calculus
derivative
function
floor function
calculus computations

Find the mistake:
$ x^2$ is $ x\plus{}x\plus{}\cdots\plus{}x$ (with $ x$ terms).
Its derivative is $ 1\plus{}1\plus{}\cdots\plus{}1$ (also $ x$ terms).
So the derivative of $ x^2$ seems to be $ x$.

I found a mistake in the line where $ x^2$ is $ x\plus{}x\plus{}\cdots\plus{}x$ (with $ x$ terms), because that's only true for nonnegative integers.
Is this the mistake they're looking for?
Or is there a mistake that has to do with taking the derivative of $ x\plus{}x\plus{}\cdots\plus{}x$ because the number of terms is a variable?

Thanks.

[quote="tornado.adv4"]...because the number of terms is a variable? [/quote]
Yes. Also the other one; derivatives of functions on $ \mathbb{Z}$ don't make sense.

Let's see what happens if we were to correct the first error by saying, say, $ x^2 = x + x + ... + x + x\lbrace {x \rbrace}$ where there are $ \lfloor x \rfloor$ terms and $ \lbrace a\rbrace$ denotes the fractional part of $ a$.  This expression is then valid for all positive numbers (check it!), but if we carry out the argument we might find that the derivative of $ x^2$ is something like $ \lfloor x \rfloor + \lbrace x\rbrace$ (at least for $ x$ away from the integers), which certainly isn't true.  This shows that though the first error is problematic, the 'essential' error (and likely the one the instructor is looking for) is the second one you pointed out.  Because the number of terms itself ($ \lfloor x \rfloor$) depends upon $ x$, we cannot simply treat it as a constant upon differentiating.

As an exercise, try to make sense of what happens if you use the definition $ x^2 = x\lfloor x \rfloor + x\lbrace x\rbrace$ along with proper use of the chain rule/product rule (you'll need to think about what the derivatives of those unusual functions are.  You'll also need to pay special attention to the integers; use the most basic limit definitions).  You should find that it [i]does[/i] yield the correct derivative for $ x > 0$! (at least 'almost' everywhere).  With just a little more effort, and another special function (hint: $ \left \vert \cdot \right \vert$), you could do something similar that works for all real numbers.


Bea usually takes the 5:00 pm train from the city and gets to her village at 6:00 pm. Her father drives to the train station from their home with a uniform speed to pick her up, and then they drive back home with the same speed. One day Bea left the city one hour early, so she got to her village at 5:00 pm. With a constant speed she started walking home on the same road her father comes to pick her up. Her father left their house to pick her up at the usual time, on his way to the station they met, he picked her up, drove back home, so they got home 20 minutes earlier than usual. (The time it took the car to turn around is too small to count.) For how many minutes did Bea walk?

[hide]
I will assume (to start) that they live 1 hour away. So Bea's father starts coming at 5:00. They will get home at 6:40, so it took 1:40 rather than 2:00 for her father. This means it was 0:50 each way instead of 1:00 each way. Thus, she walked for 10 minutes.

Easier way: They cut off 20 minutes, which equals 10 minutes each way, so Bea must have been walking for [b]10 minutes[/b] to cut 10 minutes off of the trip to and from the train station.[/hide]

[hide="attempt 2"]

oops i think i did my solution wrong...let me try again  :oops:

Let us assume the same thing that i did first. She lives 1 hour away.

She gets to her village at 6:00 on a normal day, so her father leaves at 5:00 and they get home at 7:00

If she gets home 20 minutes early, she gets home at 6:40. Because it is a round trip for her father, the trip one way is 50 minutes, meaning she gets picked up at 5:50, making her walking time [b]50 minutes[/b]. I had it being on the wrong side of the hour  :oops:

If we were to try this with other distances, it would still be 50 minutes because she is cutting off 20 minutes, which is 10 minutes one way. That means that she gets picked up 10 minutes earlier than usual. This makes it 10 minutes less than 1 hour (the hour that she gets to her village early) giving [b]50 minutes[/b][/hide]


geometry
rectangle
inradius
analytic geometry
perimeter
search
power of a point

In a 5 by 12 rectangle, one of the diagonals is drawn and then circles are inscribed in both right triangles formed. Find the distance between the centers of the two circles.

[hide]Since the circle is inscribed in a 5-12-13 triangle, the inradius is 2 (if i remember correctly). I drew the diagonal from the top left to bottom right. So the "coordinate" of the center of the bottom circle is $(2,2)$. Then since the radius of the other circle is the same, it has coordinates $(10,3)$ and by the distance formula the length is $\boxed{\sqrt{65}}$. [/hide]

Oh yeah...I forgot about that. Thanks ch1n353ch3s54a1|.

[hide]Diagonal is 13. The area of the triangles is 30. The semiperimeter is (5+12+13)/2=15. The inradius are therefore 2. The diagonal is therefore $\sqrt{(12-2*2)^2+(5-2*2)^2}=\sqrt{64+1}=\sqrt{65}$[/hide]

[quote="ch1n353ch3s54a1l"]Because (if i remember correctly) the inradius is $\frac{\text{Area}}{\text{Semiperimeter}}.$[/quote]This follows from formula that if a polygon has apothem $a$ and perimeter $p$ then area is $A=\frac12ap$. For a right triangle, $a=r$ ($r$ is inradius) and $p=2s$, so $r=\frac{A}{s}$

[hide="Answer"]$A=\frac{1}{2}(5)(12)=30=rs=r\left(\frac{5+12+13}{2}\right)=15r$, so $r=2$ on both circles. Therefore, the centers of the circles are essentially at opposite vertices of a $1\times 8$ rectangle, meaning that the distance between them is $\sqrt{8^2+1^2}=\boxed{\sqrt{65}}$.[/hide]

$A=\sqrt{\frac{a+b+c}{2}\left(\frac{a+b-c}{2}\right)\left(\frac{a-b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)}$
$A=rs\Rightarrow r=\frac{A}{s}$
$r=\frac{2A}{a+b+c}\Rightarrow 2r=\sqrt{\frac{(a+b-c)(a-b+c)(-a+b+c)}{a+b+c}}$
Multiplying out the two rightmost terms in the numerator, we have $-a^2-b^2+c^2+2ab$. Since this is a right triangle, $a^2+b^2=c^2$, so this simplifies to $2ab$. Now, multiply top and bottom by $(a+b-c)$ to get $\sqrt{\frac{(a+b-c)(2ab)(a+b-c)}{((a+b)+c)((a+b)-c)}}=\sqrt{\frac{(a+b-c)^2(2ab)}{a^2+b^2-c^2+2ab}}=a+b-c$ because $a^2+b^2-c^2+2ab=2ab$.

[quote="delta"]For a right triangle there is a simpler formula for the inradius 2r=a+b-c where c is the hipotenuse. Try to prove it![/quote]

[hide="nasty proof"]Since
$[ABC]=rs$, we can sustitute $[ABC]=\frac{ab}{2}$, and $s=\frac{a+b+\sqrt{a^2+b^2}}{2}$ to get

$\frac{ab}{2}=r(\frac{a+b+\sqrt{a^2+b^2}}{2})$.

Multiply by $\frac{2}{a+b+\sqrt{a^2+b^2}}$ and we have

$\frac{ab}{a+b+\sqrt{a^2+b^2}}=r$.

Rationalize denominator and factor to result in

$\frac{ab(a+b-\sqrt{a^2+b^2})}{2ab}=r$.

$\frac{a+b-c}{2}=r$, where $c=\sqrt{a^2+b^2}$.

Multiply this by two to get the desired result. [/hide]

*smacks head against wall for not realizing that I could have just used $\frac{ab}{2}$ to find the area of a right triangle* :wallbash_red:

By the way, you don't need to substitute $c=\sqrt{a^2+b^2}$. Just say: $A=rs\Rightarrow r=\frac{A}{s}=\frac{ab}{a+b+c}$. Multiply top and bottom by $a+b-c$: $\frac{ab(a+b-c)}{((a+b)+c)((a+b)-c)}=\frac{ab(a+b-c)}{(a+b)^2-c^2}$, and then substitute $c^2=a^2+b^2$. You can't get any better than that, really, so I don't see your solution as being nasty. Much better than mine, anyway. :P

[quote="delta"]For a right triangle there is a simpler formula for the inradius 2r=a+b-c where c is the hipotenuse. Try to prove it![/quote]

Or you could just draw radii from the center tangent to the legs and see that you get a square. Use Power of a Point on the tangent segments in the triangle and get an easy system of equations.

[quote="myc"]What's Power of a Point?

thanks  :)[/quote]
Please use SEARCH (next to MEMBERS, top right-hand side of screen) with keyword "Power of a Point" :)

I am not sure which is the right post  :? , can someone please hyperlink it?

Thx.

Also I believe that this problem was also from MIddleton Invitational 2006 on team round.  :whistling:


ratio
geometry
incenter
circumcircle
power of a point
radical axis
geometry unsolved

Here are some of my unsolved problems. Please give as many different solutions as you can give.

$ 1) ABCD$ is a parallellogram. $ X$ divides $ AB$ in the ratio $ 3: 2$ and $ Y$ divides $ CD$ in the ratio $ 4: 1$. If $ XY$ cuts $ AC$ at $ Z$, prove that $ 7AZ \equal{} 3AC$.

$ 2)$ Given four points $ A,B,C,D$ in a straight line, find a point $ O$ in the same straight line such that $ OA: OB \equal{} OC: OD$.

$ 3)$ In a given straight line $ AB$, find a point $ P$ such that the difference in the squares on $ AP$ and $ PB$ is equal to the difference between two given squares.

$ 4)$ A straight line is drawn through the incentre $ I$ if $ \triangle ABC$ perpendicular to $ AI$ meeting $ AB,AC$ in $ D$ and $ E$ respectively. Prove that $ BD \cdot CE \equal{} ID^2$.

$ 5)$ Two circles intersect at $ A$ and $ B$ ; PAQ is straight line through $ A$ meeting the circles at $ P,Q$. Find the locus of the midpoint of $ PQ$.

$ 6)$ A straight line cuts two concentric circles in $ A_1,A_2$ and $ B_1,B_2$. Prove that the four intersections of a tangent at an $ 'A'$ and at a $ 'B'$ lie on another concentric circle.

$ 7)$ A straight line of a given length subtends given equal angles at two fixed points. Prove that the straight line $ a$ always touches a fixed circle.

$ 8)$ Three equal circles pass through a given point $ H$ and meet one another two by two at $ A,B,C$. Prove that $ H$ is the orthocentre of $ \triangle ABC$.

Thanks in advance.

Hey Please reply someone?
I desperately need solutions to all of these

First problem is just simple similarity i believe.
Ok here it is,
$ AX: BX=3: 2$
and
$ CY: DY=4: 1$
Now triangles $ AXZ$ and $ YZC$ are similar.
hence
$ \frac{CY}{AX}$=$ \frac{CZ}{AZ}$
$ =>$ $ \frac{CZ}{AZ}$=$ \frac{4}{3}$
Adding 1 to both sides,
$ \frac{CZ}{AZ}+1$=${ \frac{4}{3}+1}$
$ \frac{AC}{AZ}$=$ \frac{{7}}{3}$
Done

First problem is just simple similarity i believe.
Ok here it is,
$ AX: BX=3: 2$
and
$ CY: DY=4: 1$
Now triangles $ AXZ$ and $ YZC$ are similar.
hence
$ \frac{CY}{AX}=\frac{CZ}{AZ}$

$ =>$ $ \frac{CZ}{AZ}=\frac{4}{3}$

Adding 1 to both sides,
${ \frac{CZ}{AZ}+1=\frac{4}{3}+1}$

$ \frac{AC}{AZ}=\frac{7}{3}$

Done

[quote="Rijul saini"] A straight line is drawn through the incentre $ I$ if $ \triangle ABC$ perpendicular to $ AI$ meeting $ AB,AC$ in $ D$ and $ E$ respectively. Prove that $ BD \cdot CE \equal{} ID^2.$[/quote]
Since $ \angle AIB \equal{} \frac {_1}{^2}(180^{\circ} \plus{} \angle C)$ and $\angle AIC \equal{} \frac {_1}{^2}(180^{\circ} \plus{} \angle B),$ it follows that

$ \angle DIB \equal{} \angle ECI \equal{} \frac {_1}{^2}\angle C$ and $ \angle EIC \equal{} \angle DBI \equal{} \frac {_1}{^2}\angle B$ $ \Longrightarrow$ $ \triangle IEC \sim \triangle BDI$

$ \Longrightarrow \frac {\overline{CE}}{\overline{ID}} \equal{} \frac {\overline{IE}}{\overline{BD}} \ \Longrightarrow \overline{BD} \cdot \overline{CE} \equal{} \overline{ID} \cdot \overline{IE} \equal{} \overline{ID}^2.$

[quote="Rijul saini"]Two circles intersect at A and B.  PAQ is straight line through A meeting the circles at  P,Q. Find the locus of the midpoint of PQ.[/quote]
$ \omega_1$ and $ \omega_2$ are the given circles. $ P \in \omega_1$ and $ Q \in \omega_2.$ Let $ M$ be the midpoint of $ PQ.$ Powers of $ M$ to $ \omega_1$ and $ \omega_2$ are $ p(M,\omega_1) \equal{} MA \cdot MP$ and $ p(M,\omega_2) \equal{} \minus{} MA \cdot MQ.$ Since  $ MP \equal{} MQ$ $\Longrightarrow$ $ \frac {_{p(M,\omega_1)}}{_{p(M,\omega_2)}} \equal{} \minus{} 1.$ Therefore, $ M$ moves on a circle coaxal with $ \omega_1$ and $ \omega_2$ centered at the midpoint of the segment connecting their centers.

