geometry perimeter Support ratio AMC AIME parallelogram

This geometry marathon is meant to be both entertaining, and to provide you with valuable problem solving strategies.  A wealth of problems will be gathered from texts and competitions. Each problem will stimulate your creativity and understanding of geometry, as well as problem solving in general.

Rules of the Marathon:
1) After somebody correctly answers a problem, they may post anywhere from 2-5 new problems. If they don't want to post, I will post. :P
2) Feel free to post all ideas, hints, and solutions, as long as they are hidden. We want to give everybody an equal opportunity.
3) You [b]must[/b] have fun. :D

[size=150]Problems 1-3[/size]

1) In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, find the perimeter of $\triangle AMN$.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=485&sid=17fd0de271bc4614d95373e0acbc3039[/img]

2) In right triangle $ABC$, $\angle C=90^\circ$, $CF$ is the median to the hypotenuse $AB$, $CE$ is the bisector of $\angle ACB$, and $CD$ is the altitude to $AB$. Show that $\angle DCE = \angle ECF$.

3) A line segment $AB$ is divided by points $K$ and $L$ in such a way that $(AL)^2=(AK)(AB)$. A line segment $AP$ is drawn congruent to $AL$. If $m\angle LPB=30^\circ$, find $m\angle LPK$.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=489&sid=d115a0a0047c4c41a5bd2a3b6527ea8d[/img]

[quote="vulgarfraction"][hide="1"] $\angle CBO = \angle ABO = \angle BOM$, as $BO$ bisects $\angle ABC$
Therefore $\triangle BMO$ is isoceles, and $|BM| = |MO|$

Similarly $\angle BCO = \angle ACO = \angle CON$, as $CO$ bisects $\angle ACB$. $\triangle CNO$ is isosceles, and $|CN| = |NO|$.

The perimeter of $\triangle AMN$ is equal to $|AM| + |AN| + |MN| = |AM| + |AN| + |MO| + |NO| = |AM| + |MB | + |AN| + |CN| = |AB| + |AC| = 30.$[/hide]

[hide="2"] $\triangle ABC$ is similar to $\triangle ACD (AAA), \therefore \angle ABC = \angle ACD.$

As $\triangle ABC$ is right angled, its circumcentre is the midpoint of the hypotenuse, $F$, and hence $F$ is equidistant from $A$, $B$, and $C$. Therefore, $\triangle CDF$ is isosceles.

As a result, $\angle DCF = \angle FDC = \angle ADC = \angle ADC$.

Now, $EC$ bisects $\angle ACB$ so $\angle ACE = \angle DCE$.

$\angle DCE = \angle ACE - \angle ACD$, and $\angle FCE = \angle DCE - \angle DCF$

But as proven above, $\angle ACE = \angle DCE$ and $\angle ACD = \angle DCF, \therefore \angle DCE = \angle FCE.$[/hide]

[hide="3"] $|AK||AB| = |AL|^2$

${|AB|}/{|AL}| = {|AL|}/{|AK|}$

$\angle A$ is common to $\triangle AKP$ and $\triangle APB$

$\therefore \triangle AKP$ is similar to $\triangle APB$

Let $\angle ABP = x. \angle ALP$ is an exterior angle to $\triangle BLP. \therefore \angle ALP = \angle ABP + \angle LPB = 30 + x$

But $|AL| = |AP|, \therefore \angle ALP = \angle APL = \angle APK + \angle KPL$

As $\triangle AKP$ is similar to $\triangle APB, \angle APK = \angle ABP = x \therefore \angle KPL = \angle APL - \angle APK = 30^\circ\#$[/hide][/quote]Nicely done. :)
I'll post the new problems this time, since I just changed the rules a bit and you're not online at the moment. :D

[size=150]Problems 4 and 5[/size]

4. Two telephone cable poles, 40 feet and 60 feet high, respectively, are placed near each other. As partial support, a line runs from the top of each pole to the bottom of the other. How high above the ground is the point of intersection of the two support lines?

5. In triangle $ABC$, median $AM$ is such that $m\angle BAC$ is divided in the ratio $1: 2$ and $AM$ is extended through $M$ to $D$ so that $m\angle DBA=90^\circ$. If $AC=12$, find $AD$.

[b]The next person to solve these problems may post their own geometry problems. Please try to keep them from AMC to AIME level. [/b] :)

[hide="i need a proof"]#4
i know the formula for this is $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
where x and y are the heights of the poles and z is the hight above the ground the point of intersection is. i know this is derived useind similar triangles so ill try to prove it.[/hide]

number 4:

[hide]
WOLOG, assume 40 ft pole is on the left

draw a line perpendicular to the poles and pass through the intersection point.  then we see two pairs of similar triangles.  let $x$ and $y$ be the height of the upper two triangles ($x$ on the left and $y$ on the right), then the height of the bottom two will be $40 - x$ and $60 - y$.  But we know these two are equal (because the line we drawn is parallel with the ground), so $y = x+20$. Let $d$ be the length of the line we drawn, and $l$ being the left part of the line.  Using similarity, $\frac{x}{40-x}$ = $\frac{l}{d-l}$ and $\frac{x+20}{40-x} = \frac{d-l}{l}$.  We can see that from here $\frac{x}{40-x} = \frac{40-x}{x+20}$, solving we get $x = 16$, and the point is $24$ ft off the ground.
[/hide]

[hide="4"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=490[/img]
Clearly $\triangle ACD \sim \triangle ECF$. From this, we get $\frac{z}{x}=\frac{FC}{DC}$. Also, $\triangle BCD \sim \triangle EFD$, so $\frac{z}{y}=\frac{FD}{DC}$. Notice that $DF + FC = DC$, so $\frac{z}{y}=\frac{DC-FC}{DC}=1-\frac{FC}{DC}=1-\frac{z}{x}$. Finally, \[ \frac{z}{y} + \frac{z}{x} = 1, \quad \text {or} \quad \frac{1}{x}+\frac{1}{y} = \frac{1}{z}. \] We are given that $x=60, y=40$ so we can substitute these in to find $z$.
[/hide]

[quote="4everwise"]Anybody want to give number 5 a shot?  :P

[hide="hint"]Are you sure you don't want to try yourself?  :D [hide="Real Hint"]Parallelogram[/hide][/hide][/quote]Hmm, I didn't come across that problem, but...
$m\angle BAM = x$ and $m\angle MAC = 2x$.  :)

[hide="5"]
Let $m\angle{AMC}=\theta$ and $m\angle{BAM}=x$.  Then, by the Law of Sines

\[ CM=AC\frac{\sin{2x}}{\sin{\theta}}=24\frac{\sin{x}\cos{x}}{\sin{\theta}} \]
\[ BM=AB\frac{\sin{x}}{\sin{\pi-\theta}}=AB\frac{\sin{x}}{\sin{\theta}} \]
We know $BM=CM$.  Thus,
\[ 24\frac{\sin{x}\cos{x}}{\sin{\theta}}=AB\frac{\sin{x}}{\sin{\theta}} \]
\[ \implies AB=24\cos{x} \]
But $m\angle{BAD}=x$.  Thus
\[ \cos{x}=\frac{AB}{AD} \implies AB=24\frac{AB}{AD} \implies \boxed{AD=24} \]

\[ \mathbb{QED} \] [/hide]

Are these from CPIG, by the way? :)

Reply: Yeah, some of them. I felt really good when I solved problem 5, so I decided to post it. :)

[size=150]Problem Solving Strategy #1[/size]

[hide="5 Solution"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=492&sid=5559f776b19a53699c733b8f0c6e96cc[/img]

First, we let $m\angle BAM=x$ so that $m\angle MAC=2x$. Now, we extend $AD$ to point $P$ such that $AM=MP$. We are given that $BM=MC$. Because these two lines bisect each other, quadrilateral $ABPC$ is a parallelogram.

This is where the right triangle comes in. Draw the median $BT$ to the hypotenuse of right triangle $ABD$ so that $BT=AM=MD=\frac 12 AD$.

Now, $m\angle BTP=2x$. (Exterior angle of the isosceles triangle) Recalling that $AB||PC$, we apply alternate interior angles to see that $m\angle BPA=m\angle PAC=2x$. Now, we see that triangle $BTP$ is isosceles such that $BT=BP=AC=12$. Because $AD=2BT$, $AD=24$. [/hide]
______________________________________________________________________
This leads us into Problem Solving Strategy #1. When given a triangle, its median, and asked to find the length of one of the sides or parts of the triangle, it's a good idea to try constructing a parallelogram. This doesn't always work, but can make a problem much easier in the case that it does work.

This problem also shows how helpful a prallelogram can be. :P

Here are some practice problems taken from AoPS Volume 2, Chapter 3:

Practice 1) In $\triangle ADC$, angle bisector $DB$ is drawn. If $AB=3$, $AD=6$, and $CD=8$, find $BD$.

Practice 2) Show that if $AD$ is a median of triangle $ABC$, then $4AD^2=AB^2+AC^2+2(AB)(AC)\cos \angle BAC$.

Alright, I don't want to post the next problem. 1234567890, I defer this next problem posting session to you. Please post 2-3 problems. :)

Thanks.  :P

[hide="Practice Problem 1"]By the [b]Angle Bisector Theorem[/b], $\frac{DA}{AB}=\frac{DC}{CB}$ if DB bisects $\angle ADC$.

In our case, $DA=6,AB=3,DC=8$, so we must solve for $CB$. $\Rightarrow \frac{6}{3}=\frac{8}{CB}\Rightarrow CB=4$

Now we let $\angle DBC = \theta \Rightarrow \angle DBA =180-\theta$

Applying the [b]Law of Cosines[/b] on $\triangle DBC$ and $\triangle DBA$, we have:
$DC^2=DB^2 +BC^2-2(DB)(CB)\cos{\theta}$
$DA^2=DB^2 +AC^2-2(DB)(AC)\cos{180-\theta}$

We recognize that $\cos{180-\theta}=-\cos{\theta}$ and plug in our values, giving us:
$64=DB^2+16-8(DB)\cos{\theta}$
$36=DB^2+9+6(DB)\cos{\theta}$

Solving, we see that $\boxed{DB=6}$[/hide]
[hide="Practice Problem 2"]Extend $AD$ to a point $E$ such that $ABCE$ is a parallelogram, with $AB$ parallel to $CE$, $BE$ parallel to $AC$. By the [b]Law of Cosines [/b]on $\triangle ACM$, we have:
$AE^2=CE^2+AC^2-2(AC)(CE)\cos{\angle ACE}$.

We recognize that in a paralellogram, $2AD=AE$, $\cos{\angle ACE}= \cos{180-\angle BAC}=-\cos{\angle ACE}$, and $CE=AB$.

Substituting back in, we have:
$4AD^2=AB^2+AC^2+2(AB)(AC)\cos{\angle BAC}$, as desired.[/hide]

Problems to come.................

[geogebra]bbc149f41718ee49b50119f6ee123c57ad1b3162[/geogebra]

[hide]We consider the midpoints $ P, Q, R$ of sides $ AC, BC,$ and $ AB$ respectively. Then $ O$ lies on the perpendicular bisectors of the three sides. Hence we have $ BQOR$ is cyclic, and thus it suffices to show that $ \angle QOM\equal{}\angle KOR$.

Now let the midpoint of $ AN$ be $ M_1$ and the midpoint of $ NC$ be $ M_2$.

Since $ QR$ is parallel to $ AC$, we consider the line through $ O$ parallel to $ QR$ and $ BC$.

Let this line intersect $ M_1K$ at $ Q$ and $ M_2M$ at $ T$ respectively.

Hence $ ST\equal{}\frac{1}{2}\cdot AC$ since $ AM_1\equal{}M_1N$ and $ NM_2\equal{}M_2C$ and thus $ ST\equal{}M_1M_2\equal{}\frac{1}{2}\cdot AC$.

Therefore, $ ST\equal{}QR$ and also, $ ST\parallel{}QR$.

Thus $ QRST$ is a parallelogram. Since $ KM_1\parallel{}MM_2$ and $ QT\parallel{}RS$, the angle formed by $ KM_1$ and $ RS$ is the same as that formed by $ MM_2$ and $ QT$.

This gives us $ \angle QTM\equal{}\angle KSR$. Also note that $ KOSR$ is cyclic and hence $ \angle KSR\equal{}\angle KOR$. Similarly, $ \angle QOM\equal{}\angle QTM$ and the result follows.[/hide]

New Problem...

In triangle $ ABC$, the feet of the angle bisectors of $ B$ and $ C$ are $ M$ and $ N$ respectively. The ray $ MN$ intersects the circumcircle of $ ABC$ at $ D$. Show that $ \frac{1}{BD}\equal{}\frac{1}{AD}\plus{}\frac{1}{CD}$.

That problem has already been posted, it is post #361 in this marathon.

That has been unanswered for a long time. Can you post an easier problem?

[hide="hint"]Drop perpendiculars from $ D$ to the sides of $ ABC$...

[/hide]

[hide="@resurrection`s hint"]The Simson Line!

Lots of cyclic quads, so we should be able to bash it out with Ptolemy's until we get $ \frac {1}{BD} \equal{} \frac {1}{AD} \plus{} \frac {1}{CD}\Rightarrow(AD)(CD) \equal{} (AD)(BD) \plus{} (CD)(BD)$, right?[/hide]

the simson's line implies we have to prove the  two perpendicular from D is equal to the sum of other two.how should i let the question know about the two great point M and N. i will try my best. can we get a easier problem. is my approach correct noble?

