geometry
incenter
circumcircle
analytic geometry
geometry open

Triangle ABC is not isosceles. The incenter is I, the excenter is O. The incircle touches the sides BC, CA, AB at points D, E, F, respectively. Lines FD and AC intersect at P, lines DE and AB intersect at Q. The midpoints of segments EP amd FQ are M and N, respectively. Prove that MN and OI are perpendicular.

It may be posted before but I'm not sure.

O is the excenter of angle A? Thanks in advance

I think it is the excentre of A. It is not mentioned. If it is not of it is of C.

Please, Borislav, can you check the problem more? In my first drawing (made by hand, by the way) the line which join the excenter of A and the incenetr I is manifestly non perpendicular to MN, but instead, if O is the circumcenter (as usual), then the situation is perfectly possible.
Is yours this problem?

I suspect you mean the same problem as posted at http://www.mathlinks.ro/viewtopic.php?t=185866 ...

  darij

Yes, it is the same problem as that Darij say. And O means circumcircle. With trilinear coordinates the proof is straightforward (except for computing the coordinates of the mid points), but a bit long.

I'm sorry I had no internet connection for some time. You are both right.


AMC
AIME
geometry
parallelogram
ratio
angle bisector
similar triangles

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

[hide="kinda cheap"]Assume that $ \angle{ACB}$ is right. (This is the AIME!)

Notice that $ \triangle{AMK}\sim\triangle{CPK}$ by SAS similarity.

Extend $ AM$ and $ BC$ so they meet at $ D$. Then $ \triangle{ACD}$ is a 45-45-90 triangle, and $ AC \equal{} CD \equal{} 450$ and $ BD \equal{} 450\plus{}300 \equal{} 750$.

Finally, $ \triangle{BLC}\sim\triangle{BAD} \Rightarrow \frac{LP}{AM} \equal{} \frac{LP}{180} \equal{} \frac{300}{750}$.

Solving, we get $ LP \equal{} \boxed{072}$.[/hide]

[hide]Hm this is a great example how mass points kill the problem:
Observe that $ CP\equal{}AM\equal{}180$ by SAS. Place masses $ 2, 2, 3$ on $ A, C, B$, then the mass of $ L$ is $ 5$. Therefore $ CP: PL\equal{}5: 2$, so $ LP\equal{}\frac{2\cdot 180}{5}\equal{}72$.
[/hide]

[hide]
Use Menelaus on triangle ACL with line KPB:

$ \frac {CK \cdot AB \cdot LP}{KA \cdot BL \cdot PC} \equal{} \minus{} 1$

$ \frac {CK}{KA} \cdot \frac {AB}{LB} \cdot \frac {LP}{PC} \equal{} 1$

$ 1 \cdot \frac {5}{2} \cdot \frac {LP}{180} \equal{} 1$

$ LP \equal{} 72$.[/hide]

Nice timwu, that's exactly what I did on the test.
Although you could explain your masses a little better:

Let [ ] denote the mass of a point.
$ \frac{[A]}{[C]}\equal{}\frac{AK}{CK}.$
Since $ AK\equal{}CK$, $ [A]\equal{}[C].$
$ \frac{[A]}{[B]}\equal{}\frac{BL}{AL}.$
By direct application of the Angle Bisector Theorem, we know that
$ \frac{BL}{AL}\equal{}\frac{BC}{AC}.$
We are given that $ BC\equal{}300$ and $ AC\equal{}450$, so $ \frac{BL}{AL}\equal{}\frac{300}{450}\equal{}\frac{2}{3}$.
So, the $ [A]: [B]: [C]\equal{}2: 3: 2$.
Finally, $ [L]\equal{}[A]\plus{}[B]\equal{}5$

And after that, we can do exactly what timwu did.
Thanks to Chris Juell for showing me mass points.

[hide="Similar to other solutions"]
By SAS congruency, $ \triangle MKA \cong \triangle PKC$.
Therefore, $ \angle MAK \equal{} \angle KCP$ and lines MA and CL are parallel.
Then, $ \triangle MAB \sim \triangle PLB$, so:

$ LP/MA \equal{} LB/AB$
$ LP/180 \equal{} 2/5$ (By angle bisector theorem)
$ LP \equal{} 180(2/5)\equal{} \boxed{072}$
[/hide]

I actually only noticed the solution with about 5 minutes left.  :D  I wasted over half an hour trying to use Stewart's, etc.

[hide=" solution without making trivializing assumptions"]
Draw the triangle.
Because K is midpoint of MP and midpoint of AC, AMCP is a parallelogram (diagonals are bisected).
Thus, CP(CL) is parallel to AM. Now we have similar triangles BLP and BAM. We can use angle bisector theorem to find ratio of AL: LB and thus the ratio of AB: LB. Using this ratio, we get LP = 072[/hide]

[hide="Assigning areas"]Notice that $ CP\equal{}180$ by SAS and $ BP: PK \equal{} 4: 3$ by the angle bisector theorem.
Draw segment $ AP.$ Now let $ [BCP] \equal{} 4*180k$. Employing triangles with equal altitudes,

$ [BPL] \equal{} 4(LP)k$
$ [CPK] \equal{} [APK] \equal{} 3*180k$
$ [ALP] \equal{} 6(LP)k.$

Since $ [ABK] \equal{} [CBK]$, $ 7*180 \equal{} 3*180 \plus{} 10(LP)$ and $ LP\equal{}72$.[/hide]

I used lordviking's solution with the angle bisector theorem.  I was pleased that it could be done so nicely.

[quote="timwu"][hide]Hm this is a great example how mass points kill the problem:
Observe that $ CP \equal{} AM \equal{} 180$ by SAS. Place masses $ 2, 2, 3$ on $ A, C, B$, then the mass of $ L$ is $ 5$. Therefore $ CP: PL \equal{} 5: 2$, so $ LP \equal{} \frac {2\cdot 180}{5} \equal{} 72$.
[/hide][/quote]

Why is the mass on B 3 ?

[quote="lordviking"][hide="Similar to other solutions"]
By SAS congruency, $ \triangle MKA \cong \triangle PKC$.
Therefore, $ \angle MAK \equal{} \angle KCP$ and lines MA and CL are parallel.
Then, $ \triangle MAB \sim \triangle PLB$, so:

$ LP/MA \equal{} LB/AB$
$ LP/180 \equal{} 2/5$ (By angle bisector theorem)
$ LP \equal{} 180(2/5) \equal{} \boxed{072}$
[/hide]

I actually only noticed the solution with about 5 minutes left.  :D  I wasted over half an hour trying to use Stewart's, etc.[/quote]

How did you get that MKA is congruent to PKC? Or how did you know that AK = KC?

[quote="dragon96"]How did you get that MKA is congruent to PKC? Or how did you know that AK = KC?[/quote]

AK = CK because it says so in the problem. :D

Then MK  = KP because K is the midpoint of PM.
Angles MKA and CKP are vertical angles.

Can Someone post a Diagram of this? According to my Diagram, MK = 135, making angle AMB 90 degrees, which doesnt make sense.

[quote="undefined117"][hide="kinda cheap"]Assume that $ \angle{ACB}$ is right. (This is the AIME!)

Notice that $ \triangle{AMK}\sim\triangle{CPK}$ by SAS similarity.

Extend $ AM$ and $ BC$ so they meet at $ D$. Then $ \triangle{ACD}$ is a 45-45-90 triangle, and $ AC \equal{} CD \equal{} 450$ and $ BD \equal{} 450 \plus{} 300 \equal{} 750$.

Finally, $ \triangle{BLC}\sim\triangle{BAD} \Rightarrow \frac {LP}{AM} \equal{} \frac {LP}{180} \equal{} \frac {300}{750}$.

Solving, we get $ LP \equal{} \boxed{072}$.[/hide][/quote]

I dont think you are allowed to do that since you were given AM=180. I think you just got lucky. :P. someone tell me if im wrong...

Here's a diagram, made hastily on Kig.  The numbers are inexact because I just slid $ A$ around the circle until $ AM$ was approximately 180.

abacadaea, I would say undefined117 got lucky.  I experimented with his "solution" a bit.  It turns out that it can be justified without the right angle.  Here's how:

He used the fake right angle (which, by the way, can't happen, as the conditions of the problem uniquely determine the triangles and they have no right angles) to determine that $ CD \equal{} 450$.  It turns out that this is always the case, no matter what ACB is.  That is:
By the SAS similarity (actually, congruence) he got, $ LC$ is parallel to $ AM$.  Draw the circle centered at $ C$ with radius $ 450$.  Then $ A$ lies on that circle.  By the parallelism, $ \angle{ADB} \cong \angle{LCB}$, so $ m\angle{ADB} \equal{} \frac {1}{2}m\angle{ACB}$.  Notice that $ \angle{ACB}$ is a central angle of the circle.  An inscribed angle measures half a central angle subtending the same arc, and it can be shown conversely that an angle exterior to the central angle which subtends the same arc as the central angle and whose measure is half the measure of the central angle, must be an inscribed angle.  Therefore $ \angle{BDA}$ is inscribed, so $ D$ lies on the circle and $ CD \equal{} 450$.  The rest proceeds as before.  Kind of cool, I guess.  But still, I think the angle bisector theorem solution is the simplest.

[quote="Ubemaya"]Can someone explain mass points to me? I don't get what putting masses on points means. Also, what happens if there are two points between A and B? Would it still be applicable?[/quote]

Mass points = area ratio/ceva/barycentric coordinates. I suspect its advantage is that it is easier to memorize if you are not good at geometry (?)

Connect AP and MC .now APCM is a parallelogram .now use the angle bisector theorem and menelaus and we r done ! 072

Hey, I have a quick question. Using mass points, we can get that $m_A=2, m_B=3, m_C=2,$. so $m_K=4$ and $m_P=7$. However, this results in $m_M=7-4=3,$ and $m_M$ is not equal to $m_P$ like it should. I've seen this situation multiple times, and there is a technique for split masses if it is transversal, but there is no transversal. Can anyone explain why this happens?

[quote=JustKeepRunning][quote=IAmTheHazard]I'm not entirely sure what ur solution is, but it might be because $CL$ is a cevian in $\triangle ABC$, but it's a transversal in $\triangle ABD$[/quote]

What point D are you referring to? Also, isn't mass points for lines, not triangles? That's barycentric coordinates, which I have yet to learn. Could you explain a bit more?[/quote]

The diagram that benzi made. I don't really understand ur proof (haven't rly looked at it), but it might be that.

Clearly $LC$ is parallel to $AM$, so $\triangle LPB \sim \triangle AMB$ in ratio $\frac{\frac{2}{3}}{\frac{2}{3}+1} = \frac{2}{5}$. The answer is $180 \cdot \frac{2}{5} = \boxed{72}$.

Claim: $AMCP$ is a parallelogram.
Proof: Let $AC$ and $MB$ intersect at a point $Q.$ We are given that $AQ=QC$ and $MQ=QP,$ therefore $AC$ and $MP$ bisect each other and $AMCP$ is a parallelogram. Opposite sides are equal in a parallelogram, so we have that $AM=PC=180.$

By Menelaus' Theorem on $\triangle ALC$ w.r.t. $\overleftrightarrow{BM},$ we have that $$\frac{CQ}{QA} \cdot \frac{AB}{BL}\cdot \frac{LP}{PC}=1.$$ Substituting our known values, we have $$\frac{225}{225} \cdot \frac{AB}{BL} \cdot \frac{LP}{180}=1\qquad (\diamondsuit).$$ Note that by the Angle Bisector Theorem, $$\frac{AB}{BL}=\frac{BL+LA}{BL}=1+\frac{CA}{BL}=1+\frac{AC}{BC}=1+\frac{450}{300}=\frac{5}{2}.$$ Substituting back in equation $\diamondsuit$ yields $$\frac{5}{2} \cdot \frac{LP}{180}=1 \implies LP = \boxed{72}.$$

[hide=Titu-styled proof]
$AMCP$ is a parallelogram from bisecting diagonals, implying $PC=180.$ Menelaus' gives $$1=\frac{CQ}{QA} \cdot \frac{AB}{BL}\cdot \frac{LP}{PC}=\frac{225}{225}\cdot \left(1+\frac{CA}{BL}\right)\cdot \frac{LP}{180}\stackrel{\text{ABT}}{=}\left(1+\frac{450}{300}\right)\cdot \frac{LP}{180}=1\implies LP=\boxed{72}.$$

Oh also did anyone feel like this is hard for an AIME #5, even compared with recent year AIMEs?

I (and others in the thread) have an easier solution (if your solution was the intended one, it would be much harder than a #5). Your solution is still quite innovative & impressive, though!

[quote=samrocksnature]Ah ic yeah I'm allergic to similar triangles and I always miss the easy solutions with them :/

Thx tho :)[/quote]

me too

#SimTringleBlock

Same! At first I tried to bash lengths and ended up getting something with $\sqrt{1119}$ and decided to stop and think :rotfl:

And also school math helped for once...my two-column proof PTSD reminded me of the Converse of the Bisecting Diagonals of a Parallelogram Theorem :P

First time I had to use Menelaus theorem on a AIME problem

mass pts solution:
set $A$ and $C$. Using angle bisector, you can calculate mass at $B$. Then you're basically done once you start doing some ratios on line $LC$.

Nicest AIME #5 I've seen

First, note that $MK = KP$, $KC = AK$, and $\angle MKA = \angle PKC$, so $\triangle MKA \cong\triangle PKC$. It follows that $PC = AM = 180$. Now, notice that $\frac{BP}{KP} = \frac{CK}{CB} = \frac{225}{300} = \frac{3}{4}$, so $\frac{BP}{BK} = \frac{4}{7}$. We now have
\[ \frac{[ABP]}{[ABC]} = \frac{1}{2}\frac{[ABP]}{[ABK]} = \frac{1}{2}\cdot \frac{BP}{BK} = \frac{2}{7}.\]
On the other hand,
\[ \frac{[ABP]}{[ABC]} = \frac{LP}{LC} = \frac{LP}{180 + LP}.\]
We then have $\frac{2}{7} = \frac{LP}{180 + LP}\implies LP = \boxed{72}$.

Very nice  :oops:

Mass points kill this problem. By $SAS$ triangle congruency, $\triangle AKM \cong \triangle CKP.$ Thus, $PC = AM = 180.$ By the angle bisector theorem, $\frac{CA}{AL} = \frac{CB}{BL} \implies \frac{450}{AL} = \frac{300}{BL} \implies 3BL = 2AL.$ Assign mass points $2$ on $A$ and $C.$ Since $3BL = 2AL,$ the mass of $B$ is $3,$ and thus the mass of $L$ is $5.$ Therefore, $\frac{LP}{PC} = \frac{2}{5} \implies \frac{LP}{180} = \frac{2}{5} \implies LP = \boxed{072}.$

You can also proceed by Menelaus:
Apply Menelaus on $\triangle ACL$ with line containing points $K, P, B \colon$
$$ \frac {CK}{KA} \cdot \frac {AB}{LB} \cdot \frac {LP}{PC} = 1 \implies 1 \cdot \frac {5}{2} \cdot \frac {LP}{180} = 1$$
Hence, $ LP = \boxed{072}.$

stewarts 4 times on BCK, CPM, ABM, and then APB to get LP=72


If $ a,b,c>0$ then:

\[ \boxed{ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}\ge a\plus{}b\plus{}c}\]

Try to justify your thougt  :)
Indeed, take $ a\equal{}b\equal{}2$, $ c\equal{}\minus{}1$ to get a contradiction.

