)$

$ \equal{}\cos t\left[1\plus{}\frac1{k^2}\left(\frac{\cos kt}{\cos t}\minus{}1\right)\right].$

The characteristic function of an independent sum is the product of the characteristic functions. Put in the appropriate scaling, and we have that the characteristic function of $ \frac{U_n}{\sqrt{n}}$ is

$ \cos^n\left(\frac{t}{\sqrt{n}}\right)
\prod_{k\equal{}1}^n\left[1\plus{}\frac1{k^2}\left(\frac{\cos (kt/\sqrt{n})}{\cos (t/\sqrt{n})}\minus{}1\right)\right]$

It is straightforward to show that as $ n\to\infty,$ $ \cos^n\left(\frac{t}{\sqrt{n}}\right)$ tends to $ e^{\minus{}t^2/2}.$

We now have to show that the remaining product tends to $ 1.$

Upon taking the logarithm, the product can be compared to

$ \sum_{k\equal{}1}^n\frac{1}{k^2}\left(\frac{\cos (kt/\sqrt{n})}{\cos (t/\sqrt{n})}\minus{}1\right)$

As $ n$ tends to infinity, for each $ k,$ each term of this sum tends to zero. As long as we confine our selves to a specific compact interval in $ t$ and as long as $ n$ is large enough (so that $ \cos (t/\sqrt{n})$ is bounded away from zero), then the terms of the some are bounded by a constant times $ \frac1{k^2}.$ We may invoke the Dominated Convergence Theorem, and argue that as $ n\to\infty,$ this sum tends to zero. The sum tending to zero implies the product tends to $ 1,$ which is what we need.

We now have that the characteristic function of $ \frac{U_n}{\sqrt{n}}$ tends to $ e^{\minus{}t^2/2},$ which is the characteristic function of the standard normal distribution. That's what we needed.

May be i missed something, but it seems to be pretty standart, though Lindeberg-Feller does not apply directly.
I used characteristic functions: let $ X_{k,n} = \frac {X_k}{\sqrt {n}}$, $ \phi_{k,n}(t) = Ee^{itX_{k,n}}$, then
\[ \phi_{k,n}(t) = \frac {1}{k^2}(\cos(\frac {tk}{\sqrt {n}}) - \cos(\frac {t}{\sqrt {n}})) + \cos(\frac {t}{\sqrt {n}})
\]
Next, use the well-known lemma: if $ \abs{a_i}\leq1, \ \abs{b_i}\leq1 \rightarrow |\prod_1^n a_i - \prod_1^n b_i| \leq \sum_1^n|a_i - b_i|$. All we need is to estimate
\[ |\phi_{k,n}(t) - e^{ - \frac {t^2}{2n}(2 - 1/k^2)}| = - \frac {t^2}{nk^2} + o(n)
\]
and note that $ \prod_1^n e^{ - \frac {t^2}{2n}(2 - 1/k^2})\to e^{ - t^2}$, so the limit distribution is $ \mathbb{N}(0,2)$.

P.S. Kent Merryfield is of course faster, but i still think the limit of the variance is 2.

[quote="Spasty"]Yes, the limit of the variance is 2, but $ U_n/\sqrt {n} \minus{} > N(0,1)$.[/quote]
 :?: What do you mean?

In this exercise $ \frac {U_n}{\sqrt {n}} \minus{} > N(0,1)$ and $ Var\left(\frac {U_n}{\sqrt {n}}\right) \minus{} > 2$ as $ n \minus{} > \plus{} \infty$.


LaTeX
Asymptote
Support
geometry
AMC 10
\/closed

The problem sets are so great that I just like to download the pdf version of the latest AMC 10 tests. It got only one problem.The downloaded pdf version has Latex asy text in it so I can't use it. For example it includes
text like:

(asy)
unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4;
filldraw(Arc((0,0),4,0,180)–cycle,gray,black); filldraw(Circle((0,2),2),white,black); dot((0,2));
draw((0,2)–((0,2)+2*dir(60))); label(quot;36;236;quot;,midpoint((0,2)–((0,2)+2*dir(60))),SE);
(/asy)

Why the downloaded pdf version has Latex asy text? Who should fix them?

Thanks in advance!

