Five consecutive terms of the Warloe sequence are given below.
In the Warloe sequence, each term after the third term is the sum
of the previous three terms. The 8th term has been identified.
What is the sum of the 1st term and the 13th term in the Warloe
sequence?

[asy]unitsize(1inch);
label(". . . 1, 2, 3, 6, 11, . . .",(0,0),E);
draw((0.5,0.5)--(0.5,0.1),Arrow);
label("$8^{th}$ term",(0.5,0.5),N);[/asy]

We write out the sequence backwards to get the first term.
3=2+1+0
2=1+0+1
1=0+1+(-1)
0=1+(-1)+0
1=(-1)+0+2
-1=0+2+(-1)
Therefore the first term is -1.

Then we work forwards to get the 13th term (we have up to the eleventh).
3+6+11=20
6+11+20=37.

Thus, the sum of the 1st and 13th numbers in the sequence is 36.

(Can anyone here do it in 45 seconds without a piece of paper?)

I think this is just one of those questions that aren't suited for the typical 45-second game, unless you mem it.


Human beings have discovered an inhabitable planet and soon after, they find 10 more inhabitable planets. Of these 11, only 5 are deemed "Earth-like" in its resources and the rest are deemed "Mars-like" since they lack many important resources. Assume that planets like Earth take up 2 units of colonization, while those like Mars take up only 1. If humanity mobilizes 12 total units of colonies, how many different combinations of planets can be occupied if the planets are all different from each other?

So the problem is to find the number of ways 12 may be obtained as a sum from a combination of 1s and 2s, with the limits of 6 1s and 5 2s.
You can 2 1s and 5 2s, 4 1s and 4 2s, or 6 1s and 3 2s.
There are 6C2*5C5 ways to pick the planets in the first scenario, 6C4*5C4 ways in the second, and 6C6*5C3 for the third.
These add to a total of [b]100[/b] ways.
I think.


I just answered 2 questions after algebra class last week, to try it out.
My Report says I answered 5 (and got 3 wrong) but when I look at those problems, I never saw them before.
Somehow it thinks I tried those problems.  Has anyone else had this happen to them?

Hard to say since we now show you as answering 12 problems. Before answering any more problems, can you tell us which problems the report is showing you as having completed but in reality not completed.


let $ a$ , $ b$ , $ c$ $ \in ]0,\plus{}\infty[$ , such that , $ \frac{1}{a\plus{}b\plus{}1}\plus{}\frac{1}{b\plus{}c\plus{}1}\plus{}\frac{1}{c\plus{}a\plus{}1} \geq 1$ , prove that :

$ a\plus{}b\plus{}c\plus{}1 \geq 4abc$

It reminds me about Romania 2005 unused problem (But Romania's prob is inverse with $ abc\geq 1$)
we will get that $ 1\geq abc$ from the inequality given and then it's easy to prove that $ a\plus{}b\plus{}c\plus{}1\geq 4abc$ by AM-GM

thank"s

by AM -GM we have $ a\plus{}b\plus{}c\plus{}1 \geq4 \sqrt[4]{abc}$ after that , can you explain please ?

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in ]0, + \infty[$ , such that , $ \frac {1}{a + b + 1} + \frac {1}{b + c + 1} + \frac {1}{c + a + 1} \geq 1$ , prove that :

$ a + b + c + 1 \geq 4abc$[/quote]
setting $ p = a + b + c, q = ab + bc + ca$ and $ r = abc$

$ \frac {1}{a + b + 1} + \frac {1}{b + c + 1} + \frac {1}{c + a + 1} \geq 1$

$ \Leftrightarrow 2(p + 1)\ge pq - r$

we need prove that $ a + b + c + 1 \geq 4abc \Leftrightarrow p + 1\ge 4r$

we have $ 2(p + 1)\ge pq - r \Rightarrow p + 1\ge \frac {pq - r}{2}$

thus, we need prove $ \frac {pq - r}{2}\ge 4r$

$ \Leftrightarrow pq\ge 9r$ which is true by AM-GM :wink:

nice prof
 this is my prof

$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \geq 1$

$ \Leftrightarrow 2(p \plus{} 1)\ge pq \minus{} r$

$ pq\minus{}r \equal{} (a\plus{}b)(b\plus{}c)(c\plus{}a)$

$ \implies 2(p\plus{}1) \geq (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 8abc$

$ \implies a\plus{}b\plus{}c\plus{}1 \geq 4abc$

[quote="y-a-s-s-i-n-e"]thank"s

by AM -GM we have $ a \plus{} b \plus{} c \plus{} 1 \geq4 \sqrt [4]{abc}$ after that , can you explain please ?[/quote]
Well!! we have to prove that
if $ abc\geq 1$ then $ 1\geq \sum \frac {1}{a \plus{} b \plus{} 1}$
which is true by setting $ k^3 \equal{} abc; k\geq 1$ then we have $ x \equal{} \frac {a}{k} y \equal{} \frac {b}{k} z \equal{} \frac {c}{k}$ then $ xyz \equal{} 1$
combine with
$ 1\geq\sum \frac {1}{x \plus{} y \plus{} 1}\geq\sum \frac {1}{kx \plus{} ky \plus{} 1} \equal{} \sum \frac {1}{a \plus{} b \plus{} 1}$
(for here we even got that for all $ n\in N$ and $ x,y,z>0$ if $ xyz\equal{}1$ then $ 1\geq\sum \frac {1}{x^n \plus{} y^n \plus{} 1}$ it's Baltic Way 2004 which can be proven by setting $ x^{\frac {n}{3}}\equal{}a y^{\frac {n}{3}}\equal{}b z^{\frac {n}{3}}\equal{}c$ and then it's USA 1997 which just using $ a^3\plus{}b^3\geq a^2b\plus{}ab^2$ bla bla bla and we are done!!)
so we got that if $ \sum \frac {1}{a \plus{} b \plus{} 1} \geq 1$ imply $ 1\geq abc$
so this problem is done by am-gm since $ 1\geq abc$ so $ 4\sqrt [4]{abc}\geq 4abc$
Am i right?  :lol:


How come the AoPS server is down so consistently and spontaneously? I would be doing like an alcumus problem, go check the stats, then have my whole internet freeze up. I would go to google, and it would work fine, but once I try going back to AoPS, the server is still down. Why? Is there some kind of update?

Nothing today.  I think it is probably a local problem with your computer.


Let $a,b\in R$ such that $(x-a)(x-b)+(x-a)(x-\sqrt{2})+(x-b)(x-\sqrt{2})\geq 0$ for any real $x$. Proof that $a,b\in R$\$ Q$

Z = the integers, R = the real numbers (including the integers and the rational numbers).  Perhaps you want to re-write that?

Don't know how to write those in $\LaTeX$  :ninja:

I mean, your question doesn't make any sense as written:  it's titled "Proving a and b are irrational."  Then you give a and b as integers and ask me to prove that they are real numbers?

:oops_sign:
Yes. I'm sorry for that mistake. I'm very tiered.  :fool:

[quote="Slizzel"]Let $a,b\in R$ such that $(x-a)(x-b)+(x-a)(x-\sqrt{2})+(x-b)(x-\sqrt{2})\geq 0$ for any real $x$. Proof that $a,b\in R\Q$[/quote]

I see the code now.. You meant to say at the end:

$a,b \in Q$?

If so, you might want to change the title.


Quote:What is the smallest positive integer which cannot be expressed as the sum of 8 positive cubes?

I might be wrong on this but I actually worked on this for 15 minutes until I found out how easy this was.

Okay, here's the deal: You have to find the smallest positive number that cannot be expressed as sum of 8 positive cubes. First, you don't have to find out what 8th positive cube is. All you need is 1st positive cube, which is 1. Using what we know, we get this as answer:

 [hide]1. The reason is that it's the smallest positive integer that cannot be expressed as sum of 8 positive cubes. Another hand, the smallest integer that can be expressed as 8 cubes is 8. Know why? 1+1+1+1+1+1+1+1 = 8[/hide]

I'm pretty sure that's the answer.

*thinks about how the problem would be if it was 8 or less positive cubes*

Ah yes, that is the more interesting question.  (And, of course, had I written it right, the intended question!)


[quote="N.T.TUAN"]Equation is equivalent to $\ln{2x(x+2)}=\ln{\sqrt{\frac{x+3}{2}}}(x>0)$, or $2x(x+2)=\sqrt{\frac{x+3}{2}}$. Now, it is easy! :D

You are wrong when post this problem in this box.[/quote]
Lol, i don't thinks so. I can't solve four degree equation :(
Can u  help me

No, you can, see here http://en.wikipedia.org/wiki/Quartic_equation  :wink:

Yes, you can solve a general fourth degree equation - this is a fundamental result of Galois theory and deals with field extensions, etc., etc. but which is not a topic for here.  The process is incredibly ugly and virtually useless unless you are designing a Mathematica program or whatnot.

$8x^{2}(x+2)^{2}= x+3$

$8x^{4}+32x^{3}+32x^{2}-x-3 = 0$

Solve that!

[quote]How did you find that?
[/quote]
One fairly standard way:

[hide]Notice  the first two coefficients to see that $(1/4)(32/8)=1$ so put $y=x+1$ (Or $x=y-1$ to get rid of the cubed term and get:

$8y^{4}-16y^{2}-y+6=0$
(To avoid fractions, rewrite this so that first coefficient is a square number)
or $16y^{4}-32y^{2}-2y+12=0$

Now assume it can be written as

$(4y^{2}+ay+p)(4y^{2}-ay+q)=0$
($4p+4q-a^{2}=-32$ , $a(q-p)=-2$,  $pq=12$

to get:  (a=2,p=-3,q=-4)

and you get ..

$(4y^{2}-2y-4)(4y^{2}+2y-3)= 0$

solve for y  and subtract 1 fro this to get x  :)
[/hide]

[quote]One fairly standard way: .....[/quote]
yep, that's the classic way probably... but I used Singular... it's faster  :lol:


geometry
circumcircle
geometric transformation
reflection
Euler
incenter
trigonometry

1.Denote the reflection of X w.r.t YZ is E, we will prove the EA,OI,BC are concurrent.
2.Let I1I2I3 be the excentral triangle. The lines Y Z and I2I3 are parallel because both are perpendicular to AI. Similarly, ZX//I3I1 and XY//I1I2.  Hence, the excentral triangle and the intouch triangle are homothetic and their Euler lines are parallel. Now, I and O are the orthocenter and nine-point center of the excentral triangle. On the other hand, I is the circumcenter of the intouch triangle. Therefore, the line OI is their common Euler line, contains the orthocenter H' of XY Z.
3.AI/IA1=BA/BA1  XH’=2r*cos(A/2)  XE=2h   EH’= XE- XH’
where h, r are the attitude from X to YZ and the incircle radii
   we can calculate AI/IA1= EH’/ XH’= (sinB + sinC)/sinA

Therefore, we complete the proof.

[quote="Maxim Bogdan"]Let $ ABC$ be a non-isosceles triangle, in which $ X,Y,$ and $ Z$ are the tangency points of the incircle of center $ I$ with sides $ BC,CA$ and $ AB$ respectively. Denoting by $ O$ the circumcircle of $ \triangle{ABC}$, line $ OI$ meets $ BC$ at a point $ D.$ The perpendicular dropped from $ X$ to $ YZ$ intersects $ AD$ at $ E$. Prove that $ YZ$ is the perpendicular bisector of $ [EX]$.[/quote]
Take a look at http://forumgeom.fau.edu/FG2004volume4/FG200411index.html

This is what I wrote for the RMC (and in the contest). Practically, it is a compilation of the solution given in the note April cited above. There is also another synthetic solution by Dinu Serbanescu, but I shall write it when I have more time.

[b]Theorem[/b] (Emelyanov). Let $ ABC$ be a nonisosceles triangle, for which denote by $ X$, $ Y$, $ Z$ the tangency points of its incircle with the sides $ BC$, $ CA$ and $ AB$, respectively. Let $ D$ be the intersection of $ OI$ with the sideline $ BC$, where $ O$, $ I$ are the circumcenter and incenter, respectively. The line perpendicular to $ YZ$ through $ X$ cuts $ AD$ at $ E$. Prove that the line $ YZ$ is the perpendicular bisector of $ EX$.

