Find the sum of the fractions: $\frac12+\frac16+\frac1{12}+\frac1{20}+\ldots+\frac1{9900}$

Here's what I got, and I actually got this one right when I did it.

[hide]Click [hide]On [hide]The Smiley[/hide][/hide][/hide]
[hide=":lol:"]
Okay, so if you look closer at the screen and the fractions (NOT THAT CLOSE), I can show you how to make the problem easier.

$1/2=\frac{1}{1*2}$
$1/6=\frac{1}{2*3}$
         $\vdots$
$1/9900=\frac{1}{99*100}$

Now, add them up to see a pattern
$1/2+1/6=2/3$
$2/3+1/12=3/4$
Notice how the two numbers that you multiply together are the same two numbers in the numerator and denominator.  So by continuing this pattern, you get $99/100$ because the last fraction is made by multiplying $99*100$.[/hide]

[hide]
My solution:

Just like rd5493, I changed this into a series of fractions.
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{99\cdot100}$

Buuuuut...
I used the fact that $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{(x)(x+1)}$

So it becomes
$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots-\frac{1}{100}$

Which simplifies to
$\frac{1}{1}-\frac{1}{100}=\frac{99}{100}$
[/hide]


algorithm
induction
number theory unsolved
number theory

How can one show that every integer of the form $4k+3$ has (at least one) prime factor of the same form?

I've tried some stuff(using the division algorithm to write $4k+3 = q(4k+r)$, but I haven't found any good conclusions for the case $r=1$, the other cases are quite obvious.

Edit: I think I've found an induction proof (since if $4k+r$ is of the form $4k+1$, then $q$ must be of the form $4k+3$) but arethere any other ways?

It is obviosly. If $n=4k+3$, then n is odd, therefore all prime divisores odd. If all prime divisors had form $4m+1$, then $n=p_{1}^{q_{1}}...p_{l}^{q_{l}}=1(mod \ 4)$.

My solution is:

$n=3(mod 4)$, so you can consider the powers of a generic $mod 4$ and you can discover that they can be only 0, 1, 3.
So if you want to have 3 as product, you must take at least a factor $a=3(mod 4)$. You can have it only if the prime factor $p=3 (mod 4)$.

Hope to have been clear (I'm not really good speaking english...)


'Twas last Bank Holiday, so I've been told,
Some cyclists rode abroad in glorious weather.
Resting at noon within a tavern old,
They all agreed to have a feast together.
"Put it all in one bill, mine host," they said,
"For every man an equal share will pay."
The bill was promptly on the table laid,
And four pounds was the reckoning that day.
But, sad to state, when they prepared to square,
'Twas found that two had sneaked outside and fled.
So, for two shillings more than his due share
Each honest man who had remained was bled.
They settled later with those rogues, no doubt.
How many were they when they first set out

A pound is twenty shillings (I looked it up :) ) so, if there were $n$ cyclists originally, the situation is
\[\frac{80}{n-2}=\frac{80}{n}+2\]
and I leave the rest to you :D

oh; really this problem and your solution is very nice


geometry
Polars
geometry solved
projective geometry
Pascal s theorem

Let $ \Delta ABC$ with incircle $ \Gamma$, let $ D, E$ and $ F$ the tangency points of $ \Gamma$ with sides $ BC, AC$ and $ AB$, respectively and let $ P$ the intersection point of $ AD$ with $ \Gamma$.
$ a)$ Prove that $ BC, EF$ and the straight line tangent to $ \Gamma$ for $ P$ concur at a point $ A'$.
$ b)$ Define $ B'$ and $ C'$ in an anologous way than $ A'$. Prove that $ A'\minus{}B'\minus{}C'$

there is a strong prabability that both of them are posted before,anyway,
a)use pole and polar(it directly proves this part),or similarity.
b)use menelaus for $ EFA'$ to get the ratio $ \frac{CA'}{BA'}$ and similarly for the other ratios,or use pole and polar (again it directly proves this part )

$ DEPF$ is a harmonical quadrilateral ( the tangents at $ E$ and $ F$ are concurrent at $ A$ on $ PD$), hence the tangents at $ D$ and $ P$ are concurrent on the diagonal $ EF$, hence $ A'\equal{} BC \cap EF, B' \equal{} CA \cap DF$ and $ C'\equal{} DE\cap AB$, which are collinear, done.

Best regards,
sunken rock

This problem is mine, if it´s wellknown I don´t know, the only thing I know is that it´s a very very nice result that I saw my own  :lol:
By the way very nice solution sunken rock.

For part a) just note that if $EF\cap BC=X$, then the tangent of $P$ to $\Gamma$ passes throught the point $X$ iff $P$ is in the polar of $X$ wrt circle $\Gamma$ iff $A$ is in the polar of $X$ wrt circle $\Gamma$, this is true iff $X$ is in the polar of $A$, from La hire theorem, wich is true since $X$ is in $EF$, the polar of $A$. For part b) just note that the lines $AD,BE,CF$ are concurrent in the gergonne point of the triangle $ABC$, so $A',B',C'$ are collinear  iff the polars of $A',B',C'$ wrt $\Gamma$ are concurrent (using part a)), wich is true.

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/15.%200.%20Droites%20passant%20par%20un%20point%20de%20Nobbs.pdf] here [/url]

Problem 1

Sincerely
Jean-Louis


geometry
circumcircle
Circumcenter
Triangle
IMO Shortlist
excenters
geometry solved

yes thats like the only solution  :?

by the way, how did you know that was from ISL 2003?

[[b]Moderator edit:[/b] The problem can be extended:

Let $ABC$ be a triangle, and $P$ a point in the plane. Let $D$, $E$, $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$, respectively. Furthermore, let $I_{a}$, $I_{b}$, $I_{c}$ be the excenters of triangle $ABC$. Then, prove that $AP^{2}+PD^{2}=BP^{2}+PE^{2}=CP^{2}+PF^{2}$ holds if and only if the point $P$ is the circumcenter of triangle $I_{a}I_{b}I_{c}$. ]

[quote="grobber"]Oh, I have the ISL problems :).[/quote] We (the romanian committee) gaved the problems to the students at RoMOP to be studied. :) Of course, as these problems were already used in some competitions, it is ok to put them online.

All other ISL problems are still not to be disclosed ;) :)

I only have whatever you gave to the students at the training camp, because actually, that's how I got the problems: I know quite a few of those students :).

[quote="grobber"]This is also from the ISL.

We have $PA^2+PD^2=PD^2+PE^2+EA^2=PB^2+PE^2=PD^2+PE^2+BD^2$ $\Rightarrow AE=BD$. In the same way we show that $CE=BF,\ AF=CD$, so we can see that $D, E, F$ are the projections of $I_a,I_b,I_c$ on $BC,CA,AB$ respectively.

The conclusion follows.[/quote]

I had same but didn't see why the conclusions follow, anyone can explain?

[color=darkred][b][u]PP (IMO 2003)[/u].[/b] Let $ABC$ be a triangle and $P$ a point in the interior of this triangle. Let $D$ , $E$ , $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$ respectively. Assume that $AP^{2}+PD^{2}=BP^{2}+PE^{2}=CP^{2}+PF^{2}$ . Furthermore, let $I_{a}$, $I_{b}$, $I_{c}$ be the excenters of triangle $ABC$ . Show that the point $P$ is the circumcenter of triangle $I_{a}I_{b}I_{c}$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] Observe that $PA^2+PD^2=$ $PE^2+EA^2+PD^2$ and $PB^2+PE^2=$ $PD^2+DB^2+PE^2$ $\Longrightarrow$ $AE=BD$ . In the same way we show that $CE=BF$ and $AF=CD$ . Prove easily that $D$ , $E$ , $ F$ are the projections of $I_a$ , $I_b$ , $I_c$ on $BC$ , $CA$ , $AB$ respectively, i.e. $AE=BD=s-c$ , $AF=CD=s-b$ , $BF=CE=s-a$ . Denote the projection $F'=\mathrm{pr}_{AB}I_a$ . Observe that $BF'=BD$ and $b=(s-a)+(s-c)=BF+BF'=\mathrm{pr}_{AB}[PI_a]=PI_a\cdot\sin\widehat{PI_aF'}=PI_a\cdot\sin B$ $\implies$ $PI_a=\frac {b}{\sin B}=2R$ , where $R$ is the length of the circumcenter for $\triangle ABC$ . In conclusion, $PI_a=PI_b=PI_c=2R$ , i.e. the point $P$ is the circumcenter of the triangle $I_{a}I_{b}I_{c}$ . [u][b]Otherwise[/b][/u]. Prove easily that the triangle $I_aPI_b$ , $I_bPI_c$ and $I_cPI_a$ are isosceles , i.e. $PI_a=PI_b=PI_c$ . Observe that in $\triangle I_aPI_b$ exists the relation $\frac {PI_a}{\sin\frac C2}=\frac {I_aI_b}{\sin C}$ . Therefore, $PI_a=\frac {I_aI_b}{2\cos\frac C2} =$ $\frac {r_a+r_b}{2\cos^2\frac C2}=$ $\frac {\frac {S}{s-a}+\frac {S}{s-b}}{\frac {2s(s-c)}{ab}}=$ $\frac {abcS}{2s(s-a)(s-b)(s-c)}=2R$ . In conclusion, $PI_a=PI_b=PI_c=2R$ . [b][u]Otherwise[/u][/b]. If $A'$ , $B'$ , $C'$ are the intersections of $AI$ , $BI$ , $CI$ respectively with the circumcircle of $\triangle ABC$ , then $A'B'C'\sim I_aI_bI_c$ , where $I\in A'I_a\cap B'I_b\cap C'I_c$ and $B'C'\parallel I_bI_c$ , $C'A'\parallel I_cI_a$ , $A'B'\parallel I_aI_b$ ([u]homothety[/u]). Since $2\cdot IA'=II_a$ obtain that the length of the circumradius for $\triangle I_aI_bI_c$ is equally to $2R$ , where $R$ is the length of the circumradius for $\triangle A'B'C'$ , i.e. the circumradius of $\triangle ABC$ .  See and [/color] [url=http://www.artofproblemsolving.com/blog/50478][color=red][b]here[/b][/color][/url].

[hide="Solution"]
Let $BC=a,AC=b,AB=c$, and $s$ be the semiperimeter.

Note that by the Pythagorean Theorem, $AP^2-PE^2=BP^2-PD^2\implies AE=BD$. Likewise, $AF=CD$, and $BF=CE$. It follows that $BD=s-c$. Let $D'$ be the point of tangency of the incircle with $BC$. Then $CD'=s-c=BD$, so $AD$ is isotomic to $AD'$. Since the Gergonne and Nagel points are isotomic conjugates, $D$ is the point of tangency of the A-excircle with $BC$. Thus, $I_aD\perp BC$. But $PD\perp BC$, so $P,D,I_a$ are collinear.

By similar reasoning, $P,E,I_b$ are collinear and $P,F,I_c$ are collinear. Furthermore, it is well known $\triangle{ABC}$ is the orthic triangle of $I_aI_bI_c$, so $I_aA\perp I_bI_c$. Thus,
\[ \angle{PI_bA}=90-\angle{CAI_b}=\angle{CAI_a}=\angle{BAI_a}=90-\angle{BAI_c}=\angle{PI_cA}\]
implying that $PI_b=PI_c$. Similarly, $PI_a=PI_b$, so $P$ is the circumcenter of $\triangle{I_aI_bI_c}$, as desired. // [/hide]

Suppose $P$ is the circumcentre of $\triangle I_aI_bI_c$ (which is the isogonal conjugate of the orthocentre of $\triangle I_AI_bI_c$). We have $\triangle ABC$ is the orthic triangle of $\triangle I_aI_bI_c$, so $I_aP \perp BC$ and similaarly for others. Therefore, $D, E, F$ is just the nagel triangle of $\triangle ABC$, which means $BD = AE = s-c$. So, by pythagorean theorem we have $AP^2 - PE^2 = AE^2 = BD^2 = BP^2 - PD^2$. So, done.

[quote="rizkydarmawan"]Sorry, where do you know that $D, E,$ and $F$ are the points where the excircles touch the sides of triangle $ABC$ ?[/quote]

Note that $\triangle ABC$ is just the orthic triangle of $\triangle I_AI_BI_C$. Since $P$ is the circumcentre, which is the isogonal conjugate of the orthocentre, and $\triangle ABC$ is the cevian triangle of the orthocentre, $I_AP \perp BC$ hence we deduce that it is the nagel triangle

Note that i mentioned this in my proof...

This problem reappeared in India TSTs 2004.I cannot get why they repeated the same problem......

Bye...

Sayantan....

[b]Lemma :[/b] [color=#f00]$P$ is the Bevan point of triangle $\triangle ABC$[/color]

$AP^2 + PD^2 = BP^2 + PE^2$
$\implies AE^2 + PE^2 + PD^2 = BP^2 + PE^2$
$\implies AE^2 = BD^2$
$\implies AE = BD$

Similarly, $CP=PA$ and $CE = BF$
$\implies D,E,F$ are the extouch points of $\triangle ABC$

Now since, $PD,PE,PF$ are perpendiculars to sides $BC,CA,AB$
$\implies P$ is the Bevan point, that is the circumcentre of the extriangle.

QED

Seems to be original so far based on this thread?
[hide=Solution]
By the Pythagorean theorem, $AF^2+PF^2+PD^2=PD^2+CD^2+PF^2 \implies AF=CD$. Similarly, $AE=BD, BF=CE$. We can now find that $BF=CE=s-a, FA=CD=s-b, AE=BD=s-c$. Therefore, $P-D-I_A, P-E-I_B, P-F-I_C$ are all collinear.

Let $X=I_AD \cap I_BI_C$. Then $\angle BI_AD=\frac{B}{2}, \angle CI_AD=\frac{C}{2}$ by the above collinearities. Length chase with right angle trig to get $I_AI_C=\frac{s-b}{\sin \frac{B}{2}}+\frac{s-a}{\sin \frac{B}{2}}, I_AI_B=\frac{s-c}{\sin \frac{C}{2}}+\frac{s-a}{\sin \frac{C}{2}}$. So by Ratio Lemma,
\[ \frac{I_BX}{I_CX}=\frac{s-b+s-a}{s-c+s-a}=\frac{c}{b} \]

Now suppose $O = \odot(I_AI_BI_C)$ and let $I_AO \cap I_BI_C = X'$. In reference to $\triangle ABC$,
\[ \angle I_AI_BI_C=90-\frac{B}{2} \implies \angle I_COI_A=180-B \implies \angle X'OI_C=B \]
Similarly, $\angle X'OI_B=C$. Therefore by Ratio Lemma $\frac{I_BX'}{I_CX'}=\frac{\sin C}{\sin B}$. This holds similarly for the other vertices, hence $P=O$ and we are done.
[/hide]

[hide=Storage (or whatever...)][quote=ISL 2003 G3]Let $ABC$ be a triangle and let $P$ be a point in its interior.  Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively.  Suppose that \[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\] Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$.  Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$.