[quote="new_member"]Felix,I don't think that your counterexample works  :huh:[/quote]
He probably meant $ a\equal{}b\equal{}\minus{}2,c\equal{}\minus{}1$.

What did I write^^ at first i thought $ a\equal{}b\equal{}3$, $ c\equal{}\minus{}4$ should give a couterexample (which is indeed one). Then i thought it should be possible with $ a\equal{}b\equal{}2$, $ c\equal{}\minus{}3$ too and i wrote $ c\equal{}\minus{}1$. Don't ask me why^^

[quote="moldovan"]If $ a,b,c > 0$ then:
\[ \boxed{\frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}\ge a \plus{} b \plus{} c}\]
[/quote]
\[ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}\ge a \plus{} b \plus{} c\]
\[ \Longleftrightarrow \sum{\frac {(a \minus{} b)^2}{2(a \plus{} b)}} \ge 0\]
clear.

[quote="new_member"]Hey,we know that $ a^2 \plus{} b^2 \geq \frac {(a \plus{} b)^2}{2}$ is true for every real values of $ a,b$.
Hence the left hand side will be greater or equal to:

    $ \frac {a \plus{} b}{2} \plus{} \frac {b \plus{} c}{2} \plus{} \frac {c \plus{} a}{2} \equal{} a \plus{} b \plus{} c$.

ps I don't think we need to have $ a,b,c > 0$ necessarily.In my opinion,the inequality is true for all the real values $ a,b,c$ such
that $ a \neq \minus{} b,b \neq \minus{} c,c \neq \minus{} a$.[/quote]

$ a^2 \plus{} b^2 \geq \frac {(a \plus{} b)^2}{2}$ holds for any reals $ a,b$, but $ \frac {a^2 \plus{} b^2}{a \plus{} b} \geq \frac {a \plus{} b}{2}$ is not hold for any reals $ a,b$ when $ a \plus{} b < 0$.
so if $ a,b,c < 0$, we have $ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}\le a \plus{} b \plus{} c$.

[quote="moldovan"]If $ a,b,c > 0$ then:
\[ \boxed{\frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}\ge a \plus{} b \plus{} c}\]
[/quote]
I dunno if I should post a solution, but judging from my previous posts (that are mostly wrong solutions :( )I think I will post a solution......
[hide="Solution"]
We have;
$ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}$
$ \equal{} \frac {a^2}{a \plus{} b} \plus{} \frac {b^2}{a \plus{} b} \plus{} \frac {b^2}{b \plus{} c} \plus{} \frac {c^2}{b \plus{} c} \plus{} \frac {c^2}{c \plus{} a} \plus{} \frac {a^2}{c \plus{} a}$
$ \geq \frac {4(a \plus{} b \plus{} c)^2}{4(a \plus{} b \plus{} c)} \equal{} a \plus{} b \plus{} c$ (From Titu's Lemma :lol: )
[/hide]

I have seen this problem somewhere hmmmmm :oops:  we can sole this by using titu,s lemma


function
inequalities
algebra
functional equation
algebra unsolved

Find all functions $\large f: R to R$ that satisfy the inequality:
$\large f(x+y)+f(y+z)+f(z+x) \geq  3f(x+2y+3z)$ for all x;y;z be real numbers.

I have little experience working with functional equations, but I can offer the following (rather limited) assault on the problem.

Let x = y = z. Then, we have

$f(2x) \ge f(6x)$

which is satisfied by all functions of the form

$f(x) = \frac{a}{bx^{n}+c}$ where a, b, and c are positive integers, and n is a positive even integer.

I have neither the time nor the necessary insight to find other functions or prove that all functions that satisfy the functional equation have the aforementioned form.

Given$y=z=0$we have$2f(x)+f(0)\geq 3f(x)\Rightarrow f(x)\leq f(0)$
Given$x=y=-z$we have$f(2x)+2f(0)\geq 3f(0)\Rightarrow f(x)\geq f(0)$
$\Rightarrow f(x)=f(0)\Rightarrow f(x)\equiv c \forall x \in R$


A very long conductor is turned so as to form an angle 56.There passes current I=30 A in the direction shown.Calculate the intensity of the magnetic field H in a point A, which is in the angle bisector and at a distance of 5cm from the angle vertex.

Hint: You do not need to take into account the components that form the angle, only the curved portion.

I've done it using H=1/(4pi*r)*(cosA-cosB) (applying for both curves of the conductor) and approximating some angles but I can't get the answer my prof did.

I can't completely understand where you want the point placed but if the point is at the vertex, then by Bio-Savart,
$ dB\equal{}\frac{\mu_{0}}{4\pi}\frac{I\cdot d\ell}{r^{2}}\equal{}\frac{\mu_{0}}{4\pi}\frac{I\cdot(x\ d\theta)}{x^{2}}\equal{}\frac{\mu_{0}I}{4\pi x}d\theta$
where $ x$ is the distance from the vertex to the curved part.  Integrating from $ \theta\equal{}0$ to $ \theta\equal{}\phi$ gives
$ B\equal{}\int dB\equal{}\frac{\mu_{0}I}{4\pi x}\int_{0}^{\phi}d\theta\equal{}\boxed{\frac{\mu_{0}I\phi}{4\pi x}}$.

Not sure if that is what you want.

We want H in a point which is located in the angle bisector and at the given distance from the angle vertex simultaneosly.The formula I used derives of course from biot-savart(calculates H for a conductor of certain length, i.e not very long, the angles at the formula are one acute and the other the obtuse) I guess you misunderstood it.

No, but it doesnt matter.The way is right.

I don't believe you said no, because of 1203  instead 1200.

Could you tell what is the intensity?


calculus
integration
vector
LaTeX
parameterization
real analysis
real analysis unsolved

Evaluate the integral over the curve $ C$ of $ F\cdot ds$ (what is the latex for that?), where

$ F \equal{} (e^{ \minus{} y} \minus{} ze^{ \minus{} x})i \plus{} (e^{ \minus{} z} \minus{} xe^{ \minus{} y})j \plus{} (e^{ \minus{} x} \minus{} ye^{ \minus{} z})k$, and

$ C \equal{} {\frac {ln(1 \plus{} s)}{ln2}, sin(\frac {\pi s}{2}), \frac {e^s \minus{} 1}{e \minus{} 1}}$, for $ 0 < s < 1$.

As a check that i did not typo, the curve goes from (0,0,0) to (1,1,1), and the second and third terms of F are the same as the first with the variables cycled.

I know that this problem is supposed to have an elementary solution.

(Edited to fix signs in F upon reading Fedja's observation)

[quote="Moderator"]
$ F \equal{} (e^{ \minus{} y} \plus{} ze^{ \minus{} x})i \plus{} (e^{ \minus{} z} \plus{} xe^{ \minus{} y})j \plus{} (e^{ \minus{} x} \plus{} ye^{ \minus{} z})k$
[/quote]
I'm pretty sure that it was meant to be
$ F \equal{} (e^{ \minus{} y} \minus{} ze^{ \minus{} x})i \plus{} (e^{ \minus{} z} \minus{} xe^{ \minus{} y})j \plus{} (e^{ \minus{} x} \minus{} ye^{ \minus{} z})k$
but somebody screwed it up when copying ;).

oops, you are right, but how did you know?

You're not going to do it by brute-force parametrization; integrating terms like $ \exp\left(\frac {e^s \minus{} 1}{e \minus{} 1}\right)$ is hopeless.
Now, if we'd like to avoid that, there's a tool- which we can use in the version with the $ \minus{}$ signs, but not with the $ \plus{}$ signs.

Shoot! :wallbash_red:  (kicks self)...

The tool is the fundamental theorem for line integrals. WOW! The first thing i did when i saw the problem was check to see if F was a gradient of  some f (which happens to be $ xe^{\minus{}y}\plus{}ye^{\minus{}z}\plus{}ze^{\minus{}x}$), but i wrote it wrong! ahhh!!!

Thanks to Jmerry and Fedja for helping me realize my mistake. With the minus signs this problem is nearly trivial!


Let triangle $ ABC$. $ M,N\in AC,BC$. $ L\in [MN]$. Let $ S_{ABC}\equal{}S, S_{AML}\equal{}P, S_{BNL}\equal{}Q$. Prove that
$ \sqrt[3]{S}\geq\sqrt[3]{P}\plus{}\sqrt[3]{Q}$
Have fun! :)

[quote="mufc"]Let triangle $ ABC$. $ M,N\in AC,BC$. $ L\in [MN]$. Let $ S_{ABC} \equal{} S, S_{AML} \equal{} P, S_{BNL} \equal{} Q$. Prove that
$ \sqrt [3]{S}\geq\sqrt [3]{P} \plus{} \sqrt [3]{Q}$
Have fun! :)[/quote]

$ \frac{CM}{MA}\equal{}n$,$ \frac{CN}{NB}\equal{}m$,$ \frac{ML}{LN}\equal{}k$

$ \frac{S_{MCL}}{S_{AML}}\equal{}\frac{CM}{MA}\equal{}n$  ---> $ S_{MCL}\equal{}nP$

$ \frac{S_{CLN}}{S_{NLB}}\equal{}\frac{CN}{NB}\equal{}m$  ---> $ S_{CLN}\equal{}mQ$

$ \frac{S_{MCL}}{S_{LCN}}\equal{}\frac{ML}{LN}\equal{}k$  ---> $ \frac{nP}{mQ}\equal{}k$

$ S_{MCN}\equal{}S_{CML}\plus{}S_{CLN}\equal{}nP\plus{}mQ$

$ \frac{S_{MCN}}{S_{ABC}}\equal{}\frac{MC*CN}{AC*CB}$ ---> $ \frac{nP\plus{}mQ}{S}\equal{}\frac{nm}{(n\plus{}1)(m\plus{}1)}$  ---->

$ \frac{(n\plus{}1)(m\plus{}1)}{m}*\frac{P}{S}\plus{}\frac{(n\plus{}1)(m\plus{}1)}{n}*\frac{Q}{S}\equal{}1$

$ \sqrt [3]{S}\geq\sqrt [3]{P} \plus{} \sqrt [3]{Q}$ <-> $ \sqrt[3]{\frac{P}{S}}\plus{}\sqrt[3]{\frac{Q}{S}}\leq 1$

$ \sqrt[3]{\frac{P}{S}}\plus{}\sqrt[3]{\frac{Q}{S}}\equal{}\sqrt[3]{\frac{m}{m\plus{}1}*\frac{1}{n\plus{}1}*\frac{(n\plus{}1)(m\plus{}1)}{m}*\frac{P}{S}}\plus{}\sqrt[3]{\frac{1}{m\plus{}1}*\frac{n}{n\plus{}1}*\frac{(n\plus{}1)(m\plus{}1)}{n}*\frac{Q}{S}}\leq(\frac{\frac{m}{m\plus{}1}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{(n\plus{}1)(m\plus{}1)}{m}*\frac{P}{S}}{3}\plus{}\frac{\frac{1}{m\plus{}1}\plus{}\frac{n}{n\plus{}1}\plus{}\frac{(n\plus{}1)(m\plus{}1)}{n}*\frac{Q}{S}}{3})\equal{}1$

$ \frac{1}{m\plus{}1}\equal{}\frac{n}{n\plus{}1}\equal{}\frac{(n\plus{}1)(m\plus{}1)}{n}*\frac{Q}{S}$   --> $ mn\equal{}1$

$ \frac{m}{m\plus{}1}\equal{}\frac{1}{n\plus{}1}\equal{}\frac{(n\plus{}1)(m\plus{}1)}{m}*\frac{P}{S}$ --->

$ P\equal{}\frac{mS}{(n\plus{}1)^{2}(m\plus{}1)}$

$ Q\equal{}\frac{n^{2}S}{(n\plus{}1)^{2}(m\plus{}1)}$   -----> $ k\equal{}\frac{nP}{mQ}\equal{}\frac{n*\frac{mS}{(n\plus{}1)^{2}(m\plus{}1)}}{m*\frac{n^{2}S}{(n\plus{}1)^{2}(m\plus{}1)}}\equal{}\frac{1}{n}\equal{}m$  --->

$ k\equal{}m\equal{}\frac{1}{n}$   --> $ \frac{CN}{NB}\equal{}\frac{AM}{CM}\equal{}\frac{ML}{LN}$  <-> $ \sqrt [3]{S}\equal{}\sqrt [3]{P} \plus{} \sqrt [3]{Q}$


analytic geometry
trigonometry
conics
parabola
function
calculus
derivative

A plane is inclined at an angle $\alpha$, and a projectile is to be fired from the base of the plane with velocity $v_0$.  At what angle $\theta$ should the projectile be fired to maximize its distance up the plane?

I tried to solve in an algebrical way but I obtain a negative angle....
[hide]
consider a system of coordinates with center in the point where the projectile starts. Now consider the plane as a line of equation
$r:$ $y=\tan \alpha \cdot x$. Now write the equation of the projectile in his parametric form

$x=v_0 \cos(\theta + \alpha)t$
$y=v_0 \sin(\theta + \alpha)t + \frac{1}{2}gt^2$

Now substitute  $t$ to get the equation in ordinary form

$y=\tan (\theta + \alpha)x + \frac{g}{2v_0^2 \cos^2(\theta + \alpha)}x^2$

what we want to find is the point of intersection between the line $y=\tan \alpha \cdot x$ and the parabola of the projectile in function of $\theta$.
So put in the system the 2 equations to obtain $\frac{g}{2v_0^2 \cos^2(\theta + \alpha)}x^2=(\tan\alpha-\tan(\theta + \alpha))x$. We can obviously divide by $x$, because we suppose it $\neq 0$, being $x=0$ the first intersection point between the line and the parabola. Rearrange and arrive at
$x(\theta)=\frac{2v_0^2}{g}( \tan\alpha \cdot \cos^2(\theta + \alpha)-\sin(\theta + \alpha)\cos(\theta + \alpha))$
Now we have to do derivative to find $x_{max}$

$x'(\theta)=\frac{2v_0^2}{g}(-2 \tan \alpha \cdot \cos(\theta + \alpha) \sin(\theta + \alpha)-cos^2(\theta + \alpha)+\sin^2(\theta + \alpha))$
after some boring calculus
$\theta=\frac{%Error. "atan" is a bad command.
\left(-\frac{1}{\tan \alpha}\right)}{2}-\alpha$
but it is wrong....

[/hide]

To get the object to go as far as possible up the plane, don't you just want to shoot it "straight up" the plane, so that it isn't going "sideways" at all?  Seems to me that any "sideways" component of that object's velocity would be wasted.

But maybe I'm misunderstanding the problem.

@shadysaysurspammed:I don't think the plane's rotating, but what other motion  of the plane would matter?
If you rotate the plane around back $\alpha$ degrees so it coincides with the horizontal and factor that shift into the equations accordingly, could you just use the projectile motion formulas to find an expression for distance travelled in terms of $\alpha$ and $v_0$?