Let $ A_1 , B_1 ,C_1$ be the orthogonal projections of $ D$ onto $ BC , CA , AB$ respectively.

       Then $ DB_1\equal{}DA* sin\angle DAB_1 \equal{} DA* sin \angle DAC \equal{}\frac{ DA * DC}{2R}$

        R is the circumradius of ABC.

        In the same way

        $ DA_1 \equal{} \frac{DB * DC}{2R}$ and

        $ DC_1 \equal{}\frac{DA*DB}{2R}$

        Now it is enough to prove that $ DB_1 \equal{} DA_1 \plus{} DC_1$

        Let $ m$ be the distance from $ M$ to $ AB$ or $ BC$ ... and  $ n$  the distance from $ N$ to $ AC$ or $ BC$ .

        Now put  $ x\equal{} \frac{DM}{MN}$;  $ (x > 1)$

         Then

       $ \frac{DB_1}{n}\equal{}x$   ,   $ \frac{DC_1}{m}\equal{} x \minus{} 1$  ,      $ \frac{ DA_1 \minus{} m}{ n \minus{} m } \equal{} x$

      Hence $ DB_1 \equal{} nx , DC_1 \equal{} m(x \minus{}1) , DA_1 \equal{} nx \minus{} m( x \minus{} 1) \equal{} DB_1 \minus{} DC_1$, as we wanted.

       Now it is complete.:D:D

Did you find the solution from the bulgarian math olympiad =P? oh well doesn't matter.

New Problem:

The point $ M$ inside the convex quadrilateral $ ABCD$ is such that $ MA \equal{} MC, \angle AMB \equal{} \angle MAD \plus{} \angle MCD, \angle CMD \equal{} \angle MCB \plus{} \angle MAB.$

Prove that $ AB\cdot CM \equal{} BC\cdot MD$ and $ BM\cdot AD \equal{} MA\cdot CD$.

resurrection could you please post a hint or something about your problem? I'm stuck (I'm never good at synthetic geometry)... :(
If you post your solution, here's the next problem (come on guys let's continue this marathon)

In the triangle $ ABC$ let $ X,Y$ be the tangency point of its incircle with $ BC$ and $ CA$, $ X'$ is a point on $ BC$ such that $ BX \equal{} CX'$. The straight line $ XY$ cuts the perpendicular to $ BC$ at $ X'$ in $ P$. Show that the perpendicular from $ B$ to $ AX'$ goes through the midpoint of $ PX'.$

nice problem!
[hide="simple calculation"]
Let $ BC \equal{} a, CA \equal{} b, AB \equal{} c, C \equal{} \angle ACB, \theta \equal{} \angle LBX', L\equal{}BQ \cap PX'$, and $ r\equal{}\text{inradius}$, where $ Q$ is the foot of the perpendicular from $ B$ to $ AX'$. WLOG, $ b \le c$,
Let $ X''$ be the point opposite $ X$ on the incircle centered at $ I$. It is well-known that $ X''$ lies on $ AX'$. It is clear from the given right angles that $ \theta \equal{} \angle LBX' \equal{} \angle X'BQ \equal{} \angle X'X''X$. Note now that $ LBX' \sim X'X''X$.

From triangle $ X''X'X$ we see $ \tan \theta\equal{} \dfrac{XX'}{X'X''} \equal{} \dfrac{b \minus{} c}{2r}$, so $ LX' \equal{} BX' \tan \theta \equal{} (s \minus{} c) \dfrac{b \minus{} c}{2r}$.

Now, note from $ ICX$ that $ \tan \dfrac C2 \equal{} \dfrac{r}{s \minus{}c}$. Also, $ \angle XPX' \equal{} \dfrac C2$, so $ IXC \sim XX'P$. It quickly follows that $ PX' \equal{} \dfrac{XX'}{\tan \dfrac C2} \equal{} \dfrac{(s \minus{}c)(b \minus{} c)}{r}$. Thus we are done.[/hide]

I need to practice typing some since writing this up takes longer than solving it.

Next: (much simpler)
Given a point $ P$ on the minor arc $ \stackrel{\frown}{AD}$  circumcircle of square $ ABCD$, find an expression for $ PD$ in terms of $ PA$ and $ PB$. (Hint: it's not especially pretty, but it's not anything ugly either).

I hope the question is right. Ask me for hints if needed. :)

Sorry, I meant for the lengths. It does require a bit of ingenuity to do it without much algebra.

[hide] Let $ AB\equal{}AD\equal{}x$. By Pythagoras, $ BD\equal{}x\sqrt2$.
By Ptolemy's theorem on quadrilateral $ ABDP$, we see:
$ (PB)(AD)\equal{}(PD)(AB)\plus{}(AP)(BD)$
$ \implies (PB)(x)\equal{}(PD)(x)\plus{}(AP)(2^{\frac{1}{2}})(x)$
$ \implies PB\equal{}PD\plus{}(PA)(2^{\frac{1}{2}})$
$ \implies PD\equal{}PB\minus{}(PA)(2^{\frac{1}{2}})$
[/hide]

Sorry, I can't find another decent problem.


How many even five digit numbers exist such that the first digit may not be 0 and no two consecutive digits are the same?

Can someone help me with this problem?

Editting.. misread the problem

Hmm, the no consecutive same digit and the fact that the number must be even is causing me a bit of trouble. I'm thinking about separating it into a bunch of different cases, but I'll wait to see what others come up with. Sorry couldn't help.

Hmm, ok.

So there's 9 choices for the first (1-9) , second (0-9 minus first digit) and third digits  (0-9 minus 2nd digit).

Then for the fourth digit it's easier to see that you only have 8 choices (it can't be the same as third or fifth digit). And for the fifth digit you have 5 choices (0,2,4,6,8).

Giving us 9*9*9*8*5

At least that's what I'm assuming 10000th did. If not, this way looks good to me.

5 possibilities for the last one, 9 for the 1st, 9 for the 2nd, 9 for the third, 8 for the fourth, so $5 \cdot 9 \cdot 9 \cdot 9 \cdot 8$?

I was also figuring 5*9*9*9*8, however this does not work.

You select the last number with 5 possibilities, then 10 - 1 posibilities for digits 2-4.  If the second digit is zero then there are 9 options (1-9) for the first digit, if the second digit is non-zero then there are 8 options (1-9 except the second digit)...  which makes the problem a lot harder...

EDIT:  I found it is possible to easily count up the number if you make specific assumptions about the 5th digit and add them together, but this is a very time consuming process, not very elegant, and I'm too lazy to finish doing it.

Yeah, that's what I was getting at in my first post, but since the guy that posted after me got what I had gotten the first time, I decided that I was just trying to complicate the problem.

Yeah, this looks like a pretty tough problem. FYI, it came from our first test in Discrete.

Well, for the first digit, there are 9 choices, the second, 9 choices, the third 9 choices, and this is where it gets tricky. If you say there are 9 choices for the 4th and 5 choices for the 5th, you might have the 4th digit being an even digit and that would limit the last digit to 4 choices. If you say the 4th choice has 8 options, the last digit has 5 options IFF the 4th choice is odd. If the 4th choice is even, there are 4 options for the last digit. However, the basis on the parity of the 4th digit is also based on the 3rd digit. If the 4th digit is odd, then there must have been conditions based on the third digit, etc... This is what I'm confused about.  :rotfl:

Did you eventually solve it? If so, how? Cases? A formula..?

you always choose those digits that have conditions FIRST because you can appropriately choose the middle digits. come on guys.

there are 9 choices for first digit and 5 choices for last digit. 9 choices for 2nd digit since it cannot be the same as 1st digit. 9 choices for 3rd digit by same reason. now, there are only 8 choices for 4th digit because it cannot be the same as 3rd digit and last digit. total is $9\cdot 9\cdot 9\cdot 8\cdot 5=29160$

Bleh. Hmmm

(Note that this is not a spam post, I wrote somethin then realized it was wrong, and I'm still mulling stuff over)


This clip is great.  You gotta watch it.  I hope it downloads okay on your computer, it did on mine.

http://www.funnyjunk.com/movies/49/Relaxing+Car+Drive

[i]Moderator Note: Although this clip does not have any inappropriate content, it may be deemed as "scary" for kids.[/i]

o god i hate those clips. not cause theyre bad cause therye scary! ive seen 3 of those kinds of clips already and it scared me for days afterwards. yea i scare easily.....but its still rather interesting....unpredictable and stuff


real analysis
LaTeX
real analysis unsolved

Let E be a Lebesgue measurable set with nonzero positiv measure.
Prove that for any number a less than 1, there exist an open interval I, such that the measure of the intersection of E and I is stricly greater then a times the measure of I.

Let's write it in Latex,
for every measureable set $A$ such that $\mu(A)>0$ and for every $c \in (0,1)$ there exists an interval $P$ such that $|P \cap A| > c|P|$. Proof : We can cover every measurable set $A$ with intervals for which it's true that $|A| \leq \sum_{i=1}^{+\infty}|P_i|<|A| + \epsilon$. Suppose that for every $P_i$ holds $|P_i \cap A| \leq c|P_i|$, thus
$|A|\leq \sum_{i=1}^{+\infty}|P_i \cap A|\leq\sum_{i=1}^{+\infty}c|P_i|=c\sum_{i=1}^{+\infty}|P_i|<c(|A|+\epsilon)$, so $|A|< c(|A|+\epsilon)$ thus $\frac{(1-c)|A|}{c} < \epsilon$ what's false for sufficiently small $\epsilon$ so we have a contradiction.


The numbers $1, 2, ... , 2002$ are written in order on a blackboard. Then the $1st, 4th,7th, ... , 3k+1th, ...$ numbers in the list are erased. Then the $1st, 4th, 7th, ... 3k+1th$ numbers in the remaining list are erased (leaving $3, 5, 8, 9, 12, ...$). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased?

[hide="Hint"]  Assume the numbers are doors being closed...just an idea.  It should be easy; I'll come back with a solution.[/hide]
Masoud Zargar


Have you ever completely changed your mind on an issue after an argument with someone? Have you ever changed someone else's mind?

When responding to gauss202's question, I think it is important to distinguish between a situation in which you simple argue and argue until your opponent just gives in out of weariness and a situation in which you two have a heated debate after which your opponent has truly converted his/her opinion.

I have changed my mind over this one topic over many many arguments, discussions, phychologist's appointments, and generic abuses, but not after just one argument...besides there is no one in the world who change's their mind or their view of their world with a non-exhausting short discussion, if they do that's a problem. (See the guy in Analyze This for reference; the guy that got paid like 300 bucks to get the h*** out of the shrink's office)

all the time. a lot of times, i'll argue with someone, think about it, and figure out they're right. then again, i mainly change my mind because i'm indecisive. once, my friends (so called friends) decided it'd be funny to do a science experiment on me. they argued over something where both people were right and then asked me to settle the argument for them. i didn't respond for 2 weeks.

[quote="JBL"]I have to say that I once heard someone argue very persuasively for maintaining the death penalty as a possible punishment reserved for incredibly rare and horrific cases that seriously pushed me to reconsider my own strict anti-death view.  I'm not really decided, but it hasn't come up too much since then.[/quote]

That's interesting. I used to be pro-capital punishment, and through diebates with people have since moderated my opinion. I don't believe there exists a judicial system fair enough to implement it.

The other major issue I have changed my mind about is abortion.

I've both convinced many people and been convinced.

in those cases, i think it's most comfortable to say "what do you think?" and let other people argue. i've gotta admit, it is fun to hear others argue.

[quote="gauss202"]The other major issue I have changed my mind about is abortion.[/quote]

Do you mind if I ask which way you switched?  (It seems to me that the switch away from pro-choice would be more ... normal ... but I come from a secular leftist household, so it may just be that I see pro-choice as more normal and thus more normal to be switched away from, if that makes any sense.)

By the way, this is a great question.  I'd be interested if the people who said "yes" could actually provide specific examples.

My switch was from apathetically "pro-choice" to fairly strongly "pro-life" (though I hate those terms for the two sides of the issue).

No one has ever changed my mind about an issue that I feel strongly about. I guess I'm just a very obstinate person. I don't think I'm too good at getting other people to change their minds either, so practically all arguments that I am involved in are draws, just like my chess games against 1700 players.

This is an interesting question. I don't recall (offhand) an opinion change that happened as a result of argument I had with someone. Indeed, I can think of more examples of opinion changes I have had that were prompted by noticing that people who agreed with my former position on such-and-such issue were REALLY STUPID on other issues and thus suspect in their reasoning about the issue I agreed with them on.

My most profound opinion change came about as a result of visiting [url=http://www.taipeitimes.com/News/edit/archives/2004/05/22/2003156522]Taiwan[/url], Hong Kong, and China in 1982, when the differences in economic and political systems in each place were more stark than they are today. I became much more libertarian in general and much more in favor of small government rather than big government after seeing those three culturally connected areas with their very different living conditions.

Only when I was greatly misinformed on a topic.


"My phone number is of $7$ digits, all different. If you rearrange the digits and form the largest and the smallest numbers, and subtract the latter from the former, you will get another $7$ digit number. Repeat the process (of rearranging digits, subtracting smallest from the largest number etc) enough times, and you will come back to my telephone number. "

Can you find it?