[b]Solution[/b]. We begin with a preliminary result.

[b]Lemma[/b]. The orthocenter $ H'$ of the intouch triangle $ XYZ$ lies on the line $ OI$.

We will give two different proofs for this useful result.

[i]First proof[/i]. Let $ I_{a}I_{b}I_{c}$ be the excentral triangle of $ ABC$. The lines $ YZ$ and $ I_{b}I_{c}$ are parallel because both are perpendicular to $ AI$. Similarly, $ ZX \| I_{c}I_{a}$ and $ XY \| I_{a}I_{b}$. Hence, the excentral triangle and the intouch triangle are homothetic and their Euler lines are parallel. Now, $ I$ and $ O$ are the orthocenter and the nine-point center of the excentral triangle. On the other hand, $ I$ is the circumcenter of the intouch triangle. Therefore, the line $ OI$ is their common Euler line, and thus it contains the orthocenter $ H'$ of $ XYZ$. $ \hfill \square$

[i]Second proof[/i]. Let $ A'$, $ B'$, $ C'$ be the midpoints of $ YZ$, $ ZX$ and $ XY$, respectively. Note that
\[ IA \cdot IA' = \left(\frac {r}{\sin{\frac {A}{2}}}\right) \cdot \left(r \sin{\frac {A}{2}}\right) = r^{2}.\]
Similarly $ IB \cdot IB' = IC \cdot IC' = r^{2}$. Therefore, the inversion $ \Psi(I,r^{2})$ with respect to the incircle of $ ABC$ maps the vertices $ A$, $ B$, $ C$ to $ A'$, $ B'$, $ C'$, and moreover maps the circumcircle of $ ABC$ into the circumcircle of $ A'B'C'$. But if a circle with center $ P$ is inverted in a circle with center $ Q$, then its image is a circle whose center lies on the line $ PQ$. Thus, the circumcenter $ O'$ of $ A'B'C'$ lies on the line $ OI$. Since $ A'B'C'$ is the medial triangle of $ XYZ$, its circumcenter $ O'$ is the nine-point center of $ XYZ$. Following the fact that $ I$ is the circumcenter of $ XYZ$, it becomes clear now that $ OI$ is the Euler line of $ XYZ$. Obviously, the orthocenter $ H'$ lies on this line. $ \hfill \square$

Returning to our problem, we notice that the Lemma motivates us to rewrite the last part into the following equivalent version:

[b]Claim[/b]. Let $ E$ be the reflection of $ X$ into the line $ YZ$. Denote by $ D$ the intersection of $ AE$ with the sideline $ BC$. Prove that $ D$ lies on the line $ OI$.

[i]Proof[/i]. Denote by $ T$ the intersection of $ AI$ with $ BC$. Since $ XE$ and $ AT$ are both perpendicular to $ YZ$, they are parallel and therefore the collinearity is equivalent with $ EH'/H'X = AI / IT$.

It is known that
\[ \frac {AI}{IT} = \frac {CA + AB}{BC} = \frac {\sin{B} + \sin{C}}{\sin{A}}.\]
For example, this follows from Van Aubel's theorem.

For any acute triangle, $ AH = 2R \cos{A}$. The angles of the intouch triangle are
\[ X = \frac {B + C}{2},\ \ Y = \frac {C + A}{2},\ \ Z = \frac {A + B}{2}.\]
It is clear that $ XYZ$ is always acute, and
\[ XH' = 2r\cos{X} = 2r \cos{\frac {B + C}{2}} = 2r \sin{\frac {A}{2}}.\]
Thus,
\begin{eqnarray*} \frac {EH'}{H'X} & = & \frac {EX - H'X}{H'X} = \frac {EX \cdot YZ}{H'X \cdot YZ} - 1 \\
& = & \frac {4[XYZ]}{H'X \cdot YZ} - 1\ \ \ ([\mathcal{P}] \text{denotes the area of the polygon}\ \mathcal{P}) \\
& = & \frac {2r^{2}(\sin{2X} + \sin{2Y} + \sin{2Z})}{2r\sin{X} \cdot 2r \cos{X}} - 1 \\
& = & \frac {\sin{2Y} + \sin{2Z}}{\sin{2Y}} = \frac {\sin{B} + \sin{C}}{\sin{A}}. \end{eqnarray*}
This completes our proof. $ \hfill \blacksquare$

[i]Remarks[/i]. We invite the reader to deduce the following (simple) consequences:

[b]Corollary 1[/b]. The line joining $ D$ to the projection of $ X$ on $ YZ$ passes through the midpoint of the bisector of angle $ A$.

[b]Corollary 2[/b]. The $ OI$-line is parallel to $ BC$ if and only if the projection of $ X$ on $ YZ$ lies on the line joining the midpoints of $ AB$ and $ AC$.

Since IO is the Euler line of XYZ, we can restate the enunciation as a more useful-general result.

[b]Proposition:[/b] Triangle $\triangle ABC$ is scalene with euler line $e.$ $\triangle A_0B_0C_0$ is the tangential triangle of $\triangle ABC,$ $A_0,B_0,C_0$ against $A,B,C.$ If  $D$ is the reflection of $A$ across $BC,$ then lines $e,B_0C_0, DA_0$ concur.

Using barycentric coordinates with respect to $\triangle ABC,$ we have

$A_0  (-a^2:b^2:c^2) \ , \ B_0  (a^2:-b^2:c^2) \ , \ C_0 (a^2:b^2:-c^2) \ , \ D(-a^2:2S_C:2S_B)$

$e \equiv S_A(S_B-S_C)x+S_B(S_C-S_A)y+S_C(S_A-S_B)z=0$

$\Longrightarrow \ DA_0 \equiv 2(b^2S_B-c^2S_C)x+a^2(a^2-b^2)y-a^2(a^2-c^2)z=0$

$\Longrightarrow \ B_0C_0 \equiv c^2y +b^2z=0$

Then, lines $e,B_0C_0, DA_0$ are indeed concurrent at a point $U$ with coordinates

$U \equiv (a^2b^2(a^2-b^2)+a^2c^2(a^2-c^2):-2b^2(b^2S_B-c^2S_C):2c^2(b^2S_B-c^2S_C))$

Dear Mathlinkers,
for a synthetic proof you can see
http://perso.orange.fr/jl.ayme   vol. 4 Symérique de OI... p. 22.
Sincerely
Jean-Louis

I hope I didn't mislabel any points. Here's a rather complicated solution using homothety, I wonder if someone simplify this?

Let $\triangle A'B'C'$ be the triangle such that $\triangle ABC$ is the orthic triangle of $\triangle A'B'C'$. It is well known that $I$ is the orthocenter of $A'B'C'$ and $A'$ is the excenter of $ABC$. Let $O'$ be the circumcenter of $\triangle A'B'C'$; then by Euler Line, $O', O, I$ are collinear. Let the $A$-excircle of $\triangle ABC$ touch $BC, CA, AB$ at $X', Y', Z'$, respectively. Let $AX$ intersect the $A$-excircle at $W$; reflect $W$ over $Z'Y'$ to get $W'$. Reflect $A'$ over $BC$ to get $A''$.

Note that $\triangle Z'WY'$ is homothetic to $\triangle CAB$, so $A''O' \parallel W'A'$. Let $W'A$ meet $A''O'$ again at $P$; because of parallel lines, $AP = AW'$. Since $A'$ is also the midpoint of $X'W$, and $A'A \parallel WW'$, $X'P \parallel A'A$.

Let $Q$ be the reflection of $X$ over $YZ$; then from a homothety centered at $A$ taking the incircle of $\triangle ABC$ to its $A$-excircle, $Q$ lies on $AW'$ and $XQ \parallel WW' \parallel X'P$. Since we also have $IQ \parallel W'A' \parallel PB$ and $IX \parallel A'O'$, $\triangle O'X'P$ is homothetic to $\triangle IXQ$, so $OI$, $BC$, and $AQ$ concur. Then $Q = E$, so $YZ$ bisects $EX$, as desired.

[hide]Hint : suppose that $ZY$ is perpendicular-bisector of $EX$ and $AE \cap BC = D$. it's enough to prove that $O,I,D$ are on a same line . now use sin law for $E,A,Y,Z$ and then for $O,I,B,C$[/hide]

i hope this wasn't posted above
 next $H,R,H'$ are the orthocenter of $XYZ$ and second intersections of $XI,XH$ with the circle $XYZ$.
Inversion in $I$ with radius $IY$ pictures $M,N,P$ to $A,B,C$($M,N,P$ are midpoints of $YZ,ZX,XY$) and therefore the Euler's circle of $XYZ$ into circle $ABC$. so $O$ is on the line connecting $I$ and the center of circle $MNP$(line $IH$) so $I,O,H$ are collinear $AI\cap BC=D$ then $\angle H'XR=\angle XID$ and $\angle XH'R=\angle IXD=90$ so $IXD\sim XH'R$ and $ID/XI=2*XI/XH'$ and $ID*XH'=2*XI^2$ since $\angle H'YZ=\angle HXZ=\angle HYZ$ and $HH'\perp YZ$ then $H,H'$ are symmetric wrt $YZ$.
next since $XH*AI=2*MI*AI=2*IY^2=2*IX^2$ then using the previous identity we have $XH/XH'=ID/IA=XH/EH$ so $EH=XH'$ and now $E,X$ are symmetric wrt $YZ$. so $YZ$ is the perpendicular bisector of $EX$.

Remark:
proving that $H,I,O$ are collinear can be solved without inversion considering the feet of altitudes in $XYZ$(points $P,Q,S$) and the homothethy that pictures $PQS$ into $ABC$)

[quote="plane geometry"]
3.$XH'=2r*cos(A/2)$  $XE=2h$   $EH'= XE- XH'$
where $h, r$ are the attitude from $X$ to $YZ$ and the incircle radii
   we can calculate $AI/IA_{1}= EH'/ XH'= (sinB + sinC)/sinA$

Therefore, we complete the proof.[/quote]

there is typo mistake:
$XH'=2rsin(\frac{A}{2})$ not cosine
and the rest follows.. ;)


geometry
conics
ellipse
geometry theorems

$\pi ab$, where $a$ and $b$ are the semimajor and semiminor axes.


[quote="borislav_mirchev"]I think these are interesting math problems.
Their level is not so high so many people may solve them.
Sorry if they posted before.

P.S. In the second problem relation should be CA+AE=BC.[/quote]
[hide="1"]
Consider point $ P$ on $ BC$ so that $ PB \equal{} AE$.  This means that $ CP \equal{} BC \minus{} PB \equal{} BC \minus{} AE \equal{} AC$.  Letting $ \angle ACB \equal{} 2\gamma$, we see that $ \angle CAP \equal{} 90 \minus{} \frac {\angle ACP}{2} \equal{} 90 \minus{} \gamma$.  Also, $ \angle EAP \equal{} \angle CAP \minus{} \angle CAE \equal{} 90 \minus{} \gamma \minus{} \frac {\angle BAC}{2} \equal{} 55 \minus{} \gamma$.  Furthermore, $ \angle BEC \equal{} 90 \plus{} \frac {\angle BAC}{2} \equal{} 125$ since $ E$ is the incenter of $ \triangle ABC$, so\[ \angle EBC \equal{} 180 \minus{} \angle BEC \minus{} \angle ECB \equal{} 55 \minus{} \frac {\angle ACB}{2} \equal{} 55 \minus{} \gamma \equal{} \angle PAE\]which implies that $ AEPB$ is cyclic.  Yet, $ AE \equal{} PB$, so $ AEPB$ is an isoceles trapezoid.  This means that $ \angle PAB \equal{} \angle EBA \equal{} \angle EBC \equal{} 55 \minus{} \gamma$.  Since $ \angle PAC \equal{} 90 \minus{} \gamma$, we see that $ 70 \equal{} \angle BAC \equal{} \angle CAP \plus{} \angle PAB \equal{} 90 \minus{} \gamma \plus{} 55 \minus{} \gamma \equal{} 145 \minus{} 2\gamma$.  We conclude that $ \angle ACB \equal{} 2\gamma \equal{} 75$. [/hide]
[hide="2"]
Notice that $ \angle AEB \equal{} 90 \equal{} \angle AHB$, so $ AEHB$ is cyclic, so $ \angle CAH \equal{} \angle FBH$.  Also, $ \angle FHB \equal{} 90 \equal{} \angle AHC$, so $ \triangle FBH\sim \triangle CAH$, which means that $ \frac {FH}{CH} \equal{} \frac {BH}{AH}$, so $ AH\cdot FH \equal{} HB\cdot CH$.  We also have that $ \angle HGB \equal{} 90 \minus{} \angle GBH \equal{} \angle GCB$ since $ GH\perp BC$ and $ GB\perp GC$ since $ G$ lies on the circle with diameter $ BC$.  Furthermore, notice that $ \angle GHB \equal{} 90\equal{}\angle CHG$, so $ \triangle BHG\sim \triangle GHC$, so $ \frac {GH}{BH} \equal{} \frac {CH}{HG}$, which implies that $ AH\cdot FH \equal{} BH\cdot CH \equal{} GH^2$. [/hide]

Take F’ the symmetrical of F w.r.t. BC, then ABH’C is cyclic and, by Ptolemy we have AH.F’H = BH.CH, then from the triangle BCG right angled, BH.CH = GH2.