Which PDF/problem was that? That's not supposed to happen. Regardless, asymptote support has not yet been implemented for the PDFs; that will come in the next major update, which is still a while away.

The sample pdf I downloaded is AMC 10, 2009. The problem 6 has the following text.

6 A circle of radius 2 is inscribed in a semicircle, as shown. The area inside the semicircle
but outside the circle is shaded. What fraction of the semicircle’s area is shaded?
(asy)
unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4;
filldraw(Arc((0,0),4,0,180)–cycle,gray,black); filldraw(Circle((0,2),2),white,black); dot((0,2));
draw((0,2)–((0,2)+2*dir(60))); label(quot;36;236;quot;,midpoint((0,2)–((0,2)+2*dir(60))),SE);
(/asy)

I substituted [] with () to avoid an error.

The site is so great.

Many if not most of the PDF's have bugs... you can always use the archived AMC's on the wiki, though...

[quote="vwu9"]But I can't get a decent printable version of the tests.[/quote]

Copy and paste everything, post it on Word Processor, and print it there.

For the wiki, go to Preferences -> Skins, then change your skin to Monobook. Then click on "printable version" on the sidebar to print out decent copies.

Thanks a lot. Copy and Paste method and Wiki Monobook preference both work very well. Great site and great people.


How many points of inflection does y=x^4 have?

If you think about the graph, it sort of just opens upward, so there's no inflection points.

Edit: Hah, not thinking straight. Yeah, the graph is convex everywhere, so no inflection points.

hello, that is not correct, $ y\equal{}x^4$ has no inflection points because
$ y'(x)\equal{}4x^3$
$ y''(x)\equal{}12x^2$
$ y'''(x)\equal{}24x$
$ y^{(4)}(x)\equal{}24$
the point $ x\equal{}0$ is a local minimum because $ y'(0)\equal{}y''(0)\equal{}y'''(0)\equal{}0$ and $ y^{(4)}(0)\ne 0$, this minimum is actually a global one.
Sonnhard.


I don't know if this problem fits in this category, but anyway...

In any [i]increasing geometric sequence[/i], (such as 2, 6, 18), you multiply by a [b]fixed constant greater than 1[/b] to go from any one term to the next.  If such a sequence is 7 terms long, and if all 7 terms are integers between 200 and 3000, what is the largest number in the sequence?

-----
 If the general form is x, xz, xz^2, xz^3, ... I set x = 200 and xz^6 (the last term) = 3000.  So I solved for z^6 and got 15.  I took the 6th root of 15 and got 1.5704...  This is  a fixed constant and it is greater than one, but I do not think I solved the problem correctly.

Can you solve it with a solution please?

Thanks

[hide="I think this works"]
First of all, the numbers don't have to be $ 200$ and $ 3000$, they just have to be in between.  If the first term is $ a$ and the change is $ r$, the first term is $ a$ and the last term is $ ar^6$.  We have two cases.

Case 1: $ r$ is an integer.
If this is true, then $ ar^6\le3000$.  If $ a\ge200$, we have that $ r^6\le15$, which only works for $ r\equal{}1$.

Case 2: $ r$ is a rational number.
If this is true, then $ r\equal{}\frac{p}{q}\rightarrow r^6\equal{}\frac{p^6}{q^6}$.  This suggests that $ a\equal{}kq^6, ar^6\equal{}kp^6$ for an integer $ k$.  Since $ a\equal{}kq^6\le3000$, testing values of $ q$ allows for $ q\equal{}2,3$.  If $ q\equal{}3$, since the sequence is increasing, it follows that the rate of change must be at least $ \frac{4}{3}$ since $ r$ is rational.  But this cannot occur, since then the final term, $ ar^6\ge4096$.  Thus, $ q\equal{}2$.  Now, if $ p\equal{}4$, then $ r\equal{}\frac{p}{q}\equal{}2$, which we showed cannot happen.  Thus, since we want a rational $ r$, $ p\equal{}3$.  Now we have that $ 64k\ge200$ and $ 729k\le3000$.  This can be satisfied when $ k\equal{}4$.