[b]Rijul saini[/b] wrote [quote]
4) A straight line is drawn through the incentre $ I$ of $ \triangle ABC$ perpendicular to AI meeting AB,AC in D and E respectively. Prove that $ BD \cdot CE \equal{} ID^2$. [/quote]

$ DE$ is the tangent to the circle $ BIC$ centered on $ AI$. Suppose, w.l.o.g. that $ \angle B > \angle C$ and that this circle intersects $ |AC|$ at $ B'$, there are the following equalities: $ BD \equal{} B'E$ and $ DI \equal{} EI$, hence, from the power of $ E$ to the circle $ BIC$ we get: $ IE^2 \equal{} B'E \cdot CE$, done.

Best regards,
sunken rock

[b]Rijul saini[/b] wrote: [quote]
8) Three equal circles pass through a given point $ H$ and meet one another two by two at $ A,B,C$. Prove that $ H$ is the orthocentre of $ \triangle ABC$.
[/quote]
See the attached sketch: $ O_1AO_2H$ and $ O_1HO_3C$ are rhombuses, so $ AC \parallel{} O_2O_3$, but $ O_2O_3 \perp HB$; similarly $ CH \perp AB$ and $ AH \perp BC$, done.

Additionally, it's easy to see that $ \triangle O_1O_2O_3$ has the circumcenter at $ H$ (this problem is known to us as the [b]"The 5 lei coin problem"[/b] and it's supposed it has been found by our great geometer G. Titeica during a contest, where he used the said coin to draw the circles).

Best regards,
sunken rock

Hey...all mathlinkers...especially Luis
Please post the remaining solutions fast...
Questions 2,3,6,7 are still unsolved
Tomorrow is my national olympiad... :o

[quote="sunken rock"]Good luck tomorrow then![/quote]
Thanks...

[quote="Rijul saini"][quote="Rijul saini"][quote="Rijul saini"][b][u][size=125]Unsolved Problems[/size][/u][/b]

$ 2)$ Given four points $ A,B,C,D$ in a straight line, find a point $ O$ in the same straight line such that $ OA: OB \equal{} OC: OD$.

$ 3)$ In a given straight line $ AB$, find a point $ P$ such that the difference in the squares on $ AP$ and $ PB$ is equal to the difference between two given squares.

$ 6)$ A straight line cuts two concentric circles in $ A_1,A_2$ and $ B_1,B_2$. Prove that the four intersections of a tangent at an $ 'A'$ and at a $ 'B'$ lie on another concentric circle.

$ 7)$ A straight line of a given length subtends given equal angles at two fixed points. Prove that the straight line $ a$ always touches a fixed circle.
[/quote]

Please give solutions to these... :!:  :maybe:[/quote][/quote]
Anyone? :maybe:

[quote="Rijul saini"]
$ 2)$ Given four points $ A,B,C,D$ in a straight line, find a point $ O$ in the same straight line such that $ OA: OB \equal{} OC: OD$.
[/quote]

This solution came in the geometry marathon.

[quote="oneplusone"][quote="gouthamphilomath"][b][u][i]Problem 55:[/i][/u][/b]

Given four points $ A,B,C,D$ in a straight line, find a point $ O$ in the same straight line such that $ OA: OB \equal{} OC: OD$.[/quote]
Draw circles with diameters $AD,BC$ and let their radical axis intersect the line at $O$. Then $OA\cdot OD=OB\cdot OC$ so $OA:OB=OC:OD$.[/quote]

And also, can't we directly say that choose $O$ to be at infinity?  :P

[quote="Rijul saini"]
$ 7)$ A straight line of a given length subtends given equal angles at two fixed points. Prove that the straight line $ a$ always touches a fixed circle.[/quote]

The four points consisting of the endpoints of the line segments and the two fixed points are concyclic. so, the line segment becomes a chord of fixed length and varies in any circle passing through the fixed two points.

Now, I used the help of sourbh to complete it.

We have the perpendicular from the centre to the chord to be of fixed length and so, draw a circle with centre as the centre of the circumcircle of the four points and radius as the length of the perpendicular from the centre to the chord. It is tangential to all chords possible.

So, choose the new circle to complete the proof. :blush:

[quote="Rijul saini"]
$ 3)$ In a given straight line $ AB$, find a point $ P$ such that the difference in the squares on $ AP$ and $ PB$ is equal to the difference between two given squares.

$ 6)$ A straight line cuts two concentric circles in $ A_1,A_2$ and $ B_1,B_2$. Prove that the four intersections of a tangent at an $ 'A'$ and at a $ 'B'$ lie on another concentric circle.[/quote]

Pending  :)  :)

3) Let $AP^2-PB^2=a^2-b^2$, now draw circles with radius $a,b$ centred at $A,B$, whose radical axis passes through $P$. Just choose $P$ on the radical axis.

6) Let the tangents at $A_1,B_1$ meet at $C$, $A_2,B_2$ meet at $D$, $A_1,B_2$ meet at $E$ and $A_2, B_1$ meet at $F$. Let the centre be $O$. Then $\angle CA_1O=\angle CB_1O$ so $CA_1B_1O$ is cyclic so $\angle A_1CB_1=\angle A_1OB_1=\angle A_2OB_2=\angle A_2DB_2$ so $CDFE$ is cyclic. Now note that $\angle EA_1O=\angle EB_2O=90$ so $EA_1OB_2$ is cyclic. Hence $\angle A_1CO=\angle B_2B_1O=\angle B_1B_2O=\angle A_1EO$ thus $OC=OE$ and similarly $OD=OF$ and finally $O$ is the circumcentre of $CDFE$.

Anymore? :)

[quote="Rijul saini"][quote="Rijul saini"])$8)$ Three equal circles pass through a given point $ H$ and meet one another two by two at $ A,B,C$. Prove that $ H$ is the orthocentre of $ \triangle ABC$.
[/quote]
Actually, $\triangle ABC$ is equilateral, and its circumradius is the same as the radius of the three congruent circles :)
Try to prove this.[/quote]

Hi Rijul,
This is not possible for all arc lengths formed in the circles by the $\triangle ABC$.
There is contradiction for ur statement if we take two of the arc lenghths let $\widehat{AH}\neq\widehat{BH}$.

[quote="rationalist"]
Hi Rijul,
This is not possible for all arc lengths formed in the circles by the $\triangle ABC$.
There is contradiction for ur statement if we take two of the arc lenghths let $\widehat{AH}\neq\widehat{BH}$.[/quote]
Yeah, I'm sorry, I confused it with another similar problem, but the two weren't necessarily equivalent.

[quote="Rijul saini"]$ 8)$ Three equal circles pass through a given point $ H$ and meet one another two by two at $ A,B,C$. Prove that $H$ is the orthocentre of $ \triangle ABC$.[/quote]

[u]Alternate solution:[/u]

From an inversion with centre $H$ and arbitrary radius greater than $1$, the circles map to three lines forming the triangle $A'B'C'$ where $A', B', C'$ are the images of $A, B, C$ under the inversion. Now, since the three circles are equal, $H$ is equidistant from their images $A'B', B'C', C'A'$ and so, $H$ is the incentre of $A'B'C'$.

[u]Lemma:[/u] If $O$ is the circumcentre of $HB'C'$, then $A', H, O$ are collinear.
[u]Proof:[/u] Let $B'H$ intersect $A'C'$ at $M$. In triangle $B'HC',$ we have $\angle MHC'=\angle HB'C'+HC'B'=\frac{B'+C'}{2}$. Similarly, $\angle MHA'=\frac{B'+A'}{2}$ and so, $\angle A'HC'=\angle MHA'+\angle MHC'=\frac{B'+A'+B'+C'}{2}$ which means $\angle A'HC'=\frac{\pi}{2}+\frac{B'}{2}$. In the circumcircle of $B'HC'$, we have $\angle HOC'=2\angle HB'C'=B'$ and in isosceles triangle $HOC',$ we have $\angle OHC'=\frac{\pi}{2}-\frac{B'}{2}$. So, $\angle A'HO=\angle A'HC'+\angle OHC'=\pi$ which means $A', H, O$ are collinear as required.

Continuing on the problem, we have $A'H\equiv HO$ is orthogonal to the circumcircle of triangle $B'HC'$ and so, after inverting back, due to conformity, $AH$ is perpendicular to $BC$. Similarly, $BH$ is perpendicular to $AC$ and $CH$ is perpendicular to $AB$. So, $H$ is the orthocentre of triangle $ABC$ as required.

[quote="gouthamphilomath"][quote="Rijul saini"]$ 8)$ Three equal circles pass through a given point $ H$ and meet one another two by two at $ A,B,C$. Prove that $H$ is the orthocentre of $ \triangle ABC$.[/quote]

[u]Alternate solution:[/u]
$\cdots \cdots $ Now, since the three circles are equal, $H$ is equidistant from their images $A'B', B'C', C'A'$ and so, $H$ is the incentre of $A'B'C'$. $\cdots \cdots$
[/quote]
Although I think this can be easily dealt with...but I think there are some points known as "excentres" :D which also satisfy equidistance from the sides of a triangle... :) (Don't take this too seriously...I'm only stressing this point to make sure that you write your proof rigorously)


In a certain state, the social security codes consist of at least 1 and no more than 4 letters followed by a dash and 0, 1,or 2 digits. How many different codes are possible if exactly 1 letter is repeated exactly once, but digits cannot be repeated?

Hide your solutions

Billy

[color=indigo][hide]There are 9653176 possibilities.[/color][/hide]

And you got that how? (no work)
[hide]Letters: 26*27(repeat plus no letter)*26*25*6(possitions of repeat)
Numbers: 11*10
Total: 26*27*26*25*6*11*10
26*27*26*25*660
18252*16500
>>3011580000?<<[/hide]
I think I'm wrong, but that's me.

solafidefarms... what is the correct answer?

As I changed the problem, I do not know the proper answer, but [hide]I get 26 choices for the first letter, 1 choice for the second letter, 6 ways to arrange those 2 letters, and 26*25 choices for the remaining 2 digits. However, I think I overcounted a bit for the cases in which niether of the duplicated digits is in the first slot and there is a blank before the repeated digits. [/hide]

[color=indigo]
Here is how I got it:
[hide]First I found all the possible combinations of letters:
1 letter= 0 possibilities (Because there has to be a match)
2 letters= 26 possibilities (becuse they have to match)
3 letters= 26*25 = 650 possibilities*3 ways to arrange them = 1950
4 letters= 26*25*24 = 15600 possibilities*6 ways to arrange them = 93600
So there are a total of 95576 possibilities for letters. Then for numbers:
0 numbers= 1 possibility
1 number= 10 possibilities
2 numbers= 90 possibilities
That is a total of 101. Then 101*95576 = [b]9653176 [/b] [/color][/hide]


Hi folks,

Does anyone have a good clean copy of the ARML Logo? I need something that can be imported into Adobe Illustrator. It's t-shirt time!  :)

Thanks,
Tom
Ga. ARML

there is the original one at arml.com.  I don't know if u can put it into illisturator though.

And u probably already knew that it was there :D.

Yeah, I'm looking for one with a little more resolution.

Thanks!

The logo (in AI format) can be found at http://arml.com/NewARML.otlnd.ai

Let me know if this works for you.


Let $ a_i,b_i\geq 0, i\equal{}1,2, ... ,k,$ and $ p,q,k\in N^*.$
Prove that:
$ (a_1^{p\plus{}q}\plus{}\cdots\plus{}a_k^{p\plus{}q})\cdot (b_1^{p\plus{}q}\plus{}\cdots\plus{}b_k^{p\plus{}q})\geq (a_1^pb_1^q\plus{}\cdots\plus{}a_k^pb_k^q)\cdot (a_1^qb_1^p\plus{}\cdots\plus{}a_k^qb_k^p)$

[b]Remark:[/b]
For $ p\equal{}q\equal{}1$ we get to Cauchy-Schwarz inequality.

I'm waiting for elementary proofs.

Marin Bancoș - ROMANIA

[quote="marin.bancos"]Let $ a_i,b_i\geq 0, i \equal{} 1,2, ... ,k,$ and $ p,q,k\in N^*.$
Prove that:
$ (a_1^{p \plus{} q} \plus{} \cdots \plus{} a_k^{p \plus{} q})\cdot (b_1^{p \plus{} q} \plus{} \cdots \plus{} b_k^{p \plus{} q})\geq (a_1^pb_1^q \plus{} \cdots \plus{} a_k^pb_k^q)\cdot (a_1^qb_1^p \plus{} \cdots \plus{} a_k^qb_k^p)$

[b]Remark:[/b]
For $ p \equal{} q \equal{} 1$ we get to Cauchy-Schwarz inequality.

I'm waiting for elementary proofs.

Marin Bancoș - ROMANIA[/quote]
Without loss of generality, we may assume that
$ a^{p \plus{} q}_1 \plus{} a^{p \plus{} q}_2 \plus{} \cdots \plus{} a^{p \plus{} q}_k \equal{} b^{p \plus{} q}_1 \plus{} b^{p \plus{} q}_2 \plus{} \cdots \plus{} ^{p \plus{} q}_k\equal{}k.$
The inequality becomes
$ (a_1^pb_1^q \plus{} \cdots \plus{} a_k^pb_k^q)(a_1^qb_1^p \plus{} \cdots \plus{} a_k^qb_k^p) \le k^2.$
By the AM-GM Inequality, we have
$ a^p_ib^q_i \le \frac {pa^{p \plus{} q}_i \plus{} qb^{p \plus{} q}_i}{p \plus{} q}.$
Therefore,
$ \sum^k_{i \equal{} 1}a^p_ib^q_i \le \frac {1}{p \plus{} q} \left( p\sum^k_{i \equal{} 1}a^{p \plus{} q}_i \plus{} q\sum^k_{i \equal{} 1}b^{p \plus{} q}_i\right) \equal{} k.$
Similarly, we can also prove that
$ \sum^k_{i \equal{} 1}a^q_ib^p_i \le k,$
and from multiplying these two inequalities, the conclusion follows.

It's OK [b]can_hang2007[/b]. There are a lot of nice results for this inequality. My proof is a little different from yours, but its core is the same.


inequalities Euler calculus derivative geometry parallelogram

Let ABCD be an orthodiagonal, cyclic quadrilateral. Show that:
\[ AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} AD^2\ge 8r^2
\]
and find all equality.

[i](Source: Own variation of a different problem)[/i]

EDIT: Yes, it should have been $ r^2$

[color=darkblue]Do you mean $ 8r \boxed{^2}$

I will prove indentity. In a cyclic
Denote $ E \in AC \cap BD$ and $ \left\|\begin{array}{cc} AE \equal{} a \\
BE \equal{} b \\
CE \equal{} c \\
DE \equal{} d\end{array}\right\|$.

Then we'll prove  $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4R^2$

In triangle $ ABC$ we have $ R \equal{} \frac {AB\cdot BC\cdot CA}{4S} \equal{} \frac {\sqrt {a^2 \plus{} b^2}\cdot\sqrt {b^2 \plus{} c^2}\cdot AC}{2 AC \cdot b} \Longleftrightarrow 4R^2 \equal{} \frac {(a^2 \plus{} b^2)(b^2 \plus{} c^2)}{b^2}$

We want to prove that $ \frac {(a^2 \plus{} b^2)(b^2 \plus{} c^2)}{b^2} \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \Leftrightarrow ac \equal{} bd$ which is true by power of the point $ E$.

Then $ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \equal{} 8r^2 \Leftrightarrow AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} AD^2 \equal{} 8R^2$[/color]

propab,y he means that...
another solution would be from eulers theorem for an orthodiagonial we know
$ \sum AB^2 \equal{}AC^2\plus{}BD^2\equal{}(2r)^2\plus{}(2r)^2\equal{}8r^2$

Very nice, both of you :)

Also, why do you type in blue? Many people on this forum do it and the practice makes no sense...  :huh:

[quote="Athinaios"]propab,y he means that...
another solution would be from eulers theorem for an orthodiagonial we know
$ \sum AB^2 \equal{} AC^2 \plus{} BD^2 \equal{} (2r)^2 \plus{} (2r)^2 \equal{} 8r^2$[/quote]What is Euler's Theorem for an orthodiagonal? My second question, isn't $ AC,BD<2r$? And hence $ AC^2\plus{}BD^2\equal{}(2r)^2\plus{}(2r)^2\le8r^2$, contradicts to the problem's statement.

Thanks for sharing this problem =)

[b]@ Anax:[/b] It provides color differentiation between the writing and the math, on the AoPS mode, and (to my eyes) even more in the MathLinks mode. Personally, I'd think it's too much work, though

[color=darkblue]Eulers theorem states that $ AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} AD^2 \equal{} AC^2 \plus{} BD^2 \plus{} 4EF^2$, where $ E,F$ are midpoints of diagonals. So [u][b]Athinaios'[/b][/u] solution is incorrect.[/color]

[quote="Athinaios"]ahiles my solution is correct.eulers theorem says that for every  quadrilateral
$ AB^2 \plus{} BC^2 \plus{} CD^2 \plus{} AD^2 \equal{} AC^2 \plus{} BD^2 \plus{} 4EF^2$
but for a parallelogram EF=0 :wink:[/quote]How do you prove that $ ABCD$ is a parallelogram?

o no, i misunderstood the fact that it says orthodiagonial and no orthogonial  :oops: .i am sorry


Consider the inequality $a, b, c \ge 0$

$\frac{a^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}}{c^{2}+(a+b)^{2}}\ge \frac{3}{5}$

I have a really tedious solution which I will hide at the end of my post. However, I hoped to see some solution like that of [b]arqady[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=645569#645569]here[/url]. By this I mean setting $a+b+c = 3$ and writing the ineq as

$\sum_{cyc}\left(\frac{a^{2}}{2a^{2}-6a+9}-\frac{1}{5}+k(a-1)\right) \ge 0$

where the last thing is okay because $\sum_{cyc}(a-1) = 0$. But I can't figure out how to make that work for this problem, and I would like to see if someone else can do it.

[hide="Other solution"] I will not show many steps of this because it is horrific algebra. We first let $f(a, b, c) = LHS-RHS$ and note that
$f(a, b, c)-f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right) =$

$\frac{a(b-c)^{2}(a+b+c)(a^{2}+2b^{2}+2c^{2}+3ab+3ac+2bc)}{((a+c)^{2}+b^{2})((a+b)^{2}+c^{2})((b+c)^{2}+(a+(b+c)/2)^{2})}\ge 0$

Then we also note that

$f(a, b, b) = \frac{2(a-b)^{2}(a^{2}+4ab+8b^{2})}{5(a^{2}+4b^{2})(a^{2}+(a+b)^{2})}\ge 0$

So we win. [/hide]

[hide="an easier proof"]by cauchy we can easily observe that:
$\sum_{cyc}\frac{a^{2}}{a^{2}+(b+c)^{2}}\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{a^{4}+b^{4}+c^{4}+\sum_{cyc}(ab+ac)^{2}}$
so all we need to show is:
$\frac{5}{3}(a^{2}+b^{2}+c^{2})^{2}\geq a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}+2abc(a+b+c)$
which follows from $(a^{2}+b^{2}+c^{2})^{2}\geq 3abc(a+b+c)$  :)  [/hide]
BTW I think that we can't solve this with tangent line  :wink:

[quote="Xevarion"]Consider the inequality $a, b, c \ge 0$

$\frac{a^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}}{c^{2}+(a+b)^{2}}\ge \frac{3}{5}$

I have a really tedious solution which I will hide at the end of my post. However, I hoped to see some solution like that of [b]arqady[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=645569#645569]here[/url]. By this I mean setting $a+b+c = 3$ and writing the ineq as

$\sum_{cyc}\left(\frac{a^{2}}{2a^{2}-6a+9}-\frac{1}{5}+k(a-1)\right) \ge 0$

where the last thing is okay because $\sum_{cyc}(a-1) = 0$. But I can't figure out how to make that work for this problem, and I would like to see if someone else can do it.
[/quote]

I don't think you can make it work because if you let $k=-\frac{12}{25}$

You get $\sum_{\text{cyc}}\frac{(a-1)^{2}(8a-21)}{2a^{2}-6a+9}\ge 0$

arqady is welcome to prove me wrong.

[quote="Xevarion"]Consider the inequality $a, b, c \ge 0$

$\frac{a^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}}{c^{2}+(a+b)^{2}}\ge \frac{3}{5}$

[quote]

Wow the inequality is very strict :ninja:

[quote="SplashD"]I don't think you can make it work because if you let $k=-\frac{12}{25}$

You get $\sum_{\text{cyc}}\frac{(a-1)^{2}(8a-21)}{2a^{2}-6a+9}\ge 0$

arqady is welcome to prove me wrong.[/quote]
It's also what I got. I am interested that Cauchy worked for this ineq. I will try to examine the general inequalities of this form and see which methods work when. :D

My friend noticed that we can simplify it by
$\frac{a^{2}}{a^{2}+(b+c)^{2}}\ge \frac{a^{2}}{a^{2}+2b^{2}+2c^{2}}$
So it suffices to prove that
$\sum_{cyc}\frac{a}{a+2b+2c}\ge \frac{3}{5}$
But this is a familiar kind of ineq. For example if $a+b+c = 3$ then it is
$\sum_{cyc}\frac{a}{6-a}\ge \frac{3}{5}$
Now we can apply Jensen.

[quote="SplashD"]I don't think you can make it work because if you let $k=-\frac{12}{25}$

You get $\sum_{\text{cyc}}\frac{(a-1)^{2}(8a-21)}{2a^{2}-6a+9}\ge 0$

arqady is welcome to prove me wrong.[/quote]
Yes, you are wrong, SplashD.
$\frac{a^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}}{c^{2}+(a+b)^{2}}\ge \frac{3}{5}\Leftrightarrow$
$\Leftrightarrow\sum_{\text{cyc}}\frac{(a-1)^{2}(21-8a)}{2a^{2}-6a+9}\geq 0.$ :wink:
And I confirm that my method is very weak, but sometimes it  helps.


For those who have taken the state competition already:

What rank did you get?
1st
2nd
3rd
etc

Post your Chapter Score and Ranking too.

Tell What State You're From

What problem did you miss at Chapter on the sprint round?
and
What problem did you miss at Chapter on the target round?

WI-Official Ladder Style CD

20-something at chapter, 13/8

4th at State (3rd Written)

we're allowed to discuss state scores?  [b][size=150][color=red]NO![/color][/size][/b]

california hasnt taken the test yet

[quote="jli"]we're allowed to discuss state scores?

california hasnt taken the test yet[/quote]
No, we can't discuss state scores yet.

[quote="bpms"][quote="perfect628"]Chapter-1st, 29/14-won cd
State-7th, something/something-4th cd

MA-unofficial, yet ladder, cd[/quote]
Who made it in MA?[/quote]

Leet and Daine both made it again from MA.  Perfect was 7th as he said, and the kid from Hawaii was 6th I think.  Two new kids made it to Nats.  And the coach will be ..........

[hide]
me :D

Team did awesome at all placing in top 17 in a competition with a lot of great competitors!!!!!!
[/hide]

lol top 17? now everyone knows what place the fourth place person got...

Yeah Elizabeth and I are both going back.

[quote="Fanatic"]Hey l33t, were you first?[/quote]

yeah, elizabeth was second

4th at chapter with a score of 33 and 34th at state :blush:

I have this unusual pattern:

2nd school   :winner_second:

2nd chapter   :winner_second:

2nd state   :winner_second:

I got 42 at school, 41 at chapter, -- at state.

Also, my team got first at both chapter and state.  :first:

school:2
chapter:1
state:34 [size=75][color=red]<snip>[/color][/size]

hope to see mr. frost, elizabeth, and leet (perfect, you deserve to go, you can take dan's place :P )

omg, 33 PEOPLE BEAT ME!!!

ok, so i know like 4 others from ct who should've won, but didn't
[size=75]
[color=red]<Snip.  Don't say *anything* about the States, except your own rank -- moderator rcv.>[/color][/size]

-jorian

School: 2nd
Chapter: 41 (1st)
State: Not top 10 D:

Pretty disappointed I didn't make nats, I did a lot of preparation.

[quote="mwpl11"]the indiana team only has one 8th grader! i predict our team will do really well next year, since so many ppl will be returning. It's surprising the predominance of seventh graders in Indiana mathcounts...[/quote]

This reminds me of what 1337 said last year about this year's Massachusetts team  :roll:

[quote="jhredsox"]lol, he brags without trying to

but look, only 2 of those kids made it

jorian[/quote]
When he said that only three 7th graders who made nationals lived in MA

ironically, it was actually me saying that more and i'm the only 7th grader from last year from MA who didn't make it... :|

13[color=red]th[/color] at Chapter

36[color=red]th[/color] at states

Next year Im going for 1st at chapter and in the top 4 for states.

[size=75][color=red]["th" added by moderator RCV.  If you don't make it clear if you are posting scores versus ranks, your posts may be deleted without warning.][/color][/size]

Hey Magixter, do you know someone named Abubakar?
I heard he did really well at State.

i got a score of 27 at my chapter (but got 3rd)
i don't know what i got at states but i got 3rd again
 :D SEE YOU IN FORT WORTH :D

P.S. Go Dallas Stars!!!

I got a score of 39 at chapter
then
I got 10th at State
cause i sucked

I got a score of 30 and got 4th at chapter.
I got a score of which I can't say and got 6th at states.

I hate that I had to do so well to ensure going to states. I needed a 43 for first, and about 40 for 6th(qualifying).

[quote="magixter"]In Arkansas, there was no chapter competition (lack of funding, I guess). Was there any other state like that?[/quote]

I know that Delaware and Guam are like that, I'm sure that there are others as well.

[quote="magixter"]In Arkansas, there was no chapter competition (lack of funding, I guess). Was there any other state like that?[/quote]
I think that that is more due to size than funding.

I'm positive anywhere state and defense dept. had no chapter.


[quote="vardanch"]$x,y,z$ are positive real numbers, such that $x+y+z\leq 1$. Prove that $\sum{(1-x^{3})^{2}}\geq 2$.[/quote]

Hey Vardanch :wink: !

Here is my solution.

From the problems condition it is obvious that $x,y,z\leq 1$=>

\[(1-x^{3})^{2}\geq (1-x^{2})^{2}\]
=>
\[\sum{(1-x^{3})^{2}}\geq \sum{(1-x^{2})^{2}}\]

Also you can easily see that the critical part of this problem is when $x,y$ and $z$ reache their maximum possible value,therefore, it is sufficient to prove the initial inequality in the case when $x+y+z=1$.

Now,we need to prove that
\[x^{4}+y^{4}+z^{4}+1\geq 2(x^{2}+y^{2}+z^{2})\]
<=>
\[x^{4}+y^{4}+z^{4}+(x+y+z)^{4}\geq 2(x^{2}+y^{2}+z^{2})(x+y+z)^{2}\]
<=>
\[2(xy+xz+yz)^{2}\geq x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}\]
<=>
\[x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+4xyz(x+y+z)\geq 0\]
  :P :D


Consider an elevator at rest.Now,we are in the elevator,standing at rest.If we jump up with 3m/s and immediately the elevator starts moving down with 5m/s,will we take the same time to reach the floor of the lift as it would ahve taken if the elevator at rest?

Errr... if by jumping you mean moving upwards at 3 m/s, then, NO, you will sure reach the floor in a longer time than the time you would need if the elevator was at rest. It's quite obvious actually... or did I not get the question right.

Ok,and now if the elevator was accelerating downwards and we jump up again(while the elevator is accelerating down),will it take the same time ?.Obvious it might seeem but calculations and to show how are more important ;)

If we are assuming this is classical physics, then we shall let v represent the upwards or downwards accleration of the elevator, t represent the time, and n represent the number of times we jump down, I have derrived the forumula

vt+3n. The 3 represents the downward accleration. If my equation is wrong, please don't laugh at me, since I am only in 7th grade.

I guess we could do it this way, using the elevator's frame of reference.
relative acceleration $= g-a$
where $a$ is the acceleration of the elevator relative to Earth.
relative velocity of the man $= v_i + v_l$
where $v_i$ is the initial velocity of the man relative to the Earth and $v_l$ is the initial velocity of the elevator.
Thus, time taken $= 2\frac{v_i + v_l}{g-a}$

[Edited: I guess I did not answer the question in the last post]

[quote="shadysaysurspammed"]What does your formula give?[/quote]

accleration of the elevator after n number of jumps.


I think that's because since the denominators so large that you can't the the whole block repeating digits.

[quote="kimby_102"]nope,
[/quote]

what gave you this idea? :wink: All rationals repeat sooner or later, that is the definition of a rational!

EDIT: snipped to avoid confusion :P

The definition of rational says that the number can be expressed as the ratio of two numbers (not that the number repeats eventually...or terminates but when it terminates you can just say that the 0 repeats technically).  However, all rational number [i]do[/i] repeat or terminate (even if it takes a long time).

[quote="Iversonfan2007"][quote="jb05"]Shobhit may have said this, but I can't understand anything he said in his last post... :|

Since there are arbitrarily long strings of 0s (which progressively get longer), there can be no repeating segment (because we would need more 0s for the next segment).[/quote]

 :rotfl: yea sorry, I was trying to and did say that but I didn't really say it as clear as you and added some needless stuff. I'm not very good at explaining stuff (ask anyone who went to the SC Math Camp) :blush:  :P[/quote]

SC Math Camp:
Shbhit drew a diagram and said:
"Blahblahblah....Oh wait.  Assume this is a straight line."

The line looked like the graph of sin x.

[quote="mysmartmouth"]

SC Math Camp:
Shbhit drew a diagram and said:
"Blahblahblah....Oh wait.  Assume this is a straight line."

The line looked like the graph of sin x.[/quote]

good one...

[quote="not_trig"]Why does Shobhit have a new username?  :huh:

Since all nonterminating rationals repeat, and this one doesn't, it's irrational.  :D

Not very rigorous...[/quote]

Er, basically the whole point of the problem is to show that the decimal never repeats...


This is a question from a past AMC.

Find a:

...a, b, c, d, 0, 1, 1, 2, 3, 5, 8...

How can I solve this?

Yeah, like I said, the Fibonacci Sequence.  :P

$d+0=1$
$c+d=0$
$b+c=d$
$a+b=c$
$a=c-b$
$b=d-c$
$c=-d$
$d=1$
Therefore, $a=-d-(d-c)=-d-(d-(-d))=-3d=\boxed{-3}$.

Or you could just find each term, which might be faster.

$d+0=1 \Rightarrow d=1$
$c+1=0 \Rightarrow c=-1$
$b+(-1)=1 \Rightarrow b=2$
$a+2=-1 \Rightarrow a=\boxed{-3}$

[quote="Hexaditidom"]It helps to have the entire problem:

6. In the sequence

...,a,b,c,d,0,1,1,2,3,5,8,...

[b]each term is the sum of the two terms to its left[/b]. Find a.[/quote]

[hide]Working backwards...

$1=d+0$
$d=1$

$0=c+1$
$c=-1$

$1=b-1$
$b=2$

$-1=2+a$
$\boxed{a=-3}$[/hide]

Wow, I feel like an idiot. Thanks for clarifying, it seems so obvious now.


induction
function
combinatorics proposed
combinatorics

By a [i]partition $\pi$ of an integer $n\geq 1$,[/i] we mean here a representation of $n$ as a sum of one or more positive integers where the summands must be put in nondecreasing order (E.g, if n=4 then the partitions $\pi$ are 1+1+1+1,1+1+2,1+3,2+2 and 4)
For any partition $\pi$ define $A(\pi)$ to be the nubmer of $1$'s which appear in $\pi$ and define $B(\pi)$ to be the number of distince integers which appear in $\pi$.
Prove that for any $n$ the sum of $A(\pi)$ over all partitions $\pi$ of $n$ is equal to the sum of $B(\pi)$ over all partitions $\pi$ of $n$.

i origianlly had posted this problem on the NT section. apparently a Mod moved it

i do have a NT solution for this...should i post it here? or should a Mod move this back to NT section?

Here is the correct solution of this problem
Let $G(n)$ denote the number of partition s of n and we can assume $G(0)=1$.The number of partitions $n$ in which the number m occurs at least once is $G(n-m)$.Use this result we will have $\sum A(\pi)=G(n-1)+G(n-2)+\dots+G(0)$ (you can use induction to prove this  :)  :) )
So we only need to show that:$\sum B(\pi)=G(n-1)+G(n-2)+...+G(0)$
To prove this we consider $G(n)$ as $n$ rectangular array with the rows indexed by the partitions $\pi$ of $n$ and the columns indexed by the numbers $1,2,...,n$.A checkmark is placed in row $\pi$ and column $m$ if the number $m$ appears in that partition.So we have $\sum B(\pi)=$the sum of checmarks in the array.On the other hand,the number of checmarks in the column $m$ of the array is the number of partitions in which an $m$ occurs and this is $G(n-m$.From this we easily to get the result we want to prove.So we've finished the proof  :)  :)
Do you have the same solution?I'll very happy if someone can post the solution by use generating function  :?  :?

[quote="nhat"]
I'll very happy if someone can post the solution by use generating function  :?  :?[/quote]

Using generating functions is completely straight-forward if you have seen gf's for partitions with more than one variable before, but your solution is much more appropriate for the problem. But here goes:

Let $A(q,t)= \sum_{\pi} q^{|\pi|}t^{A(\pi)}$ and $B(q,t)= \sum_{\pi} q^{|\pi|}t^{B(\pi)}$, so the exponent of $q$ is $n$ for a partition of $n$ and the exponent of $t$ counts the interesting statistics.
Now, you just have to do the three following steps:

Step 1: Prove that $A(q,t)= \frac1{1-qt}\cdot \frac1{1-q^2} \cdot\frac1{1-q^3} \cdot \frac1{1-q^4} \dots$.

Step 2: Prove that
\[ B(q,t)=(1+qt+q^2t+\dots)(1+q^2t+q^4t+\dots)(1+q^3t+q^6t+\dots)\dots\\[1ex] =\frac{1-q+q\cdot t}{1-q} \cdot \frac{1-q^2+q^2\cdot t}{1-q^2} \cdot \frac{1-q^3+q^3\cdot t}{1-q^3} \cdot \dots \]

Step 3: Prove that
$\frac{\partial}{\partial t} A(q,t) |_{t=1}=\frac{\partial}{\partial t} B(q,t) |_{t=1}.$

(Both sides turn out to be equal to $\frac q {1-q} \prod_{k=1}^{\infty} \frac 1 {1-q^k}$)

[hide="solution via generating functions"]

Let $P_k(x)=1+x^k+x^{2k}+x^{3k}+...$
$Q(x)=x+2x^2+3x^3+...$
$R(x)=Q(x)\prod_{k=2}^{\infty}P_k(x)$

After consider how the coefficients of $R(x)$ are formed, we see that that coefficient of $x^n$ equals: $\sum A(\pi)$ also:
$Q(x)=x+2x^2+3x^3+...$
$=(x+x^2+x^3+...)(1+x+x^2+...)$
$=(x+x^2+x^3+...)P_1(x)$
so that
$R(x)=(x+x^2+x^3+...)\prod_{k=1}^{\infty}P_k(x)$

now letting $\prod_{k=1}^{\infty}P_k(x)=c_0+c_1x+c_2x^2+...$
and examining how $c_i$ are formed we see that the coefficient $c_{n-k}$ is the number of partitions of $N$ which contain the integer $k$ for $1\le k\le n$.  Thus the number of partitions of $n$ in which $k$ occurs, summed for $k=1,2,3,...,n$ is $c_{n-1}+c_{n-2}+...+c_0$, and this is also the coefficient of $x^n$ in $R(x)=(x+x^2+x^3+...)\prod_{k=1}^{\infty}P_k(x)$.  But observe that the number of partitions of $\pi$ of $n$ which $k$ occurs, summed for $k=1,2,3,...,n$ is precisely the sum of $B(\pi)$ over all partitions of $\pi$ of $n$.  This completes the proof.
QED[/hide]


search
combinatorics unsolved
combinatorics

The set $ 1, 2, ..., 3n$ is partitioned into three sets $ A$, $ B$ and $ C$ with aech set containing $ n$ numbers. Determine if it is always possible to choose one number out of each set so that one of this numbers is  the sum of the other two.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=576222555&t=26762

Use the link in the thread to find a more general result.

Pierre.


Hey everyone, USNCO is coming up so I was wondering if anyone was interested in reviving the chem marathon. Ill start and next person who answers this right please post another one:
 PROBLEM 1

Three electrochemical cells were connected in series so that the same quantity of clectrical current passes through all three cells. In the first cell, 1.15g cCr metal was depositedfrom Chromium (III) nitrate solution. In the second cell, 3.15 grams of osmium was deposited from a solution of $ Os^\plus{}n$ and nitrate ions. What is the name of the salt?
In the third cell, the electrical charge paassed through a solution containing X^-2 ions caused deposition of 2.11g of metallic X. What is the electron configuration of X?

Enjoy :P

use faraday's second law of electrolysis and you now the gram equivalence of Cr from that we know it is the same for Os and X2- so we can directly find the vaalue of n for Os and knowing the moecular weight of X2- we can predict what element it is and from that write the electronic configuration !!!! do you want me to calculate and state the answer!!!! :lol:

yeah so here's my problem

http://www.mathlinks.ro/viewtopic.php?t=199186

refer this post and answer!!!! :D


Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?

First, use as many 24 packs as possible.  It's obviously 3.

90-72=18.

Now use as many 12 packs as possible.  It's obviously 1.

18-12=6.

Now use as many 6 packs as possible.  It's obviously 1.

6-6=0

3+1+1=$ \boxed5$

[quote="qballmath"]5 number of packs.[/quote]

Could you not just post answers, but post solutions too?


trigonometry
inequalities proposed
inequalities

[size=150]$ \frac {x_a}{r} \plus{} \frac {4r_a}{r_a \plus{} x_a} \minus{} \frac {4(x_a \minus{} h_a)}{x_a \plus{} r_a} \geq 5$

$ x_a \equal{} w_a,m_a,\frac {\sqrt {3}}{2}a,\frac {1}{2}(r_b \plus{} r_c),3r,\frac {3}{2}R,\frac {b \plus{} c}{2a}h_a.....$

In most cases，if a=b=c,then $ x_a \equal{} 3r$,this ineq holds.[/size]

[size=150]$ x_a\equal{}\sqrt{3}(s\minus{}a)$,does not hold,but holds in acute triangle.[/size]

[size=200]So,It is super ineq.[/size]
BQ

[size=75]we do a Simple。
let $ x_a \equal{} a\sin{A}$
<=>
$ \frac {a\sin{A}}{r} \plus{} \frac {4r_a}{a\sin{A} \plus{} r_a} \minus{} \frac {4(a\sin{A} \minus{} h_a)}{a\sin{A} \plus{} r_a} \geq 5$
$ (\sin{A} \equal{} \frac {a}{2R},r_a \equal{} \frac {sr}{s \minus{} a},..)$
<=>

$ \frac {1}{2}\frac { \minus{} a^5s \plus{} a^6 \plus{} 16a^3srR \plus{} 36r^2sR^2a \minus{} 18Rra^4 \minus{} 32R^2r^2s^2}{aRr( \minus{} a^2s \plus{} a^3 \minus{} 2srR} \geq 0$

[$ \minus{} a^2s \plus{} a^3 \minus{} 2srR \equal{} \minus{} a^2(s \minus{} a) \minus{} 2srR < 0$];
must prove:
$ \minus{} a^5s \plus{} a^6 \plus{} 16a^3srR \plus{} 36r^2sR^2a \minus{} 18Rra^4 \minus{} 32R^2r^2s^2 le 0$,
[$ R \equal{} \frac {abc}{4Rr},r \equal{} \frac {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s},s \equal{} \frac {1}{2}(a \plus{} b \plus{} c),$]

<=>
$ \frac {1}{2}\frac {a^2}{a \plus{} b \plus{} c}( \minus{} a^5 \plus{} (12bc \plus{} b^2 \plus{} c^2)a^3 \minus{} 8bc(b \plus{} c)a^2 \minus{} 5b^2c^2a \plus{} 4b^2c^2(b \plus{} c)) \geq0$
Must prove:
$ \minus{} a^5 \plus{} (12bc \plus{} b^2 \plus{} c^2)a^3 \minus{} 8bc(b \plus{} c)a^2 \minus{} 5b^2c^2a \plus{} 4b^2c^2(b \plus{} c) \geq 0;$

[a=y+z,...]

main members solution，please..[/size]

Let $ x_a \equal{} r_b$,
<=>
$ \frac {2}{(c \plus{} a \minus{} b)ac}( \minus{} ca^2 \plus{} 3cab \minus{} c^2a \plus{} a^3 \minus{} ba^2 \minus{} b^2a \plus{} c^3 \minus{} bc^2 \minus{} b^2c \plus{} b^3) \geq 0$

$ <\equal{}>\minus{}ca^2\plus{}3cab\minus{}c^2a\plus{}a^3\minus{}ba^2\minus{}b^2a\plus{}c^3\minus{}bc^2\minus{}b^2c\plus{}b^3 \geq 0$;
[a=y+z...]
<=>
$ x^2y \plus{} x^2z \plus{} xy^2 \minus{} 6xyz \plus{} xz^2 \plus{} y^2z \plus{} yz^2\geq 0$
that is easy.
BQ

[size=150]$ x_a \equal{} h_b$,

<=>
$ \minus{} 6xyz \plus{} z^3 \minus{} 13x^2y \plus{} 3x^2z \plus{} 3yz^2 \plus{} 6xy^2 \minus{} 4xz^2 \plus{} 2zy^2 \plus{} 8x^3 \geq 0$
<=>
$ (6x \plus{} 2z)y^2 \plus{} ( \minus{} 6zx \minus{} 13x^2 \plus{} 3z^2)y \plus{} (z \plus{} x)(z^2 \minus{} 5zx \plus{} 8x^2) \geq 0;$
is easy![/size]

BQ

[size=150]Cauchy inequality，
you  Beautiful and simple ,
I can not do it,although,
But I enjoy

          Come in,
          Looking for good things,
          into math
          and mathlinks every day
BQ-poem[/size]


This year's is supposed to be very good, I hear -- low risk of accidental infection or side effects.  Of course, this is coming from my college, which wants everyone to get the vaccine.  But I expect that it isn't a downright lie.

Anyhow, I think the disease is always around -- it's just that it doesn't start getting to lots of people until the weather gets colder and people start spending more time indoors.

http://www.cnn.com/2003/HEALTH/11/17/flu.season.ap/index.html
Looks like this year's is very bad.

i guess that's the reason my parents urged me to get a flu shot this year.

i'm never getting a flu shot again.  :(
stupid cold's making
me feel so stuffy. :(

I missed how those two things are related . . .
flu shots . . .
[i]prevent[/i] the flu, no?

i thought if you got the flu shot it would pervent you from getting the flu for at least a short period of time. i don't like getting colds.

a cold is not the flu -- the flu is like a cold, but worse, for longer.

yeah i guess you're right, but how much "longer" does it last. the last time i had a "cold". it lasted for a couple of months. i coughed a lot but i was feeling good enough to go to school.

[quote="Syntax Error"]yeah, when i took that duke sat thingie, i unfortunately had the flu and missed a whole week of school[/quote]

yah... me too. I got it the night before, but got worse on the day of the duke  test... that was horrible. ugh.

lets just blame astral for giving it to us even though he didnt get the flu that year

why not? Plus... we're lucky we didn't get what he was sick with that week... even though its not a flu (but he says it is... and still went to school...)

hey, i'v had the flu for the past 2 DAYS!!! reason i couldnt go to school.... at least im better now!!

[quote="Leeyuhang"]how do they come up with a vaccine before the disease comes out?
do you think it's useful?[/quote]

To answer your first question, various national disease-prevention agencies see which new strains of flu virus seem to be spreading fastest with the worst symptoms in their places of origin. Those strains are used to make up each year's flu vaccine. New strains arise all the time because of genetic drift of viruses. This year, the Fujian strain was NOT included in the Northern Hemisphere vaccine, because it wasn't grown in culture in time for the vaccine manufacture, but the Fujian strain will be included in the next Southern Hemisphere vaccine, now in production. The current Northern Hemisphere vaccine includes a Panama strain of flu that may have some cross-reactivity to the related Fujian strain.

See

http://www.cdc.gov/ncidod/diseases/flu/overview.htm

for much more about flu and flu vaccine.

To answer your other question, yes, I think getting the flu vaccine is worth it. My whole family is in a high-risk group because we care for my dad (who is paralyzed since a year and a half ago, and is now 73 years old) and we dare not spread the flu to him or to people who work where he receives therapy. I started taking flu shots after a time when I was thirty years old, in professional school, and got the flu during the semester finals of all my courses. I was so tired I could hardly remember all that I had studied so hard all semester, and I never got lower grades than I got that semester, thanks to the flu. After I take a flu shot, I usually feel MILD flu symptoms for a day or two later that week, but I go the whole season without ever catching a severe case of the flu. My children report that they don't feel anything at all after getting their shots.


Let $ a,b,c,d > 0$ prove that
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)(b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} da)(c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac)(d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd)\ge (ab \plus{} bc \plus{} ca \plus{} da\plus{}db\plus{}dc)^4
\]

[quote="encyclopedia"]Let $ a,b,c,d > 0$ prove that
\[ (a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)(b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} da)(c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac)(d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd)\ge (ab \plus{} bc \plus{} ca \plus{} da \plus{} db \plus{} dc)^4
\]
[/quote]
It's obviously true by Holder:
$ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}ac\plus{}bc)(ad\plus{}bd\plus{}d^2\plus{}ab\plus{}a^2\plus{}b^2)\cdot$
$ \cdot(ad\plus{}d^2\plus{}cd\plus{}a^2\plus{}ac\plus{}c^2)(d^2\plus{}bd\plus{}cd\plus{}b^2\plus{}c^2\plus{}bc)\geq(ad \plus{} bd \plus{} cd \plus{} da \plus{} ac \plus{} bc)^4.$

Nice, you make me to think the following can solve by Holder

\[ (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)\ge (ab\plus{}bc\plus{}ca)^3\]

Don't I?

[quote="encyclopedia"]Nice, you make me to think the following can solve by Holder
\[ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3
\]
Don't I?[/quote]
Obvious!

By Holder Inequality we have:

$ (b^2\plus{}a^2\plus{}ab)(bc\plus{}c^2\plus{}b^2)(c^2\plus{}ac\plus{}a^2) \geq (ab\plus{}bc\plus{}ac)^3$

[quote="arqady"]By the way, $ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3$ is true for all reals $ a,$ $ b$ and $ c.$ :wink:[/quote]
because we can rewrite the inequality as :
$ (ab\plus{}ac\plus{}bc)^2(a^2\plus{}b^2\plus{}c^2) \geq abc(a\plus{}b\plus{}c)^3$  :)

[quote="anas"][quote="arqady"]By the way, $ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3$ is true for all reals $ a,$ $ b$ and $ c.$ :wink:[/quote]
because we can rewrite the inequality as :
$ (ab \plus{} ac \plus{} bc)^2(a^2 \plus{} b^2 \plus{} c^2) \geq abc(a \plus{} b \plus{} c)^3$  :)[/quote]
Cool!

[quote="arqady"][quote="arqady"]By the way, $ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3$ is true for all reals $ a,$ $ b$ and $ c.$ :wink:[/quote]
The following inequality is also true.
For all reals $a$, $b$ and $c$ prove that:
\[ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3+\frac{1}{6}(a-b)^2(a-c)^2(b-c)^2\][/quote]
The following inequality is also true.
For all positive reals $a$, $b$ and $c$ prove that:
\[ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\]\[\ge \frac{1}{3}(ab \plus{} bc \plus{} ca)^2\left(2(a^2+b^2+c^2)+bc+ca+ab \right).\]

[quote=arqady][quote="arqady"]By the way, $ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3$ is true for all reals $ a,$ $ b$ and $ c.$ :wink:[/quote]
The following inequality is also true.
For all reals $a$, $b$ and $c$ prove that:
\[ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge (ab \plus{} bc \plus{} ca)^3+\frac{1}{6}(a-b)^2(a-c)^2(b-c)^2\][/quote]
http://www.artofproblemsolving.com/community/c6h622626p5935126


algebra
polynomial
function
rational function
algebra proposed

Given $k > 0$, the sequence $a_n$ is defined by its first two members and \[ a_{n+2} = a_{n+1} + \frac{k}{n}a_n  \]
a)For which $k$ can we write $a_n$ as a polynomial in $n$?

b) For which $k$ can we write $\frac{a_{n+1}}{a_n} = \frac{p(n)}{q(n)}$? ($p,q$ are polynomials in $\mathbb R[X]$).

When $k\ge0$ integer, then recurent equation can be solved $a_n=P_k(n)$, were P_k(x) is polinom with degree k.
Expessions by rational function appear when k<0 and integer.


function
number theory unsolved
number theory

Huh ? It seems to be false for $m=1$ and $n=2$ (or any other prime)  :?

yes it's not right for $m=1$  we will have $\sum_{d/n}{\mu(d)log({\frac{n}{d})=\Lambda (n)}}$

where $\Lambda$ is Mangoldt function $\Lambda (d)=logp$ if there is a prime$p$ and an integer $k$ such that : $d=p^k$ and $\Lambda (d)=0$ if not .

 we know that if $n\ge2$  we have ${\sum_{d/n} {\mu (d)}=0}$.
so $\sum_{d/n}{\mu (d)logd}=-\Lambda (n)$
 so it's not right if we take $n$ prime .


$ \ce{BH3}$ does not exist an isolated molecule. Rather, it dimerizes $ 2 \ce{BH3} \rightleftharpoons \ce{B2H6}$. Explain the bonding present in the dimer.

[hide="Hint"]

Indeed this is unusual by any localized electron model. However MO Theory explains this nicely. Ths is the for centre four electron bond structure.[/hide]

Banana bonding  :huh: . Its a 3 centre 2 electron bond or a 4 centre 4 electron bond  :)

[quote="BanishedTraitor"]Banana Bonding? What do you mean?[/quote]

Isnt that term the same everywhere then? I thought it was called banana bonding (its atleast called so in india  :D  )

I did a search. It is called the Banana Bond.

Yes that's the name of the bond and this type of bonding is called banana bonding  :D

i have a doubt does.......B4H10 exist????


On a drive from your ranch to Austin, you wish to average 64 mph. The distance from your ranch to Austin is 80 miles. However, at 40 miles (half way), you find you have averaged only 48 mph.

What average speed must you maintain in the remaining distance in order to have an overall average speed of 64 mph?

When you reach the first half you've travelled during $ \frac{40}{48}h$, suppose that the time for finishing the way (with speed $ v_{2}$) is $ k$.

Then we have the equation $ \frac{80}{\frac{40}{48}\plus{}48k}\equal{} 64$. After you have solved this you find that $ v_{2}\equal{}\frac{40}{k}mph$


The sum of the areas of two squares is 85. the length of a side of the larger square minus the length of a side of the smaller square is 1. Find  the length of a side of each square

[hide=" click to reveal hidden content"]

$ x^{2} \plus{} y^{2} \equal{} 85$

$ y\minus{}x\equal{}1$

$ x^{2} \plus{} (x\plus{}1)^{2} \equal{} 85$

$ x(x\plus{}1) \equal{} 42$

$ x\equal{}6, y\equal{}7$.

[/hide]

[quote="AndrewTom"][hide=" click to reveal hidden content"]

$ x^{2} \plus{} (x \plus{} 1)^{2} \equal{} 85$

$ x(x \plus{} 1) \equal{} 42$

$ x \equal{} 6, y \equal{} 7$.

[/hide][/quote]
I think you can't say like that because $ x$ and $ y$ may not be integer(it's not given). Finish it like that:
$ x^{2} \plus{} (x \plus{} 1)^{2} \equal{} 85$
$ 2x^2\plus{}2x\minus{}84\equal{}0$
$ 2(x\minus{}6)(x\plus{}7)\equal{}0$
Then $ x$ is equal $ 6$ of $ \minus{}7$ but we will ignore $ \minus{}7$ because the side of a square can not be negative
Then $ y$ is equal $ 7$


We basically want the largest surface area of a rectangle, square, hexagon, and circle with the smallest perimeter. This is obviously the circle, so the circular cross section of pipe will allow the greatest flow of water.


inequalities
function
inequalities proposed

I'm primarily looking for an elegant solution that doesn't use Calculus. It's for training purposes. Maybe possible with just CS / AM-GM?

For $ a,b,c$ positive reals, show:
$ \frac{6a}{9a^2\plus{}5(a\plus{}b\plus{}c)^2} \plus{} \frac{6b}{9b^2\plus{}5(a\plus{}b\plus{}c)^2} \plus{} \frac{6c}{9c^2\plus{}5(a\plus{}b\plus{}c)^2} \leq \frac{1}{a\plus{}b\plus{}c}$

Inequality is equivalent with:

  $ \sum \frac{x}{9x^2\plus{}5}\leq\frac{1}{6}$     ,  $ x\plus{}y\plus{}z\plus{}1$

  But function $ f(x)\equal{}\frac{x}{9x^2\plus{}5}$  is concave on [0,1] and follow Jensen's inequality result

  $ f(x)\plus{}f(y)\plus{}f(z)\leq3f(\frac{x\plus{}y\plus{}z}{3})\equal{}\frac{1}{6}$

[quote="hendrata01"]I'm primarily looking for an elegant solution that doesn't use Calculus. It's for training purposes. Maybe possible with just CS / AM-GM?

For $ a,b,c$ positive reals, show:
$ \frac {6a}{9a^2 \plus{} 5(a \plus{} b \plus{} c)^2} \plus{} \frac {6b}{9b^2 \plus{} 5(a \plus{} b \plus{} c)^2} \plus{} \frac {6c}{9c^2 \plus{} 5(a \plus{} b \plus{} c)^2} \leq \frac {1}{a \plus{} b \plus{} c}$[/quote]

Assume $ a\plus{}b\plus{}c\equal{}1$
\[ \sum {\frac{{6a}}{{9a^2  \plus{} 5\left( {a \plus{} b \plus{} c} \right)^2 }}}  \equal{} \sum {\frac{{6a}}{{9a^2  \plus{} 5}}}
\]
\[ \equal{} \sum {\frac{{6a}}{{\left( {9a^2  \plus{} 1} \right) \plus{} 4}}}  \le \sum {\frac{{6a}}{{6a \plus{} 4}}}  \equal{} \sum {\left( {1 \minus{} \frac{4}{{6a \plus{} 4}}} \right)}
\]
So it suffices us show that
\[ \sum {\left( {1 \minus{} \frac{4}{{6a \plus{} 4}}} \right)}  \le 1
\]

\[ \Leftrightarrow \sum {\frac{2}{{3a \plus{} 2}}}  \ge 2
\]

\[ \Leftrightarrow \sum {\frac{1}{{3a \plus{} 2}}}  \ge 1
\]

Which is true by Cauchy Schwarz.

[quote="Honey_S"][quote="hendrata01"]I'm primarily looking for an elegant solution that doesn't use Calculus. It's for training purposes. Maybe possible with just CS / AM-GM?

For $ a,b,c$ positive reals, show:
$ \frac {6a}{9a^2 \plus{} 5(a \plus{} b \plus{} c)^2} \plus{} \frac {6b}{9b^2 \plus{} 5(a \plus{} b \plus{} c)^2} \plus{} \frac {6c}{9c^2 \plus{} 5(a \plus{} b \plus{} c)^2} \leq \frac {1}{a \plus{} b \plus{} c}$[/quote]

Assume $ a \plus{} b \plus{} c \equal{} 1$
\[ \sum {\frac {{6a}}{{9a^2 \plus{} 5\left( {a \plus{} b \plus{} c} \right)^2 }}} \equal{} \sum {\frac {{6a}}{{9a^2 \plus{} 5}}}
\]

\[ \equal{} \sum {\frac {{6a}}{{\left( {9a^2 \plus{} 1} \right) \plus{} 4}}} \le \sum {\frac {{6a}}{{6a \plus{} 4}}} \equal{} \sum {\left( {1 \minus{} \frac {4}{{6a \plus{} 4}}} \right)}
\]
So it suffices us show that
\[ \sum {\left( {1 \minus{} \frac {4}{{6a \plus{} 4}}} \right)} \le 1
\]

\[ \Leftrightarrow \sum {\frac {2}{{3a \plus{} 2}}} \ge 2
\]

\[ \Leftrightarrow \sum {\frac {1}{{3a \plus{} 2}}} \ge 1
\]
Which is true by Cauchy Schwarz.[/quote]

Thank you, this is exactly what I need....


This is the first time for me to take the AMC12,
and I don't know anything about preapration for this test.
I have studied mathematics so far all by myself,
and my grades were so far so good.
I'm taking Math Analysis (Pre calculus), and I am sophomore.
But, here, after I saw some problems, I was
discouraged so badly.
I could approach to the problems in my own way, but
the many answeres came out wrong.
Here, at this point, I ve known this very wonderful website
just few days ago, and there is less than 1 month left til
the test date. And I am so anxious about it.
I want to get a high score on the AMC 12, but I'm not
that expecting too much.
If anyone of you guys have got good grades on it,
please tell me how you did. I am willing to do anything !!
 :)  :)  :)

And of course you should buy some of the wonderful books AoPS offers such as volumes 1 and 2, since school math is pretty much just memorization, whereas you need to think on the AMC.  Volume one covers pretty much everything you'd need to do well on AMC.

You say many of your answers come out wrong. There's a big difference in how you should prepare depending on whether your answers came out wrong because
[list]
[*]your method solves the problem incorrectly.
[*]you made a calculation error or other stupid mistake each time.[/list]
If it's your methods, looking at some of the AMCs on this site (see Contests above) and how people approached the harder problems should give you an idea of the ways you should be solving problems. If it's mistakes, two things: checking your work is good, and [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]this article[/url] may also be useful.

Also do as many AMC 12 problems as possible which will really help
  [url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44[/url]

[quote="Temperal"][quote="diophantient"]The pre-olympiad forum is not the place to practice for the AMC 12, for that you should go to High School Basics (the harder questions) and Intermediate Topics (the easier questions).[/quote]

Uh, don't you mean the other way around?

[[AMC 12 Problems and Solutions]] is a good resource.[/quote]

I think he means that the harder questions in High School Basics and the easier questions in Intermediate Topics are good practice for the AMC 12.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \frac{a\plus{}b\plus{}c}{ab\plus{}ac\plus{}bc}\leq\frac {a}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} ac \plus{} c^2} \plus{} \frac {c}{a^2 \plus{} ab \plus{} b^2}\leq\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{ab \plus{} ac \plus{} bc}\leq\frac {a}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} ac \plus{} c^2} \plus{} \frac {c}{a^2 \plus{} ab \plus{} b^2}\leq\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2}\]
[/quote]
LHS:*$ \frac {a}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} ac \plus{} c^2} \plus{} \frac {c}{a^2 \plus{} ab \plus{} b^2} \ge \frac {(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \equal{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$
RHS:
<=>$ \sum(\frac {a^3}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \minus{} \frac {a}{b^2 \plus{} bc \plus{} c^2}) \ge 0$
<=>$ \sum\frac {a^2}{b^2 \plus{} bc \plus{} c^2} \ge \sum\frac {bc}{b^2 \plus{} bc \plus{} c^2}$
True because
$ \sum\frac {a^2}{b^2 \plus{} bc \plus{} c^2} \ge 1 \ge \sum\frac {bc}{b^2 \plus{} bc \plus{} c^2}$


AMC
AIME
USAMTS
ARML
AMC 12 B
AMC 10
graph theory

Hey Massachusetts! My name is Ric, I'm a freshman and up until very recently, I went to New Trier in Illinois. However, next year I will be attending Andover as a 10th grader, and so I'd like to announce my arrival into the MA community. (In case you were wondering, I was also accepted to Exeter, but didn't like it nearly as much).

I love math of all kinds, in particular combinatorics. I also enjoy doing competition math. As a freshman this year, year are my math factoids:
AMC 12 B: 121.5
AIME: 5 (ew, I'm ashamed of this one, should have gotten like an 8)
USAMTS: 96 (Gold medal)
NSML: (State Math League, pretty much the NEAML or whatever equivalent)
1st place Individual (NT came in 4th)
ARML: 6 (and I rocked power and team)

I'm really excited for next year. See you there!

I hope for your continued success in solving AMC problems, power round #1's, arithmetic questions, etc. I'm sure MA will be very excited to have you as one of their new top mathletes!

I'm glad you ended up at a (safety) school that suits your intellectual abilities. :P :P :P

Exeter and Andover are both good schools.

[quote="besttate"]In case you were wondering, I was also accepted to Exeter, but didn't like it nearly as much.

[/quote]

Huh?  I thought everyone knew that Exeter was better than Andover.  I bet you just didn't get in...

hey dude nice maybe our arml team won't suck as much as we did this year

hey hey take it easy (sheesh...what with all the flaming and sarcasm)

I think we all concede that Exeter has a much stronger math program (well wtf, they have zuming). Well, Andover is a liberal arts school...

In terms of sports, I believe that Andover has the upper hand:
[code]
Fall                         Winter                     Spring
GXC - A                  G Basketball - A        Baseball - A
BXC - A                  B Basketball - E        G Crew - E
Field Hockey - A         G Hockey - A            B Crew - A
G Soccer - T             B Hockey - E            B Golf - A
B Soccer - T             G Swimming - E          G Lacrosse - E
B Football - A           B Swimming - A          B Lacrosse - E
G Volleyball - A         G Squash - A            Softball - A
B Waterpolo - E          B Squash - E            Ultimate - A
XX                       Wrestling - E           G Tennis - A
XX                       XX                      B Tennis - A
XX                       XX                      G Track - A
XX                       XX                      B Track - E
XX                       XX                      G Waterpolo - E
[/code]
Total: Andover 17, Exeter 11

Who cares about the number if they suck at them

HAHA YEAH I WAS JUST KIDDING

EXETER PWNS ANDOVER

LIKE SERIOUSLY YOU ARE TALKING ABOUT A SCHOOL THAT IS CAPABLE OF PROVIDING HALF OF THE IMO TEAM

VERSUS A SCHOOL THAT ... CAN'T

yeah, I hate Mr. Fung

he is so down on us all the time

seriously Mr. Fung, screw you

Look at Andover's matriculation data: http://www.andover.edu/cco/matrics/default.asp !  It is pretty impressive. Strangely, Exeter never published their matriculation data on their website (password required  --  I don't know why  :huh: ).

rick, this is so random...?

it is a good choice if you want any challenge in math

new trier is probably better if you want to be well rounded [thats why I am staying at NT]

is your brother moving too?

[quote="besttate"]I have to find a balance between well-roundedness and math. I am looking for the best math education that doesn't compromise my human being to much. New Trier has just been totally unaccommodating for me in math, and I intensely dislike Mr. Viktora. I pretty much started looking at schools at the beginning of this year when Mr. Viktora renigged on his word to me, but NT was still a possibility then. Exeter has a much better challenge in math, but is way too intense. Andover's math program was great, flexible, and could accommodate me.  When I went to Andover, everything just clicked. I actually think from the point of developing all around, Andover is at least equal to NT, but far surpasses it in math.

Oh, and Andy isn't moving though.[/quote]

"renigged" ?

is this supposed to be some weird variant of a racial epithet

or is it supposed to be "reneged"

Dude yeah that is exactly what I was wondering too... so eh yeah I guess all the Exeter people I know are cut-throat competitors who take performance-enhancing drugs (like uhhh Tylenol) and sabotage their opponents to win

(so hmmm that actually brings a particular person to mind, but it has nothing to do with being at Exeter)

and yeah Mr. Fung you think he has "Fun" in his name but then supposedly it is like blearghhhh

although I have never met Mr. Fung I have heard horror stories about him

dude I heard that this one time at exeter

there was this guy

who really really needed to win

ON A TEST

and then he did

uhh...the end

BUT I BET THAT NEVER HAPPENS AT ANDOVER

[quote="BFG"](1) Is this Mr. Fung's first name "Zoo-ming"?

(2) Who has more time to kill, MysticTerminator or probability1.01?[/quote]

eh their power numbers, defined as (amount of time to kill) + (mathematical ability), are the highest i've ever seen

but among middle schoolers my super number defined as (obnoxiousness) + (mathematical ability) is one of the highest. But most of that comes from the obnoxiousness, I have mediocre mathematical ability, unfortunately. :(

b[quote="bubala"][quote="besttate"]Er... OK. I admit that my post was very snobbish and I deserved to be mocked. But in all seriousness, the goal of this post was to just be informative.

I also concede that Exeter has a much stronger math program than Andover. However, Andover is still a very strong school has a stronger math program than New Trier. If my decision was based entirely on math, I would have chosen Exeter. I attended Mr. Fung's "lunchtime group" and was very impressed with both the seminar and him as a teacher. But Andover also has a totally different feel than Exeter, Exeter is a cold hard cutthroat school where the goal is simply to be the best. Like Mr. Fung was yelling at kids when they got problems wrong. Andover is a much warmer place, and is willing to be flexible and accommodating.

I put a lot of though into making my decision and I am confident that I chose the right school for me as a whole person. So, to summarize the intent of my first post, hi and I am happy to be joining this region.[/quote]

Why should it be acceptable to get problems wrong?  Am I missing something?

To reiterate:

[b]You are a traitor.

And you are "totally garbage."

And you are a liar.

And nobody likes you.[/b][/quote]

eh i kinda like him now

[quote="besttate"]I have to find a balance between well-roundedness and math. I am looking for the best math education that doesn't compromise my human being to much. New Trier has just been totally unaccommodating for me in math, and I intensely dislike Mr. Viktora. I pretty much started looking at schools at the beginning of this year when Mr. Viktora renigged on his word to me, but NT was still a possibility then. Exeter has a much better challenge in math, but is way too intense. Andover's math program was great, flexible, and could accommodate me.  When I went to Andover, everything just clicked. I actually think from the point of developing all around, Andover is at least equal to NT, but far surpasses it in math.

Oh, and Andy isn't moving though.[/quote]

Yeaaaa....the NT math program is kind of weak. However, I will say that in the past two and a half years, I have gone from getting less than 80 on the AMC 10 [yes that is pretty embarassing...freshman year... :| ] to being able to get 20ish on this year's usamo. So you don't really need school to help you with your math abilities. While I have not learned a whole lot from my classes [really I have learned almost nothing], I can still develop by reading the textbooks. Our discrete math book is [i]very[/i] good. In class, we barely learn 3 chapters. Who says that I pay any attention??? I just sit there, finish the homework, and read the book. There is a lot of good stuff in there. Who does graph theory in their precalculus class? So I guess you can just learn stuff on your own...Thats not so bad...

As for Viktora...yea... :mad: ... I remember when I was trying to double up on math classes. I literally talked to him 8 times in 3 days [yes I [i]was[/i] counting] until he stopped being evasive and let me get through the paperwork.

Also, this is so lame now. Our school with just be pathetic at math. I was thinking of doing serious stuff for math team next year for the few people who are willing to do harder math. But with John W-G gone [since he is a senior], and you leaving, we only have me, sam [cincodemayo], and like one other person who can understand something at "pre-olympiad" level.

[quote="bubala"]
You are a traitor.

And you are "totally garbage."

And you are a liar.

And nobody likes you.[/quote]

[quote="gighiuhui"]

No worries, mate.

Exeter didn't like you either.[/quote]

This Exeter guy made me feel bad for Exeter.

[quote="rnwang2"][quote="Phelpedo"]You know, with all the multiple usernames running around I'm still doubtful that gighiugiu $\neq$ bubala...[/quote]
They are obviously the same person.[/quote]
Also, trigonometry1.01 and someone else are obviously the same person.

On a related subject, is Phillips Andover planning to start competing in any local math leagues, such as MML or GBML?  MML in particular has a geographic cluster practically centered on Andover:

Westford, Chelmsford, Lowell, Tewksbury, Andover, North Andover, Haverhill

The most distant school, Westford Academy, is just 30 minutes away.

I'd love to see PA go head-to-head against some of the better Massachusetts publics, like Lexington, Canton, and ABRHS; all three compete in MML.

MML isn't possible because of scheduling conflicts (I guess its at the same time as a lot of sports on Wednesdays?). And as you know, a lot of our top mathletes people are highly needed varsity athletes :wink:

@rnwang - obviously you've never had experienced with highly advanced stalking techniques :wink:

As one of our now-retired math team coaches (at Worcester Academy) frequently pointed out, those students who are on the math team and a sports team are usually better at math.  I think that's always been obvious to our athletic coaches -- here the problem is convincing some talented students that they have a better chance of being outstanding in math.

By the way, MML competes on Thursdays, usually starting around 3:30.  How late do PA classes run?  Ours end at 3:15 but we dismiss the team around 2:15 (your [i]farthest[/i] meet would be 30 minutes away, but our [i]closest[/i] meet is 38 minutes away, per Google.)  Like you, Wednesdays and Saturdays are our athletic competition days, but we have so many early dismissals for sports on other days that a dismissal for math team doesn't seem out of the ordinary.

[quote="rjyanco"]As one of our now-retired math team coaches (at Worcester Academy) frequently pointed out, those students who are on the math team and a sports team are usually better at math.  I think that's always been obvious to our athletic coaches -- here the problem is convincing some talented students that they have a better chance of being outstanding in math.
[/quote]

hmmm but let's face it, many more mathletes do track and frisbee and tennis than basketball and baseball and football, just a note

uh yeah i would be pretty content on seeing the pa people at the state/new england meets...for i know like 1203947235 people there from arml

Why did you have to revive a year old thread...


Do any of you already play in the game ars regendi?

It is an economic simulation of your own state and quite well done apart from a few translation errors. It is NOT a war game, instead you have to make your ingrateful populace happy.

If any of you is already there or wants to go there I would be interested in having a mathematicians' alliance to prove that mathematicians are simply the best at everything.

If you follow the link below and start to play you will be counted for me as a referred gamer which will actually enable me to found an alliance:
http://www.ars-regendi.com/rid12370/

If this is not the proper forum, I apologize.

If you are interested in an alliance propose a name here or contact esmeyny on ars regendi.

Eg:
Ars Mathematica
MathLinksEveryone
Axiom


geometry
3D geometry
geometric transformation
rotation

How do you write cube problems and involving painting them? Like... I know rotation is based on them but what are some keys?

Things to keep in mind:
- How many colors do you want?
- There are two ways of picking two faces: faces that are on opposite sides or faces that are adjacent.
- There are two ways of picking three faces: faces that share a common corner or faces that make a "C" type shape
- There are two ways of picking four faces, analogous to leaving out two faces.

As for problem writing? It's pretty straightforward, just do it:
[u]Ex:[/u] [i]Bob has a cube and five colors of paint. How many ways can he paint the cube using exactly three of these?[/i]

If you aren't in the mood for trying to think up uber-original problems, you can always try octahedra  :wink:

Thanks! That was very helpful.

Can anyone also explain how [url=http://en.wikipedia.org/wiki/Burnside%27s_Lemma]Burnside's Lemma[/url] works?


group theory
abstract algebra
superior algebra
superior algebra solved

I have some problems on groups which I cant solve:

1. If $H$ is a subgroup of $G$ such that the product of two right cosets of $H$ in $G$ is again a right coset of $H$ in $G$, prove that $H$ is normal in $G$.

2. Prove that the smallest subgroup of $S_n$ containing $(1,2)$ and $(1,2,\ldots,n)$ is $S_n$, where $S_n$ denotes the symmetric group of degree $n$.

Thank you in advance

1.
it's necessary that then $Ha \cdot Hb = H(ab)$ because $e \in H$ and the spezial case $Ha^{-1} \cdot Ha = H$ gives $a^{-1}Ha \subset H$ ( use $a^{-1} \in Ha^{-1}$ ) and the same way you get $aHa^{-1} \subset H$ and so $aHa^{-1} = H$. So $H$ is normal.

2.
define $a:=(1,2) , b:= (1,2,...,n)$.
Then you get $(k+1 \mod n,k+2 \mod n)=a^{-k} b a^k$ and so all permutations of 'neighbored' numbers. Now inductively find all $(x,y)$ by conjugating the transpositions you already got and it's done

thank you very much ZetaX! I managed to solve the first question after a while too. Now I have problems with this one:

If $H$ is a maximal (i.e. there is no subgroup $K$ of $G$, s.t. $H$ is a normal subgroup of $K$) subgroup of $G$, then $G/H$ is simple (meaning that there are no normal subgroups).

I think you assume $H$ as normal again¿
The canonical homomorphism $\pi : G \to G/H$ is surjective and so for a normal subgroup $K$ of $G/H$ also $\pi^{-1}(K)$  is a (normal) subgroup of $G$ and contains $H$. We have that $\pi^{-1}(K)$ is either $H$ or $G$. Then $K=\pi (\pi^{-1}(K))$ is either $\{ eH\} \subset G/H$ or $G/H$ itself, so $K$ is no true subgroup between $G$ and $H$.


Quick question: If you're a sophomore and take the AMC 12, does the score go toward USAMO index? (I stopped after doing 14 problems)

as a sophomore, you can qualify for the USAMO through the floor

that's when you score equal to or above the lowest AIME floor among those who got in with the index

What is the floor score usually? About how many problems would I have t get right on the AIME as a sophmore to qualify for the USAMO?

this year, I would project a floor score of 7, highest 8

How difficult do you think the aime will be compared to last year?

Most likely harder, since last year's was easier than normal, but we don't have any way of knowing for sure.


[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \dfrac{5 \minus{} 3bc}{1 \plus{} a} \plus{} \dfrac{5 \minus{} 3ca}{1 \plus{} b} \plus{} \dfrac{5 \minus{} 3ab}{1 \plus{} c}\geq ab \plus{} bc \plus{} ca$.[/quote]
After expanding and homogenization we obtain
$ \sum_{cyc}(5a^5\plus{}14a^4b\plus{}14a^4c\minus{}19a^3b^2\minus{}19a^3c^2\plus{}32a^3bc\minus{}27a^2b^2c)\geq0,$ which obvious.

[quote=Inequalities Master]Let $ a,b,c\geq0$ s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \dfrac{5\minus{}3bc}{1\plus{}a}\plus{}\dfrac{5\minus{}3ca}{1\plus{}b}\plus{}\dfrac{5\minus{}3ab}{1\plus{}c}\geq ab\plus{}bc\plus{}ca$.[/quote]

After homogenization we have
\[\sum{\frac{5a^2+5b^2+5c^2+10ab-17bc+10ca}{4a+b+c}}\ge{\frac{9(ab+bc+ca)}{a+b+c}}\]
\[<=>LHS-RHS=\frac{117}{2(a+b+c)}\sum{\frac{ab(a-b)^2}{(4b+c+a)(4a+b+c)}}+\frac{45}{2(a+b+c)}\sum{\frac{(a-b)^2(a+b-c)^2}{(4b+c+a)(4a+b+c)}}\ge{0}\]

[quote=luofangxiang]arqady 一行做完[/quote]

那样的死干早过时了。不过，你想那样一行也可以。 :-D


geometry
trapezoid
3D geometry
pyramid
frustum
LaTeX
geometric transformation

Can you please solve this problem???

In isosceles trapezoid ABCD, legs AD and BC are each 5 units. Diagonal AC is 12 units and is perpendicular to leg AD. What is the length of segment AB expressed as a mixed number?

First, you draw an accurate diagram, then mark what you know. From here, you can quickly realize that triangle ADC is right, so DC (the longer base) is 13. From here, you use similarity to find out AB. Drop height AE, mark length AE as x, and the set up a proportion containing triangle ADE and triangle ADC. This proportion is

\[ \frac{5}{x}\equal{}\frac{13}{12}\]
\[ x\equal{}\frac{60}{13}\]

From here, you can find out length DE using Pythagorean theorem.

\[ DE\equal{}\sqrt{5^2\minus{}(\frac{60}{13})^2}\equal{}\sqrt{25\minus{}\frac{3600}{169}}\equal{}\sqrt{\frac{625}{169}}\equal{}\frac{25}{13}\]

Thus,

\[ AB\equal{}13\minus{}2(\frac{25}{13})\equal{}\boxed{9 \frac{2}{13}}\]

I hope this is correct...  :maybe:

Why is it 5/x = 13/12, I thought it was 5/x = 12/13 ???

Because 5 is the length of the hypotenuse of triangle ADE, so it is $ \frac{5}{x}\equal{}\frac{13}{12}$.

lol thats correct i think.....however you made it look very complicated  :D

Can you help me with one of these other problems?? I have it in another post but i want people who are actually good at solving these problems.

Georgia lives three miles from the bank of a straight river. Her friend Caleigh lives two miles from the river, on the same side, but 12 miles upstream. Georgia must get some water from the river and then go to Caleigh’s house. If Georgia jogs from her house to the river bank and then to Caleigh’s house at a rate of five miles per hour, what is the least amount of time required? Express your answer as a mixed number.

A plane parallel to the base of a right, square pyramid divides the pyramid into a smaller pyramid and a frustum that have equal volumes. If the height of the smaller pyramid is 6 cm, what is the height, in centimeters, of the original pyramid? Express your answer in simplest radical form.

For the Georgia question (the first one), the easiest route is the most direct route, which is a Pythagorean triple with 5 and 12 (do you see this?). This makes the shortest distance to Caleigh's house 13 miles. This will take $ \frac{13}{5}\equal{}\boxed{2\frac{3}{5}}$ hours.

I'm too lazy to do the second question now....  :(

Latex i dont think the pythagorean triple would apply there cuz its on the same side of the river...

For what you said below:

For the Georgia question (the first one), the easiest route is the most direct route, which is a Pythagorean triple with 5 and 12 (do you see this?). This makes the shortest distance to Caleigh's house 13 miles. This will take  hours.

How would you know that you would use 5 and 12 other than any other of the Pythagorean triple??? Also, can you guys help solve the second problem too???

Look at math's diagram. You can "reflect" Georgia's house across the river, so there is a right triangle with legs $ 5$ and $ 12$. So $ 13$ is the shortest distance.

Thanks for the help on the first problem. I finally now understand everything in the first problem. Can you please help me with the pyramid problem?? Here is the problem again:

A plane parallel to the base of a right, square pyramid divides the pyramid into a smaller pyramid and a frustum that have equal volumes. If the height of the smaller pyramid is 6 cm, what is the height, in centimeters, of the original pyramid? Express your answer in simplest radical form.

I solved the Pyramid problem and please help me with this? I think it is easy, but i am confused with the wording.

A car dealer in Yokeland advertises that customers who buy a new car will pay $ 3.99 per gallon of gas for the next three years. Gas prices in Yokeland are projected to be$5 per gallon for the next three years. If Dora drives 12,000 miles annually, and the car she buys gets 30 miles per gallon, then how much will the car dealership have to pay for this promotion for Dora???

Never mind in the previous problem, Can you help solve this???

A beetle starts at the corner of a square that measures 12 meters on each side. The beetle heads 10 meters along the diagonal, turns 45 degrees, and travels 2 more meters. How much closer to the nearest side of the square is the beetle to the second nearest side???

Please help. I am confused.

My friend helped me with the other problems. Can you help me with the ones below?

A bag contains 10 marbles. Each marble is either blue or green, and there are at least two marbles of each color. When three marbles are drawn without replacement the probability that all of them are green is the same as the probability that exactly one of them is blue. How many of the 10 marbles are blue???

A particular right prism has a rectangular octagonal base. Its base edges are 4 cm each and its height is 3 cm. Its volume can be expressed in the form (a(b + radical c)) cubic cm and c is not divisible by the square of a prime. What is that value of a+b+c???

Please answer any one of them.

nightmare i believe i and somebody else answered those two questions already in another section of this forum


ok .. aki cuekgo la parte 3 del
LIBRO DE GEOMETRIA DE LA EDITORIAL MIR..
ES LA SECCION DE AYUDAS .. PAR LA SOLUCION DE LOS PROB DE LA SECCION ANTERIOS ..

AUNQ NO ESTAN TODAS...
LES AYUDARA ,..

Carlos bRAV

[color=blue]
 Buenos dias,

les informo que por restricciones en mi
cuota de uploading...

quitare estos pdf,
por tanto, descarguenlos  antes del lunes en la noche,

Gracias[/color]


Prove that if $ p$ and $ q$ are odd primes and $ q|(a^p\minus{}1)$ then either $ q|a\minus{}1$ or $ q \equal{} 2kp\plus{}1$ for some integer $ k$.

This is equivalent to $ a^p\equiv1$ (mod q), meaning $ ord_q(a)|p$, so either $ ord_q(a) \equal{} 1$ or $ p$.  In the former case, we get $ q|a \minus{} 1$.  In the latter case, since $ a^{q \minus{} 1}\equiv1$ (mod q), $ p|q \minus{} 1$, so $ q \equal{} np \plus{} 1$ for some integer n.  p and q are odd, so n is even, say n=2k.  The result follows.

I didn't think of using orders, thank you!   :)

p/s: I think you meant $ a^{q\minus{}1} \equiv$ (mod q) instead of (mod p) right?


Fie 2 triunghiuri isoscele de laturi x,x,a respectiv x,x,b cu a diferit de b. Daca cele 2 triunghiuri au ariile egale aflatzi x. :)

Daca a=b..
Daca nu,avem $x=\frac{\sqrt{a^2+b^2}}2$. E usor de demonstrat: Fie ABC si MNP cele 2 triunghiuri cu AB=AC=MN=MN=x ,BC=a,NP=b =>$x^2sin(A)=x^2sin(M)(*)$ si cum cazul asta presupune $a\neq b \Rightarrow m(A)\neq m(M)$ rezulta conform(*) ca unghiurile BAC si NMP sunt suplementare rezulta ca avand triunghiul ABC,putem "misca" triunghiul MNP astfel incat A=M si C=N si fiindca unghiurile BAC si MNP sunt suplementare,atunci avem B-A-P coliniare..astfel se formeaza triunghiul BNP ce are A=mijlocul segm(BP) si NA=PA=BA rezulta conform reciprocei teoremei medianei ca triunghiul BNP este dreptunghic de varf N rezulta ca,conform teoremei lui pitagora,avem $4x^2=a^2+b^2$ de unde rezulta concluzia

merge si cu formula lui Heron(care e mort) ?

Probabil ca iese o ecuatie in x care da aceeasi solutie..numai ca s-ar putea sa fie un pic mai multe calcule..

nu stiu eu aveam solutia tot asa....cu ariile observai ca unghiurile dintre laturile x si x erau suplementare...si construiai primu triunghiu isoscel cu unghiu din varf y , apoi cum celalalt avea masura 180-y il faceai in continuare ca suplement :| si se obtinea un triunghiu dreptunghic...:| ca era mediana juma din ipotenuza...si iese raspunsu lu andyciup...
heron e mort ;)

pohoatzo stiam ca Heron e mort am pus intre paranteze doar ca o completare  :lol:

crek m-am prins mba:|...zi si tu kosmin sau ceva:) ca pohoatzo suna grotesc :| oricum aia nici nu era formula lu heron a fost ciordita de el..a fost a lu arhimede:|


function
integration
topology
advanced fields
advanced fields unsolved

Consider to metric spaces X = (o,1) and Y =[0,1] with the usual metric.

the sequence x_n = 1/n is Cauchy in both spaces but only convergent in Y.

Why is it not convergent in X ?

Because convergent means that the limit exists (in the metric space in question).

Look at the definition of converges:

$(x_n)$ converges iff $\exists\,L$ such  that $\forall\,\epsilon>0\,\exists\,N$ such that if $n>N$, then $d(x_n,L)<\epsilon.$

Note that that definition makes no sense unless you can produce $L.$

Now suppose we're in the metric space $(0,1)$ and looking a the sequence $\left(\frac1n\right).$ You try to tell me that this sequence converges to zero, but I say, "What's zero? I don't recognize that there is any such thing."

In principle, this is no different than the following example: Consider the metric space whose elements are the continuous functions on $[0,1]$ and whose metric is given by $d(f,g)=d_1(f,g)=\int_0^1|f(x)-g(x)|\,dx.$ Now consider the sequence $(f_n)$ where $f_n(x)$ is the piecewise linear function whose graph joins the points $(0,0),\,\left(\frac1n,1\right),\,(1,1).$

To determine whether a sequence is Cauchy, we need only consider the sequence itself, and this one passes the test - it is a Cauchy sequence. But if we ask whether it converges, we have to look for a limit function, and our space has only continuous functions in it. We can't find a continuous function which is the the limit of this sequence, so it does not converge.


The teacher asks Bill to calculate $ a\minus{}b\minus{}c$, but Bill mistakenly calculates $ a\minus{}(b\minus{}c)$ and gets an answer of 11. If the correct answer was 3, what is the value of $ a\minus{}b$?

we see that a-b+c=11,and a-b-c=3. Adding yields 2a-2b=14, so a-b=7

If we add a-b-c=3 and a-b+c=11, we have 2a-2b=14, so a-b is 7.

Dude... don't post the same answer with exactly the same solution.  It's spamming.

[quote="isabella2296"]As I said in other threads, please do not post a second answer unless

1. You have a different way of solving it.

2. There was a mistake in someone else's solution.

3. You got a different answer.

Thanks!  :)[/quote]
I have a question though:
so u use system of equations to solve it?

so you just add the two together and solve?

Yep, you just add them. This method is called elimination, because the $ c$ variable drops out.

[quote="ernie"]Yeah.  And if my brain serves me correctly (IDK where I got that one from), it is known as the [b]addition method[/b].[/quote]

I think the addition method is when you add two(or more) equations and no variables drop out, but in elimination, a variable drops out. I'm not sure if I'm right. Does anybody know?

@izzy: thanks a lot:im a total newbie at systems of equations since im not gonna be learning it till next year.
wait...but dont the B's cancel out ?

No, because in both equations they are negative. Variables only drop out when they are additive inverses. For example, $ c$ and $ \minus{}c$ or $ 2x$ and $ \minus{}2x$.

Only the C's cancel out.  You can't cancel out the B's, because you have to find the difference between A and B.

[quote="ernie"]You can't cancel out the B's, because you have to find the difference between A and B.[/quote]

Well...if they [u]were[/u] additive inverses, you kind of wouldn't have any choice.

Like: $ a \minus{} b \equal{} \text{whatever}$ and $ a \plus{} b \equal{} \text{whatever}$, you would end up with

$ a \plus{} a \minus{} b \plus{} b \equal{} 2 \times \text{whatever}$

which simplifies to

$ 2a \equal{} 2 \times \text{whatever}$

$ a \equal{} \text{whatever}$

No problem.  :wink:

Please feel free to ask if you have any other questions.  :)


From what I can see this will basically turn into some form of geometric series. This is because if $ A$ is lower or upper triangular then so is $ A^k$ where $ k\in\mathbb{N}$. And $ \det (A)$ is just the product of the entries along the diagonal.


Ok I simply am not getting this one and I have to stop wasting time; so I came to who knows best = ).

[u]sin(x)cos(y) + cos(x)sin(y)[/u]  =  [u] tan(x)+tan(y) [/u]
cos(x)cos(y) - sin(x)sin(y)         =       1-tan(x)tan(y)

P.S. How do i make a pretty image for equations or terms?

PLEEZ show individual steps (using substitutions of trig identities)

Thanks Much guys

[quote="igiul"]Ok I simply am not getting this one and I have to stop wasting time; so I came to who knows best = ).

[u]sin(x)cos(y) + cos(x)sin(y)[/u]  =  [u] tan(x)+tan(y) [/u]
cos(x)cos(y) - sin(x)sin(y)                1-tan(x)tan(y)

sorry if i botched the problem. (post #1)

PLEEZ show individual steps (using substitutions of trig identities)

Thanks Much guys[/quote]

[hide]Use identities
\begin{eqnarray*}\sin(x+y)&=&\sin x \cos y+\cos x \sin y\\ \cos(x+y)&=&\cos x \cos y-\sin x \sin y\\ \tan(x+y)&=&\frac{\tan x+\tan y}{1-\tan x \tan y}\end{eqnarray*}
and get
\begin{eqnarray*}\frac{\sin x \cos y+\cos x \sin y}{\cos x \cos y-\sin x \sin y}&=&\frac{\sin(x+y)}{\cos(x+y)}\\ &=& \tan(x+y)\\ &=& \frac{\tan x+\tan y}{1-\tan x \tan y}. \end{eqnarray*}
Done.[/hide]

Use $\text{\LaTeX}$ for pretty images.

Please don't post these problems in the intermediate forums because I think it should be in Basics.

Thanks much man, The sum difference weren't listed in the chapter; I merit that one to the author. (It was however in the front)

Will do Sirrah.

I don't think that's a good way to do this problem...
that's what I would do to DERIVE the tangent addition formula in the first place(although it could be other ways, I guess), so it's kind of a circular reasoning.
[hide="hint"]divide denominator and numerator of left hand side by $\cos x\cos y$[/hide]

btw, this kind of problem(basic trig stuff) would go in the HSB forum.


superior algebra
superior algebra unsolved

My text says
[quote]$ Gal(K/L_{i\minus{}1})/Gal(K/L_{i})$ is cyclic because $ L_i$ is a simple radical extension of $ L_{i\minus{}1}$.[/quote]

Why does the latter imply that?

$ Gal(K/L_{i\minus{}1})/Gal(K/L_i) \cong Gal(L_i/L_{i\minus{}1})$


Many people have many paths lets see which is the most poular out of these.

I was going into mathematics, phd hopefully. But then I got myself in trouble. So its over.

[quote="alkjash"]Even though I love math, I don't think it's as useful as science in the real world.[/quote]
So. Think of it as art. Then it doesnt need to be useful.  :wink:

Ha ha,  :rotfl:  who would chose desinger?
By the way, you forgot engineering. A big one.

You forgot surgeon (I want to be a heart surgeon) and musician (I want to teach music in my spare time)...oh and designing buildings would be cool (architecture) - I took designer as clothes.

[quote="sylow_theory"]Ha ha, who would chose desinger?[/quote]
That's not nice! BTW, that's not me[/quote]

[quote="sylow_theory"]By the way, you forgot engineering. A big one.[/quote]

Thank you.

[quote="alkjash"]Even though I love math, I don't think it's as useful as science in the real world.[/quote]
Nevertheless, I love math. I think "sth is loved" $\not\Rightarrow$ "sth is useful". Also, I think math is the foundation of all science, so math is very useful. (Maybe it isn't as useful as science in the real world)

why'd you put designer.. out of all the other choices anyways

I'm thinking of becoming an engineer/inventor/designer/some c.. thing (cio,cko,ceo,coo...)


How many number of $k$-tuples $(i_{1},i_{2},\ldots i_{k})$ of consecutive numbers we can count out of the given sequence $(r_{1},r_{2},\ldots r_{n})$?  :(

Just work that out:  the answer is $n-k+1$

For example, we can count $5=6-2+1$ consecutive pairs $(1,2),(2,3),(3,4),(4,5),(5,6)$ out of the sequence $(1,2,3,4,5,6)$   :lol:


Given triangle $ ABC$ with $ C$ a right angle, show that the diameter of the incenter is $ a\plus{}b\minus{}c$, where $ a\equal{}BC$, $ b\equal{}CA$, and $ c\equal{}AB$.

hello, let $ D$ the osculation point of $ \overline{AC}$ with the incircle, $ E$ with $ \overline{AB}$ and $ F$ with $ \overline{BC}$, then we have $ \overline{AD}\equal{}\overline{AE}\equal{}b\minus{}r$ and $ \overline{BF}\equal{}\overline{BE}\equal{}a\minus{}r$ and from here we get $ c\equal{}a\plus{}b\minus{}2r$, hence $ r\equal{}\frac{a\plus{}b\minus{}c}{2}$ or the diameter is given by $ a\plus{}b\minus{}c$.
Sonnhard.

By Pythagoras theorem,
c²=a²+b²
c²=(a+b)²-2ab
2ab=(a+b+c)(a+b-c)
ab={(a+b+c)(a+b-c)}/2
Area of triangle ABC=(1/2)ab
let inradius be r
rs=∆
r(a+b+c)/2=(1/2) ab
r(a+b+c)=ab
r(a+b+c)={(a+b+c)(a+b-c)}/2
r=(a+b-c)/2
2r=a+b-c
The diameter is a+b-c