Best regards,
sunken rock


geometry
3D geometry
pyramid
analytic geometry

Find the number of paths that you can make out "Georgia".

             G
           GEG
         GEOEG
       GEOROEG
     GEORGROEG
   GEORGIGROEG
  GEORGIAIGROEG

EDIT:There was a slight error when I typed this in, so arrange them in a pyramid and then solve...sorry

[quote="Arvind_sn"]Find the number of paths that you can make out "Georgia".

[code]        G
       GEG
      GEOEG
     GEOROEG
    GEORGROEG
   GEORGIGROEG
  GEORGIAIGROEG
[/code][/quote]

Problem solved. We had a similar one, but we had to spell mathematician.

yes it is...nice...now, can you prove it?

Just look at the number of paths starting with the G at the bottom left, then the number of parths starting with the next G, and so on.

It turns into a nice binomial sum.

Of course that's not a proof, but an outline  :)

[hide="better"]Work backwards, starting at the A.  You can then choose either to exclude the left or exclude the right.  You then have two choices for an I, and then two choices for an G, etc., so there are $2\cdot 2^{6}$ possibilities.  But we have, of course, counted the one which goes straight up twice.  Therefore there are in reality $2^{7}-1 = 127$ possibilities.[/hide]

[quote="Karth"][hide="One-liner  :) "]
The number of ways to go from $(0,0)$ to $(a,b)$ on a coordinate plane is $\binom{a+b}{a}$ and note that $\sum_{i=0}^{n}\binom{n}{i}= 2^{n}.$[/hide][/quote]
Yes, but mine is still better :P

[quote="mathgeniuse^ln(x)"]Karth, that isn't really "one" line.[/quote]

lol true. but it was the quickest answer. boysopranos was pretty elegant considering this was pretty simple.

[quote="mathgeniuse^ln(x)"]Karth, that isn't really "one" line.[/quote]

Well, it's the overall logic required.  :D


function
ratio
trigonometry
real analysis
geometric series
real analysis unsolved

I’m having problems with the review problems for my final. Hope somebody could help me.
a)	The set of all functions from $ N$ to $ N$ is uncountable.

I think this is true since there is a bijection from $ N$ to $ N$ ( we can show that bijection by showing that f: $ N$ into $ N$ is an injection and g: $ N$ into $ N$ is also an injection. Then by a theorem, there exists a bijection). Since $ N$ is countable, and f bijects $ N$ onto a countable set, so it must be countable.

b)	If f: $ R$ into $ R$ is differentiable, and if c in $ R$, then there exists a,b in $ R$, a<c<b such that f’(c)= ( f(b) – f(a) ) / (b-a)

I think this is false, but I can’t come up with a counter-example. My intuition tells me it’s false because it looks like the mean value theorem, but it is not.

c)	The series $ (\minus{}1)^n$/ n-th root of n converges.

I think this is true since after doing the root test on this series I got alpha=-1 , which is less than 1

d)	No at most countable subset of $ R$ other than the empty set is open w.r.t the standard metric.

I don’t get the statement here.

e)	If ($ f_n$) is a sequence of differentiable function $ f_n$: [-1,1] into $ R$ and $ f_n$ converges to f uniformly, f: [-1,1] into $ R$, then f is differentiable.

I think this is false since I remember learning that the uniform limit of differentiable need not be differentiable.
f)	If f: $ R$ into $ R$ is differentiable and f’ is continuous at all x except 0, then f’ is continuous at 0.

Can I use f(x)=1/x as a counter-example here?

g)	If f is a bounded real value function on [a,b] such that $ f^2$=f.f is Riemann integrable, then f is integrable.

I think this is true since if $ f^2$ is Riemann integrable then we just need to show that the set of discontinuities of f is equal to the set of discontinuities of $ f^2$ which is a zero set, then f is Riemann integrable ( given f is bounded).

h)If f and g are twice differentiable on (a,b) and f”=g” on (a,b), then there exists a constant c such that f(x)=g(x) +c for all x in (a,b).

I want to look at the contrapositive of this statement and prove it from there, but I just don’t know what it tells me if there is no constant c such that f(x)=g(x)+c. I want to rewrite this so that I can get more useful info, but I can’t.
i)	If ($ x_n$) is a sequence of real numbers that converges to x in $ R$ then the set {x} union {$ x_n$: n in $ N$} is compact.
j)	If the power series $ a_n$$ x^n$ has finite radius of convergence R>0, and f(x) converges for x=-R,R, then f(x)= the given power series, is Riemann integrable.
I don’t have any idea on the last two problems.

a) The set of all functions from $ \mathbb{N}$ to $ \mathbb{N}$ is $ \emph{uncountable}$.

You have misinterpreted this question. You proved that $ \mathbb{N}$ has the same cardinality as $ \mathbb{N}$. This asks about the cardinality of the set of functions. The answer is true (maybe try to show that the set of functions from $ \{0,1\}$ to $ \mathbb{N}$ is uncountable, and clearly the set in question is larger than that).

An example for b): Let $ f(x) \equal{} x^3$ and $ c \equal{} 0.$ Then $ f'(c) \equal{} 0$ but $ f$ is a strictly increasing function, so $ \frac {f(b) \minus{} f(a)}{b \minus{} a} > 0$ for all $ a\ne b.$

As for c): The root test is a [i]comparison[/i] test (a comparison to geometric series) and as with all comparison tests, it is a test for [i]absolute[/i] convergence. You must take absolute values before using the root test (or ratio test), and you can't get $ \minus{} 1$ as a limiting root or ratio. In fact, the limiting root is $ 1$ and the root test is inconclusive.

Now: what, exactly, is going on with that series?

For f): no, you can't use $ \frac1x$ as an example. You need a function that is defined [i]everywhere[/i] and differentiable [i]everywhere[/i].

There is a classic example here; if you don't already know this example, you probably wouldn't think of it:
\[ f(x)\equal{}\begin{cases}x^2\sin\left(\frac1x\right)&x\ne0\\0&x\equal{}0\end{cases}\]

For h): What is the set of all functions $ h$ such that $ h''(x)\equal{}0?$ That's a relevant question.

i): Can you show that the set is closed and bounded?

j): By Abel's Theorem, $ f$ is continuous on $ [\minus{}R,R].$ What do you know about the Riemann integrability of a continuous function on a closed interval?

Thanks for all the help Kent and Hilbert.
For h) I am able to figure out a counter-example now. It's quite simple, and I didn't get it in the first place. ( f=$ x^2$ and g= $ x^2$ + cx + b )

i) I can show that the set is closed ( union of two closed sets is closed ), and I can show it is bounded too. I forgot that in $ R^m$ , particularly, $ R$ I only need to show this to show compactness.

j) I had to look up Abel's theorem, so it is integrable.

c) I tried to take the limit of $ a_k$ and it doesn't converge to 0, so I conclude that it is not convergent.

Now, about g):

Is it possible to have $ f^2$ be a very nice function - say, for the sake of the argument, $ f^2(x) \equal{} 1$ on $ [a,b]$ - but have $ f$ itself be something really ugly?

As for d), try writing out the contrapositive of what that says.

To actually build a counterexample for e): the uniform limit of continuous functions must be continuous, so we need a continuous limit. What's a function that is continuous but not differentiable? How about $ f(x)\equal{}|x|.$ That's a nice simple example. All right then, can you build a sequence of smooth functions $ f_n(x)$ such that $ f_n(x)\to |x|$ uniformly on $ [\minus{}1,1]?$

[quote="Kent Merryfield"]Now, about g):

Is it possible to have $ f^2$ be a very nice function - say, for the sake of the argument, $ f^2(x) \equal{} 1$ on $ [a,b]$ - but have $ f$ itself be something really ugly?[/quote]
I remember that, I had a similar problem which says in general that if $ f^2$ is Riemann integrable, then $ f$ is not necessarily Riemann integrable. Show this by a counterexample:
$ f(x)$ = 1 if x is rational, and -1 if x is irrational.
I don't know if I can use this counterexample here since $ f$ is a bounded real value function? By the way, is g)  false ?

That $ f$ is most certainly bounded and real valued. All "bounded" means is that there is some $ M$ such that $ |f(x)|\le M$ for all $ x$ and we certainly have that here. So g) is false, and this is the counterexample. (And if we wanted a conclusion about $ f$ being Lebesgue integrable given that $ f^2$ is Lebesgue integrable, we'd use almost the same example, except with some non-measurable set instead of $ \mathbb{Q}.$)


I am rebuilding the current Florida Association of Mu Alpha Theta's webpage http://famat.org

http://img84.imageshack.us/my.php?image=sitesmall2bx.png

That is what the general format of the site is going to be like.  If you notice, there is a column on the right hand side that is completely empty.  I was wondering if it was possible to get ahold of the proofs so that I could include them on the site.  I was also wondering if there was a way to get the script for the fading quotes that are at the bottom of the AoPS site through an iFrame so that they can be included in this site.

The FAMAT site that I am building does have PHP functionality if necessary to run any scripts.

I of course will give credit to the Art of Problem Solving Foundation for allowing FAMAT to use the resources on its site.  I am planning on linking our forums button to the AoPS forums, and I have already added AoPS on the other math sites section.

This would all be greatly appreciated!

THANKS!
Faraz  :D

I sent you a private message about this.

What happened finally with this? Can he use the resources section of this page?


algebra
polynomial
superior algebra
superior algebra solved

Let $x_1,\ldots,x_n\in\mathbb Z,\ k_1,\ldots,k_n\in\mathbb N^*$. Show that $n!|D_n=\left|\begin{array}{cccc}x_1^{k_1}&x_1^{k_2}&\cdots&x_1^{k_n}\\x_2^{k_1}&x_2^{k_2}&\cdots&x_2^{k_n}\\\vdots&\vdots&\ddots&\vdots\\x_n^{k_1}&x_n^{k_2}&\cdots&x_n^{k_n}\end{array}\right|$.

Consider the determinant as a polynomial in n nedeterminates. This polynomial turns out to be divisible by all $ X_i-X_j $, so since all these polynomials are prime two by two, it is divisible by the vandermonde determinant (view also a polynomial in n nedeterminates). It is easy to see that this divisibility as polynomials turns out to be divisibility in integers. Also, it is hard, but nice to prove more, that $ 1!2!...n! $ divides this determinant. Hope it's correct.


AMC
AMC 10
trigonometry
geometry
MATHCOUNTS
quadratics
conics

I am thinking about holding a Getting Started Tournament.  It will be like the playoffs in most sports.  I will match ever set of two people against each other in the first round.  Winners go to the second and so on until there is a winner.  Right now I am thinking that the format will be that I will post a few(5 or so) problems on the board and those involved in the contest will pm their answers.  I might even have like a regular season i.e. have some problems that everyone does and see where people rank from there.  I haven't gotten any problems for the tourney yet but I would like to know how many people are interested in doing this.  Sorry but I don't have any prizes to give away.  You will just have to fight for glory!!!

Everyone who is interested please say so here!