No,I meant is it moving along the angle(in linear motion at an angle of alpha).Otherwise,this question is pretty straight forward.

Sorry for the confusion.  The plane is stationary, but inclined from the horizontal at an angle of $\alpha$.

Happy to meet someone from italy in this topic!

The standard method to solve this kind of problem is considering the motion along the plane, after having calculated the components of gravity along and perpendicularly to the plane.

$x=vcos\theta t-\frac{1}{2} gsen\alpha t^2$
$y=vsen\theta t-\frac{1}{2} gcos\alpha t^2$

When $y=0$ you get $t=\frac{2vsen\theta}{gcos\alpha}$

Substituting in $x(t)$ and solving $x'(\theta)=0$ you get $tg 2\theta=ctg\alpha$.


geometry
geometric transformation
invariant
inequalities
integration
topology
real analysis

This is a classic problem, although I can't seem to get a grip on it. Suppose $ A$ is measurable with positive measure. Show that $ A \minus{} A$ contains a neighborhood of $ 0$.

Prove this lemma first, it's useful:

Lemma: Suppose $ \delta\in (0,1)$ and $ E$ is a measurable set in $ \mathbb{R}^k$ of positive measure. Then there exists an open interval $ I$ so that $ \mu(E\cap I)>\delta \mu (I)$

Lemma: Let $ \epsilon > 0$ be given. Then if A is any set, then there is an open set O containing A such that $ m^{*}(O) \leq m^{*}(A) \plus{} \epsilon$.

Let $ I_{n}$ be a sequence of intervals covering A and suppose that $ m^{*}(A) \plus{} \frac {\epsilon}{2} \geq \sum_{n \equal{} 1}^{\infty} l\left(I_{n}\right)$.
If $ I_{n} \equal{} [a_{n}, b_{n})$, set $ J_{n} \equal{} \left(a_{n} \minus{} \frac {\epsilon}{2^{n \plus{} 1}}, b_{n} \right)$ so that $ A \subseteq \bigcup_{n \equal{} 1}^{\infty} J_{n}$.

Then we have $ A \subseteq O \equal{} \bigcup_{n \equal{} 1}^{\infty} J_{n}$ and $ m^{*}(O) \leq \sum_{n \equal{} 1}^{\infty} l\left(J_{n}\right) \leq m^{*}(A) \plus{} \epsilon$.

From the preceding lemma, we get $ m(O) \leq \alpha^{ \minus{} 1}m(E)$.

But $ O \equal{} \bigcup J_{n}$, is a union of disjoint open intervals.

So $ \alpha \sum l\left(J_{n}\right) < m(E) \leq \sum m\left(E \cap J_{n} \right)$.

But then $ \alpha m\left(J_{n}\right) \equal{} \alpha l\left(J_{n} \right) < m\left( E \cap J_{n} \right)$, for some n. This proves the lemma in Albanian Eagle's thread.

Think about what it means for a value $ x$ to NOT be in $ A\minus{}A$.  That's like saying that for every $ y$, the set $ A$ can contain at most one of $ y$ and $ x\plus{}y$.    If you look at an interval $ I$, you should be able to use this to imply some sort of upper bound on $ \frac{\mu(A \cap I)}{\mu(I)}$ (the smaller $ x$ is relative to the length of the interval, the better your bound).

On the other hand, we know from the suggested lemma that there must be intervals which are almost entirely contained in $ A$.   This would be a contradiction if $ x$ is too small.

Set $ A_{\delta} = A - A = \{x - y: x, y \in A \}$. Choose $ \lambda$ such that $ 2 - 2^{ - n} < 2\lambda < 1$.

By the lemma discussed above, there exists an open interval $ I$ such that $ \lambda \mu (I) < \mu (A \cap I)$.

Choose $ \delta$ to be the minimum of the lengths of the edges of $ I$, and set $ I': = \{x: |\xi_{i}| < \frac {\delta}{2}, i, ..., n\}$.

Now we want to show that $ I' \subseteq A_{\delta}$. Consider the sets $ B = A \cap I$ and $ C = \left(A \cap I\right) + x$. If for each $ x \in I'$, we have $ B \cap C \neq \emptyset$, then we are done [one may want to convince oneself here].

Now choose $ x \in I'$. By the choice of $ \delta$, $ I + x$ contains the center of $ I$.

We then have the following:

$ \mu[I \cap \left(I + x\right)] > 2^{ - n} \mu(I)$

$ \implies \mu[I \cup \left(I + x\right)] = \mu(I) + \mu(I + x) - \mu[I \cap \left(I + x \right)]$

$ < 2 \mu(I) - 2^{ - n} \mu(I) = \left(2 - 2^{ - n}\right) \mu(I) < 2 \lambda \mu(I)$ [*]

By the translation invariant property, $ \mu(B) = \mu(C) > \lambda \mu(I)$.

Hence, If B and C are disjoint, then we have:
$ 2\lambda \mu(I) < \mu(B \cup C) \leq \mu(I \cup (I + x))$ , the last inequality holds because $ B \cup C \subseteq I \cup (I + x)$.

By the [*], we have:

$ 2\lambda \mu(I) < \mu(B \cup C) \leq \mu(I \cup (I + x)) < 2 \lambda \mu(I)$, a contradiction. Hence $ B \cap C \neq \emptyset$.

Suppose $ A$ is a measureable set with $ m(A) > 0$.   We want to show that the set $ A \minus{} A$ contains an interval about the origin.

Let $ f(x) \equal{} \int 1_A(y\minus{}x)1_A(y) dy \equal{} 1_A * 1_A (x)$

Then $ f(x)$ is continuous and $ f(0) \equal{} m(A) > 0$ hence there exists $ \delta > 0$ such that  $ f(x) >0$ for $ x \in (\minus{}\delta, \delta)$

This means that $ 1_A(y\minus{}x) \equal{} 1$ for $ x \in (\minus{}\delta, \delta)$ which implies that $ y \minus{} x \in A$ for $ x \in (\minus{}\delta, \delta)$

or $ A \minus{} A$ contains an interval about the origin.


inequalities
real analysis
real analysis unsolved

If $n$ is a positive integer, let $Odd(n)$ be $\frac{n}{2^{k}}$ , where $2^{k}$ is the largest power of $2$ that divides $n$.Given two positive integers $a,b$ define a sequence by $x_{1}=a, x_{2}=b, x_{n+1}=Odd(x_{n-1}+x_{n})$.For which $a,b$ does the limit of the sequence exist?Can you describe the limit?

Since $x_{n+1}\le (x_{n-1}+x_{n})/2$ for $n\ge 4$, the sequence is bounded. Either there are two consecutive terms that are equal (and then the sequence stabilizes) or there is a cycle of length $\ge 2$. Let $A$ be the maximal element in the cycle: two terms preceding $A$ must also be equal to $A$ because of the above inequality, a contradiction.


For all x > 0  and y > 0, the radical expression $ \dfrac {\sqrt {x}}{3\sqrt {x} \minus{} \sqrt {y}}$ is equivalent to:

A. $ \dfrac {3x \minus{} \sqrt {xy}}{9x \plus{} y }$

B. $ \dfrac {3x \minus{} \sqrt {xy}}{3x \plus{} y }$

C. $ \dfrac {3x \plus{} \sqrt {xy}}{9x \minus{} y }$

D. $ \dfrac {3x \plus{} \sqrt {xy}}{3x \minus{} y }$

[hide="Solution"]First multiply by $ 3\sqrt {x} + \sqrt {y}$ on the denominator and numerator and simplify.

\begin{eqnarray*} \frac {\sqrt {x}}{3\sqrt {x} - \sqrt {y}} & = & \frac {\sqrt {x}}{3\sqrt {x} - \sqrt {y}}\cdot \dfrac{3\sqrt {x} + \sqrt {y}}{3\sqrt {x} + \sqrt {y}} \\
\frac {\sqrt {x}}{3\sqrt {x} - \sqrt {y}} & = & \frac {\sqrt {x}(3\sqrt {x} + \sqrt {y})}{9x - y} \\
\frac {\sqrt {x}}{3\sqrt {x} - \sqrt {y}} & = & \boxed{\frac {3x + \sqrt {xy}}{9x - y}} \\
\end{eqnarray*}

So the answer is $ \mathbf{(C)}$
[/hide]


geometry
3D geometry
sphere
calculus
function
integration
trigonometry

Hi all-

I have the following problem that I am trying to understand:

[b]Problem[/b]. Show that the force of attraction within a spherical shell of constant density is everywhere $ 0$.

My understanding of the statement is that, for example, in a gravitational field, the sphere would not 'cave in' on itself because the patches of the sphere are of constant density. I have the solution (below). The problem is presented in the context of advanced calculus (after talking about the implicit function theorem, surfaces and surface area). The integration is easy but I don't understand how the integral for the force is derived.

[b]Solution[/b]. Describe the shell by $ x \equal{} \sin{\phi}\cos{\theta},y \equal{} \sin{\theta}\sin{\phi},z \equal{} \cos{\phi},0\leq \phi\leq \pi, 0\leq\theta \leq 2 \pi$, and let $ P \equal{} (0,0,a)$ with $ 0\leq a \leq 1$. With $ \rho$=density (mass per unit area), the component of the force at $ P$ in the vertical direction is
\[ F \equal{} \minus{} \int_0^{2\pi}d\theta\int_0^{\pi}\frac {(\cos\theta \minus{} a)(\rho\sin\theta)}{(1 \plus{} a^2 \minus{} 2a\cos{\theta})^{3/2}} d\theta.
\]
(This may be integrated easily; for example, put $ u^2 \equal{} 1 \plus{} a^2 \minus{} 2a\cos\theta.$ One finds that $ F \equal{} 0.$)

Second, I'm interested as to why this is an advanced calculus problem. My guess is that it mathematically interesting insofar as it can be generalized to $ n \minus{}$ spheres and one must be careful with calculus. I would appreciate any help on this, since I have to give a presentation, and I have no idea what is going on!

Thanks in advance,

Los

Mhmm advanced calculus is a sketchy term. In Waterloo, Engineer's call Vector Calculus by that name but those who take "vector calculus" in the math faculty merely take Calculus 4: Vector Calculus. Also it is gone into more detail than the engineer's. So why it is called advanced calculus is not really something which could be answered I suppose. A stab at it though is it is a course taken after the normal 3 calculus' are taken and requires knowledge from each of them.

The nice way to do it is the divergence theorem.
The divergence of an inverse-square field in $ \mathbb{R}^3$ is a multiple of the density, so it is uniformly zero inside the sphere. If $ r$ is less than the radius $ R$ of the full sphere, the flux through a sphere of radius $ r$ is zero by the divergence theorem. By symmetry, the field itself is zero on the radius-$ r$ sphere.

Although the idea works in other dimensions, it's not an inverse square field in those cases.

The problem is closely related to Newton's Theorem, which states that the force of attraction felt by a particle *outside* the sphere is the same as if all of the mass of the sphere were concentrated at the center.  This can be proved with painful integration, or with a clever application of the divergence theorem.  Newton's Theorem was essential for establishing the basis for Newton's theory of gravity, since stars and planets are obviously not point-masses.

If you have to make a full-blown presentation out of it, I would recommend first doing your problem, using both multiple integration and the divergence theorem.  Then do Newton's Theorem, again using both methods and discussing the historical significance of the problem (which will requires a bit of research).  That's probably enough for a fairly lengthy presentation, so you can prune it down as needed.


A] On a 9x9 grid, we paint each square white or black. What is the maximum number of squares we can paint white without having a rectangle, sides parallel to the board, with 4 corners all white

B] Can we paint the grid without having a rectangle, sides parallel to the board, with 4 corners all white or 4 corners all black??

C]  Numbers from 1 to 100 are writen on board, the each is paint with one this colors: red, blue, orange, and grey. Can we always find two numbers painted the same color as their diference?

D]If we paint 25 numbers of each color, Can we always find two numbers painted the same color as their sum?

 :)

(edited, thanks MellowMelon and JBL)

A and B don't make sense.  Edit, perhaps?

This is a long shot as to what the problem was supposed to say

On a 9x9 grid, we paint each square white or black. What is the maximum number of squares we can paint white without having a rectangle, sides parallel to the board, with 4 corners all white, i.e. we can't pick two rows and two columns such that their four intersections are all white.

This interpretation makes the situation in B completely impossible, however, so I probably got something wrong.


algebra
polynomial
integration
absolute value
real analysis
real analysis unsolved

hello, are you sure that this sum exists?
Sonnhard.

Sonnhard, can't you please try to make contributions that have some content?  It's been so long since I've read a post of yours that had any actual ideas in it.

No, the sum doesn't exist, for the silly reason that $ 4^j \geq \binom{2j}{j}$.  The reciprocal sum also doesn't exist, for the less silly reason that $ 4^j \equal{} O\left(\binom{2j}{j} \sqrt {j}\right)$.  If we replace "4" in the given problem by anything of smaller absolute value then there is something to be done: the series $ \sum \frac {x^n}{\binom{2n}{n}}$ has some not-too-horrible form involving an arcsine, which undoubtedly has appeared on the forums before.

Sorry, here should be $ \sum_{j\equal{}0}^{n}\frac{4^{j}}{\binom{2j}{j}}$ instead of $ \sum_{j\equal{}0}^{\plus{}\infty}\frac{4^{j}}{\binom{2j}{j}}$.

[quote="Yuriy Solovyov"]Sorry, here should be $ \sum_{j \equal{} 0}^{n}\frac {4^{j}}{\binom{2j}{j}}$ instead of $ \sum_{j \equal{} 0}^{ \plus{} \infty}\frac {4^{j}}{\binom{2j}{j}}$.[/quote]

The book "A=B" from Petkovšek, Wilf and Zeilberger says that certain hypergeometric sums (which is the case of the investigated sum in this topic) in the form $ s(n): \equal{}\sum_{j\equal{}1}^{n}a_j$ can be written in the form

\[ s(n)\equal{}p(n)\cdot a_n\plus{}c,\]

where $ p(n)\in\mathbb{R}[n]$ is a polynomial and $ c\in\mathbb{R}$ is a constant. Hence, this problem can be reduced to find the polynomial $ p(n)$ such that

\[ s(n)\equal{}p(n)\cdot\frac{4^n}{{2n\choose n}}\plus{}c.\]

Moreover, we can prove that in this case $ \deg\, p(n)\equal{}1$. The rest should be obvious.

$ \sum_{j \equal{} 0}^{ \plus{} \infty}\frac {4^j}{\binom{2j}{j}}\equal{}\sum_{j\equal{}0}^{\plus{}\infty} j\cdot 4^j\int_0^1 (1\minus{}x)^jx^{j\minus{}1}dx$.


farz konid $M$ bishtarin faseleie beine 6 noghteie motamaiez  darsafhe bashad va m niz kamtarin faseleie beine in noghatsabet konid:

 $M.m\geq \sqrt 3$

manzooret ine?$M.m\geq \sqrt{3}$

agar are ke soal ghalate.