[quote="drunner2007"]i tried looking at it for a while. there must be some trick that really simplifies it. i think its safe to say that it is, at the very least, difficult to see any insight and find the solution.[/quote]

does any one care to take a shot at this?


superior algebra
superior algebra solved


inequalities
calculus
integration
logarithms
number theory unsolved
number theory

hmmm, this is frustrating, Bernoulli's Inequality has got to work, but it keeps getting smaller, can anyone help me? :)

[quote="Mathx"]find all answer of$3^n+4^n+...(n+2)^n=(n+3)^n$[/quote]
Is it $n\in N$??

Suppose $\sum_{r=3}^{n+2} r^n = (n+3)^n$.  Comparing the sum with an integral we have $\int_3^{n+3} t^n \, dt < \sum_{r=3}^{n+2} r^n$ so that $\frac{(n+3)^{n+1}}{n+1} -\frac{3^{n+1}}{n+1} < (n+3)^n$ and so $\frac{n+3}{n+1} - \frac{3}{n+1}\left(\frac{3}{n+3}\right)^n < 1$.  But $\epsilon_n = \left(\frac{3}{n+3}\right)^n \rightarrow 0$ (consider its logarithm) and so for sufficiently large $n$ we have $n+3-3\epsilon_n < n+1$, a contradiction: numerical experiment shows that this occurs for $n \ge 4$.  So the only solutions are the well-known cases $n=2$ and $n=3$.

rgep, it'd be nicer if you can provide an elementary (non-calculus) proof... :)

How about dividing both sides by $n^n$?  The right side becomes $(1+3/n)^n$, while the left becomes

$(1+2/n)^n+(1+1/n)^n +1+(1-1/n)^n+(1-2/n)^n+...+(1-(n-3)/n)^n$
$\leq e^2+e+1+e^{-1}+ \dots +e^{-(n-3)}$
$\leq {e^3 \over e-1}$

(to get from the first line to the second, use the "continuously compounded interest is better than simple interest" inequality $1+x \leq e^x$, which holds for both positive and negative $x$.  The third line is just the sum of the infinite geometric series).

$(1+3/n)^n$ is increasing in $n$ (the more frequently interest is compounded, the better), and a bit of computation shows that it is bigger than ${e^3 \over e-1}$ by the time you get to $n=7$ or so.  (a rough estimate is all you need here), and just checking the remaining cases gives you the answer.

Making the two "compound interest" inequalities rigorous may turn out to require calculus after all and there's a bit too much grunge for my taste in estimating where $(1+3/n)^n$ surpasses ${e^3 \over e-1}$ (a weak estimate would be fine here, but would require more work in checking the "small $n$" cases).


I'm curious because my profile is set to autocorrect for DST. And it's still active - despite that DST is over. In my country and every country that I know of.

Profile >> Preferences >> Internationalisation >> Summertime >> No

So the implied answer you gave to my implied question is that DST on this forum is not an auto adjust feature. i.e. it's useless.

Useless? Not quite. It's better than having to switch your time zone.


$8$ people participate in a party.

(1) Among any $5$ people there are $3$ who pairwise know each other.  Prove that there are $4$ people who paiwise know each other.

(2) If Among any $6$ people there are $3$ who pairwise know each other, then can we find $4$ people who pairwise know each other?

Well it's not hard to come up with a counterexample for (2): just take the vertices (=people) of a regular octagon, join the sides, and segments skipping 1 vertex. Removing 2 vertices still guarantees a triangle but there is no K_4... haven't finished (1) yet though

I think this may work for the first part
[hide]
Assume part (a) is false, meaning there's a group of people so that every five of them have a trio knowing each other, but no four of them does.

Lemma: In any group of three people, one pair know each other.

Proof: Suppose there were a trio of people, none of whom knew each other.  (say $a,b,c$).  Then for any two other people $x$ and $y$, the triplet who know each other amongst $a,b,c,x,y$ must include $x$ and $y$.  $x$ and $y$ were arbitrary, so this actually gets us 5 people who all know each other, and we are done.

Lemma: Each person knows at least five other people.

Proof: Consider a person $a$, and suppose that there are three people $a$ doesn't know (say $f,g,h$ WLOG).  By the first lemma $f,g,h$ all know each other.  Furthermore, there is some pair amonst $b,c,d,e$ who don't know each other (WLOG $b$ doesn't know $c$).  Now consider the quintuple. $a,b,c,f,g$.  The trio who know each other can't involve both $b$ and $c$, and therefore can't involve $a$ (who knows neither $f$ nor $g$).  Therefore either $b$ or $c$ knows both $f$ and $g$.  Similarly, either $b$ or $c$ knows both $g$ and $h$, and either $b$ or $c$ knows both $f$ and $h$.  Whichever one knows two of these pairs knows all three of $f,g,h$, and we have our quadruple, again a contradiction.

Therefore each person knows at least five others.  Say (WLOG) $a$ knows $b,c,d,e,f$.  By assumption three of those people know each other, so we again have our quadruple, and a contradiction.[/hide]

Here is a solution for $(1)$.

Inspired by the fact that $4$ is half of $8$, we decide to use the funny trick of showing that the complement of the graph is bipartite. Actually, this is not true, but fortunately the cases where the complement is not bipartite are easy to deal with.

So, supposing that the complement wasn't bipartite, we have three cases.

Case $1$: There is a triangle.

Then, there exist three people $A, B, C$, who are pairwise unacquainted. Now, for any two other people $D$ and $E$, by applying the condition on $(A, B, C, D, E)$ it is apparent that $D$ and $E$ are acquainted with each other. Therefore, we conclude that all of the other five people are pairwise acquainted, and so taking any $4$ of them yields the conclusion.

Case $2$: There is a $5$-cycle.

Then, there exist five people $A, B, C, D, E$ such that $A$ is not acquainted with $B$, $B$ is not acquainted with $C$, $C$ is not acquainted with $D$, $D$ is not acquainted with $E$, and $E$ is not acquainted with $A.$ However, applying the condition of the problem on $(A, B, C, D, E)$ yields a clear contradiction. Hence, this case is absurd and can be disposed of.

Case $3$: There is a $7$-cycle.

Let $A_1, A_2, \cdots, A_7$ be seven people such that $A_i$ is unacquainted with $A_{i+1}$ for each $1 \le i \le 7$, with indices modulo $7.$ By applying the condition on $(A_i, A_{i+1}, A_{i+2}, A_{i+3}, A_{i+4})$, it's clear that the triple of pairwise acquainted people must be $(A_i, A_{i+2}, A_{i+4}).$ This yields that $A_i, A_{i+2}$ are acquainted for each $i$, and so are $A_i, A_{i+4}.$ In particular, we know that $A_i$ and $A_j$ are acquainted if and only if $j \neq i, i \pm 1.$ Consider now the other person $V.$ We claim that $V$ must be acquainted with at least $5$ people among the $A_i$'s. Call a person $A_i$ [b] good [/b] if he/she is acquainted with $V$, and [b] bad [/b] otherwise. We desire to show that at most two people are bad. Observe that if $A_i, A_{i+3}$ are both bad, then applying the condition on $(A_i, A_{i+1}, A_{i+2}, A_{i+3}, V)$ yields a contradiction. Therefore, for each $i$ we know that one of $A_i, A_{i+3}$ is good. This immediately yields that at most half of the people are bad, so $\le 3$. Furthermore, if there are actually three people who are bad, some simple casework using the fact that one of $A_i, A_{i+3}$ is good yields that they must be $A_i, A_{i+1},$ and $A_{i+2}$ for some $1 \le i \le 7.$ However, in this case applying the condition on $(V, A_i, A_{i+1}, A_{i+2}, A_{i+3})$ yields a contradiction. Hence, we obtain that there are at most $2$ bad people. As a result, we know that there exists some $1 \le i \le 7$ such that $A_i, A_{i+2}, A_{i+4}$ are all good, because $2 < \frac{7}{3}.$ From here, we conclude that $(V, A_i, A_{i+2}, A_{i+4})$.

As we've exhausted all cases, we have resolved the problem in the case where the complement graph is not bipartite. If the complement graph is actually bipartite, then we are done by a simple application of Pigeonhole.

$\square$


Some unit cubes are stacked atop a flat 4 by 4 square. The figures show views of the stacks from two different sides. Find the maximum and minimum number of cubes that could be in the stacks. Also give top views of a maximum arrangement and a minimum arrangement with each stack marked with its height.
[asy]
string s = "1010101010111111";
defaultpen(linewidth(0.7));
for(int x=0;x<4;++x) {
for(int y=0;y<4;++y) {
if(hex(substr(s,4*(3-y)+x,1))==1) {
draw((x,y)--(x,y+1)--(x+1,y+1)--(x+1,y)--cycle);
}
}}
label("South View",(2,4),N);
s = "0101110111111111";
for(int x=0;x<4;++x) {
for(int y=0;y<4;++y) {
if(hex(substr(s,4*(3-y)+x,1))==1) {
x=x+5;
draw((x,y)--(x,y+1)--(x+1,y+1)--(x+1,y)--cycle);
x=x-5;
}
}}
label("East View",(7,4),N);[/asy]

this was in an irish programming competition!


calculus
integration
trigonometry
logarithms
complex numbers
real analysis
real analysis unsolved

Evaluate $ \int_0^\infty\frac {x^2}{\sqrt {e^x \minus{} 1}}\,dx.$

How many ways you know to tackle it?

Put $ \tan u \equal{} \sqrt{e^x \minus{} 1}$. Then $ \int_{0}^{\infty} \frac{x^2}{\sqrt{e^x \minus{} 1}} \, dx \equal{} 8 \int_{0}^{\frac{\pi}{2}} \log^2 (\sec u) \, du.$

Plugging $ \log | \sec \theta| \equal{} \ln 2 \plus{} \sum_{k\equal{}1}^{\infty} \frac{(\minus{}1)^k}{k} \cos(2k\theta)$ with some manipulation gives

$ \int_{0}^{\frac{\pi}{2}} \ln^2 (\sec u) \, du \equal{} \frac{\pi}{2} \log^2 2 \plus{} \frac{\pi^3}{24}.$

This gives the result.

Also there is a way of attacking this problem with the aid of the mean value property of $ \ln |1 \plus{} z|$ or something similar, whichi I don't fully remember.

Thanks for your answer.

[quote="sos440"]
Plugging $ \log | \sec \theta| \equal{} \ln 2 \plus{} \sum_{k \equal{} 1}^{\infty} \frac {( \minus{} 1)^k}{k} \cos(2k\theta)$[/quote]
Can you elaborate a proof for this? May be elementary or by using complex numbers? (If you know more than one way, it'd be cool.)

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176509]here[/url] one method you can use to prove that relation.


function
arithmetic sequence
algebra proposed
algebra

A function $f: N\longrightarrow N$ satisfies the following conditions.

a) $f(xy)=f(x)+f(y)-1$ for all $x,y\in\mathbb{N}$

b) There are finite numbers such that $f(x)=1$

c) $f(30)=4$

Find the value of $f(14400)$

I saw it before  :D

Let prove that $f(1)=1$ and if $x>1$ then $f(x)$ is not equal $1$ then we will have $f(14400)=4f(2)+3=11$

Thank you for your reply, ehsan2004 :)

sorry that I can't understand $f(2)=2$.Could you explain in detail? :)

kunny


What is the result when the greatest common factor of 6432 and 132 is increased by 11?

$ 132 \equal{} 2^2 \times 3 \times 11$

$ 6432 \equal{} 2^5 \times 3 \times 67$

We can get $ 2^2 \times 3$ out of both numbers, but no more, so $ 12$ is the GCF. Increasing this by $ 11$, our answer is $ \boxed{23}$


geometry
3D geometry
tetrahedron
ratio
sphere
area of a triangle
Heron\u0027s formula

In a tetrahedron all faces have the same area and all edges are tangent to a sphere. Prove that the tetrahedron is regular. Find the ratio of the volumes of sphere and tetrahedron.

Moldavia 1997

Let the tetrahedron be $ABCD$ and the sphere be $S$. Let $a,b,c,d$ be the lengths of the tangents to $S$ from $A,B,C,D$ respectively. Then $AB=a+b$, $AC=a+c,\dots$ By Heron's formula and the equal area condition for faces $ABC$ and $ABD$, $abc(a+b+c)=abd(a+b+d)$. Sice all of $a,b,c,d$ are positive, we clearly have $c=d$. Similarly, $a=b$ and $b=c$, so all of the edges are equal and the tetrahedron is regular.

The rest is just a calculation. For calculations involving regular tetrahedrons, I would embed in $\mathbb{R}^3$ with vertices at $(1,1,1),\,(1,-1,-1),\,(-1,1,-1),\,(-1,-1,1)$. The center of the sphere in question is obviously the origin, and $(1,0,0)$ is the midpoint of an edge, so the sphere in this case has volume $\frac43\pi$. The tetrahedron has volume $8-4\cdot\frac86=\frac83$, so the ratio (sphere/tetrahedron) is $\frac\pi2$.


algebra
polynomial
function
irrational number
algebra unsolved

Hello !

Would you have any clue on the following problem, I'm completely stuck !!

Let P be a polynomial with rational coefficients, and with degree greater than 2.

Aim is to prove that there exists a sequence of rational numbers (y_k) with lim y_k = + infinity, verifying: equations P(x) = y_k all admit an irrational number as their greatest solution.

Thanks a lot,

Nico

It looks like all you have to do is apply [url=http://en.wikipedia.org/wiki/Eisenstein%27s_criterion]Eisenstein's criterion[/url].  For sufficiently large $ y_k$ a real root will exist, so it suffices to make sure all roots are irrational.