I'll join as long as it is not this friday at 3:00 - 4:15 PM EST.

To try to avoid time constraints I might try to do it sort of like Rep123Max does his.  That is post the problems and give the people some time.  I probably won't give as much time as he does though.  If anyone has suggestions on how to administer the tornament (time-wise) I would be glad to hear them.

I only have a few problems so far.  I probably wont start for a week or two to let people know about the tournament and to find/make more problems.

I'll do too.

Just to be sure, the level of this should be no more higher than easy or perhaps, mid-difficult AMC10, meaning no TRIG or of course, never calculus.  And I hope when you match the people, you should match them with similar level, meaning not like this:

Me and ComplexZeta (of course, he's advancer and I'm only a getting starter)

Well I was thinking of doing a regular season type thing.  Like 10-15 problems.  From this I can seed the people.  I was thinking of putting the highest vs. lowest, second highest vs. second lowest and so on.  Or if people prefer I could try to make the matchups randoms.  The advantage of the first one is that theoretically the last round will have the two best people against each other.  I would appreciate to know which way people preferred.

Random matching is what I prefer, but it doesn't really matter to me.  As long as I'm not going against ComplexZeta... or anyone really advanced.

I am pretty sure that Simon wont compete in this.  Unless he wants to be a bully and beat up on all the getting started people  ;) .For the time issue I think this is what I'll do:  I will pm everyone and see what time they are good.  At that agreed time I will then pm them the questions and they will have one day to send me their answers.  Everyone will get the same questions but most likely at different times.  Afterwards I will post the questions and answers on the board.  This way I can work around each individuals time constraints.

Um, I'll do  it.... I'll probably be the first out though...    ;)

I am going to go ahead and post 5 warm up type problems.  The problems in the tournament will most likely be harder.  If you want to enter the tournament please [size=150]pm[/size] me answers to these questions.  So this is like an RSVP.

1. Find x if:  $ |3-4x|=10 $

2. What is the Least common multiple of 3,7,11,63, and 66?

3.  How many ways can the gold, bronze, and silver medals be given to the 1 st, 2nd, and 3 rd place finishers, respectively, in an 8 man race?

4.  Find the area of the figure made when the points (0,4), (0,5), (1,5), (5,1), (5,0), and (2,0) are connected (in that order).

5.  If the height of a projectile at time t is $ h(t)=64+16t-4t^2 $ , what is the highest point the projectile will reach?

I will post the problems after the round is over.  But if you are talking about during the round I could probably send you them if you are knocked out as long as you don't talk about them publicly.

Since you are bored check out the Mock AMC's that are going on.  They are a lot of fun( and very good practice).

I would like to participate too, at the contest... The warm up problems u proposed r very easy but very fastly chosen so there r many problems...I could also help u by proposing problems...

Those questions were a bit easy but even so it seems that you are a little too advanced for this contest.  I would appreciate your help in making problems however.

Is there any paticular reason for not using a calculator?  I would like to use a calculator  for simple arithmatic problems that I can't do in my head very fast, but It's okay If I can't  I just want to know the reason.

Well for one I don't think that you really need one. Also, it is good practice since most contests don't allow use of one anyways.  I will accept answers that aren't fully simplified as long as they are right.  E.g.  ${6\choose 3}$ is fine if you are on a time crunch or something and don't want to go any further.  Just don't give me insanely unsimplified answers and if the problem is to simplify something as much as possible it must be simplified! ;) (but that should be given)

Btw,  no more people can enter.  I cut the deadline short a day.  I will have the bracket out soon.

Mock AMCs?  I'm not familiar with contests, which is part of the reason I signed up here...the othe part is of course I need/want to practice.

The Mock AMC's are on this site also.  Go to the American Mathematics Competitions forum and look around.  I am sure you will see topics relating to the mock amc's.  Don't worry this tournament should be practice for you.  If it is not enough there is always something that will give you practice on this site.  Look around, ask questions, this is the place to learn math.

[quote="lonelytraveler8"]Perfect.  Thanks for the help, and I'm looking foward to this contest whether I do good or not.[/quote]

You should look at most contests that way. Even if you don't win, that doesn't prevent you from not taking valuable experience or new knowledge if you don't.

But, of course, winning's also very nice.  :lol:

Winning is nice of course, but I just like doing math...But I do plan on winning at least one round, However I'm really not expecting to.  But If it's good practice then it was worth signing up here.  Besides I only have to years of high school left and I'd like to learn as much before finishing as I can.

[quote="joml88"]Btw,  I want to post an attachment for the bracket but it is a word document.  Does anyone have any ideas for how I can make it accessible to everyone?
[/quote]

Aah!  I haven't been on in a while.  Did I miss anything?  I hope I didn't miss a round or anything  :blush: .

I know that Macs can open Word documents with AppleWorks.  I'm not sure if linux can.  You can always use .rtf files, though.

You could do what Aops does and use PDFs

The problem is that I got the bracket off of a site.  I was wondering if anyone knew what format I could convert it to or something like that.  Is there a way that I can get it into .pdf format?

[size=150][b]Solutions to Warm-Ups[/b][/size]

1.  $ |3-4x|=10 $

If the stuff inside the absolute value is positive then it is the same as having the equation as it is w/out the absolute value bars.  If it is negative then the absolute value will reverse its sign.  Thus we have:
\[ 3-4x=10 \ \ \ or \ \ \ -(3-4x)=10 \]

For the first equation you subtract by 3 and divide by -4 which yields $ -\frac{7}{4} $ .  For the second you multiply by -1, subtract 3 and divide by -4 to get $ \frac{13}{4} $.

2.  What is the least common multiple of 3, 7, 11, 63, and 66?

This number must be divisible by 66 so it must have 66 or 2*3*11 in its factorization.  It must also be divisble by 63.  $ 63=3^2\cdot7 $ We already have one three in the factorization so we must add another 3 and a 7 for it to be divisible by 63.  11, 7, 3 are already in the factorization so we don't have to worry about them.  So our number must contain $ 2\cdot3\cdot11\cdot3\cdot7=1386 $ in its factorization.  The LEAST common multiple will then be 1386.

3.  How many ways can the gold, bronze, and silver medals be given to the 1 st, 2nd, and 3 rd place finishers, respectively, in an 8 man race?

Let's consider how many ways we can pick the 1 st place person.  Obviously there are 8.  So once we have our 1 st place person there are 7 choices left for who gets second.  After that there are 6 people who can be 3 rd.  So there are 8*7*6=336 total ways to give out 1 st, 2 nd, and 3 rd place.

4.  Find the area of the figure made when the points (0,4), (0,5), (1,5), (5,1), (5,0), and (2,0) are connected (in that order).

Refer to the figure that is attached.  We can see that the area of our figure is equal to that big square minus those two triangles.  The big square has side length 5 and therefore area 25.  The lower left triangle (which is right)  has legs of length 2 and 4.  The area of a right triangle is bh/2.  Since it has a right triangle these two legs correspond to the base and height.  The area is (2*4)/2=4.  The other triangle's area is found similarily.  (4*4)/2=8.

The area of the figure is thus $ 25-(4+8)=13 $

5.  If the height of a projectile at time t is $ h(t)=64+16t-4t^2 $ , what is the highest point the projectile will reach?

Ok the equation we have is a quadratic equation.  If you plug in numbers and plot the corresponding points you should get a graph that looks like a parabola.  If the parabola opens up it has no maximum but has a minimum.  If it opens down it has a maximum but no minimum.  This minimum or maximum is called the vertex.

Let's look at a general parabola(quadratic equation).  \[ ax^2+bx+c=y \]

We can factor this equation by a method called completing the square.  This method takes advantage of how it is relatively easy to factor a perfect square trinomial.  First let's divide by a:
\[ x^2+\frac{b}{a}x+\frac{c}{a}=y \]

If we add $ \left(\frac{b}{2a}\right)^2 $ we get
\[ x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2+\frac{c}{a}=y+\frac{b}{2a}^2 \]
Now we have that perfect square trinomial.  We factor it to get
\[ \left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b}{2a}^2=y \]

y will be greatest or the least (depending on how the parabola opens) when $ x+\frac{b}{2a}=0 $.  From this we can find the x coordinate of the vertex.  We can see it is when $ x=-\frac{b}{2a} $.

Using this for our question we get the x coordinate to be 2.  Now 2 is the time but we want the height.  Easy fix, we just plug 2 in to get it's corresponding y (or h(t) in this case) value which comes out to be 80.

P.S. I tried to make my paint image as neat as possible  :lol: .

CutePDF or Go2PDF work, just google for them


Arrange bishops on a $ 8*8$ chess board such that no bishop controls a square on which another lie and the maximum possible number of bishops is used. Prove that: every bishops are lie on the border of the chess board.

i think 14 bishops can be used at maximum.

Let's look at the number of pairs (B,X) such that X is a square that is being threatened by the bishop B.  Assume that bishops threat the square in which they belong and there are a total of b bishops.
A bishop threats at least 8 squares and this situation occurs only when bishop is on the border of the board. (1') So there are at least 8b such pairs. (1)
It's easy to observe that a square that is a corner of the board or have got a bishop on it can be threatened by at most 1 bishop, and the others can be threatened by at most 2 bishops. Moreover, at most 2 of the 4 corners can have a bishop on it. So there are at most 2(64-b)+b-2 such pairs.(2)
From (1) and (2) 8b=<126-b so b=<14. We can easily see that b=14 is attainable with putting 8 bishops to one edge and 6 bishops to the paralel edge, without putting any to the corners in that one. Hence b must be 14 and to satisfy this, every bishop must be on the border from (1').

Some clarifications needed:

1. Can I use bishops of both colors or only one color?

2. Can I use more than 9 bishops?


Whats the product of all real numbers greater than 0? I think it is 1. My friend says it is infinity.  Am I right or wrong?

It is undefined. The fact that the reals are not well-ordered contributes a lot to this.


search
number theory unsolved
number theory

Suppose that $ m$ and $ n$ are odd integers such that $ m^2\minus{}n^2\plus{}1$ divides $ n^2\minus{}1$. Prove that $ m^2\minus{}n^2\plus{}1$ is a perfect square.

[quote=Albanian Eagle]It's been posted a few times.
here's one of them
http://www.mathlinks.ro/viewtopic.php?search_id=55820209&t=5935[/quote]
it dosen't work....

We know that the negation $-(m^2-n^2+1)$ will still divide $n^2-1$, so we rewrite it as $$(n+1)(n-1)-m^2 | (n+1)(n-1).$$

Set $n^2-1=(n+1)(n-1)=x$. We then have $x-m^2|x$. Converting this to words, we have [quote]One less than an odd perfect square is divisible by itself minus an odd perfect square[/quote]
I'm stuck here, some help? From here, I would somehow find properties of the variables then prove that the above is a perfect square...

Let's nail this guy (as far as I remember the original version involved also $m>n>1$ condition).

The idea is to set $a=\frac{m+n}{2}$, $b=\frac{m-n}{2}$ (hence $a>0$ and $b\ge 0$), and apply Vieta jumping. Specifically, after the aforementioned parameterization, it boils down $4ab+1\mid (a-b)^2-1 \implies 4ab+1\mid (a+b)^2$. To show $4ab+1$ is a perfect square, it clearly suffices to show whenever $4ab+1\mid (a+b)^2$, it is the case that $\frac{(a+b)^2}{4ab+1}$ is a perfect square.

Define the set
\[
S=\left\{k:k=\frac{a^2+b^2}{4ab+1}\quad\text{for some}\quad a,b\in\mathbb{Z}_{\ge 0}\right\}.
\]
Take any $k_0\in S$, and let $(a_0,b_0)\in\mathbb{Z}_{\ge 0}\times \mathbb{Z}_{\ge 0}$ be a pair `realizing' $k_0$ with the smallest $a_0+b_0$ sum. Clearly if $a_0=b_0$ then $k_0=0$, in which case we have the result. Assume $a_0>b_0$ without loss of generality. Interpreting this condition as a quadratic in $a$, we arrive at
\[
a^2-a\left(4k_0b_0-2b_0\right)+b_0^2-k_0=0
\]
where the roots are $a_0$ and $\overline{a}$. From Vieta's relations;
\[
a_0+ \overline{a} = b_0(4k_0-2) \quad\text{and}\quad a_0\overline{a} = b_0^2-k_0.
\]
It is evident, from the first condition, that $\overline{a}$ is integral. Moreover, the pair $(\overline{a},b_0)$ also realizes $k_0$. Now, if $\overline{a} =0$ we immediately have $k_0=b_0^2$. Assume $\overline{a}\ne 0$. In order not to have a contradiction with the minimality of $(a_0,b_0)$, one of the two cases should be valid.