[quote="saeid"]manzooret ine?$M.m\geq \sqrt{3}$

salam.
age geq bozorgtar mosavi mishe va sqrt radikal,manzooram ine.
vali fekr nakonam ghalat bashe.
mishe begin chera eshtebahe?

khob mitonim 6 ta noghte ro tori(nazdik be ham) dar nazar begirim ke $M<\sqrt[4]{ 3} \ ,\ m<\sqrt[4]{3}$ ;)  ;)

vaseye Quote bayad benevisi:
[code][quote]matn[/quote][/code]


MATHCOUNTS
trigonometry
geometry
inradius
circumcircle
angle bisector

What is the height of a regular pentagon with side length 1?

Oops! Yes, it is regular.

That makes it much harder :( If it was a house shape, then the problem would definitely be MathCounts level.

Answer should be in simplest radical form or rounded to the nearest hundredth.

All problems except the pentagon height one are solved.

[quote="anirudh"]Is the pentagon height

[hide]
$\sqrt{2}$? :maybe: [/hide][/quote]
Nope.

[quote="anirudh"]What is teh flaw in my reasoning

I created a pentagon, and the height was $2\cdot\frac{\sqrt{2}}{2}$ :maybe:  :(  :(  :maybe:

[/quote]
I don't know. Post your solution.

[quote="anirudh"]I have attached my reasoning[/quote]
How did you get that the two lengths you labeled were equal to $\frac{\sqrt{2}}{2}$? Neither of them actually are, unfortunately.  :wink:
In fact, I don't believe the problem can be solved without using trigonometry, so it isn't at the level of MathCounts.  :wink:  I believe the answer is
[hide]$\frac{\tan{54^\circ}+\sec{54^\circ}}{2}$.[/hide]
[hide="Hint"]If $r$ is the length of the inradius, and $R$ is the length of the circumradius, the height has length $r+R$.[/hide]
Since this isn't a MathCounts level problem, I'll move it to High School Basics.

[quote="i_like_pie"][quote="nebula42"]In fact, I don't believe the problem can be solved without using trigonometry, so it isn't at the level of MathCounts.  :wink:[/quote]
It CAN be solved without trig functions. That's a condtition of the problem![/quote]
If you are sure that it can, could you please post your solution without trig? (No one has been able to solve it for quite a while already, so you may as well show the solution now.)  :wink:
Also, I'm moving this to High School Basics.

nvm anirudh, i dont get either part of ur drawing

[quote="i_like_pie"]Let $x$ be the height of the pentagon and $a$ be the length of the extensions of the sides, as shown in the first diagram at the botom.

Using the Pythagorean formula we know $x^{2}+\left(a+\frac{1}{2}\right)^{2}= (a+1)^{2},$ or $x^{2}= a+\frac{3}{4}.$

Now consider the added two chords in the second diagram.

The isosceles triangle formed by these two chords and one side of the pentagon is congruent to the triangle formed by two extensions of length $a$ and the same side of the pentagon.

Thus the distance between two non-adjacent corners of the pentagon is also $a.$ Consider the chord going from either corner of the base to the top of the pentagon. We can use the Pythagorean formula, with this chord as the hypotenuse, to obtain:

$x^{2}+\left(\frac{1}{2}\right)^{2}= a^{2}$

Substituting for $x^{2}$ from above, we can solve for $a:$

$a+\frac{3}{4}= a^{2}-\left(\frac{1}{2}\right)^{2}$

$a^{2}-a-1 = 0$

$a=\frac{(1+\sqrt{5})}{2}$

Now that we know $a,$ we can solve for $x:$

$x^{2}= \left(\frac{1+\sqrt{5}}{2}\right)^{2}-\frac{1}{4}$

$x2 = \frac{5+2\sqrt{5}}{4}$

$x = \frac{\sqrt{5+2\sqrt{5}}}{2}\approx 1.5388$[/quote]
Ah. Very nice solution.  :)  :coolspeak:
This problem [i]is[/i] above the level of MathCounts (it would definitely take more than 3-4 minutes to solve this problem), but it is still a very nice problem.  :lol:

Also, I'd just like to know: what program did you use to create the diagrams? They're quite good.  :)

[hide="Hint"]If the pentagon is $ABCDE$, consider $\triangle ABD$. Find its angles, than draw the angle bisector of $\angle DAB$ and draw as much information from there as you can. Keyword: Similarity.[/hide]

[quote="nebula42"]Ah. Very nice solution.  :)  :coolspeak:
This problem [i]is[/i] above the level of MathCounts (it would definitely take more than 3-4 minutes to solve this problem), but it is still a very nice problem.  :lol:

Also, I'd just like to know: what program did you use to create the diagrams? They're quite good.  :)[/quote]
I found the problem on some site. I got the solution by myself, although it took me 10 minutes. :ninja:
I used the site's diagrams, but I think I could make something similar in Microsoft Visio.

Here is the trigonometric solution (not mine):

With some simple trig, we can express the height as $\frac{1}{2}\cdot\left(\frac{1}{\sin 36}+\frac{1}{\tan 36}\right).$

We can express $\tan 36$ as $\frac{\sin 36}{\sqrt{1-\sin^{2}36}}.$

After a few simple steps, we can express the height as $\frac{1}{2}\cdot\frac{1+\sqrt{1-\sin^{2}36}}{2\cdot\sin 36}.$

Next lets work on expressing $\sin 36.$

$\sin 180 = 16\cdot\sin^{5}36-20\cdot\sin^{3}36+5\cdot\sin36$

$16\cdot\sin^{4}36-20\cdot\sin^{2}36+5$

$16x^{4}-20x^{2}+5,$ where $x=\sin 36$

$16y^{2}-20y+5,$ where $y=x^{2}$

$y = \frac{5-\sqrt{5}}{8}$

$x = \sqrt{\frac{5-\sqrt{5}}{8}}= \sin 36$

Now put this expression for $\sin 36$ into the first equation:

$\frac{1}{2}\cdot\left(\frac{1+\sqrt{1-\frac{5-\sqrt{5}}{8}}}{\sqrt{\frac{5-\sqrt{5}}{8}}}\right)$

1/2 * [ (1+sqr(3+sqr(5))/8)) / sqr((5-sqr(5))/8) ] =

1/2 * [ sqr(8/(5-sqr(5))) + sqr((3+sqr(5))/(5-sqr(5))) ] =

1/2 * [ (sqr(8) + sqr((3+sqr(5)))/(5-sqr(5))) ] =~ 1.5389

Last few parts - I'm too lazy to LaTeX.


Find functions f and g so that f(g(x)) = H.

H(x) = (2x + 3)^4

[quote="blueshark"]Chess64,

Can you break the question down a little more?[/quote]

Um, why? :?

How about $g(x)=(2x+3)^2$, $f(x)=x^2$?

I'm still a bid confused how you got f(x) = x^4 and f(x) = x^2.
Of course, both answers are acceptable but what are the steps to finding x^4 and x^2 for f(x)?  Thank you for your patience.

basically what you do is insert parathensis into the equation so...
$(2x+3)^{4}$
$(2x+3)^{2*2}$
$((2x+3)^{2})^{2}$

I totally get it now.  Thank you all for your great help.


function
inequalities proposed
inequalities

Let $a;b;c$ be positive real numbers satisfying $1 \leq a;b;c \leq 2$.Prove that:
$a^{3}+b^{3}+c^{3}\leq 5abc$

Use convex function! That is all!  :wink:

If $f$ is convex on $[a,b]$ then $f(x)\leq\max{(f(a),f(b))}\;\forall x\in [a,b]$.  :wink:

See Darij's post [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=109418#109418]here[/url].


geometry
incenter
Euler
circumcircle
angle bisector
perpendicular bisector
congruent triangles

Prove that the incenter lies on the Euler line if and only if the triangle is isosceles.

Is the proof too hard or is it too tedious to write out?

[hide]1. If the triangle is isosceles, the incenter is on the Euler line.
Proof:
Let's say we have isosceles $\triangle ABC$, $\angle A\cong\angle B$.  If we draw the angle bisector of $\angle C$, it will be the perpendicular bisector of $\overline{AB}$ (let's say they intersect at $D$.  This is so because we create two congruent triangles: $\triangle ADC$ and $\triangle BDC$.
It is clear that $\overline{CD}$ is an altitude and a perpendicular bisector of $\triangle ABC$.  So the orthocenter and circumcenter both lie on $\overline{CD}$, so clearly, the Euler line is on $\overline{CD}$ as well.
But since $\overline{CD}$ is also an angle bisector, the incenter lies on that line, as well.

2. If the incenter is on the Euler line, the triangle is isosceles.
This one I am having trouble with.[/hide]


limit
algebra
function
domain
trigonometry
calculus
calculus computations

1.If $ f(x) \equal{} \lim_{n\to \infty}\frac {x(sinx)^n}{(sinx)^n \plus{} 1}$

Find domain and range of $ f(x)$,where $ (n\in N)$

2.Let f(x) be a bounded function.If $ L_{1}\equal{}\lim_{x\to \infty}(f'(x)\minus{}kf(x))$ and $ L_{2}\equal{}\lim_{x\to \infty}f(x)$  where $ k>0$.

   If $ L_{1},L_{2}$ both exist ,prove that $ L_{2}\equal{}\frac{\minus{}L_{1}}{k}$.

3.Using the sandwich theorem,prove that $ \lim_{n\to\infty}\frac{n!}{n^n}\equal{}0$.

[quote="pankajsinha"]1.If $ f(x) \equal{} \lim_{n\to \infty}\frac {x(sinx)^n}{(sinx)^n \plus{} 1}$

Find domain and range of $ f(x)$,where $ (n\in N)$

2.Let f(x) be a bounded function.If $ L_{1} \equal{} \lim_{x\to \infty}(f'(x) \minus{} kf(x))$ and $ L_{2} \equal{} \lim_{x\to \infty}f(x)$  where $ k > 0$.

   If $ L_{1},L_{2}$ both exist ,prove that $ L_{2} \equal{} \frac { \minus{} L_{1}}{k}$.

3.Using the sandwich theorem,prove that $ \lim_{n\to\infty}\frac {n!}{n^n} \equal{} 0$.[/quote]

[b]1)[/b] $ f(x)\equal{}\lim_{n\to \infty}\frac {x(sinx)^n}{(sinx)^n \plus{} 1}\equal{}x\minus{}x \lim_{n\to \infty}\frac {1}{(sinx)^n \plus{} 1}$

[b]For $ |\sin x| \neq 1$[/b], $ \lim_{n\to \infty} (\sin x)^n\equal{}0$, so in this case $ f(x)\equal{}x\minus{}x\equal{}0$

[b]For $ \sin x\equal{}\minus{}1$[/b], the limit doesn't exist.

[b]For $ \sin x\equal{}1$[/b], $ f(x)\equal{}x\minus{}x(\frac{1}{2})\equal{}\frac{x}{2}$

$ \sin x\equal{}\minus{}1 \Leftrightarrow x\equal{}2n \pi \plus{}\frac{3 \pi}{2}$, where $ n$ is any integer.

So the domain is $ R$ \ $ \{2n \pi \plus{}\frac{3 \pi}{2} \}$,   the range is $ \{0, n \pi \plus{}\frac{\pi}{4} \}$, where $ n$ is any integer.

[b]2) [/b] $ \lim_{n\to \infty} f(x)$ exists $ \Rightarrow \lim_{n\to \infty} f'(x)\equal{}0$

Now as $ L_1,L_2$ both exists, we have $ k L_2\plus{}L_1\equal{}\lim_{n\to \infty} f'(x)\equal{}0$, which gives the desired result.

[b]3) [/b] It's not difficult to prove that for $ n \ge 2$, $ n^{n\minus{}1} \ge n! \ge n$

So $ \lim_{n\to \infty} \frac{n^{n\minus{}1}}{n^n}\equal{}0$ and $ \lim_{n\to \infty} \frac{n}{n^n}\equal{}0$ gives the desired result by sandwich theorem.


Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a function so that: \[ f(1)=1 \]
and \[ \forall a,b \in \mathbb{R} : f(a+b) = f(a)+f(b) \]
and \[ \forall x \in \mathbb{R}_0 : f(x).f(\frac{1}{x})=1 \]
Prove that $f(x)=x$,  $\forall x \in \mathbb{R}$

We have obviously $f(q)=q$ for all $q\in \mathbb{Q}$... now how to go to the reals from there?

That's the tricky part...  :D
If you can prove that the function is strictly increasing, you are done. To prove this, here's a [hide="hint"]$\frac{2}{1-x^2}=\frac{1}{1-x}+\frac{1}{1+x}$[/hide]

You could also prove that the function is continuous.

[quote="Jan"]That's the tricky part...

You could also prove that the function is continuous.[/quote]
I am very interesting to see your proof of continuity
We have:
 $\forall x\in \mathbb{R}$      $\forall z_{n}\in\mathbb{Q}$    such that     $\lim_{n\to\infty}z_{n}=0$         $\lim_{n\to\infty}f(x+z_{n})=f(x)$
But how to prove :
 $\forall x\in \mathbb{R}$      $\forall z_{n}\in\mathbb{R}$     such that     $\lim_{n\to\infty}z_{n}=0$         $\lim_{n\to\infty}f(x+z_{n})=f(x)$

I don't have a proof for continuity, I'm just saying that proving it is sufficient to finish the question. In fact I'm also quite curious to see a proof for continuity...  :)

Well, by proving $f(x)=x$ you have proven continuity eh? ;)

I tried continuity as well but couldn't find it. The idea to prove it's increasing is rather nice.

It is not true. There are many automorfisms field $R(Q)\to R(Q)$.

But none who suffice $\forall x \in \mathbb{R}_0 : f(x).f(\frac{1}{x})=1$.

For example $f: Q(\sqrt 2 )\to Q(\sqrt 2 ), f(a+b\sqrt 2 )=a-b\sqrt 2 ,a\in Q, b\in Q.$
We have $f(x)f(\frac 1x )=1$

But that's not a $f: \mathbb{R}\rightarrow\mathbb{R}$, right? The point is exactly that the only automorphism on $\mathbb{R}$ is the identical one? :?

I think you're making this much harder than it really is. Try this:

We have $f(a+b)=f(a)+f(b)$ and $f(1)=1$. If $a=b$, then $f(2a)=2f(a)$.
Similarly, if $b=k\cdot a$, then $f((k+1)b)=(k+1)f(b)$ If we now let $b=1$, then $f(k+1)=k+1$. Let $k+1=r$. Then $f(r)=r$ for all reals.

this is because we originally define $a,b\in\mathbb{R}$, so $k\in\mathbb{R}$, so $k+1\in\mathbb{R}$, and finally $r\in\mathbb{R}$

Note that it is also true for $r\in\mathbb{C}$.

The AoPS book (volume 2) has a whole section on this type of thing.

Lol... sorry that's completely wrong. You have proven it for $n\in\mathbb{Z}$ at most. If the AoPS book has a section on it, I suggest you to read it again.