The terms of the sequence are $ 256, 384, 576, 864, 1296, 1944, 2916$, and the greatest is $ 2916$.[/hide]


Prove, that if $n$ is a composite number, $n\geq2$, then exist integers $x,y,a,b$, $x,y,a,b\geq1$ such that
$\frac{x}{a}+\frac{y}{b}=1$
$a+b=n$

[quote="dondigo"]Prove, that if $n$ is a complex number, $n\geq2$, then exist integers $x,y,a,b$, $x,y,a,b\geq1$ such that
$\frac{x}{a}+\frac{y}{b}=1$
$a+b=n$[/quote]

$a+b=n$?

And what does $\geq 2$ mean for a complex number¿

$n$ is composite so there exist $k, t \geq 2$ such that $n=kt$. And then take $a=k$, $b=n-k$, $x=1$, $y=(t-1)(k-1)$.


[quote="1234567890"].....but when does Round 3 come out? I need to redeem myself.[/quote]

Hopefully middle of next week.

[quote="zanttrang"]Excellent - that means all of winter break to mull over the problems.  Also, is there any estimate on when scores will be returned?  And how was turnout compared to the last round?  I know there was some concern over the relatively low percentage of registered students who submitted scores for the first round.[/quote]
hMm -- patience you must learn :yoda:

[quote="zanttrang"][quote="calc rulz"]Do they take the time to get out the problems because they need time to write them? Or is it because they don't want people to have too much time to do them? Or something else?[/quote]
I believe it's because they're still compiling and polishing the problems they're going to be using on the next round.[/quote]

Yep.  Round 3 should be available by the end of this week.

When will Round 2 scores come out?  :?

Thanks!


Let's say you have m things, and you need to arrange n of them, with $n>m$(obviously, you can use everything more than once). Is there a formula for that?

I don't really follow what your saying but I think you may be trying to refer to combinatorics.

Sorry if it was hard to understand. Now I'm confused as to why I put that up...arggh...moderator, could you please delete this topic?

No, I know what you are talking about.  Balls and boxes.  If you own AoPS vol. 1, read chapter 25.  But actually, if you just want a quick overview, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=balls+boxes&t=69905.

EDIT: I just reread your question, and maybe that's not what you are talking about.  It's still a useful thing to know, however.

[quote="easyas3.14159..."]THANK YOU MYSMARTMOUTH!!!I'm saved![/quote]

 :D


What is informatics?

I think i am interested in doing informatics competitions. What books would give me a good head start?

Umm.... if you're thinking informatics as in "International Olympiad in Informatics" then it pretty much means solving algorithmic problems + programming. I've never really learned anything useful from books during my high school times so I can't recommend any to you, perhaps except for [url=http://www.amazon.com/Introduction-Algorithms-Thomas-H-Cormen/dp/0262032937]Introduction to Algorithms[/url]. The best way to start doing informatics competitions is (1) learn a programming language (preferably C++ but Java would work too) and then (2) start practicing. Definitely check out [url]http://ace.delos.com/usacogate[/url], this is where we used to practice a lot, it starts out with some really really simple problems but eventually goes through everything you'll ever need.


Using on 01/1/1997 starting date, it is estimated that the population of a certain city, when completed $ t$ years after the start date, will be thousands of people $ P$, with $ P(t)\equal{}30\minus{}\frac{4}{t}$. 	The increase of population during the year 2002 will be:

a) 200 people
b) 133 people
c) 30 people
d) 4 people
e) 2 people

At the start of 2002, $ t\equal{}5$.

So we need to compute $ P(6)\minus{}P(5)$.

$ P(6)\minus{}P(5)\equal{}(30\minus{}\frac{4}{6})\minus{}(30\minus{}\frac{4}{5})\equal{}\minus{}\frac{2}{3}\plus{}\frac{4}{5}\equal{}\frac{2}{15}$

Population increase: $ 1000(2/15)\approx \boxed{133}$

So the answer is $ \boxed{\text{B}}$.

Thanks, the anwer correct is A. But I understand how to do now, thank you.

I don't like this problem because $ t \equal{} 0$ isn't the value that makes
the population zero; it's $ t \equal{} \frac {2}{15}$ of a year after the
start date.  The variable t is not allowed to be zero because of
a threat of being undefined (as opposed to the usual set-up).