[b]Case 1.[/b] $\overline{a}<0$. In this case $a_0|\overline{a}| = k_0-b_0^2>0$. Moreover, $a_0 >b_0(4k-2)$. Since $a_0\mid k_0-b_0^2$, this yields $b_0(4k_0-2)\le k-b^2$. It is evident that in this case $4kb_0+b_0^2-2b_0\le k$. If $b_0=0$, we have the condition, if $b_0=1$ we have $4k-1\le k$, forcing $k=0$; whereas for $b_0\ge 2$, we have a contradiction. Case $1$ is closed.
[b]Case 2.[/b] Now assume $\overline{a}>0$. In this case, since $(a_0,b_0)$ realizes $k_0$ with smallest $a_0+b_0$, it is the case that $\overline{a}\ge a_0$. Hence, $b_0^2-k_0 =a_0\overline{a}\ge a_0^2$. As $a_0>b_0$ (by the choice) this is impossible.

Since $k_0$ was arbitrary, the proof is finished.


summer program
MathPath
geometry
AMC
AIME
probability
LaTeX

Last year, 13375P34K43V312(now CatalystOfNostalgia) started problems of the day. He does not have the time this year so i will be doing it. The rules are the same as before, for those of you who have participated. If you have never participated, the previous Problem Of the Day thread can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168612]here[/url]. This will start tomorrow.

Solutions are not required when submitting an answer

[size=200]Day 1[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

The sum of two numbers is 2. The sum of their squares is 3. Find the sum of their cubes.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

Some of you may have seen this problem already. Note that the rules are EXACTLY the same as last time.

abacadaea... lolololololololololololololololololololololololololololololololol

at least change it a LITTLE bit.

~BOGTRO-yeah i know. Most problems will be my own.

[hide="Submissions"]9[/hide]
[hide="Correct Answers"][b]8 [/b]
27r
Bachukas
math154
lifeisacircle
FantasyLover
Textangle
pythag011
isabella2296
[/hide]
[hide="Solution"]Solution: Let a and b be the two numbers, Thus $ a\plus{}b\equal{}2$ and $ a^2\plus{}b^2\equal{}3$.
$ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab\equal{}3\plus{}2ab\equal{}4$.
So $ ab\equal{}\frac{1} {2}$.
We factor to get $ a^3\plus{}b^3\equal{}(a\plus{}b)(a^2\plus{}b^2\minus{}ab)\equal{}2(2.5)\equal{}\boxed{5}$[/hide]
[hide="Source"]Mathpath qualifying quiz 2007 #1[/hide]

[size=200]Day 2[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

In $ \Delta ABC, AB\equal{}13, BC\equal{}14,$ and $ AC\equal{}15$. Let the foot of the altitude from $ A$ to $ BC$ be $ D$ and the midpoint of $ BC$ be $ E$. Let the point where the perpendicular bisector of $ BC$ intersects $ AC$ be $ F$. $ DF$ and $ AE$ intersect at P. Find the distance from $ P$ to $ BC$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[hide="Submissions"]3[/hide]
[hide="Correct Answers"][b]2 [/b]
lifeisacircle
FantasyLover
[/hide]
[hide="Solution"]
Let $ P'$ be a point on BC such that $ PP' \parallel AD \parallel EF$
By Heros formula, $ [ABC] \equal{} \sqrt {21*6*7*8} \equal{} 84$
Thus, $ \frac {1} {2} (AD)(BC) \equal{} 84$
So $ AD \equal{} 12$
By the Pythagorean theorem, $ BD \equal{} \sqrt {13^2 \minus{} 12^2} \equal{} 5$ and $ DC \equal{} \sqrt {15^2 \minus{} 12^2} \equal{} 9$.
We have $ \angle ACD \equal{} \angle FCE$ and $ \angle CDA \equal{} \angle CEF$, Thus, by AA, $ \Delta ACD \sim \Delta FCE$
$ \frac {EC} {DC} \equal{} \frac {FE} {AD}$
We know $ EC \equal{} 7$ so $ \frac {7} {9} \equal{} \frac {FE} {12}$ so $ FE \equal{} \frac {28} {3}$

A way to get the answer is some formula in AoPS intro to geometry which gives in this case:  $ \frac {1} {PP'} \equal{} \frac {1} {EF} \plus{} \frac {1} {AD}$. Which gives $ PP'\equal{}\frac{21} {4}$

Ill leave it to you guys to prove it since i dont have the time.
For a start Use the fact that $ PP'D \sim FED$ and $ PP'E \sim ADE$.

[/hide]
[hide="Source"]own [/hide]

[size=200]Day 3[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Let S be the sum of all factors of $ 42^3$. How many factors does S have?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

When i say factor, i mean including 1 and itself(although that probably isnt what it means. Im not very good with these words).

sorry i couldnt post anything yesterday and i wont be able to do anything today. The internet at my house isnt working. Im at school right now so i have about 20 minutes on the computer. You guys can still submit answers until i post the next problem

[hide="Submissions"]5[/hide]
[hide="Correct Answers"]Everyone!
[b]5 [/b]
pythag011
1=2
Textangle
PI-Dimension
FantasyLover
[/hide]
[hide="Solution"]Solution: The number of factors of $ 42^3 \equal{} 2^3*3^3*7^3$ is $ (2^0 \plus{} 2^1 \plus{} 2^2 \plus{} 2^3)(3^0 \plus{} 3^1 \plus{} 3^2 \plus{} 3^3)(7^0 \plus{} 7^1 \plus{} 7^2 \plus{} 7^3) \equal{} 15*40*400 \equal{} 2^7*3*5^4$ Which has $ (7 \plus{} 1)(1 \plus{} 1)(4 \plus{} 1) \equal{} \boxed{80}$ factors.
If you dont understand why the first step is true, distribute it.
[/hide]
[hide="Source"]Own[/hide]

[size=200]Day 4[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Compute the sum of all the numbers in the first 10 rows of pascals triangle.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

My internet is working!!!! success! Also, if a mod could delete the above post, that would be nice. I dont want to confuse anyone

would you define '1' as the first row or the zero'th row?

also, if it is the zero'th row, then yould you take the first "10" rows (0-9) or the first "11" rows (0-10)

wikipedia says the 0th row is '1' and the first 10 rows are 0-9

[url]http://en.wikipedia.org/wiki/Pascal_triangle[/url]

[hide="Submissions"]8[/hide]
[hide="Correct Answers"]
[b]6 [/b]
Textangle
stevenmeow
chenhsi
shentang
FantasyLover
pythag011
[/hide]
[hide="Solution"]
The sum of the nth row of pascals triangle is $ 2^{n\minus{}1}$. Thus, the sum of first 10 rows is $ 2^0\plus{}2^1\plus{}...2^9\equal{}\boxed{1023}$
[/hide]
[hide="Source"]Own[/hide]

[size=200]Day 5[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A game uses a deck of $ n$ different cards, where $ n$ is an integer and $ n \geq 6$. The number of possible sets of $ 6$ cards that can be drawn from the deck is $ 6$ times the number of possible sets of $ 3$ cards that can be drawn. Find $ n$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[hide="Submissions"]7[/hide]
[hide="Correct Answers"]
[b]6 [/b]
1=2
stevenmeow
isabella2296
Textangle
FantasyLover
sup3rsolv3r
[/hide]
[hide="Solution"]
[quote="JesusFreak197"]$ \binom{n}{6} \equal{} 6\cdot \binom{n}{3}$
$ \frac {n!}{6!(n \minus{} 6)!} \equal{} 6\cdot \frac {n!}{3!(n \minus{} 3)!}\Rightarrow 6!(n \minus{} 6)! \equal{} (n \minus{} 3)!$
$ (n \minus{} 3)(n \minus{} 4)(n \minus{} 5) \equal{} 720\Rightarrow n \equal{} \boxed{13}$[/quote]
[/hide]
[hide="Source"]2005 AIME II #1[/hide]

[size=200]Day 6[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A, B, and C are randomly picked on a circle. Find the probability that $ \widehat {AB}, \widehat {AC}, \widehat {BC} > 60$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

Also, when submitting answers, make sure you put like POTD 6 instead of just POTD.

EDIT: sorry, there was an error in my latex

[hide="Submissions"]4[/hide]
[hide="Correct Answers"]
[b]0 [/b] :(
[/hide]
[hide="Solution"]
Point $ A$ can be anywhere with probability 1. Let the degree measures of B and C be the amount of degrees A must be rotated clockwise about the center to coinicide with B and C respectively. We make a graph of B and C. B and C are both between 60 and 300, and they must differ by atleast 60. The fraction of the graph that is shaded(look at the diagram) is $ \frac {60*3*60*3} {360*360} \equal{} \boxed{\frac {1} {4}}$[/hide]
[hide="Source"]Own[/hide]

[size=200]Day 7[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A circle is inscribed in an equilateral triangle with side length 6 inches. Find the area of the circle in square feet. Express your answer in terms of $ \pi$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[quote="abacadaea"][hide="Submissions"]3[/hide]
[hide="Correct Answers"][b]2 [/b]
lifeisacircle
FantasyLover
[/hide]
[hide="Solution"]
Let $ P'$ be a point on BC such that $ PP' \parallel AD \parallel EF$
By Heros formula, $ [ABC] \equal{} \sqrt {21*6*7*8} \equal{} 84$
Thus, $ \frac {1} {2} (AD)(BC) \equal{} 84$
So $ AD \equal{} 12$
By the Pythagorean theorem, $ BD \equal{} \sqrt {13^2 \minus{} 12^2} \equal{} 5$ and $ DC \equal{} \sqrt {15^2 \minus{} 12^2} \equal{} 9$.
We have $ \angle ACD \equal{} \angle FCE$ and $ \angle CDA \equal{} \angle CEF$, Thus, by AA, $ \Delta ACD \sim \Delta FCE$
$ \frac {EC} {DC} \equal{} \frac {FE} {AD}$
We know $ EC \equal{} 7$ so $ \frac {7} {9} \equal{} \frac {FE} {12}$ so $ FE \equal{} \frac {28} {3}$

A way to get the answer is some formula in AoPS intro to geometry which gives in this case:  $ \frac {1} {PP'} \equal{} \frac {1} {EF} \plus{} \frac {1} {AD}$. Which gives $ PP' \equal{} \frac {21} {4}$

Ill leave it to you guys to prove it since i dont have the time.
For a start Use the fact that $ PP'D \sim FED$ and $ PP'E \sim ADE$.

[/hide]
[hide="Source"]own [/hide]

[size=200]Day 3[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Let S be the sum of all factors of $ 42^3$. How many factors does S have?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

When i say factor, i mean including 1 and itself(although that probably isnt what it means. Im not very good with these words).[/quote]
i don't really know if heron's formula should be on here...

Yeah thats true. Sorry. I guess these problems are a bit hard for this time of the year. I'll definitely make them easier once the school year starts. I just figure since its summer, you have a bit more time to think about it.

[hide="Submissions"]6[/hide]
[hide="Correct Answers"]
[b]0 [/b] :P
[/hide]
[hide="Solution"]
I guess everyone missed the inches/feet part and got pwned. :D.
Solution 1:
The area of the triangle is $ \frac {6^2 \sqrt {3} \pi} {4} \equal{} 9\sqrt {3} \pi$ square inches. The semiperimeter is $ 9$ inches. We use the inradius formula to get $ r \equal{} \frac {A} {s}$, $ r \equal{} \sqrt {3}$ inches. Thus, the area of the incircle is $ 3 \pi$ square inches. To get the number of square feet, we must divide by $ 12^2 \equal{} 144$(12 inches in a foot), so the answer is $ \frac {1} {48} \pi$.

Solution 2:
Let $ I$ be the incenter and $ D$ be the midpoint of $ BC$. Thus, $ BD \equal{} 3$. Since $ BDI$ is 30-60-90, $ ID \equal{} \sqrt {3}$. Then we find the area using the same process as solution 1.[/hide]
[hide="Source"]Own[/hide]

[size=200]Day 8[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Calculators are permitted on this problem

At the grading of the Advanced Placement Calculus exams, 340 math teachers graded 125000 tests in 40 hours. Each test had 6 problems. What is the average number of minutes it took one teahcer to grade 15 problems? Express your answer to the nearest minute.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

For the soluion to POD 6, i forgot a diagram. :oops:

[hide="Submissions"]7[/hide]
[hide="Correct Answers"]
[b]4 [/b]
FantasyLover
chenhsi
1=2
math154
[/hide]
[hide="Solution"]
x=number of problems= 125000 * 6=750000
y= number of minutes taken by all teachers to grade all problems= 40*60=2400
z=number of minutes taken by one teacher to grade all problems= 340*2400=816000
u=number of minutes taken by one teacher to grade one problem= z/x= 816000/750000
v=number of minutes taken by one teacher to grade 15 problems =
u*15= (816000/750000)*15=16.32 which is 16 to the nearest minute.
[/hide]
[hide="Source"]Problem and solution taken from Ellis saabs website.[/hide]

[size=200]Day 9[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A positive integer $ n$ has 60 divisors and $ 7n$ has 80 divisors. What is the greatest integer $ k$ such that $ 7^k$ divides $ n$?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[hide="Submissions"]4[/hide]
[hide="Correct Answers"]
[b]2 [/b]
Bachukas
1=2
[/hide]
[hide="Solution"]
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=370839#370839
[/hide]
[hide="Source"]AMC 12B 2005[/hide]

[size=200]Day 10[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

If $ r\plus{}\frac{1} {r} \equal{} 2007 \frac{1} {2007}$ and $ s\plus{}\frac{1} {s}\equal{}2006 \frac{1} {2006}$, what is the largest possible value of $ r\minus{}s$? Express your answer as a mixed number

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

[hide="Submissions"]3[/hide]
[hide="Correct Answers"]
[b]3 [/b] Everyone
Bachukas
shentang
FantasyLover
[/hide]
[hide="Solution"]
The 2 solutions for r are(that rhymes!!) $ 2007 and \frac{1} {2007}$. The 2 for s are $ 2006$ and $ \frac{1} {2006}$. The largest possible value of r is $ 2007$, and the smallest possible value of s is $ \frac{1} {2006}$. Thus, the maximum possible value of $ r\minus{}s$ is $ 2007\minus{}\frac{1} {2006}\equal{}2006 \frac{2005} {2006}$
[/hide]
[hide="Source"]2007 state team round(#7?)[/hide]

[size=200]Day 11[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

I guess you can use a calculator on this one if you want...
A certain quadrilateral with perpendicular diagnols. The diagnols have lengths 15 inches and 4.2 feet. Find the area of the quadrilateral in square inches.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

[hide="Submissions"]7[/hide]
[hide="Correct Answers"]
[b]4 [/b] Everyone
1=2
Bachukas
FantasyLover
lifeisacircle
[/hide]
[hide="Solution"]
Some people dont learn from mistakes....lol
$ 4.2 ft \equal{} 50.4 in$
The area of any quadrilateral with perpendicular diagnols(orthogonal) is $ \frac {ab} {2}$, where a and b are the legths of the diagnols. $ 50.4*15/2 \equal{} 378$

to see why  this is true, add up the sums of the area s of the 4 small triangles created wby the diagnols
[/hide]
[hide="Source"]own[/hide]
im going to hold off on PODs until tuesday for conviniece. Its just so we can have Day X1 be monday, X2 be tuesday, ect.

[quote="abacadaea"][size=200]Day 6[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A, B, and C are randomly picked on a circle. Find the probability that $ \widehat {AB}, \widehat {AC}, \widehat {BC} > 60$.[/quote]
...
[quote="abacadaea"][hide="Submissions"]4[/hide]
[hide="Correct Answers"]
[b]0 [/b] :(
[/hide]
[hide="Solution"]
Point $ A$ can be anywhere with probability 1. Let the degree measures of B and C be the amount of degrees A must be rotated clockwise about the center to coinicide with B and C respectively. We make a graph of B and C. B and C are both between 60 and 300, and they must differ by atleast 60. The fraction of the graph that is shaded(look at the diagram) is $ \frac {60*3*60*3} {360*360} \equal{} \boxed{\frac {1} {4}}$[/hide]
[hide="Source"]Own[/hide][/quote]
Not in middle school and not doing these problems (hehe, when I wanted to do them way back when, Catalyst wouldn't let me submit. :P ), but uh... be careful when you say "random". I could find another way to randomly pick points B and C after fixing A, and it would probably give a different probability.
See http://en.wikipedia.org/wiki/Bertrand%27s_paradox _(probability) . (Can't URL tag since it has a ')

[size=200]Day 12[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Sallie earned a grade of exactly 90% for a marking period based on six equally weighted test. The four test papers she can find have grades of exactly 83%,96%,81%, adn 82%. What is the sum of the percent scores of the two missing grades?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

We are going back to easy. Problems will get harder as we approach nationals.

[hide="Submissions"]5[/hide]
[hide="Correct Answers"][b]5[/b]
Textangle
Bachukas
isabella2296
lifeisacircle
FantasyLover
[/hide]
[hide="Solution"]The sum of all her test scores is $ 90*6\equal{}540$. The sum of the ones she is missing, is therefore $ 540\minus{}83\minus{}81\minus{}82\minus{}96\equal{}\boxed{198}$[/hide]
[hide="Source"]nationals sprint 2006 #1[/hide]

[size=200]Day 13[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

When the single digit D is written at the end of the positive integer XY, with tens digit X and unites digit Y, the resulting positive 3-digit integer XYD is 619 more than the original integer XY. What is the value of the 3 digit integer XYD?

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

[hide="Submissions"]5[/hide]
[hide="Correct Answers"][b]5[/b]
Textangle
Bachukas
shentang
lifeisacircle
FantasyLover
[/hide]
[hide="Solution"]The 2 digit number XY can be expressed as $ 10X \plus{} Y$ and XYD is $ 100X \plus{} 10Y \plus{} D$. Thus, $ 100X \plus{} 10Y \plus{} D \minus{} 10X \plus{} Y \equal{} 9(10X \plus{} Y) \plus{} D \equal{} 619$. We know that $ 619 \minus{} D$ is divisible by 9, so $ D \equal{} 7$. Hence, the two digit number $ XY$ is $ 612/9 \equal{} 68$, so XYD is $ \boxed{687}$[/hide]
[hide="Source"]nationals sprint 2006 #10ish[/hide]

[size=200]Day 14[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

I randomly distribute n candies to 5 kids. No matter how i distribute it, there will always be atleast 1 kids with 9 candies. Find the minimum amount of candies i could have.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

fixed :)

message is too small blah blah blah

[hide="Submissions"]4[/hide]
[hide="Correct Answers"][b]3[/b]
Bachukas
shentang
FantasyLover
[/hide]
[hide="Solution"]This is a basic pigeonhole problem. The pigeonhole formula gives $ \lfloor \frac {n} {5} \rfloor \ge 9$. Then we get $ n \equal{} \boxed{41}$.[/hide]
[hide="Source"]own[/hide]

[size=200]Day 15[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

This is a friday. You have until Monday to do this.

A circle is inscribed in a quartercircle radius 5. Find the radius of the circle.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

I hope people realize rule 3 still exists. :)
Anyone can participate.

[hide="Submissions"]6[/hide]
[hide="Correct Answers"][b]6[/b]
Bachukas
lifeisacircle
1=2
Textangle
shentang
FantasyLover
[/hide]
[hide="Solution"]let the radius of the small circle be r. We take a look at the radius of the quarter circle that passes through the center on the circle. We see that $ r\sqrt {2} \plus{} r \equal{} 5$, so $ r \equal{} 5\sqrt {2} \minus{} 5$.
[/hide]
[hide="Source"]own[/hide]

[size=200]Day 16[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

2 dogs are 1000 feet away from each other. The run at each other at rates of 20 and 30 feet per second. Immediately when they start running, a fly starts on the nose of the dog running at 30 feet per second and starts flying towards the other dog at a rate of 50 feet per second. Once the fly reaches the nose of the other dog, it immediately reverses direction and starts flying toward the original dog again at the same speed. This process continues until the dogs collide and the fly is squished between the noses of the dogs. How far has the fly traveled?

[size=200]Day 17[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

A square is inscribed in an equilateral triangle. Find the ratio of the area of the square to the area of the triangle. Express your answer in simplest radical form as a common fraction.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u]

Exactly how is this square inscribed in this triangle? like this?
 [geogebra]4e4430be20d741f77edd6d41bc91190c1dcd84fd[/geogebra]

sorry i havent had much time to do this because of the itest.
ans to 16:1000
ans to 17:I forget...solution:use similar triangles. i will post the solution when i get home(yes i am still at school)

Day 20: f(x) is a quadratic function with f(0)=5 and f(1)=9. Find the sum of all possible values of f(2).

[size=200]Day 21[/size]

1. [b]No collaboration or public discussion[/b] until the next problem is posted.
2. PM me answers, [b]solutions not necessary.[/b]
3. No previous participation is needed to participate in any given day.
4. [b]Calculators are not permitted[/b] unless otherwise noted.
5. Answers will be accepted up until the posting of the next problem.
6. All answers should be in [b]exact[/b] form and [b]simplified[/b] as much as possible.

Two right triangles ABC and DEF have equal hypotenuses BC=EF. Triangle DEF is translated so that D=B and E=C. Given $ AB \equal{} 4$, $ AC \equal{} 3$, and $ DC \equal{} 4\frac {56}{61}$, Find $ AD\cdot BC$.

[u]Please refrain from posting anything that is not completely necessary for the purpose of this thread.[/u][/quote]

guys I know you can solve this problem. If no one pms me, I will start spamming pythag011 and Wickedestjr until they send in correct answers.

u guys skipped a few days....
Dena and Shibani each took three math tests. The median of
Dena's three scores is the same as Shibani's. Their lowest
scores are also the same. However, Dena's mean is 2 points
Jiigher than Shibani's mean. How many points higher is Dena's
"best test score than Shibani's?
[hide]16 cu feet[/hide]


You have 8 billiard balls, one of which is either lighter or heavier than the rest. How can you find the "odd ball" in two weighings using just a balance scale?

this belongs in the puzzles and brainteasers forum

This is not difficult, but this should be moved to brainteasers.

Ok, Moved:

MATHCOUNTS ---> Puzzles+Brainteasers.

[quote="thatin"]Thats old...
[hide] create 3 sets: 1 of 2 and 2 of 3 balls...[/hide][/quote]

can you explain?

If you don't know whether the odd one out is heavier or lighter, you may need three weighings...

yeah, i don't know how you can do it with 2

you weight the 2 teams of 3
if they are equal then you weight the 2
if not then you... well you should know if it is lighter or heavier; sorry I didnt noticed that there are both options. Then you need 3, there is a similar topic with coins.

[quote="Treething"]throw them all with the exact same velocity and direction vectors at somebody

whichever one hurts the most (or least) is the odd one out[/quote]

the person would be knocked out after the first ball is thrown

[quote="SplashD"][quote="bpms"][quote="box of rocks"]Actually, there could have been NINE balls with the two weighings instead of just the eight balls.

Split the nine balls into 3 groups of 3 balls apiece.  Call them A, B, and C.  Weigh A & B.  If they balance, then go to group C.
Split C into 2 balls on the balance and one left off the balance.
In the 2nd weighing, if they don't balance, there you go.  Or in the 2nd weighing, if they balance, the ball left off is the one you want.

Now go back to A & B.  What if they had not balanced after the 1st weighing?  Then choose the lettered group this time from the result of the 1st weighing to split up to use in the 2nd weighing, and it will work out in a similar way as described in the paragraph above.[/quote]
But you don't know if it's heavier or lighter and you only have 8 balls. I think treething's sol. is the best :D. You can use 8 people and no weighings.[/quote]

would you be willing to be one of the 8 people getting hit :?:  if you think his solution is so good?[/quote]
If it's for the advancement of math...
[size=59][color=white].[/color][/size]

[quote="bpms"][/quote][quote="SplashD"][quote="bpms"][quote="box of rocks"]Actually, there could have been NINE balls with the two weighings instead of just the eight balls.

Split the nine balls into 3 groups of 3 balls apiece.  Call them A, B, and C.  Weigh A & B.  If they balance, then go to group C.
Split C into 2 balls on the balance and one left off the balance.
In the 2nd weighing, if they don't balance, there you go.  Or in the 2nd weighing, if they balance, the ball left off is the one you want.

Now go back to A & B.  What if they had not balanced after the 1st weighing?  Then choose the lettered group this time from the result of the 1st weighing to split up to use in the 2nd weighing, and it will work out in a similar way as described in the paragraph above.[/quote]
But you don't know if it's heavier or lighter and you only have 8 balls. I think treething's sol. is the best :D. You can use 8 people and no weighings.[/quote]

would you be willing to be one of the 8 people getting hit :?:  if you think his solution is so good?[/quote][quote="bpms"]
If it's for the advancement of math...
[size=59][color=white].[/color][/size][/quote]

Actually, you don't need eight people at all. Once the first guy regains consciousness, hit him again. Repeat as necessary. As long as you dont whack the guy in the head (and he loses his memory) it should be fine. :D

Bpms, you're in for a wild and crazy ride through the ER. Eight times. :D

With 8 balls and not knowing whether the odd one is lighter or heavier, then there is a 1/4 chance that you will only need 2 weighings by my method although you need 3 1/2 of the time and 4 1/4 of the time.

Weigh 3 balls against 3 others. If they balance, you're lucky and only need 2 weighings. When they balance, weigh one of the 6 you've already weighed against one of the two that's left. If they balance, it's the unweighed one. If they don't balance, it's the weighed one of the 2 left.

If the 3-3 don't balance, mark one of the balls (or just keep very careful track of it) and transfer it from the heavier of the 3 piles to the 2 pile. Weigh the two new 3 piles against each other. If they balance, use the technique from before. If they don't balance, weigh 1 from the lighter pile against another from the lighter pile. If they don't balance, the lighter is the odd one. If they do balance, weigh the last one from the lighter pile with one of the ones that was in the original group of 2. If they balance, it's the marked one. If they don't balance, it's the last one from the lighter pile.

Interestingly, you can do it with 3 weighings every time with 9 balls, but I couldn't find a way to do it with 8.

[quote="K81o7"][quote="Treething"]throw them all with the exact same velocity and direction vectors at somebody

whichever one hurts the most (or least) is the odd one out[/quote]

If you could do that, you could just as easily weigh them by hand...

I have always stood by my original response to this kind of problem: beat up the person with the scale and run off with it.[/quote]

talk about brute force  :rotfl:


percent
AMC
AIME
geometry
congruent triangles

ARE YOU SMARTER THAN A 5TH GRADER

1)  There are 51 strawberries in the bowl.  Mary ate 16.  How many are left?

2) (two congruent triangles shown)  Are these figures similar or congruent?

3) A TV is on sale at Best Buy for 3000 dollars for 40 percent off.
a)  What is the sale price?
b)  What is the discount?

4) Mary has 800 dollars in her account.  She has earned 4 percent interest.  How much money is in her account at the end of the year?

5) The dinner bill is 100 dollars before tax and tip.  If tax is 6 percent and tip is 20 percent, what is the total bill after tax and tip?

6) What is 20 percent off 90 dollars?

Anyone who gets all 6 questions "correct" must go back to 5th grade.  In addition to your answer, please rate the quality of each question or ways to make the question better.  Yes, the district had a specific answer in mind for each of these questions.

There is no need to hide your solutions.  I will post the "correct" answers once enough people have weighed in.

thanks,

coach fritz

These tests always make me feel inferior  :(

2. Both!  :lol:

1. Depending on how the problem is interpreted, the answer is either 35 or 51.
2. Both - congruent triangles are similar!
3. The sale price is 3000 * 0.6=1800, so the discount is 3000-1800=1200.
4. This is 800 * 1.04=832 dollars.
5. 100 * 1.06 * 1.2=127.20 dollars.
6. 0.8 * 90=72 dollars.

Ways to make problems better:

1. Reword to "How many strawberries are left in the bowl?"
2. Make the triangles similar, so the possibility of "both" is gone.
3. For b) change it to "What is the discount, in dollars?" to clear ambiguity; one could say "40 percent"
4. State that the rate of interest is 4% per year, and she puts the money in the beginning of the year.
5. Fine
6. Fine

do what ever AIMe did.

why is he on the illinois section.

[quote="saszs"]
why is he on the illinois section.[/quote]

Because IL is beastly like that.

[quote="coachfritz"]
Anyone who gets all 6 questions "correct" must go back to 5th grade.  In addition to your answer, please rate the quality of each question or ways to make the question better.  [/quote]

What do you mean by if you get all 6 correct you have to go back to 5th grade?? Don't you mean anyone who doesn't get all 6 correct has to go back to 5th grade??  :ninja:

Haha I think he means anyone who gets all the supposed answers correct.

See how #2 is both?  So they're like trick questions.


hint:[hide]use dummy variables in both of them.[/hide]

xirtamax wrote:x^(2/3) - 7x^(1/3) + 12 = 0
9^(2x) - 2 * 9^(x) - 3 = 0

[hide]x^(2/3) - 7x^(1/3) + 12 = 0

You should see the quadratic here... the powers of x are a third what they usually are (2/3), (1/3), (0/3), so just say

x = a^3

Then using substitution:

a^2 - 7a + 12 = 0

(a - 3)(a - 4) = 0

a = 3 or 4

x = 27 or 64

9^(2x) - 2 * 9^(x) - 3 = 0

This one is a bit trickier, but there is a substitution again. Try a = 9^x

a^2 - 2a - 3 = 0

(a - 3)(a + 1) = 0

a = -1 or 3

9^x = -1 or 3

Dunno what to do from here though...[/hide]

[quote="Dune"][quote="mikewd"]
9^x = -1 or 3
[/quote]

First, 9^x can't be -1, there's no x that verify it, I think, because 9 is positive.

Solving 9^x=3 isn't really difficult since we have 3^2=9. So x=[tex]\frac 12[/tex][/quote]

Ok, thankyou. That makes sense.

The basic theme of these two problems is that equations of the form ax^2+bx+c do not necessarily have to look like that to be solved with the quadratic equation.  As long as the terms match up then you can substitute a new variable in to make it quadratic.

By the way, does anyone know if 9^x=-1 has no possible solutions, or just no real solutions?  It's just that with imaginary numbers, some weird results are possible, so I'm never sure.

It is solvable with imaginary numbers.
Good ole e^(i*x)=cos x + i*sin x.

[quote="Templar"][quote="JS1527"]I actually don't think applying De Moivre works very well.  Magnara how would you do it?[/quote]

9^x=-1
log(9^x)=log(-1)
xlog9=log(-1)

Now e^(i*pi)+1=0

xlog9=i*pi
x=i*pi/log9[/quote]

Is this an acceptable solution, though, since you're working on the premise that a logarithm with a negative anti-logarithm is acceptable?

Well geez. I plugged that into my calculator and it gave -1.  :?

I guess if working on the premise that a negative as a radicand is acceptable, then working on the premise that a negative as an antilogarithm is acceptable would work, too. :P

Was thinking of the same thing...until a friend from interstate checked it on a graphics calculator.

I think you mean ln instead of log base 10, unless it's understood that you're talking base e... is it normal to use log for base e when you're talking about De Moivre's theorem?

There was a discussion about the log issue.  Basically a lot of ppl higher up in math use log x with no subscript to mean log_e x .  IMO, most ppl use log x to mean log_10 x.  Usually you can tell from everything else that goes on in the problem which they mean.  In this case you can see what they mean quite easily.

I always took log to mean log base 10 unless otherwise specified, and ln to mean natural log, or log base e.

As has been said, it all depends on the subject. Mathematicians write log to mean base e, other people use it base 10, computer science people write log to mean base 2.

"When dealing with the logarithms to the base e, it is common especially to denote loge by ln especially if there is any likelihood that the reader might think that base 10 or base 2 logarithms might be meant. [b]In most pure mathematical work, log or ln is used to denote loge[/b]; in most engineering work, log means log10; while in information theory, log often means log2, which is traditionally written as ld (from the Latin logarithmus dualis), but also sometimes is written as lg. Whenever a possibility for ambiguity exists, this ambiguity is resolved by explicitly writing out the base."

- [url=http://en.wikipedia.org/wiki/Logarithm]Wikipedia[/url]


1.                       (2n+3)/(3n+5)          if n is a mulitple of 3
              Xn = {
                          (-3n+1)/(7n-5)         if n is not a multiple of 3

Is the sequence diverge or converge?? HOW??

2.            [(-1)^n(n cosn)]
              ______________     Converge or Diverge??? HOW??
                   (n^2 )+1

I think the problem is

Xn = ((-1)^n) n cosn/[(n^2)+1]

Can someone tell me if the sequence diverge or converge?  And please tell me how?  Because i was trying to do it but i got stuck and don't know how to proof if its converge.  Please advise

The first one diverges. Consider large $n$. When $n$ is a multiple of 3, you're adding about 2/3. Otherwise, $n$ is about -3/7. So for every three terms, you're getting $2/3 - 3/7 - 3/7 = -4/21$. So you'll get deeply negative. Divergently so.

The second one converges. Use Abel's Damped Oscillation Test (I don't know if this is the real name for the test; my analysis professor used nonstandard terminology sometimes). The theorem says that if you have a bounded oscillating sequence $a_n$ and a sequence $b_n$ that decreases to 0, then the sum $\sum_n a_nb_n$ converges. In this case, pick $a_n = (-1)^n$ and $b_n = \frac{n\cos(n)}{n^2+1}$.


Does any one know a formula/method/ect for finding the center of mass of a polygon, regular and irregular using a cordinate plane and another method not using a cordinate plane. There is one on mathworld:   [url]http://mathworld.wolfram.com/GeometricCentroid.html[/url] but that gives the geometric center which i am not sure is always the centroid. Any helps is appreciated.

Thanks

Well, on a coordinate plane, for a triangle (I don't know if it's true for all polygons), if you are given the coordinates of the vertices, the centroid is located at the average of the 3 vertices. (Sorry if that was kind of confusing)
Ex: if the points were $ (a,b), (c,d),$ and $ (e,f)$
The centroid is at: $ (\dfrac{a \plus{} c \plus{} e}{3} , \dfrac {b \plus{} d \plus{} f}{3})$

The next step is: try to find the "centroid" of a convex quadrilateral. Does it hold analogously from triangle to quadrilateral, ie the average of the coordinates of the vertices?

[b]Extension:[/b] What if the vertices have "weights?"

Im basically trying to find the center of mass for uniformed convex polygons in any method. The polygons are both regular and irregular.


MATHCOUNTS
limit
ceiling function
counting
distinguishability
LaTeX
function

[size=150]MATHCOUNTS $e$ $\displaystyle{\lim}$ $\BIG{i}$ nation![/size]

Would you like to take place in the first ever MATHCOUNTS $e$ $\lim$ $i$ nation?

If so, follow these instructions!
[hide="Instructions"]

1) PM me with "MATHCOUNTS E lim i nation!" in the title. I don't care what you put in the body; I'm not going to read it. :)

2) The competition will officially start on Tuesday, August 3, 2005, so register by then. I will post five easy problems at 12:00PM AoPS Time = 3:00PM ET.

3) Each problem will be worth its cardinal number. Problem 1 will be worth 1 point, Problem 2 will be 2 points, and so on. The total number of points possible is therefore 15.

4) To get to the next round, you must get at least $\lceil \frac{r}{2} \rceil$ of the $r$ possible points.

5) Each round will be harder and harder. We will keep having rounds until there are four people left. At that point, there will be a "countdown round."

6) Round $n$ will have $4+n$ questions.

7) In order to participate, you must be entering 9th grade or less.

[/hide]

[hide="Participants"]
jli
Boy Soprano
236factorial
Drunken_Math
math92
PenguinIntegral
nat mc
Xantos C. Guin
hwenterprise
G-UNIT
solafidefarms --> lol, solafidefarms, with his mod powers, registered himself. :P
EFuzzy
christy-metmat
mathnerd314
GoBraves
LuCy4EvA
kd5zht2
Treething
anirudh
diophantine27 --> my sister :P
nonie

That makes 21! :)
[/hide]

[hide="Final Qualifiers for Round 2"]
Drunken_Math
jli
Boy Soprano
nat mc
Treething
bob123
solafidefarms
236factorial
PenguinIntegral
EFuzzy
anirudh
christy-metmat
star99
G-UNIT
Xantos C. Guin
diophantine27

15/21=5/7 of contestants qualified.
[/hide]

[hide="Final Qualifiers for Round 3"]
PenguinIntegral
jli
solafidefarms
boy soprano
bob123
Xantos C. Guin
nat mc
236factorial
Drunken_Math

9/15=3/5 qualified.
[/hide]

[hide="Final Qualifiers for Round 4"]
bob123
Drunken_Math
PenguinIntegral
jli
Xantos C. Guin
236factorial
nat mc
Boy Soprano
solafidefarms

EVERYONE QUALIFIED! Competition is getting tough... :?
[/hide]

[hide="Final Qualifiers for Round 5"]
nat mc: 33/36
bob123: 26/36
236factorial: 23/36
Boy Soprano: 22/36
solafidefarms: 22/36
jli: 21/36
PenguinIntegral: 20/36

7/9 qualified.
[/hide]

[hide="Final Qualifiers for Countdown Round!"]
nat mc - 35/45
Boy Soprano - 29/45
236factorial - 28/45
bob123 - 25/45
jli - 16/45
[/hide]

First of all, this isn't the first ever Mathcounts Elimination thingy, because others have set up competitions with the same idea but a different name: Mathcounts Challenge, etc. (Did you take this idea from sirorange?)

What I was really here to ask, though, is when this competition will be over. Because I'm going on vacation and won't get a computer. :(

How long would we have to answer the questions?  (I'm concerned about being able to get to the computer every day.)

1) The idea of [i]a[/i] competition was from my friend sirorange. :) But the rules are completely different.

2) The competition will be over when there are only 2 people left. It depends on how good you guys are. When are you going to be gone?

3) Upon the suggestion of jli, you will have $n+2$ days for round $n$.

first of all...

im singning up

second...

it will take way to long to complete...make it n+2

I'll participate!

BTW, this should be stickied

Fine, im in.

Each of you just got an extra round ;)

Also, we should probably post what grade we are going into. Im going to 7th.

I will participate too. I guess it's going to be an interesting challenge! ! ! :)

I'll try, if I don't leave before it ends. :bye:

[quote="nat mc"]I can probablly do it after 8pm (6pm AOPS) everyday except tomorrow, i won't be home[/quote]

I can't go on at night  :blush:

[quote="hwenterprise"]You could post them tomorrow morning.

BTW: I'm his brother, so I know when he goes on the internet.

Yes, I am already out of the competition.[/quote]

Countdown round means right at the place, at one time  :roll:

It's not a round like the others.

[quote="Drunken_Math"]I'm just wondering... maybe I overlooked it... but

Can you post the answers to Round 4?

I went through the entire topic and I can't seem to find it...[/quote]

Yeah chess, i thought you said soon  :D What are round 4 solutions? How about 5?

just give me a couple dates and ill see if i can make them

[quote="chess64"]Today :D : 3:00PM ET

Thursday: 11:00 - 3:00PM ET

Friday: 11:00PM ET

That's all :( If we can't agree, there will be no countdown round.[/quote]

All is fine with me  :D

 :cool: So I'm ready....to lose  :D

wait...so the whole countdown round is in 1 session?

i might be able to make thursday 8:30-12 PCT

Can we make it AOPS time, cause I forget time zones.

Please, after 5:30 or 6 AOPS time

I can't do it then... you know what, forget it.  :( :( :(

[size=200]MATHCOUNTS Elimination! Final Results[/size]

:first: nat mc
:winner_second: Boy Soprano
:winner_third: 236factorial
fourth: bob123
fifth: jli

Congratulations to all competitors! :)

[quote="236factorial"]why not make an nearly impossible round?  :D[/quote]

I still have to write solutions.... :sleeping:


calculus
integration
calculus computations

find integral of cos(y^-0.5)/(y^0.5)?
I need specific process, thank you.

1. Make the obvious substitution.
2. Note that it's an integration by parts away from something that famously doesn't have an elementary antiderivative.
3. Check to see if you wrote it down correctly.


[hide="solution"]There are 11 possibilities for the sum, so 12.[/hide]

[quote="machack"]What is the number of tosses with 2 dices required to guarantee that you toss the same sum twice[/quote]
[hide="Answer"]The sums could be 2, 3, 4, ... 12. To make sure we have one twice, we must have them all already so the next one has to be already gotten. Thus you would roll a maximum of 11 + 1 = [b]12[/b][/hide]

[hide="Solution"]Start by finding out the greatest possible way to avoid getting the same sum. That is, you roll a $2,3,4,5,6,7,8,9,10,11,12$ in that order. Then, if you roll anything, it will have to be something you already rolled. Therefore, the answer is $\boxed{12}$.[/hide]

[hide]find the number of possible sums(11) and just add one so the answer is 12.[/hide]

This was a dumb question. not only did i get the right answer right after posting, i didn't even post the correct question... (the real quesiont is harder and different)


Will there be a gym and track? Will I have enough time to go run a few miles each day?

Also, will there be a place to buy chocolate chip cookies? Junggi owes me one.

oh snap i plan to run too...

Here was I was thinking:

I would like wake up around 7 am...go running wherever [I run xc...so I could just run around campus or w/e]. It would not take more than 45 mins. I would be back and I could go about my morning routine.

Is this acceptable? [to coordinators, etc?]

Some of the counselors led morning runs last year. I'm pretty sure they'll do so again. Also, the very early morning is probably the only time its acceptable to go running. It heats up very very quickly

unless there's an indoor track, it's gonna be probably 90 degrees or above unless you run before 7 a.m. or after dark (my personal favorite)


LaTeX
number theory unsolved
number theory

let $p$ be a prime number prove that:

$\sum_{k=1}^{\frac{p-1}{2}} \frac{(p-1)!}{k} \equiv \frac {2^p-2}{p} $  $(\mod p )$

Mahmoud, why did you pass over to $2 \sum_{k=1}^{\frac{p-1}{2}}{ \frac{(p-1)!}{k!(p-k)!} }$ at the beginning of your solution? I mean, your proof of the identity $\frac{2^p - 2}{p} \equiv 2 \sum_{k=1}^{\frac{p-1}{2}}{ (-1)^k\frac{(p-1)!}{k} } \,\,(mod\,p)$ is correct, but it took me a while to see that this identity is indeed equivalent to the identity posted by Amir2... For those who had the same difficulties with understanding this solution, I propose a better writeup:

Amir2's problem is (with a better LaTeX formatting):

[color=blue][b]Problem.[/b] If $p$ is an odd prime, prove that

$\sum_{k=1}^{\frac{p-1}{2}}\frac{\left(p-1\right)!}{k}\equiv\frac{2^p-2}{p}\mod p$.[/color]

[i]Solution.[/i] I shall use the notion of "$p$-integers" and the "congruence modulo $p$" relation on the set of $p$-integers; see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46121 post #2 for explanations of these concepts.

The binomial formula yields $2^p=\left(1+1\right)^p=\sum_{k=0}^p\binom{p}{k}=1+\sum_{k=1}^{p-1}\binom{p}{k}+1=2+\sum_{k=1}^{p-1}\binom{p}{k}$, so that $2^p-2=\sum_{k=1}^{p-1}\binom{p}{k}$. But $\binom{p}{k}=\frac{p!}{k!\left(p-k\right)!}=\frac{p\cdot\left(p-1\right)!}{k!\left(p-k\right)!}$. Now $2^p-2=\sum_{k=1}^{p-1}\binom{p}{k}$ yields

$\frac{2^p-2}{p}=\frac{1}{p}\sum_{k=1}^{p-1}\underbrace{\binom{p}{k}}_{=\frac{p\cdot\left(p-1\right)!}{k!\left(p-k\right)!}}=\frac{1}{p}\sum_{k=1}^{p-1}\frac{p\cdot\left(p-1\right)!}{k!\left(p-k\right)!}=\sum_{k=1}^{p-1}\frac{\left(p-1\right)!}{k!\left(p-k\right)!}$.

Hence, instead of proving $\sum_{k=1}^{\frac{p-1}{2}}\frac{\left(p-1\right)!}{k}\equiv\frac{2^p-2}{p}\mod p$, it will be enough to show that

$\sum_{k=1}^{\frac{p-1}{2}}\frac{\left(p-1\right)!}{k}\equiv\sum_{k=1}^{p-1}\frac{\left(p-1\right)!}{k!\left(p-k\right)!}\mod p$.

Note that all fractions in this formula are $p$-integers (since their denominators are not divisible by $p$, since $p$ is a prime and $1\leq k\leq p-1$). Thus, "congruence modulo $p$" makes sense for these fractions. So we can apply the Wilson theorem in the form $\left(p-1\right)!\equiv -1\mod p$ to the left hand side and in the right hand side, we can use

$\frac{\left(p-1\right)!}{k!\left(p-k\right)!}=\frac{\left(p-1\right)!/\left(p-k\right)!}{k!}=\frac{\left(p-1\right)\left(p-2\right)...\left(p-k+1\right)}{k!}$
$\equiv\frac{\left(-1\right)\left(-2\right)...\left(-k+1\right)}{k!}$                  (since $p\equiv 0\mod p$)
$=\frac{\left(-1\right)^{k-1}\cdot\left(1\cdot 2\cdot ...\cdot\left(k-1\right)\right)}{k!}=\frac{\left(-1\right)^{k-1}\cdot\left(k-1\right)!}{k!}=\frac{\left(-1\right)^{k-1}}{k}\mod p$.

Hence, the identity in question becomes

$\sum_{k=1}^{\frac{p-1}{2}}\frac{-1}{k}\equiv\sum_{k=1}^{p-1}\frac{\left(-1\right)^{k-1}}{k}\mod p$.

Multiply this identity with $-1$ to get

$\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k}\equiv\sum_{k=1}^{p-1}\frac{\left(-1\right)^k}{k}\mod p$.

But $\left\{1,2,...,\frac{p-1}{2}\right\} = \left\{\frac{k}{2} \mid k\in\left\{1,2,...,p-1\right\},\ k\text{ even}\right\}$, so that

$\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k}=\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{\left(\frac{k}{2}\right)}=\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{2}{k}=2\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{k}$,

while

$\sum_{k=1}^{p-1}\frac{\left(-1\right)^k}{k}=\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{k}-\sum_{1\leq k\leq p-1;\ k\text{ odd}}\frac{1}{k}$.

So we have to prove that

$2\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{k}\equiv\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{k}-\sum_{1\leq k\leq p-1;\ k\text{ odd}}\frac{1}{k}\mod p$,

or, equivalently, that

$\sum_{1\leq k\leq p-1;\ k\text{ even}}\frac{1}{k}+\sum_{1\leq k\leq p-1;\ k\text{ odd}}\frac{1}{k}\equiv 0\mod p$.

But this simplifies to

$\sum_{k=1}^{p-1}\frac{1}{k}\equiv 0\mod p$,

what is more or less obvious:

$\sum_{k=1}^{p-1}\frac{1}{k}=\sum_{k=1}^{p-1}\frac{1}{p-k}\equiv\sum_{k=1}^{p-1}\frac{1}{-k}=-\sum_{k=1}^{p-1}\frac{1}{k}\mod p$,

and thus $\sum_{k=1}^{p-1}\frac{1}{k}\equiv 0\mod p$ (in fact, since $p$ is an odd prime, the only number in $\mathbb{Z}_p$ which equals its own additive inverse is 0).

So we are done.

By the way, in the above, we have more or less proved the identity from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37352 .

  Darij