[quote="Rust"]Let $\{\sqrt D,...\}$ (D is not square of rational number) is Hamel's basis in R(Q), and $f(ax+b\sqrt D )=ax-b\sqrt D,a,b\in Q .$ Tis function satisfyed your condition.[/quote]
Take   $a=0$   &   $b>0$   then   $f(b\sqrt D )=-b\sqrt D<0$  :(

It's more interesting to transform the problem like this:

[quote="Jan"]Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a function so that: \[ \forall a,b \in \mathbb{R} : f(a+b) = f(a)+f(b)  \]
and \[ \forall x \in \mathbb{R}_0 : f(x).f(\frac{1}{x})=1  \]
Prove that $f(x)=x$,  $\forall x \in \mathbb{R}$[/quote]

[quote="Peter VDD"]That's false. $f(x)=-x$?[/quote]
You are right :blush:

We can even conclude that $f(x)=\pm x$, so it is an obvious choice of the problem creator to state $f(1)=1$. ;)

[quote="Peter VDD"]We can even conclude that $f(x)=\pm x$,  ;)[/quote]
Yes as $f(x^2)=(signf(1))f(x)^2$    &   $f(1)^2=1$

[quote="hlg"]Note that f(x) is an automorphism, and the only automorphism is the identity function.
QED[/quote]
Why is it an automorphism¿ Ans why are those the identity¿
This is more or less what you have to show... ;)

Suppose that for some $n \in \mathbb{N}$, $f(n) = n$.
Then, $f(n+1) = f(n)+f(1) = n+1$.
Thus, from our base case $n = 1$, $f(n) = n$, $\forall n \in \mathbb{N}$

Next, for any $x \in \mathbb{R}$, $f(x) = f(x+0) = f(x)+f(0)$.
Therefore, $f(0) = 0$.

Next, $\forall n \in \mathbb{N}$, $f(n-n) = f(n+(-n)) = f(n)+f(-n).$
But, $f(n-n) = f(0) = 0$.
Therefore, $f(n)+f(-n) = 0$, and therefore $f(-n) =-f(n) =-n$.
Thus, $f(n) = n$, $\forall n \in \mathbb{Z}$.

Now, $\forall m \in \mathbb{Z}$, $f(m)*f(\frac{1}{m}) = m*f(\frac{1}{m}) = 1$.
Therefore, $f(\frac{1}{m}) = \frac{1}{m}$.

Then, suppose for some $n \in \mathbb{Z}$, that $\forall m \in \mathbb{Z}$, $f(\frac{n}{m}) = \frac{n}{m}$.

Then, $f(\frac{n+1}{m}) = f(\frac{n}{m})+f(\frac{1}{m}) = \frac{n}{m}+\frac{1}{m}= \frac{n+1}{m}$.
And, $(\frac{n-1}{m}) = f(\frac{n}{m})+f(\frac{1}{-m}) = \frac{n}{m}+\frac{-1}{m}= \frac{n-1}{m}$.

Now, from our base case $n = 0$, $f(\frac{0}{m}) = f(0) = 0 = \frac{0}{m}$.
Therefore, $f(x) = x$, $\forall x \in \mathbb{Q}$.

Now, it is known that $\forall r \in \mathbb{R}$, such that r is irrational, $\exists$ a sequence $r_{n}$ of rationals converging to $r$.

Therefore, $f(r) = f(\lim_{n \to  \inf}r_{n}) = \lim_{n \to  \inf}f(r_{n}) = \lim_{n \to  \inf}r_{n}= r$, since $r_{n}\in \mathbb{Q}$.

Therefore, $f(x) = x$, $\forall x \in \mathbb{R}$.


induction
quadratics
arithmetic sequence

Extension: What is the general formula for these kinds of problems? Can you derive it? (Hint: Arithmetic Sequences!) Try proving it using induction.

The formula for the sum of the first $ n$ integers is

$ \frac{n(n\plus{}1)}{2}$.

Proving it might not be MOEMS level, and I'm quite to lazy...so...

[hide="here it is"]
nope
[/hide]

Continuing, the formula for the sum of an arithmetic sequence is:

$ a_n = a_1 + d(n - 1)$, where $ a_n$ is the $ n$th term of the sequence, $ a_1$ is the first term, and $ d$ is the common difference.

What does that weird a mean?

Isn't it $ a_n \equal{} a_1 \plus{} d(n \minus{} 1)$?

EDIT: Oh whoops, sorry xpmath.

If you look at his code, it's the same thing that you just typed. Probably some sort of error.

You could also use Arithmetic Sequences to come up with a formula, though proving it with induction probably belongs in the HSB forum.  :)

If you're curious though...

[hide="Hint: Warning - Not MOEMS Level"]

First show it works for 1. Then show it works for $ k \plus{} 1$.

[/hide]

Or to derive the formula, you can make a table and find the first few thingies.

1: 1
2: 3
3: 6
4: 10
5: 15

Then we can use quadratic sequences thingie to derive it.  ( I don't know what it is called, so i use thingie).  But since we have two primes in a row, we double it and get;

1: 2
2: 6
3: 12
.
.
.
.

2: 1*2
6: 2*3
12: 3*4

so we have $ \frac{n(n\plus{}1)}{2}$.  My explanation is weird, I didn't want to go to the time doing every single step.  And my proof is probably not MOEMS level.

INDUCTION:
[hide]
We are to prove:
$ \frac {N(N \plus{} 1)}{2}$

For $ N \equal{} 1$
$ \frac {1(2)}{2} \equal{} 1$ TRUE

Let $ N \equal{} K$ be true.
If $ N \equal{} K \plus{} 1$ is true, it is proven.

$ \frac {k(k \plus{} 1)}{2} \plus{} (k \plus{} 1)$

$ \equal{} \frac {k^2 \plus{} k \plus{} 2k \plus{} 2}{2}$

$ \equal{} \frac {k^2 \plus{} 3k \plus{} 2}{2}$

$ \equal{} \frac {(k \plus{} 1)(k \plus{} 2)}{2}$ which was to be proven.

QED[/hide]

The answer has already been posted, and this is over a month old.  Please don't revive old topics.


number theory
prime factorization
greatest common divisor
least common multiple

What is the sum of the greatest common divisor of 30 and 81 and the least common multiple of 36 and 12?

First, let's find the greatest common divisor of 30 and 81. Since the prime factorization of 30 is $ 2 \cdot 3 \cdot 5$ and the factorization of 81 is $ 3^4$, our answer is $ 3$.

The least common multiple of 36 and 12 is simply 36. The factorization of 36 is $ 2^2 \cdot 3^2$ and the factorization of 12 is $ 2^2 \cdot 3$, so it's $ 2^2 \cdot 3^2 \equal{} 36$.

Therefore, our final answer is $ 3 \plus{} 36 \equal{} \boxed{39}$.


[quote="rcv"]The product of the 9 factors $\Big(1-\frac12\Big)\Big(1-\frac13\Big)\Big(1-\frac14\Big)\cdots\Big(1-\frac{1}{10}\Big)=$

$A)\quad \frac{1}{10} \qquad B)\quad \frac{1}{9} \qquad C)\quad \frac{1}{2} \qquad D)\quad \frac{10}{11} \qquad E)\quad \frac{11}{2}$[/quote]

[hide]$\Big(1-\frac12\Big)\Big(1-\frac13\Big)\Big(1-\frac14\Big)\cdots\Big(1-\frac{1}{10}\Big)=$
$\displaystyle\left(\frac{1}{2}\right)\displaystyle\left(\frac{2}{3}\right)\displaystyle\left(\frac{3}{4}\right)...\displaystyle\left(\frac{9}{10}\right)=$
(A)$\boxed{\frac{1}{10}}$[/hide]

[hide]well, it is $1/2*2/3...9/10$. Of course, the numerators and denominators keep cancelling each other out, so A. 1/10[/hide]

[hide]The denominator cancels out the numerator in front of it, so all there is left in the numerator is 1 and only 10 in the denominators. It's 1/10.[/hide]

[hide]Most of the denominators and numerators cancel out and you're left with the numerator 1 and the denominator 10. The answer is A, 1/10[/hide]

[quote="biffanddoc"][hide]
they all cancel except the 1 and 9
$\frac{1}{9}$[/hide][/quote]
[hide]
It's $\frac12 \times \frac23 \times \frac34 \times \frac45 \times ... \times \frac89 \times \frac9{10}$.

So, everything cancels out except for the 1 and 10.  ;)[/hide]

well  :oops:
[hide]
All cancel except the 1 and 10
$\frac{1}{10}$
[/hide]
is that better :)


Due to inclement weather, many teams will not be able to make it. However, according to the webiste, there is indeed another option:

"If you can't make it, we will likely allow you to take the individual and team tests remotely, proctored by the person listed as your proctor on the registration form. E-mail hmmt-request@mit.edu with your school name and a number we can call after 5pm, and we will work something out on a case by case basis. In every case we will ask you to scan or fax your answer sheets to us on the same schedule that the contest is running in Cambridge, so plan on taking the tests from 9am-1pm EST."

thank you so much that is very helpfull!!!

So uhhh, what was the point of flying all the way to boston and spending like $300 on hotels and whatnot if we could've just taken it from home?

guts round

the guts round is like 2/3 of the fun, man

Walking around boston is a lot of the remaining. It was really nice of them to let us take it remotely, though. Glad it at least partially worked out. Unfortunately, we wanted to take the guts round unofficially but they didn't send out the problems in time. Oh well.


ceiling function
floor function
logarithms
function
calculus
integration

True or false? Let $ n$ be a positive integer. Then

$ \left\lceil{\frac{2}{2^{\frac{1}{n}}-1}\right\rceil = \left\lfloor{\frac{2n}{\ln{2}}\right\rfloor}}$.

Plugging in $ n=1$ doesn't give very good results...

Considering you have a natural log there, I feel inclined towards calculus.  That is, if we define the function $ f(x)=2^{x}$, $ \frac{2}{\ln{2}}$ is just the integral of this function from 1 to 2, while $ \frac{1}{2^{\frac{1}{n}}-1}=\frac{1+2^{\frac{1}{n}}+2^{\frac{2}{n}}+...+2^{\frac{n-1}{n}}}{n}$ is the left-hand approximation of this integral.  There's probably a way to show there's exactly one integral multiple of $ \frac{1}{n}$ between those two by determining the accuracy of approximation and then actually calculating $ \frac{2}{\ln{2}}$, but do you have a nicer solution?

The problem seems delicate to me.  By using Taylor series, we see that
\[ \frac{2}{2^{1/n}-1}= \frac{2n}{\ln 2}-1+\frac{\ln 2}{6n}+O(\frac{1}{n^{2}}) \, . \]
Thus the question boils down to whether $ \frac{2n}{\ln 2}$ can come within $ \frac{\ln 2}{6n}$ or so of an integer.

On the one hand, I checked that the problem is true for the first 10,000 values of $ n$.  That suggests that the result is true in general.  On the other hand, if the result is true, that would show (I think) that the irrationality measure of $ \ln 2$ is $ 2$, which is an open problem.

For more info on irrationality measure, see
[url]http://mathworld.wolfram.com/IrrationalityMeasure.html[/url].


geometry
incenter
circumcircle
geometric transformation
homothety
trigonometry
geometry unsolved

$ABC$ be a triangle. Its incircle touches the sides $CB, AC, AB$ respectively at $N_{A},N_{B},N_{C}$. The orthic triangle of $ABC$ is $H_{A}H_{B}H_{C}$ with $H_{A}, H_{B}, H_{C}$ are respectively on $BC, AC, AB$. The incenter of $AH_{C}H_{B}$ is $I_{A}$; $I_{B}$ and $I_{C}$ were defined similarly.
Prove that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ has all sides equal.

[quote="anonymous1173"]$ABC$ be a triangle. Its incircle touches the sides $CB, AC, AB$ respectively at $N_{A},N_{B},N_{C}$. The orthic triangle of $ABC$ is $H_{A}H_{B}H_{C}$ with $H_{A}, H_{B}, H_{C}$ are respectively on $BC, AC, AB$. The incenter of $AH_{C}H_{B}$ is $I_{A}$; $I_{B}$ and $I_{C}$ were defined similarly.
Prove that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ is regular.[/quote]

In general, it is not regular. It is just symmetric with respect to a point.

  Darij

Here is my solution. I think it's very nice.

First, denote $I$ the incenter of $\triangle ABC$.
Second, let $D_{A}$ be the foot of the bisector of $\angle A$. We define $D_{B}$ and $D_{C}$ similarly.
Finaly, let $r$ be the radius of the incircle of $\triangle ABC$.

[color=blue]Lemma 1.[/color]
Point symmetric to $H_{C}$ with respect to line $N_{B}N_{C}$ lies on line $CI$. Similarly for $H_{A}$ and $H_{B}$.
[color=blue]Proof.[/color]
Let $P$ be the projection of $B$ onto $CI$ and let $BP$ intersect $AC$ in $Q$. Obviously,

$\angle AQB=180^{\circ}-\frac{\angle A}{2}-\frac{\angle B}{2}=\angle AIB.$

So $I$ lies on circumcircle of $\triangle ABQ$, hence line $N_{B}N_{C}$ is Simson line of $\triangle ABQ$ and point $I$, therefore $P$ lies on $N_{B}N_{C}$. As

$\angle BH_{C}C=\angle BPC=90^{\circ}$

quadrilateral $BCPH_{C}$ is inscribed. So

$\angle N_{C}PH_{C}=90^{\circ}-\angle D_{C}PN_{C}-\angle H_{C}PB=$

$=90^{\circ}-\frac{\angle B}{2}-\angle H_{C}CB=\frac{\angle B}{2}=\angle D_{C}PN_{C}.$

Hence, line $PH_{C}$ is symmetric to line $CI$ with respect to $N_{B}N_{C}$. Thus the Lemma is proved.

[color=blue]Solution.[/color]
Let $\triangle A'H'_{B}H'_{C}$ be symmetric to $\triangle AH_{B}H_{C}$ with respect to line $N_{B}N_{C}$. Obviously point $A'$ lies on line $AI$. From Lemma 1. $H_{B}$ lies on $BI$ and $H_{C}$ lies on $CI$.

As $\triangle AH_{B}H_{C}\sim \triangle ABC$ we have $\triangle A'H'_{B}H'_{C}\sim \triangle ABC.$

Moreover, it is easy to verify that

$A'H'_{B}\parallel AB,$

$H'_{B}H'_{C}\parallel BC,$

$H'_{C}A\parallel CA.$

Hence, There is a homothety with center $I$ that transforms $\triangle ABC$ into $\triangle A'H'_{B}H'_{C}$. This homothety transforms incenter of $\triangle ABC$ into incenter of $\triangle A'H'_{B}H'_{C}$. But $I$ is transformed into itself.  So $I$ is the incenter of $\triangle A'H'_{B}H'_{C}$. Therefore, $I$ is symmetric to $I_{A}$ with respect to line $N_{B}N_{C}$. So

$I_{A}N_{B}=I_{A}N_{C}=r.$

We prove similarly for the remaining segments. Thus I have proved that the hexagon $I_{A}N_{B}I_{C}N_{A}I_{B}N_{C}$ has all sides equal to $r$.