Hence we have this discrepancy between gauss1181's expected
solution and the OP's stated answer.


geometry
geometric transformation
reflection
geometry proposed

Construct a quadrilateral $ABCD$ given $AB = a$, $BC = b$, $CD = c$,$DA = d$.it is also knwn that the diagonal through $AC$ bisects angle $A$.You are allowed to use an unmarked ruler and a compass.

This problem was given to $9^{th}$ graders at a junior olympiad and be assured none could solve it.

In the segment $AD$ (or prolongation) construct the point $B'$ such that $AB'=a$. Now, draw the circunferences with centres $B'$ , $D$ and radius $b$, $c$, respectively. Let $C$ be one of the points of intersection of these circunferences. Finally, construct the reflection point $B$ of $B'$ with respect to $AC$.

It's not that tough, in my opinion.

See the attached diagram. Let $B'=\mathcal{S}_{AC}(B)$. Then $A-B'-D, CB'=b$ and $AB'=a$, hence with fixed point $A$, $\triangle B'CD$ is immediately constructible. After we find $C$, we just reflect $B'$ back to $B$ and we're done.

EDIT: Beaten by a second.

probably we felt it was tough because here reflections are quite an underused topic. i tried to construct a circle whish touches both $AB$ and $AD$ though without much success.


Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[f(x+y)f(x-y)=(f(x)f(y))^{2}\;\;\;\forall x,y\in\mathbb{R}.\]


For every subset of $ S\equal{}\{1,2,...,32\}$ with $ k$ elements, if there exists distinct $ a,b,c$ such that $ a|b$ and $ b|c$, what is the least possible value of $ k$?

A) 17 B) 24 C) 25 D) 29 E) None

[hide]
From $ \{1,2,4,8,16,32\},\{3,9,27\},\{5,10,20\},\{6,12,24\},\{7,14,28\}$ we must remove at least $ 8$ elements. For example, removing $ 1,2,3,4,5,6,7,8$ gives us a subset of size $ 24$ without divisibility triples, hence answer is $ \textbf{(C)}$.
[/hide]

edit; ah crud  :roll:

[quote="azjps"]For example, removing $ 1,2,3,4,8,10,12,28$ gives us a subset ... without divisibility triples[/quote] $ \{5, 15, 30\}$ ;)  Why didn't you just remove the obvious $ 1,2,3,4,5,6,7,8$?  Then $ 9 \cdot 2 \cdot 2 > 32$ prevents any triples.


While Steve and LeRoy are fishing $ 1$ mile from shore, their boat springs a leak, and water comes in at a constant rate of $ 10$ gallons per minute. The boat will sink if it takes in more than $ 30$ gallons of water. Steve starts rowing toward the shore at a constant rate of $ 4$ miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?

$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 10$

It takes $ \frac{1}{4}\text{hr}$ or $ 15\text{min}$ to get back to shore. Let $ x$ be the rate that LeRoy bails water out. Then we have that they do not sink, so \[ 30\ge 15\cdot (10\minus{}x)\implies x\ge 8\] (and equality obviously occurs) so our answer is $ 8$, or $ \boxed{D}$.

[hide]i took a slightly different approach to this problem. Since it will take them 15 minute to row back, it should take the water 15 minutes to get 30 gallons. So only 2 gallons can come in per minute and so 10 g/min - 2 g/min = $ \fbox{8 gal/min}$. i took me like 30 seconds to do this problem in the actual contest.[/hide]

Number 11.  The fastest way would probably be cognos'

I skipped this one because I thought that there were 100 min in an hour so I got $ 30\geq25(10\minus{}x)$ and the answer I got wasn't one of the choices  :blush:

[quote=sugar_rush][hide=solution]It will take $15$ minutes to row the $1$ mile back to shore while traveling at $4$ miles per hour. After these $15$ minutes, the total amount of water that will come into the boat is $150$ gallons, and if LeRoy bails at $x$ gallons per minute, then $15x$ gallons will be bailed out. The boat will sink if $150-15x\geq 30\rightarrow 15x\leq 120\rightarrow x\leq 8$. The answer is $\boxed{\textbf{(D)} ~8}$.[/hide][/quote]

ok bumping 1 old lowend amc question was fine
2 was ok

but this is getting annoying

[quote=OTMS_MATHCLUB][quote=sugar_rush]@above [quote=rohan.sp]Hi, RFT can bump these posts as long as he posts a solution.