[i]Proposed by C.R. Pranesachar, India [/i][/quote]
[b][color=#000]Solution:[/color][/b] \begin{align*} AP^2+PD^2=BP^2+PE^2=CP^2+PF^2 \implies AE=BD=x, DC=AF=y, BF=CE=z \end{align*}
From here, $2(x+y+z)=a+b+c$ $\implies$ $x=s-c, y=s-b, z=s-a$ $\implies$ $D,E,F$ are foots from $I_A,I_B,I_C$ to $BC,AC,AB$. Note that, in $\Delta BI_AC$, $I_AI,I_AD$ are Isogonal WRT $\angle BI_AC$. Similar goes for $\Delta CI_BA$, $\Delta AI_CB$. Since, $I$ is orthocenter WRT $\Delta I_AI_BI_C$, $P$ becomes the circumcenter $\qquad \blacksquare$

[/hide]

Storage.
[quote=ISL 2003 G3]Let $ABC$ be a triangle and let $P$ be a point in its interior.  Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively.  Suppose that \[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\] Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$.  Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$.

[i]Proposed by C.R. Pranesachar, India [/i][/quote]

$$\begin{cases} AP^2+PD^2=BP^2+PE^2\implies AE=BD \\ AP^2+PD^2=CP^2+PF^2\implies AF=CD \\ BP^2+PE^2=CP^2+PF^2\implies BF=CE\end{cases}$$ So from here we get that $AD,BE,CF$ bisects the perimeter of the triangle $ABC$. So, $\{AD,BE,CF\}$ are the $A,B,C\text{ Nagel Cevians}$ respectively. Hence, $\overline{P-D-I_A},\overline{P-F-I_C},\overline{P-E-I_B}$. After that a bit of angle chasing shows that $PI_A=PI_B=PI_C\implies P$ is the $\text{Bevan Point}$ or Circumcenter of the Excentral Triangle of $\triangle ABC$ i.e $I_AI_BI_C$.

This seemed too easy for G3  ;)

[hide=Solution]Firstly, note that $$AP^2 + PD^2 = BP^2 + PE^2 \implies AP^2+BP^2-BD^2=BP^2+PE^2=AP^2-AE^2 \implies BD=AE$$Similarly, we get : $CD=AF$ and $BF=CE$.

Now, $a=AE+AF=b-CE+c-BF \implies 2BF=b+c-a=2s-2a \implies BF=s-a$. Similarly, we get : $BD=s-c$ and $CE=s-b$. Hence, we conclude that $D,E$ and $F$ are the $A-\textbf{excircle}, B-\textbf{excircle}$ and $C-\textbf{excircle}$ touch points. Hence, $\overline{P-D-I_A},\overline{P-F-I_C},\overline{P-E-I_B}$.

Now, $\angle BI_AD=90-\frac{B}{2}=BI_CF$. Hence, $PI_A=PI_C$. Also, $\angle CI_AD=90-\frac{C}{2}=CI_BE$. Hence, $PI_A=PI_B$. Thus we conclude that $PI_A=PI_B=PI_C$. In other words, $P$ is the circumcenter of $I_AI_BI_C$, as desired. $\blacksquare$   [/hide]

In fact, in this solution, I will not even refer to $I_a, I_b, I_c,$ as the problem is equivalent to demonstrating that $P$ is the reflection of $I$ over $O,$ where all centers are defined wrt $\triangle ABC$. It is easy to see by doing pairwise subtraction that $AE=BD, DC=AF, EC=FB$. Notice that this essentially shows that the points $D,E,F$ are the extouch points, or that $PF$ is the reflection of the perpendicular from $I$ to $AB$ over the perpendicular from $O$ to $AB$, in other words, they are isotomic conjugates of the intouch points(aka the extouch points). A similar thing hold for $PE,$ and therefore, because the corresponding perpendicular of $P$ are reflections of the perpendiculars from $I$ over the perpendiculars from $O,$ we get that $P$ is the reflection of $I$ over $O$.

[hide]Note that the equations given in the problem statement give \[AF=DC,AE=BD,EC=FB\] Now it is well known that the circumcenter of the excentral triangle is the reflection of $I$ over $O$ where $I$ is the incenter and $O$ is the circumcenter of $ABC$.

Now to show that the reflection of $P$ over $O$ is $I$ (let this be $P'$), note that $O$ lies on the perpendicular bisectors of all sides. Thus if we let $D',E',F'$ be the feet to the sides from $P'$ analogously, then we have $AF'=BF$ and similarly for $D',E'$. This solves the problem, as we now have $BF'=BD'$ and similarly for $A,C$ so $P'$ is the incenter and we are done. [/hide]

[quote=AnonymousBunny]That's because most of the countries use problems from the ISL for their TST's (which is why the ISL is kept strictly confidential until the next IMO).

[hide=" Solution"]
By Pythag, we get that
\[PA^2+PD^2= PB^2+PE^2 \implies PD^2+PE^2+AE^2 = PD^2+PE^2+BD^2 \\ \implies AE=BD.\] Similarly, $BF=CE$ and $AF=CD,$ so $D,E,F$ are the points of tangency of the excircles opposite $A,B,C$ with $BC,CA,AB$ respectively, so $I_aD \perp BC, I_bE \perp CA, I_cE \perp AB,$ and as a consequence, lines $I_aD, I_bE, I_cF$ are concurrent at $P.$

Let the line passing through $A$ and perpendicular to $I_bI_c$ meet $BC$ at $X.$ Since $\angle ACI_b = \angle ABI_c,$ $\angle XAC = 90^{\circ} - \angle ACI_c= 90^{\circ} - \angle ABI_c = \angle XCA,$ so $X$ lies on the internal angle bisector of $\angle BAC,$ which implies $I_aA \perp I_bI_c.$ Similarly, $I_bB \perp I_cI_a$ and $I_cC \perp  I_aI_b.$

Then, note that $\angle I_aAC = \angle I_aAB \implies \angle AI_bP = \angle AI_cP \implies PI_b=PI_c.$ Similarly, $PI_c=PI_a$ and we're done. $\blacksquare$ [/hide][/quote]

Confidential to the common people and contestants :).
Since trainers know the problems.

My Sol : (I was dumb tho)
Bit of length chasing to get $AE=BD,BF=CE,AF=CD$.
Thus they are extouch points and now angel chase.

Lemma1 : Let incircle touch BC at D' and A-excircle touch BC at D. CD = BD'.
 Proof : Let AC = b, AB = c, BC = a, AB+BC+CA/2 = p. Let A-excircle meet AC and AB at S and K.
 AS = AS+AK/2 = AB+AC+CD+BD/2 = p
 CD = AS - AC = p - b
 BD' = p - b

Lemma2 : Let A-excircle touch BC at D, B-excircle touch AC at E and C-excircle touch AB at F. AF = CD, AE = BD and BF = CE.
 Proof :  Let incircle touch BC at D', AC at E' and AB at F'
 CD = BD' = BF' = AF
 BD = CD' = CE' = AE
 BF = AF' = AE' = CE.

Lemma3 : Only 3 points like D,E,F exists on segments of ABC such that AF = CD, AE = BD and BF = CE.
 Proof : let assume there are also points X,Y,Z such that AX = CY,AZ = BY and BX = CZ.
 AX = AF + n so CY = CD + n so BY = BD - n so AZ = AE - n so CZ = CE + n so BX = BF + n witch gives contradiction.

AP^2 + PD^2 = AE^2 + PE^2 + PD^2 and BP^2 + PE^2 = BD^2 + PD^2 + PE^2 so AE = BD. same way we can prove AF = CD and BF = CE. From lemma2 and lemma3 we have D,E,F are extouch points in triangle ABC so PDIa, PEIb and PFIc are collinear.
∠PIaIb = 90 - (90 - ∠A/2) = ∠PIbIa ---> PIa = PIb.
same way we can prove PIb = PIc P is circumcenter of IaIbIc.
we're Done.

P.

[img width=50]https://media.discordapp.net/attachments/821919868332015627/955272968357937202/Screen_Shot_2022-03-20_at_6.13.37_PM.png?width=1242&height=1170[/img]

Since $AP^2+PD^2=BP^2+PE^2,$ we have $AP^2-PE^2=BP^2-PD^2.$ This simplifies to $AE=BD.$ Similarly, $CD=AF,BE=FB.$

Note that letting $AB,BC,CA$ be constants, we have unique values $AE,CD,FB.$ This fixes $D,E,F.$ Note that when $D,E,F$ are exactly the $A,B,$ and $C$ excenter touchpoints on $BC,CA,AB$, respectively, these equalities before satisfy. Thus, $I_BE\perp AC,I_AD\perp BC,I_CF\perp AB.$

This implies that $I_BEP,I_CFP,I_ADP$ are all collinear. Note that $ABC$ is the orthic triangle of $\triangle I_AI_BI_C$ so \[\angle CI_BP=90^\circ-\angle I_BCE=90^\circ-\angle I_BI_CI_A=\angle BI_BI_C.\] This is true for the other vertices of $\triangle I_AI_BI_C$ as well. It follows that $P$ is the isogonal conjugate of $I,$ so while $I$ is the orthocenter, $P$ is the circumcenter.

First, we will prove that the circumcircle of $I_AI_BI_C$ satisfies this property. We will then show that there exists only one point satisfying this property, finishing the problem.
Let $\angle I_BI_AI_C = A$, $\angle I_AI_CI_B = C$, $\angle I_CI_BI_A = B$. Then $\angle I_API_B = 2C$, so $\angle I_BI_AP = 90 - C$, while since it is well-known that $ABC$ is the orthic triangle of $I_AI_BI_C$, $I_CI_BCB$ is cyclic, so $\angle BCI_A = C$ and $\angle CI_AD = 90 - C$. Therefore, $I_A$, $D$ and $P$ are collinear. The same argument works to prove that $I_B$, $E$ and $P$ are collinear and that $I_C$, $F$ and $P$ are collinear. Now, note that $I_AC = \frac{DC}{cos(C)}$, $I_CI_A = \frac{I_AC}{cos{A}}$, so
$DC = I_AI_Ccos(A)cos(C)$. In the same way, $I_CA = AFcos(A)$, $I_CI_A = I_CAcos(C)$, so $AF = I_AI_Ccos(A)cos(C)$. This implies that $AF = DC$.
Note that $(AF)^2 = (DC)^2$ implies that $(AF)^2 + (PD)^2 = (PC)^2$, so $(AF)^2 + (PF)^2 + (PD)^2 = (CP)^2 + (PF)^2$, so since $(AF)^2 + (PF)^2 = (AP)^2$,
$(AP)^2 + (PD)^2 = (CP)^2 + (PF)^2$. A similar argument works to prove the other two equalities, so the circumcircle of $I_AI_BI_C$ satisfies this property.

Now, note on the other hand that by doing the previous argument backwards we are forced to have $AF = DC$, $BF = CE$, $AE = BD$. Note that this implies that $AF - AE = AB - AC$, $BF - BD = BA - BC$, $CD - CE = CB - CA$. Then, note that $BD = BF + BC - BA$, $CD = BC - (BF + BC - BA) = BA - BF$, $CE = CA + BA - CB - BF$, $AE = CA - (CA + BA - CB - BF) = CB + BF - BA$, $AF = BA - AC + CB + BF - BA = CB + BF - AC$, $BF = AB - (CB + BF - AC) = AB + AC - CB - BF$, so $2BF = AB + AC - BC$, and $BF = \frac{AB + AC - BC}{2}$.
Therefore, the position of $F$ on line $AB$ is fixed. In the same way, the positions of $D$ and $E$ are fixed, so $P$, the intersection of the perpendiculars to the sides that the three points are on at those points, is fixed, as required.

Clearly, we have $$BF = \sqrt{BP^2 - PF^2} = \sqrt{CP^2 - PE^2} = CE.$$ Analogously, we find $CD = AF$ and $AE = BD$. Thus, the $A$-excircle touches $BC$ at $D$, the $B$-excircle touches $CA$ at $E$, and the $C$-excircle touches $AB$ at $F$, implying $\overline{I_ADP} \perp BC$ and similarly for the other two excenters.

Now, let $k_a, k_b, k_c$ denote denote the tangents to $(I_AI_BI_C)$ at $I_A, I_B, I_C$ respectively. It's well-known that $ABC$ is the orthic triangle of $I_AI_BI_C$, so $I_BI_CBC$ is cyclic with diameter $I_BI_C$. Hence, $k_a \parallel BC \perp \overline{I_ADP}$ follows from Reim's. Similarly, we deduce $k_b \perp \overline{I_BEP}$ and $k_c \perp \overline{I_CFP}$, which finishes. $\blacksquare$

[b]Remark:[/b] This problem is way too easy for a G3.

i'm sorry, what?

Note that
\[
AE^2 = AP^2 - PE^2 = BP^2 - PD^2 = BD^2 \implies AE = BD
\]
and similarly $BF = CE$, $CD = AF$ - so $D$, $E$, $F$ are the corresponding excircle touchpoints on $BC$, $CA$, $AB$ respectively. In particular, we deduce $I_AD$, $I_BE$, $I_CF$ concur at $P$. Hence $P$ is the [i]Bevan Point[/i] of triangle $ABC$, which is known to be the circumcenter of $I_AI_BI_C$. $\square$

Use Pythagorean Theorem to remove $P$ from the image.

\[BD^2-AE^2=CD^2-AF^2\]
\[CE^2-BF^2=AE^2-BD^2\]
\[AF^2-CD^2=BF^2-CE^2\]

which implies all sides are $0$. So $P$ is the Bevan point of $\triangle ABC$, done.

$\color{magenta} \boxed {\textbf{SOLUTION G3}}$

$$AP^2 - PE^2 = BP^2 - PD^2 \implies AE = BD=y$$
Similarly, $AF=CD=x$ and $BF=CE=z$

Let, $\textbf{A-excircle}$ meets $BC$ at $D'$

So, $$CD'=s-AC=s-y-z=x=CD \implies D' \equiv D$$
$\implies I_A,D,P$ are collinear and $I_AP \perp BC$

Simlarly, $I_BP \perp AC \equiv E$ and $I_CP \perp AB \equiv F$

We need to show, $PI_A=PI_B=PI_C$

$$\angle PI_AI_C=\angle DI_AB=\frac{1}{2} \angle ABC=\angle BI_CF=\angle I_AI_CP \implies PI_A=PI_C$$
Similar work gives desired $\blacksquare$


Select 4 numbers from 0-9 randomly and using the 4 selected numbers (in order of selection) to form a 4 digit number, is the probability of the 4 digit number having all four digits the same (ie. 1111, 2222, 3333) smaller than the probability of the 4 digit number not having all four digits the same? why or why not?

[hide]
Since each number has equal probability of being chosen, the probability of having all the digits being the same is the same as having the digits differ.

Unless it is meant that the probability of getting one of the ten possible combinations in which all of the digits are the same, then the probability of obtaining such a thing is ,$p$ for possible outcomes, $\frac{10}{p}$, which is less than obtaining something other than those ten with complement the probability of $\frac{p-10}{p}$.
[/hide]

lol this problem should be moved to the beginners forum let alone intermediate..

[quote="AstroPhys"][hide]
Since each number has equal probability of being chosen, the probability of having all the digits being the same is the same as having the digits differ.
[/hide][/quote]

Having four digits the same and having no digits are not equal in proability.  Four of the same digit is $\frac{10\cdot1\cdot1\cdot1}{10\cdot10\cdot10\cdot10}$, while having four digits differ is $\frac{10\cdot9\cdot8\cdot7}{10^{4}}$.  Clearly they are not equal.

For that explaination, it was assumed that the four times a number is selected are independent of each other.  Since it is not a biased selection the probability of any number chosen is the same as another.  Obtaining a $4444$ has the same probability as if obtaining $4356$, since each individual number is $\frac{1}{10}$, each of the groups of the final four digits have the same chance of occuring.

This was then countered in the second explaination stating that there are more ways not obtaining all the digits being the same, therefore it has a higher probability of occuring.


There is not. If you want to talk about sports, do so in the Fun Factory.


[hide]I don't believe that there are any solutions: drawing a graph, it is clear x must be positive, since $2^{x}$ is always positive, and for positive $x$, $2^{x}>1$, so $x>1$, but for $x=1$ $2^{x}=2$, and the $2^{x}$ rapidly pulls away from the line as x increases...

that was unclear, but anyway, I think it's right [/hide]

I think you're right...

I'm wondering how you would solve an equation such as
$2^{x}= x+1$
even though the solutions (for this one) are obvious.

How about:

For which real numbers $a$ are there solutions to the equation $x=a^{x}$?

and solving the more general equation $x=a^{x}+b$...

[quote="lotrgreengrapes7926"][quote="tim1234133"]How about:

For which real numbers $a$ are there solutions to the equation $x=a^{x}$?[/quote]

[hide] Any $a<\sqrt[e]{e}\approx1.44466786101$.

We will find the value of $a$ where there is one solution.
$\frac{d}{dx}a^{x}=a^{x}\log(a)$
$\frac{d}{dx}x=1$
$a^{x}\log(a)=1$.

Since $x=a^{x}$, we can keep substituting to get an infinite "power tower."
$a^{a^{a^{...}}}\log(a)=1$
$a^{a^{a^{...}}}=\frac{1}{\log(a)}$
$a^{\frac{1}{\log(a)}}=\frac{1}{\log(a)}$
$a=\frac{1}{\log(a)^{\log(a)}}$
$\frac{1}{a}=\log(a)^{\log(a)}$

From here, I used a graphing calculator.  :D [/hide][/quote]

Ya have to love $e$  :P

(Oh btw isnt $\frac{d}{dx}a^{x}= a^{x}\ln{a}$?)

:D

[quote](Oh btw isnt $\frac{d}{dx}a^{x}= a^{x}\ln{a}$ ?)[/quote]

I think in this kind of context $\ln{a}$ and $\log{a}$ generally mean the same thing... though I suppose using $\log$ is kinda ambiguous


geometry
inradius
circumcircle
inequalities proposed
inequalities

Let $ ABC$ be a triangle with $ 2p\equal{}a\plus{}b\plus{}c$ , inradius $ r$ and circumradius $ R\ .$  Prove that

$ \boxed {\ 8p\left(5Rr\minus{}r^2\right)\ \le\ \prod (b\plus{}c)\plus{}\prod (b\plus{}c\minus{}a)\plus{}\sum a (a\minus{}b)(a\minus{}c)\ \le\ 8p(R\plus{}r)^2\ }\ .$

My solution is calculation, we can calculate
\[ \prod (b \plus{} c) \plus{} \prod (b \plus{} c \minus{} a) \plus{} \sum a (a \minus{} b)(a \minus{} c) \equal{} 2p(p^2 \plus{} 4Rr \plus{} r^2)
\]
so we must prove
\[ 8p\left(5Rr \minus{} r^2\right) \le 2p(p^2 \plus{} 4Rr \plus{} r^2) \le\ 8p(R \plus{} r)^2
\]

\[ \Leftrightarrow 16Rr \minus{} 5r^2\le p^2\le 4R^2 \plus{} 4Rr \plus{} 3r^2
\]

Which is true.

Yes, from this nice identity $ \prod (b \plus{} c) \plus{} \prod (b \plus{} c \minus{} a) \plus{} \sum a (a \minus{} b)(a \minus{} c) \equal{} (a\plus{}b\plus{}c)(bc\plus{}ca\plus{}ab)$ ...


integration
trigonometry
calculus
function
logarithms
calculus computations

Find $\int \frac{x^2}{(x\cos x - \sin x)(x\sin x + \cos x )}dx$

Here's a solution I thought of:

Take out $\cos^2x$ out of the denominator and make the substitution $t = \tan x$ and this integral will become $\int \frac{(\arctan t)^2}{(\arctan t - t)(t\arctan(t) + 1)} dt = \int \frac{\frac{(\arctan t)^2}{(t\arctan t + 1)^2}}{\frac{\arctan t - t}{t\arctan t + 1}} dt$

Note that this integral is in the form $\int \frac{f'(t)}{f(t)}dt$, so the primitive becomes $\ln|\arctan t - t| - \ln|t\arctan t + 1|$. Substituting back we get our answer:

\[ \ln|x - \tan x| - \ln|x\tan x + 1|  \]

Afterthoughts: There are of course some conditions on x so that neither the original function nor this primitive will be undefined, but I won't bother figuring them out precisely. I'm also not sure why it works out to say that $\arctan\tan x = x$ when this is not true on every interval. Maybe someone can else can explain that to me. It seems like there is a lot of "cheating" in the way I've calculated this integral. The substitution $x = \arctan t$ only describes x in the interval $]-\pi/2, \pi/2[$, yet I get a primitive that works in any interval (as long as it's not undefined somewhere)? I don't understand these things.

$\frac{x^2}{(x\cos x - \sin x)(x\sin x + \cos x)}=\frac{x\cos x}{x\sin x+\cos x}+\frac{x\sin x}{x\cos x-\sin x}$

$=(\ln |x\sin x+\cos x|)'-(\ln |x\cos x-\sin x|)'.$ Integrate this,

$\int \frac{x^2}{(x\cos x - \sin x)(x\sin x + \cos x )}dx=\ln |x\sin x+\cos x|- \ln |x\cos x-\sin x|$

$=\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+Const.$

While the same thing could be asked about my solution (I had a 'feeling' it would be the way I did it), how did you come up with the decomposition of the integrand? Is there a method for these kind of expressions, or did you also just intuitively expect such a decomposition to be possible?

The denominator of $\frac{x^2}{(x\cos x - \sin x)(x\sin x + \cos x)}$ is two degree for $\cos x$ and $\sin x$, so

I remarked that the numerator of the fraction could be transformed into $x^2=x^2(\sin ^2 x+\cos ^2 x).$
In this way, I made the numerator and denominator the same degree.

As another example, $\cos x=\frac{\cos ^ 2 \frac{x}{2}-\sin ^ 2 \frac{x}{2}}{\cos ^ 2 \frac{x}{2}+\sin ^ 2 \frac{x}{2}}.$

[i]Kunny[/i], the your solution is identically with the my solution !. Here is our solution:

$u(x)=x\cos x-\sin x\Longrightarrow u'(x)=-x\sin x\ ;$
$v(x)=x\sin x+\cos x\Longrightarrow v'(x)=x\cos x\ ;$
$x^2=x\cos x\cdot (x\cos x-\sin x)+x\sin x\cdot (x\sin x+\cos x)=u(x)\cdot v'(x)-u'(x)\cdot v(x).$

$I\equiv \int \frac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}\ dx\Longrightarrow$

$I=\int \frac{u(x)\cdot v'(x)-u'(x)\cdot v(x)}{u(x)\cdot v(x)}\ dx\Longrightarrow$

$I=\int \frac{v'(x)}{v(x)}\ dx-\int \frac{u'(x)}{u(x)}\ dx\Longrightarrow$ $I=\ln |v(x)|-\ln |u(x)|+C\Longrightarrow$

$I=\ln\left | \frac{v(x)}{u(x)}\right| +C\Longrightarrow$ $I=\left| \frac{x\sin x+\cos x}{x\cos x-\sin x}\right| +C.$

Here is a more clear drawing up of the Kunny's solution (he is the first !). I like to write in LaTeX.


LaTeX
superior algebra
superior algebra unsolved

Just a tip: you're likely to get more responses if you don't force people to open Word attachments. (and strange as it seems, not everybody [i]has[/i] Word)

Another tip: learning to use embedded LaTeX (you can start with the link on the left) is fun!  And it will let you put formulae directly into your posts.

There is no "link on the left" for ML skin users, but there is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]Learn LaTeX topic[/url].

MS Word is not an alternative to LaTeX, as you will notice as soon as you try to write a longer text with the Word formula editor and after 4 pages, half of the formulas will instantly be scaled up and another third of the formulas will disappear. ;)

  darij

[quote="arosu"]Determine the status of each of the following real numbers among the following options:
•	Option 1: A number in Q.
•	Option 2: A number in Q[2] but not in Q.
•	Option 3: A number not in Q[2].

The numbers are inside the attachment.[/quote]
Prove that
sqrt(1+sqrt(2)+sqrt(1+sqrt(2))) is not rational nor in the field of Q(sqrt2).

You don't need to say "not rational nor in the field of Q(sqrt2)". All rationals lie in $ \mathbb{Q}\left(\sqrt2\right)$, so it would be enough to say "not in $ \mathbb{Q}\left(\sqrt2\right)$".

Simplify a bit. If $ \sqrt{1\plus{}\sqrt{2}\plus{}\sqrt{1\plus{}\sqrt2}}$ would lie in $ \mathbb{Q}\left(\sqrt2\right)$, then what simpler number would also lie in $ \mathbb{Q}\left(\sqrt2\right)$ which can be easily shown [b]not[/b] to lie there (at least more easily than $ \sqrt{1\plus{}\sqrt{2}\plus{}\sqrt{1\plus{}\sqrt2}}$ itself)?

  darij

[quote="darij grinberg"]You don't need to say "not rational nor in the field of Q(sqrt2)". All rationals lie in $ \mathbb{Q}\left(\sqrt2\right)$, so it would be enough to say "not in $ \mathbb{Q}\left(\sqrt2\right)$".

Simplify a bit. If $ \sqrt {1 \plus{} \sqrt {2} \plus{} \sqrt {1 \plus{} \sqrt2}}$ would lie in $ \mathbb{Q}\left(\sqrt2\right)$, then what simpler number would also lie in $ \mathbb{Q}\left(\sqrt2\right)$ which can be easily shown [b]not[/b] to lie there (at least more easily than $ \sqrt {1 \plus{} \sqrt {2} \plus{} \sqrt {1 \plus{} \sqrt2}}$ itself)?

  darij[/quote]
Is sqrt(1+sqrt(2)), because I already proved this one of not being in Q(sqrt2)?

Yes, if we assume that $ \sqrt {1 \plus{} \sqrt {2} \plus{} \sqrt {1 \plus{} \sqrt2}}\in\mathbb{Q}\left(\sqrt2\right)$, then we get $ \sqrt{1\plus{}\sqrt2}\in\mathbb{Q}\left(\sqrt2\right)$ (because if an element lies in $ \mathbb{Q}\left(\sqrt2\right)$, then we can square it and subtract $ 1\plus{}\sqrt2$ from it and the result will still lie in $ \mathbb{Q}\left(\sqrt2\right)$), and thus you get a contradiction.

  darij


A digital music library contains 12 days of music and takes up 16,000 megabytes of disk space. On average, how many megabytes of disk space does one hour of music in this library take up? Express your answer to the nearest whole number.


vector
real analysis
real analysis unsolved

$ u \in L(E)$ where $E$ is normed vector space over $K=R$ or $C$

1/ Suppose $|||u||| \leq 1 $ prove that $ker(u-Id)\cap Im(u-Id)=(0)$

2/ Suppose $|||u||| = 1 $ prove that $ker(u-Id)\cap Im(u-Id)=(0)$

looks like this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=40279


Let $a, b, c$ three positive reals such that
\[\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1. \]
Show that
\[a+b+c\geq ab+bc+ca. \]

$\sum \frac{1}{a+b+1}\geq 1$ $\leftrightarrow$ $\sum (1-\frac{a+b}{a+b+1})\geq 1$ $\Leftrightarrow$ $2\geq \sum \frac{a+b}{a+b+1}=\sum \frac{(a+b)^{2}}{(a+b)^{2}+a+b}\geq \frac{4(a+b+c)^{2}}{2(\sum a^{2}+\sum ab+\sum a)}$ $\Rightarrow$
$\sum a^{2}+\sum ab+\sum a \geq (a+b+c)^{2}$ $\Leftrightarrow$ $a+b+c\geq ab+bc+ca$.qed

I have anther one.

Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$. Prove that
\[\frac1{1+a+b}+\frac1{1+b+c}+\frac1{1+c+a}\leq1.\]

?? That's in contradiction with the posted one...?

By applying the Cauchy-Schwarz inequality, we obtain \[ (a+b+1)(a+b+c^{2})\geq(a+b+c)^{2},\]thus we have \[\frac{1}{a+b+1}\leq\frac{c^{2}+a+b}{(a+b+c)^{2}}.\]Now by summing ciclically we obtain \[\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq\frac{a^{2}+b^{2}+c^{2}+2(a+b+c)}{(a+b+c)^{2}}\]
But from the condition, we can see that \[ a^{2}+b^{2}+c^{2}+2(a+b+c)\geq (a+b+c)^{2},\]and therefore \[ a+b+c\geq ab+bc+ca.\]
Equality occurs if and only if $a=b=c=1$.The ineq which pvthuan posted is also in my notes, and the solution is the same as above.

Andrei Ciupan.

[u]Otherwise:[/u] Let $a+b=x;b+c=y;c+a=z$. The inequality becomes:
\[\sum\dfrac{1}{m+1}\ge 1\Rightarrow \sum (m+1)(n+1)\ge (m+1)(n+1)(p+1)\Rightarrow m+n+p+2\ge mnp\]
We back at the variables $a,b,c$:
$2(a+b+c)+2\ge (a+b)(b+c)(c+a)$.
We have the identity : $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$.
Note $a+b+c=m;ab+bc+ca=n$ and $abc=p$. By $AM-GM$, $mn\ge 9p$. But $2m+2\ge mn-p\ge \dfrac{8mn}{9}$.
If $m<n$, we use the inequality : $m^2\ge 3n>3m\Rightarrow m>3\Rightarrow 9m+9\ge 4mn>12n>12m\Rightarrow m<3$, contradiction !

[b]Romania [/b]Junior Balkan Team Selection Tests 2007
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698&p=2914767#p2914767
The following inequalities are also true.
Let $a, b, c$ three positive reals such that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ Show that\[a+b+c\geq 3. \]

[quote="sqing"][b]Romania [/b]Junior Balkan Team Selection Tests 2007
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698&p=2914767#p2914767
The following inequalities are also true.
Let $a, b, c$ three positive reals such that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $ Show that
\[a+b+c\geq 3. \][/quote]
Prove :$1\ge \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge \frac{9}{2(a+b+c)+3}$

$abc=1\Rightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $

$a+b+c \geq ab+bc+ca \Rightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1. $
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2924523
Positive real numbers $a,b,c$ satisfy$\frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c} = 1$Prove that$ a+b+c \geq ab+bc+ca$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2137595

Leta, b, c three positive reals such that
1/(a+b+1)+1/(c+b+1)+1/(a+c+1)≥ =>a+b+c≥a2b+b2a+c2b+b2c+c2a+a2c
show  that
a+b+c≥ ab+bc+ac  => a+b+c ≥a2b+b2a+a2c+c2a+b2c+c2b≥ab+bc+ac

a2b+b2a≥2ab
b2c+c2b≥2bc
c2a+a2c≥2ac
2ab+2ac+2bc≥ab+bc+ca
ab+bc+ac≥0

My solution:
$  a,b,c>0,\sum\frac{1}{a+b+1}\ge 1\Leftrightarrow 2+2(a+b+c)+abc\ge (a+b+c)(ab+bc+ca) $
But, by Am-Gm, we have:$ abc\le\frac{(a+b+c)(ab+bc+ca)}{9} $; result
$ 4(a+b+c)(ab+bc+ca)\le 9+9(a+b+c) $
It is sufficient that $ 9+9(a+b+c)\le 4(a+b+c)^2\Leftrightarrow a+b+c\ge 3 $
So for $ a+b+c\ge 3 $ inequality is true!
If $ a+b+c\le 3 $ we have:
$ 3(a+b+c)\ge (a+b+c)^2\ge 3(ab+bc+ca)\Rightarrow a+b+c\ge ab+bc+ca\Leftrightarrow QED $
The proof is ended!
________
Marin Sandu


Euler
quadratics
Gauss
number theory proposed
number theory

Suppose coprime positive integers a, b, c satisfy $a^2 - ab + b^2 = c^2$.
Prove that if d is a divisor of c, then d = 1 (mod 6).

The source is Euler or Lagrange or something of that age.

[hide="Sketch of solution"]a/b or b/a is a primitive 6th root of 1 modulo any prime divisor of d, so 6 | (p-1)[/hide]

Something more elementary? Please? (What about quadratic reciprocity and the following theorem: if p = 1 (mod 6) is prime then there exist integers x, y such that p = x^2 + 3y^2.)

From $p|c$, $p$ prime we get $(2a-b)^2\equiv -3b^2(p)$,

This means $\Big({-3\over p}\Big)= 1$

Using Gauss quadratic reciprocty theorem you can finish.


logarithms
ratio
LaTeX
calculus
calculus computations

test the series

$ \sum^{\infty}_{n\equal{}2} (\frac {1}{\ln n})^{\ln n}$

for convergence or divergence

$ \left(\frac1{\ln n}\right)^{\ln n}\equal{}e^{(\ln n)(\minus{}\ln\ln n)}\equal{}n^{\minus{}\ln \ln n}.$

Since $ \ln\ln n>2$ eventually, we have $ n^{\minus{}\ln \ln n}<\frac1{n^2}$ eventually.

The series converges by comparison to $ \sum\frac1{n^2}.$

Note: the root test and ratio test are inconclusive for this series.

It converges. Use cauchy condensation test. Any sereis, a_n converges iff the sereis 2^ka_(2^k) converges.

Applying the above,

your sereis converges iff the sereis (ak/e)^-ak converges, where a=ln2. This sereis surely converges, for its terms decrease faster than those of a geometric sereis.

[quote="bharathhm"]Any sereis, a_n converges iff the sereis 2^ka_(2^k) converges.[/quote]
It's not quite "any series." The condensation test has one additional hypothesis, namely that the sequence $ a_n$ is positive and (eventually) decreasing.

That hypothesis does hold here and this is a correct use of the condensation theorem. (It would have been nice if bharathhm had used $ \text{\LaTeX}$.)

what's cauchy condensation test?

you can solve it like this:

since $ \ln n$ is increasing, we can find some number p such that

$ \ln n$ $ >$ $ e^{p}$ $ \forall$ $ n > N$

$ \implies$ $ (\ln n)^{\ln n}$ $ >$ $ e^{p\ln n}$

$ \implies$ $ (\ln n)^{\ln n}$ $ >$ $ n^p$

$ \implies$$ \frac {1}{(\ln n)^{\ln n}}$ $ <$ $ \frac {1}{n^p}$

choosing n sufficiently great, we can make the given series less than any desired $ p$-series. hence the given series converges.

There is a test that may work and it is called Abel-Dini scale.


Factor these:

1. $ 6x^{12} \plus{} 35x^7 \plus{} 36x^2$

2. $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$

3. $ (ab \plus{} cd)^2 \plus{} (ac \minus{} bd)^2$

Finding roots:

a. Find all the real roots of $ x^4 \plus{} (2 \minus{} x)^4 \equal{} 34$.

b. If x is real and $ x^3 \plus{} 4x \minus{} 8 \equal{} 0$, find all possible values of $ x^7 \plus{} 64x^2$.

hello, we have
1.) $ 6x^{12} \plus{} 35x^7 \plus{} 36x^2 \equal{} x^2(3x^5 \plus{} 4)(2x^5 \plus{} 9)$
2.) $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} xz \minus{} yz)$
3.) $ (ab \plus{} cd)^2 \plus{} (ac \minus{} bd)^2 \equal{} (b^2 \plus{} c^2)\cdot(a^2 \plus{} d^2)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have
2.) $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} xz \minus{} yz)$
Sonnhard.[/quote]

How did you pull that out of thin air?

[quote="Dr Sonnhard Graubner"]
3.) $ (ab \minus{} cd)^2 \plus{} (ac \minus{} bd)^2 \equal{} (b^2 \plus{} c^2)(a^2 \plus{} d^2)$   . . [I deleted the extra multiplication dot.]
Sonnhard.[/quote]

Dr Sonnhard Graubner,

no, for the third one, you typed out the wrong problem.

You were supposed to factor this one instead:

$ (ab \plus{} cd)^2 \plus{} (ac \minus{} bd)^2$   . . . [This still needs to be done.]

Had you looked farther, it would be clear that he only typoed--he did solve the problem.

[quote="Smartguy"]Factor these:

3. $ (ab \plus{} cd)^2 \plus{} (ac \minus{} bd)^2$

[/quote]
I quoted this to show what it was/is in case it is edited.

[quote="Arrange your tan"]Had he looked farther, he could have made his correction in a follow-up post instead of letting an error sit there to confuse.[/quote]

No, he linked it back to the wrong expression, so he did not
show it correctly, although he knows how to do it correctly.
If you're grading on partial credit, he gets something.  If you're
grading on all or nothing, he gets nothing, because it all has
to be correct.

Arrange your tan, what exactly is your point here?

Dr Sonnhard Graubner got correct solution, and clearly only made mistake while copying from his notes. Instead of being a smartass, why didn't you write whole solution correctly yourself?

arrange your tan, why don't you post the "correct solution" yourself?

Thanks flame, you are a genius!


conics
parabola
symmetry
limit
calculus
calculus computations

[url=http://bildr.no/view/487869][img]http://bildr.no/thumb/487869.jpeg[/img][/url]
[url=http://bildr.no/view/487870][img]http://bildr.no/thumb/487870.jpeg[/img][/url]

1.  why `the sequence diverges when $ x\leq \minus{}1/2$ and $ x\geq 3/2$ ?

2. why `n' in the last formula ?

Thanks

The solution is indeed lacking rigor, and even incorrect at some point. This is a classical case when an image is worth a thousand words: the graph of $ f$ is an "inverted" parabola, with the line $ x\equal{}\frac {1} {2}$ as symmetry axis, point $ V(\frac {1} {2}, \frac {1} {2})$ as its apex, passing through points $ A(\minus{}\frac {1} {2}, \minus{}\frac {1} {2})$ and $ B(\frac {3} {2}, \minus{}\frac {1} {2})$, and intersecting the first bisector $ y\equal{}x$ in points $ V$ and $ A$.

The solution errs by not recognizing that for $ x \equal{} \minus{}\frac {1} {2}$ and $ x \equal{} \frac {3} {2}$, the sequence becomes stationary equal to $ \minus{}\frac {1} {2}$ from term $ x_1$ on, thus convergent to this value.

For $ x < \minus{}\frac {1} {2}$ and for $ x > \frac {3} {2}$, the sequence is decreasing, with $ x_1 < \minus{}\frac {1} {2}$. Now, if the sequence was lower bounded, then it would be convergent (Weierstrass), and we can find its limit by passing to the limit (like in the solution), only that the roots of that equation are $ \pm \frac {1} {2}$ and thus they are not convenient ($ x_1 < \minus{}\frac {1} {2}$ forces the limit to be also less than $ \minus{}\frac {1} {2}$). Therefore the sequence is lower unbounded, hence divergent, with $ \lim_{n \to \infty} x_n \equal{} \minus{}\infty$.

Finally, for $ x \in (\minus{}\frac {1} {2}, \frac {3} {2})$, we have $ |x \minus{} \frac {1} {2}| < 1$, and moreover, by its definition, $ \left | x_{n\plus{}1} \minus{} \frac {1} {2} \right | \equal{} \left | x_n \minus{} \frac {1} {2} \right |^2$. By iterating this relation we get $ \left | x_{n\plus{}1} \minus{} \frac {1} {2} \right | \equal{} \left | x \minus{} \frac {1} {2} \right |^{2^{n\plus{}1}} \to 0$ as $ n \to \infty$.


How would you produce allyl alcohol from ethanol?

This is just a suggestion. Not sure how valid the procedure is. :blush:

[hide=" Synthesis"]

First thing to notice is that $ \text{ethanol}$ has one less carbon atom than $ \text{allyl alcohol}$, so we want to add one more.

1. Convert ethanol to ethanal (using $ \ce{PCC}, \ce{CH2Cl2}$).
2. Using phosphorous ylide ($ \ce{P(Ph)3CH2}$) convert ethanal into propene
3. Propene can be converted into allyl alcohol using the reagent: $ \ce{SeO2}$.
[/hide]

That synthesis look perfectly ok to me. Can you propose other routes?

Thanks! I did almost [hide="the same"]$ \ce{CH3CH2OH}\xrightarrow{\ce{Cr2O7^2\minus{}},\ce{H^\plus{}}}\ce{CH3CHO}\xrightarrow{\ce{CH3MgBr},\ce{H2O}}\ce{(CH3)2CHOH}\xrightarrow{\ce{H2SO4},\Delta}\ce{CH3CH\equal{}CH2}\xrightarrow{\ce{HO^\minus{} Br^\plus{}},\ce{R2O2}}\ce{CH3CHBrCH2OH}\xrightarrow{\ce{KOH(alc)}}\ce{CH2\equal{}CHCH2OH}$[/hide], but my process is a bit too long. :oops:

And [hide="this"]$ \ce{EtOH}\rightarrow\ce{C2H2}\xrightarrow{\ce{HCN},\ce{Ba(CN)2}}\ce{H2C\equal{}CHCN}\xrightarrow{\ce{NaBH4}}\ce{H2C\equal{}CHCH2NH2}\xrightarrow{\ce{HNO2}}\ce{H2C\equal{}CHCH2OH}$[/hide] is what my friend suggested.

Are the above OK? :maybe:

Well it will be nice if you can clarify your friend's first step: $ \ce{EtOH} \rightarrow \ce{C2H2}$.

Actually when he proposed his method to me he wrote "do yourself" lol. But I think we can do it like this: dehydrate ethanol to ethene, add Br2 in CCl4 to get $ \ce{BrH2C \minus{} CH2Br}$ and finally heat the dihalide with alcoholic KOH.

And thanks Banished, I got to learn about some new reagents from your post above. :)

The last step (amine ---> alcohol with nitrous acid) would also fail miserably. Why?

Nayel, you could browse through this forum: there are several nice problems in organic synthesis that were solved some time ago.

Um... because it would react with the double bond?

Yes I have searched and found old and nice problems. But most of them are too advanced for my level. :(


how do we check for all usernames? i just want to know how. i kinda forgot because i haven't been in this forum for over a year.

I think that would be in Memberlist, which is in the toolbar that contains the shiny "AoPS Blogs" and "Contests"

[quote="A+MATH"]Over a year? You just joined Nov 4. 2005.  :P[/quote]
Uh... I think what he meant was he had an account, but his last use of it was over a year ago (he doesn't remember the account username)

That's kind of why his account name is "fake account"...

To fakeaccount, I don't think there's much you can do. If you remember some of your posts, or where you might have posted, you can try the search, and try to find what your username was, but otherwise, just create/stick with a new acount.

ah thanks! i remember now. my username was gameworld7.

its locked now because i spammed too much 2 years ago...


conics
hyperbola
rotation
function
geometry solved
geometry

Given an arbitrary triangle, can it be translated, rotated and scaled such that all three vertices lie on the rectangular hyperbola xy=k??

can it be only translated and rotated such that all three vertices lie on the rectangular hyperbola xy=k??

For the first part:

The rectangular hyperbolas are similar, so we can assume k=1.

Any triangle ABC has at least one acute angle. Suppose that angle is A. We translate the triangle so that A is the point (-1,-1).

Consider B(x)=(x,1/x). Let C(x) be the one point such that AB(x)C(x) is similar to ABC. The point C(x) is unique because we must keep the orientation, because we can't make symmetries.

Assume ABC is oriented clockwise or else change B and C.
It is clear that C(x) is a continuous function on (0, \infty). What is important is that C((0,\infty)) is a curve D.
If we tend x to 0 on the positive side, the angle between AB and Oy tends to 0. Because A is acute, AC cuts Ox in a point on the right side of (-1,0) and thus cuts the right side of the hyperbola and C gets inside the right side of the hyperbola.

If we tend x to \infty we get a point C outside the right side of the hyperbola.
So the curve D has points on either side of the right side of the hyperbola, so D cuts the right side of the hyperbola at a point C(x_0).
AB(x_0)C(x_0) is the triangle we were looking for.

For the second part I think the answer is no.
For k=1 take a small equilateral triangle such that it cannot have vertices on both sides of the hyperbola.
I state, but I'm too lazy to check, that on one side of the hyperbola cannot lie an equilateral triangle.

I checked:

Suppose we have positive distinct reals a,b,c such that (a,1/a),... are the vertices of an equilateral triangle.
Then (a-b)^2+(1/a-1/b)^2=...
These lead to
a+c-(1/a+1/c)/ac-2b+2/abc=0,...
By summing these we get
6/abc=sum1/a^2b \geq 6/abc, so a=b=c (a contradiction).


What is the greatest two-digit prime number with the property that the product of its digits is also a prime number?

We see that each of our two digits are factors of this prime-numbered product.  Because the only factors of a prime are $ 1$ and itself, we know that one of our digits must be a $ 1$.  We'll put $ 1$ in the one's place to maximize our number.  The largest 1-digit prime is $ 7$.  So that gives us our ten's digit. Now we just need to check to see if our number is prime.  And indeed, our answer, $ \boxed {71}$ is prime.


If you drop a class, does the message board go away? If it doesn't, are you still allowed to answer problems?

If you drop a class, you will no longer have access to that class's message board or transcripts.

What happens if you drop a class before it starts? Will you get a refund? I have never taken an AoPS class and I am considering taking one.

You get a full refund if you drop the class [b]before[/b] the third class starts.

[quote="rrusczyk"][quote="myyellowducky82"]I still have the message board there (Advanced Mathcounts/AMC8)[/quote]

You won't in a week or so; we cut off access to the message board for dropped students after the 3rd or 4th class ends.[/quote]

you won't what?

can't someone just enroll in every class and drop all of them and get 3 weeks free?

I think the AoPS Staff would notice a trend after the 3rd class or so in a row that has been dropped. And usually, the first three weeks don't go into anything that complex so the person won't get the real benefit from the class. From my experience in the Intro/Interm classes, the first few weeks aren't very challenging.


calculus
derivative
limit
calculus computations

Here it is - use limits to find the derivative of $\frac{\sqrt{x-1}}{\sqrt{x+1}}$

The derivative, using limits is just $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$.

Plugging our equation into this formula, we obtain:

$\lim_{h \to 0}\frac{\frac{\sqrt{x+h+1}}{\sqrt{x+h-1}}-\frac{\sqrt{x-1}}{\sqrt{x+1}}}{h}$.

Based on this, how would you simplify the numerator? I've multiplied both sides by $\sqrt{x+h-1}\cdot \sqrt{x+1}$, but then I still have to deal with the $h$ on the bottom. Can anyone show me to how proceed?

You need to rationalize the numerator by multiplying numerator and denominator by the conjugate. That will get you a factor of $y$ in the numerator.

But what is the conjugate?

Would I have to do this?

$\left(\frac{\sqrt{x+h+1}}{\sqrt{x+h-1}}-{\frac{\sqrt{x+1}}{\sqrt{x-1}}\right) \cdot \left(\frac{\sqrt{x+h+1}}{\sqrt{x+h-1}}+{\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)}}$


Prove, that the equation
$3x^{2}-7y^{2}+1=0$
has an infinite number of solutions in natural numbers.

We quickly find one solution $(x,y)=(3,2)$.

By the theory of Pell's equation, there are infinitely many solutions $(a,b)$ to $a^{2}-21b^{2}=1$.  Then if

$3u+v\sqrt{21}= (3\cdot3+2\sqrt{21})(a+b\sqrt{21})$,

i.e., $u=3a+14b$, $v=2a+9b$, we have $9u^{2}-21v^{2}=-3$, or $3u^{2}-7v^{2}=-1$.


floor function
limit
integration
calculus
calculus open

Does anyone know good estimations for the quantity $\sum_{k=1,n}{\frac{1}{f(tk)}}$ where t is irational? Here f is the fractional part.

What does the term "fractional part" mean?

The fractional part of a real number $x$ is $x-\lfloor x\rfloor$, where $\lfloor x\rfloor=\text{floor}(x)=\text{int}(x)$ is the greatest integer less then or equal to $x.$

As for the sum, it won't be very well-behaved. Here's one estimate:

$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{f(tk)}=\int_0^1\frac1x\,dx=\infty.$

So the sum harazi asked for is not $O(n).$ It's bigger than that.


show that if $ a_i >0$ and $ \sum_{1}^n{a_i^6}\equal{}n$ then $ \sum_1^n{a_i^2}\le \sum_1^n{a_i^5}$

By Holder:
 $ (\sum_1^n{a_i^2})^4\le (\sum_1^n{a_i^5})(\sum_1^n{a_i})^3$ if $ \sum_1^n{a_i}\leq \sum_1^n{a_i^2}$ we are done.

This last is true iff: $ \sum_1^n{a_i^2 \minus{} a_i}\geq 0$ consider $ f(x) \equal{} x^2 \minus{} x$ $ f''(x) \equal{} 2$ then $ f$ is convex. If $ \sum_1^n{a_i} \equal{} S$ and WLOG the $ a_i$ is a decreasing series, then $ [a_1,\dots ,a_n] > [\frac {S}{n},\dots,\frac {S}{n}]$. Applying Karamata, we get the result.

Use	RCF_LCF_and_EV theory .
And  here  solution.
I hope  to see  other solution.
 Occur  if and only if  $ a_1\equal{}a_2\equal{}...\equal{}a_{80} \equal{}\sqrt[3]{3/4}$and$ a_{81} \equal{}\sqrt[3]{6}$


floor function
function
ceiling function

when n is even, (n/2)!
when n is odd, ((n+1)/2)!
For the soln refer to the pdf attached

Why you atached a .pdf? You could simple wrote that   :lol:

[quote="shyong"]shouldnt it be something like this ?

[hide]

If $N$ is even , break it into two groups (each with $\frac{N}{2}$ digit ) . Since the first digit cannot start with $0$ , we have $9$ ways . Then the second digit onward to $N/2$th digit have $10$ ways each . So total is $9\times 10^{\frac{N}{2}-1}$

If $N$ is odd , break it into three groups , two with $\frac{N-1}{2}$ digit while the third one with $1$ digit . Now ,similarly we will get $9\times 10^{\frac{N-1}{2}}$ .

Is it correct ?

 :?

[/hide][/quote]

I agree, but you can combine your two end results into:

[hide]
$9\times10^{\lfloor\frac{N-1}{2}\rfloor}$

[/hide]

[quote="aidan"]What are those symbols? The code gives it as "floor"....[/quote]

they are floor symbols. $\lfloor x \rfloor$ means the greatest integer less than or equal to x. The opposite is also a function $\lceil x \rceil$ is ceiling and it means the least integer greater than or equal to x.

[hide]If $N$ is even, then the first and second half must be the same, so $p(N)=9 \cdot 10^{N/2-1}$

If $N$ is odd, then the first and second half, not including the middle # must be the same, so $p(N)=9 \cdot 10^{\lfloor (N-1)/2 \rfloor }$ is the general solution for all $N$. [/hide]


number theory
combinatorics proposed
combinatorics

Let $ p$ be a prime number and $ s$ be the number of bijections $ f: \{1,2,\cdots,2p\} \to \{1,2,\cdots,2p\}$ that satisfies: $ f^p$ is an identity, where $ f^p \equal{} f(f(f(...(f(p))...)))$ ($ p$ times). Prove that $ s$ is not a divisor of $ 2p$.

Do you mean $ s$ is not divisible by $ 2p$?

Hope we are supposed to prove that $ 2p\nmid s$ as t0rajir0u pointed out. Writing out the permutation map in disjoint cycle notation, we observe that $ \ell$ ,length of a cycle , then $ \ell \mod p \mid p$. Hence cycles can be of length $ 1$ or $ p$. Computing, $ s \equal{} 2{2p\choose p} \plus{} 1$

That's not right.  The number of permutations with two cycles of length $ p$ is $ \frac{(2p)!}{2p^2}$ and the number of permutations with one cycle of length $ p$ is $ \frac{(2p)!}{p}$.

[quote="t0rajir0u"]That's not right.  The number of permutations with two cycles of length $ p$ is $ \frac {(2p)!}{2p^2}$ and the number of permutations with one cycle of length $ p$ is $ \frac {(2p)!}{p}$.[/quote]

That's still not quite right I think. I get

$ s \equal{} \frac {(2p\minus{}1)!}{p} \plus{} \frac {2(2p\minus{}1)!}{p!} \plus{} 1$

By Wilson's theorem this is $ s \equiv (\minus{}1)^2 \plus{} 2.(\minus{}1) \plus{} 1 \equiv 0 \bmod{p}$

but that's fine because its odd unless $ p \equal{} 2$ which gives $ s \equal{} 10$, so its not a multiple of $ 2p$  :huh:

My result is $ s \equal{} 1 \plus{} {2p \choose p}(p \minus{} 1)! \plus{} {2p \choose p}((p \minus{} 1)!)^2$. I think there is some thing wrong with the problem, since $ s > 2p$. I will ask my teacher again. Thanks all of you :lol:


[quote="nayel"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ \frac{1}{8abc+(a+b-c)^{3}}+\frac{1}{8abc+(b+c-a)^{3}}+\frac{1}{8abc+(c+a-b)^{3}}\le \frac{1}{3abc}\]
[/quote]
Nice inequality, nayel! :lol:
$ \sum_{cyc}\frac{1}{8abc+(a+b-c)^{3}}\leq\frac{1}{3abc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{1}{8abc}-\frac{1}{8abc+(a+b-c)^{3}}\right)\geq\frac{3}{8abc}-\frac{1}{3abc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a+b-c)^{3}}{8abc+(a+b-c)^{3}}\geq\frac{1}{3}.$
But by Holder's inequality we obtain: $ \sum_{cyc}\frac{(a+b-c)^{3}}{8abc+(a+b-c)^{3}}\geq$
$ \geq\frac{(a+b+c)^{3}}{3(24abc+(a+b-c)^{3}+(a+c-b)^{3}+(b+c-a)^{3})}=\frac{1}{3}.$

thank you arqady your solution is nice too. :)  it's similar to my solution.


Yeah, so our snow has all melted away, and I'm afraid that there will not be any snow for all of Hanukkah, which ends tomorrow :mad: . What is up with this stupid weather? Does anyone else have snow right now?

No, but I'm what? A mile away from you?

Snow would be nice. I'm rather tired of this fog.

Probably a bit more than a mile, but I'm not actually really sure. I'm going to Chicago for the weekend, so I'm sure that it will snow while I'm gone just to spite me :dry:

Wow. It rained all day yesterday until 3 PM when it switched over to snow and didn't quit until 10. Some of my relatives have lost power. We ended up getting about 6 inches of absolutely heavy stuff. Shoveling was a nightmare! We will get a white Christmas after all!

[quote]
PostPosted: Today, at 4:15 pm
mustafa
Probably a bit more than a mile, but I'm not actually really sure.[/quote]

Actually, I know which middle school you attended, so I can find an upper and lower bound for the distance between our houses.

You were correct about it being more than a mile; you live between 1.067 and 5.667 miles from me. I'm guessing by neighborhoods that it's not close to either of those, though.

Merry Christmas people I am glad it snows for some!(and rains for others)

We have not had snow in 2 years since I was a mere 6th grader.

Everytime we go under 20 degrees, no snow comes, even though weather predictions have predicted snow numerous times.  :(

i think that we haven't gotten snow for 2 years maybe 3...hay did you guys get that bad power outage that one year? it was like several years ago.

Well, I'm in Chicago right now, and around eleven at night, it sprinkled snow for about two minutes, so I guess I kind of technically had a white Christmas...


algebra
polynomial
Vieta
quadratics
number theory theorems
number theory

I am confused with the validity of the vieta-jumping method. The idea of the method is to assume that the existence of a solution in the statement in question is wrong. Does it imply that if you can show that the existence of a solution is
wrong then the problem is true? why is that so?

Vieta-jumping can be used either to construct an infinite family of solutions to a quadratic Diophantine or prove that none exist (by showing that you can find one with contradictory properties).  I'm not sure what you're asking.

Most probably [color=red]queue_zniv16[/color] is asking about the method of showing contradiction. Yes, "the method of contradiction" woks this way. Because if u can show by contradiction that $ x\not > y$ then $ x\le y$ (and about other cases contradiction works almost in the same way)
For learning more about vieta jumping see here [url]http://reflections.awesomemath.org/2007_5/vieta_jumping.pdf[/url]


Amanda has a 2-cm rod, a 5-cm rod and a 9-cm rod.  She will randomly choose a fourth rod from a box containing rods of lengths 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.  What is the probability that Amanda's four rods will be able to form a convex quadrilateral, if the rods are to be connected endpoint to endpoint?

Express you answer as a common fraction.

[hide]I got $\boxed{\frac{3}{5}}$
Ummm i just made the rods given almost form a striaght line. So the 16,18,and 20 can't form a line since 2+5+9=16. Then the two can't either because it's too short (Draw a diagram). Not sure if this is right... but yeah. [/hide]

[hide]Basically, you have to see if they are able to form a quad. with each angle less than 180.
Just like in a triangle, the 3 sides sides have to be [b]greater[/b] than the fourth.
2+5+9= 16 so 16,18, and 20 won't work.
2+2+5 = 9 so 2 won't work
that leaves 6 others out 10 so [b]3/5[/b][/hide]


[quote="AndrewTom"]$ PQRS$ is a rectangle. $ PQ \equal{} 2QR$. $ E$ is a point such that $ PQE$ is an equilateral triangle overlapping rectangle $ PQRS$. $ M$ is the mid-point of $ EQ$. Find the size of angle $ QMR$.[/quote]
i think it will be 30
if true i will send complet solution
mr : mohamed

It seems easy.
[hide]$ QM\equal{}QR\equal{}\frac{1}{2}PQ$, so $ \triangle QMR$ is isoceles.
But $ \angle MQR\equal{}\angle PQR \minus{} \angle PQE\equal{}90\minus{}60\equal{}30$
So since $ \angle QMR\equal{}\angle QRM, \angle QMR\equal{}75$.[/hide]

the answer should be 75 degrees since QM =MR and <MQR=90-60=30 degrees


geometry
trapezoid
circumcircle
ratio
geometry solved

On the side BC of trapezium ABCD (AB||CD) exist a point K, such that triangle ADK is equilateral triangle. Proove that on the side AD exist a point L, such that triangle BCL is equilateral.

Suppose that the rays $AD,BC$ intersect at $P$

Construct the circle passing through $C,D,K$.
The circle intersects the line $DA$ again at $F$.

The triangles $PFC, PKD$ are similar, so $\frac{PC}{PD} = \frac{PF}{PK} = \frac{FC}{DK}$

From the parallel lines $AB,CD$ we get:

$\frac{PC}{PD} = \frac{PB}{PA} = \frac{BC}{AD}$

Finally $\frac{BC}{AD} = \frac{FC}{DK}$

It is $AD=DK$ because $\triangle ADK$ is equilateral , so $BC=FC$

Also, $\widehat{BCF} = \widehat{KCF} = \widehat{KDF} = 60^0$

Hence, the triangle $BCF$ is equilateral and $F$ is on the line $AD$.

If the circumcircle of $CDK$ is tangent to the line $DA$, then just rename point $F$ as $D$ and the above ratios are still true.

We can draw the circumcircle of $\triangle CDK$ [u]only if[/u]  there is such a triangle.
Since $K \in BC$, the three points can't be collinear.
The only problem is when $K \equiv C$.
Then $\triangle ADC$ is an equilateral triangle

In this case, we draw a circle which passes through $D$ and is tangent to $BC$, at the point $C$.
This circle intersects again the line $AD$ at a point $F$.

The triangles $PFC, PCD$ are similar, so $\frac{PC}{PD} = \frac{PF}{PC} = \frac{FC}{CD}$

$\Rightarrow  \frac{BC}{AD} = \frac{FC}{CD} \Rightarrow  BC=FC$ again

The angle $\widehat{FCB}$ is $60^0$ because it is equal to $\widehat{FDC}$

Here is a slightly shorter solution, probably shorter because I am using angles only (no ratios) and don't really care about degenerate cases as Pontios does.

[color=blue][b]Problem.[/b] Let ABCD be a trapezium with AB || CD. Assuming that there exists a point K on the line BC such that triangle ADK is equilateral, prove that there exists a point L on the line AD such that triangle BCL is equilateral.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°. Assume that the equilateral triangle ADK is negatively directed, i. e. we have < ADK = < DKA = < KAD = 60° (and not -60°).

Let the circumcircle of triangle CDK intersect the line AD at a point L (apart from D). [If the points C and K coincide, then the circumcircle of triangle CDK has to be understood as the circle passing through the points C and D and touching the line BC at the point C. If the circumcircle of triangle CDK touches the line AD at the point D, then simply let L = D.] Then, < LCK = < LDK; in other words, < LCB = < ADK. Since < ADK = 60°, we thus get < LCB = 60°.

On the other hand, since the point L lies on the circumcircle of triangle CDK, we have < DLK = < DCK. But AB || CD yields < DCK = < ABK. Thus, < DLK = < ABK. In other words, < ALK = < ABK. Thus, the points A, B, L and K lie on one circle, so that < KBL = < KAL. In other words, < CBL = < KAD. Since < KAD = 60°, we thus get < CBL = 60°.

Since < LCB = 60° and < CBL = 60°, the triangle BCL is equilateral, and the problem is solved.

  Darij


The exterior angles of a regular polygon each measure $ 20^\circ$. How many sides does the polygon have?

The sum of all the exterior angles of a regular polygon is 360. Thus the answer is $ 360 \div 20 \equal{} 18$.


I was wondering if I can get some feedback on whether or not my answer is correct
If an observer is 200 feet from a building and the angle of elevation to the top of a building is 60 degrees, what is the height of the building?
For this problem I got 400ft is that right

[quote="Altheman"]no, that is the distance from him to the top is 400 ft, the answer is $200\sqrt{3}$, this is not a place for homework by the way...[/quote]

I would not discourage people asking questions of this type.  I think that posting a whole homework set should be highly discouraged.  But if someone is not understanding a concept, even if it's from homework or class, that should be OK to ask...it's just a little too basic for Intermediate.  I moved it to Getting Started but a moderator there can move it lower (I'm not very familiar with what kind of material should be posted in the lower levels so I'll just leave it in GS).

With this problem, you basically have a 30-60-90 triangle with the side between the right angle and the 60 degree angle being 200. You can either use trigonometry (specifically tangents) to find the height of the building, or you can apply basic principles of 30-60-90 triangles to get 200*sqrt3 as an answer.

Nope. 400 is the length of the hypotenuse, and you're tring to find the leg opposite 60 degree angle. The correct answer is $200\sqrt{3}.$


algebra
polynomial
limit
functional equation
algebra unsolved

Find  polynomial P(x) with real coefficients such that for all $ x \in R \minus{} {0}$, the following equality holds

$ (P(x))^2 \plus{} (P(\frac {1}{x}))^2 \equal{} P(x^2) \cdot P(\frac {1}{x^2})$

\[ P(x)^2 \plus{} P\left(\frac {1}{x}\right)^2 \equal{} P( \minus{} x)^2 \plus{} P\left( \minus{} \frac {1}{x}\right)^2 \\
\Rightarrow \left(P(x) \plus{} P( \minus{} x)\right)((P(x) \minus{} P( \minus{} x)) \equal{} P\left( \minus{} \frac {1}{x}\right)^2 \minus{} P\left(\frac {1}{x}\right)^2\]
The right hand side approaches $ 0$ as $ x\to \infty$ (by continuity at $ 0$), while the left hand side cannot approach $ 0$ unless it is identically $ 0$ (since $ P(x) \plus{} P( \minus{} x)$ and $ P(x) \minus{} P( \minus{} x)$ are polynomials and therefore are both unbounded unless they are constant).  Therefore, $ P$ is either odd or even.

Suppose $ P$ is of order $ k$, and let $ a_k \equal{} \lim_{x\to \infty}\frac {P(x)}{x^k}\neq 0$, and $ a_0 \equal{} P(0)$.  Then,
\[ \lim_{x\to 0}x^{2k}\left(P(x)^2 \plus{} P\left(\frac {1}{x}\right)\right) \equal{} a_k ^2 \\
\lim_{x\to 0}x^{2k}P(x^2)P\left(\frac {1}{x^2}\right) \equal{} a_0 a_k \\
\Rightarrow a_0 \equal{} a_k\]
But if $ P$ is odd then $ a_0 \equal{} a_k \equal{} 0$ and we have a contradiction, so $ P$ is even.  It follows that $ P(\sqrt {|x|})$ is a polynomial satisfying the original functional equation, and by repeating the entire argument $ n$ times it follows that $ P\left(|x|^{2^{\minus{}n}}\right)$ is a polynomial satisfying the functional equation for each $ n\in \mathbb{N}$.  This is a absurd unless the order of $ P$ is $ 0$, i.e. $ P$ is constant.  Therefore, $ P\equiv 0$ is the only solution.


What do you want? Please post clearly to anybody can understand

The dot product is distributive.  Using this fact, we can quickly prove that

$\left< a_{1}, a_{2}, ... a_{n}\right> \cdot \left< b_{1}, b_{2}, ... b_{n}\right> = a_{1}b_{1}+...+a_{n}b_{n}$

Which provides a very succinct statement (and proof!) of the Cauchy inequality.

[quote="jgao"][quote="Leao"]Actually, I have a few questions about vectors involving cross and dot products that I'm confused about...
1. Why is $(\vec{a}$x$\vec{b})$x$\vec{c}\cdot\vec{c}$ equal to $(\vec{a}$x$\vec{b})\cdot\vec{c}$x$\vec{c}=0$ ?
In addition can somebody please show me a proof of Lagrange's Formula, that is:
2. $(\vec{a}$x$\vec{b})$x$\vec{c}=-(\vec{b}\cdot\vec{c})\vec{a}+(\vec{a}\cdot\vec{c})\vec{b}$[/quote]Dot product and cross product are different[/quote]
I'm not really sure what you mean, could you please elaborate?  :blush:

[quote="jgao"]I i remember correctly

$a \dot b=$abosolute value{a}abosolute value{b}cos \theta
$a \times b=$abosolute value{a}abosolute value{b}sin \theta

I dont know have to write vectors in latex so assume a, b are vectors[/quote]
But how does this answer my question? I know the definitions of dot and cross products...


1. Karen knows her old analog clock (one that only gives the hours 1-12) is fast. By coincedence when she left for a 20 day vacation, the clock's time was accurate, and when she came back after exactly 20 days the clock's time was also accurate. Given that the clock is fast, what is the least amount of minutes its minute-hand could have moved in one hour?

How do I start on this problem?

Thanx.

since it's an analog clock, the time could run 12 hours in 20 days, or 24 hours in 20days or 36 in 20days....

and make a ratio..It could help tremendesly..(did I spell it right?) :D

[quote]it's spelled tremendously...you were pretty close.[/quote]

He he...

thanks...

I'm only in 7th grade, so I'm not so good at spelling......

 :blush:


$ x^{3}\minus{}3x\plus{}1\equal{}0$ roots $ a,b,c$ $ (a<b<c)$ . Prove that $ b^{2}\minus{}a\equal{}c^{2}\minus{}b\equal{}a^{2}\minus{}c\equal{}2$

[quote="zaya_yc"]$ x^{3} \minus{} 3x \plus{} 1 \equal{} 0$ roots $ a,b,c$ $ (a < b < c)$ . Prove that $ b^{2} \minus{} a \equal{} c^{2} \minus{} b \equal{} a^{2} \minus{} c \equal{} 2$[/quote]

It's immediate to see that if the three roots are $ a < b < c$, we have $ a < \minus{} \sqrt 3 < 0 < b < 1 < c < 2$, so $ c^2 > b^2$ and so $ c^2 \minus{} 2 > b^2 \minus{} 2$
Since $ \minus{} 1 < c < 2$ we also have $ (c \plus{} 1)(c \minus{} 2) < 0$ and so $ c > c^2 \minus{} 2$ and so $ c > c^2 \minus{} 2 > b^2 \minus{} 2$.
Then $ a<\minus{}\sqrt 3$, so $ a^2>3$ and $ a^2\minus{}2>1>b$

Then, since $ c$ is a root of $ x^3 \minus{} 3x \plus{} 1$, then $ (c^2 \minus{} 2)^3 \minus{} 3(c^2 \minus{} 2) \plus{} 1 \equal{} c^6 \minus{} 6c^4 \plus{} 9c^2 \minus{} 1$ $ \equal{} (c^3 \minus{} 3c \plus{} 1)(c^3 \minus{} 3c \minus{} 1) \equal{} 0$

So $ u\equal{}c^2\minus{}2$ is also a root  and so $ v\equal{}u^2\minus{}2$ is also a root and so $ w\equal{}v^2\minus{}2$ is also a root and, since we have  $ c > c^2 \minus{} 2 > b^2 \minus{} 2$, we get $ b \equal{} c^2 \minus{} 2$ and $ a \equal{} b^2 \minus{} 2$

And so either $ a^2\minus{}2\equal{}b$, either $ a^2\minus{}2\equal{}c$ but we saw that $ a^2\minus{}2>b$ so $ a^2\minus{}2\equal{}c$

Hence the result.

I have nice soln.

[quote="zaya_yc"]$ x^{3} \minus{} 3x \plus{} 1 \equal{} 0$ roots $ a,b,c$ $ (a < b < c)$ . Prove that $ b^{2} \minus{} a \equal{} c^{2} \minus{} b \equal{} a^{2} \minus{} c \equal{} 2$[/quote]
Take $ x \equal{} 2\cos \phi$, we solve trig equation and get solutions $ a \equal{} 2\cos \frac {2\pi}{9}$, $ b \equal{} 2\cos \frac {4\pi}{9}$, $ c\equal{}2\cos \frac {8\pi}{9}$. With some calculuses of trig, we are done!


calculus
integration
geometry
rectangle
rotation
3D geometry
sphere

Consider a right circular cylinder with radius $ r$ of the base, hight $ h$. Find the volume of the solid by revolving the cylinder about a diameter of the base.

maybe i dont understand the question.. if you revolve a cylinder around some diameter, then doesnt the rectangle above the diameter catch all of the points in the revolution?

When the cylinder rotates about the diameter, you will see the solid looks swell outside?

I get a really messy formula here:

If $ h\le r,$ then we simply get the sphere of radius $ r$ and volume $ \frac{4\pi r^3}3.$

So the interesting case is $ h>r,$

Define $ x_0\equal{}\frac{r(h^2\minus{}r^2)}{h^2\plus{}r^2}.$ I get

$ V\equal{}2\pi\int_0^r\max\left(\sqrt{r^2\minus{}x^2},h\left(1\minus{}\frac{x}r\right)\right)^2\,dx$

$ \equal{}2\pi\int_0^{x_0}(r^2\minus{}x^2)\,dx\plus{}2\pi\int_{x_0}^rh^2\left(1\minus{}\frac{x}r\right)^2\,dx$

$ \equal{}2\pi\left(r^2x_0\minus{}\frac{x_0^3}3\plus{}\frac{h^2}{3r^2}(r\minus{}x_0)^3\right).$

I don't have the heart to put in the formula for $ x_0$ to see if it simplifies any further.

Why do we need to classify the cases? or does my poor English in the context lead you misunderstand?

By the way, my answer is wrong; I'll have to work on the fix. This is a seriously messy problem. (In particular, I think I'm integrating the wrong thing on each side of $ x_0.$)

We have to classify the cases because it does matter.

Let the cone be given by $ z\equal{}h\minus{}\frac{h}r\sqrt{x^2\plus{}y^2}.$ We will rotate around the $ x$ axis.

Fix $ x$ and WLOG assume $ x>0.$ Slice through the cone at with a plane at that fixed $ x.$ We get a hyperbola. The radius at that particular $ x$ of the solid of rotation is determined by the point furthest from the $ x$ axis. To find that, we must maximize $ \sqrt{y^2\plus{}z^2}$ subject to the restriction $ z\equal{}h\minus{}\frac{h}r\sqrt{x^2\plus{}y^2}.$ It turns out that this maximum happens at one of the extremes - either when $ y\equal{}0,$ in which case the distance is $ z\equal{}h\left(1\minus{}\frac xr\right),$ or when $ z\equal{}0,$ in which case the distance is $ y\equal{}\sqrt{r^2\minus{}x^2}.$ If $ h\le r,$ the second value - the rim of the circle - is always the maximum, so we rotate to get the sphere. If $ h>r,$ then each term is maximized over some portion of the interval in question.

Now I'm going to go away for a while and try to track down my errors.

OK, I think I can try again. Assume $ h > r.$ As above, I think the volume is

$ V \equal{} 2\pi\int_0^r\max\left(r^2 \minus{} x^2,\frac {h^2}{r^2}(r \minus{} x)^2\right)\,dx.$

As above, let $ x_0 \equal{} \frac {r(h^2 \minus{} r^2)}{h^2 \plus{} r^2},$ which is where the maximum trades sides.

Note that $ r \minus{} x_0 \equal{} \frac {2r^3}{h^2 \plus{} r^2}.$

$ V \equal{} 2\pi\left[\int_0^{x_0}\frac {h^2}{r^2}(r \minus{} x)^2\,dx \plus{} \int_{x_0}^r(r^2 \minus{} x^2)\,dx\right]$

$ \equal{} 2\pi\left[\frac {h^2}{3r^2}\left(r^3 \minus{} (r \minus{} x_0)^3\right) \plus{} \frac {2r^3}3 \minus{} r^2x_0 \plus{} \frac {x_0^3}3 \right]$

Now, if I've done the polynomial multiplications correctly (a big if), I get

$ V \equal{} \frac {2\pi(h^8r \plus{} 3h^6r^3 \plus{} 3h^4r^5 \plus{} 13h^2r^7 \minus{} 4r^9)}{3(h^2 \plus{} r^2)^3}.$

I won't swear that that's right but it at least gives the correct answer for $ h \equal{} r.$

For $ h\gg r,$ this looks asymptotically like $ \frac {2\pi h^2r}3.$ In that configuration, we have approximately a double cone of height $ r$ (doubled) and base radius $ h.$

I see your mistake, Kent: it's a cylinder, not a cone.

[hide="Not so bad after all"]I'll work with knowledge of how plane figures revolve; we want to find a plane figure that revolves to form the same solid.

Let the cylinder be $ 0\le \sqrt {x^2 \plus{} y^2}\le r$, $ 0\le z\le h$ and the axis of rotation be the $ x$-axis. We want to "project" onto the $ xz$-plane. On a slice $ x \equal{} x_0$, the solid's cross-section is determined by the point with the largest distance $ \sqrt {y^2 \plus{} z^2}$. Since $ 0\le y^2\le r^2 \minus{} x_0^2$ and $ 0\le z\le h$, the maximum distance is $ \sqrt {r^2 \minus{} x_0^2 \plus{} h^2}$, achieved on the rim of the top circular face.

The figure in the $ xz$-plane that rotates for the same solid is $ \minus{} r\le x\le r$, $ 0\le z\le \sqrt {r^2 \plus{} h^2 \minus{} x^2}$, and the solid is in fact a sphere with two end caps cut off. Integrating by the disk method, the volume is $ \int_{ \minus{} r}^r \pi\cdot (r^2 \plus{} h^2 \minus{} x^2)\,dx \equal{} \frac43\pi r^3 \plus{} 2\pi rh^2$.
[Edit- silly mistake fixed][/hide]

Ack! I just realized that I've been working a different problem than the one asked. I was working with a [b]cone[/b] when the problem clearly says [b]cylinder[/b].

This is easier. At location $ x,$ the radius of the piece that we revolve is the distance from the $ x$ axis to the rim of the opposite circle, which is the hypotenuse of a right triangle with legs $ h$ and $ \sqrt{r^2\minus{}x^2},$ namely $ \sqrt{r^2\plus{}h^2\minus{}x^2}$

That makes the volume

$ V\equal{}\pi\int_{\minus{}r}^r(r^2\plus{}h^2\minus{}x^2)\,dx\equal{}2\pi\int_0^r(r^2\plus{}h^2\minus{}x^2)\,dx$

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

That is a much easier problem than the one I was trying to work.


[i]A student gave me this problem , but I have not time to try it :[/i]

  If $ k \in \mathbb N$, prove that the equation $ x \plus{} y \plus{} \frac {1}{x} \plus{} \frac {1}{y} \equal{} 3k$

 has not  positive rational solutions. He told me , he found the problem in an english book.

 Babis

Hmm it reminds me something (it was a problem from Baltic way and from Archimidis 2006)

I don't have a link so I will present the solution

Let $ x\equal{}\frac{k}{l}$ and $ y\equal{}\frac{m}{n}$ . After cearing the denominators we have that

$ mn(k^2\plus{}l^2)\plus{}lk(m^2\plus{}n^2)\equal{}3klmn$  . From this we have that mn divides lk and lk divides mn so $ kl\equal{}mn$

so the relation  becomes $ k^2\plus{}l^2\plus{}m^2\plus{}n^2\equal{}3k$

Now note that a square is 0 or 1 mod 3 and we can't have k=l (mod 3)  or m=n(mod 3)  since they are relative prime

so $ k^2\plus{}m^2\plus{}l^2\plus{}n^2\equal{}2(\mod 3)$ which is absurb and we are done


An integer x is not the square of another integer. Show that x is not the square of a rational number.

Assume (p/q)^2 = x for p,q in Z and gcd(p,q) = 1. Then p^2/q^2 = x and p^2 = xq^2. I know this is relatively easy, but where do you get the contradiction?

I know that if x is a prime number, then sqrt(x) is irrational. But it's weird working with x being a composite number.

Well for a square to be ..well a square, all of the powers of each prime factor must be even. So the left hand side ($ p^2$) has all even powers.
The right hand side: $ (q^2)$ has all even powers. Therefore $ x$ must have prime factors whose exponents are all even. But it is given that this cannot be the case (An integer x is not the square of another integer). So at least one prime factor, lets say $ p_i$ has an odd exponent (WLOG of course :P). So the exponent count on the left side does not equal the count on the right side. Contradiction, etc..
As you can see, prime or composite x is irrelevant.


How many regions can you cut a circle into with $ n$ line segments?

I have the formula, with an algebraic proof (this is stolen from AIME15's Tournament, by the way) here:

[hide]
The explicit formula to find $ r$ is $ \boxed{f(n) = \binom{n}{4} + \binom{n}{2} + 1}$.

We can prove this by starting with $ n$ points, all connected, and adding one more point. Connecting it to its two neighbors adds $ 1$ region each. Connecting it to their neighbors adds $ 1(n - 2) + 1$regions each. Connecting it to their neighbors adds $ 2(n - 3) + 1$ regions each. Connecting it to their neighbors adds $ 3(n - 4) + 1$ each. Each line from the new point to an old vertex crosses $ k(n - k - 1)$ lines connecting the $ k$ vertices on one side of the line to the remaining $ n - k - 1$ vertices on the other side, and crossing $ m$ lines results in $ m + 1$ new regions. That means the number of regions added is:
\begin{align*} f(n + 1) - f(n) & = \sum_{k = 0}^{n - 1}k(n - k - 1) + 1 \\
& = \sum_{k = 0}^{n - 1}1 + nk - k(k + 1) \\
& = n + n^2\times\left(\frac {n - 1}{2}\right) - n(n - 1)\left(\frac {n + 1}{3}\right) \\
& = \frac {n^3 - 3n^2 + 8n}{6} \\
& = \frac {n(n + 1)(n + 1)}{6} - 2n\times\left(\frac {n + 1}{2}\right) + 2n \end{align*}
Interrupting our proof, our recursive equation can now be found: $ \boxed{f(n + 1) = f(n) + \frac {n(n + 1)(n + 1)}{6} - 2n\times\left(\frac {n + 1}{2}\right) + 2n}$.

Continuing: Now we have:
\begin{align*} f(n) & = f(0) + \sum_{k = 1}^{n}f(k) - f(k - 1) \\
& = 1 + \sum_{k = 1}^{n}\frac {(k - 1)(k)(k + 1)}{6} - \frac {2k(k - 1)}{2} + 2(k - 1) \\
& = 1 + \frac {(n - 1)(n)(n + 1)(n + 2)}{24} - \frac {(n - 1)(n)(n + 1)}{3} + \frac {2n(n - 1)}{2} \\
& = 1 + \frac {n(n - 1)(n^2 - 5n + 18)}{24} \\
& = 1 + \frac {n(n - 1)}{2} + \frac {n(n - 1)(n - 2)(n - 3)}{24} \\
& = 1 + \binom{n}{2} + \binom{n}{4} \end{align*}
as desired.
[/hide]

However, since the formula is quite concise, I'm wondering if there's a nice combinatorial argument.  If you know it, please post it.  Thanks!

The question is posed incorrectly. That would just be the same as cutting the plane with $ n$ line segments. You probably mean how many regions, at most, are formed by connecting every pair of points of an $ n$-gon inscribed in a circle. For that, refer to the last hide tag [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1399387#1399387]here[/url], ignoring the fact that I mistakenly put "regular" there.

Your solution solves a slightly different problem than the problem you actually gave.  Here's a solution to the problem as you stated it:

[hide="Combinatorial Solution"]First note that in an optimal configuration, no three lines are concurrent.  If they were, we could translate one of them by an arbitrarily small distance to get another region.  Similarly, no two lines intersect on the circle.

Now choose an arbitrary orientation so that none of the lines are horizontal.  Then look at the lowest point of each region.  These lowest points can only be intersection points of two lines, intersection points of a line and the circle, and the lowest point of the circle.

Each intersection point of two lines gives exactly one lowest point, and there are $ \binom{n}{2}$ intersection points.  Each line intersects the circle twice, and exactly one of these two points is a lowest point.  The lowest point of the circle gives one more lowest point.

Now the number of regions is the same as the number of lowest points, which is $ \binom{n}{2}\plus{}n\plus{}1$.[/hide]

If you have Engel, his enumerative combinatorics chapter does this and other similar problems as examples, I believe.


inequalities
geometry
circumcircle
incenter
inradius
inequalities proposed

In any triangle the inequality holds
$\frac{R}{r}\geq 2+3\sum(\frac{b-c}{b+c})^{2}$.

do $a=y+z$...then the inequality is equivalent to:
$\iff \frac{(x+y)(y+z)(z+x)}{4xyz}\ge 2+\sum\frac{3(y-z)^{2}}{(2x+y+z)^{2}}$
$\iff \frac{x\sum(y-z)^{2}}{4xyz}\ge \sum \frac{3(y-z)^{2}}{(2x+y+z)^{2}}$
$\iff \left(\sum{\frac{1}{4yz}-\frac{3}{(2x+y+z)^{2}}\right)(y-z)^{2}\ge 0}$
$\iff \left(\sum{\frac{(2x+y+z)^{2}}{4yz}-3\right)\frac{(y-z)^{2}}{(2x+y+z)^{2}}\ge 0}$
$\iff \left(\sum{x(2x+y+z)^{2}-12xyz\right)\frac{(y-z)^{2}}{(2x+y+z)^{2}}\ge 0}$
Wlog:$x \ge y \ge z$
Let $S_{x}=x(2x+y+z)^{2}-12xyz$...
Because:$\frac{(x-z)^{2}}{(x+2y+z)^{2}}\ge \frac{(x-y)^{2}}{(x+y+2z)^{2}}\iff \frac{x-z}{x-y}\ge \frac{x+y}{x+z}\ge \frac{x+y+z+y}{x+y+z+z}$(So notice that we can do it like SOS.)
$S_{y}\ge 0 \iff (x+2y+z)^{2}-12zx \ge (x+3z)^{2}-12zx= (x-3z)^{2}\ge 0$
$S_{x}\ge 0$ is obvious.
$S_{y}+S_{z}\ge 0 \iff y(x+2y+z)^{2}-12xyz+z(x+y+2z)^{2}-12xyz$
$=y(x^{2}+4y^{2}+z^{2}+4xy+2xz+4yz)+z(x^{2}+y^{2}+4z^{2}+2xy+4xz+4yz)-24xyz$
$=x^{2}y+x^{2}z+4xy^{2}+4xz^{2}+5yz^{2}+5y^{2}z+4y^{3}+4z^{3}-20xyz \ge 0$
But we have:
to prove the last inequality here we may let $y=z$($AM \ge GM$)
and also let $y=z=1$(if not,mulpi $\frac{1}{y^{3}}$ and replace $\frac{x}{y}$ by $x$)
So we should only prove
$x^{2}+9-6x=(x-3)^{2}\ge 0$ which is trivial .
So we solved the whole problem :)
Very nice and stronger inequality :)

[quote="zhaobin"]
Let $S_{x}=x(2x+y+z)^{2}-12xyz...$
Because:$\frac{(x-z)^{2}}{(x+2y+z)^{2}}\ge \frac{(x-y)^{2}}{(x+y+2z)^{2}}\iff \frac{x-z}{x-y}\ge \frac{x+y}{x+z}\ge \frac{x+y+z+y}{x+y+z+z}$(So notice that we can do it like SOS.)
$S_{y}+S_{z}\ge 0 \iff y(x+2y+z)^{2}-12xyz+z(x+y+2z)^{2}-12xyz$
[/quote]
Nice!

Thank you two:lol: .And Vasc what about your solution?

Thanks Vasc for nice inequality,I see a nice corollary of it
Given are $\triangle ABC,O,I$ is circumcenter and incenter, $r,s$is inradius and semiperimeter then
$OI\ge (s-a)sin\frac{|B-C|}{2}+(s-b)sin\frac{|C-A|}{2}+(s-c)sin\frac{|A-B|}{2}$
 :D  :D

My solution is also SOS.
Let $S_{x}=\frac 1{12yz}-\frac 1{(2x+y+z)^{2}}$,...
For $x \geq y \geq z$, we have $S_{x}\geq 0$, $S_{y}\geq 0$ and $S_{y}+S_{z}\geq 0$.
For the last, denote $s=x+y+z$ and assume that $y+z=2$. Since $x=s-2$, ineq. becomes $F(s,yz) \geq 0$.
We have
$F(s,yz) \geq F(s,1) \geq 0$.

Thanks for show your solution :)
[quote="Vasc"]Nice solution, zhaobin.  :lol:
Equality occurs for $a=b=c$ and also for $a=b=2c$ and any permutation thereof.[/quote]
I think the inequality occurs for  $x=y=z$ and also for $x=y=2z$ and any permutation thereof :D

[quote="Vasc"][quote="zhaobin"]$S_{y}\ge 0 \iff (x+2y+z)^{2}-12zx \ge (x+3z)^{2}-12zx= (x-3z)^{2}\ge 0$
and also let $y=z=1$(if not,mulpi $\frac{1}{y^{3}}$ and replace $\frac{x}{y}$ by $x$)
So we should only prove
$x^{2}+9-6x=(x-3)^{2}\ge 0$ which is trivial .
[/quote]
Therefore, equality holds for $x=y=z$, and also for $y=z=x/3$ and any permutation thereof. That is, for $a=b=c$, and also for $b=c=2a$ ...[/quote]
 :blush: ..I am thinkin wrong


[quote="hungkhtn"]I think you should think about it before you post here. Don't post many problem of Crux of some other TST in short time.[/quote]
I'm sorry. These ineq are from my friend and I don't know they're from Crux

But the last ineq is mine ,please solve it.

[quote="spanferkel"]I guess for the last one, the slightly stronger $\sum(a^4b+a^4c)+3\sum(a^3b^2+a^3c^2)+12\sum ab^2c^2\ge20(a^2+b^2+c^2)abc$ should also hold.[/quote]
But how do you solve this ineq?


[quote="Tony2006"]Este es un problema que me marcaron en una tarea de entrenamientos de platas nacionales:

Las circunferencias C1 y C2 se cortan en dos puntos distintos A y B. Una linea arbitraria que pasa por A corta de nuevo a C1 en C y a C2 en D. Sea M un punto arbitrario en el segmento CD. la paralela a BC por M corta al segmento BD en K y la paralela a BD por M corta al segmento BC en N. La perpendicular por N a BC corta al arco BC de C1 que no contiene a A en el punto E. La perpendiular por K a BD corta al arco BD de C2 que no contiene a A en F. Muestra que angulo EMF =90º.

Dejen sus ideas y soluciones  :P !![/quote]

Que problema feito! muchas letras...   lo hice en el Cabri y daba, o sea que ahora falta probarlo.... jejeej, esa tarea se las dejo a uds.... si quieren posteo un dibujo.....

Si puedes por favor pon el dibujito!!  :)

Disculpen, pero no sé que tiene esta compu que no me deja ponerlo al dibujo!
Si quieren avísenme y se lo paso por mail a alguien así lo postean

Estas seguro que estan bien puestos los nuevos puntos, por que lo estoy haciendo y como que no da!!! ;)

[quote="Tony2006"]
Las circunferencias C1 y C2 se cortan en dos puntos distintos A y B. Una linea arbitraria que pasa por A corta de nuevo a C1 en C y a C2 en D. Sea M un punto arbitrario en el segmento CD. la paralela a BC por M corta al segmento BD en K y la paralela a BD por M corta al segmento BC en N. La perpendicular por N a BC corta al arco BC de C1 que no contiene a A en el punto E. La perpendiular por K a BD corta al arco BD de C2 que no contiene a A en F. Muestra que angulo EMF =90º.
[/quote]

Bueno lo primero es marcar más cosas:
Sea $\angle CBD = \theta$
Sea $O_1 y O_2$ los centros de $S_1 y S_2$  respectivamente.
El segundo punto de corte de EN en $S_1$ se llama H.
Sea G el punto de corte de EN y FK.
Sean $X_1$ el corte de MK con EN y $X_2$ el corte de FK con NM.

$\angle O_2O_1B= 1/2 \angle AO_1B = \angle ACB$,
$\angle O_1O_2B= 1/2 \angle AO_2B = \angle ADB$
Los triangulos $CBD$ y $O_1BO_2$ son semejantes.
En general toda la figura tiene una dilatación espiral:
 -Con centro en B
 -Con angulo $\theta$
 -De razón $\frac{CB}{BD}$
                Esta transformación me relaciona:
                       -$S_2 con S_1$
                       -La recta BD con la recta BC

Sean K' y F' las imagenes de K y F después de la dilatación espiral, respectivamente.
por lo anterior y usando las razones de la transformación:
$\frac{K'B}{KB}=\frac{CB}{DB} \Rightarrow K'K es paralela a CD$
Esto me da que $\angle DCN= \angle KK'B$ y es fácil ver que $\angle CNM = \angle K'BK$
Como NMBK es paralelogramo tengo que $MN=KB$.
Todo lo anterior me da que los triangulos CNM y K'BK son congruentes con CN=K'B

$\angle BKF= \angle BK'F'$, por las propiedades de la dilatación espiral y F' cae sobre $S_1$
CN=K'B entonces $NO_1=K'O_1$ como $HO_1=F'O_1$ entonces HF' es paralela a NK, NK'F'H es un paralelogramo ( rectangulo en realidad ) y NH=K'F'.

Por potencia de puntos en N con el circulo $S_1$: $EN\cdot NH =CN \cdot NB$:
Como:
-NH=K'F'
-CN=K'B
$EN\cdot K'F' =K'B \cdot NB \Leftrightarrow \frac{EN}{NB}=\frac{K'B}{K'F'}$.
Otra vez usando las proporciones de la rotación dilatada:
$\frac{KB}{KF}=\frac{K'B}{K'F'} \Rightarrow \frac{KB}{KF}=\frac{EN}{NB}$
Usando que NMBK es paralelogramo KB=MN y NB=MK:
1)
$\frac{MN}{KF}=\frac{EN}{MK}$ o $\frac{MK}{KF}=\frac{EN}{MN}$

Sabemos que $90=\angle GX_1K = \angle GX_2N$
Entonces es fácil ver, ya que comparten un angulo que los triangulos $GX_1K$ y $GX_2N$ son semejantes:
$\angle GKX_1 = \angle GNX_2$
Esto nos da que $\angle FKM = \angle ENM$
con el resultado 1) tenemos que los triangulos MKF y ENM son semejantes.

$\alpha = \angle MEN = \angle FMK$
$\beta = \angle NME = \angle KFM$
$\angle FKM = 180- \beta - \alpha$
$\angle MKG = \beta + \alpha$
$\angle NGK = 90-\beta - \alpha$
($\angle GX_1M = \angle GX_2M= 90$)
$\angle NMK = \angle X_1MX_2= 90+\beta + \alpha$
$\angle EMF = 90+\beta + \alpha - \angle NME - \angle FMK = 90$

QED.

Porfavor revisenlo y me avisan si está completo.

[quote="Tony2006"]que es una dilatacion espiral es algo asi como homotecia explicame porfavor :) !!![/quote]

También se conoce como Semejanza espiral.
Es la combinación de una homotecia central y una rotación con el mismo centro.

Como dato curioso, si la rotación se hace en otro centro el resultado es también una dilatación espiral con centro diferente a los dos escogidos. En general cualesquiera figuras semejantes se relacionan con una dilatación espiral o por una reflección. ( Se considera que una translación es una dilatación con centro en el infinito o algo así)

[quote="Tony2006"]disculpa me podrias decir propiedades de la dilatacion espiral que conoces como las que nombraste en la solucion del problema, por favor :huh: !!!![/quote]

Creo que las más importantes son las siguientes.

Si O es el centro de la dilatación espiral, $\alpha$ es el angulo, $\omega$ es la razón y A' es la imagen de A después de la transfomación:

-$\angle A'OA = \alpha$ (Mantiene angulos)
-$A'O/AO=\omega$ (Mantiene razones)
-Toda figura después de la transformación es semejante a su original.

De estas dos se tienes que para cualesquiera dos puntos P y su imagen después de la transformación P'
el triangulo POP' es semejante al triangulo AOA'.


We just blew some air into a spherical balloon and doubled its volume. By how much did we multiply the surface area ?

(A) 2 (B) 2^1/3  (C) 2^2/3    (D) 2^3  (E) none of these

[hide="Solution"]If the volume increases by $ k^3$, the radius must increase by $ k$, so the surface area must increase by $ k^2$. Therefore, an increase of $ k$ in the volume induces an increase of $ k^\frac{2}{3}$ in the surface area, so let $ k\equal{}2$. Therefore, the answer is $ \boxed{C}$.

[/hide]

darkdieuguerre's way is probably the best, but on a problem like this, remember you can make up your own spheres and numbers and come to the same answer.


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

a) Let $ G=\cup_{i=1}^{n}H_{i}$, which $H_{i}\geq G$. Prove that if $[G: H_{n}]=\infty$, then $ G=\cup_{i=1}^{n-1}H_{i}$.
b) Prove that if group $G$ is union of finite number of its abelian subgroups, then $[G: Z(G)]<\infty$.

Just a comment: although no proof has been presented, the first problem appeared at least twice, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63717]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=covering+difficult&t=80745]here[/url]. As for the second one, it's an immediate consequence of the first one: according to (a), we can eliminate those subgroups which have infinite index, so we may as well assume that all our abelian subgroups have finite index. Their intersection will then be a subgroup of finite index contained in the center, so we're done.

Here is one (long) way to proceed.
First you can assume that $\forall i \in \{1, \ldots n-1 \}, H_{i}\not\subset \cup_{j\not= i}H_{j}$.
Now, assume that $G\not= \cup_{i=1}^{n-1}H_{i}$, and let $h \in H_{n}\setminus \cup_{i=1}^{n-1}H_{i}$
Let's show, by induction over $k \in \{1, \ldots n-1 \}$, the following lemma:

[i][b]Lemma:[/b][/i] there are $k$ distinct subgroups $H_{i_{1}},\ldots,H_{i_{k}}\in \{ H_{1},\ldots, H_{n-1}\}$ and $(x_{p})_{p\in \mathbb{N}}\in G^{\mathbb{N}}$ s.t.
\[\forall p \geqslant 0, x_{p}\in \left( \bigcap_{j=1}^{k}H_{i_{j}}\right) \setminus H_{n}\quad \mbox{and}\quad \forall m,p \in \mathbb{N}, m \not=p \Rightarrow x_{m}H_{n}\not= x_{p}H_{n}\]
  - If $k=1$. $[G: H_{n}]=\infty$, so there is $(x_{p})_{p\in \mathbb{N}}\in G^{\mathbb{N}}$ s.t. $\forall m,p \in \mathbb{N}, m \not=p \Rightarrow x_{m}H_{n}\not= x_{p}H_{n}$. $x_{p}H_{n}$ is at most once equal to $H_{n}$, so we can assume that $\forall p \geqslant 0, x_{p}H_{n}\not= H_{n}$, which means that $x_{p}\not\in H_{n}$.
$G \setminus H_{n}\subset \cup_{i=1}^{n-1}H_{i}$, so let $i_{1}\in \{1, \ldots n-1 \}$, s.t. there is an infinity of $x_{p}$ belonging to $H_{i_{1}}$. We can extract from $(x_{p})$ a sequence $x_{\phi(p)}$ s.t. $\forall p\geqslant 0, x_{\phi(p)}\in H_{i_{1}}$, which proves the lemma for $k=1$.

  - If $k\in \{1, \ldots n-1 \}$, assume that the lemma is true for $k$ and let's show that it is true for $k+1$.
Let $H_{i_{1}},\ldots,H_{i_{k}}\in \{ H_{1},\ldots, H_{n-1}\}$ and $(x_{p})_{p\in \mathbb{N}}\in G^{\mathbb{N}}$ s.t.
\[\forall p \geqslant 0, x_{p}\in \left( \bigcap_{j=1}^{k}H_{i_{j}}\right) \setminus H_{n}\quad \mbox{and}\quad \forall m,p \in \mathbb{N}, m \not=p \Rightarrow x_{m}H_{n}\not= x_{p}H_{n}\]
$hx_{p}\in G \setminus \left( \bigcap_{j=1}^{k}H_{i_{j}}\cup H_{n}\right)$
So let $i_{k+1}\in \{1, \ldots n-1 \}\setminus \{i_{1}, \ldots i_{k}\}$ and $\phi ~: ~\mathbb{N}\to \mathbb{N}$ which defines an extract sequence $(x_{\phi (p)})$, s.t. $\forall p\geqslant 0, hx_{\phi (p)}\in H_{i_{k+1}}$.
Now define $y_{p}= x_{\phi (0)}^{-1}x_{\phi (p)}$ for $p \geqslant 1$, and we have $y_{p}\in \left( \bigcap_{j=1}^{k+1}H_{i_{j}}\right) \setminus H_{n}$
Finally, assume we had $m\not=p$, $m,p \geqslant 1$, s.t. $y_{p}H_{n}=y_{m}H_{n}$. Define $g \in H_{n}$ s.t. $x_{\phi (0)}^{-1}x_{\phi (p)}g=x_{\phi (0)}^{-1}x_{\phi (m)}$, then $x_{\phi (p)}g=x_{\phi (m)}$, which implies $x_{\phi (p)}H_{n}=x_{\phi (m)}H_{n}$ which is impossible. As a result $\forall m,p \in \mathbb{N}^{*}, m\not=p \Rightarrow y_{m}H_{n}\not= y_{p}H_{n}$. That proves the [i][b]lemma[/b][/i] for $k+1$.

Let's [u][b]conclude[/b][/u] by using that lemma for $n-1$:
Let $x\in \left( \bigcap_{i=1}^{n-1}H_{i}\right) \setminus H_{n}$, $x\not\in H_{n}$, $hx\not\in H_{n}$, and $hx \not\in H_{i}$, for all $i\in \{1, \ldots, n-1 \}$. So $hx \in G \setminus \left( \bigcup_{i=1}^{n}H_{i}\right)$, which is impossible.[list][/list][list][/list]


Solve the system of the equations :)
 $ (x \plus{} y)^{3} \equal{} 3z$
 $ (y \plus{} z)^{3} \equal{} 3x$
 $ (z \plus{} x)^{3} \equal{} 3y$

let $ x \plus{} y \equal{} c,y \plus{} z \equal{} a$ and $ x \plus{} z \equal{} b$ .
so the system $ \Longrightarrow a^3 \equal{} 3\frac {b \plus{} c \minus{} a}{2}$
                      $ b^3 \equal{} 3\frac {a \plus{} c \minus{} b}{2}$
                     $ c^3 \equal{} 3\frac {b \plus{} a \minus{} c}{2}$
so                  $ a^3 \plus{} b^3 \equal{} 3c$
                      $ b^3 \plus{} c^3 \equal{} 3a$
                      $ a^3 \plus{} c^3 \equal{} 3b$
so                  $ (a^3 \plus{} b^3) \minus{} (b^3 \plus{} c^3) \equal{} 3(c \minus{} a)$ so $ (a \minus{} c)(a^2 \plus{} ac \plus{} c^2 \plus{} 3) \equal{} 0$ so $ a \equal{} c$ because $ a^2 \plus{} ac \plus{} c^2 \plus{} 3 \equal{} (a \plus{} c/2)^2 \plus{} \frac {3c^2}{4} \plus{} 3 > 0.$ and with the same way $ a \equal{} b \equal{} c$ so $ x \equal{} y \equal{} z$ so $ 8x^3 \equal{} 3x$ so $ (x,y,z) \equal{}${$ (0,0,0),\sqrt {\frac{3}{8}},\sqrt {\frac{3}{8}},\sqrt {\frac{3}{8}}),( \minus{} \sqrt {\frac{3}{8}}, \minus{} \sqrt {\frac{3}{8}}, \minus{} \sqrt {\frac{3}{8}})$}.


Given that $f(x) = 3x + 1$, what is the value of $f(3) + f(5)$?

if $f(x) = 3x + 1$ then $f(3) = 10$ and $f(5) = 16$, and thus all u do is add $10$ and $16$ which gives u $26$