i learned it from Geometry Unbound by Kiran Kedlaya...

[url]http://www-math.mit.edu/~kedlaya/geometryunbound/[/url]

its used to find the area of a polygon given the coordinates for its vertices...

Given points $(x_1, y_1), (x_2, y _2), ... (x_n, y_n)$ (in either clockwise or counterclockwise order) that define a polygon in the plane, the area of the polygon can be found by the following procedure:

Create a table with $x$-coordinates in one column and $y$-coordinates in the other, starting with $(x_1, y_1)$ and going to $(x_n, y_n)$, and then repeating $(x_1, y_1)$ at the end.  Now, draw lines from $x_1$ to $y_2$, from $x_2$ to $y_3$, and so forth down to $y_1$ again.  Multiply along the lines and add together - this is $A$.

Draw lines going the other way from $y_1$ to $x_2$, from $y_2$ to $x_3$, and so on down to $x_1$ again.  Multiply along those lines and add together - this is $B$.

Then the area of the polygon is $\frac{A - B}{2}$.

(This is called the Shoelace Theorem because the pattern of lines resembles shoelaces.)

This result can be understood as a sum of cross products, but for now just take it on faith that it works.  :P

Example:  Find the area of the triangle defined by $(3, 4), (1, 7), (17, 20)$.

The points, in counterclockwise order starting from $(3, 4)$, are $(3, 4), (17, 20)$, and $(1, 7)$, so our first column is $3, 17, 1, 3$ and our second column is $4, 20, 7, 4$.

$A = 3 \times 20 + 17 \times 7 + 1 \times 4 = 183$
$B = 4 \times 17 + 20 \times 1 + 7 \times 3 = 109$
$\frac{A-B}{2} = 37$

shobhit, your 'general' form seems a bit too specific.. :huh:

is it necessary to list the numbers in counterclockwise order?

clockwise seems to produce a negative area

good call

one more question; does this theorem apply to concave polygons as well?

It seems like it should, but it would take someone who knows more about it that I do to tell you for sure...

Yes.  As long as the polygon is "simple" (does not cross itself), the shoelace theorem will work as long as you use the points in the right order.


geometry
circumcircle
Euler
geometry proposed

Find the locus of points $ P$ such that the lines joining the vertexes $ A$, $ B$, and $ C$ with the circumcenters of $ PBC$, $ PAC$ and $ PAB$ resp. are concurrent.

I can prove that the lines joining the vertexes $ A$, $ B$, and $ C$ with the circumcenters of $ \triangle{PBC} , \triangle{PCA}, \triangle{PAB}$ are concurrent if and only if: $ P \in (ABC)$ or $ \frac {PA^2 \minus{} b^2}{PA^2 \minus{} c^2} \cdot \frac {PB^2 \minus{} c^2}{PB^2 \minus{} a^2} \cdot \frac {PC^2 \minus{} a^2}{PC^2 \minus{} b^2} \equal{} 1 \ \ (*)$
But I don't know the locus of points $ P$ which satisfies $ (*)$. I guess that it's a high-degree curve. Anybody help me, please ? :(
P/s:
Coincidentally, I also proved that the Euler lines of $ \triangle{PBC} , \triangle{PCA}, \triangle{PAB}$ are concurrent if and only if $ P \in (ABC)$ or $ \frac {PA^2 \minus{} b^2}{PA^2 \minus{} c^2} \cdot \frac {PB^2 \minus{} c^2}{PB^2 \minus{} a^2} \cdot \frac {PC^2 \minus{} a^2}{PC^2 \minus{} b^2} \equal{} 1$
( The proof of two above conditions are separate, but they have the same ideas )
It means that the lines joining the vertexes $ A$, $ B$, and $ C$ with the circumcenters of $ \triangle{PBC} , \triangle{PCA}, \triangle{PAB}$ are concurrent if and only if the Euler lines of $ \triangle{PBC} , \triangle{PCA}, \triangle{PAB}$ are concurrent. I posted this problem on Mathlinks about six months ago, but no one interested: http://www.mathlinks.ro/Forum/viewtopic.php?p=953497
I'm quite busy right now, so I'll post my solution in a few days later  :)

It can be done by the same method, which you used here: http://www.mathlinks.ro/viewtopic.php?t=18424 .
Very nice, thank you.

Yes, the methods I used for the these problems are same :) ( although that I think the two problems of this topic are a easier, especially the first one ). Thanks April very much for reminding me. This may helps me not have to post the solution, as I said  :D  :P


geometry
circumcircle
trapezoid
geometric transformation
rotation
perpendicular bisector
geometry proposed

Let $ A,B,C$ be collinear points on the plane with $ B$ between $ A$ and $ C$. Equilateral triangles $ ABD,BCE,CAF$ are constructed with $ D,E$ on one side of the line $ AC$ and $ F$ on the other side. Prove that the centroids of the triangles are the vertices of an equilateral triangle, and that the centroid of this triangle lies on the line $ AC$.

Let $ (O_1)\equiv(ADB),$ $ (O_2) \equiv(BCE)$ and $O\equiv (O_1) \cap (O_2).$ It is enough to see that $ \angle AOC \equal{} 120^{\circ}$ $\Longrightarrow$ $ O$ lies on the circumcircle $ (O_3)$ of  $\triangle CAF.$ $ OA , OB$ and $ OC$ are radical axes of $ (O_1) \cup (O_3),$  $ (O_1) \cup(O_2)$ and $ (O_2) \cup (O_3)$ $\Longrightarrow$ $OA \perp O_1O_3,$ $OB \perp O_1O_2$ and $OC \perp O_2O_3.$ As a result, we have $ \angle O_2O_1O_3 \equal{} \angle AOB \equal{} 60^{\circ}$ and $\angle O_1O_2O_3 \equal{} \angle BOC \equal{} 60^{\circ}$ $ \Longrightarrow$ $\triangle O_1O_2O_3$ is equilateral.

[hide] Indeed, $ \triangle O_1O_2O_3$ can be seen as the  Napoleon triangle of the degenerate $\triangle ABC$[/hide]

Centroid of $ \triangle O_1O_2O_3$ lies on the line $ AC$ $\Longleftrightarrow$ Sum of the oriented distances from $ O_1,O_2,O_3$ to $ AC$ equals zero. Letting $ M,N,L$ be the midpoints of $ AB,BC,AC$

$ O_1M \equal{} \frac {_1}{^3}DM \ , \ O_2N \equal{} \frac {_1}{^3}EN \ , \ O_3L \equal{} \frac {_1}{^3}FL \ \Longrightarrow$

$ O_1M \plus{} O_2N \minus{} O_3L \equal{} \frac {_1}{^3}(DM \plus{} EN \minus{} FL) \equal{} \frac {_1}{^6}\sqrt {3}(AC \minus{} AC) \equal{} 0$

$\Longrightarrow$ The centroid of $ \triangle O_1O_2O_3$ lies on the line $ AC,$ as desired.

using cos law
we acquire G1G2=G2G3=G1G3
we can also prove G1AX G2CY are both equilateral triangles
G1G2G3XY are coocyclic  further, G1G2YX is an isosceles trapezoid thus the circumcenter lies on the perpendicular bisector of XG1,YG2
done

Denote $ G_1,G_2,G_3$ respectively by the centroid of $ \triangle DAB$, $ \triangle EBC$, $ \triangle FAC$. Consider the rotation around $ G_i$, $ i\in \{1,2\}$ with the angle $ 120^{\circ}$, we have $ \mathcal {R}^{120^{\circ}}_{G_1}: A\mapsto B$. Similarly, we also have $ \mathcal {R}^{120^{\circ}}_{G_2}: B\mapsto C$ $ \Longrightarrow \mathcal {R}^{120^{\circ}}_{G_1}$ $ \circ$ $ \mathcal {R}^{120^{\circ}}_{G_2}: A\mapsto C$. But ${ \mathcal {R}^{120^{\circ}}_{G_1}}$ $ \circ$ $ \mathcal {R}^{120^{\circ}}_{G_2}$ $ = \mathcal {R}^{240^{\circ}}_{G}$ $ = \mathcal {R}^{ - 120^{\circ}}_{G}$. Therefore $ \mathcal {R}^{ - 120^{\circ}}_{G}: A\mapsto C$ $ \Longrightarrow G\equiv G_3$. But we also have $ \angle GG_1G_2 = \angle GG_2G_1 = \frac {120^{\circ}}{2} = 60^{\circ}$ $ \Longrightarrow \triangle G_3G_1G_2$ is an equilateral triangle.
Now, let $ \mathcal {Z}\left (B, - 120^{\circ},\frac {BE}{BA}\right ): A\mapsto E$, $ D\mapsto C$, $ \triangle BDA\mapsto \triangle BCE$. Hence $ G_1\mapsto G_2$. Therefore, $ \angle G_1BG_2 = 120^{\circ}$. Let the circumcircle $ (O)$ of $ \triangle G_1BG_2$ intersects $ AC$ at $ G_4$ as the second point. We have $ \angle G_1OG_2 = G_1BG_2 = 120^{\circ}$, but $ \angle G_4G_1G_2$ $ = \angle G_2BG_4$ $ \equiv$ $ \angle G_2BC$ $ = 60^{\circ}$, $ \angle G_4G_2G_1 = \angle {G_1BA} = 60^{\circ}$. Then it is followed that through the rotation with center $ G_4$, angle $ - 120^{\circ}$, we have $ \mathcal {R}^{ - 120^{\circ}}_{G_4}: G_1\mapsto G_2$. Hence $ G_4$ must be the centroid of $ \triangle G_1G_2G_3$. The result is lead as follow. $ \square$

[quote="Napoleon"][color=darkred]Let $ A$ , $ B$ , $ C$ be three points on the plane. Equilateral triangles $ ABD$ , $ BCE$ , $ CAF$ are constructed outside of given triangle.

Prove that $ AE \equal{} BF \equal{} CD$ , $ AE\cap BF\cap CD\ne\emptyset$ , the centroids $ G_c$ , $ G_a$ , $ G_b$ of  the triangles $ ABD$ , $ BCE$ , $ CAF$

are the vertices of an equilateral triangle and the triangles $ ABC$ and $ G_aG_bG_c$ have a common centroid $ G$ and ... another properties.[/color]

[hide="Indication."]
[color=darkblue]You can prove easily using the complex numbers.[/color][/hide][/quote]

[color=darkblue][b][u]Particular case[/u].[/b] When $ B\in (AC)$ obtain the proposed problem in which the common centroid of the triangles $ ABC$

(degenerated !) and $ G_aG_bG_c$ is the point $ G\in (BM)$ , where $ M$ is the midpoint of $ [AC]$ and $ BG \equal{} 2\cdot GM$ .[/color]


vector
geometry
geometric transformation
reflection
geometry proposed

Let $\textbf{u}, \textbf{v}, \textbf{w}$ be three vectors in plane, each of lenght 1.
a) Prove that we can choose the signs $+,-$ so that $|\textbf{u}\pm\textbf{v}\pm \textbf{w}|\leq 1$;
b) Give an example of three vectors $\textbf{u}, \textbf{v}, \textbf{w}$ such that no matter we choose the signs $+,-$ we have  $|\textbf{u}\pm\textbf{v}\pm \textbf{w}|\geq 1$

[quote="N.T.TUAN"]If $ABC$ is a triangle for center of $(ABC)$ is $O$ and its radius equal to $1$, then $|OA+OB+OC|\leq 1$. In fact, $|OA+OB+OC|=3|OG|$ and $9OG^{2}=9-AB^{2}-BC^{2}-CA^{2}\leq 9$. From this, we have a).[/quote]

But then you get $OG^{2}\le 1$, so $|OA+OB+OC| = 3|OG| \le 3$.  The quantity $|OA+OB+OC| = 3$ when all three vectors point in the same direction.

Choose the signs so that triangle $ABC$ is acute (or right).  (Draw a line through $O$.  If $A$, $B$, and $C$ all lie on one side of the line, then reflect the "middle point" through $O$.)  Then the orthocentre $H$ lies inside (or on) triangle $ABC$, and the part (a) follows.

For part (b), just take three vectors at $120^\circ$, i.e. triangle $ABC$ is equilateral.

[quote="nsato"]For part (b), just take three vectors at $120^\circ$, i.e. triangle $ABC$ is equilateral.[/quote]

Actually, this doesn't work :( .  Instead, take two vectors pointing in opposite directions, then take the third vector perpendicular to the first two.

Three colinear vectors "-->-->-->" works too


Normally, I try to focus a bit more, but that's been failing, so I think I'll be playing around with my attitude a little bit for AMC 12B.

None of the above. My attitude swings on a moment's notice from giddy to anxious to afraid to concentrating.

Mostly just in the zone...I try to block out outside noise and concentrate all my energy on the problem.

Am I the only person that voted normal?
Hmph.

I stay completely relaxed.  I try to work on the problems as I would normally work on problems, and maybe hurry up a little if I'm running out of time.  There are a few things I like to remind myself:

- Read the problems really slowly.
- Work slowly.
- I'm capable of solving every problem.

Seems to work pretty well for me.

Yeah, I was blasting Kashmir (Led Zeppelin) and the Rocky theme on the morning of 12A, so I was feeling *epic* when I was taking it... that probably explains my stupid mistake on #18. Now that I know I'm already safe for AIME, I'll try to be more mellow for 12B... after I listen to stairway.

[quote="123456789"]Usually on the AMC, I get really excited before the contest. Then I sit down, with my heart racing, and speed through the first 15 ish problems. As I work farther into the contest, I become calmer. (Oh, and I probably made one or two stupid mistakes in those first 15.) Then I do the next 5 problems rather quickly again, so I have like over 35 minutes to do the last 5 problems. Then I skip to number 25, spend way too much time on it, start panicking and looking at the clock, and then work backwards from 25. Then I realize I have like 10 minutes left, and I solve one or two of the last five, rush on another one, and then after the test, I think I owned the test.