Hello !

Errr... :-) !

Actually, starting from P with rational coefficients, I can more or less guess how to multiply it by a rational so that the problem is stated in terms of a polynomial Q with integers coefficients.

Then, provided the dominant coefficient of Q is positive, given any large enough y_k rational number, there will be a unique solution to Q(x) = y_k (intermediate value theorem), and the associated x will necessarily be the greatest solution of this equation (since Q will be an increasing function at some point).

Now, the problem is to prove this solution will necessarily be irrational, and here, I don't really understand how to apply Eisenstein... Whence would come the prime number to use, and why should the coefficients of Q - y_k verify the Eisenstein criteria conditions ?

If this step solved, then the polynom Q - y_k will be irreductible, hence with irrational roots, hence the x solution of Q(x) = y_k will be irrational.

But Eisenstein kills me here... All the more since for every y_k, it is needed to prove that Q - y_k will be irreducible...

Cheers,

Nico


number theory
greatest common divisor
algorithm
Euclidean algorithm

[hide="solution"]Here is the euclidian algoithm:

(589,236)
(236,353)
(236,117)
(117,119)
(117,2)
(175,2)
.
.
.
(3,2)

The GCF is one[/hide]

What do you mean "linear combination"?

[quote="inom"]Write gcd of 236 and 589 as lineer combination of two numbers.[/quote]

[hide]
what do you mean by linear combination of 2 numbers??

by using http://mathworld.wolfram.com/GreatestCommonDivisor.html,

the GCD is 1[/hide]

Linear combination of x and y is any nuber of the form kx+ly.
Our task is to find for x=589 and y=236 a pair (k,l), such that
k.589+l.236=gcd(589,236)=1. As my previous post shows (maybe not clearly enough), the solution is k=117 and l=-292. Actually, there are infinitely many solutions, but this is the smallest one (the others are k=117+236m, l=-292-589m where m is aribitrary integer).


prove,

$ 2sin(\dfrac{\pi}{n})cos(\dfrac{\pi}{n})sin(\dfrac{2\pi}{n})sin(\dfrac{3\pi}{n})... sin(\dfrac{(n \minus{} 1)\pi}{n}) \equal{} \dfrac{n}{2^{n \minus{} 1}}$

good luck!

Hello,
It's pretty well-known. But the only related post I can find is this one: http://www.mathlinks.ro/viewtopic.php?p=231507#231507
In the above link joml88 gave the solution for the special case $ n\equal{}5$ but you can basically use the same argument for any $ n$.

I have a question about joml88's method:

From the identity $ |1\minus{}e^{i\theta}|\equal{}2\sin\left(\frac{\theta}{2}\right)$,
how do you get $ \prod_{k\equal{}1}^4 \sin\left(k \pi/5\right) \equal{} \frac{1}{16}\prod_{k\equal{}1}^4 |1\minus{}e^{ik\pi/5}|$? Shouldn't it be $ \frac{1}{16}\prod_{k\equal{}1}^4 |1\minus{}e^{i2k\pi/5}|$ instead? I checked that they both evaluate to the same thing, but why are they the same?

[quote="kcn2rivers"]I have a question about joml88's method:

From the identity $ |1 \minus{} e^{i\theta}| \equal{} 2\sin\left(\frac {\theta}{2}\right)$,
how do you get $ \prod_{k \equal{} 1}^4 \sin\left(k \pi/5\right) \equal{} \frac {1}{16}\prod_{k \equal{} 1}^4 |1 \minus{} e^{ik\pi/5}|$? Shouldn't it be $ \frac {1}{16}\prod_{k \equal{} 1}^4 |1 \minus{} e^{i2k\pi/5}|$ instead? I checked that they both evaluate to the same thing, but why are they the same?[/quote]it should be $ \frac {1}{16}\prod_{k \equal{} 1}^4 |1 \minus{} e^{i2k\pi/5}|$.
The intended product evaluates to:
$ \prod_{k \equal{} 1}^4 |1 \minus{} e^{i2k\pi/5}|\equal{}\prod_{k \equal{} 1}^4 \sqrt{(1 \minus{} e^{i2k\pi/5})(1 \minus{} e^{\minus{}i2k\pi/5})}$ $ \equal{}\prod_{k \equal{} 1}^4 (1 \minus{} e^{i2k\pi/5})\equal{}1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\Big|_{x\equal{}1}\equal{}5$
To explain the last equality, the monotonic polynomial that contains all the 5th roots of unity is $ x^5\minus{}1$, and the polynomial that contains the 5th roots of unity (except 1) is $ \frac{x^5\minus{}1}{x\minus{}1}\equal{}1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4$.
In the meanwhile, the other product evaluates to:
$ \prod_{k \equal{} 1}^4 |1 \minus{} e^{ik\pi/5}|\equal{}\prod_{k \equal{} 1}^4 \sqrt{(1 \minus{} e^{ik\pi/5})(1 \minus{} e^{\minus{}ik\pi/5})}$ $ \equal{}\sqrt{\prod_{k \equal{} 1}^4(1 \minus{} e^{ik\pi/5})(1 \minus{} e^{\minus{}ik\pi/5})}\equal{}1\plus{}x^2\plus{}x^4\plus{}x^6\plus{}x^8\Big|_{x\equal{}1}\equal{}5$
To explain the last equality, the monotonic polynomial that contains all the 10th roots of unity is $ x^{10}\minus{}1$, and the polynomial that contains the 10th roots of unity (except 1 and -1) is $ \frac{x^{10}\minus{}1}{(x\minus{}1)(x\plus{}1)}\equal{}1\plus{}x^2\plus{}x^4\plus{}x^6\plus{}x^8$.


Let $ a,b,c>0,$ such that $ \displaystyle\frac{a\plus{}b}{bc}\equal{}\frac{b\plus{}c}{ca}\equal{}\frac{c\plus{}a}{ab}.$ Prove that $ a\equal{}b\equal{}c.$

[hide]assume $ a > b,c$

$ \frac {a \plus{} b}{bc} > \frac {b \plus{} c}{ca}$

contradiction

Assuming any one of the three is the largest leads to a similar contradiction.

So $ a \equal{} b \equal{} c$[/hide]

[hide="solution"]
Clear all the denominators to get the set of equations \begin{align*}a^2+ab&=b^2+bc\\b^2+bc&=c^2+ac\\a^2+ab&=c^2+ac.\end{align*} Bring all the terms to one side and factor each equation to get \begin{align*}(a-b)(a+b)+b(a-c)&=0\\(b-c)(b+c)+c(b-a)&=0\\(a-c)(a+c)+a(b-c)&=0.\end{align*} Since $ a,b,c>0$ and each expression equals 0, \begin{align*}a&=c\\b&=c\\a&=b\end{align*} so $ a=b=c.$[/hide]


There exist non-negative integers $a$ and $b$ such that for any $a < x < b$, it is true that $\sqrt{x} > x$.  Find the sum $a + b$.

[hide]Isn't it just a=0, b=1, and a+b=1?  :huh: Not like a number 30 to me. [/hide]

Yes, that was the intended answer.  I would argue, however, that any nonnegative integer should be accepted as an answer.  For example, we could choose $a = 0$ and $b=0$, or $a = 1$ and $b= 1$, or even $a = 3$ and $b = 2$.  Let's look at the statement that results when $a = 0$ and $b = 0$:

"For any $0 < x < 0$, it is true that $\sqrt{x} > x$."

I would say that this statement is true.  The hypothesis $0 < x < 0$ is always false, which means, by the usual definitions of logic, that the statement as a whole is true.

Do you agree that there is an error in the problem?

I don't think they're expecting that kind of logic from 6th-8th graders. It does seem ridiculously easy for a State #30, though.

THey probably should've specified that x must exist....


[color=blue]Let n be a natural number and m in {1,2,...,n}.
Prove that S_m( n )=sum(k=1,n) k^m (n,k) can be divided by n*2^(n-m).
where (n,k)=n!/(k!(n-k)!). [/color]

cheers! :D :D

We can easy step-by-step calculate sum fm(x) = \sum xk-1km(n,k). We will use equality Sm(n)=fm(1).

For m=1 : f1(x) = ((1+x)n)'=n(1+x)n-1 ==> S1(n)=n.2n-1.

Next step : fm+1(x)=(x.fm(x))' ==> fm(x) is a sum of n.M.(1+x)sxt where M is integer, s \geq n-m ==> n(1+x)n-m | fm(x) ==>  n.2n-m | Sm(n).


Can we have some type of ARML Seminar, kind of like we did with the AIME?

i was wondering about a seminar like we did with the AIME

[quote="masamune2387"]There is an ARML math jam, I believe.[/quote]

One is scheduled for Mid-May iirc.


The points of a circle are colored in green and yellow, such that every equilateral triangle inscribed in the circle has exactly 2 vertices colored in yellow. Prove that there exist a square inscribed in the circle which has at least 3 vertices colored in yellow.

Consider any regular $12$-gon inscribed in the circle. Its vertices can be divided into $4$ groups of three which are vertices of an equilateral triangle. Thus, there are $8$ yellow points among the $12$.
Now divide the $12$ vertices into $3$ groups of four vertices which ar the vertices of a square. There are $3$ squares and $8$ yellow points so that, from the pigeon-hole principle, there is at least one square with $3$ yellow vertices.

Pierre.


search
geometry
geometric transformation
rotation
geometry proposed

In following figure, three triangles $ OPP'$,$ OQQ'$,$ ORR'$ are equilateral. Let $ P''$,$ Q''$,$ R''$ be midpoints of $ P'Q$,$ Q'R$,$ R'P$ respectively. Prove that the triangle $ P''Q''R''$ is equilateral.

I search a few synthetic proofs, so happy to your help. Thanks for reading my post.

1.	Q’’A=1/2P’Q’=1/2PQ=BP’’
2.	P’R’=PR  => ∠ABR’=60  BR''=BA  => BR’’=AR’’=BA
3.	<Q’’A  BP’’>=60  <AR’’  BR’’>=60    => ∠P’’BR’’=∠Q’’AR’’
4.	△Q’’AR’’=△P’’BR’’    ∠Q’’R’’P’’=∠AR’’B=60
5.	P’’Q’’R’’ is an Equilateral triangle

[quote="77ant"]Thanks for your reply, plane geometry.
I have a few odds.
What is the point $ A$ ?  Why $ PR$=$ P'R'\Longrightarrow\angle ABR'$=60 and why $ BR'$=$ R''A$ ?
Why are $ B$,$ A$,$ Q''$ collinear ? Why $ \angle{(Q''A,BP'')}$=60 and $ \angle{(AR'',BR'')}$=60 ? :([/quote]
A is the midpoint of P'R
sorry I made a tope    it's BA=BR''  not  BR’=R’’A
notice the angle between PR and P'R'   =   the angle between BA and BR'' =60

<Q’’A BP’’>=60 <AR’’ BR’’>=60  , use  rotation  you will find the conclusion

ps:I don't see B,A,Q'' are collinear  and they are not collinear

See: http://www.mathlinks.ro/Forum/viewtopic.php?p=1009087#1009087

Best regards,
sunken rock

We can prove easily this problem with complex numbers. Choose the origin $ O$ for  the complex plane and denote $ X(x)$ - the point $ X$ with the affix $ x$  .  Suppose  $ P(p)$ , $ Q(q)$ , $ R(r)$ . Then $ P_1(wp)$ , $ Q_1(wq)$ , $ R_1(wr)$ , where $ w^3\plus{}1\equal{}0$ , i.e. $ w^2\minus{}w\plus{}1\equal{}0$ . Observe that $ 2p_2\equal{}wp\plus{}q$ , $ 2q_2\equal{}wq\plus{}r$ , $ 2r_2\equal{}wr\plus{}p$ . The triangle $ P_2Q_2R_2$ is equilateral iff $ p_2^2\plus{}q_2^2\plus{}r_2^2\equal{}p_2q_2\plus{}q_2r_2\plus{}r_2p_2$ iff $ \left(w^2\minus{}w\plus{}1\right)\cdot\left(\sum p^2\minus{}\sum pq\right)\equal{}0$ , what is truly.


Let $ a/b$, $ c/d$, $ e/f$ be three fractions with $ b,d,f\in{\mathbb{N}}$,  and let $ \abs{ad - bc} = 1$, then $ \exists{x,y\in{\mathbb{Z}}}$ s.t. $ ax + cy = e, bx + dy = f$ and that $ e/f$ lies (properly) between $ a/b$ and $ c/d$, when these are positive.

[quote="kunny"]Don't you forget the condition something else, say, relatively prime.[/quote]

nope, that's all the conditions you need

Now, it's the time nsato's explain should come.

$ ad \minus{} bc \equal{} 1$ gives the relative primality you want.  But what is the problem asking for?  Are we given $ \frac{e}{f}$ and asked to find $ x, y$ (this is clearly not in general possible)?  Are we asked to prove that $ \frac{e}{f}$ satisfying the condition is always between the two given fractions (this is trivial)?

Yes, this problem doesn't make sense. :wink:

This is a similar problem.

Let $ a\equal{}5x\plus{}4y,\ b\equal{}6x\plus{}5y$ for $ x,\ y\in\mathbb{N}$, prove the following contexts.