[quote="derezins"][quote="nxxx"]Why is $A'H_{B}'$ parallel to $AB$?[/quote]
Dear nxxx,
line symmetric to $AC$ with respect to $N_{B}N_{C}$ is parallel to AB, because $\angle AN_{B}N_{C}=\angle AN_{C}N_{B}$.[/quote]
Thank you very much!
Now I understand this perfectly well! :rotfl:

The triangles $\triangle AH_{B}H_{C}\sim \triangle ABC$ are oppositely similar with similarity coefficient $\cos A$ and the bisector $AI \equiv AI_{A}$ of $\angle A$ is common. Let the incircle $(I_{A})$ of the $\triangle AH_{B}H_{C}$ touch $AH_{C}\equiv AB$ at $T_{B}.$ Then $I_{A}T_{B}= IN_{C}\cos A = r \cos A,$ $AT_{B}= (s-a) \cos A = AN_{C}\cos A,$ so that the other leg $T_{B}N_{C}$ of the right triangle $\triangle I_{A}T_{B}N_{C}$ is

$T_{B}N_{C}= AN_{C}-AT_{B}= (s-a) (1-\cos A) = \frac{2 r \sin^{2}\frac{A}{2}}{\tan \frac{A}{2}}= r \sin A.$

The hypotenuse is then $I_{A}N_{C}= \sqrt{I_{A}T_{B}^{2}+T_{B}N_{C}^{2}}= r \sqrt{\cos^{2}A+\sin^{2}A}= r$ and similarly, $I_{A}N_{B}= I_{B}N_{A}= I_{B}N_{C}= I_{C}N_{A}= I_{C}N_{B}= r.$


algebra
function
domain
advanced fields
advanced fields unsolved

Where can I found some theory about saturated fields? Namely, I have to prove that

1. If $ S_{i},S_{i}^{\prime}$ are ring of cardinality at most the continuum for $ i\in\mathbb{Z}_{>0}$ and for every non-principal ultrafilter $ D$ on $ \mathbb{Z}_{>0}$ we have
\[ \prod S_{i}/D\equiv\prod S_{i}^{\prime}/D,\]
and $ A\equal{}\prod S_{i}/\bigoplus S_{i},A^{\prime}\equal{}\prod S_{i}^{\prime}/\bigoplus S_{i},$ then $ A$ and $ A^{\prime}$ are saturated.

2. Elementarily equivalent saturated systems of the same cardinality are isomorphic.

[quote="puuhikki"]

2. Elementarily equivalent saturated systems of the same cardinality are isomorphic.[/quote]

This follows as a trivial corollary of the following theorem:
Let T be a complete theory, if M, N are saturated models of T of the same cardinality, then M is isomorphic to N.

If M is elementary equivalent to N, then Th(M)=Th(N), which is of course a complete theory. I guess there is no need for further explanation.

The proof to the above theorem should be found in any standard model theory book. The basic idea is that if |M|=|N|=k, we construct a series of k elementary embeddings from M to N (and vice versa). By carefully restricting the domain of each f_a, it's possible to see that the union of those functions is an isomorphism between N and M.

Thanks. My thesis advisor said that I can use the fact without proving it. I have already done my thesis so unfortunately I can't add the proof to my thesis.  :wink:


calculus
integration
trigonometry
calculus computations

Using the well-known identity $\sin 2x = \frac{2 \tan x}{1+\tan^{2}x}$, we get that
\begin{eqnarray*}I &=& \int_{0}^\frac{\pi}2 \frac{1-\frac{2 \tan x}{1+\tan^{2}x}}{\left( 1+\frac{2 \tan x}{1+\tan^{2}x}\right)^{2}}\, dx \\ \ &=& \int_{0}^\frac{\pi}2 \frac{\left( 1-\tan x \right)^{2}}{\left( 1+\tan x \right)^{4}}\cdot \left( 1+\tan^{2}x \right) \, dx .\end{eqnarray*}

Substitute $y = \tan x, \, dy = \left( 1+\tan^{2}x \right) \, dx$, so \[I = \int_{0}^\infty \frac{(1-y)^{2}}{(1+y)^{4}}\, dy .\] One final substitution, $z = 1+y$, so \[I = \int_{1}^\infty \frac{(2-z)^{2}}{z^{4}}\, dz = \left.-\frac{3z^{2}-6z+4}{3 z^{3}}\right|_{1}^\infty = \frac13 .\]


hey

Can someone help me with this, thanks in advance!

Prove that if p is not prime then $(p-1)! \not\equiv-1 (mod p)$ or in other words $p \nmid (p-1)!+1$

[hide]
For $p > 4$, we can always write $p = a \cdot b$ with distinct $a, b < p-1$ unless $p = k^{2}$ for some prime $k$. So in all of those cases $(p-1)! \equiv 0 \pmod{p}$. If $p = k^{2}$ for prime $k$, then $k \cdot (2k) \equiv 0 \pmod{p}\Rightarrow (p-1)! \equiv 0 \pmod{p}$ anyway. Lastly, for $p = 4$ we have $3! \equiv 2 \pmod{4}$. So none of them are $-1$.[/hide]

thank you but does someone has any other solution.

this is what i'm doing

Prove by contradiction, let's say p is not prime and $(p-1)! \equiv-1 \pmod{p}\Rightarrow p \mid (p-1)!+1$. If p is not the prime then $p = s \cdot r$ for some inetgers r and s such that $1 < r \le s < p$. Therefore, $r \mid (p-1)!+1, s \mid (p-1)!+1$ but $r \nmid (p-1)!+1, s \nmid (p-1)!+1$. Since r and s are factors of p then we should have $(p-1)! \equiv 0 \pmod{r}$ and $(p-1)! \equiv 0 \pmod{s}$ but we have $(p-1)! \equiv-1 \pmod{r}$ and $(p-1)! \equiv-1 \pmod{s}$. Hence it contradicts our assumption, therefore if p is not prime then $(p-1)!\not\equiv-1 \pmod{p}$

Someone let me know if this makes sense, thank you!

That's basically what paladin did.

I don't understand the need to find multiple solutions to a problem that depends only on one application of one concept.  This problem doesn't really have that kind of depth.


Jane and her brother each spin this spinner once. The spinner has five congruent sectors. If the non-negative difference of their numbers is less than 3, Jane wins. Otherwise, her brother wins. What is the probability that Jane wins? Express your answer as a common fraction.

[asy]size(101);
draw(scale(2)*unitcircle);
for(int i = 0; i<5; ++i)
{
draw((0,0)--2dir(90+i/5*360));
label(string(i+1),1.3dir(45-i/5*360));
}
draw((0,0)--1.5dir(75),EndArrow(4));[/asy]

the probability that jane's brother wins : $ \frac{6}{25}$
1 and 4, 1 and 5, 2 and 5 and vice versa (so 6 ways)

$ 1\minus{}\frac{6}{25}\equal{}\frac{19}{25}$

answer : $ \frac{19}{25}$


We slove the inversion problem so that. From number $ 1; 2; 3; 4; 5; 6; 7; 8; 9$. How many four digit number, in that it have non odd number OR it have one odd number.

Let $ \bar{abcd}$ is number we find, with $ a, b, c, d \in$ $ \{1; 2; 3; 4; 5; 6; 7; 8; 9\}$.

$ \bullet$ If $ \bar{abcd}$ have non odd number:
$ a, b, c, d \in \{ 2; 4; 6; 8 \}$ $ \Longrightarrow$ we have $ 4^4$ number found.

$ \bullet$ If $ \bar{abcd}$ have one odd number:
At this we have five way so:

+) $ a, b, c, d \in \{1;2;4;6;8\}\Longrightarrow$ we find $ 5^4$ number.

+) $ a, b, c, d \in \{3;2;4;6;8\}\Longrightarrow$ we find $ 5^4$ number.

+) $ a, b, c, d \in \{5;2;4;6;8\}\Longrightarrow$ we find $ 5^4$ number.

+) $ a, b, c, d \in \{7;2;4;6;8\}\Longrightarrow$ we find $ 5^4$ number.

+) $ a, b, c, d \in \{9;2;4;6;8\}\Longrightarrow$ we find $ 5^4$ number.

$ \bullet$ But we have $ 9^4$ four digit number.

So we find $ \boxed {9^4 \minus{} 4^4 \minus{} 5.5^4 \equal{} 3180}$ number.

in my opinion,,,If abcd  have one odd number:
there are 5(4^4) numbers
thus

we find ((9^4) - 6(4^4)) numbers....

is it right?

EDIT: nevermind....

[quote="garry_guitarist"]in my opinion,,,If abcd  have one odd number:
there are 5(4^4) numbers
[/quote]

but wouldn't it be more like $ 4^3$ since we are required to choose the odd number, and there are 4x4x4 ways to get the other three?

[quote="pi^2/6"][quote="garry_guitarist"]in my opinion,,,If abcd  have one odd number:
there are 5(4^4) numbers
[/quote]

but wouldn't it be more like $ 4^3$ since we are required to choose the odd number, and there are 4x4x4 ways to get the other three?[/quote]

hmm
abcd, element of 1,2,4,6,8
if we put a=1,, there will be 64 numbers(bcd element of non odd numbers)
if we put b=1,, there will be 64 numbers(acd element of non odd numbers)
if we put c=1,, there will be 64 numbers(abd element of non odd numbers)
if we put d=1,, there will be 64 numbers(abc element of non odd numbers)
so there are 4^4 numbers

3,5,7,9 is the same way,,,

so if abcd have only one odd number,,,there will be 5(4^4) numbers,,,,, in my opinion....

ohh yeah you're right. I forgot to pay attention to the order. It should be 4^4.

Why even break it into cases?  5 odd numbers in the set, 4 even.  For 2 even and 2 odd, we have $ {}_4C_2 \cdot 5^2 4^2 \equal{} 6 \cdot 400 \equal{} 2400$ possibilities, for 3 odds and 1 even we have $ {}_4C_3 \cdot 5^3 4 \equal{} 4 \cdot 500 \equal{} 2000$ possibilities and for all odd digits we have $ 5^4 \equal{} 625$ possibilities.  Adding these up, we have a total of $ 5025$ possibilities.

Anyway we are all assuming that we are allowed to reuse digits.  If we aren't, the whole interpretation changes.

thanhnam2902, you overcounted the number of possibilities with one odd because doing $ 5^4$ presumes that each odd can be chosen multiple times.  Instead you should have $ 4 \cdot 4^3 \equal{} 4^4$ - four ways to choose the spot for the odd and 4 ways to choose each remaining digit.

garry_guitarist, you are right. I suppose what I did is just a more complex way to solve it.

[quote="garry_guitarist"]from number 1,2,3,4,5,6,7,8,9 we want to make  4 digit number at least there are 2 odd number.....how many number is possible?[/quote]
[hide="My attempt"]3 cases(because I can't do combinatorics D:
first case is that there are 2 odd numbers in this 4 digit number. _ _ _ _  we can re-arrange this in $ 4C2$ ways. $ 5\cdot4\cdot4\cdot3\equal{}360\cdot6\equal{}2160$
next case is that there are 3 odd numbers. _ _ _ _ we can re-arrange this 4 ways. $ 5\cdot4\cdot3\cdot4\equal{}360\cdot4\equal{}1440$
Final case is that all 4 digits are odd numbers. And there is only one way to arrange it. $ 5\cdot4\cdot3\cdot2\equal{}120$
$ 2160\plus{}1440\plus{}120\equal{}3720$[/hide]

[quote="facis"]ckck, $ 5\cdot4\cdot4\cdot3 \equal{} 240$, not $ 360$.  But other than that, your work is correct (just recalculate your numbers) [i]assuming[/i] that we can only use each digit once.  If we can reuse digits, there are $ 5025$ possibilities as I showed above.[/quote]
Lol I knew I did something wrong somewhere...Thats what happens when you try to do a problem in your head as you start reading the topic... -_-
I can't edit it anymore D:


this is posted in the wrong forum...it belongs to the calculus section.

Have a mod move it. :)

Wow, I can actually understand the Chinese text here. Too bad I don't know how to type it. :D

[quote="4everwise"]Have a mod move it. :)
[/quote]
Ask Valentin.

I think it's not good to answer them. Many Hong Kong students post their homework problems on internet and ask for answers, which is disgusting. :mad: Luckily this is not happening in mathlinks, but that has occured in quite a number of forums I have been.

Well, it is really disgusting if the students post their homework problems on internet and just for getting the answers.
I wish they could learn much by using the way.

[quote="Dreamer"][quote="BoomHead"]A curve y=x^2and the point R(-1,0).P(s,0) is a variable point on the x-axis ands is increasing at the rate of 5 units per second . Q is a point on the the curve such that PQ is perpendicular to the x-axis

Calculate the rate of increasing of the length of QR when s=2 units.[/quote]

As PQ is perpendicular to the x-axis,
the rate of increasing of the length of QR is dl=sqrt(dx*dx+dy*dy),
y=x^2
so,  dy=2xdx
s=2 and dx=5 units/sec
we get:  dy=20 units/sec.
so,dl=sqrt(5*5+20*20) units/sec[/quote]

I got a different answer from Dreamer for this question. Below is what I did:
$l=sqr((x+1)^2+x^4))$
$dl/dt=(dl/dx)*(dx/dt)=(2(x+1)+3x^3)/(2*sqr((x+1)^2+x^4)*(dx/dt)=(6+24)/(2*5)*5=15$

Is there somthing wrong:?:


geometry
LaTeX
modular arithmetic
algebra
binomial theorem

There are some maths questions i need help from everybody.Thanks.

1)When each side of a square increased by 5 feet,the area is increased by 175 square feet.Then the dimensions of the original square is?

2)a man 40 years old,his son 10 years old.in how many years will the father be three times as old as the son?

3)James tore out several successive pages from a book.The number of the 1st pages he tore out was 183,and it is known that the number of last page is written with the same digits in some order.So,how many pages did James tear out?

4)Ahmad,Ben and Chole with their and their wives Amy,Barbie and Candy each purchased at least one animal from a farm.each of six spent an exact number of dollars for each animal equal to the number of animals he or she purchased.Ahmad purchased 23 animals more than Barbie did,and Ben spent 11 dollar per animal more than Amy.Also each husband spend a total of 63 dollar more than his wife.Then,who was married to whom?

5)John owns a pasture on which the grass has been cut to a uniform height of two inches. The grass in the pasture grows uniformly at a constant rate. He also owns a cow, a horse, and a sheep. The grass existing and growing in the pasture is sufficient to feed all three animals grazing together for 20 days. It would feed the cow and the horse alone for 25 days, the cow and the sheep alone for 33  days,and the horse and the sheep alone for 50 days. How long would it sustain
(i)	the cow alone?
(ii)	the horse alone?
(iii)	the sheep alone?

6)A ball projected vertically upward with initial speed v0 ft/sec is at time t sec at a distance s ft from the point of projection as given by the formula s = v0t – 16t^2.
(i)	If the ball is given an initial upward speed of 128 ft/sec, at what time it be 100 ft above the point of projection

(ii)	Plot the graph of s against t where v0 = 128.