A few minutes/hours later, I realize I made a ton of stupid mistakes and will have to retake.[/quote]

THIS is exactly what happened to me when i was taking the test.


you have sixteen games to pick either home win or visitor win each game has a box you must answer all 16 correct how many different combinations are there to ensure you get perfect. ex

1.home_vs visitor_
2.home_vs visitor_
you have 16 teams you have to select you have to get 16 correct put a checkmark in the space. how many combinations will you have so you will answer all 16 correctlyby making these combos you will have every possible outcome covered how many combos are there

I still don't get it...

Does the question ask to find the number of "winning" combinations?

You need to get all 16 matches "correct" so there is just 1 possibility?

Or am I missing the point of the question?


Prove the following theorem:

If $\{x_{1}, x_{2},...,x_{\phi(m)}\}$ is a reduced residue system modulo m and $(a,m)=1$, then the set $\{ax_{1}, ax_{2},...,ax_{\phi(m)}\}$ is a reduced residule system modulo m, too.

Note: $\phi(m)$ is the totient function...number of numbers less than $m$ that are relatively prime to $m$.

Sorry if I'm just not thinking right, but how do you know that $\{ax_{1}, ax_{2},...,ax_{\phi(m)}\}$ is a reduced residule system modulo $m$?

Any reduced residue system $\bmod m$ is isomorphic to any other.

The set we are currently interested in has the correct number of members, and each member is itself a residue relatively prime to $m$, meaning each member is a reduced residue $\bmod m$.

Hence it only remains to show that the $\varphi(m)$ residues are distinct.

[quote="t0rajir0u"][hide="Solution"] Of course, all of the resulting elements are relatively prime to $m$, and they are all members of $U_{m}$, so it remains to show that they are distinct.

But this is obvious.  Suppose $ax_{i}\equiv ax_{j}$.  Since $(a, m) = 1$ we have a multiplicative inverse; hence

$x_{i}\equiv x_{j}$,

Contradiction. [/hide][/quote]

wait, what do you mean by all the elements are relatively prime to $m$?

Are we discussing the same concept here?

[b]Definition:[/b]  A reduced residue system $\bmod m$ (denoted, in my experience, $U_{m}$), is the finite multiplicative group

$\{ x_{1}, x_{2}, ... x_{ \varphi(m) \}}$

Under modular multiplication.  (In saying it is a group, I stipulate that it must be closed under modular multiplication, that it contains $1$, and that multiplicative inverses are defined for each element (no multiplicative inverse of $0$, of course))

The criterion for multiplicative inverse is the most important; in order that $a^{-1}\bmod m$ exists it is required that

$\exists b : a \cdot b \equiv 1 \bmod m$

In other words,

$\exists x, y: ax+my = 1$

Which requires that $gcd(a, m) = 1$.

Now, if our first set $\{ x_{1}, x_{2}, ... x_{ \varphi(m) \}}$ is a reduced residue system, then $gcd(x_{i}, m) = 1 \forall i$.  Furthermore, if $gcd(a, m) = 1$ it follows that

$gcd(ax_{i}, m) = 1 \forall i$

And therefore that the set

$\{ ax_{1}, ax_{2}, ... ax_{ \varphi(m)}\}$

Only contains elements that are relatively prime to $m$.  In order that this be isomorphic to our first set, all of its elements need to be distinct, which is easy to prove since the multiplicative inverse of $a$ exists.

[b]Remark:[/b]  This is true of any finite group.

yes that makes sense. i got confused earlier.

Normally it's not a field (not closed under addition).

Btw, this theorem is often used to prove the existence of inverses (and generalises in some ways to other rings/subsets of those).

Oh, sorry.  I meant it was a group under multiplication.   :(

[quote="t0rajir0u"]it follows that

$gcd(ax_{i}, m) = 1 \forall i$

And therefore that the set

$\{ ax_{1}, ax_{2}, ... ax_{ \varphi(m)}\}$

[/quote]

Thanks, that's what I was looking for, so they do not necessarily have to be in order, but contain all the elements of $\{ x_{1}, x_{2}, ... x_{ \varphi(m)}\}$.


I was just wondering if anyone here does indeed debate? or do speech?

I debat, but I'm a novice. In fact, I barely know what novice means and what difference it makes.

i participated in JSA but not so much anymore

I do it, well my school only does speech. I'm only a novice though, and have only been to like 2 tournaments  :D

i can prove and argue well, but i'm not really good with debating infront of a crowd :blush:.


geometry
circumcircle
incenter
angle bisector
geometry solved

Let $ABC$ be a triangle. The its incircle $i=C(I,r)$ touches the its sides in the points $D\in (BC),E\in (CA),F\in (AB)$ respectively. I note the second intersections $M,N,P$ of the lines $AI,BI,CI$ respectively with the its circumcircle $e=C(O,R)$. Prove that the lines $MD,NE,PF$ are concurrently.

[u]Remark.[/u] If the points $A',B',C'$ are the second intersections of the lines $AO,BO,CO$ respectively with the circumcircle $e$ then the points $U\in MD\cap A'I,V\in NE\cap B'I,V\in PF\cap C'I$ belong to the circumcircle $w$.

I forgot to say that the remark is a independent problem.

Here's my proof.

$Lemma)$ Consider two triangle $ABC$, $DEF$ which resemble the other and $AB||DE, BC||EF, CA||FD$.

Then $AD, BE, CF$ are concurrent.

$Lemma-pf)$ Let's think just triangle $ABC$ and the segment $DE$ (erase $F$ in your brain).

Let's denote $K$ by $AD\cap BE$. Then It's self-evident that $AD: DK= BE: EK$.

Now, denote $F'$ by a point on line $KC$ which satisfies $AD: DK= CF': F'K$.

Then we get triangle $DEF'$ which resembles $ABC$. or $F=F'$. -> OK!

Now, Let's prove the problem!

We already know that $FD\perp BI$. Here, Let $\angle A= 2 \alpha, \angle B= 2 \beta, \angle C= 2\gamma$.

Then $\angle IPM= \alpha$, and $\angle PIB = |\angle BIF - \angle PIF| = |(\gamma + \alpha) - |\angle AIF - \angle AIP| | = \beta + \gamma$.

So, we get $MP \perp BI$. Therefore, $MP||DF$.

Similarly, $MN||DE$, $NP||EF$.

Finally, we get two similar triangle $DEF$, $MNP$ with corresponding sides parallel.

By the Lemma we proved above, $MD, NE, PF$ are concurrent.

$Q.E.D.$

Note: If we go ahead a little more, we get $O, I, K$ ($K$ is the intersection of $MD, NE, PF$) are collinear.

$PB=PI,\ MB=MI\Longrightarrow PM\perp BI;$
$BF=BD\Longrightarrow FD\perp BI.$
$PM\perp BI,\ FD\perp BI\Longrightarrow PM\parallel FD$
(analogously $MN\parallel DE,\ NP\parallel EF$).
$MN\parallel DE,\ NP\parallel EF,\ PM\parallel FD\Longrightarrow$
([u]Desarques's theorem[/u]) $MD\cap NE\cap PF\ne \emptyset.$

[u]Remark.[/u] I note $S\in MD\cap NE\cap PF$. It is very interesting that the point
$\Gamma \in AD\cap BE\cap CF$ (Gergogne's point) belongs to the line $OI$.

[quote="Ph-An"]Let $ABC$ be a triangle. The its incircle $i=C(I,r)$ touches the its sides in the points $D\in (BC),E\in (CA),F\in (AB)$ respectively. I note the second intersections $M,N,P$ of the lines $AI,BI,CI$ respectively with the its circumcircle $e=C(O,R)$. Prove that the lines $MD,NE,PF$ are concurrently.[/quote]

Since I is the incenter of triangle ABC, the line AI is the angle bisector of the angle CAB. Thus, < BAM = < CAM. But since the point M lies on the circumcircle of triangle ABC, we have < BAM = < BCM and < CAM = < CBM, so that this becomes < BCM = < CBM; thus, the triangle BMC is isosceles, so that BM = CM. Hence, the point M is the midpoint of the arc BC of the circumcircle of triangle ABC. Similarly, the points N and P are the midpoints of the arcs CA and AB of this circumcircle, respectively. Now, the problem becomes equivalent to the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=46118 .

Do you find this difficult? I think the main observation is that the triangles MNP and DEF are homothetic, what is quite straightforward to see by angle chasing (or just well-known: the sidelines of both triangles are perpendicular to the lines AI, BI, CI, respectively)...

[quote="Ph-An"][u]Remark.[/u] If the points $A',B',C'$ are the second intersections of the lines $AO,BO,CO$ respectively with the circumcircle $e$ then the points $U\in MD\cap A'O,V\in NE\cap B'O,V\in PF\cap C'O$ belong to the circumcircle $w$.[/quote]

I guess you mean $U=MD\cap A^{\prime}I$, $V=NE\cap B^{\prime}I$ and $W=PF\cap C^{\prime}I$.

Nice problem!! Here is a proof:

Since the three assertions are cyclical permutations of each other, we will prove only one of them. So the problem reduces to:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle. The incircle of triangle ABC has center I and touches its side BC at a point D. Let the line AI meet the circumcircle of triangle ABC at the point M (apart from A). Let O be the circumcenter of triangle ABC, and let the line AO meet the circumcircle of triangle ABC at the point A' (apart from A). Prove that the lines MD and A'I intersect on the circumcircle of triangle ABC.[/color]

[i]Proof of Theorem 1.[/i] We will work with directed angles modulo 180°.

As I showed above, the point M is the midpoint of the arc BC of the circumcircle of triangle ABC. Now, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20815#p20815]http://www.mathlinks.ro/Forum/viewtopic.php?t=6095 post #2[/url], the point M is the circumcenter of triangle BIC. In other words, if we denote by $\mu$ the circumcircle of triangle BIC, then the center of this circle $\mu$ is M. The radius of the circle $\mu$ is obviously MB = MC = MI.

Now, let R be the image of the point D under the inversion with respect to the circle $\mu$. Then, the point R lies on the ray MD and satisfies $MD\cdot MR=MB^2$. Hence, $\frac{MD}{MB}=\frac{MB}{MR}$, what, together with < DMB = - < BMR, yields that the triangles DMB and BMR are inversely similar. Thus, < BRM = - < DBM. Similarly, < CRM = - < DCM. Hence,

  < BRC = < BRM - < CRM = (- < DBM) - (- < DCM) = < DCM - < DBM
        = < (BC; CM) - < (BC; BM) = < (BM; CM) = < BMC.

Thus, the point R lies on the circle through the points B, M and C; this is obviously the circumcircle of triangle ABC.

Let G be the image of the point A' under the inversion with respect to the circle $\mu$. Then, the point G lies on the ray MA' (since M is the center of $\mu$).

Obviously, the inversion with respect to the circle $\mu$ maps the points B, C and I to themselves (since these points lie on the circle $\mu$); hence, it maps the circumcircle of triangle ABC to the line BC (in fact, the circumcircle of triangle ABC passes through the center M of the inversion circle $\mu$, and thus is mapped to a line; and this line is the line BC, since the points B and C are mapped to themselves by the inversion). Hence, as the point A' lies on the circumcircle of triangle ABC, its image G under our inversion must lie on the line BC.

As the point G is the image of the point A' under the inversion with respect to the circle $\mu$, the point A', in turn, is the image of the point G under this inversion.

Now, since the point D is the point of tangency of the incircle of triangle ABC with the side BC, while the point I is the center of this incircle, we have $ID\perp BC$, so that < IDG = 90°. On the other hand, since the chord AA' of the circumcircle of triangle ABC passes through the center O of this circumcircle, it is a diameter of this circumcircle; hence, < AMA' = 90°, what rewrites as < IMG = 90°. Thus, < IDG = < IMG, what yields that the points I, G, D and M lie on one circle.

Now, the inversion with respect to the circle $\mu$ maps this circle to a line (in fact, every circle through the inversion center M gets mapped to a line by this inversion), and this line must pass through the images of the points I, G and D under this inversion, i. e. through the points I, A' and R. Hence, the points I, A' and R lie on one line, i. e. the point R lies on the line A'I.

Altogether, we see that the point R lies on the circumcircle of triangle ABC and on the lines MD and A'I. Thus, the lines MD and A'I intersect on the circumcircle of triangle ABC, and Theorem 1 is proven.

  Darij

Yes, I corrected  the remark. This remark is a well-known property and it appears in the book: [b]Gheorghe Titeica-Culegere de probleme de geometrie[/b] [u](in the Rumanian language),[/u] [b]1937[/b].

[u]A solution of the problem from the remark.[/u] I note $S\in MD\cap A'I$.

$\widehat {DIM}\equiv \widehat {IAA'},\ \frac{ID}{IM}=\frac{AI}{AA'}\ (2Rr=ID\cdot AA',\ IA\cdot IM=p(I)$ -
the power of the point $I$ w.r.t the circumcircle $w$)$\Longrightarrow$
$\triangle DIM\sim \triangle IAA'\Longrightarrow \widehat {IMD}\equiv \widehat {AA'I}\Longrightarrow S\in w$.


geometry
incenter
circumcircle
geometry unsolved

[b][size=167]I is the incenter of triangle ABC.⊙I cuts the segment BC,CA,AB at A',B',C'.Proof that the circumcenter of AIA',BIB',CIC' are collinear.[/size][/b]

The reason I don't draw the picture is the picture seems really "huge" that will be unclearly for you to see.

Dear Mathlinkers,
see : http://www.mathlinks.ro/Forum/viewtopic.php?t=283185
Sincerely
Jean-Louis


Find all functions that satisfy $ f(f(x)\plus{}y) \equal{} f(x^2\minus{}y) \plus{} 4yf(x)$ for all real $ x,y$.