(1) $ (a,\ b)\equal{}(x,\ y)$

(2) If we choose $ x,\ y$ appropriately, any rational numbers $ r$ with $ \frac{4}{5}<r<\frac{5}{6}$ can be expressed as $ r\equal{}\frac{a}{b}$.

1982 Gihu University entrance exam/medical


[quote="José"]Supposing that I want to propose a problem for an IMO, how can I do it??????.[/quote]You talk to the IMO Team Leader from your country and present him/her (in your case it's a her :) ) the problem, with solution. The team leader can afterwards send up to 6 problems to the IMO committee as proposals from his/her country.

Thanks!!!

[quote="Valentin Vornicu"]You talk to the IMO Team Leader from your country and present him/her (in your case it's a her :) ) [/quote]

Do you know her???

Of course I know her :) I know almost all the Leaders ;)


trigonometry
limit
calculus
calculus computations

Find constants a and b in order that

$\frac{1+acos2x+bcos4x}{x^{4}}$ may have a finite limit as x approaches 0,and find the value of this limit

[quote="benorin"]Nicely done, FireyHydra.[/quote]
Thanks! :)

[I did not use proper limit notation in my solution. To be sure, $L$ is evaluated for $x\rightarrow 0.$]

[quote="Michael Niland"]$L\equiv\frac{1+a\cos2x+b\cos4x}{x^{4}}\in R$ $\Longrightarrow$ Find the constants $a$ and $b\ .$[/quote]
$y: =2x\Longrightarrow$ $L=16\cdot \lim_{y\to 0}\frac{1+a\cos y+b\left(2\cos^{2}y-1\right)}{y^{4}}=$ $16\cdot\lim_{y\to 0}\left(\frac{1-\cos y}{y^{2}}\right)^{2}\cdot \frac{2b\cos^{2}y+a\cos y+1-b}{(1-\cos y)^{2}}=$ $4\cdot \lim_{y\to 0}\frac{2b\cos^{2}y+a\cos y+1-b}{(1-\cos y)^{2}}\ .$
$z: =\cos y\Longrightarrow$ $L=4\cdot\lim_{z\to 1}\frac{2bz^{2}+az+1-b}{(z-1)^{2}}\ .$ I used the well-known limit $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}\ .$
$t: =z-1\Longrightarrow$ $4\cdot\lim_{t\to 0}\frac{2bt^{2}+(a+4b)t+a+b+1}{t^{2}}$ $\Longrightarrow$ $a+b+1=0$ and $a+4b=0$ $\Longrightarrow$ $a=-\frac{4}{3}$ , $b=\frac{1}{3}$ and $L=\frac{8}{3}\ .$

Yes, my solution is same as yours, Virgil Nicula.
This is a standard solution for Japanese high school students. :D


[quote="gigaman"]Given x,y,z positive such that xyz=1. Prove

$ \frac {3\sqrt {xy} \plus{} \sqrt {z}}{\sqrt {(x \plus{} y)(z \plus{} 1)}} \plus{} \frac {3\sqrt {yz} \plus{} \sqrt {x}}{\sqrt {(y \plus{} z)(x \plus{} 1)}} \plus{} \frac {3\sqrt {zx} \plus{} \sqrt {y}}{\sqrt {(z \plus{} x)(y \plus{} 1)}}\ge 6$[/quote]

it's not true for $ x\equal{}2$,$ y\equal{}\frac{1}{2}$, $ z\equal{}1$

[quote="SaYaT"][quote="gigaman"]Given x,y,z positive such that xyz=1. Prove

$ \frac {3\sqrt {xy} \plus{} \sqrt {z}}{\sqrt {(x \plus{} y)(z \plus{} 1)}} \plus{} \frac {3\sqrt {yz} \plus{} \sqrt {x}}{\sqrt {(y \plus{} z)(x \plus{} 1)}} \plus{} \frac {3\sqrt {zx} \plus{} \sqrt {y}}{\sqrt {(z \plus{} x)(y \plus{} 1)}}\ge 6$[/quote]

it's not true for $ x \equal{} 2$,$ y \equal{} \frac {1}{2}$, $ z \equal{} 1$[/quote]

Sorry, I made a little mistake in calculation

But that one is true :D

Given x,y,z positive such that xyz=1. Prove

$ \frac {3xy \plus{} 1}{\sqrt {(x \plus{} y)(z \plus{} 1)}} \plus{} \frac {3yz \plus{} 1}{\sqrt {y \plus{} z)(x \plus{} 1)}} \plus{} \frac {3zx \plus{} 1}{\sqrt {(z \plus{} x)(y \plus{} 1)}}\ge 6$

[quote="gigaman"]

Given x,y,z positive such that xyz=1. Prove

$ \frac {3xy \plus{} 1}{\sqrt {(x \plus{} y)(z \plus{} 1)}} \plus{} \frac {3yz \plus{} 1}{\sqrt {y \plus{} z)(x \plus{} 1)}} \plus{} \frac {3zx \plus{} 1}{\sqrt {(z \plus{} x)(y \plus{} 1)}}\ge 6$[/quote]

This ineq can solve by Am-GM in one line :D

Yes , only one line
$ LHS \ge 2 \sum \frac{ \frac{3}{z}\plus{}1}{ \sum \frac{1}{x}\plus{}1 }\equal{}6$

[quote="Vincent Gilbert"]Yes , only one line
$ LHS \ge 2 \sum \frac {\frac {3}{z} \plus{} 1}{\sum \frac {1}{x} \plus{} 1 } \equal{} 6$[/quote]
Can you solve clearly? Please... :blush:

Ofcourse,
$ \sqrt{(x\plus{}y)(z\plus{}1) }\equal{} \sqrt{ ( \frac{1}{x}\plus{}\frac{1}{y})(1\plus{}\frac{1}{z}) } \le \frac{  \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}1}{2} (xyz\equal{}1)$
$ \equal{}\frac{1}{2}( xy\plus{}yz\plus{}zx\plus{}1)$


Prove that every positive integer which is not a member of the infinite set below is equal to the sum of two or more distinct numbers of the set \[\{ 3,-2, 2^{2}3,-2^{3}, \cdots, 2^{2k}3,-2^{2k+1}, \cdots \}=\{3,-2, 12,-8, 48,-32, 192, \cdots \}.\]

lets split that set in two series:
$ a_i \equal{} 2^{2i}3 \equal{} 4^i 3$ for $ i \ge 0$ and
$ b_i \equal{} \minus{} 2^{2i \plus{} 1} \equal{} \minus{} 4^i 2$ for $ i \ge 0$
Defining:
$ c_i \equal{} a_i \plus{} b_i \equal{} 4^i$ for $ i \ge 0$
$ d_i \equal{} b_i \equal{} \minus{} 2 \times 4^i$ for $ i \ge 0$
$ e_i \equal{} a_i \equal{} 3 \times 4^i$ for $ i \ge 0$

Every positive integer can be written as a sum of powers of 4 using the digits $ \{0,1, \minus{} 2,3\}$
by replacing in finite steps the pair of digits $ \cdots (x)(2) \cdots$ with $ \cdots (x \plus{} 1)( \minus{} 2) \cdots$
then every positive integer is equal to the sum of distinct numbers of the original set.


[quote="mathgeniuse^ln(x)"]Wow!  Okay, but is this considering the fact that it could be arranged odd odd even even.  I think you have only shown even even even even.[/quote]
OK, I guess my initial explanation is somewhat lengthy. Here is a much simpler way of saying the same:

We want to roll one of the even numbers twice and one of the odd numbers twice. There are 3 even numbers. So the number of ways to choose one of the even numbers is $\binom{3}{1}= 3$. Similarly the number of ways to choose one of the odd numbers is $3$.  The probability of getting the chosen even number in any of the two rolls is $\frac{1}{6}\cdot \frac{1}{6}$. Similarly the probability of getting the chosen odd number in any of the two rolls is $\frac{1}{6}\cdot \frac{1}{6}$. The two even and two odd numbers can be rolled in $\binom{4}{2}= 6$ permutations.

So the desired probability is, \[6 \times 3 \times 3 \times \frac{1}{6^{4}}= \frac{1}{24}\]

Oh, okay.  Thank you for your explanation.

[hide]
I guess...
$\frac{{6 \cdot 1 \cdot 3 \cdot 1 \cdot 3}}{6^{4}}$

Extra 3 is for the different ways you can place 2 cubes in 4 spots (with the 6 always in front).[/hide]

[quote="mthiyaga"][quote="mthiyaga"]Four number cubes, with the numbers 1 to 6, one per face, are rolled. What is the probability that two of the numbers are the same even number and that the other two numbers are the same odd number? Express your answer as a common fraction.[/quote]
[hide]
Method 1 - Using counting
--------------------------------

There are three even numbers, 2, 4, 6. The number of ways to choose two 2's from 4 cubes is $\binom{4}{2}= 6$. So we see that there are $3 \cdot 6 = 18$ ways to choose two of either 2, 4, or 6 from the 4 cubes.

Having chosen two even numbers, there are two cubes left. We have three odd numbers, 1, 3, and 5. The number of ways to choose two 1's from 2 cubes is $\binom{2}{2}= 1$. So we see that there are $3 \cdot 1 = 3$ ways to choose two of either 1, 3, or 5 from the remaining 2 cubes.

This gives us a total of $18 \cdot 3 = 54$ ways to choose two even numbers that are the same and two odd numbers that are the same from 4 cubes. Since there are $6\cdot 6 \cdot 6 \cdot 6 = 1296$ possibilities with 4 cubes, we find that the probability that two of the numbers are the same even number and that the other two numbers are the same odd number is, \[\frac{54}{1296}= \boxed{\frac{1}{24}}.\] Method II - Using Probability
-----------------------------------

We want two same even numbers on any of the two cubes and two same odd numbers in the other two cubes. The number of ways to choose any two cubes from four cubes is $\binom{4}{2}= 6$. Probability of obtaining an even number in the first cube is $\frac{1}{2}$. Having obtained an even number in the first cube, obtaining the same number in the second cube is $\frac{1}{6}$. So this gives us a total probability for the getting the same even numbers on any of the two cubes is $6 \times \frac{1}{2}\times \frac{1}{6}= \frac{1}{2}$.

Now there only two cubes left. The probability of obtaining an odd number in the first cube is $\frac{1}{2}$. Having obtained on odd number in the first cube, the probability of obtaining the same number in the second cube is $\frac{1}{6}$. So the probability of obtaining the same odd numbers in any of the two cubes is $\frac{1}{2}\times \frac{1}{6}= \frac{1}{12}$. Note that we could have very well chosen the odd numbers first and even numbers after. It doesn't matter.

So the probability of obtaining two same even numbers and two same odd numbers with four cubes is \[\frac{1}{2}\times \frac{1}{12}= \boxed{\frac{1}{24}}.\] [/hide][/quote]Wow that is a neat answer

[hide="Slight Variant"]Step 1:  The probability of rolling exactly two evens and two odds is 6/16.  [This is the same as flipping exactly two heads out of four coin tosses.]

Step 2:  Given we are known to have exactly two evens, arbitrarily label one of the even dice as "A" and the other as "B".  Die "A" is guaranteed to be right.  Since we know die "B" is even, it matches die "A" with probability 1/3.

Step 3:  Given we are known to have exactly two odds, arbitrarily label one of the odd dice as "C" and the other as "D".  Die "C" is guaranteed to be right.  Since we know die "D" is odd, it matches die "C" with probability 1/3.

Answer:  $\frac{6}{16}\times\frac13\times\frac13=\frac{1}{24}$[/hide]


geometry
complex numbers
geometry proposed

One constructs the equilateral triangles BCD, CAE and ABF externally, on the sides of the triangle ABC, and names M, N and P their centres.
Let's M' be the symmetrical of M w.r.t. BC, N' and P' being analogously obtained.
Then the following equality holds:
S[MNP] - S[M'N'P'] = S[ABC]

[i]S[XYZ] means the area of the triangle XYZ.[/i]

Best regards,
sunken rock

Hello!
The triangles $ MNP$ and $ M'N'P'$ are equilateral.
$ AN = AN' = \frac {b\sqrt {3}}{3}$ and $ AP = AP' = \frac {c\sqrt {3}}{3}$

$ NP^{2} = \frac {1}{3}\cdot[b^{2} + c^{2} - 2bc\cdot cos(\angle A + 60^{\circ})]$
$ S[MNP] = \frac {NP^{2}\cdot\sqrt {3}}{4} = \frac {\sqrt {3}}{12}\cdot[b^{2} + c^{2} - 2bc\cdot cos(\angle A + 60^{\circ})]$

$ N'P'^{2} = \frac {1}{3}\cdot[b^{2} + c^{2} - 2bc\cdot cos(\angle A - 60^{\circ})]$
$ S[M'N'P'] = \frac {N'P'^{2}\cdot\sqrt {3}}{4} = \frac {\sqrt {3}}{12}\cdot[b^{2} + c^{2} - 2bc\cdot cos(\angle A - 60^{\circ})]$

$ S[MNP] - S[M'N'P'] =$
$ = \frac {\sqrt {3}}{12}\cdot[b^{2} + c^{2} - 2bc\cdot cos(\angle A + 60^{\circ}) - b^{2} - c^{2} + 2bc\cdot cos(\angle A - 60^{\circ})] =$
$ = \frac {2bc\sqrt {3}}{12}\cdot[cos(\angle A - 60^{\circ}) - cos(\angle A + 60^{\circ})] =$
$ = \frac {bc\sqrt {3}}{6}\cdot 2\cdot sin \angle A\cdot\frac {\sqrt {3}}{2} = \frac {bc\cdot sin\angle A}{2} = S[ABC]\Rightarrow$
$ \Rightarrow S[MNP] - S[M'N'P'] = S[ABC]$ :)

Alternatively, one can use complex numbers (I won't write the computations) to obtain that $ [MNP] \equal{} \frac {1}{2}[ABC] \plus{} \frac {\sqrt {3}}{24} \left(a^{2} \plus{} b^{2} \plus{} c^{2}\right)$ and $ [M'N'P'] \equal{} \frac {1}{2}[ABC] \minus{} \frac {\sqrt {3}}{24} \left(a^{2} \plus{} b^{2} \plus{} c^{2}\right).$

Do you happen to have a synthetic solution Stan?

Hello, everybody,

Attached, a half-synthetic solution.
For the 2nd relation there I could not find a proper synthetic calculation, since the triangles BCM' and ABP' (see my drawing) have the area M'BP' overlapping.

Best regards,
sunken rock


Let me start with Physics i will post an unanswered question here from the Physics Forum
[size=150][color=blue] Problem 1(P)[/color][/size]
There are two discs rotating with angular speed $ \omega$ in oppsite direction.Both of them have some number of mirrors attached to them in such a way that a light ray coming from a source kept in between them gets reflected n times at each mirror in one second and only when each mirror is vertical.The ray started Horizontally
1. Find the angular speed $ \Omega$ of such a ray.
2.The speed $ v_{n}$ of the nth image

X plays the following game.
He generates a sequence where he initially puts down (deliberately) the values 1 and 4 as the 1st 2 terms of the sequence.
The rules of the game are :
1. Whenever you have a perfect square which is deliberately put down by you as the last term of a sequence, then the next term of the sequence is got by adding it and the previous term of the sequence.
2. If the last term of the sequence is a non perfect square
Case 1 : not divisible by 3
Put deliberately the next perfect square of the sequence (next to the last perfect square in the sequence) and follow step1.
Case 2 : divisible by 3
Subtract 5 to get the next term of the sequence
3. If the last term of the sequence is a perfect square got by 2 (i.e. not deliberately added) then add to the sequence the next perfect square and follow step 1.
For Example the sequence proceeds as :
1,4
1,4,5
1,4,5,9
1,4,5,9,14
1,4,5,9,14,16
1,4,5,9,14,16,30
1,4,5,9,14,16,30,25
1,4,5,9,14,16,30,25,36
1,4,5,9,14,16,30,25,36,61
1,4,5,9,14,16,30,25,36,61,49
1,4,5,9,14,16,30,25,36,61,49,110
1,4,5,9,14,16,30,25,36,61,49,110,64
1,4,5,9,14,16,30,25,36,61,49,110,64,174
1,4,5,9,14,16,30,25,36,61,49,110,64,174,169
...............................................................

The question is : While playing such a game, how many perfect squares does X miss till 1,00,00,000.

Hint
[hide]
Follow the pattern by generating the sequence.
For example you will find that  are missing from the sequence first.
Try to generate a pattern for no. of nos missing at a go each time and hence try to deduce the pattern

[/hide]

as a side issue no doubt Rohit and shreyas are working hard for  INOI  :D

Not me da only Shreyas and his Friend Kartik Prashant.  :P
Those two are hell a lot good in Computers.
And they seem to be goin' on  :wow:

i have a friend of mine who wrote that even last year and he is also thrugh this year hope he gets through


integration
logarithms
function
calculus
calculus computations

\[ 2xyy^{\prime}\equal{}4x^{3}\plus{}3y^{2}\]
put in in the form
\[ y^{\prime}\plus{}p(x)y\equal{}q(x)y^{n}\]
which is
\[ y^{\prime}\minus{}\frac{3}{2x}y\equal{}2x^{2}y^{\minus{}1}\]
so we see that $ n \equal{}\minus{}1$. Now substitute $ v \equal{} y^{1\minus{}n}$

which is $ v \equal{} y^{1\minus{}(\minus{}1)}\equal{} y^{2}$ so that $ \frac{dv}{dx}\equal{} 2yy^{\prime}$. Now substitute this into the original DE:
\[ xv^{\prime}\equal{}4x^{3}\plus{}3v\]

\[ v^{\prime}\minus{}\frac{3}{x}v\equal{}4x^{2}\]

which is linear in $ v$, so use the integrating factor $ \mu\equal{}e^{\minus{}\int\frac{3}{x}dx}\equal{}e^{\minus{}3\ln |x|}\equal{}x^{\minus{}3}$ to get

\[ \left( x^{\minus{}3}v\right)^{\prime}\equal{}4x^{\minus{}1}\]

\[ \int\left( x^{\minus{}3}v\right)^{\prime}dx\equal{}4\int x^{\minus{}1}dx\]

\[ x^{\minus{}3}v\equal{}4\ln |x|\plus{}C\]

solve for $ v$ and backwards substitute $ v\equal{}y^{2}$ to get

\[ v\equal{}4x^{3}\ln |x|\plus{}Cx^{3}\]

\[ \boxed{y^{2}\equal{}4x^{3}\ln |x|\plus{}Cx^{3}}\]

Put $ z\equal{}y^{2}$.

We have linear DE $ xz'\equal{}4x^{3}\plus{}3z$.

Another method:

let $ y \equal{} u\cdot v$, where u and v are both functions of x that we will determine. Substituting in the Bernoulli differential equation and rearranging terms we get

$ 2xuv(u'v\plus{}uv') \equal{} 4x^{3}\plus{}3u^{2}v^{2}$

$ \Rightarrow  2xuu'v^{2}\plus{}2xu^{2}vv'\minus{}3u^{2}v^{2}\equal{} 4x^{3}$

$ \Rightarrow  v^{2}(2xuu'\minus{}3u^{2})\plus{}2xu^{2}vv' \equal{} 4x^{3}$.    (1)

Without loss of generality, let's now suppose that our function u is such that $ 2xuu'\minus{}3u^{2}\equal{} 0$, and so

$ 2x\frac{du}{dx}\equal{} 3u\Rightarrow \frac{du}{u}\equal{}\frac{3}{2}\cdot\frac{1}{x}dx\Rightarrow  u \equal{} x^{3/2}$.

Now substituting our function u into (1) we get

$ 2x\cdot x^{3}\cdot vv' \equal{} 4x^{3}\Rightarrow  vv' \equal{}\frac{2}{x}\Rightarrow  vdv \equal{}\frac{2}{x}dx$

$ \Rightarrow \frac{v^{2}}{2}\equal{} 2\ln|x|\plus{}C\Rightarrow  v^{2}\equal{} 4\ln|x|\plus{}C$

and so

$ y^{2}\equal{} u^{2}v^{2}\equal{} x^{3}(4\ln|x|\plus{}C)\Rightarrow  y^{2}\equal{} 4x^{3}\ln|x|\plus{}Cx^{3}$.


I believe this question has been posed before, but due to the increasing number of responses and posts from users outside of the state, many of which are not really math related, I was wondering whether this forum should remain an open forum where anyone can post, or a more private group.

Please let me know what you think.

does it really matter that much if people from other states post in here? if someone really thinks that's a problem, then maybe we could close it, but i don't see anything wrong with alabama people posting in here.

Yeah, I think it's kind of fun to have the Vestavia kids drop in every once in a while.

By the way, Howard, are you still a moderator of this forum?  (Your username isn't at the top any more.)  We don't usually need a mod, but recently there was some annoying spam in a thread which was happily deleted.


How many terms are there in the expansion $ (2n\plus{}3o\minus{}4)^5$ after like terms are combined? I've seen a lot of problems like this, and im pretty sure that Pascal's triangle helps. Can anyone explain the method to me?

[hide="Hint"] Each term is of the form $ n^a o^b 1^c$, where $ a, b, c \ge 0, a \plus{} b \plus{} c \equal{} 5$.  [/hide]

[hide]Using what t0rajir0u said, we need to find the number solutions to $ a\plus{}b\plus{}c\equal{}5$ in nonnegative integers.

Using [url=http://mathforum.org/library/drmath/view/52269.html]balls and urns[/url], we have $ \binom{5\plus{}2}{2}\equal{}\binom{7}{2}\equal{}\frac{7 \cdot 6}{2} \equal{} \boxed{21}$.[/hide]

[quote="123456789"][quote="madness"][quote="123456789"][hide]Using what t0rajir0u said, we need to find the number solutions to $ a \plus{} b \plus{} c \equal{} 5$ in nonnegative integers.

Using [url=http://mathforum.org/library/drmath/view/52269.html]balls and urns[/url], we have $ \binom{5 \plus{} 2}{2} \equal{} \binom{7}{2} \equal{} \frac {7 \cdot 6}{2} \equal{} \boxed{21}$.[/hide][/quote]

[hide]Or we could do it by generating fns  :) [/hide][/quote]
[hide]
What would the function look like?[/hide][/quote]

coefficient of $ x^5$ in the expansion of $ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\plus{}x^5)^5$

[quote="madness"][/quote][quote="123456789"][quote="madness"][quote="123456789"][hide]Using what t0rajir0u said, we need to find the number solutions to $ a + b + c = 5$ in nonnegative integers.

Using [url=http://mathforum.org/library/drmath/view/52269.html]balls and urns[/url], we have $ \binom{5 + 2}{2} = \binom{7}{2} = \frac {7 \cdot 6}{2} = \boxed{21}$.[/hide][/quote]

[hide]Or we could do it by generating fns  :) [/hide][/quote]
[hide]
What would the function look like?[/hide][/quote][quote="madness"]

coefficient of $ x^5$ in the expansion of $ (1 + x + x^2 + x^3 + x^4 + x^5)^5$[/quote]

Oh yea.


Let $ a$, $ b$ and $ c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Then prove that
\[ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}
\]

i multiplied out but cant finish: here waht i have:

Obviously denominators are all positive.

Clear denominator:  we have $\sum 2(1-ab)(1-bc) \geq 9\prod (1-ab)$

this is:

$6 + 2abc(a+b+c) - 4(ab+bc+ca) \leq 9 + 9(abc)^2 - 9(ab+bc+ca) + 9abc(a+b+c)$

or

$5(ab+bc+ca) \leq 3 + 7abc(a+b+c) + 9(abc)^2$

Maybe I did some mistakes, but after homegenizing I couldnt do it just with Shur and Muirhead
I also think, that you've got wrong sign near $9(abc)^2$.

This inequality was publised in the last issue of Crux Mathematicorum Journal.

After homogenizing the inequality becomes

$2\sum a^4bc+9a^2b^2c^2+9\sum a^4b^2+3\sum a^6\geq 3\sum a^3b^2c+10\sum a^3b^3+5\sum a^5b$

This seems very strong.  Even using Schur ($\sum a^6+\sum a^4bc\geq \sum a^5b$) I can't kill the $\sum a^5b$ on the RHS.

my solution is ugly, but i hope it's correct:
The initial ineq is equivalent to
$S_a (b-c)^2 + S_b (c-a)^2 + S_c (a-b)^2 \ge 0$, where $S_a = 5(b^3 a + c^3 a) + 6(b^2c^2 + bca^2) - a^3(b+c)$
WLOG assume $a \ge b \ge c$
Now it's apparent that $S_c \ge 0$, $S_b \ge 0$, $S_b + S_a \ge 0$, $a-c \ge b-c$, so we are done.

after multiplying with the common denominator you can continue as follows.

http://www.mathlinks.ro/Forum/viewtopic.php?t=56979

 :rotfl:

really bad isn't it?

[quote="silouan"][quote="Soarer"]The initial ineq is equivalent to
$S_a (b-c)^2 + S_b (c-a)^2 + S_c (a-b)^2 \ge 0$, where $S_a = 5(b^3 a + c^3 a) + 6(b^2c^2 + bca^2) - a^3(b+c)$
[/quote]

How did you reach in this form ?[/quote]
It is equivalent to $\sum \frac{1-3ab}{1-ab} \ge 0$, as $a^2+b^2+c^2=1$, it's equivalent to $\frac{2(a-b)^2+2(c^2-ab)}{1-ab} \ge 0$, now use the identity $c^2-ab = (c-a)(c+a)+(c-b)(c+b)+(a-b)^2$, and then use the old method to group the terms.

It show that S.o.S is very useful for this kind.

[quote="dannyikc"]Hint: Use 1 - c^2 = a^2 + b^2 >= 2ab and make use of the convexity of 1/(1 + x^2).[/quote]
it's only convex on an interval, not on the whole of [0,1]

[quote="Detective"][quote="Soarer"]now use the identity $c^2-ab = (c-a)(c+a)+(c-b)(c+b)+(a-b)^2$, and then use the old method to group the terms.[/quote]

You made a little mistake there, it should be

$2(c^2-ab) = (c-a)(c+a)+(c-b)(c+b)+(a-b)^2$[/quote]
Thanks. That's only a typo here, not a mistake in my calculations.

Prove that if $a,b,c\geq0$ and $a^2+b^2+c^2=1$, then
\[ \frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\leq\frac{9}{2}.  \]

My solution. WLOG: assume $a\geq b,c$ and write $b=Xa,c=Ya$. So $0\leq X,Y\leq1$. Then the inequality becomes
\[ \frac{X}{1+X^2+Y^2-X}+\frac{Y}{1+X^2+Y^2-Y}+\frac{XY}{1+X^2+Y^2-XY}\leq\frac{3}{2}.  \]
But this inequality is easy to prove.   :)

[quote="silouan"]Vasc this is your solution too ?[/quote]
Yes, this is.

You can see here .... http://www.mathlinks.ro/Forum/topic-61915.html

There's also  some open idea here.

I have two other solutions to your problem, Vasc. The first is using the substitutions $x=a+b+c$, $y=ab+bc+ca$, and $z=abc$. The last and critical step in my proof is quite different from the one in CRUX MATHS; it is simpler and nice.

The second solution is to use
\[ \frac1{1-xy}+\frac1{1-yz}+\frac1{1-zx}\leq\frac{27}8,\;\; x,y,z\geq0, \;\; x+y+z=1,  \]
which reduces to proving that
\[ 3-11(xy+yz+zx)+18xyz\geq0.  \]
This is solved by supposing that $z=\min\{x,y,z\}$ so that $z\leq 1/3$. Then we transform it into $(3z-1)^2(2z+1)\geq0$.

Then we note that
\[ \frac1{1+ab}+\frac1{1+bc}+\frac1{1+ca}\geq\frac94,\;\;\; a^2+b^2+c^2=1.  \]
Your inequality follows.

[quote="Vasc"]I don't understand your solution, pvthuan![/quote]

In the first proof, I also use some manipulations as published in CRUX MATHS, but then I did not use Schur, and completing squares $(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2$. Instead, I used just AG and and then turn it into an inequality of single variable.

For the second proof, let me give more details:
After solving
\[ \frac1{1-xy}+\frac1{1-yz}+\frac1{1-zx}\leq\frac{27}{8},\;\;\; x+y+z=1,  \]
(note that this also implies
\[ \frac1{1-x^2y^2}+\frac1{1-y^2z^2}+\frac1{1-z^2x^2}\leq\frac{27}{8},\;\;\; x^2+y^2+z^2=1  \]

we will have
\[ \frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}+\frac1{1+ab}+\frac1{1+bc}+\frac1{1+ca}\leq\frac{27}{4},\;\;\;\; a^2+b^2+c^2=1.  \]
The left hand of this is
\[ \frac1{1-ab}+\frac1{1+ab}=\frac2{1-a^2b^2}\,\;\;, ...  \]
Since $a^2+b^2+c^2=1$.
\[ \frac1{1+ab}+\frac1{1+bc}+\frac1{1+ca}\geq\frac9{3+ab+bc+ca}\geq\frac94.  \] :lol:  :lol:  :lol:
So that your inequality is
\[ \frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}\leq\frac{27}{4}-\frac94=\frac92.  \]

Vasc, can you post the solution for the more general case you claimed in CRUX?

A sketch of the proof:

$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} ac} \plus{} \frac {1}{1 \minus{} bc}\leq\frac {9}{2}\Leftrightarrow$
$ \Leftrightarrow3 \plus{} 7abc(a \plus{} b \plus{} c) \minus{} 5(ab \plus{} ac \plus{} bc) \minus{} 9a^2b^2c^2\geq0.$
But $ abc(a \plus{} b \plus{} c)\geq9a^2b^2c^2$ (the vasc's hint!)
Hence, remain to prove that $ 3 \plus{} 6abc(a \plus{} b \plus{} c) \minus{} 5(ab \plus{} ac \plus{} bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4 \minus{} a^3b \minus{} a^3c \plus{} a^2bc \plus{} (a \minus{} b)^4)\geq0,$ which obviously true.

Let $x,y,z$ are nonnegative numbers such that $x^{2}+y^{2}+z^{2}=1$, then

$\sum\frac{1}{1-yz}\leq \frac{4\sum{x}\sum{yz}}{\prod{(y+z)}}\leq\frac{9}{2}$;

$\sum\frac{1}{1-yz}\leq\sum\frac{27x^{2}}{2(2x+y)(2x+z)}\leq\frac{9}{2}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=66660

this was one of the problems in mathematics magazine too...

[quote="ityaa"]Let $ a$, $ b$ and $ c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Then prove that
\[ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}\]
[/quote]
Let $ \{a,b,c\}\subset\mathbb R$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
and let assume $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ ( $ v$ can be imaginary number ) and $ abc \equal{} w^3.$
Hence, $ \sum_{cyc}\frac {1}{1 \minus{} ab}\leq\frac {9}{2} \Leftrightarrow3 \plus{} 7abc(a \plus{} b \plus{} c) \minus{} 5(ab \plus{} ac \plus{} bc) \minus{} 9a^2b^2c^2\geq0\Leftrightarrow f(w^3)\leq0,$
where $ f(w^3) \equal{} w^6 \minus{} 7u(3u^2 \minus{} 2v^2)w^3 \plus{} 15v^2(3u^2 \minus{} 2v^2)^2 \minus{} 9(3u^2 \minus{} 2v^2)^3.$
But $ f$ is  convex function.
Thus, $ f$ gets a maximal value when $ w^3$ gets an extremal value
id est, when two numbers from $ \{a,b,c\}$ are equal, which easy to check.


geometry
Pythagorean Theorem
Pascal\u0027s Triangle

OK again I'm running out of topics, so far I've covered
basic number terminology
basic geometry terminology
Pythagorean theorem
How to do multiplication quickly
exponents and radicals
and things that were suggested in the older thread.

Any more topics that I can cover? (and it can't be too hard although half of the kids are turning out to be very hopeful)

6th graders, 4 of them (well +1 unofficially)

what about Pascal's triangle??? you can talk about combinations along with that and then have that lead into permutations...it's amazing how many kids in my Algebra class don't know what Combinations/Permutations are...

Yeah I talked about that too. (Under the category of things mentioned before)

Combinatrics/Factorials might be good for my second half of the class, great idea!

thanks...you can go into a lot of good stuff with factorials...such as question 7 of the chapter sprint last year...which was counting the number of zeros at the end of: (100!)(200!)(300!)...fun huh...

Coming at ya, Tare. Read through this. You might find some stuff of interest for the youngin's.

http://www.thiel.edu/mathproject/atps/PDF/default.htm

Aaaah, great resource, time to sneak into the ole math room again...

Thank you so much, that should cover months!


[hide]
Pretend that you are distributing -1's to a, b, c, d, & e. First, you must give a single -1 to a, b, & c to ensure that they're negative. Because you gave away three -1's, you're left with twelve -1's.

Distributing the 12 remaining -1's among 5 variables can correspond to placing 4 dividers in 12 + 4 spaces and putting the -1's in the remaining spaces.

So the total number of solutions is $\binom{12+4}{4}=\boxed{1 820}$
[/hide]

[hide="a suggestion"]Just multiply both sides by -1 and equate them. Then eliminate one from each of d and e and apply the standard method of balls and urns[/hide]

[quote="nutz_for2.718281828"]How many solutions are there to

a+b+c+d+e=-15

where a, b, and c are negative numbers, and d and e are nonpositive numbers(...-3, -2, -1, 0)?[/quote]

[clearly he means integers]

if we rewrite it as:

$(-a-1)+(-b-1)+(-c-1)+(-d)+(-e)=12$

each bracket is a nonegative integer...applying the ball and urn formula...we get

$\binom{12+5-1}{4}=1820$


function
geometry
geometric transformation
Support

I've started this topic because I'm quite sure that the South Tyrolean question is not well known around the world, especially in the USA.
For example: do you know where it is?
Do you know something about his history, his traditions?
What do you think about self-determination?

isn't that the part in italy mostly populated with germans and austrians. france and united kingdom gave it to the italy after the war (it was promised to be given to italy in order that italy abandon her previous alliance)

am i right?

Well, Südtirol is mostly populated with german speaking people. Most of them are of austrian origin, even if it's better to say Tyrolean origin because in the past it was quite indipendent from the Habsburg empire, because of its conformation and traditions.

In fact Südtirol has suffered lots of insults and damages in his recent hystory because of the arrogance of the great powers of Europe and of the USA with the complicity of president Wilson during the first and second world war. In fact they denied basilar principle of democracy as the self-determination. I would say it's more or less the same situation as in the Euskadi.

The central government oppresses the minority and, if in the past it used the brute force and torture, now the oppression is cultural, but not less humiliating.
And I think bilinguism is the principal means of oppression.
As far as italian government wouldn't recognize the self-determination principle for absurd reasons but in the end of september Austria will insert in its new constitution a sort of principle that give it a tutelar function with respect to Südtirol in front of Italy.

Deimos, you do seem to know a bit about everything don't you :wink: .

No, I hadn't heard about it, it's an interesting topic you started here.
As I understand it, the region of Sudtirol was at first Austrian, became Italian after the first world war, and remained Italian soil during and after World War II, when the Nazis even acted against the interest of Germanspeaking people in order not to strain relations with Mussolini.

How is the situation now?  What languages do you speak?  I assume German and English, and Italian? (as your second language?)  What kind of school did you go to?

Living in a trilingual country I can understand the significance of a difference in languages (here it has a major influence on politics)

Oh, I don't live in Südtirol but in the north of Italy in a region near to it. In fact I'm italian-speaking but for some reasons I'm keeping in contact with the reality of that beautiful land.
As I've already said I think that the bilinguism is the modern mean of oppression with the new concept of unite Europe, because they try to hide the problem and to give it an antiquated description. It's common to hear people saying: "Why do they ask indipendence when the borders no longer exist?" These people haven't lived in their native land with a foreign that impose language, traditions, laws that don't belong to they!

If you think about it, in Europe there are lots of Nations without State!  :(
The rights of self-determination, as the civil rights are universal values that belong to everyone.

Here are some words of Eva Klotz, the daughter of Georg, a combatant for freedom, but consider as a terrorist by Italy: "What is the native land? The Heimat, Native Land, for me is the little church, the hill, the stream with the stones, the trees of my country. It is a spiritual place, where is preserved the ideal of my father. All of these was historically austrian and must return to Austria if South Tyrolean people would. It's a question of democracy."

I don't know a lot about the situation, but as I said, I am familiar with language problems.

But tell me, how many native speakers of Italian live there, and how many speak German?  Is there (affordable) access to education in the native languages of both groups?  Are road signs in both languages?  Do police officers, employees in hospitals etc... speak both languages. Is Südtirol properly represented in Italian parliament and are Germanspeaking parties allowed to participate in the elections.

All those things are sensitive, in Belgium nearly every piece of land near the border is disputed, and there are many complex intermediate language situations.

However, I am referring to the Dutch/French situation, there are also about 70.000 German speakers.  They live in territory that Belgium took from Germany in the first World War (just like Südtirol).  It may seem obvious we should give it back to Germany... but what about those who have settled in territories and don't speak German (in Eupen it's 10%)?

[quote="fredbel6"]
But tell me, how many native speakers of Italian live there, and how many speak German? [/quote]
70% are German-speakers, 25% Italian-speakers, 5% Ladin.
[quote]Is there (affordable) access to education in the native languages of both groups? [/quote]
In general, yes.
[quote]Are road signs in both languages?  Do police officers, employees in hospitals etc... speak both languages.[/quote]
Road signs are in both languages, but the translations into italian are completely false. For example, Südtirol is Alto Adige, instead of Sudtirolo.
Most of German-speaking people refused to speak in Italian becuase they consider this as an insult respect to their history.
[quote]Is Südtirol properly represented in Italian parliament and are Germanspeaking parties allowed to participate in the elections.[/quote]
There are at most 5 parliamentary all of the same party, the Südtiroler Volks-Partei (the German-speaking people mass party), which since 1950 obtain more than 70% every election.
[quote]... but what about those who have settled in territories and don't speak German (in Eupen it's 10%)?[/quote]
They haven't any fault, but this is not a sufficient argument to leave all things as now!

I'm sorry but although I'm interested I simply don't know enough about it.

What do you mean with "bilinguality" being evil?  Are you talking about individual bilinguality or the bilinguality of a region?

But what about all other public jobs?  Bus drivers, ambulance drivers, nurses, doctors at your local hospital, farmacists, police officers, JUDGES (can a legal dispute be settled in German?),...

What about the Italianspeakers?  Do they go to school in a local school?  Do they have to learn German?
You say they don't want to speak German, but .... is it because they simply can't?

I'm especially interested because I want to compare the situation to ours in Flanders.  (Flemish towns with an ever increasing Frenchspeaking population, demanding and receiving political representation, demanding education...)

[quote="fredbel6"]
What do you mean with "bilinguality" being evil?  Are you talking about individual bilinguality or the bilinguality of a region?[/quote]
I mean a language caracterize a nation and a culture. If you are not allowed to speak only your own language you are deprived of something
very important.

[quote]But what about all other public jobs? Bus drivers, ambulance drivers, nurses, doctors at your local hospital, farmacists, police officers, JUDGES (can a legal dispute be settled in German?),...[/quote]
In the services that in most percentage depend from the State, people must speak mostly in Italian (police, judges). But consider that those kind of works are acomplish mostly by Italians. In the other cases people normally speak in German. You have to consider that the Italians live mostly in Bozen, which is the chief town. This fact is important. You have to known that last year, 113 Mayors of all the 115 South Tyrolean municipalities sent a letter to the Austrian Parliament in which they demanded a sort of protection. One of the two municipalities that didn't subscribe the letter was in fact Bozen.

[quote]What about the Italianspeakers?  Do they go to school in a local school?  Do they have to learn German?
You say they don't want to speak German, but .... is it because they simply can't?[/quote]
No, Italians don't have to learn German. They must to learn it only if they would enter in an administrative body managed by the Province.
It's a reciprocal thing. Italians don't want to speak German and South Tyroleans don't want to speak Italian.

As I understand it, you speak Italian, your parents or grandparents aren't Austrian either, but you are sympathetic to their cause?
Would you be willing to accept the consequences for you if something would change?

[quote]I mean a language caracterize a nation and a culture. If you are not allowed to speak only your own language you are deprived of something
very important.[/quote]
But you said they were allowed to speak and do most (like education) in their own language? Please describe the depriving better if you like.

[quote]In the services that in most percentage depend from the State, people must speak mostly in Italian (police, judges). But consider that those kind of works are acomplish mostly by Italians.[/quote]
But that can't be good! :o   If for instance one Germanspeaker beats up another, will the trial be in German or Italian?
If you get hit by a car and taken into an ambulance, will the people be able to understand you in both languages? (This is very important for children who couldn't have learnt another language anyway because they are simply too young).  I know that this can KILL (language problems have caused a train accident here in 2001)

[quote]No, Italians don't have to learn German. They must to learn it only if they would enter in an administrative body managed by the Province.[/quote]
Well that's something!  But is it enforced?  Is there some sort of a test for those people to check their German?
It is confusing though when you speak of Italians.  Do you mean new people from the rest of Italy, people who have lived in Sud Tirol all the time.... technically everyone in Sud Tirol is Italian, right?

[quote]It's a reciprocal thing. Italians don't want to speak German and South Tyroleans don't want to speak Italian[/quote]
That makes the situation completely different from the one in Belgium.  Flanders is crawling with French teachers, and most people know a fairly good amount of English and French, it's an asymmetric situation sustaining itself out of convenience...

Yes, you are right. I'm an Italian speaking but I am for the South Tyrolean cause, I support German-speaking because it's clear to me the difficult situation in which they live.
When I go there, I refuse to speak my own language even if I know only few words in German.

Yes, technically everyone is Italian in Südtirol, I mean Italian-speking.

For the autonomy satute I hope you'll find some answers [url=http://www.provinz.bz.it/lpa/publ/publ_getreso.asp?PRES_ID=1922]here[/url].


A motorist made a 60 mile trip averaging 20 miles per hour.  On the return trip, he averaged 30 miles per hour.  What was the motorist's average speed for the entire trip?

[hide]the harmonic mean of 20 and 30 is 2/(1/20 + 1/30) or 24 mph. [/hide]

[hide] 120/ 5hrs= [b]24 miles per hour[/b][/hide]

[hide]Use the harmonic mean to get $\frac{1}{\frac{(\frac{1}{30}+\frac{1}{20}}2)}$=1/1/24=24 for the whole way

I apologize for the glitched LaTeX and for reviving a year-and-a-half old topic[/hide]

Just remember it is D=RT, so therefore it is D/T=R, in this case 120/5=24.

What is the point of a harmonic mean? When would you need to use it?


Prove that there are an infinity of numbers that are both square and triangular, e.g. 1, 36, etc.

Basically we need to show that there are an infinite number of solutions $\left(m,n\right)$ to the equation $n^{2}+n-2m^{2}=0$.

consider the sequence $1$, $8$, $a_3$, $a_4$,...

with $a_{n+2}=(a_{n+1}-a_n)^2$

where the trianglular numbers are $\frac{a_n(a_n+1)}{2}$, checking the first few terms works, try induction to prove

EDIT: woops, this only works up to $a_4$


geometry
circumcircle
parallelogram
perpendicular bisector
geometry proposed

Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$.
The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$.
The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$.
The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$.
The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$.
Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$.

Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$.

[i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular:

[b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$.

[b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$.

[i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos .

I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions).

  Darij

Obviously $ANHM$ is a parallelogram, therefore points $N, D, M$ are collinear. Consider the circle with diameter $BC$(centered at $E$) and point $A$ (1). It can be easily shown that points $M, N, U, V$ belong to the radical axe of the abovementioned circles and the conclusion follows.
(a)	from the fact that $D, A', D', E$ ($A'$ is the foot of the altitude from $A$) belong to the Euler’s circle, we can conclude that ${DD'}\perp {AE}$.
(b)	Well, $X$ also belongs to the radical axe from (1) and we are done.

[quote="darij grinberg"]

[b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$.

It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions).

  Darij[/quote]

If O is the circumcenter of ABC, then (is easy to prove that) OE = AD and from this, as OE//AD it follows that AO//DE so ED is orthogonal to AX. But this means that D is the orthocenter of AEX, from which we have the thesis.

[quote="darij grinberg"]Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$.
The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$.
The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$.
The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$.
The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$.
Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$.

Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$.
 It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions).

  Darij[/quote]

As NAH  (MAH) is similar-orthogonal to ABC then the conrespondents medians AE and ND (ND) are orthogonal too.

If F is the intersection between AN and CU and B' the midpoint of AC, as B'U//AF,  it holds that FU=UC. Therefore U is colinear with ND. Similarly V is coliner with MD and we are done.


number theory
greatest common divisor
algebra proposed
algebra

Given two natural numbers $ a$ and $ b$.
We know that $ \dfrac{a^2\minus{}1}{b\plus{}1}$ $ \plus{}$ $ \dfrac{b^2\minus{}1}{a\plus{}1}$ is integer
Prove that $ \dfrac{a^2\minus{}1}{b\plus{}1}$ and $ \dfrac{b^2\minus{}1}{a\plus{}1}$ are integers too.

Since $ a,b$ are integers, let $ \frac{a^2\minus{}1}{b\plus{}1} \equal{} \frac p q$ where gcd$ (p,q)\equal{}1$.
Then eliminating $ b$, we have $ \frac{a^2\minus{}1}{b\plus{}1}\plus{}\frac{b^2\minus{}1}{a\plus{}1} \equal{} \frac {q(q(a^2\minus{}1)\minus{}p)^2 \minus{}qp^2 \plus{}p^3(a\plus{}1)}{qp^2(a\plus{}1)}.$
Let $ a\plus{}1 \equal{} q^tk$ where $ t \geq 0$ is an integer, and $ k$ is not divisible by $ q$.
Then $ q^{t\plus{}1}$ divides $ qp^2(a\plus{}1)$ and then divides $ q(q(a^2\minus{}1)\minus{}p)^2 \minus{}qp^2 \plus{}p^3(a\plus{}1) \equal{} q^3(a\plus{}1)^2(a\minus{}1)^2\minus{}2q^2p(a\plus{}1)(a\minus{}1)\plus{}p^3kq^t$. Since $ q^3(a\plus{}1)^2(a\minus{}1)^2\minus{}2q^2p(a\plus{}1)(a\minus{}1)\plus{}p^3kq^t \equal{} kp^3q^t \mod[q^{t\plus{}1}$, it follows that $ q$ divides $ kp^3$, but $ q$ is coprime with $ p$ and does not divide $ k$, therefore $ q\equal{}1$ and $ \frac{a^2\minus{}1}{b\plus{}1}$ is an integer.
It easily follows that $ \frac{b^2\minus{}1}{a\plus{}1}$ is also an integer.

Pierre.


i guess this problem was too hard for pre-olympiad

let the radius of the black circle, $O$, be $r$. point $p$ is selected on the circumfrence of the black circle. and a blue circle, $Q$, with radius $R$ is drawn (centered at $P$).
the area of circle $Q$ (which is outside $O$), is equal to area of circle $O$.
find $R$ in terms of $r$.

[quote="henry23"]hard? i think not... .[/quote]

If you try to solve it with trigonometry you end up with an equation which involves $\theta$ and $cos\theta$.

Is there a pure geometric solution?

a similar problem was posted on the University of Washington Challenge of the week for last week! (its over now).

they posted the solution, in that case, they had given a value for r and solved for R

try googling it...it should be somewhere on the web! ;)


Prove that:
$3(ab+bc+ca)(1+52abc)\geq 79abc$
where $a;b;c$ are positive real numbers and satisfying:$a+b+c=1$

The stronger inequality holds: If $a,b,c$ are positive real numbers such that $a+b+c=1,$ then \[3(ab+bc+ca)(1+48abc)\geq 75abc. \] According to EV, it is enough to prove the inequality when $a=b:$ \[3\left(a^{2}+\frac{2}{a}\right)\left(60+8a^{3}+\frac{6}{a^{3}}+\frac{1}{a^{6}}\right)\geq 75\left(4a^{2}+\frac{4}{a}+\frac{1}{a^{4}}\right).\] But \[\mathrm{LHS}-\mathrm{RHS}=\frac{6(1-3a^{3}+2a^{6})^{2}}{a^{7}}\geq 0\]
which proves our inequality.

[quote="bodan"]
According to EV, it is enough to prove the inequality when $a=b:$[/quote]

What is EV?? Is it some cool way of solving inequalities??

[quote="Myszon11"]

What is EV?? Is it some cool way of solving inequalities??[/quote]
That is a Vasc's theorem!

[quote="arqady"][quote="bodan"]The stronger inequality holds: If $a,b,c$ are positive real numbers such that $a+b+c=1,$ then
\[3(ab+bc+ca)(1+48abc)\geq 75abc. \]
[/quote]
I think, if $3(ab+bc+ca)(1+52abc)\geq 79abc$ is true then it's stronger
than $3(ab+bc+ca)(1+48abc)\geq 75abc$ . :wink:[/quote]

Then the original inequality isn't true, because $(48,75)$ is optimal. Indeed, when taking $a=0.554943,\;b=0.222529,\;c=0.222528$ we have
\[\mathrm{LHS}-\mathrm{RHS}=3(ab+bc+ca)(1+52abc)-79abc\approx-0.0103647.\]


please...or better yet, does anyone have the link with the past arml problems? Like ones before 1996?

What link are you referring to?  A ton of ARML problems can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arml+practice&t=30264]here[/url]

For copyright purposes, we weren't allowed to post whole competitions at a time.  Also, that means that we put very few Power round stuff in there (or relays).

that's fine...the link I was referring to was the one you pm ed me a while back with ARML questions from 2005,2004, etc.

well...okay i found the link i was talking about
http://www.math.sc.edu/~filaseta/contests.html

If Tony has 100,000,000 ISP addresses, what is the sum of his IP addresses, assuming each IP address is 8 digits?

[IM CLEANING UP SPAM....DELETE THIS POST PLEASE]

um...how are ip addresses coded...doesn't each ip address have its own number of digits, you have to specify these things in the problem!

and btw, delete that post!


It is perhaps well known but it is interesting (and easy):

Let $G$ be a group and $X$ a set. Consider an action of
$G$ on $X$ that has only one orbit(a transitiv action).
Show that there exists $a \in G$ such that $a.x \neq x$
for every $x \in X$.

Here's a solution for the finite case (finite $X$ and $G$). I haven't thought about the infinite case. Is it still true?

We assume $|X|>1$, of course. By Burnside's formula for the number of orbits, we have $1=\frac 1{|G|}\cdot\sum_{g\in G}|X^g|\ (*)$, where $X^g$ is the number of elements of $X$ which are fixed by $g\in G$. $1\in G$ fixes all the elements of $X$, so $|X^1|>1\ (**)$. If $|X^g|\ge 1,\ \forall g\in G$, from $(*)$ and $(**)$ we get a contradiction (the RHS in $(*)$ becomes strictly larger than the LHS).

I did the same  :)
and I forgot to say that it is a finite group  :D


geometry
trigonometry
parallelogram
AMC
AIME
cyclic quadrilateral
power of a point

I created this problem myself - comments are welcome!

In cyclic quadrilateral ABCD with no sides parallel, BA and CD are extended to intersect at point R. AC and BD intersect at point P. Let [tex]\angle APR = \alpha[/tex] and [tex]\angle DPR = \beta[/tex]. Prove

[tex]\displaystyle \sqrt{
{
 {
  K_{\Delta APR}K_{\Delta CPR}
 }\over{
  K_{\Delta BPR}K_{\Delta DPR}
 }
}} =
{
 {
  \sin \alpha
 }\over{
  \sin \beta
 }
}[/tex],

where [tex]K_{\mathrm{figure}}[/tex] represents the area of that figure.

Beautiful problem():

[hide]K_(APR)=(AP)(PR)sin(alpha)/2, becuase construct a congruent triangle that shares side AR and therefore we have a parallelogram. Height of the parallelogram is APsin(alpha). therefore area is (AP)(PR)sin(alpha)/2.

Same reason

K_(DPR)=(DP)(PR)sin(myself)/2

K_(CPR)=(CP)(PR)sin(180-alpha)/2

K_(BPR)=(BP)(PR)sin(180-myself)/2

since sin(180-alpha)=sin(alpha),

therefore

K_(APR)K_(CPR)=(AP)(CP)(PR)^2sin^2(alpha)/4

K_(DPR)K_(BPR)=(BP)(DP)(PR^2)sin^2(myself)/4

divide them we have

(sin^2(alpha)/sin^2(myself))(AP)(CP)/(BP)(DP)

since by power of a point theorem

(AP)(CP)/(BP)(DP)=1

square root give us the desired result[/hide]

What time limit would you recommend for this problem if it were part of a real contest?

Well, it depends what kind of contest it's on.  If it were an olympiad question, it would be a very easy olympiad question.  However, I could see it occupying a medium position on the AIME and probably only as one of the very last AMC 12 problems, if at all.  I sort-of agree with Beta that if someone has the appropriate mathematical tools, the solution pops to mind rather quickly.  A very nice question.