Thanks,
rohan.sp[/quote]

Therefore, I can also bump as long as I have a solution. Hope this helps.[/quote]

I think he meant "a distinct solution" than the ones in that thread[/quote]

[quote=v_Enhance]
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.
[/quote]

This is from the pinned C&P Posting recs.

Got this question wrong while mocking because I think the inequality was strict (> instead of $\geq$)

My reasoning was at exactly 8 minutes, they would sink, so while they reached the shore they have sinked. Why is this wrong?


Sum the series
$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^m(n3^m+m3^n)}$


A block is projected up an inclined plane making an angle of 30 degree with the horizontal at a speed of 12m/s. The coefficient of friction is 0.4. Compute:
a) The speed of the block as it returns to its starting point
b)The time required for the block to ascend and descend the plane

Hi, here is the solution of your problem.
 For the upward joureny along the incline, the acceleration  is directed down the incline and has a magnitude $ a_1\equal{}g(\sin\alpha \plus{} \mu \cos\alpha)$ where $ \mu$ is the friction coefficient and $ \alpha$ is the angle of the incline. It comes to rest (instantaneously)  after a time $ t_1 \equal{} \dfrac{v_0}{g(\sin\alpha \plus{} \mu \cos\alpha)}$ covering  a distance $ s \equal{} \dfrac{v_0^2}{2g(\sin\alpha \plus{} \mu \cos\alpha)}$. Here, $ v_0$ is the initial projection speed.

For the return journey, the acceleration is still down, but its magnitude becomes $ a_2\equal{} g(\sin\alpha \minus{} \mu \cos\alpha)$. Going down the incline, it must cover the same distance $ s$ as it did while going up. Accordinly, the final speed of the block as it returns to its starting point is given by
\[ v\equal{}\sqrt{2a_2s} \equal{} v_0 \sqrt{\dfrac{\sin\alpha \minus{} \mu \cos\alpha}{\sin\alpha \plus{} \mu \cos\alpha}}\]
Also, the time $ t_2$ for the downward jourey can be obtained from the relation $ t_2 \equal{} v/a_2$ as
\[ t_2 \equal{} \dfrac{v_0}{g(\sin\alpha \minus{} \mu \cos\alpha)}\sqrt{\dfrac{\sin\alpha \minus{} \mu \cos\alpha}{\sin\alpha \plus{} \mu \cos\alpha}} \equal{} \dfrac{v_0}{g} \sqrt{\dfrac{1}{\sin^2\alpha \minus{} \mu^2 \cos^2\alpha}}
\]
Plugging in the values you can easily obtain the required answers.


Give the greatest common divisor of $ 6^3$ and $ 3^6$.

I think it's $ \fbox{27}$. Did anyone get something else?

$ 6^3\equal{}2^3 \cdot 3^3$
$ 3^6\equal{}3^6$.

Taking the smallest powers in common, $ 3^3$.

Or, use the euclidian algorithm.

$ (216,729)\equal{}(216,81)\equal{}(54,81)\equal{}(54,27)\equal{}(27,27)$.

So yes, you are correct.


When I build it, there are two errors,

1. LaTex Error: File `fancyhdr.sty' not found.
Type X to quit or <RETURN> to proceed,
or enter new name. (Default extension: sty)
Enter file name:

2. Emergency stop.
<read *>

1.29 \pagestyle
                  {fancy}
No pages of output.
Transcript written on solutions.log.

There must be a lot of people who use the template, so I didn't expect an error. Does anyone know what those errors mean or what I should do?

Edit:Never mind, I just found another thread that asked the same thing

The problem is that you don't have fanyhdr package installed on your TeX system.
Please read [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]this[/url].


Se stie ca $a_n= 1 + 1/2 + 1/3 + ... + 1/n$ are limita :inf:.
Fie p :ge: 1 arbitrar.
$|a_n_+_p - a_n| = | 1/(n+p) + 1/(n+p-1) + ... + 1 / (n + 1)| < p/n -> 0 $
Deci oricare ar fi  :varepsilon: > 0 si p :ge: 1 exista N a.i. $|a_n_+_p - a_n| <$ :varepsilon: pt. orice n :ge: N, deci $a_n$ convergent(ceea ce e absurd)
Deci unde gresesc?

p nu trebuie fixat in definitia sirului Cauchy.