To tired to post the entire solution so there it is a [hide="almoust solution hint  :D "]take $ y \equal{} \minus{} f(x)$ and then $ y \equal{} x^2$.Then you have to prove  that $ f(x) \equal{} x^2$[/hide]

[hide="partial solution"]
Plugging in values:
\[ f(0) \equal{} f(x^2 \plus{} f(x)) \minus{} 4f(x)^2 \\
f(f(x) \plus{} x^2) \equal{} f(0) \plus{} 4x^2f(x)
\]
Subtracting, we get $ 4x^2f(x) \equal{} 4f(x)^2$, which leaves us with two possibilities for each $ x$: $ f(x)\equal{}0$ or $ f(x)\equal{}x^2$. In each case, obviously, $ f(0)\equal{}0$ and $ \forall x, f(x)\ge0$. From here we have two cases: $ \forall x, f(x)\equal{}0$ or $ \exists a \ne 0$ such that $ f(a)\equal{}a^2$. Thus, $ a^2>0$ as well. In the latter case, setting $ x\equal{}a$, we have $ f(a^2\plus{}y)\equal{}f(a^2\minus{}y)\plus{}4a^2y$. $ \forall y  \in\mathbb{R}^{\plus{}}$, if $ f(a^2\plus{}y)\equal{}0$, then $ f(a^2\minus{}y)\equal{}\minus{}4a^2y<0$. However, $ f(a^2\minus{}y)\ge 0$, a contradiction, so $ \forall y\in\mathbb{R}^{\plus{}}, f(a^2\plus{}y)\equal{}(a^2\plus{}y)^2$.
[/hide]

[quote="JBL"][quote="worthawholebean"]either $ f(x) \equal{} 0$ or $ f(x) \equal{} x^2$[/quote]

No.  We have $ \forall x, (f(x) \equal{} 0) \vee (f(x) \equal{} x^2)$, not $ (\forall x, f(x) \equal{} 0) \vee (\forall x, f(x) \equal{} x^2)$.  There's still more work to do.[/quote]Oops. Fixing now.

[quote="JBL"][quote="worthawholebean"]either $ f(x) \equal{} 0$ or $ f(x) \equal{} x^2$[/quote]

This means $ \forall x, (f(x) \equal{} 0) \vee (f(x) \equal{} x^2)$, not $ (\forall x, f(x) \equal{} 0) \vee (\forall x, f(x) \equal{} x^2)$.  There's still more work to do.[/quote]
yes you're right,let $ f(x)\equal{}0$ for some $ x\in\mathbb{R}$ then we must have:

$ f(y)\equal{}f(x^2\minus{}y)$

for every $ y\in\mathbb{R}$

now let $ y\not \equal{}0$,we know that $ f(y)\equal{}0$ or $ f(y)\equal{}y^2$.

if $ f(y)\equal{}y^2$ then we would have:

$ y^2\equal{}f(x^2\minus{}y))\not \equal{}0$

thus we must have $ f(x^2\minus{}y)\equal{}(x^2\minus{}y)^2$ thus:

$ y^2\equal{}x^4\plus{}y^2\minus{}2x^2y\Rightarrow x^4\equal{}2x^2y$

for every $ y\in\mathbb{R}\minus{}\{0\}$,thus we must have:

$ x\equal{}0$

so in this case $ f(x)\equal{}0$ implies $ x\equal{}0$.

--------------------------------------

now if $ f(y)\equal{}0$,if there exists a real number $ z$ such taht $ f(z)\equal{}z^2$ then according to the last case we would have $ y\equal{}0$.
now if there was'nt any $ z\in\mathbb{R}\minus{}\{0\}$ such that $ f(z)\equal{}z^2$ means that for every $ z\in\mathbb{R}\minus{}\{0\}$ we have $ f(z)\equal{}0$ hence $ f\equiv 0$.

thus we have these cases:

$ f\equiv 0$

and $ \forall x\in\mathbb{R} : f(x)\equal{}x^2$


Is there a positive integer $ n$ such that among 200th digits after decimal point in the decimal representations of $ \sqrt{n}$, $ \sqrt{n\plus{}1}$, $ \sqrt{n\plus{}2}$, $ \ldots,$ $ \sqrt{n\plus{}999}$ every digit occurs 100 times?

[i]Proposed by A. Golovanov[/i]

So sorry for my spam post, but I'm really interested in this problem. I hope a soltuion for it.

Consider $ n\equal{}10^{400}\plus{}1$, then $ \sqrt{n\plus{}k}\equal{}10^{200}\sqrt{1\plus{}\frac{k\plus{}1}{10^{400}}}$. Note that \[ 1\plus{}\frac{k}{2\cdot 10^{400}}<\sqrt{1\plus{}\frac{k\plus{}1}{10^{400}}}<1\plus{}\frac{k\plus{}1}{2\cdot 10^{400}},\] where $ 0\leq k\leq 999$. It follows that 200th digit in $ \sqrt{n\plus{}k}$ equals last digit of $ [k/2]$.

[quote="orl"]Is there a positive integer $ n$ such that among 200th digits after decimal point in the decimal representations of $ \sqrt {n}$, $ \sqrt {n \plus{} 1}$, $ \sqrt {n \plus{} 2}$, $ \ldots,$ $ \sqrt {n \plus{} 999}$ every digit occurs 100 times?

[i]Proposed by A. Golovanov[/i][/quote]

Yes, there is.

I suggest to look for a natural number $ a$ such that : $ a \plus{} p10^{ \minus{} 200}\le\sqrt {a^2 \plus{} 100p \plus{} k} < a \plus{} (p \plus{} 1)10^{ \minus{} 200}$ $ \forall p\in[0,9]$ $ \forall k\in[0,99]$

If such "$ a$" exists, then $ n \equal{} a^2$ is an answer to the required problem.

[u][b]First part i[/b][/u]s $ a \plus{} p10^{ \minus{} 200}\le\sqrt {a^2 \plus{} 100p \plus{} k}$

$ \iff$ $ 2ap10^{ \minus{} 200} \plus{} p^210^{ \minus{} 400}\le 100p \plus{} k$

$ \iff$ $ a\le 50\cdot 10^{200} \plus{} \frac {k10^{200}}{2p} \minus{} \frac 12p10^{ \minus{} 200}$ and so, since this must be true $ \forall p\in[0,9]$ $ \forall k\in[0,99]$, it would be enough to have  :

$ a\le 50\cdot 10^{200} \minus{} \frac 9210^{ \minus{} 200}$

and so $ a < 5\cdot 10^{201}$

[u][b]Second part is [/b][/u]$ \sqrt {a^2 \plus{} 100p \plus{} k} < a \plus{} (p \plus{} 1)10^{ \minus{} 200}$

$ \iff$ $ 100p \plus{} k < 2a(p \plus{} 1)10^{ \minus{} 200} \plus{} (p \plus{} 1)^210^{ \minus{} 400}$

$ \iff$ $ 50\cdot10^{200} \plus{} \frac {k \minus{} 100}{2(p \plus{} 1)}10^{200} \minus{} \frac 12(p \plus{} 1)10^{ \minus{} 200} < a$ and so, since this must be true $ \forall p\in[0,9]$ $ \forall k\in[0,99]$, it would be enough to have  :

$ a > 50\cdot10^{200} \minus{} \frac 1210^{199} \minus{} \frac 1210^{ \minus{} 200}$

And so $ a\ge 50\cdot 10^{200} \minus{} \frac 1210^{199}$ $ \equal{} 4995\cdot 10^{198}$

So we got $ 4995\cdot 10^{198}\le a < 5000\cdot 10^{198}$

And so, for example, $ \boxed{n \equal{} 4999^210^{396}}$ would be an answer.


In an acute-triangle $ABC$,Prove
\[ \sin^2 A+\sin^2 B+\sin^2 C>2  \]

$a = 2R\sin A$, so we just need $\sum a^2 > 8R^2$.

Since $( \sum a^2 ) - 9R^2 = -OH^2$, it becomes $R^2 > OH^2$.  Since H is inside the triangle, it is obviously true.

I think you are right.
Can I ask how do you $R>OH$


analytic geometry
vector
invariant
combinatorics unsolved
combinatorics

Show that any given nonequilateral triangle can be cut in to 3 parts satisfying:
1)One of the parts must be similar to the given triangle.
2)From the other two parts a triangle that is similar to the given can be formed.

Does there always exist affine transformation mapping arbitrary triangle to an isosceless right triangle?

This question may seems unrelated but I think it actually solves the problem (as far as it is quite easy in the above mentioned case to do the cutting we are asked to).

Or do I think wrong?   :?

[quote="hardsoul"]Show that any given nonequilateral triangle can be cut in to 3 parts satisfying:
Kyrgyzstan MO 2005,final round,day 2.[/quote]

Do you send us all questions from Kyrgyzstan MO 2005, final round?

Thanks.

[quote="Megus"]Does there always exist affine transformation mapping arbitrary triangle to an isosceless right triangle?[/quote]

Yes. Just take a skew coordinate system where the vertices of the triangle have the coordinates (0, 0), (0, 1) and (1, 0) (in other words, two sides of the triangle are the unit vectors along the axes of this skew coordinate system). Now apply an affine transformation which maps this skew coordinate system to a rectangular, Cartesian one. Then, the triangle becomes an isosceles right triangle.

[quote="Megus"]This question may seems unrelated but I think it actually solves the problem (as far as it is quite easy in the above mentioned case to do the cutting we are asked to).[/quote]

Don't forget that similitude of triangles is [i]not[/i] an invariant under affine transformations, so the above transformation [i]does not[/i] solve the problem.

  Darij

[quote="darij grinberg"]
Don't forget that similitude of triangles is [i]not[/i] an invariant under affine transformations, so the above transformation [i]does not[/i] solve the problem.
[/quote]

Oh...  :blush: silly me - I don't even know how could I think of preserving similtude - sorry for bothering

valeriano wrote:
Do you send us all questions from Kyrgyzstan MO 2005, final round?

Thanks.
No I didn't.
Betwean how do you write this command
"hardsoul wrote: ..." in different color....    I think you understand me.[b][/b]


Is your unfinished post supposed to be a question or what? Should it be deleted?


analytic geometry
calculus
derivative
linear algebra
matrix
function
topology

Show that grad(u) is independent of the cartesian coordinate system.

Some other related question I asked myself: "the total derivative (df)(a)... is this is also independent of the basis we choose? The definition of the derivative does not mention a fixed orthogonal basis anywhere... Must we choose an orthogonal basis? Or is (df)(a) similar (in the sense of conjugate) to some other matrix if we change of basis?"

That's not really enough for this to make sense.

Here's what I think you mean:
There is a general notion of the gradient of a scalar-valued function on a smooth manifold. This gradient is coordinate-independent, in the sense that (a) it lives in a space which doesn't necessarily have coordinates, and (b) it transforms nicely under changes of coordinates; if $ \phi$ and $ \psi$ are parametrizations of a neighborhood of $ p$, with $ \psi=\phi\circ F$ near $ p$, $ \nabla(u\circ\psi)(x)=\nabla(u\circ \phi)(F(x))\cdot F'(x)$.
If this is an $ m$-dimensional manifold, the gradients (in coordinates) are $ 1\times m$ row vectors, and $ F'$ is an $ m\times m$ matrix. In the manifold world, the gradient lives in the cotangent space, and it's one of the first examples of the "$ d$" operator.

Derivatives in general also work without reference to an explicit basis. There's one set of coordinates in the domain, and another in the codomain. Both change-of-basis matrices come into the transformation, one as an inverse.

[Edit- fixed missing symbols]

Sorry. I don't know what manifolds are :(

as a hint there was given that change of cartesian coordinate system is done by translation + rotation of basis.

if someone can do it for R^2, I'll try to generalize it myself to R^n

This is how I understand the question: the gradient $ \nabla u=\langle \partial u/\partial x_{1},\dots \partial u/\partial x_{n}\rangle$ is defined in terms of coordinates $ x_{1},\dots,x_{n}$. Why is the vector $ \nabla u(x)$ the same no matter which orthogonal coordinates we choose? [url=http://planetmath.org/encyclopedia/Gradient2.html]PlanetMath[/url] explains this reasonably well.

Ok, thanks, mlok. That's what I meant

So if I take planetmath's definition, i can just say that the scalar product is invariant under orthogonal transformations, and that suffices?

Yannick, this is what I did to answer this question a long time ago. Probably a bit complicated, but oh well.

Nice. Thanks Arne  :coolspeak:  This was exactly what I was looking for :)

All right- my last post was pitched at the wrong level. This one doesn't mention manifolds, but it's still heavy on the linear algebra.

A note of caution: if we think of the gradient as a vector in the ordinary sense, it's [b]not[/b] really coordinate-independent; the change-of-basis matrix is transposed. If we stick to orthonormal coordinate systems, we can read this transpose as an inverse, and all we're left with is the contravariant action (the order of composition is reversed).

The real fix is to interpret the gradient as something other than a vector: it's an element of the dual space instead. This notion can be made coordinate-independent, and it makes something like this easy.

The PlanetMath article effectively writes the gradient as a column vector, which leads to the transpose in that chain rule entry. I prefer writing it as a row, which allows me to put the multiplication on the other side instead.


Construct a Borel set $ E\in \mathbb{R}$ such that $ 0 < m(E\cap I) < m(I)$ for every nonempty segment $ I$. Is it possible to have $ m(E) < \infty$ for such a set?

this problem had been discussed on the forum,
I know that Grober, Fedja , ....also commented on some one's ( I cannot remember) nice solution but I cannot find it out.
You should try to look for it  :lol:


Two particles with positive charges  q_1= 0.540nC  and  q_2= 7.70nC  are separated by a distance of 1.18m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

hey use coulombs law here
which is
$ F  \equal{} 1/4  PIE  Eo (Q1  *  Q2/R^2)$

Let the distance of the desired point from charge $ q_{1}$ be $ x$ m.
Thus, from charge $ q_{2}$, the distance is $ (1.18\minus{}x)$ m.

