Euler
geometry
circumcircle
geometry proposed

The circle inscribed in a triangle $ ABC$ touches the sides $ BC, CA, AB$ in $ D, E, F$, respectively. Prove that $ \triangle ABC$'s Euler line is equal to triangle $ \triangle DEF$'s Euler line.

In an arbitrary triangle $ ABC$, denote by $ I$ the circumcenter of the triangle and by $ D$, $ E$, $ F$ the intersection points of the lines $ AI$, $ BI$, $ CI$ with the sides $ BC$, $ CA$ and $ AB$, respectively.

In a problem posted already on the forum, it is stated that the Euler lines of $ \triangle{ABC}$ And $ \triangle{DEF}$ coincide or are parallel. Well this is WRONG!

The correct statement should be:

If $ H_{1}$ is the orthocenter of $ \triangle{DEF}$, then the line $ IH_{1}$ is parallel with the Euler line of $ \triangle{ABC}$.

[color=red][EDITED by pohoatza][/color]

If $ D$, $ E$, $ F$ are the tangency points of the incircle  with the sides $ BC$, $ CA$ and $ AB$, respectively of an arbitrary triangle $ ABC$, and $ H_{1}$ is the orthocenter of $ \triangle{DEF}$, then the line $ IH_{1}$ (the Euler line of $ \triangle{DEF}$) contains the circumcenter of $ \triangle{ABC}$.

[color=red][EDITED by pohoatza][/color]

[color=darkred][b]Remark.[/b] Nu va suparati ([b]Sorry[/b]) Domnule profesor ([b]Mr. teacher[/b]), incercati ([b]try[/b]) sa scrieti ([b]write[/b]) in limba engleza ([b]in English language[/b]), va rog ([b]please[/b] ) !
Cred ([b]I think[/b]) ca nu ati bagat de seama unde sunteti. Si mie mi s-a intamplat pe la inceput .
Va multumesc pentru interventie. Oportuna si la obiect. Foarte interesante observatiile Dvs. ![/color]

Welcome to the forum, Mihai Miculita! As Virgil Nicula pointed out that in this forum the only language is English, please edit and translate your above messages. Thank you!

@Euler-ls: You probably meant: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=21311#21311[/url], so this topic is locked.


geometry trigonometry function circumcircle AMC USA(J)MO USAMO

In $ \triangle ABC$, $ \angle ABC \equal{} 45^\circ$. Point $ D$ is on $ \overline{BC}$ so that $ 2 \cdot BD \equal{} CD$ and $ \angle DAB \equal{} 15^\circ$. Find $ \angle ACB$.
[asy]
pair A, B, C, D;
A = origin;
real Bcoord = 3*sqrt(2) + sqrt(6);
B = Bcoord/2*dir(180);
C = sqrt(6)*dir(120);
draw(A--B--C--cycle);
D = (C-B)/2.4 + B;
draw(A--D);

label("$A$", A, dir(0));
label("$B$", B, dir(180));
label("$C$", C, dir(110));
label("$D$", D, dir(130));
[/asy]
$ \textbf{(A)} \ 54^\circ \qquad \textbf{(B)} \ 60^\circ \qquad \textbf{(C)} \ 72^\circ \qquad \textbf{(D)} \ 75^\circ \qquad \textbf{(E)} \ 90^\circ$

I'm really bad at geometry and eh.. same thing applied to this problem. It took me so long time to figure it out but when I did, I was glad that I got it right.   :D (I have this thing for later geometry problems that if I don't get it, I try and try till I get the right answer --> don't matter if I have to do for couple of days)

[hide="Go TI-89 calculator"]
First, set $AB = 1$. This is something I learned that if you set ONLY one length to be a number you want, you'll still get a right answer (note: this only applies for angles) because from this, you can form a similar triangle that has your length and satisfy the problem (making two lengths won't necessarily satisfy this!). Anyway, this makes my life a lot easier. Then, draw $CK$ where $K$ is on $\overline{AB}$ such that $\triangle BCK$ is a right triangle. Since we know $\angle ABC = 45^\circ$, it's easy to find that $\angle BCK = 45^\circ$. Before using this, note that in $\triangle ABD$, we have, from Law of Sine:

$\frac{BD}{\sin 15^\circ} = \frac{AB}{\sin 120^\circ}$

From this, we get $BD = \frac{AB \cdot \sin 15^\circ}{\sin 120^\circ}$. Now, back to our right triangle $BCK$, we have, again from Law of Sine:

$\frac{CK}{\sin 45^\circ} = \frac{CB}{\sin 90^\circ}$

From this and substituting $CB$ with $\frac{3 AB \cdot \sin 15^\circ}{\sin 120^\circ}$, we get $CK = \frac{3 AB \sin 15^\circ \sin 45^\circ}{\sin 120^\circ}$. From here, rest is pretty self-working with calculator.. Using Pythagoren Theorem, we'll get $BK$ (we know what $CB$ and $CK$, remember?). Also, by $AB = 1$, we now have $AK$ as well. Again using Pythagorean Theorem, we can find what $AC$ is. Take this value and use Law of Sine:

$\frac{AC}{\sin 90^\circ} = \frac{AK}{\sin \angle KCA}$

If you didn't round anything, you should get $\angle KCA = 30^\circ$. But since we're looking for $\angle BCA$, the final answer is $\angle BCA = \angle BCK + \angle KCA = 45^\circ + 30^\circ = 75^\circ$. D is the correct answer. [/hide]

No Calc Solution:

[hide]Let . We see that . By Law of Sines in , we can find  to be . By another use of Law of Sines in , we can see that  Rearranging, we find that .

To find , we must eliminate ; we do this by noting that . Plugging this back in, we find

We find this by expanding and canceling the  from the right term in the numberator and making it  on the left term in the numerator. Thus:

, which obviously implies . D.[/hide]

[quote="FMako"]In response to Iversonfan2005's post:

What is R?[/quote]

Sorry, it was a typo, I meant $B$ :blush:

[quote="Iversonfan2005"][quote="Silverfalcon"]Just wondering, how did you evaluate $\frac{\sin 15^\circ}{\sin 45^\circ}$? Did you use half-angle formula (I can't remember it so show if you did) or something else?[/quote]

He probably used $\sin{15}=\sin{(45-30)}$ and did it that way, I think that's easier than doing $\sin(\frac{30}{2})$ since you don't get caught up in the radicals[/quote]

Oh okay. Both solutions are indeed very nice!

Rarely; the only time it really ever is needed is when the question specifically asks for a half-angle of something (usually something like $22.5^{\circ}$) or if you're dealing with some angle-bisector problem (and even then, Mass Points, Angle Bisector Theorem, and Stewart's is usually more than enough to kill the problem). Out of all the half angle formulas the I'd say the $\tan$ half-angle formula is the one that I've had to use the most, if at all, and I usually use the $\frac{\sin x}{1+\cos x}$ form of it anyways.

[quote="Silverfalcon"][quote="Elemennop"]$\sin 15^{\circ}$ is simply one of the values we were required to memorize in our Algebra II/Trig class. We often had 'timings' where the teacher wrote 10 random trig functions (ie. $\sin 1095^{\circ}$), and we had three minutes to do all 10. By the end of the year we could call off the top of our head the value of all six trig functions on any multiple of $15^{\circ}$, as well as values that would simplify down to $\sin 18^{\circ}$ and $\cos 36^{\circ}$.

So I just immediately knew that $\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$ without really thinking about it. If you aren't able to remember those values quickly you should probably practice them some, because they come up very often in alot of trig problems (plus it expands how much you can do with geometry problems using trig).[/quote]

True. But I have other things to learn so I'll probably not "memorize" this (I know how to find it using $\sin(\alpha - \beta)$). Just wondering, how often do you use Half-Angle Formula? I don't really use it at all so I don't know if it's much of use at all (well... I guess it should but I never really used them on math contest)[/quote]

You should use half angle formula only for numbers where you can't use the $(a+-b)$ formula, like $35$ for instance. I just hate doing the radicals because sometimes, you can get a radical underneath a radical which gets messy and complicated.

[quote="1234567890"]How do you get 35? :?

I mean, we only know multiples of 30, 45, 60, etc and we can't add/half/subtract in any way to get 35.[/quote]

No, I'm just giving an example, the first thing that came on the top of my head.... :blush:

[quote="breez"]This problem is easy with trig.[/quote]

ok...?

[hide="I guess you want a clarification of my opinion"]
One application of law of sine to get the length of the cevian and you then one application of law of cosines to get the side opposite the 60 degrees. You end up with a triangle with 3 sides: $\frac{\sin 45^{\circ}}{\sin 15^{\circ}}-2-\sqrt{6}$. Now use law of cosine to solve for $\cos \theta$ where $\theta$ is the angle of interest. Take $\arccos (.258819...) = \boxed{75^\circ}$[/hide]

No, I just said that because I thought it was random that this really old thread was brought back to life, and there are a couple of solutions using trig already...

[hide]Draw altitude $AX$, and let $DX=y$.  Then $AX=BX=y\sqrt{3}$, and $BD=BX-DX=y\sqrt{3}-y$.  We have $BC=3\cdot BD=3y\sqrt{3}-3y$.  Then $CX=BC-BX=2y\sqrt{3}-3y$.  Finally, $\tan C=\frac{AX}{CX}=\frac{y\sqrt{3}}{2y\sqrt{3}-3y}=2+\sqrt{3}=\tan 75^{\circ}$.[/hide]

[quote="Iversonfan2005"]Non Trig Solution
[hide]
Let's call $DC=2x$ and $BD=x$ and then let's draw in $CX$ which is perpendicular to $DA$ and draw in $BX$ too. From here, we know that $\angle{XDC}=60$ so $\angle{CXD}=30$ Since $DX=DB$ we see that since $\angle{XDB}=120$ then $\angle{DXB}$ and $\angle{DBX}$ are equal to $30$ which means that $\angle{BXA}$ and $\angle{BAX}$ are both equal to $15$ and so $\angle{AXB}=150$ and so $\angle{CXA}=90$. We also know that $AX=BX=CX$ and so $\angle{XAC}$ and $\angle{XCA}$ are both equal to $45$, so $\angle{ACB}=30+45=\boxed{75}$[/hide][/quote]

I don't see how you obtained $AX=BX=CX$ without trig. It follows from law of sines, but how else do you get it?

I think he did it by isosceles triangles.

oh man that was from the 2005 era so I don't remember the solution and don't really feel like doing it again. I must've used iscoceles triangles though cuz I didn't know any trig back then.

[color=darkblue][b]A synthetic ("slicing") method.[/b] Denote the point $E\in (AD)$ so that $EA=EB$ and the middlepoint $M$ of the segment $[DC]$.
Prove easily that $DB=DE=DM=ME=MC$, $CE\perp AD$, $EA=EB=EC$, $m(\widehat{ACE})=45\Longrightarrow \boxed{\ m(\widehat{ACB})=75\ }\ .$

[b]A metrical method 1.[/b] Suppose w.l.o.g. $BD=1\ (DC=2)$ and denote $AD=x$, $m(\widehat{CAD})=\phi .$ Apply the generalized Pithagoras' theorem :
$AB^{2}=DB^{2}+DA^{2}-2\cdot DB\cdot DA\cdot\cos 120\Longrightarrow AB=\sqrt{x^{2}+x+1}\ .$
$AC^{2}=DC^{2}+DA^{2}-2\cdot DC\cdot DA\cdot \cos 60$ $\Longrightarrow AC=\sqrt{x^{2}-2x+4}\ .$
$AD^{2}=BD^{2}+BA^{2}-2\cdot BD\cdot BA\cdot\cos 45$ $\Longrightarrow$ $x+2=\sqrt{2(x^{2}+x+1)}$ $\Longrightarrow$
$x^{2}-2x-2=0$ $\Longrightarrow$ $x=1+\sqrt 3$ $\Longrightarrow$ $AC^{2}=(1+\sqrt 3)^{2}-2(1+\sqrt 3)+4$ $\Longrightarrow$ $AC^{2}=6\ .$
$DC^{2}=AD^{2}+AC^{2}-2\cdot AD\cdot AC\cos\phi\Longrightarrow$ $4=(1+\sqrt 3)^{2}+(\sqrt 6)^{2}-2(1+\sqrt 3)\sqrt 6\cos\phi$ $\Longrightarrow$
$\cos \phi=\frac{\sqrt 2}{2}$ $\Longrightarrow$ $\phi=45$ $\Longrightarrow$ $\boxed{\ m(\widehat{ACB})=75\ }\ .$

[b]A metrical method 2.[/b] Denote the point $E\in (AD)$ so that $EA=EB$ and the point $F\in AD$ so that $FB\perp AB$. Prove easily that $DB=DE$, $m(\widehat{AFB})=75$ and $EA=EB=EF=2\cdot DB\cdot \cos 30=\sqrt 3$, i.e. the point $E$ is the circumcenter of the triangle $ABF$. The line $BD$ meets again the circumcircle of the triangle $ABF$ in the point $C_{1}$. Therefore, $C_{1}F\perp C_{1}A$ and $C_{1}F=C_{1}A$, $m(\widehat{AC_{1}B})=75$, $DF=\sqrt 3-1$, $DA=\sqrt 3+1$ and $DB\cdot DC_{1}=DF\cdot DA=(\sqrt 3-1)(\sqrt 3+1)=2=DB\cdot DC$, i.e. $C\equiv C_{1}$. In conclusion, $\boxed{\ m(\widehat{ACB})=75\ }\ .$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=107194

[b]A trigonometrical proof.[/b] $m(\widehat{ACB})=x$, $m(\widehat{DAC})=120-x$, $2=\frac{DC}{DB}=\frac{AC}{AB}\cdot \frac{\sin\widehat{DAC}}{\sin\widehat{DAB}}=\frac{\sin 45}{\sin x}\cdot\frac{\cos(30-x)}{\sin 15}$ $\Longrightarrow$
$\boxed{\ 2\sin x\sin 15=\sin 45\cos (30-x)\ \ (1)\ }$ $\Longleftrightarrow$ $2\sin x\cdot 2\sin 15\sin 45=2\sin^{2}45\cos (30-x)$ $\Longleftrightarrow$ $2\sin x(\cos 30-\cos 60)=\cos (30-x)$   $\Longleftrightarrow$
$\sqrt 3\sin x-\sin x =\cos (30-x)$ $\Longleftrightarrow$ $\sqrt 3\sin x=\sin x+\sin (60+x)$ $\Longleftrightarrow$ $\sin x=\sin (30+x)$ $\Longleftrightarrow$ $x+(30+x)=180$ $\Longleftrightarrow$ $\boxed{\ x=75\ }\ .$
[b]Otherwise ("ugly")[/b], the relation  $(1)\Longleftrightarrow$ $2\sin 15\tan x=\sin 45(\cos 30+\sin 30\tan x)$  $\Longleftrightarrow$ $\tan x=\frac{\sin 45\cos 30}{2\sin 15-\sin 45\sin 30}=$
$\frac{\frac{\sqrt 2}{2}\cdot\frac{\sqrt 3}{2}}{2\cdot\frac{\sqrt 2}{2}\cdot\frac{\sqrt 3-1}{2}-\frac{\sqrt 2}{2}\cdot\frac{1}{2}}=$ $\frac{\sqrt 3}{2(\sqrt 3-1)-1}=\frac{1}{2-\sqrt 3}=2+\sqrt 3$ $\Longleftrightarrow$ $\tan x=\tan 75$ $\Longleftrightarrow$ $\boxed{\ x=75\ }\ .$
[b]Remark (maybe somebody doesn't know !).[/b] $\tan 75=\tan (45+30)=\frac{\tan 45+\tan 30}{1-\tan 45\tan 30}=$ $\frac{1+\frac{\sqrt 3}{3}}{1-\frac{\sqrt 3}{3}}=$ $\frac{3+\sqrt 3}{3-\sqrt 3}=$ $\frac{\sqrt 3+1}{\sqrt 3-1}=$ $\frac{(\sqrt 3+1)^{2}}{2}=2+\sqrt 3\ .$

[b]Generalization.[/b] $\boxed{\ \triangle ABC\ ,\ m(\widehat{ABC})=x\ ;\ D\in (BC)\ ,\ m(\widehat{BAD})=y<x\ ,\ \frac{DB}{DC}=\frac{\sin 2y}{\sin 2x}\Longrightarrow m(\widehat{ACB})=90-y\ }\ \ .$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=107292 [/color]

[quote="Virgil Nicula"]In the my previous message I offered several proofs and a generalization for the proposed problem.[/quote]

jeez dude, u just slaughtered that amc problem...MURDERER

good job  :D

I know this is old, but damn am I glad I got this! (first 24 I got right by myself...I know I am one of the USAMO hopefuls who doesnt have the years of problem solving experience since my school started this only last year).

I drew a triangle ABC with the B being on the lower left hand side, for AD (labeled a) I just extended D to A in such a way that the angle ADB looked to be about 120. I then drew an altitude from A to the DC and called that altitude c as well as the point of intersection E. I labeled DE as p and then found CE and AE in terms of a and did arctan(CE/AE) to get 75.

Is this a sound way to do things or did I just get lucky?

Another non-trig solution:
Let E be the reflection of B through AD. Then $ DE \equal{} DB \equal{} \frac{1}{2}AC$ and $ \angle{EDC} \equal{} 60.$ Thus triangle $ EDC$ has $ E\equal{}90$ and $ C\equal{}30$.
Now we have $ \angle{BCE} \equal{} 30 \equal{} \angle{BAE},$ so $ ABEC$ is cyclic. Therefore: $ \angle{ACB} \equal{} \angle{AEB} \equal{} 75.(q.e.d)$

[quote="dgreenb801"]I did it Iversonfan 2007's way. However, at first, I had a really cool idea that I wonder if someone can complete. Let ABCDEFGHIJKL be a regular 12 gon inscribed in a circle. Then, the given triangle is triangle ADH with cevian AG. Can someone prove AG trisects DH?[/quote]

This gave me an idea to reverse the problem:
Given triangle $ ABC$ with $ A\equal{}60, B\equal{}45, C\equal{}75$. Point $ D$ lies on $ BC$ such that $ \angle{DAB} \equal{} 15$. Prove that $ 2DB \equal{} CD$

My proof for the reversed one:
We have: $ \frac{CD}{BD} \equal{} \frac{AC.sinCAD}{AB.sinBAD} \equal{} \frac{sin45.sin45}{sin75.sin15} \equal{} \frac{2(sin45)^2}{2sin75.cos75} \equal{} \frac{1}{sin150} \equal{} 2 (q.e.d)$

My proof(slightly different from Elemenop's proof):
By Law of Sines in triangle ABD, we have: BD/AB=sin15/square root(3)/2.Since 2BD=BC, we have: BC/AB=4sin15/square root(3)
Moreover, BC/AB=sin(135-C)/sinC
From both equations we get: sin(135-C)/sin(C)=4.sin15/square root(3)
Solving C by exactly the same way as Elemenop did but I got a wrong answer::cotgC=1- 2/square root(3).
PS:I went over my solution 5 times but I couldn't find out any mistakes in it. Can anybody help me by pointing out which part was wrong in my solution? I really need to know where's my mistake. I already checked by plugging in the result I guess at first, which is the same as the correct answer(75 degree) and it doesn't satisfy the equation: sin(135-C)/sin(C)=4.sin15/square root(3). But I don't understand how it's wrong. :(

[quote="ghjk"]Since 2BD=BC, we have: BC/AB=4sin15/square root(3)
[/quote]

It is not $ 2BD\equal{}BC$; it is $ 2BD \equal{} CD$, thus $ 3BD \equal{} BC$

There's supposed to be a diagram included with the contest problem (for the purpose of keeping the Contest Collections updated). Here it is:
[asy]
pair A, B, C, D;
A = origin;
real Bcoord = 3*sqrt(2) + sqrt(6);
B = Bcoord/2*dir(180);
C = sqrt(6)*dir(120);
draw(A--B--C--cycle);
D = (C-B)/3 + B;
draw(A--D);

label("$A$", A, dir(0));
label("$B$", B, dir(180));
label("$C$", C, dir(110));
label("$D$", D, dir(130));
[/asy]

[quote=agbdmrbirdyface]There's supposed to be a diagram included with the contest problem. Here it is:
[asy]
pair A, B, C, D;
A = origin;
real Bcoord = 3*sqrt(2) + sqrt(6);
B = Bcoord/2*dir(180);
C = sqrt(6)*dir(120);
draw(A--B--C--cycle);
D = (C-B)/3 + B;
draw(A--D);

label("$A$", A, dir(0));
label("$B$", B, dir(180));
label("$C$", C, dir(110));
label("$D$", D, dir(130));
[/asy][/quote]

i think u can still draw the diagram intuitively

Guys, this thread is over 10 years old...

Nah it's fine; I told him to post his diagram in this thread so that I could copy it over to the main post.
[quote=Mathscienceclass]i think u can still draw the diagram intuitively[/quote]
True, but that's not the point.  If the actual AMC 2001 contest pdf includes a diagram, then so should the entry in the Contests page.

Below is a really, really straightforward solution if you know some basic trigonometry and the Ratio Lemma (generalized Angle Bisector Theorem).
[hide=Solution: Mild Trig Bash (No Calculator Needed)]Let the measure of $\angle DAC=\lambda$. The answer we are looking for is given by $120^\circ-\lambda$. Angle chasing gives $\angle ADB=120^\circ$ and $\angle CDA=60^\circ$. We are now motivated to use trigonometry since these are all angles we know trignometric functions of. Let $AB=x$ and let $AC=y$ and let $CD=a$. Using the Law of Sines, we have $\frac{x}{\sin 120^\circ}=\frac{a}{\sin 15^\circ}$ and $\frac{2a}{\sin \lambda}=\frac{y}{\sin 60^\circ}$. The sine of 15 degrees (we actually don't end up needing this!!) and the sine of 120 degrees are known. By the Ratio Lemma, $\frac{x \sin 15^\circ}{y \sin \lambda}=\frac{1}{2}$. We can also use Law of Sines again and get $\frac{y}{\sin 45^\circ}=\frac{3a}{\sin (\lambda + 15^\circ)}$. Some manipulation shows $\frac{\sin (\lambda + 15^\circ)}{\sin \lambda}=\frac{\sqrt{6}}{2}$. At this point, we can try out each of the answer choices, which are different options for the value of $120^\circ-\lambda$. The only answer choice that works is $\boxed{\text{D}}$.
[/hide]

Put point X on DA and let DX=DB. There will be some nice symmetries that quickly lead to AX=BX=CX, $\angle{ABX}=\angle{BAX}=15$, and $\angle{ACX}=\angle{CAX}=45$, $\angle{BCX}=30$, so $\angle{C}=75$

I'm not sure whether the diagram provided on the physically printed exam was accurate. Regardless, it is easy to draw our own accurate diagram. We have a protractor, so we can just measure $\angle ACB$ and get $75^{\circ}$. Besides $72^{\circ}$, the answer choices are far enough apart that there is no doubt, and enough precision in our drawing can convince us it's not $72^{\circ}$. Just something strange to note about this problem.

The diagrams don't [i]always[/i] have to be accurate.  Just be grateful you have a rough sketch, instead of a paragraph-long description of one.

[hide=yeah nice trig praktice] mark $BD=a$ and $DC=2a$from Law of sine's in $\triangle{ABD}$ we get:\\

$AD=\frac{2a}{\sqrt{3}-1}$\\

and $AB=\frac{2a}{\sqrt{3}-1}\cdot \sqrt{2}\cdot \frac{\sqrt{3}}{2}\implies AB=\frac{a\sqrt{6}}{\sqrt{3}-1}$\\

also from stewart's we have $AC=a\sqrt{6}$\\

so from Law of sine's  in $\triangle{ADC}$ we get:\\

$\sin{\angle{ACB}}=\frac{1}{\sqrt{6}-\sqrt{2}}\implies \sin{\angle{ACB}}=\frac{\sqrt{3}+1}{2\sqrt{2}}=\sin(75^{\circ})\implies \angle{ACB}=\boxed{75^{\circ} (D)}$ [/hide]


Good ol' telescoping series  :lol: I would do it myself, but it would look exactly the same as Kunny's solution.

Edit: actually, you seem to have made some sort of plug-in error on your last step.  The sum should be

$S = \frac{2005 \cdot 2006}{2(2005^2 + 2005 + 1)}$

If you evaluate your sum, you'll see that it's different.

As the result, your answer is the same mine. ;)

Sorry, this is wrong. :blush:

[quote="kunny"]As the result, your answer is the same mine. ;)[/quote]
No, try plugging your answer and mine into a calculator - the two numbers are different.

Edit: okay, you corrected it now, and our answers agree ;-)


integration
calculus
calculus computations

Oh ,You  are right , result of you same as me  :)

[quote="IndoChina"]fantastic! :rotfl:[/quote]
 You do it very good and clear ( :play_ball:   )


geometry
3D geometry
ratio
quadratics
algebra
factorization
difference of cubes

This is from AoPS Vol I:

Evaluate: $ \frac{1}{1^{ \frac{1}{3} } + 2^{ \frac{1}{3} } + 4^{ \frac{1}{3} } } + \frac{1}{4^{ \frac{1}{3} } + 6^{ \frac{1}{3} } + 9^{ \frac{1}{3} } } + \frac{1}{9^{ \frac{1}{3} } + 12^{ \frac{1}{3} } + 16^{ \frac{1}{3} } }$

Basically the books says you need to recognize the fact that $ {1^{ \frac{1}{3} } + 2^{ \frac{1}{3} } + 4^{ \frac{1}{3} }}$ is the same thing as $ 2^{ \frac{3}{3} } - 1^{ \frac{1}{3} }$.  I can follow the solution from there (also noting similar things in the other fractions).  My question is how do you recognize that the fractions are in fact the difference of cubes?  Is it something that you just "see", or is there something in particular that you should look for?  I really have problems recognizing neat factorizations with fractions and or fractional exponents.

$ 2 \minus{} 1 \equal{} 1 \equal{} (2^{\frac {1}{3}} \minus{} 1^{\frac {1}{3}})(4^{\frac {1}{3}} \plus{} 2^{\frac {1}{3}} \plus{} 1^{\frac {1}{3}})$

$ (4^{\frac {1}{3}} \plus{} 2^{\frac {1}{3}} \plus{} 1^{\frac {1}{3}}) \equal{} \frac {1}{2^{\frac {1}{3}} \minus{} 1^{\frac {1}{3}}}$

It's just things you notice after you learn how to apply it to different problems. Just looking at the cube roots gives me the idea.

So now you just apply this to each fraction and the answer follows. :)

Usually when you see radicals in the denominator, it's a signal that you will get something simpler if you rationalize it.

You probably don't want to do that. Notice that the fraction is in the denominater, therefore, it gets rid of the fraction.

$ \frac{1}{1^{\frac{1}{3}}\plus{}2^{\frac{1}{3}}\plus{}4^{\frac{1}{3}}} \equal{} \frac{1}{\frac{1}{2^{\frac{1}{3}}\minus{}1^{\frac{1}{3}}}} \equal{} 2^{\frac{1}{3}}\minus{}1^{\frac{1}{3}}$

See?

$ 4, 6, 9$ is the simplest example of a geometric progression of integers where the common ratio isn't an integer, and it's a useful example to recognize.  More generally the quadratic form $ a^2 \plus{} ab \plus{} b^2$ has a lot of nice properties.

Rationalization actually shouldn't be the first thing on your mind when you see this problem.  The first thing on your mind should be to make the substitution $ x \equal{} \sqrt[3]{2}, y  \equal{} \sqrt[3]{3}$, since numbers tend to obscure the structure of a problem.

Oh, I see. $ a^2 \plus{} ab \plus{} b^2 \equal{} \frac{a^3 \minus{} b^3}{a\minus{}b}$. Very nice.


inequalities
search
inequalities proposed

Let $f(x)=x^2-x+1$, the prove that:

For $x,y,z \in \mathbb{R}$ :

$f(x)f(y)f(z) \ge \frac{(xyz)^2+xyz+1}{3}$

It's probably been posted here before, but you can either expand it / complete the square (yeah right) or... here's a hint:

[hide]Try it for positive reals first.  What does proving $3(x^2-x+1)^3 \geq x^6+x^3+1$ remind you of?  Now, finish with some sign casework...[/hide]

Btw, I thought the RHS was $(xyz)^2-(xyz) \cdots$?

This inequality can be also found in my book of inequalities. It is problem A.P.19. :D
A connection to this problem is A.P. 14 ;)  from my book. ;)

Can anyone post a clearly sols for it please ?
  I has expand it already and it is so terrible. It looks extremely nasty.
  Please give me a right puzzle or  cezar lupu's book (what is its name?)
Thank you! :!:

http://www.mathlinks.ro/Forum/viewtopic.php?t=140
http://www.mathlinks.ro/Forum/viewtopic.php?t=43486
http://www.mathlinks.ro/Forum/viewtopic.php?t=13467

(Two of these links were the result of a search for the string *x^2-x+1* in the Inequalities Section. Very great effort... ;) )

  darij


Does anyone else here like kenken puzzles? They are kind of like sudoku, for those that don't know.

Yes, they are fun.

I think there are some on the new york times website

yeah... kenken is a lot of fun... especially the $ 6\times6$'s :)
idk if the New York Times has them, but I know that the London times has them on their website.

Don't be confused. Kenken is a great game to improve your thinking.

I've made some free videos about how to start solving a basic 4x4 Kenken, as well as an advanced 9x9 KenKen in which the operation signs are not given.

You can check it out at:
[url]http://mathmojo.com/chronicles/kenken/[/url]

i love these puzzles, especially the bigger ones, like 8X8 or 9X9

I know Kenken those are easy. Except for the 9's those are hard.

If you subscribe Reader's Digest, you get one every month!
very effective!

http://www.kenken.com/playnow.html

They update daily and go from 4x4 to 9x9. In my opinion, a really well done website.

i agree, tried it for today, a bit harder then the ones i usually do.. i have a kenken book! lolz

Yeah, will shortz has a few books out... i have the hardest one, and it's really fun, but hard (obviously). Anyway, I'd recommend it.


What's next in the sequence:

$7, 41, 28, 65, 31, 26, 25, 5, 1, 2, 4, 8 ...$

If you think you see it, you're seeing it correctly.  Just give a reasonable explanation for those 3 weird terms... (you'll know what I'm talking about when you figure it out).

[hide]From what I see.

Every other term the number is in the reverse order. The next term is possibly 16? Each term is doubled the previous, but is then grouped so that it's a two digit number (and reversed if necessary)[/hide]

the next term is not 16, but it is close.

[hide]I see what you do.  You double the term before it, then take the last two digits and reverse them.  The next term is 61.[/hide]

[quote="Hamster1800"][hide]

It's 61. Each tem is the last term, doubled and reversed. (remove all 0s)

[/hide][/quote]
Ha! beat you.  Kinda funny we posted at the same time.  When he said doubled I noticed you just reverse the digits

I thought every even term is reversed, and the 13th isn't even, so it shouldn't be reversed.

yes it is 61.  you double and then you read the number from right to left (and therefore get rid of any 0's that come in the front).


function
logarithms
real analysis
real analysis unsolved

Let $U=B(0,1)\subset\mathbb{R}^{2}$. I want to show that $H_{0}^{1}(U)\nsubseteq C(U)$.
I need an example of a function in $u\in H_{0}^{1}(U)=W_{0}^{1,2}$ that is not continuous.
Can I use $u(x)=log(|x|)$?

No, your function is not in $H^{1}$. But $v(x)=\log\log(e/|x|)$ works.


irrational number
advanced fields
advanced fields unsolved

Find an uncountable chain of subsets of natural numbers, ordered by inclusion. By "chain of subsets of $ \mathbb{N}$" I of course mean a set $ A\subset\mathcal{P}(\mathbb{N})$ such that $ A$ is totally ordered by "$ \subset$". Show that we can find an $ A$ such that $ |A|=2^{\aleph_{0}}$.

[hide="my solution"]For each real number $ \alpha$, we can define the map $ F\colon\mathbb{R}\rightarrow\mathcal{P}(\mathbb{Q})$ by $ F(\alpha)=\{r\in\mathbb{Q}\colon r\leq\alpha\}$. It is easily seen that $ F$ is an injection, and that $ F(\mathbb{R})$ is a chain with the required properties (provided that we identify each rational number with some natural number via a bijection $ G$.)[/hide]

Hello.

Take a bijection $ \varphi:{\mathbb N}\to{\mathbb Q}$. For $ x$ any irrational number, let $ A_{x}$ be the set $ \{n\in{\mathbb N}: \varphi(n) < x\}$. Clearly, for $ x<y$ we have $ A_{x}\subset A_{y}$ whence $ \{A_{x}\}_{x\in{\mathbb R}\setminus{\mathbb Q}}$ is linearly ordered. Further, for $ x<y$ we have $ A_{x}\neq A_{y}$ since there exists a rational number $ z\in (x,y)$ and so $ \varphi^{-1}(z)\notin A_{x}$, but $ \varphi^{-1}(z)\in A_{y}$.

Hanno


It's equivalent to $(x-1)^{3}-8 = 16 \sqrt{x}$
Sketching quickly, we can see that there is exactly one solution.
By inspection, I find that it's between $x=4$ and $x=5$. I don't know how to find the exact value.

[quote="chien than"]Solve of equation:
$x^{3}-3x^{2}+3x-16\sqrt{x}-9=0$[/quote]
Maybe $x^{3}-3x^{2}+3x-16\sqrt[3]{x}-9=0$ $?$ :wink:
If so the hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149399


Hi,

I'm trying to prove this, but I can't really find a way:

If f is a continuous function and f[0] = f[2], prove that there is an y in [0,1] so f(y) = f(y+1).

Thanks for any hints,

RicOh

define $ g(y)\equal{}f(y\plus{}1)\minus{}f(y)$. then $ g(1)\plus{}g(0)\equal{}0$, so one of the values $ g(1),g(0)$ is non-negative, the other non-positive. now use the intermediate value theorem.


The function $ f(x)$, for which the set of all real numbers is field, assumes only positive values  and $ f'(x)\equal{}2f(x)$ for everyone $ x \in R$. To appoint all points in the period $ x \in [\minus{}\pi, \pi]$, which the function $ g(x)\equal{}f(x)\cdot cos^2{x}$ has the minimum in.

I think that the answer is [hide="this one"]$ \minus{}\frac{\pi}{2},\frac{\pi}{2}$[/hide].
Is it correct?

Greetings
GioMott


If $ a,b,c,d\in{R}$
and $ |a|\plus{}|b|\equal{}1;
      |c|\plus{}|d|\equal{}1;
      |a\plus{}b|\plus{}|c\plus{}d|\equal{}2;
      |a\minus{}b|\plus{}|c\minus{}d|\equal{}2$;
Prove that $ |a\plus{}b|\equal{}|a\minus{}b|\equal{}|a|\plus{}|b|$ and $ |c\plus{}d|\equal{}|c\minus{}d|\equal{}|c|\plus{}|d|$

$ |a| \plus{} |b| \le |a \plus{} b|$
$ |c| \plus{} |d| \le |c \plus{} d|$

However: $ 2 \equal{} |a| \plus{} |b| \plus{} |c| \plus{} |d| \equal{} |a \plus{} b| \plus{} |c \plus{} d| \equal{} 2$ then
$ |a| \plus{} |b| \equal{} |a \plus{} b|$
$ |c| \plus{} |d| \equal{} |c \plus{} d|$

idem for $ a$, $ \minus{} b$ and $ c$ and $ \minus{}d$


An a x b x c box has half the volume of an (a+2) x (b+2) x (c+2) box, where a ≤ b ≤ c. What is the largest possible c?

I believe that $ c$ can be an arbitrarily large real number.

Is there some extra condition such as requiring that $ a$, $ b$, and $ c$ be integers?


[b]how many possibilities are there?[/b]

if you havent heard of su doku, it is a puzzle consisting of a 3*3 grid, where each square itself is a 3*3 grid so you have in total a 9*9 grid which is seperated into 9 sections. one number goes in each of the small squares such that each row has the numbers 1-9, each column also has the numbers 1-9, and each of the large squares consisting of 9 smaller ones also has the numbers 1-9, so that no number is repeated.

If that was a bad explanation then this website is very clear in explaining it:

http://www.sudoku.com/

you can check the following site for all that information:
 http://en.wikipedia.org/wiki/Sudoku

as you might see, the $n\times n$ Sudoku problem is not at all easy. Even counting the number of sudoku grids needs a good computer search.


induction
number theory
Diophantine equation
Diophantine Equations

(Leo Moser) Show that the Diophantine equation \[\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}+\frac{1}{x_{1}x_{2}\cdots x_{n}}= 1\] has at least one solution for every positive integers $n$.

[quote="silvergrasshopper"]Great problem, and also great solution. Where does the idea of defining $y_i = x_i$ for $i=1,2,...,n$ and $y_{n+1} = x_1 x_2 ... x_n + 1$ comes from?[/quote]

Thanks!
The idea comes from trying to use induction from smaller cases, defining $y_i = x_i$ for $i=1,2,...,n$ and then getting the value of $y_{i+1}$


calculus algorithm search topology probability videos function

This question has been coming up more and more in my mind recently, and AoPS user Scrambled wrote a Facebook note on this a while ago. But how useful is the math that most researchers are doing these days? Let me clarify - we're not talking about things like math applied to things like population genetics, computational linguistics, etc., that allow us to [i]directly[/i] understand what's going on in the real world. And we're not talking about elementary mathematics that's applied to any scientific field, calculus being a perfect example.

Perhaps I am too naive in my understanding of research mathematics to do justice to this topic, forgive me, but are we getting carried away with the kinds of mathematical research we're doing? I sat in on a Siemens regional competition presentation a few weeks ago, and someone had a project on the logarithmic number derivative. It was a perfectly rigorous study of mathematics, but his conclusion at the end was that his work [i]confirmed[/i] something that was already pretty widely speculated about encryption (again, forgive me, I don't recall much of the mathematical detail). Is this really useful?

We developed pretty heavy machinery over several centuries to prove Fermat's Last Theorem. But I mean, after all these centuries, we were pretty convinced anyway that it was true. What, really, is the value in the real world of a proof of this theorem? Even if for some reason Fermat's Last Theorem came up in real life, wouldn't the numbers we cared about be small enough such that advances in computer science already would have kept us safe? I guess all the stuff we did with elliptic curves and algebraic number theory did allow us to develop faster factoring algorithms for encryption, yes. This brings me to my next point - we justify doing mathematics by saying that it is the language of the sciences, which it of course is. But it seems to me that most mathematicians (pure mathematicians, at least), don't often do mathematics with this in mind. Please forgive and correct me if I'm wrong. But I've always justified my love of math to skeptical friends by saying things like "who knows if we're doing to need this contrived mess in 50 years?". Is this [i]really[/i] a good enough justification for doing math the way we do it now, with such a high level of abstraction?

Sorry for the jumbled-ness of all of these ideas, I just kind of spit them out. Hopefully it's enough to have some kind of discussion.

This old [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1542504543&t=180878]discussion[/url] may be relevant to this topic.

I've always felt I was good at math, and that math "was my thing", but when I think of a future me, I never seem to picture a math professor, or a man who spends his time proving obscure theorems and formulae. And while that may not necessarily be what a mathematician does, I've always seen myself as going into another field, and if I do mathematics, likely a practical math field like engineering. Pure math, while entertaining, just doesn't seem to be that beneficial. I want to die having made a difference in the world, remembered for helping people and changing lives, not proving a 500 year old theorem assumed to be true.

How useful is art, really?  I mean, it's entertaining and pleasing and all, but it doesn't put food on anyone's table, right?

Maybe the parallel isn't perfect, but I tend to think of mathematics more as art than science.  It's true that some mathematics is useful (including some originally considered very much "pure", e.g. big chunks of very abstract functional analysis are nowadays indispensable in engineering and computational mathematics), but I don't really see why the perceived usefulness of this or that branch of math is relevant to whether or not we should pursue it.  Or rather, fundamentally, I guess I disagree with the definition of "useful" implied here.  Pure math gives us the tools to imagine and manipulate a universe full of wonderful abstract objects limited only by our own creativity, which I think is just as useful as most other human endeavors -- even if it doesn't add much to the GDP.

One might retort that while other art can be appreciated by the general public, pure mathematics cannot.  I don't agree.  At least in my personal experience, I've found that of my friends who know nothing about mathematics, most are more interested in hearing about, say, topology than about how probability can be used to valuate stock options.  I think the reason many people don't appreciate the subject -- moreover, [i]detest[/i] it -- has more to do with the way it's taught in school than its innate value.  Most people merely haven't been introduced to math as art, but it need not be that way, and maybe will not always be.  But that's a digression for another thread.

As already mentioned in the other topic, a lot of times a use surfaces decades or centuries after it was developed for pure joy or just out of curiosity with Number Theory as an example. But even more, there is in fact no reason to favour modern physics (which is in some sense more absurd than math: the latter doesn't claim to have to do with reality) or art (which has even less usage) or most parts of engineerings (which primarily don't help the poor, but the rich, or who do you think buys the cool new gaming consoles, builds 500 metres tall buildings, uses nukes or or just shoots people to the moon¿) over math. Or even more abstract: why is "thinking" worse than "doing"¿ Was man made for playing video games or for conquering the mysteries of logic and the universe¿

Joe: If you haven't read it, you should take a look at http://www.maa.org/devlin/LockhartsLament.pdf .

I'm not sure about a lot of things being said.  (Disclaimer :  I'm studying math and cello performance, so I have some view of both math and "art", although I'm an undergrad, so some other people here probably have better perspective on these things than I do, although the opposite is possibly also true.)

In particular, I am suspicous of the parallel between math and art.  It seems to me (although I could be mistaken) that a lot of math is not about solving specific, contrived problems, but rather about understanding what's happening in a particular situation.  Also, a lot of math applies to number theory, where (these days) it looks like there is often a short path from theoretical results to practical applications (cryptography, etc.), at least as far as computing goes.  So it seems to me like math often has direction---there are goals beyond simply "have fun" and "solve this random problem".  Perhaps I am being optimistic, but it seems to me like there are certain questions that are begging to be answered, and fortunately, some of us happen to enjoy pursuing the answer.  To me, at least, it seems like fun is a secondary aspect of math.  Sure, fun is possibly a big reason why many of us do math---but that's not (usually) the reason that people are paid to do it.  Researchers aren't paid so that people outside the profession can be awed by the beauty in math journals.  Rather, I suspect that ultimately they are paid because math has a pretty strong history of turning out to be useful.

By contrast, artists (musicians) are paid solely so that amateurs can have art (music) to admire.  I am not sure about the proposition that art is useless, either---in its many forms, it can brighten people's lives, inspire them, give them an escape from the hassles of everyday life, and so forth.  That is not an easy thing to do!

Sometimes, science can serve the same function as art.  I think that the lunar landings are a prime example of this.  Unfortunately, the applied sciences seem to be better at providing inspiration to the general public than math does.  To the layman, proving that the equation $ x^n \plus{} y^n \equal{} z^n$ has no nontrivial solutions probably just doesn't seem nearly exciting as seeing a human being walk on the moon.

By the way, I'm curious about the assertion that most parts of engineering don't help the poor.  I get the impression that lots of engineers work in designing things like buildings and vehicles, which seem to me like they would favor the poor about as much as most other things.

[quote="ZetaX"][i](line breaks added)[/i]
... there is in fact no reason to favour:
- modern physics (which is in some sense more absurd than math: the latter doesn't claim to have to do with reality)
- art (which has even less usage)
- engineerings (which primarily don't help the poor, but the rich, or who do you think buys the cool new gaming consoles, builds 500 metres tall buildings, uses nukes or or just shoots people to the moon)

over math. Or even more abstract: why is "thinking" worse than "doing"¿ Was man made for playing video games or for conquering the mysteries of logic and the universe?[/quote]

I'm not entirely sure what you're trying to prove with all of your parenthetical comments. It seems like you're saying that because physics is absurd, art is useless, and engineering doesn't help the poor (all of which are highly questionable statements), they should not be favored over pure mathematics by society. I'll expand on my objection to the third statement, in particular, after my final exam period is over (if it is still relevant).

The question you pose about "thinking" vs "doing" is an interesting one. I am of the opinion that an individual should choose, within their means, how to value each. However, I don't think unlocking the mysteries of the universe and playing video games are the necessarily most relevant examples of thinking and doing, respectively, and it's [i]certainly[/i] not a useful comparison.

For me, a precondition to doing serious math is recognizing that, in all probability, none of it will be useful for years, decades, or millennia, or never will be. You can never know what possible uses things will have. The best example of this is number theory - for 2500 years, the thought that it would have practical application was preposterous. Our world is fundamentally continuous, not discrete. You probably won't see the results of your work in your lifetime, but maybe some lemma you prove will be key to a new model of quantum mechanics that allows for huge improvements in world quality of life... 500 years later.

G.H. Hardy's [i]A Mathematician's Apology[/i] is a really interesting read. It talks about this issue some.

Many aspects of math are meant for enjoyment, not purpose. Of course, there are some really elegant applied mathematical branches, but I personally feel that mathematics in its purity is truly beautiful.

[quote="worthawholebean"]Our world is fundamentally continuous, not discrete.

...

You probably won't see the results of your work in your lifetime, but maybe some lemma you prove will be key to a new model of quantum mechanics that allows for huge improvements in world quality of life... 500 years later.[/quote]

Do you know what 'quantum' means?

The vast majority of professional mathematicians not employed by the NSA work at universities or colleges; the vast majority of these jobs require some moderate to large amount of teaching.  I think justifying mathematics in terms of its direct positive effects on people is extremely difficult*, and justifying mathematics in terms of its potiential future positive effects on people is so speculative as to be silly, but justifying the project of education is not very difficult at all (at least if you believe in any version of the enlightened individual or enlightened citizen).  In fact, I'm quite content that my mathematical output should have no consequences beyond mathematics -- I'm off the hook for all the various sorts of unpleasant things that people inflict on each other, directly and indirectly.

* This is also true of the vast majority of other human endeavors, including most of scientific research and engineering and whatever else.  People just genuinely don't spend very much time improving the lives of other people.

[quote="new_member"]According to wikipedia: http://en.wikipedia.org/wiki/Quantum :P[/quote]

Wah, Wah, Great! You do possess an awesome wit-- I'm bedazzled by it. I have a query: Were you trying to be witty or were you trying to give some authentic information pertaining to 'What constitutes the definition of a Quantum'?

OK, so up until the 1900s we thought our world was continuous... my point stands.


Let $ ABC$ is a triangle, let $ O$ is a point in triangle $ ABC$. Let $ AO, BO,CO$ meet $ BC, CA, AB$ at $ A', B', C'$ respectively. Prove that: $ \frac {OA}{OA'} \plus{} \frac {OB}{OB'} \plus{} \frac {OC}{OC'}\ge9$.

The statement is false, because for an equilateral triangle with $ O$ (or $ M$  :P ) at the center, we have $ \frac{OA}{OA'}\plus{}\frac{OB}{OB'}\plus{}\frac{OC}{OC'} \equal{} 6$.

[quote="Ahiles"]As [u][b]archimedes1[/b][/u] mentioned, the corect problem is $ \frac {OA}{OA'} \plus{} \frac {OB}{OB'} \plus{} \frac {OC}{OC'} \ge 6$, with equality if $ O \equal{} G$.

[b][color=darkblue]Add $ 3$ to both sides:

$ \bigg( \frac {OA}{OA'} \plus{} 1\bigg) \plus{} \bigg(\frac {OB}{OB'} \plus{} 1\bigg) \plus{} \bigg(\frac {OC}{OC'} \plus{} 1\bigg) \ge 9$

$ \frac {AA}{OA'} \plus{} \frac {BB}{OB'} \plus{} \frac {CC}{OC'} \ge 9$

By Gergonne  $ \frac {OA'}{AA} \plus{} \frac {OB'}{BB} \plus{} \frac {OC'}{CC} \equal{} 1$.

If we substitute: $ \frac {OA'}{AA} \equal{} x, \ \frac {OB'}{BB} \equal{} y, \ \frac {OC'}{CC} \equal{} z$.

Our inequality is equivalent to $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge 9$ with $ x \plus{} y \plus{} z \equal{} 1$.

But by Cauchy: $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge \frac {9}{x \plus{} y \plus{} z} \equal{} 9$, and we are done.[/color][/b][/quote]
[color=darkblue]Thank you very much.[/color]


LaTeX
combinatorics unsolved
combinatorics

In a company of $ 2N$ boys and $ 2005$ girls for every pair of girls exactly $ N$ boys know just one of this girls. Prove that the number of boys who know all the girls don’t exceed $ \frac{N}{1003}$

[quote="outback"]for every pair of girls exactly $ N$ boys know just one of this girls. [/quote]
Can you say more precisely what this means? Is there one girl in the pair such that $ N$ boys know her and not the other? Or are there $ N$ boys who know either girl but not the other?

[quote="randomgraph"]
Or are there $ N$ boys who know either girl but not the other?[/quote]

It's this one..

Off-Topic@OP: You are posting lots of problems but not always with the most specific sources, i.e. only Russia and the year but no grade or problem number. I ask you to compile these problems links sorted by competition in an organized fashion to add them to the resource section.

[quote="gauravpatil"]does it mean that that for a pair of girls  if both of them know a boy then  the boy is not counted among the $ N$[/quote]

If you mean both of them is known by a boy, yes.

If i understand properly, that's my solution:
Suppose otherwise, and we have $ m>\frac{N}{1003}$ boys who knows all girls.
Let $ a_{i}$ is number of girls who know boy number "i" . For every pair of girls we have N boys who knew just one of them. Let's count the sum of number boys for every pair of girls.It will be $ \left(\begin{array}{c}2005\\2\end{array} \right)*N$
Let's count this sum second time. Every boy know just one of the girl in $ (2005\minus{}a_{i})*a_{i}$ pairs. So we have
$ \left(\begin{array}{c}2005\\2\end{array} \right)*N\equal{}\sum_{i\equal{}1}^{2N}{(2005\minus{}a_{i})*a_{i}}$
The max of $ (2005\minus{}a_{i})*a_{i}$ is 1002*1003. Then we have
$ \left(\begin{array}{c}2005\\2\end{array} \right)*N\equal{}\sum_{i\equal{}1}^{2N}{(2005\minus{}a_{i})*a_{i}}\leq 1002*1003*(2N\minus{}m)<1002*1003*(2N\minus{}\frac{N}{1003})\equal{}1002*2005N\equal{}\left(\begin{array}{c}2005\\2\end{array} \right)$
Which is contrudiction.
So number of boys who know all the girls don’t exceed $ \frac{N}{1003}$  :!:
 :roll:

Nice proof (though you dropped an $ N$ in your last equation).  Three latex suggestions: \binom{top}{bottom} gives $ \binom{top}{bottom}$ and \cdot gives a nice multiplication dot $ \cdot$.  Also, the align* enviroment makes for nice sequences of equations:

\begin{align*}
1 + 2 & = (5 - 4) + (5 - 3) \\
& \leq (5 - 4) + (5 - 4) \\
& = 10 - 8
\end{align*}

\begin{align*}
1 + 2 & = (5 - 4) + (5 - 3) \\
& \leq (5 - 4) + (5 - 4) \\
& = 10 - 8
\end{align*}


If $ a,b,c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ \sqrt{3a\plus{}\frac 1{b}}\plus{}\sqrt{3b\plus{}\frac 1{c}}\plus{}\sqrt{3c\plus{}\frac 1{a}} \ge 6$.

Although Inequalities are not interested me much as before, I am still very pleased to solve your nice problem, VASC. For this one, we have to prove that
\[ \sum \frac {1}{a} + 2\sum \sqrt {(\left (3a + \frac {1}{b}\right)\left(3b + \frac {1}{c}\right)} \ge 27.
\]
Applying Cauchy-Schwarz Inequality
\[ (\left (3a + \frac {1}{b}\right)\left(3b + \frac {1}{c}\right) = \left (a + 2a + \frac {1}{b}\right)\left(\frac {1}{c} + 2b + b\right) \ge \left(\sqrt {\frac {a}{c}} + 2\sqrt {ab} + 1\right)^2.
\]
So it remains to prove that
\[ \sum \frac {1}{a} + 2\sum \sqrt {\frac {a}{c}} + 4\sum\sqrt {ab} \ge 21.
\]
Recall two following quite simple (and quite easy to prove)

\[ \sum \frac{1}{a}+2\sum \sqrt{ab} \ge 9\\]
\[ \sum \frac{a}{b}+\frac{3(ab+bc+ca)}{a^2+b^2+c^2} \ge 6\]

We have the desired result.

Notice that both inequalities (1) and (2) are very slight. We can change it, for example

\[ \sum \frac{1}{a}\plus{}\frac{5}{2}\sum \sqrt{ab} \ge \frac{21}{2}\\]

\[ \sum \frac{a}{b}\plus{}\frac{9(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \ge 12 \\]

So we can make something stronger than VASC's initial problem.

[quote="arqady"][quote="hungkhtn"]
\[ \sum \frac {a}{b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge 6
\]
[/quote]
I think, it's wrong. Try $ a \equal{} 3$ and $ b \equal{} c \equal{} 2.$ :wink:
But the following inequality:
$ \sum \frac {a}{b} \plus{} \frac {(4\sqrt2 \minus{} 3)(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge 4\sqrt2$ ends the proof by your beautiful idea, hungkhtn.[/quote]

Oh yes... you are true ... I am really forgetful... oh the old-age  :blush:

Thank you very much for editing this idea  :lol:  Very nice idea  :lol:

The following inequality also holds

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{(3\sqrt[3]{4}\minus{}2)(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}\geq 3\sqrt[3]{4}\plus{}1$

Nice your idea, Hungkhtn!  :lol:
[quote="arqady"]silouan, but for my inequality there is a proof by one line. :P[/quote]
Indeed, the Arqady's inequality is equivalent to
$ \sum a(b^2\minus{}ab\minus{}\sqrt 2 bc\plus{}\sqrt 2 ac)^2 \ge 0$.  :o

hungkhtn, you saw here it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84160&start=20 $ ?$  :)

Dear arqady your inequality is so nice and togheter with

$ \sum \frac {1}{a} \plus{} \frac {5}{2}\sum \sqrt {ab} \ge \frac {21}{2}$ ...$ (1)$

solve the Vasc problem.

I solved (1) using mixing variables method and using your hint.

Thank you very much, Arqady. :)

[url= http://www.artofproblemsolving.com/community/c6h549607p3188769]Generalization:[/url]
Let $a,b,c>0,a+b+c=3,$ and $n\in N^*, $ prove that
\[\sqrt{3a^n+\dfrac{1}{b}}+\sqrt{3b^n+\dfrac{1}{c}}+\sqrt{3c^n+\dfrac{1}{a}}\ge6.\]
Let $a,b,c>0,a+b+c=3 $ . For $\lambda \ge \frac{1}{3}$ ,prove that\[ \sqrt{a+\frac{\lambda}{b}}+\sqrt{b+\frac{\lambda}{c}}+\sqrt{c+\frac{\lambda}{a}}\ge 3\sqrt{1+\lambda}.\]

Let $ a,b,c$ are positive numbers such that $ a+b+c=3.$ [url=https://artofproblemsolving.com/community/c6h2646266p22899241]Prove that[/url]
$$\sqrt{a+3b+\frac2c}+\sqrt{b+3c+\frac2a}+\sqrt{c+3a+\frac2b}\ge3\sqrt 6$$
( IMOC 2017)


What are some good textbooks for high school geometry and algebra II? And by this I mean standard school curriculum stuff, not contest problem solving. Thanks guys.

For Geometry, it depends on what you are looking for. The book I used (in CTY) that emphasized on proofs, is Geometry by Jurgenson; a small, orange, book... I think this book is pretty hard in level, but it's helpful for learning.

yea my school used Geometry by Jurgenson when i was in Burma (well my school is international-american school :) ) and for algebra 2 my school used Glencoe Algebra 2 by Holiday but that book is too easy and nowhere near the level of easy AHSME problems. hmm i suggest Challenging Problems in Algebra by Posamentier after you have learned the basics of algebra materials, that way you can get out of routine and boring algebra problems from your classes  :D  my next book for geometry is Euclid's Elements! i gotta try and see the real deal of this big masterpiece

Well, actually I want stuff that's easier than normal. I know this kid with a learning disability who I'm trying to help.

ah, this makes it much clearer... let me think about some of the early geo and alg books i used to use in 8th and 9th grade...

in 9th grade i used: Geometry by Jurgenson, and Glencoe Algebra 2 by Holiday

in 8th grade i used: Beginning Algebra but for forgot author and it did not have geometry

hey i just found this in amazon.com, name is Passport to Algebra and Geometry (assuming that student has "learning disability")
[url=http://www.amazon.com/exec/obidos/tg/detail/-/0395879884/ref=pd_sbs_b_6/002-9189145-9840054?%5Fencoding=UTF8&v=glance]www.amazon.com/exec/obidos/tg/detail/-/0395879884/ref=pd_sbs_b_6/002-9189145-9840054?%5Fencoding=UTF8&v=glance[/url]

If you're looking for easy, the [url=http://www.keypress.com/catalog/products/key_to/]Key To books[/url] by Key Curriculum Press are pretty good, and mathematically accurate.


A circle with radius R is inscribed into an equilateral triange. Three smaller circles, each with radius r are inscribed into the corners of the triangle so that each little circle touches two sides of the triangle and also touches the large circle.

Find the value R/r

[quote="Iversonfan2005"]Let's pick sides for the equilateral triangle, let's say each side is $2$, thus, the area of the triangle would be $3^{1/2}$. Thus $R(3)=3^{1/2}$, so $R=(3^{1/2})/3$ Now, we know that each of the little circles are inscribed in little equilateral triangles also, each with side length of $1/2$, thus the area of each of those triangles would be $(3^{1/2})/16$ so then $r(3/4)=(3^{1/2}/16),  thus r=(3^{1/2})/12.
Sorry, I don't know what's up with my latex but the answer is $4$.[/quote]

answer is not 4, and you missed a dollar sign

...
It's actually 3...here's a short proof...
Once again, let the length of each side of $ABC$ be 2.  Let the circle touch $AB$, $AC$, and $BC$ at $D$, $E$ and $F$ respectively.  Let the incenter be $I$.  Draw $AI$, and then draw $ID$.  $ID$ is perpendicular to $AB$, and $AI$ bisects angle $A$, so $AID$ is a 30-60-90 triangle.  Thus, with $AI=\frac{2\sqrt{3}}{3}$, we know that the radius of the circle is half of that, or $\frac{\sqrt{3}}{3}$.  The height of the triangle is $\sqrt{3}$, so the radius of the circle is a third of the height of the triangle.  Now draw one of the smaller circles in the section nearest vertex $A$.  Draw the line tangent to both $I$ and this smaller circle (call it $C_1$).  Let the common point of tangency be $Z$.  Just assume that this line is parallel to $BC$...i'm too lazy to prove that right now...  Let it intersect $AB$ and $AC$ at $X$ and $Y$ respectively.  $XY||BC$, so $AXY$ is equilateral.  The radius of $I$ is one third of the height of $ABC$, so $AXY : ABC=\frac{1}{3}$.  Therefore, ratio is 3.

Btw, how did you get that the side length of each of the smaller triangles is 1/2?  just wondering  :D

EDIT: i messed up my latex by missing dollar sign i always do that  :D

OHH!!! It is $3$, I see what I did. the smaller equilateral trianges would have side length of $2/3$! not $1/2$. Thus the area of the the smaller triangles would be $(3^{1/2}/9$ which would also be the radius of the smaller circles since the semiperimeter is $1$, so then the ratio would be $3$. Foolish mistakes :blush:  :blush: Atleast I fixed it though ;)

[quote="calculuslover800"]Sorry if this sounds stupid, but how did you get the lengths of the smaller triangles equals 2/3?  Or rather, how do you know that the lengths of the triangles are one third the length of the big triangle?  Everything else makes sense.[/quote]

Because make the circle inscribed in the bigger triangle be inscribed in a hexagon, by drawing the three equilateral triangles that the circle is indscribed in. That hexagon contains 6 equilateral triangles all the same size as the 3 smaller equilateral triangles, thus the bottom sides of those three triangles make up one side of the big equilateral triangle. I don't know how I could attach a drawing but you should be able to see it.

I think that's why each of the little equilateral triangles would have 1/3 of the length of the big equilateral triangle...am I right?

[quote="Iversonfan2005"][quote="K81o7"]Using the fact that the perpendicular from $I$ to the base is half of $AI$ (you can make a 30-60-90), you know that the radius of the big circle is 1/3 the height...
If you wanted to, you could say that the altitudes, medians and angle bisectors overlap, and from there it's easy...[/quote]

Hmm...I wouldn't have really thought of that :oops: I just saw the hexagon and since it was an equilateral triangle, I thought the Hexagon was regular so the triangles it was composed of were congruent to the ones that held the three circles.[/quote]

That way works too.  You just have to show that it is a regular hexagon (which isn't too hard.)


Given $B>0$ that $\frac{x^3}y\geq A(x-y-B),\text{ }x,y\in\mathbb{R}_+$. Find $\max A$.

If $x < y+B$, then the RS is negative and the ineq must be true.  This is the motivation for the substitution $x = y+B+k$.  Assuming that the RS is positive, then k is positive.

Then $\frac{(y+B+k)^3}{yk} \ge A$(*), so we want to find the minimum of the LS of (*).    We have y,B,k positive.

Considering as a function in y, the derivative has a sign equal to the sign of $2y - B - k$.  Then $y = \frac{B+k}{2}$ for the minimum.

Similarily, $k = \frac{B+y}{2}$.  Solving, $y=k=B$.  So the minimum of the LS of (*) is $27B$.  This is the maximum for A.

----

ps this might be bad, i didnt check it


Sean $a$ y $n$ enteros tales que $n$ divide a $(a^{2}+1)$. Demostrar que existe un entero $b$, tal que $n(n^{2}+1)$ divide a $(b^{2}+1)$

[b]UNA FORMA[/b]

$b= n+a (n^{2}+1)$ satisface lo pedido.

[b]OTRA FORMA[/b]

Usando el siguiente enunciado (que debe ser un teorema)

Sea $n>1$ un numero entero.
Existe un entero $a$ tal que $n$ divide a $a^{2}+1$ si y solamente si existen enteros $\alpha$, $m$ tales que $n= 2^{\alpha}.m$, $\alpha=0, 1$ y todo divisor primo $p$ de $m$ es de la forma $p \cong 1 mod(4)$

Disculpa jonfv, como s te ocurrio la primera solucion??? Es decir, usaste alguna congruencia, teorema chino, o nada mas se te ocurrio de la nada...

Bueno, fue de una manera inocente.
$n$ divide a $(xn+a)^{2}+(yn+1)^{2}$ para cualesquiera enteros $x$, $y$.
Entonces $n(n^{2}+1)$ divide a
$((xn+a)^{2}+(yn+1)^{2})(n^{2}+1) = ((xn+a)n-(yn+1))^{2}+((xn+a)+(yn+1)n)^{2}$
tomando $y=xn+a$, la última expresión queda
$1+(n+(xn+a)(n^{2}+1))^{2}$
y para $x=0$, obtenemos el $b$ que di al comienzo.


superior algebra
superior algebra unsolved

[quote="ciprian"]Let $(A,+,\cdot)$ a commutative ring such that the elements $1+1+1$ and $1+1$ are invertible and such that $x^{3}=y^{3}\Rightarrow x=y$. Prove that if $a,b,c\in A$ and $a^{2}+b^{2}+c^{2}=ab+bc+ca$ then $a=b=c$.[/quote]

$a^{2}+b^{2}+c^{2}=ab+bc+ca$ yields $\left(a^{2}+b^{2}+c^{2}\right)-\left(ab+bc+ca\right)=0$. But some algebra shows that

$\left(c-a\right)^{3}-\left(a-b\right)^{3}=\left(\left(c-a\right)-\left(a-b\right)\right)\cdot\left(\left(a^{2}+b^{2}+c^{2}\right)-\left(ab+bc+ca\right)\right)$
$=\left(\left(c-a\right)-\left(a-b\right)\right)\cdot 0=0$,

so that $\left(c-a\right)^{3}=\left(a-b\right)^{3}$. Since $x^{3}=y^{3}\Rightarrow x=y$, this yields $c-a=a-b$, and similarly we find that $a-b=b-c$. Thus, the trivial equation $\left(c-a\right)+\left(a-b\right)+\left(b-c\right)=0$ becomes $\left(a-b\right)+\left(a-b\right)+\left(a-b\right)=0$, or, equivalently, $\left(1+1+1\right)\left(a-b\right)=0$. Since $1+1+1$ is invertible, this yields $a-b=0$, so that $a=b$. Similarly, $b=c$. Thus, $a=b=c$, and we are done.

Note that we have used the invertibility of $1+1+1$ only, not that of $1+1$.

  darij

Thx for the fast reply darij :)  I tried the same idea but I think I was too tired to do the computations well :blush:  This problem was used some years ago at the regional stage of the Romanian Olympiad.


Hi!

I found (Q19) in Exercise 4.3 on Page 137 of Challenges and Thrills Of Pre-College Mathematics an interesting question for research!!!

This is the question "STATED WORD BY WORD" as in the book:-

"X is any point on AB and the median AD of triangle ABC meets XC at Y. Prove that XY/YC = AX/XB"

If u have doubt regarding the authenticity of the ques, plzz turn to the above-mentioned page and chapter of the book.

Now the task is to research this question, come up with findings and find an appropriate solution!! :lol:

This may be a sitter for RMO/INMO/IMO guys :D  but try it out,guys!!!

I think that the problem is either wrong or incompletely stated.Let X be the midpoint of AB.Then Y is centroid and AX/BX=1
But XY/YC=1/2

@apratimgr

good :D u r going in the the right direction but complete the research!!!!

It is probably a printing mistake. $ \frac{XY}{YC}\equal{}\frac{AX}{AB}$ holds. (easy to prove using ceva's theorem)

[quote="fbaggins"]
"$ X$ is any point on $ AB$ and the median $ AD$ of triangle $ ABC$ meets $ XC$ at $ Y$. Prove that $ \frac{XY}{YC} \equal{} \frac{AX}{AX}$"
[/quote]
Hi..well this IS really a sitter...and I've corrected your question...
Just apply MENELAUS' THEOREM on $ \triangle XBC$ for the line $ AYD$...
We get,
\[ \frac {BD}{DC}\cdot\frac {CY}{YX}\cdot\frac {XA}{AB} \equal{} \minus{} 1\]
Since, $ \frac {BD}{DC} \equal{} 1$
\[ \Longrightarrow \frac {CY}{YX}\cdot\frac {XA}{AB} \equal{} \minus{} 1\]

\[ \Longrightarrow \frac {CY}{YX} \equal{} \minus{} \frac {AB}{XA}\]
We are Done...!!!
 :lol:


Let $ \alpha,$ $ \beta,$ $ \gamma$ and $ \delta$ are real numbers such that $ \alpha\plus{}\beta\plus{}\gamma\plus{}\delta\equal{}\pi.$ Prove that
\[ \cos\alpha\plus{}\cos\beta\plus{}\cos\gamma\plus{}\cos\delta\leq2\sqrt2\]

[quote="nateharman1234"][hide]
Well if they are all less than pi/2 jensens finishes the job as the cosine is concave on (0,pi/2).  And if one of them is greater than pi/2 well the cosine is negative, and that is clearly bad but i don't feel like rigorously justifying that.[/hide][/quote]

Note $ \alpha, \beta, \gamma, \delta \in \mathbb{R}$ not $ \mathbb{R}^{ \plus{} }$ so you would also need to justify the case of when some are negative.

[quote="nateharman1234"]  And if one of them is greater than pi/2 well the cosine is negative, and that is clearly bad but i don't feel like rigorously justifying that.[/quote]
There is a one-line solution for all $ \alpha,$ $ \beta,$ $ \gamma$ and $ \delta$ are real numbers such that $ \alpha\plus{}\beta\plus{}\gamma\plus{}\delta\equal{}\pi.$ :wink:

I am not sure what the one-liner is but I think we can make use of the fact $ \cos{x} \equal{} \cos{ (\minus{} x)}$.

[quote="madness"]Can't we directly apply Jensen's inequality?
 :D[/quote]

Problem is due to the fact that the concavity of $ f(x) \equal{} \cos{x}$ changes as $ x$ varies.

$ \frac {\cos(\frac {\alpha \plus{} \beta}{2}) \cos(\frac {\alpha \minus{} \beta}{2}) \plus{} \sin(\frac {\alpha \plus{} \beta}{2}) \cos (\frac {\gamma \minus{} \delta}{2})}{\sqrt {2}} \leq \frac {\cos( \frac {\alpha \plus{} \beta}{2})}{\sqrt {2}} \plus{} \frac {\sin( \frac {\alpha \plus{} \beta}{2})}{\sqrt {2}} \equal{} \sin(\frac {\alpha \plus{} \beta \plus{} \frac {\pi}{2}}{2}) \leq 1$
so we have
$ \frac {\sum \cos \alpha}{2\sqrt {2}} \equal{} \frac {\cos(\frac {\alpha \plus{} \beta}{2}) \cos(\frac {\alpha \minus{} \beta}{2}) \plus{} \sin(\frac {\alpha \plus{} \beta}{2}) \cos (\frac {\gamma \minus{} \delta}{2})}{\sqrt {2}} \leq 1$
so done

Yes, pardesi! It is that idea!:lol:
[quote="pardesi"]$ \frac {\cos(\frac {\alpha \plus{} \beta}{2}) \cos(\frac {\alpha \minus{} \beta}{2}) \plus{} \sin(\frac {\alpha \plus{} \beta}{2}) \cos (\frac {\gamma \minus{} \delta}{2})}{\sqrt {2}} \leq \frac {\cos( \frac {\alpha \plus{} \beta}{2})}{\sqrt {2}} \plus{} \frac {\sin( \frac {\alpha \plus{} \beta}{2})}{\sqrt {2}}$
[/quote]
But it can be wrong. :(
I think, the following reasoning better:
\[ \sum_{cyc}\cos\alpha\leq2\sqrt {\cos^2\frac {\alpha \minus{} \beta}{2} \plus{} \cos^2 \frac {\gamma \minus{} \delta}{2}}\leq2\sqrt2
\]


vector
calculus
function
derivative
calculus computations

Hello, I am stuck at this problem doing some independent calculus study  :blush:  Can anybody give me some hints? Please no full solution. Thank you.

Let $f: \mathbb{R}\rightarrow\mathbb{R}^n$ be a differentiable mapping with $f'(t)\ne\vec{0}$ for all $t\in\mathbb{R}$. Let $\vec{p}$ be a fixed point not on the image curve of $f$. If $\vec{q}=f(t_0)$ is the point of the curve closest to $\vec{p}$, that is, $|\vec{p}-\vec{q}|\le|\vec{p}-f(t)|$ for all $t\in\mathbb{R}$, show that the vector $\vec{p}-\vec{q}$ is orthogonal to the curve at $\vec{q}$. Hint: Differentiate the function $\varphi(t)=|\vec{p}-f(t)|^2$.

I tried using the hint but I keep getting $\varphi'(t)=-2f'(t)(\vec{p}-f(t))$

PS: Is there a condition for 'closest point' on the first derivative? I'm not sure...

Be sure you remember what is a vector and what is a scalar, and remember that we can write $|v|^2$ as $v\cdot v.$

Now $\varphi(t)=|\vec{p}-f(t)|^2=(\vec{p}-\vec{f}(t))\cdot(\vec{p}-\vec{f}(t))$

Differentiate that to get $\varphi'(t)=-2\vec{f'}(t)\cdot(\vec{p}-\vec{f}(t)).$

That's a dot product; it can be zero without either factor being zero, if the factors are orthogonal.

I'm sorry for late reply, Mr. Merryfield. I was having dinner   :blush:
I hope I'm getting this right: $\varphi$ is a mapping $\varphi: \mathbb{R}\rightarrow\mathbb{R}$

Since $\vec{q}$ is the closest point of the curve to $\vec{p}$, $\varphi(t_0)$ attains minimum, and so $\varphi'(t_0)=-2f'(t_0)\cdot(\vec{p}-f(t_0))=0$. But $f'(t)\ne 0$ for all $t\in\mathbb{R}$ and $f'$ is the tangent vector, so vector $f'(t_0)$ is orthogonal to $\vec{p}-\vec{q}$ or $\vec{p}-\vec{q}$ is orthogonal to the curve at $\vec{q}$


This forum has grown large enough that this may be of interest:

Post here if you went to Mathcounts nationals at some point.  If you want.

--

Me:

I went to nats in 2002 and 2003.  Team [hide]Nevada.[/hide]

 8-)

All the way back in 1990 and 1991.  Yes, I am ancient.

I went in 2001, 2002 and 2003.  It helps to be from a small "state".

[quote="MathFiend"]I hope to go this year, but I doubt it will happen. My team has never gone. Oh well, a girl can always dream...[/quote]

It might happen.  Online you can now study with top MATHCOUNTS students from all over the country.  Maybe that could make the difference.

[quote="Tare"]You're not ancient. You're younger than most adult people at AoPS.

Where can I find MATHCOUNTS problems?[/quote]

You can order books of MATHCOUNTS problems from the mathcounts.org website (I believe).  I remember my MATHCOUNTS teacher (who was still coaching the state team this past year) having stacks of manuals of MATHCOUNTS problems.

yeah, there are lots of people on this forum whove gone to nats. i havent before, one last year. hopefully i will accompany chinaboy 8-)

2001 I was in 6th grade; my school didn't let me. (:()
2002 I had a scheduling conflict; couldn't go
2003 I went. and sucked. darn.

Even so, I doubt I'll ever get to go...oh well, I can at least watch it on TV


I don't know if this is true, but I hear that the new countdown round will be single elimination w/ seeds 1-4 a bye the first round and seeds 5-12 competing.  Is this really going to happen?  If so, any other details?

[quote="yif man12"]and also, some of the people who are really high seeds might be mad if they're not that great at countdown and go from 1 to like 8.[/quote]

Yeah that was what I was concerned about.  I can't do countdown at all.

all in all, it would probably take longer than the one we have now, because of all the lower positions. i'd enjoy a round where they just pick 8 random people, do a just for fun countdown round and give the winner a little plaque. it'd be like the fun round that actually gets you something

Yes, this is true about the new format (according to my state director, anyway).  I believe the rules state that ties will not be broken, thus 3rd and 4th place are "semifinalists", 5th through 8th are "quarterfinalists", and everyone else is a "participant".
In response to yifman, remember that in the last few years, the top ten at nationals have been separated by very little in terms of score.  So this may actually be more fair, in that the written round winner has no advantage over the 4th place student, when they were probably only separated by one point, and likewise with 5th through 12th places.  I personally think that the most fair way to do it would be to have the countdown round somehow be part of each student's score (assuming the student is in the top ten or so), but that has other disadvantages (such as unfair pairings and the impossibility of having a ten-way countdown match).

I think MathCounts should just be harder so the scores will vary more (nationals only, of course)  :twisted:.  That way we don't have to mess with the countdown thing, in which speeds usually vary by very little.

[quote="Rep123max"]Even though MATHCOUNTS is over for me, I think its a good idea. The countdown round didnt have enough weight before and now it will be way more important to people[/quote]

Is this necessarily a good thing?  Countdown puts quick readers and speedy arithmeticians (is that a word?) at an advantage, as opposed to those who are better at actually understanding and doing math..

[quote="Rep123max"]Sprint round puts those who are immune to making mistakes winners. [/quote]

Doesn't Countdown sort of do the same thing?  And also, for sprint, the majority of competitors can't answer more than 25 of the questions, so it does require mathematical talent to be able to win.. I think ;)

Of course, this is coming from a person who hates Countdown because she can't read quickly, can't do speedy computations, and also doesn't do well under pressure/one-on-one :)

I think all of MC puts those who are immune to mistakes in top ranks...unless countdown is official...

although countdown does kind of put fast readers/thinks at an advantage, it is a good way to see "who really belongs at the top"...at least, it does that better than the other two individual rounds...

My problem in MC was always (except in my 6th grade year) that the problems were too easy and I made too many stupid mistakes

Maybe the best thing to do would be to have countdowns harder and trickier, so that it's less a thing of speed.

mainly, i just fool around during the countdown round in states. when i'm waiting for others to finish, i'll randomly guess answers and see if they're right. usually i get 2 or 3.
Back on topic, yes, it is good for mathcounts to be evening out the distribution of skills. But the truth is, no matter what, you can't make everybody happy. Whatever you do, there'll be some people hating you for it.

[quote="yif man12"]That would be pretty cool. but then how would they decide places 5-12 after the countdown round was done? and also, some of the people who are really high seeds might be mad if they're not that great at countdown and go from 1 to like 8.[/quote]

How can you go down more than one spot?  I thought you could only move down 1 spot at the most....

By the new system, it's a single elimination.  Look at the first few messages.

Personally, I don't like the new countdown style. I thought the ladder format was ingenious, because it made a difference, yet the better you did in the written round, the better you would start in countdown, so there was a lot of emphasis put on the written rounds. I also liked it because you could only go down one place, but you could move up 9 places potentially (10th to 1st), so there was always hope, but it would be difficult. I also don't like how with the new forma, there isn't 3rd through 12th place, but rather there's "semi-finalist," "quarterfinalist," and "Countdown round participant." So now, instead of someone saying they were 9th in the nation, they will have to explain what it means to be a "participant." What I'm concerned about more than anything else, is how this is the second year that ESPN is covering the countdown round on TV, and it is "coincidentally" also the year they move to a sports-bracket-type countdown round -- similar to those seen on ESPN.

Now that I've said the negatiges, I will mention a positive. It has to do with the fact that the top finishers at nationals have written round scores that are so close together, so the person who finishes first with a score of 46 shouldn't have such an advantage over the person who finished eighth with a score of 45 (okat, so that's a bit of an exaggeration). So, in that respect, maybe it was a good change.

But I still don't think so much emphasis should be put on countdown round. Here's an example of why: Take a look at Nick Wage from Wisconsin in 2002, who, in the countdown round, didn't just fall back a spot at Nationals and State, but also at Chapter. And yet, he was in the top ten in the country. If there was a bigger stress put on the countdown round, he wouldn't have done very well, even though he's one of the best in the country.

Which brings be to my final point. This bracket system may be good at Nationals, but they should not expand it to state (As far as I am aware, they have no plans to). They should not force the State written round Champion to win in the Countdown round to go to Nationals. It would be possible for none of the top four to make it to Nationals if the top four was bad at countdown, like Nick Wage was.

[quote="yif man12"]lol, remember when jeff thought it was alphabetical order?[/quote]

I still think that. It would explain why I got 17th out of about 20 tied with the same score. Also, someone whose last name begins with Y who had the same score ended up last in the tiebreak.

That's actually strangely possible!  I lost the 4-way tie for 2nd place at Nats, and our last names just happened ("happened"??) to be:

2nd -- Chen
3rd -- Chien
4th -- Kishore
5th -- Zhai

Also, the tie at states for 2nd:

2nd -- Gauthier
3rd -- Iglesias

Last year's tie at states:

1st -- Gauthier
2nd -- Zhai

Interesting...

Last year I got 5th and tied with 4th place at state.  probability1.01 what was his or her last name?

I think that in the old days tiebreakers were more complicated: they broke ties according to some weird subset of problems instead of the last n.

Oh, and way back in '99, when JBL and I tied for 2nd at states, that was also alphabetical :-)  (He won, Lewis before Miller.)

[quote="JMI"]I know that they use actual tiebreakers (whoever gets the most sprint round questions correct, is one of them) for the state team. I was just wondering if the tie I was in was also in alphabetical order.[/quote]

No, it was Nayak.  But still, it's rather unlikely to have 4 in alphabetical order (1/24, or, MC style, 0.04 :-)).

[quote="JSRosen3"]I'm not sure if that's what they do all the time, but I'm pretty sure that they used it at Nationals last year past the first 57 or so. But at states, to determine who gets on the team or not, I think they use some real tiebreakers.[/quote]
lol, you're still convinced that it's alphabetical order? well, it might be possible. our whole team tried to prove you wrong. ahhh, good times

Ok here is what I think the rankings were, and I placed (with bias) possible scores for ranks 6-13.

44:
1 -- Gauthier
40:
2 -- Chen
3 -- Chien
4 -- Kishore
5 -- Zhai
39:
6 -- Chu
38:
7 -- Benjamin
8 -- Bernstein
9 -- Hu
10 -- Zhu
37:
11 -- Mao
36:
12 -- Elder
13 -- Huang
14 -- Nahlik

Since Nahlik was on my team, I know his score was 36, which makes everything fit right.  Of course, I exercised total bias in grouping the ranks, but it is still a possibility...

i know its definetely not in alphebetical order (at least at chapter and state), because at chapter, i got a 42 and got 4th, and somebody from my school (with the same last name) got 42 and placed 6th and i know that the person who got 5th definetly did not have the same last name as me, so ha!!!

at state, somebody with 42 (last name Yang) got 4th, and somebody else with 42 (last name Mao) got 5th...

I think I heard somewhere Elder got a 37, but that list is very strange indeed...

Last year, i got a 33 and 69th. 62nd place also got a 33. my last name begins with C. i kinda doubt the alphabetical order thing a little bit, but that was last year.

[quote="mathdool"][quote="probability1.01"]
7 -- Benjamin
8 -- Bernstein
9 -- Hu
[/quote]

umm.. You mixed 7 and 8 because I remember Hu against Benjamin in round 1... that was entertaining.[/quote]
o yeah! bernstein definitely didn't go up against Hu.


A 12-hour clock stopped at 11:00 AM Friday morning. From noon Friday through midnight Sunday, how many times did this clock show the correct time?

It will show the correct time at 11 AM and 11 PM, since it is a 12 hour clock. We easily see that this time occurs $ \boxed{5}$ times.

Is this testing whether we interpret "midnight Sunday" as the end of Sunday (answering 5) or the start of Sunday (answering 3)?

It seems to me that the question is unusually ambiguous, even for a warm-up.  I doubt mathcounts would put this in a countdown or other contest round.


linear algebra
matrix
linear algebra unsolved

[b]Let[/b] $ A$ be a real ,symmetric ,$ nxn$ , tridiagonal matrix (All entries not on the main diagonal or the diagonals just above and below the main one are zero.) Assume $ b_j \neq 0 ,\forall j$
  $ A\equal{}\left[\begin{array}{ccccccc} a_1 & b_1 & 0 &  0 & ... & 0 & 0 \\ b_1 & a_2 & b_2 &  0 & ... & 0 & 0 \\ 0 & b_2 & a_3 &  b_3 & ... & 0 & 0 \\ ... & ... & ... &  ... & ... & ... & ... \\ 0 & 0 & 0 &  0 & ... & a_{n\minus{}1} & b_{n\minus{}1} \\  0 & 0 & 0 &  0 & ... & b_{n\minus{}1} & a_{n} \end {array} \right]$
Prove :
1)$ Rank(A) \geq n\minus{}1$
2) $ A$ has $ n$ distinct eigenvalues.
[b]Thanks ![/b]

1.) We observe the lower $ (n\minus{}1)\times (n\minus{}1)$ submatrix of $ A$:

$ A_{sub}\equal{}\left[\begin{array}{ccccccc}
b_1 & a_2 & b_2 & 0 & ... & 0  \\
0    & b_2 & a_3 & b_3 & ... & 0 \\
...  & ...    & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & b_{n\minus{}2}& a_{n\minus{}1}  \\
0 & 0 & 0 & 0 & 0 & b_{n\minus{}1}  \end {array} \right]$

$ \det {A_{sub}}\equal{}\prod^{n\minus{}1}_{k\equal{}1} b_k \neq 0 \Rightarrow A_{sub} \; \mathrm{is\; not \;singular} \Rightarrow rank{A_{sub}}\equal{}n\minus{}1$

Since $ rank A \geq rank A_{sub} \Rightarrow rank A \geq n\minus{}1$.

2) follows directly from 1). If $ \lambda \in \mathbb{R}$ is an eigenvalue of $ A$, then $ \mathrm{rank}(A\minus{}\lambda I_n) \geqslant n\minus{}1$, hence $ \dim \ker (A\minus{}\lambda I_n) \equal{}1$, thus $ A$ has $ n$ eigenvalues because it is diagonalizable.


[quote="chien than"]Given $ a, b, c >\frac{1}{\sqrt{6}}$ satisfying:
$ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$.
Prove that:
$ \frac{1\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{1\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{1\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ca\minus{}b^{2}}}\geq 2(a\plus{}b\plus{}c)$[/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155354
Why you do it, chien than?


The number r can be expressed as a four-place decimal 0.abcd, where a, b, c, and d represent digits, any of which could be zero. It is desired to approximate r by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to r is 2/7. What is the number of possible values of r?

[hide]The fractions with numerator $1$ or $2$ closest to $\frac{2}{7}$ are $\frac{1}{4}=\frac{7}{28}$ and $\frac{1}{3}=\frac{7}{21}$.  Since $r$ must be closer to $\frac{2}{7}$ than either of these other two fractions, we have $\frac{15}{56}<r<\frac{13}{42}$, or $0.2679\le r \le 0.3095$  Therefore, the number of possible values for $r$ is $3095-2679+1=\boxed{417}$[/hide]

[hide="Answer"]$r\approx \frac{2}{7}\approx 0.285714...$
We have limited options for other fractions to use because our only possible numerators are $1$ and $2$. We have $\frac{1}{3}\approx 0.3333...$ as a boundary on the upper side, and $\frac{1}{4}=0.2500$ on the lower side. Since we want $\frac{2}{7}$ to be the closest, we have $0.2679\le r\le 0.3095$, which gives us a total of $\boxed{417}$ possible values.[/hide]


calculus
derivative
inequalities
real analysis
real analysis unsolved

Here is my question:

[img]http://img12.imageshack.us/img12/8159/clipimage002r.jpg[/img]

I have a solution, but I didn't use "Log sum Inequality". Did I miss sth. in my sol.?
Also, I tried to use "Log sum Ineq" but I couldn't figure a way out. If anyone solves it by using this, please
send the sol.
Thanks :)

[hide="My Solution"]$ d_p(\bold{x},\bold{y}) = (\sum_{i = 1}^{n}{|x_i - y_i|^p})^{1/p} = z$

Take ln of both side for derivation

$ d_p(\bold{x},\bold{y}) = \frac {ln(\sum_{i = 1}^{n}{|x_i - y_i|^p})}{p} = ln z$

Take derivative of both sides w.r.t p

$ \frac {z'}{z} = - \frac {1}{p^2}ln(\sum_{i = 1}^{n}{|x_i - y_i|^p}) + \frac {1}{p}\frac {1}{\sum_{i = 1}^{n}{|x_i - y_i|^p}}\sum_{i = 1}^{n}{ln(|x_i - y_i|)|x_i - y_i|^p} = \frac {1}{p^2}(\frac {\sum_{i = 1}^{n}{ln(|x_i - y_i|^p)|x_i - y_i|^p}}{\sum_{i = 1}^{n}{|x_i - y_i|^p}} - ln(\sum_{i = 1}^{n}{|x_i - y_i|^p}))$

Let $ a_i = |x_i - y_i|^p$, then

$ z' = \frac {z}{p^2}(\frac {\sum_{i = 1}^{n}{ln(a_i)a_i}}{\sum_{i = 1}^{n}{a_i}} - ln(\sum_{i = 1}^{n}{a_i}))$

We need
$ z' \le 0 \Longrightarrow (\frac {\sum_{i = 1}^{n}{ln(a_i)a_i}}{\sum_{i = 1}^{n}{a_i}} - ln(\sum_{i = 1}^{n}{a_i})) \le 0$
$ \Longrightarrow \sum_{i = 1}^{n}{ln(a_i)a_i} \le (\sum_{i = 1}^{n}{a_i})ln(\sum_{i = 1}^{n}{a_i})) \quad(1)$

Proof of (1) without "Log Sum Inequality":

Use the second inequality with $ \alpha_i = \frac {a_i}{\sum_{j = 1}^{n}{a_j}}$

$ \sum_{i = 1}^{n}{\frac {a_i}{\sum_{j = 1}^{n}{a_j}}} = 1 \quad ,$       So, it's OK.

We have,

$ \sum_{i = 1}^{n}{\alpha_i ln(\alpha_i)} \le 0$
$ \sum_{i = 1}^{n}{\alpha_i ln(\alpha_i)} = \sum_{i = 1}^{n}{\frac {a_i}{\sum_{j = 1}^{n}{a_j}} ln(\frac {a_i}{\sum_{j = 1}^{n}{a_j}})} = \frac {1}{\sum_{j = 1}^{n}{a_j}}(\sum_{i = 1}^{n}{a_i}(ln(a_i) - ln\sum_{j = 1}^{n}{a_j} ) ) \le 0$

$ \sum_{i = 1}^{n}{a_i}ln(a_i) \le \sum_{i = 1}^{n}{a_i}ln\sum_{j = 1}^{n}{a_j} = (ln\sum_{i = 1}^{n}{a_i}) \sum_{i = 1}^{n}{a_i}$

It is Inequality (1)[/hide]

I do not see any problem with your solution and have no idea how to use the hint either. As a matter of fact, you can prove everything without computing any derivatives. The shortest solution just uses the inequality $ \sum_j\beta_j^r\le 1$ whenever $ \beta_j>0$, $ r\ge 1$, and $ \sum_j \beta_j\equal{}1$. Can you find it? ;)


[quote="Virgil Nicula"]Let $ m$ , $ n$ be two natural numbers and let  $ x$ , $ y$ , $ z$ be three real positive numbers. Prove that

$ \frac {x^{n \plus{} 1}}{y^n} \plus{} m\ge \frac {(x \plus{} m)^{n \plus{} 1}}{(y \plus{} m)^n}$ Is truly the inequality $ \frac {x^{n \plus{} 1}}{y^n} \plus{} z\ge \frac {(x \plus{} z)^{n \plus{} 1}}{(y \plus{} z)^n}\ \ ?$[/quote]
By Holder inequality, we have:
$ \left(\frac {x^{n \plus{} 1}}{y^n}\plus{}z\right)(y \plus{} z)^n \ge \left(\sqrt[n\plus{}1]{\frac{x^{n\plus{}1}}{y^n}\cdot y^n}\plus{}\sqrt[n\plus{}1]{z\cdot z^n}\right)^{n\plus{}1}\equal{}(x\plus{}z)^{n\plus{}1}$ :)


Γεια σας παιδιά, παρακολουθώ αρκετό καιρό αυτό το forum και αποφάσισα να γίνω και εγώ μέλος.  Για αρχή παραθέτω μια άσκηση γεωμετρίας που μπήκε στην Ινδία το 2003 για την επιλογή της ομάδας για την IMO.  Η άσκηση είναι η εξής:
Έστω Α', Β', C' να είναι μέσα των πλευρών BC, CA, AB αντίστοιχα, ενός οξυγωνίου, μη ισοσκελούς τριγώνου ABC και έστω D, E, F τα ίχνη των υψών από τις κορυφές A,B,C πάνω σ' αυτές τις πλευρές αντίστοιχα.  Θεωρούμε το τόξο DA' του κύκλου των εννιά σημείων του τριγώνου ABC και έστω σημείο A'', που τριχοτομεί αυτό το τόξο, το οποίο βρίσκεται πλησιέστερα στο Α'.  Ορίζουμε ανάλογα τα σημεία Β'' (πάνω στο τόξο ΕΒ') και C'' (πάνω στο τόξο FC'). Να δείξετε ότι το τρίγωνο Α''Β''C'' είναι ισόπλευρο.

Καλωσόρισες [b][size=100]demcharis[/size][/b]  στην παρέα μας και καλή συνέχεια.

Υπάρχουν κάποιες ιδέες για το όμορφο πρόβλημα που έβαλες και θα τις μοιραστώ μαζί σας, αν και δεν είναι πλήρως επεξεργασμένες. Νομίζω όμως ότι οδηγούν σε λύση και ελπίζω ότι δεν κάνω λάθος.

Όπως ξέρουμε ο κύκλος $ (N),$ των εννέα σημείων και ο περιγεγραμμένος κύκλος $ (O)$ του  $ \bigtriangleup ABC,$ είναι ομοιόθετοι, με κέντρο ομοιοθεσίας το ορθόκεντρο $ H$ και λόγο $ 1: 2$

Έτσι οι ευθείες $ HA'',$ $ HB'',$ $ HC'',$ τέμνουν τον κύκλο $ (O)$ στα σημεία έστω $ A'_{1},$ $ B'_{1},$ $ C'_{1}$ αντιστοίχως και αρκεί να αποδείξουμε ότι είναι ισόπλευρο το τρίγωνο $ \bigtriangleup A'_{1}B'_{1}C'_{1}.$

Αν τώρα θεωρήσουμε τα σημεία $ D'\equiv (O)\cap HD,$ $ E'\equiv (O)\cap HE$ και $ F'\equiv (O)\cap HF,$ μπορούμε εύκολα να δούμε ότι τα σημεία $ A'_{1},$ $ B'_{1},$ $ C'_{1},$ τριχοτομούν αντιστοίχως τα τόξα $ \widetilde{A_{1}D',}$ $ \widetilde{B_{1}E',}$ $ \widetilde{C_{1}F'}$ αντιστοίχως $ ($ συγνώμη για το αδόκιμο σύμβολο του τόξου, αλλά δεν μπόρεσα να βρω το σωστό $ ),$ όπου $ A_{1}\equiv (O)\cap HA',$ $ B_{1}\equiv (O)\cap HB',$ και $ C_{1}\equiv (O)\cap HC'.$

Μπορούμε τώρα να δούμε κάποιες ισότητες που νομίζω ότι βοηθάνε.

Από $ D'A_{1}\parallel BC$ έχουμε $ \widetilde{BD'} \equal{} \widetilde{CA_{1}}$ $ ,(1)$ και παρόμοια $ \widetilde{CB_{1}} \equal{} \widetilde{AE'}$ $ ,(2)$ και $ \widetilde{AC_{1}} \equal{} \widetilde{BF'}$ $ ,(3)$

Επίσης, επειδή τα σημεία $ D',$ $ E',$ $ F',$ είναι συμμετρικά του $ H,$ ως προς τις πλευρές $ BC,$ $ AC,$ $ AB$ αντιστοίχως, έχουμε τις ισότητες $ \widetilde{CD'} \equal{} \widetilde{CE'}$ $ ,(4)$ $ \widetilde{AE'} \equal{} \widetilde{AF'}$ $ ,(5)$ και $ \widetilde{BD'} \equal{} \widetilde{BF'}$ $ ,(6)$

Από τις ισότητες $ (1),$ $ (2),$ $ (3),$ $ (4),$ $ (5),$ $ (6),$ αν δεν έχω κάποιο λάθος στις πράξεις, προκύπτει ότι για να είναι $ \widetilde{A'_{1}B'_{1}} \equal{} \widetilde{B'_{1}C'_{1}} \equal{} \widetilde{C'_{1}A'_{1}}$ αρκεί να αποδειχθεί ότι ένα από τα τρία τόξα $ \widetilde{A_{1}D'},$ $ \widetilde{B_{1}E'},$ $ \widetilde{C_{1}F'},$ ισούται με το άθροισμα των δύο άλλων.

Στο σχήμα μου όπου $ AB \equal{} 6,7 < BC \equal{} 10,0 < CA \equal{} 11,2,$ αρκεί να αποδειχθεί ότι $ \widetilde{A_{1}D'} \equal{} \widetilde{B_{1}E'} \plus{} \widetilde{C_{1}F'},$ που νομίζω αποδεικνύεται εύκολα, αν λάβουμε υπόψη μας ότι τα σημεία $ A_{1},$ $ B_{1},$ $ C_{1},$ είναι αντιδιαμετρικά των $ A,$ $ B,$ $ C$ αντιστοίχως.

Δεν έχω πολύ χρόνο τώρα, θα το ξανακοιτάξω όμως και αν δεν υπάρχει λάθος, θα βάλω αργότερα περισσότερες λεπτομέρειες και σχήμα.

Κώστας Βήττας.


Worth k points:
1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6... + 1/n = 10
Find the value of n, to the nearest whole number.

Worth 3k points:
1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6... + 1/n = v
Develop an equation that relates v and n

Worth 2k points:
Explain your answer.

On a scale of e --> pi, how hard was this for you?
(e being the easiest, pi being the hardest)

What's the source?  The typical answer to this problem is straightforward depending on how much calculus you know:  there is a pretty standard notion that $ 1 \plus{} \frac{1}{2} \plus{} ... \plus{} \frac{1}{n}$ is about $ \ln n$ and some pretty standard ways of making that approximation precise enough for you to figure out roughly how $ n$ and $ v$ are related, but there's not much in the way of an exact formula for one in terms of the other.

There is no such value of $ n$.  I think I know what you're looking for, but my answer to the actual question posed is that there is no $ n$.

And you didn't even specify what metric to use when determining our difficulty rating!  :roll:

Also, why does it say "Find the value of n to the closest whole number?"

Ah.  What you want to say is "find the whole number $ n$ that minimizes the difference between the LHS and the RHS."

Ah, there is no solution, because if you use the integral, you get e^10, which doesn't fit the series.


Let $ V$ be a vector space over $ \mathbb{C}$ with $ M_{2}(\mathbb{C})$.

Let $ f:\ V\rightarrow V$ be a map such that $ f(X) \equal{}\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)X\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right),\ \ \ \ X\ \in V.$

(1) Prove that $ f$ is a linear mapping.
(2) Find the matrices representation of $ A$ [color=blue]for $ f$[/color] related to bases of $ V$, $ \left(\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right),\ \left(\begin{array}{cc}0 & 1\\ 0 & 0\end{array}\right),\ \left(\begin{array}{cc}0 & 0\\ 1 & 0\end{array}\right),\ \left(\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right).$
(3) Find $ det A$.

Let $ a,b\in\mathbb{C}$ and $ X,Y\in V$. Then

$ f(aX\plus{}bY)\equal{}\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)\left(aX\plus{}bY\right)\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)$

 $ \equal{}\left(\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)(aX)\plus{}\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)(bY)\right)\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)$

$ \equal{}\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)(aX)\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)\plus{}\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)(bY)\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)$

$ \equal{}a\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)X\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)\plus{}b\left(\begin{array}{cc}3 & 4\\ \minus{}2 &\minus{}3\end{array}\right)Y\left(\begin{array}{cc}1 & 2\\ \minus{}1 &\minus{}1\end{array}\right)$

$ \equal{}af(X)\plus{}bf(Y)$

So it is linear.

Sorry, but I don't understand what is $ A$.

Sorry for I made a typo. :oops:

Please correct the text as follows.

(2) Find the matrices representation of $ A$ for $ f$ related to bases ...... [color=red][Done. -mathmanman][/color]

kunny


trigonometry
integration
logarithms
real analysis
real analysis unsolved

Prove :
for all $ a>1$ series converge
$ \sum\frac{1}{n^{a}}\frac{1}{|\sin(\pi\sqrt{2}n)|}$

The crucial point is that the distance from $ \sqrt 2 m$ to the nearest integer is at least $ c/m$ with some absolute $ c>0$. Let us now look at the numbers $ n$ from $ N$ to $ 2N$ with $ |\sin\pi\sqrt 2 n|< 1/t$ for $ t>0$. There are no such numbers for $ t>CN$ and the distance between two such consequtive numbers is at least $ ct$ for $ t<CN$, so we have at most $ CN/t$ of them. Thus
\[ \sum_{N\le n<2N}\frac {1}{|\sin\pi\sqrt 2 n|}\le\int_{0}^{CN}\min\{N,CN/t\}\,dt\approx N\log N\]
The rest is standard: just split your sum into dyadic blocks from $ 2^{k}$ to $ 2^{k\plus{}1}$ and apply this estimate to each block.

yes  :) you are right.
I note
this problem also can be generalizait. if we instead $ \sqrt{2}$ write some algebraic number and use http://en.wikipedia.org/wiki/Liouville_number


analytic geometry
function
calculus
derivative
vector
geometry proposed
geometry

Let $A,B,C,D$ be points in space which are not on a same plane and no any 3 points are not colinear.
Suppose that the sum of the segments $AX+BX+CX+DX$ is minimized at $X=X_0$ which is different from $A,B,C,D.$ Prove that $\angle{AX_0B}=\angle{CX_0D}.$

[quote="kunny"]Let $ A,B,C,D$ be points in space which are not on a same plane and no any 3 points are not colinear.
Suppose that the sum of the segments $ AX \plus{} BX \plus{} CX \plus{} DX$ is minimized at $ X \equal{} X_0$ which is different from $ A,B,C,D.$ Prove that $ \angle{AX_0B} \equal{} \angle{CX_0D}.$[/quote]
Let $ A;B;C;D$ and $ P$ have coordinates $ (x_1;y_1;z_1;),....,(x_4;x_4;x_4)$ and $ (x;y;z)$
Then the function being minimized is
$ f(P)\equal{}\sum \sqrt{(x\minus{}x_i)^2\plus{}(y\minus{}y_i)^2\plus{}(z\minus{}z_i)^2}$
At its minimum,its three partial derivatives are zoro,but these are precisely the three coordinates of $ u_a\plus{}u_b\plus{}u_c\plus{}u_d$,where $ u_a$ is the unit vector$ \frac{P\minus{}A}{\parallel{}P\minus{}A\parallel{}}$ and so on
Thus this sum is zoro,and so $ u_a.u_b\equal{}u_c.u_d$ for $ P\equal{}X_o$ ,which proves the claim.


The number $\ N$ is a perfect square whose last three digits are $\ x25$.  Find all possible values for $\ x$/

[hide]A $25$ is retained in a perfect square any time a $5$ is the unit digit of the squared number.

The pattern follows: $2,6,2,0,0$ [from $15^2$].

Hence the only numbers possible for "x" are $\boxed{2,6,0}$[/hide]

Just by looking at it 2,6, and 0 obviously work and I think those are the only ones...

[hide]
We want $100x+25=25(4x+1)$ to be a perfect square, so $4x+1$ ($0\le x\le 9$) has to be a perfect square (because $25$ is a perfect square).

A quick check shows that $x=0$, $x=2$ and $x=6$ work.
[/hide]


Let for all $|x|\leq1$ holds $|ax^{2}+bx+c|\leq1.$
Prove that for all $|x|\leq1$ and $0\leq\alpha\leq1$ holds $|ax+\alpha b|\leq2.$

Quite same problem is the following one .
For $-1\leq x\leq 1$ and  $-1\leq ax^{2}+bx+c\leq 1$
then $-4\leq 2ax+b\leq 4$


If any 2 are equal, it is trivial, so assume $ x,y,z$ are distinct.

$ \sum_{cyc} |\frac {x}{y \minus{} z}| \ge 2$.

We claim that we can either increase $ x,y,z$, or decrease $ x,y,z$ by the same amount, so as 1 of the variables is 0 and the L.H.S. does not increase.

To do this, we notice that by increasing $ x,y,z$ by an amount which does not change any of their signs changes the L.H.S. in the opposite direction than if we had decreased $ x,y,z$ by that amount. Therefore, we can either increase or decrease all of the variables until 1 of them hits 0 and the L.H.S. does not increase (if none ever hit 0, then we can clearly see that $ |\frac {x}{y \minus{} z}|$ is unbounded, and we are done).

Therefore, we only need to check the inequality when 1 of the variables is $ 0$, WLOG let $ z$ be 0.

$ |\frac {x}{y}|\plus{}|\frac {y}{x}|\ge 2$,

$ \frac {|x|}{|y|} \plus{} \frac {|y|}{|x|} \ge 2$, which is just simple AM-GM.

Q.E.D.

Cheers,

Rofler

[quote="fjwxcsl"]Let $ {x,y,z}\in{R},$prove that
$ |x(x \minus{} y)(x \minus{} z)| \plus{} |y(y \minus{} z)(y \minus{} x)| \plus{} |z(z \minus{} x)(z \minus{} y)|\ge{2|(x \minus{} y)(y \minus{} z)(z \minus{} x)|}.$[/quote]

Really nice inequality...

If any 2 are equal, it is trivial.
when $ x,y,z$ are distinct, we can assume that $ x < y < z$ and $ x \equal{} y \minus{} t, z \equal{} y \plus{} u,t,u > 0$, then
\[ |x(x \minus{} y)(x \minus{} z)| \plus{} |y(y \minus{} z)(y \minus{} x)| \plus{} |z(z \minus{} x)(z \minus{} y)|\ge{2|(x \minus{} y)(y \minus{} z)(z \minus{} x)|}\]<=>\[ \frac {|y \minus{} t|}{u} \plus{} \frac {|y|}{t \plus{} u} \plus{} \frac {|y \plus{} u|}{t} \ge 2.\]
when $ y \ge t$,
$ \frac {|y \minus{} t|}{u} \plus{} \frac {|y|}{t \plus{} u} \plus{} \frac {|y \plus{} u|}{t} \equal{} \frac {y \minus{} t}{u} \plus{} \frac {y}{t \plus{} u} \plus{} \frac {y \plus{} u}{t} \ge \frac {t \minus{} t}{u} \plus{} \frac {t}{t \plus{} u} \plus{} \frac {t \plus{} u}{t} \equal{} \frac {t}{t \plus{} u} \plus{} \frac {t \plus{} u}{t}\ge 2$
inequality holds;
when $ t \ge y \ge 0$,
$ \frac {|y \minus{} t|}{u} \plus{} \frac {|y|}{t \plus{} u} \plus{} \frac {|y \plus{} u|}{t} \equal{} \frac {t \minus{} y}{u} \plus{} \frac {y}{t \plus{} u} \plus{} \frac {y \plus{} u}{t}$
let $ f(y) \equal{} \frac {t \minus{} y}{u} \plus{} \frac {y}{t \plus{} u} \plus{} \frac {y \plus{} u}{t}$,
because $ f(0) \equal{} \frac {t}{u} \plus{} \frac {u}{t} \ge 2$ and $ f(t) \equal{} \frac {t}{t \plus{} u} \plus{} \frac {t \plus{} u}{t}\ge 2$
we get $ f(y) \ge 2$, inequality holds;
when $ 0 \ge y \ge \minus{} u$ and $ \minus{} u \ge y$ are similar.

See also http://www.mathlinks.ro/viewtopic.php?t=20869  (Rofler has my solution).

  darij


Let $a;b;c$ be real numbers  satisfying $a^{2}+b^{2}+c^{2}=1$.Find min $A=a^{3}+b^{3}+c^{3}$


I had post again the problem in separated thread:

Find all $x,y,z$ are  positive integer,$(x,y,z)=1$ ,$y$ is odd integer and:
$\frac{1}{x^2}+\frac{1}{y^8}=\frac{1}{z^2}$

and:
Find all integer roots of:
$y^2=x^3+(x+4)^2$

I need elementary solution.

The first one is really easy :
Just clear the denominators, it leads to a sum of two odd integers which has to be odd...
The hypothesis about $\gcd (x,y,z)$ is useless.

Pierre.

W.Sierpinski, 250 Problems In Elementary Number Theory (5/177)
I am tired of writing solution (easy but too long) :?

I haven't W.Sierpinski, 250 Problems In Elementary Number Theory.
Problem (2) is in Crux.A remainder question  is find $x,z$(Not have $(x,y,z)=1$) IF $y=15$.


calculus
derivative
function
algebra
polynomial
number theory open
number theory

Does there exist a derivation on functions $ f : \mathbb{Z} \to \mathbb{Z}$ that coincides exactly with the [url=http://en.wikipedia.org/wiki/Formal_derivative]formal derivative[/url] in the case that $ f$ is an integer polynomial?  How about functions $ f : \mathbb{Z}_{\ge 0} \to \mathbb{Z}$?

[hide="Thoughts"] In the latter case, we can write $ f(n) \equal{} \sum_{k\equal{}0}^{\infty} a_k {n \choose k}$ where $ a_0 \equal{} f(0), a_1 \equal{} f(1) \minus{} f(0), ... a_k \equal{} (\Delta^k f)(0)$.  Note that when $ f(n)$ is not a polynomial $ a_k$ does not terminate.  It is then possible to find the formal derivative of each Newton polynomial; in particular,

$ {x \choose k} \equal{} \sum_{i\equal{}0}^{k} \frac{s(k, i)}{i!} x^i \implies$
$ \frac{d}{dx} {x \choose k} \equal{} \sum_{i\equal{}1}^{k} \frac{s(k,i)}{(i\minus{}1)!} x^{i\minus{}1}$.

We then have

$ f(n) \equal{} \sum_{k\equal{}0}^{\infty} a_k \sum_{i\equal{}1}^{k} \frac{s(k,i)}{(i\minus{}1)!} n^{i\minus{}1} \implies$
$ f(n) \equal{} \sum_{i\equal{}1}^{\infty} \left( \sum_{k\equal{}i}^{\infty} a_k \frac{s(k,i)}{(i\minus{}1)!} \right) n^{i\minus{}1}$.

Unfortunately, when $ a_k$ is arbitrary it seems rather unlikely that the coefficients of the above converge.  I realize this is a lot more work than simply using the forward difference (which is sufficiently analogous), but I have been wondering about this question for some time now. [/hide]

We can define $ f'(x)\equal{}f(x\plus{}1)\minus{}f(x), x^{|n}\equal{}x(x\minus{}1)...(x\plus{}1\minus{}n)$, then $ f\to f'$ linear operator and $ (x^{|n})'\equal{}nx^{|n\minus{}1}$.
But Leibnic rule must be modefied.

I know that; I was asking for, among other things, an operator that exactly respects the product rule.

[hide="More thoughts"] The reason the approach I tried doesn't work is likely because polynomial interpolation is very bad at approximating arbitrary data; in particular, interpolating polynomials can wildly fluctuate between the interpolation points, so the non-convergence of the derivative I tried to define is a consequence of this (probably).

If an approach based on interpolation is going to work, then perhaps it'll be based on a Fourier series instead? [/hide]


inequalities
number theory proposed
number theory

Show that $ 3^{2008} \plus{} 4^{2009}$ can be expressed as product of two positive integers each of which is larger than $ 2009^{182}$ .

I remember seeing this problem recently in the forum :D .

sorry for double posting....didnt notice ...
See ...http://www.mathlinks.ro/viewtopic.php?p=1695943#1695943

No problem with new topic
Use sophie germain identity as you told. I will tell you the inequality part later.(Hard to type)