Wow! I can understand "$p$ is not a fixed number in Cauchy criteria". Right?


AMC
USA(J)MO
USAMO
algebra proposed
algebra

Prove that the product of the 2 real roots of
$ x^{4}\plus{}x^{3}\minus{}1\equal{}0$ is a root of $ x^{6}\plus{}x^{4}\plus{}x^{3}\minus{}x^{2}\minus{}1\equal{}0$
 :)

One of possible solutions can be found on:

[url]http://www.kalva.demon.co.uk/usa/usoln/usol773.html[/url]


ARML
geometry
AMC
AIME
ratio
angle bisector
Pythagorean Theorem

1. Triangle ABC is inscribed in a circle and $BP$ bisects angle $ABC$. If $AB=6$, $BC=8$, $AC=7$, compute $BP$.

For this one, I first used the angle bisector theorem to get $BP = 3$ and $PC = 4$. Then I [hide]used a wierd formula from AOPS Volume 2 that gave directly gave me the length of $BP$ as
$6$, but the correct answer should be $8$. How come? [/hide]

2. Given triangle $ABC$, with $AB =15$ and $CA=12$. The median from $B$, the altitude from $C$, and the angle bisector from $A$ are all concurrent inside the triangle. Compute the length of $BC$.

For number 1 the problem should stipulate that the point P is on the circle, not the triangle.

Problem number 2 is one of my favorite geometry problems. I think it was posted originally by JBL a few years ago. It's harder than a typical ARML problem - its probably around the level of a late AIME question. It requires you to pull together a lot of the theorems you know about triangles.

2
[hide]By the angle bisector theorem, $BC$ is split into segments that are in a ratio of 4 to 5. Let's call these lengths $4x$ and $5x$. Let $a$ and $b$ denote the length of the section of $AB$ closest to $A$ and $B$ respectively. Then by Pythagorean theorem,

$144 - a^2 = 81x^2 - b^2$

$x = \frac {\sqrt {144 - a^2 + b^2}}{9}$

Since $b = 15 - a$ we substitute to get

$x = \frac {\sqrt {369 - 30a}}{9}$

By Ceva,

$\frac {4 \sqrt {369 - 30a}}{9} \cdot 6 \cdot (15 - a) = \frac {5 \sqrt {369 - 30a}}{9} \cdot 6 \cdot a$

$4(15 - a) = 5a$

$a = \frac {20}3$

Since $BC = \sqrt {369 - 30a}$ we substitute $a = \frac {20}3$ to find that $BC = 13$[/hide]

[quote="gauss202"]
Problem number 2 is one of my favorite geometry problems.[/quote]

I personally do not like this problem very much (though it is alright, and fairly creative).  It basically just requires straightforward application of a lot of theorems.  Basically if you know Ceva, you'll get this problem and if you don't you won't. But for testing knowledge of these theorems it is alright.

The main thing that I like about the problem isn't so much the problem itself (though I like that too), but the triangle involved. In particular, the fact that the third side happens to turn out so nice.

If you solve the problem generally with sides b and c, you'll see that it seems quite unusual for the third side to come out so nice. When the problem was originally posted I think we had some discussion about how the proposer of the problem came up with such a triangle, and if there were any others like it.

Yes, I still have the mathematica program on which I calculated the smallest solutions to it.  The general problem does not have a nice-looking solution at all!

I am actually quite curious where you (Keta) came across this question.

I think he came across it from a set of Geometry problems that I put online for SC ARML Team practice - which in turn mostly came from this site.

Problem 2 is very similar to a problem from Poland Math Olympiad 1961. I think there is a post about that in the Olympiad section.

[quote="gauss202"]For number 1 the problem should stipulate that the point P is on the circle, not the triangle.
[/quote]

Umm, ok, what do you do then?  I think I might be missing something obvious because no one has mentioned this problem... :blush:

[hide]I used angle bisector, stewarts and then power of a point on that one.[/hide]

oh yeah, power of a point, duh!  thanks, i was completely lost. :lol: