geometry
circumcircle
geometric transformation
reflection
geometry solved

Let $ABC$ be a triangle with circumcenter $O$. Let $P$ and $Q$ be points on the segments $AB$ and $AC$, respectively, such that $BP : PQ : QC = AC : CB : BA$.
Prove that the points $A$, $P$, $Q$ and $O$ lie on one circle.

[i]Alternative formulation.[/i] Let $O$ be the center of the circumcircle of a triangle $ABC$. If $P$ and $Q$ are points on the sides $AB$ and $AC$, respectively, satisfying $\frac{BP}{PQ}=\frac{CA}{BC}$ and $\frac{CQ}{PQ}=\frac{AB}{BC}$, then show that the points $A$, $P$, $Q$ and $O$ lie on one circle.

[quote="JG"]Let $ABC$ be a triangle with circumcenter $O$. Let $P$ and $Q$ be points on the segments $AB$ and $AC$, respectively, such that $BP : PQ : QC = AC : CB : BA$.
Prove that the points $A$, $P$, $Q$ and $O$ lie on one circle.[/quote]

Nice problem! Here is my (probably not optimal) solution:

[b]Lemma 1.[/b] Let ABC be a triangle with circumcenter O. If X, Y, Z are three points on the sidelines BC, CA, AB of this triangle ABC, and if the triangle XYZ is directly similar to the triangle ABC, then the circumcircles of triangles AYZ, BZX and CXY pass through the point O.

[i]Proof of Lemma 1.[/i] We will use directed angles modulo 180°.

Let S be the orthocenter of the triangle XYZ. Then, the lines YS and ZS are altitudes of the triangle XYZ, and thus perpendicular to the corresponding sides ZX and XY of this triangle. Hence, < (YS; ZX) = 90° and < (ZS; XY) = 90°. Hence,

  < YSZ = < (YS; ZS) = < (YS; ZX) + < (ZX; XY) + < (XY; ZS) = < (YS; ZX) + < (ZX; XY) - < (ZS; XY)
            = 90° + < (ZX; XY) - 90° = < (ZX; XY) = < ZXY.

But since the triangle XYZ is directly similar to the triangle ABC, we have < ZXY = < CAB, and thus < YSZ = < CAB = < YAZ. Consequently, the points A, Y, Z and S are concyclic. Therefore, < SAY = < SZY. In other words, < SAC = < SZY. But

  < SZY = < (ZS; YZ) = < (ZS; XY) - < (YZ; XY) = 90° - < (YZ; XY) = 90° - < ZYX.

Since the triangle XYZ is directly similar to the triangle ABC, we have < ZYX = < CBA, and thus < SZY = 90° - < CBA. Since < SAC = < SZY, we thus get < SAC = 90° - < CBA.

On the other hand, consider the circumcenter O of the triangle ABC. This circumcenter O is the center of the circle through the points A, B, C. Thus, by the central angle theorem for directed angles modulo 180°, we have < OAC = 90° - < CBA.

Comparing this with < SAC = 90° - < CBA, we get < SAC = < OAC. Thus, the point S lies on the line AO. Similarly, the point S lies on the lines BO and CO. Thus, the point S, lying on the lines AO, BO and CO, must coincide with the point O. Consequently, our former result that the points A, Y, Z and S are concyclic can be restated as follows: The points A, Y, Z and O are concyclic. In other words, the circumcircle of triangle AYZ passes through the point O. Similarly, the circumcircles of triangles BZX and CXY also pass through the point O. This proves Lemma 1.

Now the solution of the problem:

We will use non-directed angles in the following part of the solution.

Let R be a point in the plane such that the triangle RQP is directly similar to the triangle ABC. Then, < PRQ = < CAB and RP : PQ : QR = AC : CB : BA. Comparing this with the condition BP : PQ : QC = AC : CB : BA, we get RP : PQ : QR = BP : PQ : QC. Thus, RP = BP and QR = QC. In other words, the triangles RPB and RQC are isosceles.

The isosceles triangle RPB yields < BRP = < PBR, and the isosceles triangle RQC yields < QRC = < RCQ. Furthermore, we know that < PRQ = < CAB. Thus,

  < BRC = < BRP + < PRQ + < QRC = < PBR + < CAB + < RCQ = < ABR + < CAB + < RCA.

On the other hand, the sum of angles in the quadrilateral ABRC yields < CAB + < ABR + < BRC + < RCA = 360°, so that

  360° = < CAB + < ABR + < BRC + < RCA = < BRC + (< ABR + < CAB + < RCA)
          = < BRC + < BRC = 2 < BRC,

and $\measuredangle BRC=\frac{360^{\circ}}{2}=180^{\circ}$. Consequently, the point R lies on the line BC. Since we also know that the points Q and P lie on the lines CA and AB, respectively, and since the triangle RQP is directly similar to the triangle ABC, we can apply Lemma 1 to the points X = R, Y = Q and Z = P, and see that the circumcircles of triangles AQP, BPR and CRQ pass through the point O. The only of these three circumcircles we are interested about is the circumcircle of triangle AQP. So we have obtained that the circumcircle of triangle AQP passes through the point O. In other words, the points A, P, Q and O are concyclic. $\blacksquare$

[b]EDIT:[/b] This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=49106 .

  darij

Interesting solution Darij, thanks!. pestich, I was also trying to complete a solution in that way but i don´t know how to handle that PQ-CB. There´s a nice solution by pascual, i hope he posts it.

Ok here it goes:

Let $P'$ be the reflection of $P$ trought the midpoint of $AB$. Let $Q'$ be the reflection of $Q$ trought the midpoint of $AC$.

We easily get $AP'Q'$ is similar to $ACB$, so $P'Q'=PQ$. This implies that $OPQ$ is congruent to $OP'Q'$. And from here and an easy angle chasing, it follows that the cuadrilateral is conciclic ( I dont have a paper here so i hope you can finish the angle chasing by yourselves)

BTW Jose, visit the spanish forum! :D


It's quite easy ... First we prove a lemma:

[b]Lemma 1:[/b]  For $\alpha\geq8$ we have $\sqrt{\alpha+1}\geq1+\sqrt[3]{\alpha}.$
[b]Proof:[/b]  It's obvious because $x^{1/2}$ grows faster then $x^{1/3}$ and $\sqrt{8+1}=1+\sqrt[3]{8}.$ We can also do it by squaring it.

Now let's assume $abc=1.$ Then the original inequality is rewritten as \[ \sqrt{3+\sum a^3+\frac{1}{a^3}}\geq 1+\sqrt[3]{2+\sum a^3+\frac{1}{a^3}} \] which is obviously true because of lemma 1 for $\alpha=2+\sum a^3+1/a^3\;\;(\alpha\geq2+2+2+2=8.)$

Your solution is very good,an this one is a Korea's problem in 2000(or 2001)

Dividing by abc, it becomes
$\sqrt{(\frac{a}{c}+\frac{b}{a}+\frac{c}{b})(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})} \ge 1+\sqrt[3]{(\frac {a^2}{bc}+1)(\frac {b^2}{ac}+1)(\frac {c^2}{ab}+1)}$

Let $x=\frac {a}{b} , y=\frac {b}{c} , z=\frac {c}{a}$ , we obtain the constraint xyz = 1.

It takes the form

$\sqrt{(x+y+z)(xy+yz+zx)}\ge 1+\sqrt[3]{(\frac{x}{z}+1)(\frac{y}{x}+1)(\frac{z}{y}+1)}$

$(\frac{x}{z}+1)(\frac{y}{x}+1)(\frac{z}{y}+1)=(z+x)(y+x)(z+y)$

and $(x+y+z)(xy+yz+zx) = (z+x)(y+x)(z+y)+xyz=(z+x)(y+x)(z+y)+1$

Letting $(z+x)(y+x)(z+y)=p^3$ the inequality now becomes $\sqrt {p^3+1}\ge1+p$

$\sqrt {p^3+1}\ge1+p \Longleftrightarrow p(p+1)(p-2)\ge0 \Longleftrightarrow p\ge2$

$p^3=(x+y)(y+z)(z+x)\ge8xyz=8$ (well-known inequality) so $p\ge2$ and the proof has finished


calculus
derivative
function
trigonometry
calculus computations

Prove that if F is a twice differentiable function with F(0) = 0 and F(1) = 1 and F'(0) = F'(1) = 0, then |F''(x)| >= 4 for x in (0,1).

Hint: Prove that either F"(x) >= 4 for some x in (0,1/2) or else F"(x) <= -4 for some x in (1/2,1)

Then, show that we actually have |F"(x)| > 4.

 I think you'll have to use the Mean Value Theorem for this, but I'm not sure how

Thanks for reading

[quote="pluristiq"]doesn't seem right.

consider $ f(x) \equal{} \minus{} \frac {1}{2}\cos{\pi x} \plus{} \frac {1}{2}$.

$ f'(x) \equal{} \frac {\pi}{2}\sin{\pi x}$,

$ f''(x) \equal{} \frac {\pi^2}{2}\cos{\pi x}$,

so $ f$ satisfies the conditions of the problem, but $ f''(0) \equal{} \frac {\pi^2}{2} > 4$, so in any neighborhood of 0, $ f''(x) > 4$, whereas $ f''(\frac {1}{2}) \equal{} 0 < 4$, so there are values of $ f''$ both bigger than and less than 4 in absolute value.[/quote]

 Thank you, I appreciate your reply.

 Oops, I think that was my fault, it should be:

 Prove that if F is a twice differentiable function with F(0) = 0 and F(1) = 1 and F'(0) = F'(1) = 0, then |F''(x)| >= 4 for SOME x in (0,1).

 Sorry for the ambiguity

This has appeared before, more than once.

A hint:
Consider $ f(x)\equal{}\begin{cases}2x^2&0\le x\le\frac12\\ 1\minus{}2(1\minus{}x)^2&\frac12<x\le 1\end{cases}$.

[quote="jmerry"]This has appeared before, more than once.

A hint:
Consider $ f(x) \equal{} \begin{cases}2x^2 & 0\le x\le\frac12 \\
1 \minus{} 2(1 \minus{} x)^2 & \frac12 < x\le 1\end{cases}$.[/quote]

 Thanks for the hint, I'll try  it again.. but could you direct me to a similar thread? I do not know what I would be searching for..

Thanks again

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123255]Here[/url] is one such topic, and there's another topic in a link inside the hint in my first post there.

thanks, that answer sounds like an answer I'd be looking for.. but his constructed function g(x) = 2x^2 - f(x)  seems strange to me.. how would I have come up with that? what's the thinking process behind that - and how do I know that it'll behave just like f(x)?

 I'd appreciate an answer very much, this is something I truly wish to know :S

Thanks


I found this problem in my 92-93 Mathcounts book which is in the National Competition.  How do you solve it??? :?  :?

Given that a,b,c and d are positive integers, and: a+1/(b+1/(c+1/d))=181/42,  find a+b+c+d

Notice that [hide]1/(b+1/(c+1/d)) < 1. Therefore, a = floor(181/42) = 4. You can then get b, c, d in the same manner.

a+b+c+d = 4+3+4+3 = 14.[/hide]

Yeah, you're just trying to find the 4th convergent for the continued fraction for 181/42.

Now I understand!! :o   With 1/(b+1/(c+1/d))<1, that makes a={181/42} which is the value of the largest integer less than 181/42 which 168/42 or 4=a.  Then I subtract that from the original number and take the reciprocal to get b+1/(c+1/d)=42/13.  {42/13} equals 39/13 or 3=b.  Then do the same pattern over middle steps over again to get c+1/d=13/3.  {13/3}=12/3 or 4 and with the steps, makes d=3.  So 4+3+4+3 is 14.

Ergo, a+b+c+d is 14.  Thank you, wyra!


geometry
analytic geometry
algebra
polynomial
Euler
3D geometry
function

Nobody doubts about your extraordinary ability to solve the most difficult problems one can imagine, but saying that a problem famous for its difficulty (by the way, AMM it's not its first origin, it was developped in France and quite many mathematicians struggled with it for some time)  it's not difficult is a little bit too much, IMHO.

Hi harazi, good to see you on this forum :) I didn't know about this problem's history. I didn't say it was easy. And I didn't even say that I could solve it -- I never tried to (and of course, you know that I don't have any "extraordinary" abilities). By saying  that it's not as difficult as it might seem, I meant that it's not impossible, and it does admit a very short and beautiful solution (although finding that solution would be an extreme challenge; still, Sperner's lemma is standard in such problems, so the solution is not such a "big surprise").

A very engaging discussion of the problem's history and
a beautiful solution (involving not only Sperner's lemma,
but some p-adic analysis also) can be found in

Algebra and Tiling
Homomorphisms in the Service of Geometry
Sherman Stein and Sándor Szabó

It's an MAA book.

Tom

[quote="vess"]This is a jewel (neither as easy nor as difficult as it might seem):

Prove that a square cannot be dissected into an odd number of triangles of equal area.

(Generalize as much as possible)

--Vesselin[/quote]

This problem was posted first here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=464692&highlight=#464692

please continue the discussion in one topic only. Thanks

Ok, I didn't use either Sperners Lemma nor 2-adic analysis, but some facts about resultants and I need to work in the algebraic closure of $\mathbb{F}_2$. You may decide what's more elementary ;).

In the whole proof, we assume the square to be unit square with vertices $(0,0), (0,1), (1,1), (1,0)$.

So, first the plan of the proof: We consider all points which are vertices of some triangle of the subdivision. Then we observe that the necessary conditions for the areas give exactly one equation more than coordinates of these points. This means that a necessary condition for the existence of these points is the vanishing of the resultant of these polynomials(The resultant of $k+1$ polynomials in $k$ variables vanishes iff these polynomials have a common root in $\mathbb{C}$). But now, we see that in the case that $n$ is odd, there is no solution in $\overline{\mathbb{F}_2}$, meaning that the resultant of these polynomials does not vanish $\mod 2$ and therefore does not vanish at all. This last argument works since the resultant of integer polynomials is again an integer and the reduction $\mod 2$ of it is the resultant for the reduced polynomials $\mod 2$ in $\overline{\mathbb{F}_2}$.

Let the points which are vertices of some triangles in the subdivision and are not vertices of the square be $p_1=(x_1,y_1),...,p_k=(x_k,y_k)$. We can now form a planar graph by looking at our subdivision as being a graph with the points of the triangles as vertices and the edges of the traingles as vertices. The only problem occurs for collinear points. In that case we move some point on that line slightly away from the line such that the graph stays planar; if the point is connected to points on both sides of that line, we "split" the point into two points and move each one of them to one side of the line; this gives at most more variables.

Now, every area in that planar graph gives an equation; in fact, either it comes from an "old" triangle which has to have area $\frac 1n$, or it comes from a perburted line which has to have area 0. Denoting by $V$, $E$, $A$ the number of vertices, edges and areas of that planar graph, respectively, we know on the one hand that $V-E+A=1$ since Euler told us, and on the other hand $E=\frac 32 A+2$ since every area has exactly three edges around it and every edge has exactly two areas next to it, except those four border edges of the square. This gives $V-3=\frac 12 A$. It follows that $k\leq V-4=\frac 12 A - 1$ or $2k\leq A-2$. Now, one of the equations is redundant as the sum of triangles must clearly be 1, therefore there are actually $A-1\geq 2k+1$ equations, one more than the number of variables.

Now, assume the triangle of the triangle with vertices $(x_i,y_i),(x_j,y_j),(x_k,y_k)$ has area $A$. This means that $x_iy_j-x_jy_i+x_jy_k-x_ky_j+x_ky_i-x_iy_k=\pm 2A$ by well-known formulas for the area. But reducing $\mod 2$, this means that $x_iy_j-x_jy_i+x_jy_k-x_ky_j+x_ky_i-x_iy_k=0$ for all those triangles which need to have area $0$ or area $\frac 2n$ in the case $n$ odd.
Now assume there was a solution in $\overline{\mathbb{F}_2}$. The above condition, however, means, that $(x_i,y_i),(x_j,y_j),(x_k,y_k)$ are collinear. Therefore we may start with 2 vertices of the square and then some point $(x_i,y_i)$ which forms a triangle of the subdivision with those 2 vertices(or is collinear with them) and we get that it is collinear to them. Continuing that way, we see that all the points need to be collinear and therefore also the vertices of the square. But this is clearly a contradiction.

I just wanted to add that my proof gives a much more general result:

If we have some polyhedron $P$ in $\mathbb{R}^n$ whose vertices are integer coordinates whose reductions $\mod p$ do not lie in any plane for a prime $p$, then there is no subdivision of $P$ into $n+1$-simplices such that for every volume $V$ of such a simplex, we have $p|n!V$.

I have to add that my prof misses some argument: If the resultant is 0, only a projective solution is guaranteed, but my proof only handles the case of an affine solution; I couldn't really handle that problem until now...
For affine solutions the following argument can replace the collinearity argument:
Colour a point in colour 1 if has both coordinates 0, in colour 2 if only its second coordinate is 0 and in colour 3 if its second coordinate is not 0. By Sperners Lemma(finally!) there is a triangle with vertices of colour 1, 2 and 3. But it is immediate to see that this triangle has not area 0 in $\overline{\mathbb{F}_2}$.

Peter

Some of the asked generalizations:

[b]a)[/b] For $ n>4$, a regular $ n$-gon can be partitioned into $ k$ triangles of equal area if and only if $ n|k$.

[b]b)[/b] An $ n$-dimensional cube can be partitioned into $ k$ simplices of equal hypervolume if and only if $ n! | k$.

Also note that Peter's solution uses Sperner's lemma, which cannot be applied that easiliy as it needs a triangulation, not just a subdivision into triangles (in fact, one needs a slight generalisation of the lemma, then everything works out).

As promised, the $p$-adic proof and the generalisation of Sperners lemma:

Let $P$ be a polygon (nondegenerated, no self intersections).
A [i]triangular subdivision[/i] $T$ of $P$ is a subdivision of $X$ into finitely many triangles, that is: we have finitely many triangles $t \in T$ such that $X = \bigcup _{t \in T}t$ and distinct $t$ share at most some boundary points. A [i]subpolygon[/i] of $T$ is any of the triangles or $P$ as a polygon itself. A [i]vertex of $T$[/i] is any point that is a vertex of any of the triangles (note that this automatically includes the vertices of $P$). A [i]segment[/i] of $T$ is any segment (with it's endpoints) that is completely in $T$; it is called a [i]border segment[/i] if it's exactly a side of a subpolygon and it's called a [i]short segment[/i] if it contains no other vertices of $T$ than it's endpoints. A [i]subdivision of a segment[/i] $s$ of $T$ is defined to be a subdivision of $s$ into into segments of $T$ that share at most their endpoints.

A coloring of  the vertices of $T$ with three colors ($A,B,C$) is called [i]flat[/i] if every segment of $T$ contains at most two colors (counting all endpoints and interior points); in fact, for what we will do, a more general definition will suffice: it is flat if all border segments contain at most two colors. We call a segment [i]odd[/i] if it's endpoints are colored $A,B$ (both occuring, in any order). We call $P$ [i]odd[/i] if the number of odd border segments is odd. We call a triangle of $T$ [i]colorfull[/i] if it's vertices have all three colors.

[u][b]Generalized Sperner's Lemma in two dimensions (GSL2):[/b][/u]
Let a polygon $P$ and a triangular subdivision $T$ of $P$ and a flat coloring of $T$ be given.
Then $P$ is odd if and only if the number of colorfull triangles is odd. Especially, there is at least one colorfull triangle then.

[u]Proof:[/u]
After all that introduction, the proof is rather easy. We need one little lemma (in fact, it would be GSL1, see further below), which to prove is left to the reader (hint: consider two cases, one where the color $C$ appears and one where it doesn't, remember that we don't have all three colors on a segment, and count changes of color):
Let $s$ be a segment of $T$, subdivided into segments $s_i$. Then $s$ is odd if and only if an odd number of the $s_i$ is odd.

For any subpolygon $t$ of $T$ we define $n_t$ to be the number of odd short segments on the border of $t$ and $N_t$ as the number of odd segments that are sides of $t$. Using the lemma, we find that $n_t \equiv N_t \mod 2$ for every subpolygon $t$.
As every short segment borders exactly two subpolygons (note that this is true if one of them is $P$), we get that $\sum_{t \in T} n_t + n_P \equiv 0 \mod 2$ or equivalently $\sum_{t\in T} n_t \equiv n_P \mod 2$.

Now a triangle $t$ is colorfull if and only if $N_t$ is odd (if all colors are present or a color different from $C$ is missing, this is trivially true; if $C$ is the only missing color, then w.l.o.g. the triangle is colored $ABA$ and the property is again true). As a result, if we let $c$ be the number of colorfull triangles, we get $c \equiv \sum_{t \in T} N_t \equiv \sum_{t \in T} n_t \equiv n_P \equiv N_P \mod 2$, which is what we wanted to prove.

[b]Lemma:[/b]
For a given prime $p$, there is a function (called an [i]additive valuation[/i] at/over $p$) $v_p : \mathbb C \to \mathbb C \cup \{\infty\}$ such that:
- $v_p(x)=\infty \Leftrightarrow x=0$
- $v_p(xy) = v_p(x)+v_p(y)$ (multiplicativity)
- $v_p(x+y) \geq \min\{v_p(x),v_p(y)\}$ (nonarchimedean property)
- $v_p(p)=1$.
[b]Corollary:[/b]
- $v_p(1)=0$
- $v_p(-x)=v_p(x)$
- if $v_p(x)<v_p(y)$ then $v_p(x+y) = v_p(x)$ (weakest wins)
- if $p \nmid m,n \in \mathbb Z$ then $v_p(p^s \frac mn)=s$, $s \in \mathbb Z$.

[u]Proof of lemma:[/u]
On $\mathbb Q$, we can use the $p$-adic additive valuation, that is: write a given nonzero rational as $p^s \frac mn$ with $m,n,s \in \mathbb Z$ and $m,n$ not divisible by $p$ and set $v_p\left(p^s \frac mn\right) = s$ (and additionally $v_p(0):= \infty$); this construction automatically fulfills the last statement of the corollary, but we will not use this (you can prove them directly from the axioms on $v_p$). Now we consider the $p$-adic numbers $\mathbb Q_p$, to which this valuation extends uniquely, and as I pointed out in another thread ( http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1951482#p1951482 ), their algebraic closure (to which that valuation extends, too) is isomorphic to $\mathbb C$. Thus we have an isomorphism of fields $i: \mathbb C \to \bar{\mathbb Q}_p$ and we can define $v_p(x) := v_p(i(x))$ for $x \in \mathbb C$.
Alternatively, we will only need this lemma for finitely generated extensions of $\mathbb Q$ (contained in $\mathbb R$), so another and less AC-using way would be to show:
If $K$ is a field having some additive valuation $v$ and if $L=K(a)$ is a field generated by adding either a single algebraic or a single transcendental element, then we can extend $v$ to a valuation on $L$.
[u]Proof of corollary:[/u]
$v_p(1) = v_p(1^2)=2v_p(1)$, thus $v_p(1)=0$. Similar, $0=v_p(1) = v_p((-1)^2) = 2 v_p(-1)$, giving $v_p(-x)=v_p(x)+v_p(-1)=v_p(x)$.
If $v_p(x)<v_p(y)$, we have $v_p(x+y) \geq v_p(x)$ by the nonarchimedean property. We also get $v_p(x) = v_p((x+y)+(-y)) \geq \min\{v_p(x+y),v_p(y)\}$, but it can't be $\geq v_p(y)$, thus we get $v_p(x) \geq v_p(x+y)$ and thus the result. The last line is left to the reader (or follows from the construction).

[b]Division lemma:[/b]
If $m,n$ are integers such that $v_p(m) \leq v_p(n)$ for all primes $p$ then $m|n$. (we are choosing some $v_p$ for each $p$, but as we have seen in the corollary, they all agree on the integers.)
[u]Proof:[/u]
Easy and left to the reader.

If $x=(x_1,x_2)$ is any point in the plane, we define it's $p$-adic color $c(x)$ to be:
- $A$ if and only if $v_p(x_1) < v_p(x_2)$ and $v_p(x_1) \leq 0$
- $B$ if and only if $v_p(x_1) \geq v_p(x_2)$ and $v_p(x_2) \leq 0$
- $C$ if and only if $v_p(x_1),v_p(x_2)>0$
Note that this is defined for all points (and I recommend drawing a picture).

[b]Triangular coloring lemma (TCL):[/b]
Let $x,y,z$ be a (maybe degenerate) triangle in the plane with area $a$. If the triangle is colorfull, then $v_p(a) \leq - v_p(2)$.

[u]Proof:[/u]
Assume it is colorfull, w.l.o.g. assume that $c(x)=A,c(y)=B,c(z)=C$.
We first show that $c(x-z)=A$, $c(y-z)=B$ and (trivially) $c(z-z)=C$, so that we may assume $z=0$ afterwards. We have $v_p(x_1) < v_p(x_2)$ and $v_p(x_1) \leq 0 < v_p(z_1)$, which by the the weakest wins leads to $v_p(x_1-z_1)=v_p(x_1) \leq 0$; if $v_p(z_2)>v_p(x_2)$ then we get $v_p(x_2-z_2)=v_p(x_2)$ and are done, otherwise we get $v_p(x_2-z_2) \geq v_p(z_2) > 0 \geq v_p(x_1-z_1)$, so in this case we get color $A$, too. Showing $c(y-z)=B$ is similar and left to the reader.
Now we have $z=0$, and it's well known that the area of a triangle with vertices $0,x,y$ is given (up to sign, which we don't care about) by $\frac 12 (x_1y_2-x_2y_1)$. As $c(x)=A$ and $c(y)=B$, we get $v_p(x_1) < v_p(x_2)$ and $v_p(y_1) \geq v_p(y_2)$, in combination yielding $v_p(x_1y_2) = v_p(x_1)+v_p(y_2) < v_p(x_2)+v_p(y_1) = v_p(x_2y_1)$. But the weakest wins says $v_p(x_1y_2-x_2y_1) = v_p(x_1y_2) = v_p(x_1)+v_p(y_2) \leq 0$, the later inequality by $c(x)=A,c(y)=B$. In total, we get $v_p(2)+v_p(a) =  v_p(2a) \leq 0$.

[b]Corollary:[/b]
If we color the full plane this way, we get at most two colors on each line.
[u]Proof:[/u]
Three points on a line form a triangle of area $0$. Noting that $v_p(0)=\infty > -v_p(2)$ we get the result.

Let's finish the original problem with our new armada of stuff:
Now we are given the square $S$, w.l.o.g. with vertices $(0,0),(1,0),(1,1),(0,1)$ and some triangular subdivision $T$ into $m$ triangles. We color all it's vertices as told above, especially $c(0,0)=C$, $c(1,0)=A$, $c(1,1)=A$, $c(0,1)=B$. Then $S$ is obviously odd, and the GSL2 (note that the condition on segments is fulfilled by the aforementioned corollary) gives us a colorfull triangle $t$. Call it's area $a$, by the TCL, we get that we have $v_p(a) \leq -v_p(2)$. If all triangles are of equal area (this hasn't been used until now), we get $a=\frac 1m$ and thus $-v_p(m) = v_p(a) \leq -v_p(2)$, in other words $v_p(m) \geq v_p(2)$, so by the division lemma $2|m$.

Well, how to generalize this¿
Let's start defining the coloring in dimension $n$:
We are given $n+1$ colors which we call $0,1,2,...,n$.
Let $x=(x_1,...x_n$), we define the [i]color[/i] $c(x)=0$ if and only if all $v_p(x_i)>0$, and otherwise, we choose the color to be the index $i$ for which $v_p(x_i)$ is minimal (and we take the largest such $i$ if there are more of them).
A simplex (generalization of tetrahedron for higher dimensions) is [i]colorfull[/i] if it's $n+1$ vertices have all $n+1$ possible colors.

Then we get the same properties, just in higher dimensions:

[b]Simplicial coloring lemma (SCL):[/b]
Let $x_0,...,x_n$ be a (maybe degenerate) $n$-dimensional simplex in $\mathbb R^n$ with volume $a$. If the simplex is colorfull, then $v_p(a) \leq - v_p(n!)$.
[b]Corollary:[/b]
Any hyperplane contains at most $n$ colors. In fact any $m$-dimensional affine subspace contains at most $m+1$ colors.

[u]Sketch of proof:[/u]
W.l.o.g., let the color of $x_i$ be $i$. As in the two dimensional case, we may assume $x_0=0$. Then the volume of the simplex is $a=\frac{1}{n!} \det(x_1,...,x_n)$. Let $x_{i,j}$ be the $j$-th component of $x_i$, then Leibniz' formula says $ \det(x_1,...,x_n) = \sum_{\sigma \in S_n} \pm x_{1,\sigma(1)}\cdot ... \cdot x_{n,\sigma(n)}$, and we readily check that $v_p(x_{1,\sigma(1)}\cdot ... \cdot x_{n,\sigma(n)})$ is minimal only for $\sigma$ being the identity (use that $v_p(x_{i,i}) \leq v_p(x_{i,j})$ and equality cannot hold if $j>i$, both by $x_i$ having color $i$). Then weakest wins gives $v_p(\det(x_1,...,x_n)) = v_p(x_{1,1}\cdot ... \cdot x_{n,n}) = v_p(x_{1,1})+...+v_p(x_{n,n}) \leq 0$, the last inequality by definition of the coloring. Thus you get $v_p(n!a) \leq 0$.

Now the harder part: Sperners lemma. For this we need to define even more stuff than we did before.
A [i]simplicial subdivision[/i] $T$ of a polytope $P$ is a subdivision into finitely many simplices, having at most their boundaries shared. A [i]polytopical subdivision[/i] is defined the same, but the smaller objects may be more complex polytopes. A [i]triangulation[/i] is a simplicial subdivision such that two simplices are either disjoint or share a complete face; every polytope has a triangulation (unlike some other definitions, this one allows additional interior points!). A [i]subpolytope[/i] of $T$ is any simplex/polytope of $T$ or $P$ itself. A coloring of $T$ with $0,...,n$ is called [i]flat[/i] if every $k$-face of every subpolytope contains at most $k+1$ colors (on it's boundary and interior; this is true for triangulations if and only if it is true for the faces of $P$). The standard $p$-adic coloring is always flat by the corollary.

Given a $m$-dimensional ($m\leq n$) simplex $s$ with colored vertices, we define $C(s)=0$ if not every color $0,...,m$ occurs exactly once (this may either happen by having a color twice or by having a color $>m$); otherwise we set $C(s)=1$. Given a $m$-dimensional polytope $P$ with colored vertices, we take any triangulation $T$. Then we define $C(P) := \sum_{s\in T} C(s) \mod 2$; this is well defined $\mod 2$ as we will see later and will not depend on $T$ (for now, just fix a triangulation for any possible polytope). We can even generalize this to complexes of polytopes (that is: polytopes that share only their boundary and whose coloring is flat; in our cases: polytopical subdivisions and boundaries of polytopes), which defines us two other numbers: $c(P) := \sum C(Q)$ where the sum goes over all $(m-1)$-dimensional faces $Q$ of $P$, and $C(T)=\sum_{s\in T} C(s)$ which is defined for any polytopical complex $T$.

Some remark: The definition of $c$ is asymmetrical, as it uses the $(n-1)$-dimensional version of $C$ that only considers simplices with colors $0,1,...,n-1$, ignoring the color $n$. We could easily have defined it ignoring another color, which as we will now see would have given nothing different.

[b]Main proposition:[/b]
a) Given a flat colored $n$-dimensional polytope $P$, the definition of $C(P)$ (and thus of $c(P)$) does not depend on the triangulation.
b) We have $c(P) \equiv C(P) \mod 2$ (especially, the definition of $c$ does not really depend on the distinguished color).

[u]Proof:[/u]
Start by checking b) for a $n$-dimensional simplex with no further subdivision. This again contains three cases: if a color other than $n$ is missing, then $c(P)=0=C(P)$; if no color is missing, $c(P)=1=C(P)$; if $n$ is the only missing color, then $c(P)=2,C(P)=0$. In all cases $c(P) \equiv C(P)\mod 2$.

For a general polytope take a triangulation $T$ of $P$. Then checking $c(P) \equiv C(P) \mod 2$ (both defined by this subdivisions!) is done exactly the same way than in two dimensions, using the simplicial case: every interior triangular face is counted exactly twice and every outside face is counted once in $c$ and once in $C$.
As $c(P)$ only depends on $(n-1)$-dimensional simplices, we may by induction on $n$ assume GSL to be proved for them; actually, this works only for faces $Q$ having the colors $0,...,n-1$, but if the color $n$ occurs on $Q$, $c$ and $C$ are $0$ because another color is missing as our coloring was flat and we are happy again. In both cases $c(P)$ is independent from any subdivision, and thus $C(P)$ is invariant if we change our triangulation, too. This shows a) and b).

Now the general Sperner lemma can be written down in it's full glory:
[u][b]Generalized Sperner lemma :[/b][/u]
For any polytopical subdivision $Q$ of $P$ we have $c(P )\equiv C(P) \equiv C(Q) \mod 2$ (and everything is well-defined).

We have two ways to procede in the proof (details left to the reader):
- just subdivide $Q$ even further until it is a triangulation of $P$ and every single polytope in $Q$ (yes, this is possible); then having proved invariance of all $c,C$ from the chosen triangulations, we immediatly get the theorem.
- procede again by induction on the dimension, noting that a polytopical subdivision induces one on each of the faces of each $Q_i$.

[u]Remark:[/u]
As we have seen, all that $c,C$ are the same ($\mod 2$). This gives an efficient way to compute this number (other than the obvious "subdivide into simplices as you wish and count"-algorithm). Given a polytope $P$ and some coloring, we want to compute $C(P)$. Instead, we may also compute $c(P)$, which reduces our problem to computing $C(Q)$ for faces $Q$ of $P$. But we already know that $C(Q)=0$ if any of the vertices of $Q$ contains the "last" color we distinguished from counting in $c$ (in our case, taking the most frequent color may be the best choice), so we don't need to count this faces. Now counting the remaining $C(Q)$, we may again distinguish a color and instead count $c(Q)$, and so on.

From that point, proving the hypercube case of the original question is easy: Assume that the unit hypercube $H_n=[0,1]^n$ is subdivided into $m$ simplices. Identify the lower dimensional hypercube $H_k$ by $[0,1]^k \times {0}^{n-k}$ as a face of $H_n$. We distinguish the color $n$, implying that $C(Q)=0$ for all faces $Q$ of $H$ that have a vertex of color $n$. The vertices of color $n$ are exactly those such that the last coordinate is $1$, and every face except $H_{n-1}$ contains at least one point of it. So we are reduced to compute $C(H_{n-1})$ (having the same coloring as if we would have directly started in dimension $n-1$), which inductively reduces us the compute either $C(H_1)$ or $C(H_0)$, both turning out to be $1$. Thus $C(H_n) \equiv 1 \mod 2$, thus our simplicial subdivision contains an odd number of colorfull simplices (especially at least one, call it $s$), and by SCL we get $v_p(m) \geq v_p(n!)$ and finally $n!|m$.

Now to tackle some of the more difficult problems, it gets necessary to have an oriented version of Sperner's lemma. For this, one distinguishes between the two possible orientations a colorfull simplex may have and counts it either $+1$ or $-1$ accordingly (more formal definition: if we are given a simplex $s$ with vertices $a_0,...,a_n$ such that $a_i$ has color $i$, we translate it such that $a_0=0$ and define $C(s) = \mathrm{sgn} \det(a_1,...,a_n)$) (it is still defined to be $0$ if not all colors occure exactly once). Similar, the boundary simplices $s^\prime$ with vertices $a_1,...,a_n$ are counted positively or negatively according to how they "look from the outside" (one possible definition: add a point $a_0$ such that the full-dimensional simplex $s$ with vertices $a_0,...,a_n$ is completely inside of $P$, then define $C(s^\prime) = C(s)$). Again, we extend this definition to arbitrary polytopes and complexes of them.
Then we get the same results by the same proofs (which one can use to solve problem a) in my previous post):

[u][b]Oriented generalised Sperner lemma (OGSL):[/b][/u]
For any polytopical subdivision $Q$ of $P$ we have $c(P)=C(P)=C(Q)$ (an equality of integers), especially these numbers do not depend on the chosen triangulation or distinguished color.


Rationalize the following expression:

$ \frac {1}{\sqrt [3]{x} \plus{} \sqrt [3]{y} \plus{} \sqrt [3]z}$

Okay, I get a really nasty answer.
We first find that
$ \frac {1}{a \plus{} b \plus{} c} \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}$, but then we're left with $ x \plus{} y \plus{} z \minus{} 3 \sqrt [3] {xyz}$ in the denominator, so we have to use $ (m \minus{} n)(m^2 \plus{} mn \plus{} n^2) \equal{} m^3 \minus{} n^3$ to get rid of it. There's probably an easier way.

[quote="t0rajir0u"]There shouldn't be.  Generically, three extensions of degree $ 3$ give you an extension of degree $ 27$, so division (rationalization) is going to involve a lot of terms.[/quote]
Hmm, I don't really see the reasoning here... could you elaborate a little more?

$ \mathbb{Q}[x, y, z, a, b, c]/(a^3 \minus{} x, b^3 \minus{} y, c^3 \minus{} z) : \mathbb{Q}[x, y, z] \equal{} 27$.  Roughly, this means that it requires $ 27$ terms to do arithmetic:  every term of the form $ \sqrt[3]{x^i y^j z^k}$ where $ i, j, k \in \{ 0, 1, 2 \}$ is necessary in general.


how do i do this question?

Find the exact value of   $\cos \frac{5}{12}\pi$

cos(5/12pi) = cos(1/4pi + 1/6pi)

then use cos(a+b) rule.

Or use half-angle formulas for ${5\pi\over 6}$

i believe neelnal's method contains larger amounts of sexiness and asskickage:

cos(pi/6)cos(pi/4) - sin(pi/6)sin(p/4)
(rt3/2)(rt2/2) - (1/2)(rt2/2)
(rt6 - rt2)/4

too lazy to write everything out in latex so maybe: $\frac{\sqrt6-\sqrt2}{4}$


On average, how many students qualify to USAMO via the AMC 10 per year? And what USAMO index do you need for USAMO qualification if you did the AMC 10?

I got a 129 on the AMC 10A, so what score do I need for AIME?

The USAMO index is anywhere around 200-210 for 10th graders and below. Given that you got 129, you need at least an 8 on the AIME to qualify for USAMO. (I suggest you strive for more than 8, though :))

Frequently Asked Question USAMO #5 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

I forget exactly where I saw this, but a few months ago I was browsing the AMC wesite trying to figure out whether I should take the AMC 10 or 12 to have the best chance of qualifying for the USAMO.  I found out that since the AMC 10 is easier, my USAMO index (AMC score + 10*AIME score) would be (significantly) boosted if I took the AMC 10.  In addition, I remember seeing some statistic about 10th graders and below who qualified for the USAMO a few years ago, and the good majority of them took the AMC 10.

Hope this helps!

4. The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value. If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.

This is taken from the USAMO #5 FAQ.

Q : Let's say I take both the 10 and the 12. If I don't make the top 330 USAMO index of the amc12 people, but I do make the next 160 from the AMC10, will I be able to take the USAMO? Or do the people qualifying on the 10 have to only take amc10's?

Q: Also, is there a maximum number of USAMO students that can be selected from a given state or province? I noticed that in the past two years there were exactly 25 people from NJ who made USAMO, and I thought this might not be a coincidence.

dang they make it too hard on amc 10 takers
i should have gone for amc 12b instead of amc10 again

to make usamo, i need a 7ish? i got a 144 on 10a, and im taking 12b.

8 to be safe.  Last year 132/7 missed usamo by 0.5 points, but this year is supposedly really really easy.


we have the problem of hanoi but not with 3 sticks to put discs on them,we have n sticks.
we have $n\geq 3 ;m>0$ where n number of sticks and m number of discs.
how could i go about solving ?

I think finding the optimal algorithm is an open problem even for $4$ sticks, but I'm not sure. Try googling it. I remember searching for something like this and finding some articles a long time ago.


A fair coin is tossed 10 times. Compute the probability that heads does not occur on consecutive tosses.

The answer is (6C5+7C4+8C3+9C2+10C1+11C0)/1024 = 9/64. I guess you can probably find a neat formula for the top on account of how it works out so nicely, but the numbers are small enough so that the will of Allah alone suffices.


geometry
3D geometry
combinatorics proposed
combinatorics

The edges of a cube are labeled from 1 to 12 in an arbitrary manner. Show that it is not possible to get the sum of the edges at each vertex the same.
Show that we can get eight vertices with the same sum if one of the labels is changed to 13.

Hi

Perhaps ur 1st part can be readily solved if u keep this in mind.

[b]Every number on an edge shows up twice[/b]. So adding all the numbers u get at the vertices (which is obtained by adding numbers on the 3 edges meeting at the vertices) u get a sum s where

         s= sum of numbers on all vertices (8*num at one vertex)

           = 2*(1+2+3+.....+12) = 2*78

 [b]Now for all the vertices to have the same number the total sum of numbers on the vertices should be divisible by 8
[/b]

i.e 8 sud divide 2*78 which is not so.

Hope it is clear....

[quote="Johann Peter Dirichlet"]
Show that we can get eight vertices with the same sum if one of the labels is changed to 13.[/quote]
Clearly we must delete an odd. It works for 3 and 11. So we can get eight vertices with sum 11 and 10 respectively.


Can someone please explain the solution for this in detail. I don't understand why it's not $ \frac{1}{3}$.
Thank you very much.

6. Participation in the local soccer league this year is 10% higher than last year.
The number of males increased by 5% and the number of females increased
by 20%. What fraction of the soccer league is now female?
(A) $ \frac{1}{3}$  (B) $ \frac{4}{11}$   (C) $ \frac{2}{5}$ (D)  $ \frac{4}{9}$  (E)$ \frac{1}{2}$

[hide="write equations"]
$ \frac{105}{100}m\plus{}\frac{120}{100}f\equal{}\frac{110}{100}(m\plus{}f)$
Multiply on both sides by 100
$ 105m\plus{}120f\equal{}110m\plus{}110f$
$ 10f\equal{}5m$
$ 2f\equal{}m$ Probably where you got it wrong, it says how many females are [b]now[/b] in the league?
So we can just assume that there were (last year) 10 females and 20 males.
After the increase it is 12 and 21.   So the fraction of females is $ \frac{4}{11}$

[/hide]


[hide]
I think that the answer should be $-2^{x+6}$.[/hide]


geometry
3D geometry
sphere
topology
Functional Analysis
linear algebra

When you have to answer or explain something, you can say "basically that" or "essentially that". It may happen that the matter is not completely clear for you, so you use these words to mask your lack of knowledge, or laziness of thinking.
What is your opinion? Do you or others use these words to cover what you don't know well?

Really?  I usually use those words to avoid talking about technicalities that would distract from a beginner's understanding of a new concept.  For example, one might say that "a Hilbert space is essentially the most general setting in which the notion of angle makes sense."  This is not strictly true since this definition says nothing about completeness (so really we're talking about pre-Hilbert spaces), but it's a useful way to think about them nonetheless.

I use those to skip trivial steps, like why the number of ways you can pick 3 people out of 7 to form a comittee is $ \binom73$. I don't use them a lot, though.

I don't say that's the rule but it might happen, especially non-intentionally.
Thank you for your answers.

As far as words used to ignore issues you don't understand I find the word "clearly" to be much guiltier.

I very rarely use 'clearly,' it has the potential to frustrate someone to whom it isn't so clear.

I use 'essentially' when I'm summarizing an explanation and give people the important information. I find this especially useful when I'm explaining physics -- many people get too bogged down in the mathematics to step back and get a feel of the physics at work, so it is important to to give a physical explanation that sort of bypasses the math, aside from just deriving it mathematically.

If I'm explaining something that I'm not quite sure of, in most cases, I just honestly say that I'm not quite sure, but I think that such and such is true.

This is probably wrong of me, but when someone says essentially, I kinda find them a bit...pretentious(that might be too strong of a word).
But when Im trying to sound smart I say essentially  :P

I agree with t0rajir0u when he said that he uses those words for explaining things that may not be so clear to beginners.  It's a good way for beginners to get a better, somewhat clearer picture of something they may not have either grasped before or have not seen before. Ish.

I don't really act pretentious with any specific words that currently come to mind, but sometimes I do use, "You're completely inane if...", but only to those who annoy me and can't understand 2+3 is not equal to 7.
...sorry, math tutoring issues.

My math teacher uses it to illustrate the general overthought of stuff, like ta-daing the magic equation that we've worked so hard to derive and such.
Or when he starts ranting about the beauty of Fibonacci Numbahs. Stuff like that is also when I've heard it used.

I use "basically" all the time when tutoring or helping friends with homework.  It tends to make them feel more capable of grasping the concept that otherwise scares them.  Usually I say it when there's a relatively messy formula or equation that people just take for granted: time to say "Look, basically what this does/means is...".

Sorry, t0rajir0u, I don't mean to nitpick, but I really dislike your characterization of a Hilbert space as "essentially the most general setting in which the notion of angle makes sense."  (For example, is a sphere a Hilbert space?)

No, but the tangent space at every point can be given a Hilbert space structure.  This idea extends to Riemannian manifolds in general, but the point is that the dot product is a local construction and takes place explicitly in the tangent space.  I agree that my statement wasn't maximally general; if you're talking to a high school student who has at best a passing familiarity with linear algebra, would you want to start talking about Riemannian manifolds?

I might have said "inner product space" instead of "Hilbert space." That serves two purposes. Technically, you can talk about angles even if you don't have completeness. At a much lower level of explanation, an inner product is an almost-familiar object that you can pull out and look at.

Not to be argumentative, but I didn't say anything about Riemannian geometry.  I just mentioned the sphere, an object that is well-known to most high school students.

In any case, that's just one of many things I didn't like about that pithy description of Hilbert space.  Here's another: Inner products provide a notion of angle *and* length.  (For the interested reader: Geometry with angles but not lengths is called "conformal geometry," as opposed to Riemannian geometry which has both angles and lengths.)


function
real analysis
real analysis unsolved

Let P = {x0,x1,x2, ... ,xn} be a regular partition of the interval [a,b]. (Show that if f is continuous and decreasing on [a,b], then...

Uf(P) - Lf(P) = [f(a) - f(b)]▲x

We have that $ U(f,P)\minus{}L(f,P)\equal{}\sum_{i\equal{}1}^{n}M_{i}(x_{i}\minus{}x_{i\minus{}1})\minus{}\sum_{i\equal{}1}^{n}m_{i}(x_{i}\minus{}x_{i\minus{}1})$.  The fact that $ f$ is decreasing means that $ m_{i}\geq M_{i\minus{}1}$, and the fact that $ f$ is continuous means that in fact they are equal.  Also note that $ M_{1}\equal{}f(a)$ and $ m_{n}\equal{}f(b)$.

A very visual proof. The nickname I've heard for the proof idea is "stack the boxes." Do draw the picture for yourself.

The hypothesis that $ f$ is monotone is everything to this argument; that $ f$ is continuous is unimportant, although if we dropped the hypothesis of continuity, we'd have to modify the conclusion to
\[ U(f)(P)\minus{}L(f)(P)\le [f(a)\minus{}f(b)]\Delta x,\]
which is still sufficient to prove that $ f$ is (Riemann) integrable.

For an increasing function and a regular partition, that would be
\[ U(f)(P)\minus{}L(f)(P)\le [f(b)\minus{}f(a)]\Delta x.\]


Let $ p$ and $ q$ be two consecutive terms of the sequence of odd primes. The number of positive divisor of $ p \plus{} q$, at least

$\textbf{(A)}\ 2  \qquad\textbf{(B)}\ 3  \qquad\textbf{(C)}\ 4  \qquad\textbf{(D)}\ 5  \qquad\textbf{(E)}\ 6$

[hide]Since $ r\equal{}p\plus{}q$ is even, it must have at least $ 4$ divisors: $ 1,2,{r\over 2},r$. For $ p\equal{}3,q\equal{}5$ we get $ r\equal{}8$ which has exactly $ 4$ divisors. Hence the answer is $ C$.[/hide]


inequalities
trigonometry
geometry
circumcircle
perpendicular bisector
geometry proposed

$\triangle$ [b]ABC : $2a^2=b^2+c^2\ ,$ centroid[/b] $G\Longrightarrow \sin \widehat {ABG}+\sin \widehat {ACG}\le 2\cos A\ .$

[color=darkred][b]An extension.[/b] Given are the triangle $ABC\ ,$ the middlepoint $P$ of the side $[AB]$ and the point $M\in (AC)$ so that $MA=k\cdot AB$, where $k\in (0,1)\ .$ Prove that :
$2a^2=2(1-k)b^2+c^2\Longrightarrow\boxed {\sin \widehat {ABM}+\sin \widehat {ACP}\le 2\cos A}$
$($ with equality if and only if the triangle $ABC$ is isosceles in the vertex $C\ )\ .$[/color]
[u]Remark.[/u] For $k=\frac 12$, i.e. and the point $M$ is the middlepoint of the side $[AC]\ ,$ we obtain the proposed problem $(\ G\in CP\cap BM\ )\ .$

Let A', B', C' be the midpoints of a = BC, b = CA, c = AB. The A-median of the triangle $\triangle ABC$ is $AA' = \frac{\sqrt{2b^2 + 2c^2 - a^2}}{2} = \frac{a \sqrt 3}{2}.$ Thus for fixed B, C, the locus of A is a circle (A') centered at the midpoint A' of BC and with radius $AA' = \frac{a \sqrt 3}{2}.$ Let the perpendicular bisector of BC meet the locus circle (A') at $A_0.$ $A_0A'$ is the $A_0-$altitude of the isosceles triangle $\triangle A_0BC.$ From $BC = a,\ A_0A' = \frac{a \sqrt 3}{2},$ this triangle is equilateral, $\angle CA_0B = 60^\circ.$ The circumcircle $(O_0)$ of this equilateral triangle is centered on the perpendicular bisector of BC and $A_0O_0 < A_0A',$ hence the circles $(A'),\ (O_0)$ touch at $A_0$ and $(O_0)$ is inside of $(A').$ Any other point A on $(A')$ is outside of $(O_0)$, hence $0 < \angle A \le 60^\circ.$ By the cosine theorem for the triangle $\triangle ABC,$

$\cos \widehat A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{a^2}{2bc}$

The B-, C-medians of the triangle $\triangle ABC$ are $BB' = \frac{\sqrt{2c^2 + 2a^2 - b^2}}{2} = \frac{c \sqrt 3}{2},$ $CC' = \frac{\sqrt{2a^2 + 2b^2 - c^2}}{2} = \frac{b \sqrt 3}{2}.$ By the sine theorem for the triangles $\triangle ABA',\ \triangle CBA',$

$\frac{\sin \widehat{BAA'}}{\sin \widehat B} = \frac{BC}{2 AA'} = \frac{a}{a \sqrt 3} = \frac{1}{\sqrt 3}$

$\frac{\sin \widehat{CAA'}}{\sin \widehat C} = \frac{BC}{2 AA'} = \frac{a}{a \sqrt 3} = \frac{1}{\sqrt 3}$

We have $AG = \frac 2 3 AA' = \frac{a \sqrt 3}{3},$ $BG = \frac 2 3 BB' = \frac{c \sqrt 3}{3},$ $CG = \frac 2 3 CC' = \frac{b \sqrt 3}{3}$ and $\angle BAG \equiv \angle BAA',$ $\angle CAG \equiv \angle CAA'.$ By the sine theorem for the triangles $\triangle ABG,\ \triangle ACG,$

$\frac{\sin \widehat{ABG}}{\sin \widehat{BAG}} = \frac{AG}{BG} = \frac a c$

$\frac{\sin \widehat{ACG}}{\sin \widehat{CAG}} = \frac{AG}{CG} = \frac a b$

Consequently,

$\sin \widehat{ABG} + \sin \widehat{ACG} = \frac a c \cdot \frac{\sin \widehat B}{\sqrt 3} + \frac a b \cdot \frac{\sin \widehat C}{\sqrt 3} = \frac{a(b \sin \widehat B + c \sin \widehat C)}{bc \sqrt 3} =$

$= \frac{a(b^2 + c^2)}{2bcR \sqrt 3} = \frac{a^3}{bcR \sqrt 3}$

where R is circumradius of the triangle $\triangle ABC.$ We want to show that this is less than or equal to $2 \cos \widehat A = \frac{a^2}{bc}.$

$(?)\ \ \ \frac{a^3}{bcR \sqrt 3} \le \frac{a^2}{bc}\ \Longleftrightarrow\ a \le R \sqrt 3$

But since A is either identical with $A_0$ or lies outside of the circumcircle $(O_0)$ of the equilateral triangle $\triangle A_0BC$ with circumradius $R_0 \le R,\ a = R_0 \sqrt 3 \le R \sqrt 3,$ this is obviously true.


An equilateral triangle shares a common side with a square as shown. What is the number of degrees in $ m\angle CDB$?
[asy]pair A,E,C,D,B; D = dir(60); C = dir(0); E = (0,-1); B = C+E; draw(B--D--C--B--E--A--C--D--A); label("D",D,N); label("C",C,dir(0)); label("B",B,dir(0));[/asy]

With an equilateral triangle on a square, angle BCD is 90+60=150 degrees. $ \overline{DC}$ and $ \overline{CB}$ are equal, making triangle BCD isosceles, meaning angle CDB is $ \frac{180\minus{}150}{2} \equal{} \boxed{15}$

[quote=kevinmathz]wrong! 30[/quote]

Ebob's solution is correct.


Just like to point out that AHSME forbids calculators. :D

[quote="13375P34K43V312"]I am trying to use the AHSME to prepare myself for AMC 12. The way I am planning to do it is to skip the first 3 questions, and the last 2, and take it as a 25 question test under current AMC 12 rules. Is that the best way to do it?[/quote]

Who knows. Personally I would just do all 30 in the normal 90 minutes.

[quote="lotrgreengrapes7926"]Just like to point out that AHSME forbids calculators.[/quote]
This is only true before 1994. The tests from 1994 on allowed calculators.

I'd recommend following the rules at the time for taking it (90 minutes). Trying to match the current model exactly isn't worth it.

[quote="13375P34K43V312"]So just use the rules that were set forth then?

But the modern AMC 12s, and 10s for that matter, I'm using the new scoring system. Am I correct for doing that?[/quote]
Actually, however many points you give yourself for each problem has practically no effect on how much you get out of the practice.  If you want a predictor of how well you would do on the AMC if you took it tomorrow, my advice would be to take a recent test (like last year's or something) and score it using the scoring system that will take effect this year.  (But if you find it likely that you will qualify for the AIME, my advice is really to work on past AIMEs.)


function
parameterization
algebra
domain
inequalities
vector
advanced fields

Hi guys:

Can somebody tell me if there is any software which can tell me if a particular function (input by user) is CONVEX or NOT ?

Thank you..

[quote="erpg"] .. .  there is any software which can tell me if a  function (input by user) is CONVEX or NOT ? [/quote]
Nice !
I think that it's  better to write your own routine.  Perhaps may be used one of following proposed  methods:
Consider $I: =[a,b]$ and $f: I\to {\mathbb R}$. As [b] input parameters : [/b]
---the analytic expression of  function $f$ ,
---the domain $I$ .
---a small  number $\varepsilon\; ,\; \varepsilon \in (0,1)$  , e.g. $\varepsilon={10}^{-8},$ which simulates the   "  precision "   .
---An integer $N$  which represents the number of "selections"  $(x,y,z)$ in case (1.1)  or $(x,h)$ in cases
(1.2) or (2).
--- A positive integer $L, L\in \{0,1,2\}$ which represents a " level" ; for instantce if we use (1.2) and consider

$N=500 , L=0$ then  the significane of L is that (1.2) is satisfied on  100$\%$  selections.
 $N=500 , L=1$ means that   (1.2) is verified only at 80 $\%$ from the selections.
As [b] out parameter [/b] consider a positive number, called for instance  KOD , having (for example) the sigificance:
KOD=0 means that the function is "convex (in compure), and KOD=1 if  the function is not convex (see below)
[color=red] [b] [1] [/b] [/color]  For     $x,y,z$ distinct points from $I$ define
\[ (1.1) \; \; \; \; D[x,y,z; f] : = \frac{f(x)}{(x-y)(x-z)} + \frac{f(y)}{(y-x)(y-z)} + \frac{f(z)}{(z-y)(z-y)} \; ,  \]
or when the points   $x,y,z , y=x+h, z=x+2h \; h\ne 0 ,$ it's possible to work with
\[ (1.2) \; \; \; \; D_h(f;x ): = f(x) - 2f(x+h) + f(x+2h) \; \; \; ,\; \; \; \frac{a-x}{h} \le h \le \frac{b-x}{2} \; x\in I .  \]
[color=green] [b] Denote by $D$  the right term in (1.1)  or in (1.2) .
Step ($\pounds$) :
Test if    $(*) \; \; \; \; D> -\varepsilon$  for  some admissible variables  (In case (1.1) for some  $x,y,z$ , in case (1.2)  for $h,x$ admissible ,
  i.e. the points  $x ,x+h,x+2h$  must belong to $I .$
 If (*) holds, consider that $f$ is  "convex" (in computer) on its domain and  KOD=0 and L= ... .  [/b] [/color]
[color=red] [b][2] [/b] [/color]  Consider that instead of right-members from (1.1) or (1.2) we take
\[ (2)\; \; \; \; D: = 7f\left(x-\frac{h}{3}\right)+27f\left(x+\frac{h}{3}\right) - 2f\left(x+\frac{2h}{3}\right)) - 32f\left(x-\frac{h}{6}\right)  \]
where the variable $x,h$ are admissible.  [color=green] [b] Follow Step ($\pounds$) .  [/b] [/color]
[size=75] [b]Note :[/b] 1) When  $f: I\to {\mathbb R}$ is convex on $[a,b] ,$  then  $D[x,y,z;f]\ge 0$ and also $D_h(f,x) \ge 0$.
2)  If  $f^{\prime\prime}(x)$ exists, then    $f^{\prime\prime}(x) \approx h^2 D ,$  $D$ being as in $(2)$  . It's a " (very) good" approximation.
[/size]

can the above procedure be applied to the function taking only descrete values (both input and output) ? My function takes 3 inputs(say 'd','L' and 'm') and one depends on another (say 'm' depends on 'L') ???

[quote="erpg"]can the above procedure be applied to the function taking only descrete values (both input and output) ?[/quote]
Perhaps following help :
[b]1.[/b]Consider a set of [color=red][b] equidistant  data  [/b][/color] $\begin{array}{||c|c|c|c|c||} \hline\hline p & p+1 &p+2&\ldots &p+m\\ \hline y_0&y_{1}&y_2&\ldots &y_{m}\\ \hline\hline \end{array}\; ,$ where  $y_0,y_1,....,y_{m}$   are the values of a
certain (unknown)  function $F$ $: D\times U^{\nu}\to {\mathbb R}$ or $y(x)=F\left(x;\alpha_1,\alpha_2,\ldots,\alpha_{\nu}\right) \; ,\; x\in D,$ at the points   ([color=blue] [b] "mesh" [/b][/color])  $x_0,x_1,...,x_m$
 with  $x_j=p+j\in D ,$  that is  $y_j=F\left(x_j;\alpha_1,\alpha_2,\ldots,\alpha_{\nu}\right)$ where $\alpha_1,....,\alpha_{\nu}$ are possible other variables. Further consider these
 $\alpha_j\in U$  as  [b]parameters[/b] . Assume that the [color=blue] [b] aim [/b] [/color] is to investigate the convexity to respect to   $x$ , i.e. convexity of  $y=F\left(\cdot ;\alpha_1,\alpha_2,\ldots,\alpha_{\nu}\right) .$
Then we have as (unique)  method (according to the fact that [color=blue][b] the mesh is equidistant [/b][/color])  we must test if following inequalities  are true
\[ (*)\; \; \; \; \; \begin{array}{|c|} \hline y_{i+2}-2y_{i+1}+y_i \ge -\varepsilon \\ \hline \end{array} \; \; ,\; \; \; \mbox{for}\; \; \; i \in \{ 0,1,..., {\mathbf k},\ldots , m-2 \}  \]
where $\varepsilon \in (0,1)$ is a small number (e.g.$\varepsilon : ={10}^{-8}$) .
[color=green]
As [b]input [/b] consider:
---   the data $p, m , (m\ge 2),$
--- the $m+1$- dimensional vector $(y_0,y_1,....,y_{m} ),$
---an integer ${\mathbf k}$ satisfying $0\le {\mathbf k} \le m-2$  which help to simulate a "level" .
[/color]

 [color=blue]
[b]Outputs [/b] must contains an KOD$\equiv$ [b]error -cod [/b].
  For instance if in (*)  all inequalities  are satisfied, then KOD:=0.
 Suppose that  (*) are true only for $i\in \{0,1,...,k\} .$ Then KOD:=1  .
[/color]

[size=75]
The situation of [color=red] [b]non-equidistant mesh [/b][/color] is different , but possible to investigate.
[/size]

[quote="flip2004"]
[size=75] The situation of [color=red] [b]non-equidistant mesh [/b][/color] is different , but possible to investigate.
[/size][/quote]
[b]2.[/b]Consider a set of [color=red][b]non- equidistant  data  [/b][/color] $\begin{array}{||c|c|c|c|c||} \hline\hline x_0 & x_1 &x_2&\ldots &x_m\\ \hline y_0&y_{1}&y_2&\ldots &y_{m}\\ \hline\hline \end{array}\; .$  In order to investigate the convexity of $F(\cdot;\alpha_1,....,\alpha_{\nu} )$
 (according to the fact that [color=red][b] the mesh is non- equidistant [/b][/color])   we  test  following inequalities  :
\[ (**)\; \; \; \; \; \begin{array}{|c|} \hline \\ \displaystyle \displaystyle \frac{y_i}{(x_i-x_{i+1})(x_i-x_{i+2})}+ \frac{y_{i+1}}{(x_{i+1} - x_{i})(x_{i+1}-x_{i+2})}+\\ \\ + \displaystyle \frac{y_{i+2}}{(x_{i+2}-x_{i})(x_{i+2}-x_{i+1})} \ge -\varepsilon \\ \\ i \in \{ 0,1,..., {\mathbf k},\ldots , m-2 \} \\ \\ \hline \end{array}  \]
[color=green]
As [b]inputs [/b] consider:
---$m , (m\ge 2),$
--- the $m+1$- dimensional vectors $(x_0,x_1,...,x_{m})\; ,\; (y_0,y_1,....,y_{m} ),$
---the integer ${\mathbf k}$ , $0\le {\mathbf k} \le m-2$  which  simulates a "level" . [/color]   [color=blue]
[b]Outputs [/b]  contains  KOD$\equiv$ [b]error -cod [/b].
[/color]


I do not know if this is the approppriate place for made this warning, but it seems impossible to reach the web page of the journal KVANT and also that of the Internet Biblioteka in russian. Could maybe some body help all of us about this?
Best regards

Professor, I have tried to open both the [url=http://kvant.info/]site of Kvant[/url], and the [url=http://kvant.mirror1.mccme.ru/]Kvant archive[/url] and both these sites seem to be working.

As far as the Internet Biblioteka is concerned, I am not sure which of the online book repositories in Russian are you referring to.

It seems to be a local problem of my machine, because other colleagues say they can open the pages.
I means http://ilib.mccme.ru
Many thanks


Will ice melt faster in fresh water or salt water (completely saturated), and why?  My first thought was salt water, but my textbook says fresh water.

Thanks in advance!

The salt makes it take longer for phase changes.

[quote="feynman_wannabe"]The NaCl in the water cause something known as freezing point depression. So basically, in the salt water, the melting/freezing process will occur at a lower temperature. Because its at a lower temperature, I suspect that the kinetic evergy of each ice atom is lowered, and hence, the phase transition will take a longer time.[/quote]

That's a nice idea.  However, I was wondering if there's anyway to test that with an experiment.

Thanks!

[quote="JRav"]While this is true, the difference is not really that great unless you start working with high quantities.  You should be able to perform this experiment (to some extent) in your chemistry class.[/quote]

I can obviously test to see that ice melts faster in fresh water than in salt water.  However, it there a way to confirm that the salt really affects the freezing point depression which makes the phase change take longer?

Thanks!

I have performed this experiment at home. It needs a narrow beaker, a marker pen, a stopwatch, and four ice cubes...(All these things are rarely found, so do order timely!!! :D )
Pour the beaker upto $ \frac{1}{6}th$. Now, drop gently one ice-cube inside the beaker water, mark the level of the water and start your stopwatch(better if you can do all of them simultaneously!). Count for 2-3 minutes, and mark the water level, and pick up the remaining portion of the cube. [b]Don't let it melt completely.[/b]
Next, pour two teaspoons of salt in the water, and drop a second ice cube, mark the new level, start your stopwatch.........mark the level, pick up the remaining portion of the cube.
Next, pour two more teaspoons of salt in the water, and repeat the actions mentioned above. Finally, pour two more teaspoons of salt and end your experiment up with the forth ice-cube.
Time for analysis! Measure the distances between each two marks for the four icecubes. Let the distances be $ d_{1}, d_{2}, d_{3}$ and $ d_{4}$ respectively. If you accurately measure the times equal for the four ice-cubes, you will obtain $ d_{1}< d_{2}< d_{3}<d_{4}$. You will also see that if you continue the experiment likewise, the distance between the marks will gradually tend to a constant value. When the water is saturated with the salt, you will always obtain the same distances under identical conditions.
Hope the experiment will be of some use to you. Thank u.
 :)

[quote="Sunkern_sunflora"]I have performed this experiment at home. It needs a narrow beaker, a marker pen, a stopwatch, and four ice cubes...(All these things are rarely found, so do order timely!!! :D )
Pour the beaker upto $ \frac {1}{6}th$. Now, drop gently one ice-cube inside the beaker water, mark the level of the water and start your stopwatch(better if you can do all of them simultaneously!). Count for 2-3 minutes, and mark the water level, and pick up the remaining portion of the cube. [b]Don't let it melt completely.[/b]
Next, pour two teaspoons of salt in the water, and drop a second ice cube, mark the new level, start your stopwatch.........mark the level, pick up the remaining portion of the cube.
Next, pour two more teaspoons of salt in the water, and repeat the actions mentioned above. Finally, pour two more teaspoons of salt and end your experiment up with the forth ice-cube.
Time for analysis! Measure the distances between each two marks for the four icecubes. Let the distances be $ d_{1}, d_{2}, d_{3}$ and $ d_{4}$ respectively. If you accurately measure the times equal for the four ice-cubes, you will obtain $ d_{1} < d_{2} < d_{3} < d_{4}$. You will also see that if you continue the experiment likewise, the distance between the marks will gradually tend to a constant value. When the water is saturated with the salt, you will always obtain the same distances under identical conditions.
Hope the experiment will be of some use to you. Thank u.
 :)[/quote]

This tells me that salt indeed makes the ice melt slower, but how do I know that the salt affects [i]affects the freezing point depression, lowering the kinetic energy of each ice atom, which makes the phase change take longer? [/i].  Is there a way that I can test this?

Thanks!

I see.....I'll try to provide u a better experiment....like the one u want.
 :)

Classic examples of freezing point depression labs include the creation of ice cream. you need to put salt in ice in order to super cool the ice so that the ice cream can be formed at room temperature. This procedure is fairly common and can be found online.

Yes....you are absolutely right feynman_wannabe! Thank you!
 :)


[hide]Expand the left and simplify the right to get

$169a^2+13ab+13ac+bc=169a^2+169a+bc$.

Cancel out like terms to get

$13ab+13ac=169a$.

$13a(b+c)=169a$.

$b+c=13$. [/hide]

[quote="Fermat -Euler"]$a , b , c$ are  réels $(a\neq 0 , b\neq 0 ,c\neq 0)$ satisfaies $(13a+b)(13a+c)=169a(a+1)+bc$
[color=red]Find [/color]  $b+c$[/quote]

[hide]
Expanded:
$169a^2+13ac+13ab+bc=169a^2+169a+bc$
The $bc,169a^2$ can be subtracted.
$13ac+13ab=169a$
$13a(c+b)=169a$
$b+c=13$

[/hide]


If x,y,z,u,v are real numbers so that:
sqrt(x)+sqrt(y)+sqrt(u)+sqrt(v)+2sqrt(z-2)=x+y+u+v+z. Find x;y;u;v and z.

unknown source

[b]Solution[/b]

Easy computations gives u to:
(sqrt(x)-1/2)^2+(sqrt(y)-1/2)^2+(sqrt(u)-1/2)^2+(sqrt(v)-1/2)^2+(sqrt(z-2)-1)^2=0 so the sum of sqaures if 0 iff (sqrt(x)-1/2)^2=(sqrt(y)-1/2)^2=(sqrt(u)-1/2)^2=(sqrt(v)-1/2)^2=(sqrt(z-2)-1)^2=0
z=3 and the rest are 1/4

cheers

damn .. I am too slow.. seems that harazi was faster for the first three problems posted by Georg! :(

I was fast because the problems are trivial. If you were to see me spending up to three hours thinking about a certain problem, then you wouldn't say I'm fast.

well... yeah they are easy .. but you were faster.. I guess that everyone has "problems with some problems", which require them some time to finish a solution to a certain hard problem! :D :D


linear algebra
matrix
vector
LaTeX
linear algebra unsolved

Prove that if $ A$ is nonnegative definite $ nxn$ matrix and if $ B$ is any $ nxm$ matrix, then $ B^{T}AB$ is nonnegative definite. Find the analogus result for positive definite matrices.

Since you're using transposes, I'm going to assume you're working with real matrices. (In complex matrices, we could use the conjugate transpose or Hermitian adjoint.)

If $ C=B^TAB,$ we want to show that $ x^TCx\ge 0$ for all vectors $ x.$

$ x^TCx=x^T(B^TAB)x=(Bx)^TA(Bx)\ge 0$ since $ A$ is positive semidefinite.

Following the same argument, is having $ A$ be positive define enough to conclude that $ C$ is positive definite? No, not by itself; you would also need to know that $ Bx\ne 0.$ Hence the result is this:

If $ A$ is a positive definite $ n\times n$ real matrix and $ B$ is an $ n\times m$ real matrix of rank $ m$ (that is, the columns of $ B$ are independent), then $ B^TAB$ is positive definite.

$ \text{\LaTeX}$ tip: say "n \times n" rather than "n x n".


geometry
integration
calculus
rotation
calculus computations

I'm having trouble finding the volume by rotating an area bouded by curves, etc, and rotating about one of the axes or a line.  Here are some of the problems I'm really stuck on!

Find the volume formed by rotating the region enclosed by:
x = 8 y and y^3 = x with y <(or equal) 0
about the y-axis

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves
x=0, y=1, x=y^{12} , about the line y=1.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x = -4

y=x^2, \ x = y^2

Those are just a few that I'm having problems with.  For some reason I understand the stuff where you use cylindrical shells, but my prof isn't exaclty the greatest, and I don't think I have all the necessary equations either.  Thanks a bunch

1. I'll try to integrate in $y$, which amounts to slicing the volume by planes perpendicular to the $y$-axis. Cross-sections are washers with radii $8y$ and $y^{3}$ (up to minus sign), and with area $\pi((8y)^{2}-(y^{3})^{2})$. To find the limits for $y$, I must find where the curves $x=8y$ and $x=y^{3}$ intersect: $8y=y^{3}$ when $y=0$ or $y=\pm \sqrt{8}$. Since we are interested in $y\le 0$, the limits for $y$ are $-\sqrt{8}\le y\le 0$. So,
\[V=\int_{-\sqrt{8}}^{0}\pi((8y)^{2}-(y^{3})^{2})\,dy\]
2. Now I will integrate with respect to $x$, which amounts to slicing the volume by planes perpendicular to the $x$-axis. Cross-sections are disks with radius $1-x^{1/12}$ and with area $\pi((1-x^{1/12})^{2})$... Or maybe I don't want to integrate in $x$ after all. Integrating in $y$, I will slice the region of integration by lines perpendicular to the $y$-axis and rotate each slice around  $y=1$, obtaining a cylindrical shell. It's very helpful to draw a sketch of one shell and identify its dimensions: height $y^{12}$, radius $1-y$, thickness $dy$. So,
\[V=\int_{0}^{1}2\pi (1-y)y^{12}dy\]
Shells would work better for #3 too.


In triangle $ABC$, $D$ is the midpoint of side $AB$. $E$ and $F$ are points arbitrarily chosen on segments $AC$ and $BC$, respectively. Show that $[DEF] < [ADE] + [BDF]$.

A sketch: Let $x = CE/AC$ and $y = CF/BC$.  The inequality becomes

$1-\frac{1}{2}(1-x)-\frac{1}{2}(1-y) - xy < \frac{1}{2}(1-x)+\frac{1}{2}(1-y)$
$\iff 0 < xy-x-y+1 \iff 0 < (1-x)(1-y)$ which holds...


algebra
polynomial
LaTeX
algebra unsolved

hey can't u use \[ La Tex\], I can't see the problem clearly :mad:

Problem $ \text{\LaTeX}$ed for those who can't read it.
[quote="anorithdialga"]Let $ P(z) = z^n + a_{n - 1}z^{n - 1} + \cdots + a_0$ be a monic polynomial with complex coefficients.
Choose $ j \in \{0, \cdots , n \}$ so that the roots of $ P$ can be labeled $ \alpha_1 , \cdots , \alpha_n$ with
$ |\alpha_1|, \cdots , |\alpha_j| > 1$
$ |\alpha_{j + 1}|, \cdots , |\alpha_n| \le 1$

Prove that
$ \prod_{i = 1}^{j}|\alpha_i| \le \sqrt {|a_0|^2 + \cdots + |a_{n - 1}|^2 + 1}$
[/quote]


algebra
function
domain
integration
calculus
calculus computations

(dy/dx) = [(x^2)*e^((-x)^3)]/y

1. find the particular solution passing through(0,-1)
2. What is the domain of the solution?
3. Is the solution an increasing or decreasing?  Justify.
4. What is the range of the function?
5. FInd the equilibrium solution.

$ \frac{dy}{dx}\equal{}\frac{x^2 e^{(\minus{}x)^3}}{y}$

$ y(x) y'(x)\equal{}e^{\minus{}x^3} x^2$

$ \int_{x_0}^t y(x) y'(x) \, dx\equal{}\int_{x_0}^t e^{\minus{}x^3} x^2 \, dx$

$ \frac{1}{2} \left(y(t)^2\minus{}y\left(x_0\right){}^2\right)\equal{}\frac{1}{3} \left(\minus{}e^{\minus{}t^3}\plus{}e^{\minus{}x_0^3}\right)$

$ y(t)\equal{}\pm \sqrt{y\left(x_0\right){}^2\plus{}\frac{2}{3} \left(e^{\minus{}x_0^3}\minus{}e^{\minus{}t^3}\right)}$

And you can do the rest of your homeword  :wink:


Stanford
college
ratio
Stanford Math Tournament
AMC

Source: Stanford Math Tournament 2006

Poles $ A,B,$ and $ P_1, P_2, P_3,...$ are vertical line segments with bases on the x-axis. The tops of poles $ A$ and $ B$ are (0,1) and (200,5), respectively. A string $ S$ contains (0,1) and (200,0) and intersects another string connecting (0,0) and (200,5) at point $ T$. Pole $ P_1$ is constructed with $ T$ as its top point. For each integer $ i>1,$ pole $ P_i$ is constructed so that its top point is the intersection of $ S$ and the line segment connecting the base of $ P_{i\minus{}1}$ (on the x-axis) and the top of pole $ B$. Find the height of pole $ P_{100}$.

Hint: Can you find the height for $ P_n$?

[hide="hint"]If you have not already seen this theorem, prove that $ \frac{1}{a}\plus{}\frac{1}{b} \equal{} \frac{1}{P_1}$.[/hide]
[hide]Let the base of $ A$ be $ A_0$, and the top be $ A_1$.  Similarly for $ B$.  Finally, for $ P_1$, let the top be $ T_1$ and the bottom be $ T_0$.  Then we have that the triangles $ A_1 A_0 B_0$ and $ T_1 T_0 B_0$ are similar with ratio $ \frac{A_0 B_0}{T_0 B_0}$.  So, $ \frac{P_1}{A} \equal{} \frac{T_0 B_0}{A_0 B_0}$.

Similarly, $ \triangle B_1 B_0 A_0 \sim \triangle T_1 T_0 A_0$ with ratio $ \frac{A_0 B_0}{T_0 A_0}$.  So, $ \frac{P_1}{B}\equal{}\frac{A_0 T_0}{A_0 B_0}$.
Hence, $ \frac{P_1}{A}\plus{}\frac{P_1}{B}\equal{}\frac{T_0 B_0 \plus{} A_0 T_0}{A_0 B_0} \equal{} \frac{A_0 B_0}{A_0 B_0}\equal{}1$.
ao, $ \frac{1}{A}\plus{}\frac{1}{B}\equal{}\frac{1}{P_1}$.

Applying this theorem, we thus have $ \frac{1}{B}\plus{}\frac{1}{P_{n\minus{}1}}\equal{}\frac{1}{P_n}$.  As $ B\equal{}5$,we can write this into losed form as $ \frac{1}{P_n} \equal{} 1 \plus{} n \cdot \frac{1}{5}$.  Hence, $ \frac{1}{p_{100}}\equal{}1\plus{}20\equal{}21$, so the height of the 100th pole is $ \frac{1}{21}$. Note that the 200 is irrelevant![/hide]

Hmm, interesting. I used same idea except I found general form for the height of $ P_n$ involving $ a$ and $ b$.

I think your solution is neater though. Well done!  :)


The point $ C$ is on segment $ AB$. A straight line through $ C$ intersects the circle with diameter $ AB$ at $ E$ and $ F$, the circle with diameter $ AC$ again at $ M$, and the circle with diameter $ BC$ again at $ N$. Prove that $ MF\equal{}EN$.

P- the intersection point of circle with diameter AB and line AM
Q- the intersection point of circle AB with line BN
BQ||AP => MF=EN

Why does $ BQ\parallel{}AP$ imply $ MF\equal{}EN$?  :maybe:

AB is diameter so AP and QB are symmetric through the midperpendicular of MN.


I'm doing an independent drawing with a cannonball bouncing from 40ft. If it bounces back up 2/3 of what it falls, (following the sequence of Down 40ft, Up 26.66, down 26.66, up 17.77, etc.), Can anyone tell me how much this cannonball would weigh and when gravity would cause it to stop bouncing and come to rest? Thank you. (I'll post the pic when I get it done.)

theoretically, wont it be an infinite gp sort of? theoretically i think the time to come to complete rest is $ \infty$ though in real life, maybe friction makes it come to rest. Also, i think practically, we can neglect the higher order terms  :)

It is, but since gravity acts upon it, it will eventually stop. I want to find at what point the sequence would end.

Now Thats quite easy. Its quite interesting that it will stop in finite time.
I will solve it generaly.
If we drop it from height $ h$, then the time elapsed until, the first bounce is obviously:
\[ t_0\equal{}\sqrt{\dfrac{2h}{g}}\]
Here $ g$ is the usual. :D
Let the coefftient $ k<1$ be the quotient of the incoming and after the bounce velocities. That is:
\[ k\equal{}\dfrac{v_{n\plus{}1}}{v_n}\]

So just before the first bounce the balls velocity is
\[ v_0\equal{}\sqrt{2gh}\]
 after the first bounce the velocity is
\[ v_1\equal{}k\sqrt{2gh}\],
 so the time elapsed until the second bounce is
\[ t_1\equal{}2k\sqrt{\frac{2h}{g}}\].

The velocity after the second bounce is then:
\[ v_2\equal{}k^2\sqrt{2gh}\]

ANd so on.

So the total time is, the sum of these elapsed times:

\[ T\equal{}\sqrt{\dfrac{2h}{g}}\plus{}2k\sqrt{\dfrac{2h}{g}}\sum_{n\equal{}0}^{\infty}k^n\]
As we see this is an infinite geometric series, and we know that its sum is finite, hence the total time elapsed until the ball stops is:

\[ T\equal{}\dfrac{1\plus{}k}{1\minus{}k}\sqrt{\dfrac{2h}{g}}\]

Hmm. I just noticed that you give the quotient of the bouncing height not the velocity. In phyiscs the usual parameter used for describing this inelastic bouncing bouncing is the number I used.
But no worries the only change is that for your quotien $ n$ that is the quotient of the heights, my quotient of velocities is $ k\equal{}\sqrt{n}$. You can figure this out easily I am sure. Just check the begining :D

So now in your case according to the above :

\[ k\equal{}\sqrt{\dfrac{2}{3}} \; \; \; \; h\equal{}40\text{cm}\equal{}0.4\text{m}\]

Hence the total time until the ball stops is (I take $ g\equal{}9.81m/s^2$) :

\[ T\approx 2.827s\]

just to add to the good solution posted above mine -->

these two easily derivable formula should be remembered in such cases -->

if e is the coefficient of restitution and n collisions have occurred

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/b/a/fbab90684c50fbedc3927b08d66aacc3400be3e8.gif[/img]

H_n = e^2n * h_0

now using this formula the question becomes a child's play -->

t_0 = root(2h/g)

t_1 = 2 root ( e^2 * 2h/g)

t_2 = 2 root ( e^4 * 2h/g)

and so on

we have multiplied by 2 to include the time of going up and coming down

hence

total time = root (2h/g) + 2 root(2h/g) [ e + e^2 + e^4 +.....]

              = root(2h/g) + 2 root(2h/g)[e/(1 - e)]

             [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/c/3/4c3350e68eb37ec3361a6b88f2120bd24c04413f.gif[/img]

Alright guys! I'll get that worked out a bit, and I'll finish my drawing. Really it's nothing too mathematical, I'm more of an artist than a mathematician, so you helped me a lot. ^_^)v

But wait, Thaakis, you used Cm. I wanted it in Feet. I thought that 3 seconds was bit quick. Ha ha. Sorry I didn't make it more clear. ^_^

Ah yes I see, well then $ 40ft\equal{}12.192m$.

Hence the time, you just substitute it into the formula:

$ T\approx  15.6s$

Alright. I'm going to re do the entire drawing with these new calculations, and post it on my [url=http://mini-skirtus.deviantart.com/]DeviantArt[/url] and then come back here and post it here too. Thanks for the help all!


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Find  f:Q->Q if for any q
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f07f69b636080c5a6ba83885a4ff2093.gif[/img]
is infinite.

I don't understand the $|q|$ part...
Since there is a bijection between the positive integers $\mathbb{N}$ and $\mathbb{Q}$, it's enough to find such function on the new set.
Take any bijection (surjection suffices) from $\mathbb{N}$ to $\mathbb{N}_0[x]$, the polynomials with nonnegative integers as coefficients (or more concise since it's the only thing needed: to the set of sequences of nonnegativ integers with as good as all elements $=0$).
Then map any polynomial $\neq 0$ to the sum of it's coefficients and $0$ to $1$. This is the desired surjection.

Well, does someone remember the problems text¿ I don't...  :blush:


For this problem, an expression such as $XC$ represents the two-digit number with a tens digit $X$ and units digit $C$. In the equations below, each letter represents a unique digit, and $C=1$. Wat is the value of $R^{D}$?

$M^{L}=XC$      $L^{M}=RL$      $M^{R}=UFA$      $L^{P}=CDFL$

[hide]We look at the first equation, and we have $M^{L}=X1$

81 is the only 2-digit number ending in 1 that can be epressed as $a^{b}$. ($3^{4}$)

So M = 3, L = 4, X = 8.

Then we can use that in the 2nd equation and get $4^{3}= 64.$

Now we know R = 6.

$M^{R}= 3^{6}= 729$

So U = 7, F = 2, A = 9.

Last equation now: $L^{P}=CDFL$

The numbers used are 1, 2, 3, 4, 6, 7, 8, 9

Since P can't be 0 (Because then CDFL would have to be 1, which is impossible), P is 5.

$L^{5}= 4^{5}= 1024$, so D = 0.

And since we're looking for $R^{D}=6^{0}$,  the answer is 1.[/hide]


A solid sphere of ice with radius 4 is placed on a hollow cone (vertex pointed down) of radius 4 and height 5. When the ice melts, how much of the water will overflow the cone? Express your answer as a common fraction in terms of $ \pi$.

The volume of the cone is $ \frac{16\pi\times 5}{3}\equal{}\frac{80}{3}\pi$
The volume of the sphere is $ \frac{4}{3}\pi\times (4)^3\equal{}\frac{256}{3}\pi$

The amount overflowed is $ \frac{256}{3}\pi\minus{}\frac{80}{3}\pi\equal{}\frac{176}{3}\pi$


topology
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real analysis unsolved

Say we have open sets from two topological spaces. A is an open subset of T1 and B is an open subset of T2. So for these two open sets, A, B. Is A X B open itself? I see this is the case in R x R, where R is the real number line. I am wanting to say that this is just true in general... If so, anyone know a good proof? Or can anyway tell me how to show this?

By the definition the topology on $ T_1\times T_2$ is generated by $ \{A\times B: A\subset T_1, B\subset T_2, A,B\text{ are open}\}$. So $ A\times B$ is open just because we define the topology on the product space in such way.

Update: added that "$ A$ and $ B$ are open" in the definition.

That isn't true? just because A is in T1 and B is in T2 doesn't mean that they are open... and you certainly can't just define AxB to be open since it is in T1 X T2....there are closed sets in a topological space...and just because they are open doesn't mean that they are the things generating the topology. These aren't necessarily members of the given basis.... Perhaps I am misinterpreting your post though...

There are open sets that aren't part of that product base. Products of open sets are still open*; that's an essential part of the definition.

(*: The definition is somewhat different for Cartesian products of infinitely many sets.)

It may follow from the definition, but the definition of a topological space says nothing about the products... if just demands the the empty set and the whole space be open...and that there is closure amongst unions of open sets.. and finite intersections of open sets... So, how would I go about showing that is follows is the question...

It follows immediately from the definition of the [b]product topology[/b] not from the definition of the topological space.

We use the [b]standard[/b] definition of the topology on the product space. Certainly, you can define another topology on the product space, e.g. trivial topology. Then $ A\times B$ is not necessarily open.


In atriangle ABC we have 3a=(2 \sqrt 3)m_a.Prove that
bm_c+cm_b=2am_a

This one is somewhat easy.
  First, you can prove that AB'GC' are cyclic (by proving cotg(BGC)+cotgA = 0).(B',C' are midpoints of AC,BA, G is the centroid of ABC).
  and finally by using Ptolemee theorm, we finish it. :D


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Aliens, Armageddon, when Kim invites Bush over for tea ( in that order).

I don't see a supervolcanic eruption destroying life on earth: they've happened before.

Black hole/gamma ray burst seems so amazingly unlikely (I don't know the figures, but the chance of this destroying us any time soon must be, like one in a billion)

I think 'machines overtaking humans' is also a bit overrated, actually making a machine that actively wants to kills its creators [u]and[/u] has the power to do so would be well beyond current capability.

Global Warming is a real threat, however I don't think that it will destroy life on earth, we might just have to live nearer the poles.

[quote="G-UNIT"][quote="PenguinIntegral"]Nukes. One guy gets trigger happy and it's the end of all human life.

Seriously though, way too much power for one person.[/quote]

  I agree.  Instead, the government should distribute family-size portions of uranium to each household across the country.  This way, in order to wage war with Weapons of Mass Destruction, a country would need [i]all[/i] its citizens backing them.[/quote]

That's a brilliant idea. Nuking another country would [i]require[/i] a referendum -- you need to get enough people to turn in their uranium and plutonium in order to make any nukes! The only problem is that a large group of people could get together and make an illicit nuke with government-provided materials. Sure, you could make it illegal, but if it was that easy to have plutonium, well, consider how easy it is to have guns now in the USA, and how many people get shot...

What are supervolcanoes and how could those or global warming ever destroy life¿  :huh:
(and the more realistic factor "sun" is missing instead)

[quote="Amir.S"][quote="ragnarok23"]I recently watched a documentary on the 7 most likely ways life on Earth will be destroyed. Any opinions on this?
#7- Black Hole/ Gamma Ray Burst
#6- Machines overtaking humans
#5- Supervolcanoes
#4- Asteroid
#3- Nuclear War
#2- Pandemic
#1- Global Warming
If you don't know any of these terms, you can search for them on google.[/quote]

none of them.[/quote]

yeah, what if you divide it by zero.  :)

joke. I' think that some people are going to create a supervirus who is going to kill most of the mankind. only few humans will survive, then an enormous asteroide will colide with earth and eventually, destroy whole life on earth.

(btw, by "life" you mean only humans or animals and plants and ... too)

[quote="Ignite168"]You forgot basically ALL of the most likely possibilities.

1. Sun evaporates all water on Earth and life cannot function.

2. Water does not all evaporate but Earth burns up in the Sun.

3. There are no more resources left.

4. An outside force in space, such as meteor, asteroid, etc. like the ones that destroyed the dinosaurs.

5. Sun starts slowing down first and not enough energy.

6. Sun slows down and Earth is pulled into another gravitational orbit from something killing it in some way.[/quote]
I watched a documentary on this and I took the top seven reasons the top scientists all over the world say...

[quote="Ignite168"]You forgot basically ALL of the most likely possibilities.

1. Sun evaporates all water on Earth and life cannot function.

6. Sun slows down and Earth is pulled into another gravitational orbit from something killing it in some way.[/quote]
1) Um, hells no. Water doesn't magically dissapear.

6) Gravity doesn't automagically decide to change.

[quote="SplashD"]How come evolution isn't on there?[/quote]
How would evolution destroy life?

An amazing lack of information, or good information.

"Where will all the water go?"  Without the extreme low temperature in the atomosphe blocking the water from reaching the ionizing solar radiation at the top of the atomosphere, the hydrogen (produced from the captured decay protons and the major part of the solar system) will be seperated from the oxygen and gain enought lift to pass the Van Allen belts and be blown away by the solar wind (as is all free hydrogen -- not much with 20% oxygen and helieum).

Global Climate Change  Can you say brainwashed?  First, half of all energy comes from inside the earth and seems to be monotically decreasing.  Second, half of the carbon on the earth is involved in the inorganic carbon cycle (dissolving into oceans, being part of carbonate rocks and providing the explosive force behind volcanos).  Third, before the earth cooled enought to have liquid water, there was no inorganic carbon cycle, so all the carbon in the inorganic carbon cycle was in the atomosphere as carbon dioxide.  Fourth, before the earth cooled enough to have liquid water, there was no organic carbon cycle, no oil, no coal, so all the carbon in the current organic carbon cycle was in the atomosphere.  Fifth, the sun has been decreasing its energy output monotically over time.  Thus, if the earth didn't have a runaway greenhouse effect when all factors were far more favorable than they seem to ever be able to be reached today -- it won't happen now.

Let's think about this rationally.  The oil companies can only get about half of the oil that they find out of the ground.  If it is all burned and placed into the atomosphere, we only have half as much atomospheric carbon as was there when the oil precursor was taken from the carbon in the atomosphere, clearly not a rational earth destruction fear.  Next these same fear-mongers tell us that we are filling up the world with our garbage.  (This is just sequestered carbon -- all the paper comes from trees that remove carbon from the atomosphere.  While not on a huge scale, this tends to reduce atomospheric carbon.  Studies show that after 50 years in a landfill, newspaper hasn't even started to yellow yet.  It doesn't decay, it just sits.)

Next, the sun will not blow up (prediction only, we haven't run the experiement).  It will expand, past earth's orbit.  This means that the earth will be destroyed by fire, if something else doesn't take it first.  (Note that this is also what the Bible says, so we don't hear about the agreement between science and the Bible.  The Bible was ridiculed until science came to the same view then the critics became quiet on this point.)

Atomic wars, give me a break.  While mankind does have the power to kill many, kill all life on the planet isn't nearly possible.  Note that in killing a population or a sample one has to deal with distributions of individuals resistance to the selected method.  (Bio-monitoring of water uses the concentration that kills half of the sample group.)  This shows that science accepts the fact that it is relative easy to kill the first and relative hard to kill the last of a population or sample.

If we consider atomic wars, the weapons will not be delivered in a way to maximize the death.  Those surviving will find a rich bounty of food and things when the radiation and weather have been reduced to more normal levels.  Note that the North and South pole switch periodically, so the earth is left with no Van Allan belts for considerable periods of time.  This significantly increased radiation hasn't scoured the earth of life.

So many bombs were dropped in WWI and WWII that they are still being retreived from many fields that are in active farming.  We tried very hard to destroy everything and while many were hurt, is wasn't close to what was feared.

The Nobel prizes were funded by the invention of dynamite, which Albert thought would forever tarnish his memory and was feared to all the complete destruction of the earth.  (I know that just because someone was wrong before doesn't mean that we are wrong now, but it does suggest a certain caution of adopting an arguement that has persisted for a long time, just being updated with the current most deadly weapons.)

Chemicals -- not a chance.  Things are so much better in most of the world than they were a short time ago and those areas poisoned did get sick, but recovered.

Diseases -- not a chance.  The Black Death in the middle ages in Asia and Europe when no effective treatment was availabe only killed 1/3 of the populations at best.  This was, and would be a great disaster.  However, the population was not as robust as we currently are and it isn't even close to earth destruction.  The flue of 1913 wasn't effectively treated, since we were so concerned about killing the soldiers on the other side of WWI, only killed a score of millions.  Mao and Stalin did far worse.

It goes on and on, but try to think constructively and critically and don't just follow the position someone says because "everyone else does" or "you're an idiot if you don't beleive this."  Even if an idiot, as some I'm sure would call me, says something, the idea should be evaluated on its own merits.  Note that even a stopped clock (with hands) is right twice a day.

Thanks for reading, hope you are also thinking.

I think life will be destroyed by Global Warming.

According to the [url=http://en.wikipedia.org/wiki/Maya_calendar#2012]Mayan Calendar[/url], the end of the world will be on December 12, 2012.

[quote="kstan013"]North Korea is probably going to go all crazy and kill every single one of us. :maybe:[/quote]
That's pretty likely. :|

[quote="i_like_pie"]According to the [url=http://en.wikipedia.org/wiki/Maya_calendar#2012]Mayan Calendar[/url], the end of the world will be on December 12, 2012.
12/12/12?

No government is stupid enough to launch a nuke as part of any attack, terrorists maybe, governments no.

I doubt any of the above reasons will be it.
We'll probably have a central automation system (by 2012????)
and some idiot will forget to write code as to how to handle a division by zero error, the ensuing chaos will probably anhialate everything.

Reminding me somehow about The Terminator :D

[quote="Padmavathi"]I got this weird email which says that NASA has predicted that some giant asteroid will collide with earth destroying life...and February 1st 2019 is D-Day. so no more life on earth from 2nd Feb 2019....
 I wonder if this is true but then it looks pretty probable. i've attached 1 pic...will do the next one 2morrow but that looks morphed and unreal...
this one really scares me...d'ya think it will happen? :([/quote]

i'm pretty sure that it is like a .001 chance of that happening. and even if it hits it'll probably kill about a couple billion (all though that is so much that I don't want it 2 happen) even though there will probably be about 10 billion by then, it wont wipe out humans.

Yah, it might not wipe out humans, but it could wipe out nortamerica(by by jack :( ) Also, I was serious about that planks constant thing, its my theory of how the world will end.

[quote="Padmavathi"]I got this weird email which says that NASA has predicted that some giant asteroid will collide with earth destroying life...and February 1st 2019 is D-Day. so no more life on earth from 2nd Feb 2019....[img]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=5561&thumb=1[/img]
 I wonder if this is true but then it looks pretty probable. i've attached 1 pic...will do the next one 2morrow but that looks morphed and unreal...
this one really scares me...d'ya think it will happen? :([/quote]

yeah i think i saw a NOVA ScienceNOW on that. i think NASA has developed some interesting ideas to stop that. here is one i remembered:

$\cdot$ Launching a rocket system to the incoming asteroid. SMall rockets are attatched to specifit points on the asteroid. At the precise time, the rockets will fire, therefore, propelling the asteroid on a different course.

Yeah that's an easy way to get rid og the Big Bad Asteroid!
So i guess that brings us back to square 1-nuclear wars...
but a faster way to end the world(If there is anyone evil enough to do such a horrible thing) is to take out math and science from schools!! lol

[quote="Ihatepie"]like I said there is so small a chance of that happening that people don't have to think of it as a problem[/quote]
If you're talking about taking out math and science from schools...it is sickening to hear this but in my class of 200, about 150 hate math and science and have opted for FINE ART!!!!!!
Scary, isn't it?  :lol:

[quote="ursula"]Nathan, the sun will expand over Mars in the case it expands most, never will make it to Jupiter(there's a long way from Mars to Jupiter). Anyway, the temperature on Earth will rise so much before the Sun reaches the Earth that life as we know it will disappear in no time. We'll be a long way from earth in any case. In a Vogon  ship or something

Ursula[/quote]Uhhh... and you think that the earth will not be destroyed before then? I am not saying that that is necessarily the wrong side of the argument, it's just one thing you might possibly not have considered :wink:

Assuming the existence of an enormous quantity of dark matter, life will end in the big crunch  :) Nothing can stop that


A bicylcle tire has a 30 inch diameter.  If the tire spins at a rate of 80 revolutions per minute, what distance is traveled in one minute?

The circumference of the tire is $ 30\pi$.  Since there are $ 80$ revolutions in a minute, the answer is simply $ 30\pi \times 80$ or $ \boxed{2400\pi \text{ inches per minute}}$ or $ \boxed{200\pi \text{ feet per minute}}$.

That would be $ 2400\pi$  :wink: inches per minute.

SOMEONE NEEDS TO EDIT THEIR POST!
and what's up with all the trig questions from math92?


Consider the set in $\mathbb{R}^{2}$: \[\mathcal{C}: = \{(x,y)|Ax^{2}+Bxy+Cy^{2}+Dx+Ey=0\}\] where $D^{2}+E^{2}\ne 0$ is placed on a cartesian plane. Clearly the point $(0,0)$ lies on $\mathcal{C}$. Find the equation of the tangent to the graph at $(0,0)$.

[hide="Note"]There is an intuitive argument. Of course one can use calculus, but that is less interesting in my opinion. However, when broken down, they essentially use the same idea. I will present the solutions if nobody gets them. [/hide]

[quote="mffmartinelli"]Deriving both sides of the equation in x we get that the equation of the tangent is $y=-\frac{D}{E}x$.[/quote]

yes, that is clearly correct, but this book uses no calculus (i suppose that their proof is not completely rigirious, but it is intuitive)

[color=darkblue][b][u]Generalization.[/u][/b]$f(x,y)\equiv Ax^{2}+2Bxy+Cy^{2}+2Dx+2Ey+F=0$ and $f(x_{0},y_{0})=0$ $\Longrightarrow$
the equation of the tangent $TT$ in the point $T(x_{0},y_{0})$ to the graph $G_{f}$ is :
$\boxed{Axx_{0}+B(xy_{0}+x_{0}y)+Cyy_{0}+D(x+x_{0})+E(y+y_{0})+F=0}\ \ (1)\ .$

[b][u]Proof.[/u][/b] $f'_{x_{0}}(x,y)=2Ax_{0}+2B\left[y_{0}+x_{0}y'(x_{0})\right]+2Cy_{0}y'(x_{0})+2D+2Ey'(x_{0})=0$ $\Longrightarrow$
$y'(x_{0})=-\frac{Ax_{0}+By_{0}+D}{Bx_{0}+Cy_{0}+E}$. The equation of the tangent $TT$ is $y-y_{0}=y'(x_{0})(x-x_{0})$ $\Longleftrightarrow$
$(y-y_{0})(Bx_{0}+Cy_{0}+E)+(x-x_{0})(Ax_{0}+By_{0}+D)=0$ $\Longleftrightarrow$
$Axx_{0}+B(x_{0}y+xy_{0})+Cyy_{0}+Dx+Ey=$ $Ax^{2}_{0}+2Bx_{0}y_{0}+Cy^{2}_{0}+Dx_{0}+Ey_{0}=-Dx_{0}-Ey_{0}-F$
because $f(x_{0},y_{0})=0$. Therefore, the equation of the tangent $TT$ is $(1)$.

[b]Remark.[/b] In the our particular case $f(x,y)\equiv ax^{2}+bxy+cy^{2}+dx+ey=0$ and $f(0,0)=0$,
the equation of the tangent $TT$ in the point $T(0,0)$ to the graph $G_{f}$ is $\boxed{\ dx+ey=0\ }\ .$[/color]


Prove that the Cauchy problem \[ (x')^{2}-2xx'+x^{2}(-e^{t}+1)=0 \] where $x(0)=1$, has at least two distinct solutions.

ecuatia se mai scrie $(\frac{x'-x}{x})^2=e^t$. de aici avem $\frac{dx}{x}=(1+e^\frac{t}{2})dt$ si respectiv $\frac{dx}{x}=(1-e^\frac{t}{2})dt$, adica solutiile $x=e^{t+2e^\frac{t}{2}-2}$ si $x=e^{t-2e^\frac{t}{2}+2}$


A $ 4\times 4$ block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?
\[ \setlength{\unitlength}{5mm} \begin{picture}(4,4)(0,0) \multiput(0,0)(0,1){5}{\line(1,0){4}} \multiput(0,0)(1,0){5}{\line(0,1){4}} \put(0,3){\makebox(1,1){\footnotesize{1}}} \put(1,3){\makebox(1,1){\footnotesize{2}}} \put(2,3){\makebox(1,1){\footnotesize{3}}} \put(3,3){\makebox(1,1){\footnotesize{4}}} \put(0,2){\makebox(1,1){\footnotesize{8}}} \put(1,2){\makebox(1,1){\footnotesize{9}}} \put(2,2){\makebox(1,1){\footnotesize{10}}} \put(3,2){\makebox(1,1){\footnotesize{11}}} \put(0,1){\makebox(1,1){\footnotesize{15}}} \put(1,1){\makebox(1,1){\footnotesize{16}}} \put(2,1){\makebox(1,1){\footnotesize{17}}} \put(3,1){\makebox(1,1){\footnotesize{18}}} \put(0,0){\makebox(1,1){\footnotesize{22}}} \put(1,0){\makebox(1,1){\footnotesize{23}}} \put(2,0){\makebox(1,1){\footnotesize{24}}} \put(3,0){\makebox(1,1){\footnotesize{25}}} \end{picture}
\]$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

[hide]
The top-left to bottom-right diagonal will be 1+10+17+22=50
The top-right to bottom-left diagonal will be 4+9+16+25=54

The difference is 4. B
[/hide]

I have no idea how I got 8...
My mathematical skills aren't very good...this is what the calculator was for!!!

[b]Bonus[/b]

The answer does not depend on what block of dates you choose.

(this explains why they do the couple weird steps)