So,that's all my questions.Hope someone can help me to solve the questions.I really need your help.Very thanks you.thanks you.[/b]

[color=darkblue]Please solve for me the question 4,5 and 6,please.Thanks.[/color]

7)solve equations.
   3x+2y-z=-1
   x+y+z=6
   3x+y+2z=15

8)solve equations.
   3x+2y=18
   xy=12

So,until now the questions are.thanks,buddies,u guys are really helped me.thanks you very much

$1.$
[hide]$A_{\text{square}}=s^{2}$

Thus an area where we increase each side by $5$ is:
$A=(s)^{2}$
$A+175=(s+5)^{2}$

Plus $A=s^{2}$ into the second equation to produce:
$s^{2}+175=(s+5)^{2}$
$s^{2}+175=s^{2}+10s+25$
$150=10s$
$s=15$

Thus it was a $\boxed{15 \text{ x }15}$ sqaure.[/hide]

$2.$
[hide]Man $=40$, Son $=10$

Let $n$ be the number of years after the current time.

$40+n=3(10+n)$
$40+n=30+3n$
$10=2n$
$n=5$

It will take $\boxed{5}$ years for the father to be three times the age of his son.[/hide]

$4.$ Try to avoid using dollar signs.  These produce latex (ie the bold text that appears around your dollar signs).  If you must follow the directions when you just type in one dollar sign to produce just one.

[hide="3"]
$\frac{318-184}{2}=67$
[/hide]
[hide="6 HINT"]
Plug in the given values to get the answer.
[/hide]
The fourth problem does not make sense.

[hide="5."]
Let C, H and S represent how many inches of grass the cow, horse, and sheep eat each day. Also, let G be the number of inches the grass grows each day. It follows that $20G-20C-20H-20S =-2$, $25G-25C-25H =-2$, $33G-33C-33S =-2$, and $50G-50H-50S =-2$. After dividing the equations, subtracting them and simplifacation $G=16/825$, $C=3/50$, $H=13/330$, and $S=1/50$. Put these into $X(G-Y)=-2$ where Y is C, H, or S and X is the answer for the corresponding animal to get:
i. about 49.25 (I used a graphing calculater and got this but didn't find an interger when I squared it :what?: )
ii. 100
iii. 3300
[/hide]

thanks all guys,i need 2 solve out all this questions 2nite.your help are really really need.Because my maths skill is lousy that i even can only know some basic maths only.all tis ques are too hard for me.So now,because of pushing my brain so much,now i get the headache.Thanks you for all your help.

[quote="vishalarul"][quote="Train"]7)find the remainder of when 9^2007 devided by 8

9)solve equations.
   3x+2y-18
   xy=12
[/quote]
[hide="7"]
We note that $9^{2007}\equiv (8+1)^{2007}\pmod{8}$. Applying the binomial theorem, we notice that all but one of the terms is canceled by the $\pmod{8}$. That term is $1$. So the answer is $1$.[/hide]

Is the first equation in 9) supposed to be $3x+2y=18$?[/quote]

yes,it is 3x+2y=18
thanks for your help

EDIT: Don't delete the original problems you posted and change the numbering.. just leave them.  Now my numbers are out of whack.

$7.$
[hide]You could just look at it as:

$9 \equiv 1 \mod 8$
$9^{2007}\equiv 1^{2007}\mod 8$
$9^{2007}\equiv 1 \mod 8$

$\boxed{\text{Remainder}=1}$[/hide]

$8.$
[hide]$3x+2y-z=-1$
$x+y+z=6$
$3x+y+2z=15$

Take the first two equations and add them to produce $\rightarrow 4x+3y+0=5$.
Take the second equation, multiply it by $2$ and then subtract it from the third equation $\rightarrow x-y=3$

Now we have another system:
$4x+3y=5$
$x-y=3 \implies 3x-3y=9$

Adding the latter part of the second to the first produces $\rightarrow 7x=14 \iff \boxed{x=2}$.  Substitute this into $x-y=3 \iff 2-y=3 \iff \boxed{y=-1}$.  Now substitute both of these into one of your original equations $\rightarrow x+y+z=6 \iff 2-1+z=6 \iff \boxed{z=5}$

Solution: $\boxed{(2,-1,5)}$[/hide]

$9.$
[hide]$3x+2y=18$
$xy=12$

Taking the second equation $xy=12 \rightarrow y=\frac{12}{x}$ and plug it into the first.

$3x+2 \left( \frac{12}{x}\right)=18$
$3x^{2}+24=18x \implies x^{2}-6x+8=0 \implies (x-4)(x-2)=0 \implies x=2,4$

Plugging $x=2$ and $x=4$ back into the original equation $xy=12$ we see our solution become $\boxed{(2,6), (4,3)}$.[/hide]

$10.$
[hide]Note that $2002 \equiv 0 \mod 7$. Thus:

$2003 \equiv 1 \mod 7$
$2004 \equiv 2 \mod 7$
$2005 \equiv 3 \mod 7$
$2006 \equiv 4 \mod 7$

Thus:
$2003 \cdot 2004 \cdot 2005 \equiv 1 \cdot 2 \cdot 3 \mod 7 \iff 2003 \cdot 2004 \cdot 2005 \equiv 5 \mod 7$ and

$2006^{3}\equiv 4^{3}\mod 7 \iff 2006^{3}\equiv 64 \mod 7 \iff 64 \equiv 1 \mod 7$
$\iff 2006^{3}\equiv 1 \mod 7$.

Adding these two remainders produces a remainder of $\rightarrow 1+5=\boxed{6}$[/hide]

thanks so much.anybody help?the question haven't complete.and i need to solve it now in 6 hours from now.thanks if helping.realyy thankful for all your help

You should be able to do $6.$ with the hint provided.. its not hard.  $4.$ I don't think anyone understands what you mean.. I don't at least.  I'm not accusing but as long as you're doing these for self improvement its okay but this isn't a "do your homework for you" forum.

what is the hint?
it's ok,i dun mean wan 2 tis forum do homework for me.
i juz ask,i really really cannot solve tis problem,so i ask at here,hope someone can help me.
anyway,thanks.
the 4th question,i really give up with it,or mayb u mean the 4th ques oso has the hint?

[quote="AstroPhys"][hide="6 HINT"]Plug in the given values to get the answer.[/hide][/quote]
AstroPhys gave it to you.

but i juz saw tis sentence onli:
Plug in the given values to get the answer.
so,i dunno...
can u explain 2 me,pls
thanks

Why is it so difficult? Just do what it tells it to do :arrow:  $v_{0}=128$ and $s=100$. Substitute and solve for $t$. Then graph it.

[quote="Train"]it's ok,i dun mean wan 2 tis forum do homework for me.
i juz ask,i really really cannot solve tis problem,so i ask at here,hope someone can help me.
[/quote]

Okay, the best thing is to let people know what you need help on unless you know how to solve it and want to see how others would solve it like many people do. :)


vector
linear algebra
linear algebra unsolved

Show that a set of Vectors, which contains a set of linearly dependent vectors, is linearly dependent. What is the anlogous statement about  linearrly independent vectors ?

There is no "analogous statement." There is only the contrapositive: any subset of an independent set is independent.


The digits of a positive two-digit number $N$ are interchanged to form an integer $K$.  Show that $N-K$ is an integral multiple of 9.

Let $t$ be the tens digit of $N$ and $d$ the ones digit.

Thus, $N-K = (10t+d)-(10d+t)=9t-9d=9(t-d)$


number theory
least common multiple
relatively prime
prime numbers

In a certain football league, the only ways to score points are to kick a field goal for 3 points or score a touchdown for 7 points. How many positive scores cannot possibly be obtained by a team in this league?

is there any fast way to do this w/o brute force? :huh:

By the Chicken McNugget Theorem, we know the largest score that cannot be obtained is $ 7 \times 3\minus{}(7\plus{}3)\equal{}11$.  Brute force from here isn't tough.  $ 1$, $ 2$, $ 4$, $ 5$, $ 8$, and $ 11$ are the only ones that cannot be attained, for a total of $ \boxed{6}$.

[quote="AIME15"]By the Chicken McNugget Theorem, we know the largest score that cannot be obtained is $ 7 \times 3 \minus{} (7 \plus{} 3) \equal{} 11$.  Brute force from here isn't tough.  $ 1$, $ 2$, $ 4$, $ 5$, $ 8$, and $ 11$ are the only ones that cannot be attained, for a total of $ \boxed{6}$.[/quote]
What's the Chicken McNugget Theorem?

For two relatively prime numbers $ a$ and $ b$, the largest sum that cannot be attained using just $ a$ and $ b$ is $ ab\minus{}(a\plus{}b)$.

Isn't that larger than $ ab\minus{}(a\plus{}b)$ at least in some cases?  Then we can achieve $ ab\minus{}(a\plus{}b)\plus{}1$ with just $ a$ and $ b$, so that doesn't work.

So with three numbers it's just brute force?  Or can you use the McNugget for 2 choices together and find LCM or something like that..?

[quote="007math"]For three or more numbers, is it simply:

$ abc \minus{} ( a \plus{} b \plus{} c)$

?[/quote]

7,5,3

Try Frobenius number.


what is the greatest possible value of $f(y)$ if $f(y)=\frac{22}{3y^2-4y+5}$

I'll give hint for this one.

[hide]Note that the MAX of the whole thing occurs at the MIN of the bottom. Now, how can we minimize the bottom then? Well.. we have a parabola opening upward..  :)[/hide]

[hide]The minimum of the denominator is $-\frac{b}{2a}=\frac23$. So when $y=\frac23$, $f(y)=6$.[/hide]


Let $ ABC$ be a triangle and incenter touch $ BC,CA,AB$ at $ A',B',C'$, $ A'',B'',C''$ is projection of $ A',B',C'$ to $ B'C',C'A',A'B'$, $ Ge$ is Gergonne point of $ ABC$ assume $ B''Ge\cap CC''\equal{}\{M\}$, $ B''C''\cap AC\equal{}N$ prove that quadrilateral $ B''MNC$ is cyclic.

hi,[b]encyclopedia[/b],can you post your idea?thank you[/b]


Given two polynomials: $f(x)=x^{2}+px+1$, $g(x)=x^{2}+qx+1$.

$\alpha, \beta$ are the roots of $f(x)$. $\gamma, \delta$ are roots of $g(x)$. Show that:

$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^{2}-p^{2}$

[quote="vmsvl"][hide=" :rotfl: "]use viete.th.[/hide]<====[/quote]

 :rotfl:  I know. It's always good to have a laugh. Anyone with a full solution?


function
number theory
relatively prime
number theory unsolved

Prove that, for relatively prime integers $ a$ and $ b$, $ a^{\phi (b)} \plus{} b^{\phi (a)} \equiv 1 \: \rm{mod} \: ab$.

EDIT: Solved. Please ignore.

we prove that $ a^{\phi (b)}\plus{}b^{\phi (a)}\equiv 1\: \rm{mod}\: a$ and $ a^{\phi (b)}\plus{}b^{\phi (a)}\equiv 1\: \rm{mod}\: b$.
Equivalent $ b^{\phi (a)}\equiv 1\: \rm{mod}\: a$ and $ a^{\phi (b)}\equiv 1\: \rm{mod}\: b$.
This is true.(Euler)


Here are problems from our national contest last year. I hope you will find something interesting for yourselves. The average success on these problems was under 50%.

You can make separate topics for each problem and post the links here ;)
It would be very appreciated.  :)

Ok
Here's where you can discuss problems from this contest.

Problem9.1. http://www.mathlinks.ro/Forum/viewtopic.php?t=30654
Problem9.2. http://www.mathlinks.ro/Forum/viewtopic.php?t=30655
Problem9.3. http://www.mathlinks.ro/Forum/viewtopic.php?t=30655
Problem9.4. http://www.mathlinks.ro/Forum/viewtopic.php?t=30655
Problem10.1. http://www.mathlinks.ro/Forum/viewtopic.php?t=30647
Problem10.2. http://www.mathlinks.ro/Forum/viewtopic.php?t=30648
Problem10.3. http://www.mathlinks.ro/Forum/viewtopic.php?t=30649
Problem10.4. http://www.mathlinks.ro/Forum/viewtopic.php?t=30653
Problem11.1. http://www.mathlinks.ro/Forum/viewtopic.php?t=30645
Problem11.2. http://www.mathlinks.ro/Forum/viewtopic.php?t=30644
Problem11.3. http://www.mathlinks.ro/Forum/viewtopic.php?t=30641
Problem11.4. http://www.mathlinks.ro/Forum/viewtopic.php?t=30652
Problem12.2. http://www.mathlinks.ro/Forum/viewtopic.php?t=30646
Problem12.3. http://www.mathlinks.ro/Forum/viewtopic.php?t=30650

Good luck!


The problem is that you have to find how many solutions and what they are to the following problem.

  abc
+def
____
  ghi

Where (a,b,c,d,e,f,g,h,i)  represent one of the integers from 1 to 9 distinctly.

Ex.  735
    +246
   _____
      981

Each integer 1 to 9is used once.  Is there a way to solve this problem without brute forcing it? I'm pretty stuck... Thanks for the help.

Use that ghi has to be a multiple of 9 to reduce the brute force checking.

[quote="barce11"]How do we know it is a multiple of 9?[/quote]

$ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\equal{}45$ is a multiple of 9. Denote by $ Q(n)$ the digit sum of $ n$; it is well-known that $ Q(m)\plus{}Q(n)\equiv Q(m\plus{}n)\pmod 9$, so \[ 2Q(ghi) \equiv Q(ghi)\plus{}Q(abc)\plus{}Q(def) \\\equiv a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g\plus{}h\plus{}i\equiv 45\equiv 0 \pmod 9.\] So $ ghi$ has to be a multiple of 9.


integration
linear algebra
linear algebra theorems

My prof derived fourier series using $ T \equal{} {d^{2}}/{dx^{2}}$ as linear operator.
He defined $ <f|g> \equal{} \int_c^{c\plus{}2L} f^{*}g dx$
And then he found eigenvalues...and so on.
But my question is, before finding eigenvalues, he checked $ <Tf|g>\equal{}<f|Tg>$. What's the point of doing this?
I've got to ask this prof directly, but I missed his office hour...

Self-adjoint operators have nicer spectral properties than general operators since the [url=http://en.wikipedia.org/wiki/Spectral_theorem]spectral theorem[/url] holds for them.


probability
function
probability and stats

A dice is thrown $ n$ times. Let $ A_i\in \{\{1\},\{2\},\{3\},\{4\},\{5\},\{6\}\}$ denote the value you get at the $ i$th throw. What is the probability that $ |\cup_{1\leq i < n} A_i| \equal{} 5$ and $ |\cup_{1\leq i\leq n} A_i| \equal{} 6$.

let $ X_n\equal{}|\bigcup_{i\equal{}1}^n A_i|$

so $ 1\le X_n \le 6$ for each n=1,2,...
given $ X_n\equal{}k<6$ lets find $ P(X_{n\plus{}1}\equal{}k)$

so we roll a die, and the chances of getting one of our k already rolled is $ \frac{k}{6}$

giving $ P(X_{n\plus{}1}\equal{}k|X_n\equal{}k)\equal{}\frac{k}{6}\equal{}1\minus{}P(X_{n\plus{}1}\equal{}k\plus{}1|X_n\equal{}k)$
let $ p_k(n)\equal{}P(X_{n\plus{}1}\equal{}k)$  for n=0,1,.. so,
$ p_1(n)\equal{}P(X_{n\plus{}1}\equal{}1|X_n\equal{}1)P(X_n\equal{}1)\equal{}...\equal{}\frac{1}{6^n}$
and
$ p_k(n)\equal{}\frac{k}{6}p_k(n\minus{}1)\plus{}\frac{7\minus{}k}{6}p_{k\minus{}1}(n\minus{}1)\equal{}\frac{7\minus{}k}{6}\sum_{j\equal{}1}^n \frac{k^{j\minus{}1}}{6^{j\minus{}1}}p_{k\minus{}1}(n\minus{}j)$

$ p_2(n)\equal{}\frac{5}{6}\sum_{j\equal{}1}^n \frac{2^{j\minus{}1}}{6^{j\minus{}1}}\frac{1}{6^{n\minus{}j}}\equal{}\frac{5(2^n\minus{}1)}{6^n}$

$ p_3(n)\equal{}\frac{4}{6}\sum_{j\equal{}1}^n \frac{3^{j\minus{}1}}{6^{j\minus{}1}}\frac{5(2^{n\minus{}j}\minus{}1)}{6^{n\minus{}j}}\equal{}\frac{20}{6^n}\sum_{j\equal{}1}^n 3^{j\minus{}1}(2^{n\minus{}j}\minus{}1)$
$ \equal{}\frac{20}{6^n}\left[ 2^{n\minus{}1}\sum_{k\equal{}0}^{n\minus{}1}\left( \frac{3}{2}\right)^j\minus{}\sum_{j\equal{}0}^{n\minus{}1} 3^j \right]\equal{}\frac{20}{6^n}\left[ 2^{n\minus{}1}\cdot\frac{ \left(\frac{3}{2}\right)^n\minus{}1}{\frac{3}{2}\minus{}1}\minus{}\frac{ 3^n\minus{}1}{3\minus{}1} \right]$
$ \equal{}\frac{10}{6^n}[3^n\plus{}1\minus{}2^{n\plus{}1}]$

$ p_4(n)\equal{}\frac{3}{6}\sum_{j\equal{}1}^n \frac{4^{j\minus{}1}}{6^{j\minus{}1}}\frac{10}{6^{n\minus{}j}}[3^{n\minus{}j}\plus{}1\minus{}2^{n\minus{}j\plus{}1}]\equal{}\frac{30}{6^n}\sum_{j\equal{}1}^n 4^{j\minus{}1}(3^{n\minus{}j}\plus{}1\minus{}2^{n\minus{}j\plus{}1})$
$ \equal{}\frac{30}{6^n}\left[ 3^{n\minus{}1}\sum_{j\equal{}0}^{n\minus{}1} \left(\frac{4}{3}\right)^j\plus{}\sum_{j\equal{}0}^{n\minus{}1} 4^j \minus{}2^n\sum_{j\equal{}0}^{n\minus{}1} 2^j\right]$
$ \equal{}\frac{30}{6^n}\left[ 3^{n\minus{}1}\frac{\left(\frac{4}{3}\right)^n\minus{}1}{\frac{4}{3}\minus{}1}\plus{}\frac{4^n\minus{}1}{4\minus{}1} \minus{}2^n(2^n\minus{}1)\right]$
$ \equal{}\frac{10}{6^n}\left[ 4^n\minus{}3^{n\plus{}1}\plus{}3\cdot2^n\minus{}1\right]$

$ p_5(n)\equal{}\frac{2}{6}\sum_{j\equal{}1}^n \frac{5^{j\minus{}1}}{6^{j\minus{}1}}\frac{10}{6^{n\minus{}j}}(4^{n\minus{}j}\minus{}3^{n\minus{}j\plus{}1}\plus{}3\cdot 2^{n\minus{}j}\minus{}1)$
$ \equal{}\frac{20}{6^n} \left[4^{n\minus{}1}\sum_{j\equal{}0}^{n\minus{}1}\left(\frac{5}{4}\right)^j\minus{}3^n\sum_{j\equal{}0}^{n\minus{}1}\left(\frac{5}{3}\right)^j\plus{}3\cdot 2^{n\minus{}1}\sum_{j\equal{}0}^{n\minus{}1} \left(\frac{5}{2}\right)^j\minus{}\sum_{j\equal{}0}^{n\minus{}1}5^j\right]$
$ \equal{}\frac{20}{6^n} \left[4^n(\left(\frac{5}{4}\right)^n\minus{}1)\minus{}\frac{3}{2}\cdot 3^n(\left(\frac{5}{3}\right)^n\minus{}1)\plus{}2^n(\left(\frac{5}{2}\right)^n\minus{}1)\minus{}\frac{1}{4}(5^n\minus{}1)\right]$
$ \equal{}\frac{5}{6^n}\left[ 5^n\minus{}4^{n\plus{}1}\plus{}2\cdot3^{n\plus{}1}  \minus{}2^{n\plus{}2} \plus{}1\right]$

$ p_6(n)\equal{}\frac{1}{6}\sum_{j\equal{}1}^n \frac{6^{j\minus{}1}}{6^{j\minus{}1}}\frac{5}{6^{n\minus{}j}}\left[ 5^{n\minus{}j}\minus{}4^{n\minus{}j\plus{}1}\plus{}2\cdot3^{n\minus{}j\plus{}1}  \minus{}2^{n\minus{}j\plus{}2} \plus{}1\right]$
$ \equal{}\frac{5}{6^n}\left[5^{n\minus{}1}\sum_{j\equal{}0}^{n\minus{}1} \left(\frac{6}{5}\right)^j\minus{}4^n\sum_{j\equal{}0}^{n\minus{}1} \left(\frac{6}{4}\right)^j\plus{}2\cdot 3^n \sum_{j\equal{}0}^{n\minus{}1} \left(\frac{6}{3}\right)^j\minus{}2^{n\minus{}1}\sum_{j\equal{}0}^{n\minus{}1} \left(\frac{6}{2}\right)^j\plus{}\sum_{j\equal{}0}^{n\minus{}1} 6^j\right]$
$ \equal{}\frac{5}{6^n}\left[5^{n\minus{}1}\frac{\left(\frac{6}{5}\right)^n\minus{}1}{\left(\frac{6}{5}\right)\minus{}1}\minus{}4^n\frac{\left(\frac{6}{4}\right)^n\minus{}1}{\left(\frac{6}{4}\right)\minus{}1}\plus{}2\cdot 3^n \frac{ \left(\frac{6}{3}\right)^n\minus{}1}{\left(\frac{6}{3}\right)\minus{}1}\minus{}2^{n\minus{}1}\frac{ \left(\frac{6}{2}\right)^n\minus{}1}{\left(\frac{6}{2}\right)\minus{}1}\plus{}\frac{ 6^n\minus{}1}{6\minus{}1}\right]$
$ \equal{}\frac{1}{4\cdot 6^n}\left[ 19\cdot 6^n\minus{}4\cdot 5^{n\plus{}1}\plus{}10\cdot4^{n\plus{}1}\minus{}40\cdot 3^n\plus{}5\cdot 2^n\minus{}4\right]$

The probability that the face 1 appears $ k_1$ times, the face 2 appears $ k_2$ times, …, and the face 6 appears $ k_6$ times in $ n$ repeated throws is equal to
\[ \frac {{n!}}{{n_1 !n_2 ! \cdots n_6 !}}\left( {\frac {1}{6}} \right)^{n_1 } \left( {\frac {1}{6}} \right)^{n_2 } \cdots \left( {\frac {1}{6}} \right)^{n_6 } \equal{} \frac {{n!}}{{n_1 !n_2 ! \cdots n_6 !}}\left( {\frac {1}{6}} \right)^n\]
where$ n_1 \plus{} n_2 \plus{} \cdots \plus{} n_6 \equal{} n$.

Let $ X_n \equal{} \left| {\bigcup\limits_{i \equal{} 1}^n {A_i } } \right|$.
In order to have$ X_n \equal{} 6$, each face must appear at least once. We take every possible combination of $ n_1 ,n_2 , \ldots ,n_6$ such that $ n_1 \plus{} n_2 \plus{} \cdots \plus{} n_6 \equal{} n$ and$ n_1 ,n_2 , \ldots ,n_6 \ge 1$. Hence,
\[ P\left( {X_n \equal{} 6} \right) \equal{} \left( {\frac {1}{6}} \right)^n \sum\limits_{\scriptstyle n_1 \plus{} n_2 \plus{} \cdots \plus{} n_6 \equal{} n \atop \scriptstyle n_1 , \ldots ,n_6 \ge 1} {\frac {{n!}}{{n_1 !n_2 ! \cdots n_6 !}}}\]
If$ X_n \equal{} 5$, then we obtained 5 different faces, one face not appearing at all. Then, one of the numbers $ n_1 ,n_2 , \ldots ,n_6$ must be zero. Hence,
\[ P\left( {X_n \equal{} 5} \right) \equal{} \left( {\frac {1}{6}} \right)^n \sum {\frac {{n!}}{{n_1 !n_2 ! \cdots n_6 !}}}\]
where the sum is taken over all 6-tuples of positive integers $ n_1 ,n_2 , \ldots ,n_6$ such that $ n_1 \plus{} n_2 \plus{} \cdots \plus{} n_6 \equal{} n$ and exactly one integer is zero. The face that does not appear can be chosen in 6 ways. Then,
\[ P\left( {X_n \equal{} 5} \right) \equal{} 6\left( {\frac {1}{6}} \right)^{n} \sum\limits_{\scriptstyle n_{i_1 } \plus{} n_{i_2 } \plus{} \cdots \plus{} n_{i_5 } \equal{} n \atop \scriptstyle n_{i_1 } , \ldots ,n_{i_5 } \ge 1} {\frac {{n!}}{{n_{i_1 } !n_{i_2 } ! \cdots n_{i_5 } !}}}\]
Now let’s compute$ P\left( {X_n \equal{} k} \right)$. Of 6 faces, $ k$ will appear and $ 6 \minus{} k$ will not. The faces which do not appear can be chosen in $ \left( {\begin{array}{*{20}c} 6 \\
{6 \minus{} k} \\
\end{array}} \right) \equal{} \left( {\begin{array}{*{20}c} 6 \\
k \\
\end{array}} \right)$ ways. Hence,
\[ P\left( {X_n \equal{} k} \right) \equal{} \left( {\begin{array}{*{20}c} 6 \\
k \\
\end{array}} \right)\left( {\frac {1}{6}} \right)^n \sum\limits_{\scriptstyle n_{i_1 } \plus{} n_{i_2 } \plus{} \cdots \plus{} n_{i_k } \equal{} n \atop \scriptstyle n_{i_1 } , \ldots ,n_{i_k } \ge 1} {\frac {{n!}}{{n_{i_1 } !n_{i_2 } ! \cdots n_{i_k } !}}}\]
[u][b]Remark[/b][/u]. The sum
\[ \sum\limits_{\scriptstyle n_{i_1 } \plus{} n_{i_2 } \plus{} \cdots \plus{} n_{i_k } \equal{} n \atop \scriptstyle n_{i_1 } , \ldots ,n_{i_k } \ge 1} {\frac {{n!}}{{n_{i_1 } !n_{i_2 } ! \cdots n_{i_k } !}}}\]
equals the number of onto (surjective) functions
\[ f: \left\{ {1,2, \ldots ,n} \right\} \to \left\{ {1,2, \ldots ,k} \right\}\]
The number of such functions is also equal to $ k!S\left( {n,k} \right)$, where $ S\left( {n,k} \right)$ is the Stirling number of the second kind and counts the number of partitions of an $ n$-set into $ k$ cells. We have the following formula for $ S\left( {n,k} \right)$
\[ S\left( {n,k} \right) \equal{} \frac {1}{{k!}}\sum\limits_{j \equal{} 0}^k {\left( { \minus{} 1} \right)^{k \minus{} j} \left( {\begin{array}{*{20}c} k \\
j \\
\end{array}} \right)j^n }\]
The final result is
\[ P\left( {X_n \equal{} k} \right) \equal{} \frac {{\left( {\begin{array}{*{20}c} 6 \\
k \\
\end{array}} \right)\sum\limits_{j \equal{} 0}^k {\left( { \minus{} 1} \right)^{k \minus{} j} \left( {\begin{array}{*{20}c} k \\
j \\
\end{array}} \right)j^n } }}{{6^n }},\forall n,k \in \mathbb{N},1 \le k \le 6,n \ge k\]

[u][b]Generalization.[/b][/u] If instead of 6 possible outcomes, you have $ s$ possible outcomes (each with probability $ 1/s$ of occurrence), then
	\[ P\left( {X_n  \equal{} k} \right) \equal{} \frac{{\left( {\begin{array}{*{20}c}
   s  \\
   k  \\
\end{array}} \right)\sum\limits_{j \equal{} 0}^k {\left( { \minus{} 1} \right)^{k \minus{} j} \left( {\begin{array}{*{20}c}
   k  \\
   j  \\
\end{array}} \right)j^n } }}{{s^n }},\forall n,k \in \mathbb{N},1 \le k \le s,n \ge k\]


A rectangular prism $77 \times 81 \times 100$ is cut into cubes of edge 1 by planes parallel to the faces of the prism. A given internal diagonal pierces how many of these cubes?

Source: Australian Math Competition, S28, 2002

Is this problem too hard for Getting Started?

There is a certain formula for this problem (something like the chicken mcnugget formula, except extended), but I cannot recall it.

[hide]
Principle of Inclusion-Exclusion

http://www.artofproblemsolving.com/Forum/topic-73967-120.html
There's an previous AIME problem exactly like this; there's a good explanation of the solution in the above thread :)
[/hide]


Let $ p_1\equal{}q_1\equal{}1$\[ p_n\equal{}p_{n\minus{}1}\plus{}2q_{n\minus{}1}\\q_n\equal{}q_{n\minus{}1}\plus{}p_{n\minus{}1}\]
Prove:\[ \left|\begin{bmatrix}p_n^2 & 2q_nq_{n\plus{}1}\\2q_nq_{n\minus{}1} & p_n^2\end{bmatrix}\right|\equal{}1
\]

[hide="Discussion"] Here's a general method for tackling mixed recursions:  Let $ \mathbf{s}_n \equal{} \left[ \begin{array}{c} p_n \\
q_n \end{array} \right]$ and let $ \mathbf{M} \equal{} \left[ \begin{array}{cc} 1 & 2 \\
1 & 1 \end{array} \right]$; then the given recursion is

$ \mathbf{s}_n \equal{} \mathbf{M} \mathbf{s}_{n\minus{}1}$

Find the eigenvectors/eigenvalues of $ \mathbf{M}$ and write $ \mathbf{s}_1$ in terms of the eigenvectors.  Then you have closed forms for $ p_n$ and $ q_n$.  The rest is algebra.  :) [/hide]


Solve the equation $ 7^{x}\minus{}3^{y}\equal{}4$  in positive integers.

Bad title, locked!
http://www.mathlinks.ro/Forum/viewtopic.php?t=135914