Applying the intensity expression,

$ \frac{q_{1}}{4 \pi \epsilon x^2}\equal{}\frac{q_{2}}{4 \pi \epsilon (1.18\minus{}x)^2}$

$ \Rightarrow \frac{0.54}{x^2}\equal{}\frac{7.70}{(1.18\minus{}x)^2}$

$ \Rightarrow \left(\frac{1.18}{x} \minus{}1 \right)\equal{}3.776$

$ \Rightarrow x\equal{}\frac{1.18}{4.776}\equal{}0.247$

So the distance is $ 0.247$ m from charge $ q_{1}$.
 :)

oops i got wrong while i wrote
i realised it later but could u all tell me how to delete ones post when one has already sent it!
thanx :)

You have to contact the moderators....
 :)


Is the answer b? I'm not familiar with sets.  :(

Yes; $A\cup B=\{1,2,3\}\cup \{2,4,6\}=\{1,2,3,4,6\}$, and $(A\cup B)\cap C=\{1,2,3,4,6\}\cap \{1,3,5\}$, or the elements that those two sets have in common, which are $\{1,3\}$, and our answer is $\boxed{B}$.


geometry perimeter parallelogram trigonometry rectangle

A parallelogram has a height of 5, and area of 50, and angles of 150 and 30. What is the perimeter of this parallelogram?

[hide]
(10 x square root 3)/3 + 20
[/hide]

[hide]um...this is prolly wrong...but is the perimeter 30?[/hide]

[hide]\[\frac{10}{\sqrt{3}}\] because if you use the properties of a 30 60 90 degree triangle you get 5 divided by \[\frac{\sqrt{3}}{2}\] That equals \[\frac{5\times2}{\sqrt{3}}\] The perimeter is$\frac{20}{\sqrt{3}}+20$(Add in the other 2 sides of side length 10 since the area is 50 and the height is 5)I could explain this better if i could use a diagram![/hide]

[quote="B4k4ka"][quote="nonie"][hide]\[\frac{10}{\sqrt{3}}\] because if you use the properties of a 30 60 90 degree triangle you get 5 divided by \[\frac{\sqrt{3}}{2}\] That equals \[\frac{5\times2}{\sqrt{3}}\] I could explain this better if i could use a diagram![/hide][/hide][/quote]
[hide]
yeah same here,  a diagram would be easier to use.

BTW, he asked for the perimeter so you have to add on the other 2 sides as well[/hide][/quote]

oops ill fix that!

[quote="anirudh"]doyou use trig for tis?[/quote]
I didn't. At least i don't think it counts as trig.

[quote="jaden"][hide]You create a rectangle moving the imaginarium triangle to the right. Then you add 5x2 + 10x2=30[/hide][/quote]

Interesting solution. but what about this?

[hide]The 30-60-90 triangle you see implies that the slant has length of 10. So 10(2)+10(2)=40? [/hide]

I just want to tell you guys, I first posted this in classroom math  :blush: , one of the mods/admins bumped it here.  :blush:

[quote="Holier Than Thou"][quote="nonie"] The perimeter is$\frac{20}{\sqrt{3}}+20$(Add in the other 2 sides of side length 10 since the area is 50 and the height is 5)I could explain this better if i could use a diagram![/quote]

I think nonie and B4k4ka already solved it and got $20+\frac{20}{\sqrt3}$. The diagonal sides aren't $10$, but $\frac{10}{\sqrt3}$.[/quote]
Well, according to 236factorial, who actually posted the problem...
[hide]
[quote="236factorial"]The 30-60-90 triangle you see implies that the slant has length of 10. So 10(2)+10(2)=40?[/quote]
And I agree with him that the slant has length of $10$, not $\frac{10}{\sqrt{3}}$...
[/hide]

[hide="trig for those who cant recall algebra I"]
$\sin{30}=\frac{5}{s}$, where s is the slant.

$0.5=\frac{5}{s}$

$s=\frac{5}{0.5}$

$s=10$  ;) [/hide]

Here is my image with solution. Everything is included in the image below:
[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=239]Click here to see full size[/url]

I think what people are doing, is when they're drawing the diagram, they don't realize that the figure can sometimes not be drawn to scale, for the triangle on the left, the $30$ degree angle is on the BOTTOM and not on the top, therefore, the hypotenuse is twice the base which is $5$, also the height of the parallelogram. Then since the height is $5$, the bases of the parallelogram are $10$, so $10+10+10+10=\boxed{40}$. People, this wasn't a hard problem ;)

[quote="236factorial"]I made this up in about 5 seconds, i thought it was easy enough for classroom, not a discussion in high school...  :roll:[/quote]

I can't believe people are missing this, this is an sat problem at hardest.

[quote="riddler"][quote="frt"][quote="riddler"]My solution[/quote]
40 is correct, but can you explain your answer a bit more clearly?[/quote]
well you get 2 triangles similiar in this thus we can find the hypotenuse.

P.S: COOOL avatar Anirudh![/quote]

It's Mozilla Firefox, the internet browser

[quote="riddler"][quote="Iversonfan2005"]People, this wasn't a hard problem ;)[/quote]
nobody said it was. :P[/quote]

Then why did so many people miss it? ;)

I thought the slant bizness was conjecture :?


function
algebra
domain
inequalities unsolved
inequalities

If f is a convex function and x1, x2, x3 lie in its domain, then

f(x1)+f(x2)+f(x3)+f((x1+x2+x3)/3) \geq 4/3*(f((x1+x2)/2)+f((x1+x3)/2)+f((x2+x3)/2)).

[i]via Titu Andreescu.[/i]

[i][b]A<B(A is less than B)[/b], Kiran Kedlaya, based on notes for the Math Olympiad Program, Version 1.0[/i], page 7.

in this book, convex funtions are defined as follows:

f is said to be convex if
{(x,y)\in R2, y\geq f(x)}
is convex.


Number one:

in three weeks, Jeremy will receive a 15% raise which will bring his yearly salary to (dollers)25,300. What is Jeremy erning now? include a diagram in your solution.

Number Two:

The average of the first of Bob's Algebra 1 test scores was 90, and his average for the five was 82. What wa Bob's overall average?

Heres the first one, didn't really understand the second one. :huh:
[hide]1) 426.3  a week, sorry to lazy for a diagram[/hide]

[quote]The average of the first of Bob's Algebra 1 test scores wa 90, and his average for the five wa 82. What wa Bob's overall average?[/quote]
For this problem, did you mean that the average of the first _ tests is 90?

Yeah i think its the first test was 90 and other five wore 82

[hide="1"]Let $x$ be what Jeremy is earning now. This gives $1.15x=25300$, so he earns $\$22,000$ a year right now.[/hide]

[hide="2"]$\frac{90+5\cdot82}{6}=83\frac13$[/hide]


You can use power of a point because of the tangent line and the secant line can you not?

edit: nevermind i thought they gave you AD.

Actually, I don't think Power of a Point would work in this case because you have two variables.

Notice that the triangle inscribed in the circle has two sides of length 6 and 8. That's enough info for me to guess that the third side is 10, so $AB=18-10=8$.   :P

It turns out that $AD=12$ based on that info, so the numbers are fairly nice too.  :D

[quote="Bictor717"][quote="Altheman"][hide="hint"]find a pair of similar triangles[/hide][/quote]

[hide="Solution"]We have $\angle BDA\cong\angle ACD$ (because both intercept $\widehat{BD}$) and $\angle A\cong\angle A$, so $\triangle ABD\sim\triangle ADC$.
$\frac{AC}{AD}=\frac{CD}{BD}\Rightarrow AD=\frac{27}{2}$
$\frac{CD}{BD}=\frac{AD}{AB}\Rightarrow AB=\frac{81}{8}$ (or use PPT)[/hide][/quote]

AD should be 12 . Therefore, AD^2=AB*AC  AB is 9


geometry
3D geometry
tetrahedron
puzzles

You have 2 identical copies of a fair 4-sided die, each with one of the numbers
4, 4, 4, and 9 on one of the faces of each die.  They have 2 identical copies of a fair 4-sided die, each with one of the numbers 5, 5, 5, and 1 on one of the faces of each die.  These 4 dice are all that are present and available.

You and they start out with 110 dollars apiece. You rolled 22 times in a row, and for each time (round) that a person's roll beat the other person's, the winner collected 5 dollars from the loser.

At the end of the 22 rolls in this instance, you have 100 dollars, and they have 120 dollars.

Now, they offer you a chance to gain the most money back by choosing
2 dice for yourself (from what's present) to roll 22 times against the remaining
2 dice for them (from also what's present).  The rolls of 2 or more dice are added together to give their composite number to compare against each other.

Question:

Which 2 dice out of the 4 dice would you choose to maximize your chances of gaining the most money back from them?

Note:  If you look straight down on a 4-sided die that is resting on a flat surface, 3 faces will show (at a slant), so it is the face on the farther part of the tetrahedron that is out of view that is actually the side that counts for the die's number.

you should keep the same two dice u used for the first 22 rolls


[quote="SUPERMAN2"]Let $ P_k(x) = 1 - x + x^2 - x^3 + ... + ( - 1)^{k - 1} x^{k - 1}$,where $ k \in N$*.Prove that $ n^\sum_{k = 1}\binom{n}{k}P_k(x) = 2^{n - 1}P_n(\frac {x - 1}{2})$[/quote]

I think, the statement is not true. Take x=0. Then $ P_k(0) = 1$. Hence,
$ \sum_{k = 1}^{n}{n\choose k}P_k(0) = - 1 + \sum_{k = 0}^{n}{n\choose k} = 2^n - 1$. On the RHS wqe get for x=0 the expression
$ 2^{n - 1}P_n\left ( - \frac {1}{2}\right ) = 2^{n - 1}\sum_{j = 0}^{n - 1}\frac {1}{2^j} = 2^{n - 1}\cdot\frac {1 + \frac {1}{2^n}}{\frac {3}{2}}.$
But in this case LHS equals not RHS for a general positive integer $ n$.

The general method for the solution of your problem can be done with the help of the fact that for $ x\neq - 1$ we have
$ P_k(x) = \frac {1 + x^k}{1 + x}.$
This implies that
$ \sum_{k = 1}^{n}{n\choose k}P_k(x) = \frac {1}{1 + x}\left (2^n + (1 + x)^n - 2\right ).$
Similarly,
$ P_n\left (\frac {x - 1}{2}\right ) = \frac {2}{1 + x}\left (1 + \left (\frac {x - 1}{2}\right )^n\right ).$


AMC USA(J)MO USAMO MATHCOUNTS geometry HMMT ratio

Here is my annual update of high schools with the most USAMO qualifiers.  I counted all qualifiers from 1996 to 2006.  The schools are listed below in descending order.

To conserve space, I used the code A = 10, B = 11, C = 12, D = 13, E = 14, F = 15, G = 16, and H = 17.

Let me know if you see any errors in this list.

[code]
                      YEAR
               9 9 9 9 0 0 0 0 0 0 0
SCHOOL   STATE 6 7 8 9 0 1 2 3 4 5 6 TOTAL
TJ          VA 5 4 3 4 B E H E E B F 112
Stuyvesant  NY 7 5 1 4 5 9 8 8 4 3 9  63
IMSA	     IL 6 5 3 3 5 8 9 6 6 5 5  61
AAST        NJ 5 2 F 2 2 3 8 5 6 5 7  60
Homeschool  -- 1 3 4 4 9 6 7 5 7 5 8  59
Exeter      NH 1 2 2 1 1 1 4 9 B 8 F  55
Blair       MD 2 3 2 0 5 2 9 8 5 7 8  51
ICAE        MI 1 3 4 4 4 4 5 2 2 2 9  40
Lexington   MA 2 2 1 1 5 6 4 4 3 3 7  38
Gunn        CA 4 3 2 2 2 3 3 3 5 2 5  34
Lynbrook    CA 1 2 1 0 1 4 3 4 3 5 7  31
W. Windsor  NJ 0 0 1 1 4 2 2 4 2 2 7  25
P. Andover  MA 2 2 1 0 2 3 4 4 0 1 6  25
Monta Vista CA 1 1 1 2 5 4 3 2 0 2 3  24
Naperv. N.  IL 1 0 1 3 2 6 2 3 3 1 2  24
Saratoga    CA 0 0 0 0 3 2 2 3 2 2 6  20
Hunter      NY 0 1 2 1 3 4 4 2 2 0 1  20
J. Madison  WI 0 1 1 2 3 2 2 2 2 2 1  18
Mad. West   WI 0 0 2 2 4 3 2 2 1 0 0  16
NCSSM       NC 3 3 0 2 1 1 1 1 1 1 1  15
St. Mark's  TX 0 0 1 2 2 4 3 1 1 0 0  14
Rox. Latin  MA 0 0 0 2 1 3 2 3 0 1 1  13
LB Johnson  TX 4 2 1 1 0 1 0 1 1 2 0  13
Troy        MI 0 0 0 1 2 2 2 1 0 1 3  12
[/code]

Ravi, you have to include the 12 schools so that mine can go on there  :D

Aww, if only you went back one more year our school would have Mathew Crawford's year and we'd have 14 overall :(

What about the qualifiers:total student population ratio? I know TJ has like 2 or 3 times the student population of Exeter.

[quote="chess64"]Blair is magnet.

So Exeter had the most qualifiers this year...[/quote]

But you gotta add on Daniel Li from VA, as most likely he will go to TJ next year.

13-13  :roll:

[quote="filletwho"]Troy High School has had 12 since 1999...I'm sure we've had more from 96-99...oh, tetrahedr0n replied before me. We're talking about the same school.[/quote]Troy High School in Michigan didn't have any qualifiers in 1996 or 1997.  Unfortunately, I don't have the results book for 1998.  From 1999 to 2006, Troy did have 12 qualifiers (specifically 1, 2, 2, 2, 1, 0, 1, and 3).

So, for now, Troy doesn't make the cut. Sorry!

Wow, I just got accepted into Exeter, and will be attending my junior and senior years... How intimidating.  I'm used to being top in my school in most things, but I didn't even qualify for the USAMO this year (If only I could read my own handwriting  :mad:).

What's really impressive is Exeter's depth. They have a bunch of qualifiers going to USAMO, but completely different people who did well at the HMMT. Sherry Gong was there but she was not one of their top scorers! This is good praise coming from someone who goes to Andover ;)

Well Rickards has had 11 since 2002, thats when we first started getting qualifiers (Shaanan Shetty).

I suppose we're just about there with the scoreboard.

2002: 1
2003: 2
2004: 1
2005: 4
2006: 3

We're getting there lol.

Wheeler High School GA -- Yes, we're a magnet school.
2003: 1
All other years: 0

I almost qualified this year. Oh well, I still get a plaque and I can quadruple our school's total.

Because the AMC has invited 57 more students to participate in this year's USAMO, I have updated the list of top schools on the first page of this thread.

In Homeschool, did you count Andrew Ardito? He is listed as Albany Area Math Club, but he's homeschooled. (Certain groups sometimes do not like homeschoolers, thus sometimes it is necessary to go under the auspices of a math club or such.)

[quote="krustyteklown"]Lexington is possibly the best non magnet school.[/quote]
Probably so, but note the cluster of neighboring non-magnet schools from the Silicon Valley just below: Gunn (Palo Alto), Lynbrook (San Jose), Monta Vista (Cupertino), Saratoga.

And not far below them is the Madison, WI school listed as "J. Madison," also known as Madison Memorial, with 18. Of those 18, at least 12 (and likely more) were students named Po-_ Loh. I doubt that you'll find another single family to match that.

[quote="Ravi B"]Oh yeah, thanks, I had forgotten to count Andrew.  I have now added him in to this year's count of homeschoolers.  The funny thing is that I will be Andrew's coach (representing New York) at the MathCounts national competition this year; I should have remembered.

Here are the 8 homeschoolers that I have recorded as USAMO qualifiers this year:  Derrick Sund (AZ), Michael Viscardi (CA), Jeffrey Manning (CA), Mark Pritt (MD), Thomas Snell (NJ), Andrew Ardito (NY), Daniel Li (TX), and Nicholas Zehender (VA).

By the way, TJ and Exeter are now tied for most qualifiers this year: 15.[/quote] :roll:

[quote="Xantos C. Guin"][quote="MCrawford"][quote="Kent Merryfield"]
And not far below them is the Madison, WI school listed as "J. Madison," also known as Madison Memorial, with 18. Of those 18, at least 12 (and likely more) were students named Po-_ Loh. I doubt that you'll find another single family to match that.[/quote]

Don't speak too soon.  The Benjamin family in Indiana is just getting started.[/quote]

Do David and Nathan have a younger sibling?[/quote]

2 of them, as far as I know; Anna and Seth I think, or is it Sean? (thus DNAS)

[quote="randomdragoon"][quote]Is there any explanation for the "clumping" that you sometimes see in regular public schools though? Is it possible that a strong coach or one very strong player can lead to several students qualifying for USAMO? [/quote]

Statistical flukes.[/quote]

I think that schools that perform well attract smarter parents with smarter kids, who perform well/beter, and the cycle continues. Lexington is an example, and borders my town, where there is no math team, but the standardized test scores are better. Also, 90% of the students in my school participate in sports, and 75% year round.  :huh:  :huh:  :huh:

[quote="rrusczyk"][quote="Ravi B"][quote="ICAEmichigan"]ICAE is a weekend enrichment program in Micihgan. The students learn at ICAE during weekends but compete in AMC through their home schools. ICAE has put 40 students in USAMO since 1996. (http://www.icae.org)

1 3 4 4 4 4 5 2 2 2 9

However, the only student showing up as an ICAE student is this year is - Randy Jia, a 7th grader who could not compete from his home school.[/quote]Congratulations!  I am an admirer of ICAE, and am impressed at the success rate of its students. [/quote]

We at AoPS are ICAE fans, too.  Anyone who is very interested in building a community program should take a close look at how ICAE did it.  They've had phenomenal success, and made a big impact on a lot of students in their area.  (I had the good fortune of visiting ICAE last year during National MATHCOUNTS in Detroit.)[/quote]

Why is he listed under ICAE though? If it's just a weekend program, shouldn't he be listed under his middle school?

The Benjamins have been very impressive so far, but with just 4 of them they can't match the Po _ Loh legacy :P

[quote="solafidefarms"][/quote][quote="Xantos C. Guin"][quote="MCrawford"][quote="Kent Merryfield"]
And not far below them is the Madison, WI school listed as "J. Madison," also known as Madison Memorial, with 18. Of those 18, at least 12 (and likely more) were students named Po-_ Loh. I doubt that you'll find another single family to match that.[/quote]

Don't speak too soon.  The Benjamin family in Indiana is just getting started.[/quote]

Do David and Nathan have a younger sibling?[/quote][quote="solafidefarms"]

2 of them, as far as I know; Anna and Seth I think, or is it Sean? (thus DNAS)[/quote]

Oh. I was wondering what DNAS stood for and never figured it out.

[quote="Phelpedo"][/quote][quote="rrusczyk"][quote="Ravi B"][quote="ICAEmichigan"]ICAE is a weekend enrichment program in Micihgan. The students learn at ICAE during weekends but compete in AMC through their home schools. ICAE has put 40 students in USAMO since 1996. (http://www.icae.org)

1 3 4 4 4 4 5 2 2 2 9

However, the only student showing up as an ICAE student is this year is - Randy Jia, a 7th grader who could not compete from his home school.[/quote]Congratulations!  I am an admirer of ICAE, and am impressed at the success rate of its students. [/quote]

We at AoPS are ICAE fans, too.  Anyone who is very interested in building a community program should take a close look at how ICAE did it.  They've had phenomenal success, and made a big impact on a lot of students in their area.  (I had the good fortune of visiting ICAE last year during National MATHCOUNTS in Detroit.)[/quote][quote="Phelpedo"]

Why is he listed under ICAE though? If it's just a weekend program, shouldn't he be listed under his middle school?

The Benjamins have been very impressive so far, but with just 4 of them they can't match the Po _ Loh legacy :P[/quote]

How many Po_ Loh's are there?

[quote="JSteinhardt"][quote="krustyteklown"]Lexington is [i]possibly[/i] the best non magnet school :D

(pending the status of Stuyvesant and Blair...)[/quote]

I think we can go ahead and say that it [i]is[/i] the best non magnet school. I was actually gonna go there except instead we moved to Virginia so I ended up going to TJ. Really though, I'm surprised they don't win more things like HMMT, etc.; I have a lot of friends there and they (the school) definitely should be doing well in stuff like that.[/quote]

One of the bigger reasons we didn't win HMMT this year was because our registration got botched...
We're looking to make a comeback next year tho :D

(who do you know at LHS? PM)

[quote="krustyteklown"]

One of the bigger reasons we didn't win HMMT this year was because our registration got botched...
We're looking to make a comeback next year tho :D[/quote]

hahaha yeah that's always somewhat of an impediment


hi,sorry,it's my first time on this and i couldn't find superscript,I want to find Maclaurin's series for e to the power of -kh up to and including the term h to the power of 3,hope this makes it a little clearer,thanks..Stuart.

From $ e^x\equal{}\sum_{n\equal{}0}^{\infty}\frac{x^n}{n!}$ it follows that $ e^{\minus{}kh}\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}k)^n h^n}{n!}$. If we keep only the terms up to $ h^3$, the answer is $ 1\minus{}kh\plus{}\frac{k^2}{2}h^2\minus{}\frac{k^3}{6}h^3$.


trigonometry
inequalities unsolved
inequalities

Let $n$ be natural number. Prove that: $\displaystyle \sum_{i=1}^{3n}|Sin(i)| >\frac {8n}{5}$

prove this first:

$\forall x$ it holds that $|\sin x| + |\sin (x+1)| + |\sin (x+2)| > 2\sin 1$


[u]Prove[/u]
Let $ f: X\rightarrow Y$ and $ g: X\rightarrow Y$ be continous functions from a topological space $ X$ into a Hausdorff space $ Y$.
Then  $ A\equal{}\{x: f(x)\equal{}g(x)\}$ is a closed subset of $ X$

we have to show that $ X\minus{}A$ is open. let $ x\in X\minus{}A$. then $ f(x)$ and $ g(x)$ are distinct points in $ Y$. so by the fact that $ Y$ is Hausdorff, there exist open subsets $ U$ and $ V$ of $ Y$ with $ f(x)\in U, g(x) \in V$. then $ f^{ \minus{} 1}(U)\cap g^{ \minus{} 1}(V)$ is an open neighborhood of $ x$ and is contained in $ X\minus{}A$. thus every point in $ X\minus{}A$ has an open neighborhood contained in $ X\minus{}A$, hence $ X\minus{}A$ is open.

A very useful alternate definition of $ X$ being Hausdorff is that $ \Delta : \equal{} \{(x,x) : x \in X\} \subset X \times X$ is a closed subset. (You should prove these are equivalent.) Then $ A \equal{} (f \times g)^{\minus{}1}(\Delta)$, where $ (f \times g)(x) \equal{} (f x, g x).$ Therefore $ A$ is closed.

(You should also search the forum before posting; this has definitely been asked before.)


How many ordered triples $ (x_{1},x_{2},x_{3})$ are there such that $ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}10$ given that $ x_{1},x_{2}$ and $ x_{3}$ are whole numbers?

Let's say we have $ 10$ balls in a line.  We can place $ 2$ dividers among the balls so each space between the dividers represents the value of $ x_1, x_2$ and $ x_3$.  We have $ 12$ spots where we can put the two dividers, and we need to choose $ 2$ of them.  Our answer is thus $ \boxed{\binom{12}{2}}$.


Let $ a_n$ be a sequence defined by $ a_1\equal{}3,\ a_{n\plus{}1}\equal{}2a_n^2\minus{}1\ (n\equal{}1,\ 2, \cdots)$.

Find $ \lim_{n\to\infty} \left(1\plus{}\frac{1}{a_1}\right)\left(1\plus{}\frac{1}{a_2}\right)\cdots \left(1\plus{}\frac{1}{a_n}\right).$

Well, the limit equals $ \sqrt{2}$, but I got this experimentally.

[hide]
Let $ \alpha=3+2\sqrt{2}$ so that $ a_0=\frac{1}{2}( \alpha +\frac{1}{\alpha})$. Then $ a_n=\frac{1}{2}( \alpha^{2^n}+\alpha^{-2^n})$ for all positive integers $ n$, since
\[ 2[\frac{1}{2}( x+\frac{1}{x})]^2-1=\frac{1}{2}( x^2+\frac{1}{x^2})\]
As a result,
\[ 1+\frac{1}{a_n}=1+\frac{2\alpha^{2^n}}{\alpha^{2^{n+1}}+1}=\frac{( \alpha^{2^n}+1)^2}{\alpha^{2^{n+1}}+1}\]
\[ =\frac{\alpha^{2^{n+1}}-1}{\alpha^{2^{n+1}}+1}\cdot \frac{\alpha^{2^n}+1}{\alpha^{2^n}-1}\]
So the product telescopes, and we get
\[ \prod_{k=1}^n ( 1+\frac{1}{a_n}) =\frac{\alpha^{2^{n+1}}-1}{\alpha^{2^{n+1}}+1}\cdot \frac{\alpha+1}{\alpha-1}\]
Since $ \alpha >1$, the first fraction tends to 1 as $ n\rightarrow \infty$. The limit is then
\[ \frac{\alpha+1}{\alpha-1}=\frac{4+2\sqrt{2}}{2\sqrt{2}+2}=\boxed{\sqrt{2}}\]
[/hide]


Does anyone know where I can find the [b]official[/b] solutions to some of the USAMO problems? I have solutions for 2002 to 2004 only. I am looking for solutions to the problems from 1989 to 2001 if possible. Thanks a lot..

Same here... if anybody knows, please post. Thanks.


geometry
3D geometry
tetrahedron
induction
combinatorics unsolved
combinatorics

Let $ ABC$ be an equilateral triangle with side length equal to $ N \in \mathbb{N}.$ Consider the set $ S$ of all points $ M$ inside the triangle $ ABC$ satisfying

\[ \overrightarrow{AM} \equal{} \frac{1}{N} \cdot \left(n \cdot \overrightarrow{AB} \plus{} m \cdot \overrightarrow{AC} \right)\]

with $ m, n$ integers, $ 0 \leq n \leq N,$ $ 0 \leq m \leq N$ and $ n \plus{} m \leq N.$

Every point of S is colored in one of the three colors blue, white, red such that

[b](i) [/b]no point of $ S \cap [AB]$ is coloured blue
[b](ii)[/b] no point of $ S \cap [AC]$ is coloured white
[b](iii)[/b] no point of $ S \cap [BC]$ is coloured red

Prove that there exists an equilateral triangle the following properties:

[b](1)[/b] the three vertices of the triangle are points of $ S$ and coloured blue, white and red, respectively.
[b](2)[/b] the length of the sides of the triangle is equal to 1.

[i]Variant:[/i] Same problem but with a regular tetrahedron and four different colors used.

I think the following solution is correct but I'm not 100% sure.

We prove by induction. It's easy to show for small values of $N$ that it must be true.

Suppose $X$ is any point in $S$ such that $Y = X + \vec{AB}/N$ and $Z = X + \vec{AC}/N$ are both in $S$. We call the triangle enclosed by $XYZ$ a "good" triangle. Note that all good triangles are unit triangles and they all have the same orientation.

We now assume that there is no unit triangle that has 3 differently colored vertices. So that means for each unit triangle, there is a "majority" color among the vertices. For instance, 2 blue and 1 red (which means the majority color is blue) or 3 white (which means the majority color is white) and so on. For each GOOD triangle, color the triangle according to its majority color among the vertices.

Note that in order for a unit triangle to have a majority color blue, at least two vertices must be colored blue. Thus, for any good triangle with a side on $AB$, its majority color cannot be blue because the two vertices on $AB$ cannot be blue. Its majority color is either red or white. Similarly, for any good triangle with a side on $AC$, its majority color is red or blue, and on $BC$, blue or white.

If we know create a new triangular lattice with points on the CENTERs of the good triangles, and we color those centers according to the majority color of the corresponding unit triangle, we get a new tri-colored lattice. This new lattice satisfies the conditions of the problem for $N-1$. That is, on $AB$, the vertices are colored red or white, and so on. The inner vertices are freely colored. Due to induction hypothesis, there must be a tri-colored unit triangle here, and we shall show that it leads to our tri-colored unit triangle in the original lattice.

Suppose in the new lattice, the tri-colored unit triangle is $PQR$ where $P$ is blue, $Q$ is white, $R$ is red. Consider three good triangles in the original lattice whose centers are $P,Q,R$, and one non-good triangle in the middle. There are 6 points total in the original lattice. Since we need at least 2 blue points in the good triangle corresponding to $P$, at least 2 white points in the good triangle corresponding to $Q$, and 2 red points in the good triangle corresponding to $R$, thus among 6 points, there are exactly 2 blue 2 white and 2 red points. There's a limited way of doing it. It's hard to explain here but you can try it on paper and see what I mean. Eventually there's only two possible configurations such that it's possible, and in both cases, the inner non-good triangle has tri-colored vertices.

[quote="hendrata01"]
If we know create a new triangular lattice with points on the CENTERs of the good triangles, and we color those centers according to the majority color of the corresponding unit triangle, we get a new tri-colored lattice. This new lattice satisfies the conditions of the problem for $N-1$. [/quote]

You have to be careful because the triangle lattice formed on the centers of the triangles isn't a lattice of the same form as $S$.

This problem is a slightly weaker form of [url=http://en.wikipedia.org/wiki/Sperner's_lemma]Sperner's lemma[/url]. Wiki gives a nice proof.

[quote="ocha"]

You have to be careful because the triangle lattice formed on the centers of the triangles isn't a lattice of the same form as $S$.

This problem is a slightly weaker form of [url=http://en.wikipedia.org/wiki/Sperner's_lemma]Sperner's lemma[/url]. Wiki gives a nice proof.[/quote]

I think it actually is. We only consider the centers of the good triangles, that is, the unit triangles that have same orientation as the big triangle. We don't consider all centers of unit triangles.


If I roll a fair, regular six-sided die four times, what is the probability that I will roll the number $ 1$ exactly three times?

There are four rolls, out of which you can pick 3 to have 1, so 4C3. Multiply that by 1/6^3(probability of rolling 3 1's) and 5/6(probability of not rolling a 1) to get 5/324

[hide=Solution]The answer is simply just
$$\frac{5 \cdot \binom{4}{3}}{6^4}=\frac{5}{324}$$


Heights are all same for all triangles on height AH (same goes for height CH) so therefore we need only to calculate base ratios.

BR=1.5RS, SH=0.5RS. Thus BH=3RS, or 33.3%.


inequalities
inequalities solved
Sides of a triangle
3-variable inequality
Symmetric inequality
Japan
2001

We have $\left(\sum a\right) \left(\sum a^3\right)\ge\left(\sum a^2\right) ^2$. So we have to prove that if $x,y,z$ are sides of a triangle (eventually degenerate) then $(x+y+z)^3\ge 4(x^3+y^3+z^3)$. So, we need to prove that if $a,b,c\ge 0$ then $(2a+2b+2c)^3\ge 4\sum (a+b)^3$, or $2\left(\sum a\right) ^3\ge \sum (a+b)^3$, or $6(a+b)(b+c)(c+a)\ge 3\sum ab(a+b)$, which is obvious.

[hide="Solution"]Without loss of generality, suppose that $a=\max\{a,b,c\}$. This implies $a\ge b, a\ge c,$ and $b^2+c^2\ge a^2$.

Now, $(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)(a^2+b^2+c^2)^2\ge 4a^4(a^2+b^2+c^2)$$=4(a^6+a^4b^2+a^4c^2)\ge 4(a^6+b^6+c^6)$, as desired. $\blacksquare$[/hide]

[quote=nickolas]Three nonnegative real numbers satisfy $a,b,c$ satisfy $a^2\le b^2+c^2, b^2\le c^2+a^2$ and $c^2\le a^2+b^2$. Prove the inequality
\[(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3)\ge 4(a^6+b^6+c^6).\]
When does equality hold?[/quote]
[b]When does equality hold?[/b]

In sqing's solution, the last two line holds only when $a=b=c=0$, so that is the equality case.

Assume $\sum a^2 =2$. Then $a,b,c \leq 1$ by given conditions. We have:
$$\left( \sum a \right) \left( \sum a^2 \right) \left( \sum a^3 \right) \geq \left( \sum a^2 \right)^3 = 4 \left( \sum a^2 \right) \geq 4 \left( \sum a^6 \right)$$

With equality iff $a, b, c =0,1$ and $\sum a^2 =2$, combining which and normalising out gives $(a,b,c) = (0,k,k)$ and permutations :)


So, who is planning to go to HMMT?

I might be able to get a ride there if (a) there are enough people going to convince the driver it's worth it and (b) there's some convenient time for us to all meet up (lunch?).

I'm going!  Let's see...there's me, Russell, Emil, Nick, Gabe, Eddie, Benno, Rachel, Sarah, Josh, Chetak, Elizabeth going on the Yellow Pig and Chlorine (I think) teams.  Then there's some NYC people going I would assume, presumably Adam S, Danny Y, Kevin, Jin, David...I assume AAST will be there, so Ian and Mark...yeah, I think that's about it.  So close to 20 people from HCSSiM 2007 (I can't speak for any other year).  I'd say that's worth it  :lol: Lunch would definitely be a convenient time, from what I remember of last year there's not that much to do during lunch.

Alas, we Chicago folk have math team regionals on that day and thus will be unable to attend. Best of luck to everybody.

Okay, I'll be there---so where should we all meet?  And are folks wanting to get lunch out somewhere, or doing the pizza order thing?  (I want to know ahead of time so I know whether to bring food with or not.)  I don't know how many people are reading this thread at the moment, so can someone propagate whatever decisions are made through the HCSSiM teams?

Photos from yesterday are posted at http://picasaweb.google.com/portablepig
but there were a couple of people there not in any of the photos, sadly.


geometry AMC AIME trigonometry USA(J)MO USAMO

I have AoPS Volume I,II, Challenging Problems in Geometry, and Mathematics of Choice. I covered most of the AoPS Volume I and bit of Volume II but almost no part of Challenging Problems in Geometry and Mathematics of Choice. I know my weakest parts are geometry and counting (I'm pretty strong at algebra. and NT.. I'm able to answer most of the questions in AMC-12, at least more than amount I can do in geometry and counting).

So, here is my question: For those who studied with at least some or all of the books above, how would you recommend studying those books? Key chapters are also preferred and if possible, the way to attack the books (especially last two) in quick time. I like both books CPIG and MC but they take awful long time to solve it and I want to cover important parts before AMC so I can do well on it.

Or general advice on developing geometry and counting are also welcome..

[quote="probability1.01"]Hmm... I have CPIG, and I think it's too hard for AMC. It's suitable more for AIME level geometry. I don't think AMC tends to have much geometry though, so I'm not sure exactly what you're trying to study. Most AMC geometry can either be bashed by trig or is relatively simple.[/quote]

Well, even few geometry can make my score higher. My goal is to get at least 120 in AMC-12. I'm currently getting around 112 and among them, I barely solve any geometry. I even got two questions on 21-25 which were NT and algebra..

I don't think counting will immediately help me since I'm not naturally good counter.  :D But I have many geometry problems so I need tip on doing geometry.

I think I'll work on AoPS volume I again thoroughly and start going on AMC geometry with CGIP. I don't know how much I'll get done but hey, it's a worth a try.  :D

Any advice? Like from beta who I believed got most of the AIME geometry problems? :D

Volume 1 and Chapters 2, 3, 4, 5 (5 probably wont show up much though) should be sufficient, content wise, to cover [i]any[/i] AMC problem.  After that just go through and practice old AMC questions.  After that you should have mastered AMC geometry...move up to AIME geometry, etc.

is the geometry is aops 2 sufficient for amc 12 or aime geometry, or do i need to get more books?

The basic tools needed for the AMC/AIME is covered in the AoPS books.  For the AIME however, you will need to practice additional problems.  Challenging Problems in Geometry is one good option.  My favorite option is just going through old AIME's :D

Joe, just quick question.

I remember "most" of the concepts in AoPS Volume I. I don't know if it will do me any good if I read over the basic sections. So, should I just focus on problem part? I don't understand why it takes so much time for me to solve AMC geometry even I know the most of the concepts. Does this mean that I should simply get a lot of geometry problems and work on them? Or should I read the lectures?

 :?

[quote="joml88"][quote="DPopov"]I would say practice, practice, practice! You are probably familiar with most if not all of the concepts but you probably don't have enough experience in applying them when it comes to problems, so the more problems you attempt the better you will become in seeing what needs to be used.[/quote]

I agree completely.[/quote]

Thanks. I guess I was asking an easy way to raise my geometry and counting skills without actually practicing them.

Um, geometry can be solved that way.. But I have really really bad counting skills. Is there anyway I can develop it? I mean, if I know the concepts, I can practice problems. Yet I have really low comprehension on counting.

Any idea?..

[quote="Silverfalcon"]Thanks. I guess I was asking an easy way to raise my geometry and counting skills without actually practicing them.
[/quote]

heh wouldn't we all like to improve without practice...

[quote="Silverfalcon"]Yet I have really low comprehension on counting.

Any idea?..[/quote]

Take Intro Counting. It really helps. I was in the same position as you so I took it as my first AoPS class last summer. (Don't worry about the "intro". We solved a USAMO problem in that class.)

If you don't want to learn by solving a lot of problems, I've found that being with friends who share a similar interest in math and are better or worse in certain areas can be most advantageous to learning.  Depending on your school or environment, this might be difficult.


Let $ {x,y,z}\in{R^\plus{}},k\in{N},$prove (or disprove) :

   \[ \sum{\sqrt{\frac{x^{k\plus{}1}}{(2^{k\plus{}1}\minus{}3)x^k\plus{}(y\plus{}z)^k}}}\leq{\sqrt{\frac{x\plus{}y\plus{}z}{2^k\minus{}1}}}.\]

  Note. The inequality holds for k=1,2,3,4,5.

[quote="fjwxcsl"]Let $ {x,y,z}\in{R^ \plus{} },k\in{N},$prove (or disprove) :
\[ \sum{\sqrt {\frac {x^{k \plus{} 1}}{(2^{k \plus{} 1} \minus{} 3)x^k \plus{} (y \plus{} z)^k}}}\leq{\sqrt {\frac {x \plus{} y \plus{} z}{2^k \minus{} 1}}}.
\]
Note. The inequality holds for k=1,2,3,4,5.[/quote]
I think it is hard.


What is the value of the expression
\[ \frac {1}{\log_2 100!} \plus{} \frac {1}{\log_3 100!} \plus{} \frac {1}{\log_4 100!} \plus{} \cdots \plus{} \frac {1}{\log_{100} 100!}?
\]$ \textbf{(A)}\ 0.01 \qquad \textbf{(B)}\ 0.1 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 10$

[hide="Solution"]

Yay!

$\log_2 100! = \frac{\log{100!}}{\log{2}}$

$\frac{1}{\frac{\log{100!}}{\log{2}}} = \frac{\log{2}}{\log{100!}}$

Continue this. The denominator is always $\log 100!$. $\log a + \log b = \log{ab}$. Thus, the numerator is $\log 100!$.

The answer is 1.

[/hide]

Ha.  They put this exact question on this year's MMPC.

[quote="JesusFreak197"][hide="Answer"]$\frac{1}{\log_2100!}=\frac{\log 2}{\log 100!}$
Therefore, the expression sums to $\frac{\log 100!}{\log 100!}=1\Rightarrow \boxed{C}$.[/hide]

That's a little too easy for a #22 for anyone who knows anything about logs. :huh:[/quote]
I agree - the fact that I was able to do it in my head after glancing at it means that it was a bit too easy  :P But I suppose logs are considered an "advanced" concept, and maybe back in 1998 they weren't taught nearly as much.

Hmm...I can't understand why they would make this a number 22 :o

[quote="Iversonfan2005"][quote="beta"]Remember it was on #22 on AHSME, and it has 30 questions instead of 25. So this is like a #18 on current AMC12.[/quote]

Still...I feel AHSME problems (20-30) are MUCH harder than AMC problems 18-25...[/quote]

Agreed. I can do pretty much every problem after about 2000 but I find ones I can't do from before that point.

[quote="Iversonfan2005"]

Still...I feel AHSME problems (20-30) are MUCH harder than AMC problems 18-25...[/quote]

Huhhh??? Well the AHSME is supposed to really similar to AMC12(and it is).

Haha this problem literally takes a second if you know that [hide]$\frac{1}{\log_a b}=\log_b a$.[/hide]

[quote="Pakman2012"]Cmon you guys.. of course this problem was a 22.
I bet half the problem was figuring out what a log was in math terms.
Square root of tree...

Everyone knows logs weren't invented until 2000.

(this is a joke)

But yeah, good problem, fun, easy stuff.  Maybe it was just a freebie.[/quote]Not to spam, but this is too funny not to comment on.

 :rotfl:  :rotfl:  :rotfl:  :rotfl:

Edit: The Math League number 6 today was harder than this problem. ;)

[quote="4everwise"]

Edit: The Math League number 6 today was harder than this problem. ;)[/quote]
That was a fun problem!  I'm not sure if I'm allowed to post it here yet, but it's definitely taking a look at.

The only downside is if you have a TI-89, you could just plug in the sum and do the "approximate" command, and it would give you the correct answer; I feel like they should ban that calculator from the Math League exams, since some problems can be brute-forced using the 89's features.

[quote="zanttrang"]The only downside is if you have a TI-89, you could just plug in the sum and do the "approximate" command, and it would give you the correct answer...[/quote]

:blush: Yeah, that's what I did...I got a perfect score. I still had about 15 minutes left when I finished so I could have figured it out the real way if I wanted to, but I had a ping-pong tourney to go to... :roll:

Guys, this isn't really an easy problem if you're beginner (all people above who posted obviously know math well and members of AoPS).

If you give this to regular student, you can't guarantee that he/she will definitely get it. $\log_a b = \frac{1}{\log_b a}$ isn't very hard to notice but it's certainly not obvious to everyone.

I remember seeing #21 or something in 2002 AMC-12 which used same logic.

Well, it's a #22. An average student isn't [i]supposed[/i] to get it. On the other hand, it's too easy to be among the last five, because those are supposed to be more difficult. I mean, you shouldn't be able to solve it in less than 30 seconds, but you can solve this in 1 second. ;)

EDIT: fixed typo

Personally, I thinks the AMC thinks logs are too hard. In 9th grade, I somehow managed to qualify for the AIME. I solved a total of 2 problems: #1, which I actually brute forced by dividing the numbers, and #11, the log problem, which I managed to solve in 5 minutes (and I had only learned geometric series in math class a few days before the AIME). So, yeah, I think the log problems are way too easy.

The AIME was 2004 I, I think.

Edit: It was #12 I solved, not #11.

Well, I wouldn't call that question very easy (although it's easy for a number 12), but it's still pretty difficult for many aime participants

[quote="4everwise"]Edit: The Math League number 6 today was harder than this problem. ;)[/quote]
That was a nice problem. I started out the same way for this problem, then I realized F(1) is very very similar to F(2005).

I remember this problem from a practice AMC 12... and I remember having extra time and staring at it to see if there was a catch somewhere in there

I'm paranoid like that  :?


function
number theory proposed
number theory

[quote="spx2"]let $q$ be an integer
$S=\sum \left(i+\frac{q}{i}\right)$ , $(\forall)i $ so that$ i|q$ .
how do i find a formula for S ?[/quote]
Let $q=p_1^{e_1}...p_k^{e_k}$.
$\sum i = \prod_{i=1}^{k} (1+..+p_i^{e_i})$
$\sum \frac{1}{i} = \prod_{i=1}^k (1+...+\frac{1}{p_i^{e_i}})$
Put them in and we can do some simplification i bet.

i just made a correction
sorry i didn't put it in the first place
i'm really tired so this may be the reason
this is a supplimentary condition
$i\leq \left[ \frac{n}{i}\right]$
i figgured this would also be equivalent to saying $i\leq \sqrt{n}$

and also when does this happen ? -> $S(n)=s(n)$
where $S(n)$ is the sum described above and $s(n)$ is the sum of
all divisors of $n$

Did I miss something or $S = 2 \sigma (q)$?
where $\sigma (q)$ is the sum of positive divisors of $q$.

Pierre.

You wrote it during I was writing my message...so yes $S = \sigma (q).$

Pierre.

[quote="spx2"]$siuhochung$ i really dont understand how what you wrote can help me ...
why did you wrote $S$ as beeing a factorisation ?
how does that help me ?
and the last product i really dont understand[/quote]
Recall the divisor function so actually $S = 2 \sigma(q)$ as Pierre has said. Pierre, do you mean $S = 2 \sigma (q)$ in the last post?

p.s. i've modified my last post. There were too many errors.

[quote="pbornsztein"]Not anymore, since he adds the condition $i \leq \sqrt n.$

Pierre.[/quote]

oh, i didn't read his second post...
Your original expression $S= 2 \sigma (q)$. Adding $i < \sqrt{n}$ divide the sum into two equal and symmetric parts, so the new sum (the one you wanted actually) equals $\sigma (q)$, or the s(n) you defined.

[quote="DCo"]You're ignoring the interesting case: if q is a square, you have to add a root q to Siu's formula.  Otherwise his formula is correct and best.[/quote]
why?

[quote="DCo"]Say q = 9.  We have s = (1+9)+(3+3) =/= 1+3+9 = S(n).
The three gets repeated, as does, in general, $q^{.5}$ when it is an integer.[/quote]
What is your definition of s(n)? I think when we evaluate $\sigma _1 (n)=\sum_{d|n} d$ we won't count same factor twice.

"and also when does this happen ? -> S(n)=s(n)
where S(n) is the sum described above and s(n) is the sum of
all divisors of n"

I confused S and s.  But still, S(n) != s(n) when n is a square (because S is bigger)


[b]If you love Physics, I'm sure you're searching for books and articles and journals that are regarded as the best in the world in their respective fields![/b]
[hide]( I know this coz I am in this category!!!)[/hide]
So come on. [b]List all the best books you know in various fields[/b]. Also please include a link(possibly at amazon) as well as a grading - [b]Basic(Elementary), Intermediate and Advanced. Problem books [/b]and any other software etc. related to Physics are most welcome!!!
Possible Categories are (these can be changed later!):
[b]General/Expository
Mechanics
Thermodynamics and Statistical Physics
Electrodynamics
Optics
Modern Physics( Relativity, fundamental paricles)
Quantum World( QMechanics, QField Theory, QElectrodynamics.. etc)
Problem Books
Experiment Books[/b]

[size=150][b]Please suggest the best!!![/b][/size]

No one has any suggestions or queries???

Problem Books

I think this is nice :

"200 Puzzling Physics Problems" by Peter Gnadig, G. Honyek, K. F. Riley

[quote="eminem"]ok SHADYSAYSURSPAMMED YOU START !!, tell your books!! :lol:  :lol:  :rotfl:  :rotfl:

hey rushil how about you tell some of YOUR OWN books which you use  :lol:  ;)  before we start telling?[/quote]

Cmon eminem,why the $giveandtake$ attitude,the guy needs help...and i'll tell my sources right away rushil,as soon as eminem tells me where he got those 2 sums he posted in intermediate forum recently :D  ;)

Stop playing games!!! And I'm not asking for books like HC Verma or Arihant or Resnick. I am asking for topic specific advanced books.

Hey , ineqlover. I already have the excellent book "200 Puzzling Physics Problems" It's truely wonderful, isn't it!!! Thanks!

hey shady , our books would be toooooooooooooo $LOWLY$ for our friend Rushil here, ya know ! :P  ;)  ;)  :P  Or maybe you  do have some weird $\" TOPIC SPECIFIC ADVANCED BOOKS\"$ for him ........ :rotfl:  :rotfl:  :rotfl:

as for the intermediate questions (HEHEeheehhehehehe) ,forget it !

if u like irodov's problem book(i know i do),you can get his theory books on mechanics and electromagnetics which deal with the subject matter at almost the JEE level(if not higher).
Most 'advanced ' books do not really contain stuff that would be useful to ppl preparing for JEE or olympiad. Though some books have problems that bear strong resemblances to actual JEE and olympiad questions. if ur interested look up in ur local library.

I like Physics: Calculus by E. Hecht. It's the only physics book I have and I know it's too easy for you :D I'm using it to self-study for the AP Physics B exam.

[quote="Rushil"]I knew I'd have such a response from you!!!! I fyou were offended, sorry!!!!

My basic point was that if you were going to recommend those, better not fro I already know ! Also, those books are mainly(as far as I know) limited to India.  Of course, all your recommendations are welcome!!![/quote]

ok rushil, truce

hey do you have

morrison boyd - organic chemistry
robinson odom holtzclaw - general chemistry
tipler- physics

I have morrison - the standard organic text!!!

Never heard of holtzclaw --- please give more details!!!!

Tipler- saw it , too basic, almost equivalent to Resnick -- already have Resnick and also Young!!

Keep the suggestions coming!!!

ok, how about these???

fundamentals of physics-  Ivanov

college physics - Miller

do you use MTG magazines and books ??

Haven't seen Miller and seen Ivanov but not much impressed. There was another Physics book by some Russian I don't remember that was very good!!

No, I don't use MTG stuff. I saw a Math magazine once but ... um... dunno , didn't quite get up to the mark. How are they???

Rushil,where u got that book(200 puzzling)?
I can't get it here in Australia.
My friend in Malaysia also can't buy it..

I got that book from India, It's available here  - though you need to know where to find it. Australia should have some good online stores. Why don't you try those? Maybe your friend in Malaysia can get it from Singapore - I think most books are available there!

Oh!!! Of course, yo can never forget those!!!

I think this is quite nice :

OPTICS by Eugene Hetch.

Just a correction * Eugene Hecht is the author.

I have that too . Its good - concepts and new ideas are well explained. Though at times, the talk of history and repeating the same thing over and over to make it 'clear' sort of bores me, on the whole , keeping in mind the exercises and explanations etc. , Hecht for Optics is a very good book ( by the way, it is recommended by MIT and Princeton!) . A comparable book in India (don't know elsewhere) is Ajoy Ghatak's Optics. This book takes a more mathematical approach to things rather than explaining more of their 'physical meaning' ( sort of like Griffiths' Electrodynamics!) . It's got mainly derivations and difficult differential equations. on the other hand, Hecht takes up a more 'explanatory ' role while not sacrificing the math a teeny bit (that's why it is quite big!) Interesting concepts like the actual mechanism behind slowing of light in a medium are explained to great depths and clarity!

Hey, IneqLover, which class are you in??? If in college, which one!

Rushil,please tell me where to get the "200 physics problms book" PLEAs!!E

Preparing for IPhO
Do u think "Fundamantals of Physics" by Halliday is enough?
Or,"Feynman Lectures On Physics" better?

Krane is better than halliday though i don't know the level of ipho,any site for ipho problems?


inequalities
LaTeX
inequalities proposed

Prove for positive $a_i$s

$\frac{a_1}{a_2+a_3+...+a_n} + \frac{a_2}{a_1+a_3+a_4+...+a_n} + ... + \frac{a_n}{a_1+...+a_(n-1)} \geq  \frac{n-1}{n}$

P.S. the last one is supposed to be a_(n-1) but I'm not sure how to edit that in LaTeX!  :D

[quote="georgesand"]
P.S. the last one is supposed to be a_(n-1) but I'm not sure how to edit that in LaTeX!  :D[/quote]

a_{n-1} -> $a_{n-1}$ ;)

Let $S = \sum a_i$.

$\sum \frac{a_i}{S-a_i} =\sum \frac{S}{S-a_i} - n \geq \frac{n^2}{n-1} - n = \frac{n}{n-1}$ by AM-HM, qed.

You can show this inequality ie. by Jensen, too.


AMC
AIME
modular arithmetic
arithmetic sequence
number theory

[hide]if you add one and take the square roots of these terms, we get the sequence: 2, 4, 5, 7, 8, 10, 11, i.e. all positive integers :ge:2 excepting multiples of 3. Well, the 1994th term of this will be the 997th term of 4, 7, 10, ... (every second term) which is a normal arithmetic sequence with a=4 and d=3. a+(n-1)d=4+(996)3=2992, so the answer is 2992:^2:-1, which is definitely not an AIME answer.[/hide]

Sorry, it's the remainder when divided by 1000 to get an AIME answer.

[hide="Solution"]
We have $3m = n^2 - 1 = (n-1)(n+1)$, so $m\equiv 1 \text { or } 2 \pmod {3}$. So if we let $f(x) = x^2 - 1$, we get the sequence $\{a_n\} : = \{ f(2), f(4), f(5), f(7), \ldots \}$. It will be easier to deal with if we place $f(1)$ in the beginning of the sequence, even though it's not in the sequence because $f(1) = 0$, which isn't positive. So then we get $\{b_n\} : = \{f(1), f(2), f(4), f(5), f(7), f(8), \ldots \}$, which we need the 1995th term of. If we split this into 2 sequences, we get $\{c_{n}\} : = \{ f(1), f(4), f(7), \ldots\}$ and $\{d_n\} : = \{ f(2), f(5), f(8), \ldots \}$. The 1995th term will be in $\{c_n\}$, which has all the odd-numbered terms of $\{ b_n\}$. Since 1995 is the $(1995+1)/2=998$th odd number, it is the 998th term of $\{ c_n\}$, which is $1+3(997)$. Hence we need $f(1+3\cdot 997) \pmod {1000}$, or

\begin{eqnarray*} (1+3 \cdot 997)^2 -1 &\equiv & (1+3\cdot (-3))^2 -1 \pmod {1000}\\ &\equiv & \boxed {63} \pmod {1000}. \end{eqnarray*}
[/hide]

[hide="Answer"]Every perfect square $n^2\equiv 0,1\bmod{3}$. Those $\equiv 1\bmod{3}$ are all of the ones where $n$ is not divisible by $3$. However, we do not count $n=1$, since we want positive multiples of $3$. Therefore, the $1994^{th}$ term is $2992^2-1=(2992-8)(2992+8)+63=3000(2984)+63$, so our answer is $\boxed{63}$.[/hide]

Start listing all of the possible ways in which this is possible in order...

2^2 - 1, 4^2 - 1, 5^2 - 1, 7^2 - 1, 8^2 - 1, 10^2 - 1, 11^2 - 1, 13^2 - 1, ...

There is a pattern here... the squares are, in order, 2, 4, 5, 7, 8, 10, 11, 13, and so on. We can break this sequence up into separate partitions as follows:

| 2, 4| 5, 7| 8, 10| 11, 13| ...

As you can see, each number above in boxes (|) have a difference of 2, while adjacent numbers have a difference of 1. Using a bit of algebra and logic, one can derive that the 1st number of the nth partition is 3n-1 and the 2nd number of the nth partition is 3n+1. Since 1994 is in the 997th partition (1997/2 = 997), the number is (997 x 3 + 1)^2 - 1. We need to take this to the mod 1000.

Thus we are searching for (997 x 3 + 1)^2 - 1 (mod 1000). Multiplying out, we get (2992)^2 - 1 (mod 1000) lol don't multiply it ALL out that'd take too much work... it would work, but you could save some time, not to mention save yourself of possibly multiplying incorrectly  :)

so, we have (2992)^2 - 1 (mod 1000)... this is the same thing as (-8)^2 - 1 (mod 1000) (AoPs Volume 1 if you aren't familiar with modular arithmetic), which equals 64 - 1 (mod 1000), which equals 63 (mod 1000). Therefore, the answer is [b]63[/b].[/b]

[quote="jeffrey"]The increasing sequence $3, 15, 24, 48, \ldots$ consists of those positive multiples of 3 that are one less than a perfect square.  What is the remainder when the 1994th term of the sequence is divided by 1000?[/quote]

[hide="My Solution"]
Notice that for $a \equiv 1,2 \pmod{3}$, $a^2 - 1 \equiv 0$, and for $a \equiv 0$, $a^2 - 1 \equiv 2 \not\equiv 0$. Therefore, our sequence is the sequence $2^2 - 1$, $4^2 - 1$, $5^2 - 1$, $7^2 - 1$, $\cdots$. Let us temporarily include $1^2-1$. Notice the first 1994 terms of this sequence are the first $1994 \cdot \dfrac32 = 2991$ terms of the sequence $x^2 - 1, 1 \le x \le 2991$, without the multiples of three. Therefore, the 1995th term of this sequence is $2992^2 - 1$, which is the 1994th term of our original sequence. Therefore, our answer is $2992^2 - 1 \equiv (-8)^2 - 1 \equiv \boxed{063}$.
[/hide]

Note that in order for $n^2-1 = (n+1)(n-1) \equiv 0 \pmod 3,$ $3 \nmid n.$ Let us temporarily include $1^2-1$ in our sequnce. Then the $1994$th number in our sequnce has $n = \frac{3 \cdot 1994}{2} = 2991.$ However, we have to subtract one since $2991$ is a multiple of $3,$ and shift our number again by adding $2$ because we included $1^2-1$ in our sequnce at the beginning. Thus, $n = 2992.$ Therefore, $$2992^2 - 1 \equiv (-8)^2 - 1 \equiv \boxed{063} \pmod{1000}.$$


how many integer solutions for the given conditions. m.n>=0  and   m^3 + n^3 +99mn = 33^3

I believe this is an old ARML problem.  We have the factorization

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} \frac{1}{2} (a \plus{} b \plus{} c)( (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)$,

and letting $ a \equal{} m, b \equal{} n, c \equal{} \minus{}33$ we have $ (m \plus{} n \minus{} 33)( (m \minus{} n)^2 \plus{} (n \plus{} 33)^2 \plus{} (m \plus{} 33)^2 )$.  Clearly we need $ m \plus{} n \equal{} 33$, hence there are $ 34$ solutions.


Let $ a,b,c \in R$ and let $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Find maximum of $ P\equal{}a\plus{}b\plus{}c\minus{}abc$.

using the inequility,

$ \frac{a^2 \plus{}b^2 \plus{}c^2 }{3}\ge\ (\frac{a\plus{}b\plus{}c}{3})^2$

given $ a^2 \plus{}b^2 \plus{}c^2 \equal{}3$

we get,

$ a\plus{}b\plus{}c\le\ 3$..............(i)

now, using AM-GM inequility :

$ abc\le\ (\frac{a\plus{}b\plus{}c}{3})^3$

or, $ a\plus{}b\plus{}c\ge\ 3(\sqrt[3]{abc})$..................(ii)

from (i) & (ii)

$ abc\le\ 1$...................(iii)

subtracting (i) & (iii) we get,

$ a\plus{}b\plus{}c\minus{}abc\le2$

so, maximum value of $ a\plus{}b\plus{}c\minus{}abc\equal{}2$

                                     * * * * *

[quote="dimensions"]

$ abc\le\ (\frac {a \plus{} b \plus{} c}{3})^3$

[/quote]
Are you sure?

yes, it is simple AM-GM inequility

$ \frac{a\plus{}b\plus{}c}{3}\ge\sqrt[3]{abc}$               :idea:

[quote="dimensions"]yes, it is simple AM-GM inequility

$ \frac {a \plus{} b \plus{} c}{3}\ge\sqrt [3]{abc}$               :idea:[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} \minus{} 9.$
By the way, [quote="dimensions"]
so, maximum value of $ a \plus{} b \plus{} c \minus{} abc \equal{} 2$
[/quote]
Try $ a\equal{}b\equal{}\sqrt{1.5}$ and $ c\equal{}0.$ :wink:

there is an important condition for AM-GM inequility,

a,b,c must be positive real numbers,

we have to assume it,as all inequilities are solved only for a,b,c belongs to positive real no's................................[/b]

Yes, fjwxcsl! It's also my result.
If $ a\equal{}\min\{a,b,c\}\geq0$ then, easy to check that $ a\plus{}b\plus{}c\minus{}abc\leq2.5.$
If $ a\equal{}\min\{a,b,c\}\leq0$ then,
$ a\plus{}b\plus{}c\minus{}abc\leq a\plus{}\sqrt{2(b^2\plus{}c^2)}\minus{}a\cdot\frac{b^2\plus{}c^2}{2}\equal{}a\plus{}\sqrt{2(3\minus{}a^2)}\minus{}\frac{a(3\minus{}a^2)}{2}\equal{}$
$ \equal{}\frac{1}{2}\left(a^3\minus{}a\plus{}2\sqrt{2(3\minus{}a^2)}\right)$ and we can show that $ a_{max}\equal{}\minus{}\sqrt{\frac{4\minus{}\sqrt{13}}{3}}.$

[color=darkblue]Hi every One!

You know that  $ \boxed{a, b, c \in R}$. So you can't use AM-GM inequality.[/color]

But $ \sqrt {\frac {b^2 \plus{} c^2}{2}} \ge \frac{b \plus{} c}{2}$ and $ b^2 \plus{} c^2 \ge 2bc$, so arqady's solution is still valid.

arqady, can you explain better why $ a \plus{} b \plus{} c \le abc \plus{} 2.5$ for $ a, b, c \ge 0$?

[quote="feliz"]
arqady, can you explain better why $ a \plus{} b \plus{} c \le abc \plus{} 2.5$ for $ a, b, c \ge 0$?[/quote]
It's easy, but ugly:
$ a \plus{} b \plus{} c \minus{} abc\leq2.5\Leftrightarrow12\left(\sum_{cyc}(a^3 \plus{} a^2b \plus{} a^2c \minus{} abc)\right)^2\leq25(a^2 \plus{} b^2 \plus{} c^2)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(13a^6 \minus{} 24a^5b \minus{} 24a^5c \plus{} 39a^4b^2 \plus{} 39a^4c^2 \minus{} 48a^3b^3 \plus{}$
$ \plus{} 48a^4bc \plus{} 24a^3b^2c \plus{} 24a^3c^2b \minus{} 10a^2b^2c^2)\geq0.$
Id est, enough to prove that $ \sum_{cyc}(13a^6 \minus{} 24a^5b \minus{} 24a^5c \plus{} 39a^4b^2 \plus{} 39a^4c^2 \minus{} 48a^3b^3 \plus{} 5a^4bc)\geq0,$
which is true by SOS:
$ \sum_{cyc}(13a^6 \minus{} 24a^5b \minus{} 24a^5c \plus{} 39a^4b^2 \plus{} 39a^4c^2 \minus{} 48a^3b^3 \plus{} 5a^4bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(13a^6 \minus{} 48a^5b \plus{} 78a^4b^2 \minus{} 48a^3b^3 \plus{} 5a^4bc)\geq0\Leftrightarrow$
$ \Leftrightarrow13\sum_{cyc}(a^6 \minus{} a^5b \minus{} ab^5 \plus{} b^6) \minus{} 35\sum_{cyc}(a^5b \minus{} a^4b^2 \minus{} a^2b^4 \plus{} ab^5) \minus{}$
$ \minus{} 5\sum_{cyc}(c^4a^2 \minus{} 2c^4ab \plus{} c^4b^2) \plus{} 48\sum_{cyc}(a^4b^2 \minus{} 2a^3b^3 \plus{} a^2b^4)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(13a^4 \minus{} 22a^3b \plus{} 26a^2b^2 \minus{} 22ab^3 \plus{} 13b^4 \minus{} 5c^4)\geq0.$
Let $ a\geq b\geq c.$ Hence, $ S_c\geq0,$ $ S_b\geq0$ and $ (a \minus{} c)^2b^2\geq(b \minus{} c)^2a^2.$
Hence, $ \sum_{cyc}(a \minus{} b)^2S_c\geq(a \minus{} c)^2S_b \plus{} (b \minus{} c)^2S_a$ and
$ (a \minus{} c)^2S_b \plus{} (b \minus{} c)^2S_a\geq0\Leftarrow a^2S_b \plus{} b^2S_a\geq0.$
But $ a^2S_b \plus{} b^2S_a\geq0\Leftrightarrow13(a^2 \plus{} b^2)c^4 \minus{} 22(a^3 \plus{} b^3)c^3 \plus{} 10(a^4 \plus{} b^4)c^2 \plus{}$
$ \plus{} 16(a^4 \plus{} b^4)c^2 \minus{} 22(a^5 \plus{} b^5)c \plus{} 13a^6 \minus{} 5a^4b^2 \minus{} 5a^2b^4 \plus{} 13b^6\geq0,$ which is obvious
because $ (a^3\plus{}b^3)^2\leq(a^2\plus{}b^2)(a^4\plus{}b^4)$ and $ (a^5\plus{}b^5)^2\leq(a^4\plus{}b^4)(a^6\plus{}b^6).$

Thank you!

I want to observe I [i]really[/i] don´t think it is easy (I spent hours trying to understand).

Anyway, where can I find a text about SOS?

[quote="arqady"]Yes, fjwxcsl! It's also my result.
If $ a \equal{} \min\{a,b,c\}\geq0$ then, easy to check that $ a \plus{} b \plus{} c \minus{} abc\leq2.5.$
If $ a \equal{} \min\{a,b,c\}\leq0$ then,
$ a \plus{} b \plus{} c \minus{} abc\leq a \plus{} \sqrt {2(b^2 \plus{} c^2)} \minus{} a\cdot\frac {b^2 \plus{} c^2}{2} \equal{} a \plus{} \sqrt {2(3 \minus{} a^2)} \minus{} \frac {a(3 \minus{} a^2)}{2} \equal{}$
$ \equal{} \frac {1}{2}\left(a^3 \minus{} a \plus{} 2\sqrt {2(3 \minus{} a^2)}\right)$ and we can show that $ a_{max} \equal{} \minus{} \sqrt {\frac {4 \minus{} \sqrt {13}}{3}}.$[/quote]

[color=darkblue]Thank arqady. Your solution is very good.
My idea as [b]feliz[/b]'idea. What can we done this problem by S.O.S only.[/color]

feliz, thanhnam2902, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127


Let be given $a<b$ are positive integer numbers and $p$ is old prime.
Find  $k$ great such that $p^k|C_{pb}^{pa}$.
($C_{n}^{k}=\frac{n!}{k!(n-k)!}$).

Forgive my rusty combinatorics, but if I understand you question correctly, what I want to do is to write out b and (a-b) in base p. Then I find for how many i (from 1 onwards) this is true: The last i digits for both b and (a-b) add up to more than p^i.
(100...(i '0's)...0 in base p)

Then that would be the largest k.
Can't think of a closed form.


algebra
polynomial
calculus
integration
algebra unsolved

Let $P$ be polynomial with real coefficients with $n$ different zeros $z_1, z_1, \dots, z_n$. Complex number $t$ is different from $\frac{z_i}{z_j}$, $i \neq j$. Prove that
\[ 0 = \sum_{k = 1}^{n} {z_k \left [ \left ( \frac{1}{P (tz_k)} + \frac{1}{t^2 P (\frac{z_k}{t})} \right ) \prod_{j \neq k} {\frac{1}{z_k - z_j}} \right ] } \]

alexilic

If I didn't do nonsense again as I'm doing the whole day today, then it should follow from the residue theorem for $f(x)=\frac{x}{P(x)P(tx)}$. Integrating around a large circle, the integral approaches 0, on the other hand, it is given by $2\pi i$ times the sum of the residues which is exactly your sum.

Peter

Can you explain what do you mean?
I am not familiar with residue theorem!


Prove that $ 13$ is the largest prime that can divide two successive integers of the form $ n^{2}\plus{}3$.

[hide="solution"]Suppose that there is some prime $ p$ so that $ p|n^2\plus{}3$ and $ p|(n\plus{}1)^2\plus{}3$.  Then:
$ p|(n\plus{}1)^2\plus{}3\minus{}n^2\minus{}3\equal{}2n\plus{}1\Rightarrow p|2n\plus{}1, n^2\plus{}3\Rightarrow p|2n^2\plus{}n, 2n^2\plus{}6\Rightarrow$
$ p|2n^2\plus{}n\minus{}2n^2\minus{}6\equal{}n\minus{}6\Rightarrow p|2n\plus{}1, n\minus{}6\Rightarrow p|2n\plus{}1\minus{}2(n\minus{}6)\equal{}13\Rightarrow p|13$.  Hence $ 13$ is the largest, and only, prime that this works for.  An example would be $ 13|6^2\plus{}3\equal{}39$ and $ 13|7^2\plus{}3\equal{}52$.[/hide]