Let $ x$ be positive real number,then find the minimum value of

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

(2)$ x^n\plus{}\frac{1}{x}$  ($ n\in \mathbb{N}$)

hello, to 1) we have $ (x\minus{}1)^2(1\plus{}2x)\geq0$ for all real $ x$, this is equivalent with
$ 2x^3\minus{}3x^2\plus{}1\geq0$
$ 2x^3\minus{}3x^2\plus{}1\geq0$
$ 2x\plus{}\frac{1}{x^2}\geq3$ since $ x>0$.
The equal sign holds if and only if $ x\equal{}1$.
Sonnhard.

hello, in (2) we have $ x^n\plus{}\frac{1}{x}\geq\frac{n\plus{}1}{n^{\frac{n}{n\plus{}1}}}$.
Sonnhard.

Alternatively, with AM-GM,...

[hide="1"]
$ 2x\plus{}\frac1{x^2}\equal{}x\plus{}x\plus{}\frac1{x^2}\ge3\sqrt[3]{x\cdot x\cdot\frac1{x^2}}\equal{}3$, with equality when $ x\equal{}1$.
[/hide]
[hide="2"]
$ x^n\plus{}\frac1x\equal{}x^n\plus{}\underbrace{\frac1{nx}\plus{}\frac1{nx}\plus{}\cdots\plus{}\frac1{nx}}_n\ge(n\plus{}1)\sqrt[n\plus{}1]{x^n\cdot\underbrace{\frac1{nx}\cdot\frac1{nx}\cdots\frac1{nx}}_n}$
$ \equal{}\frac{n\plus{}1}{\sqrt[n\plus{}1]{n^n}}\equal{}\frac{(n\plus{}1)\sqrt[n\plus{}1]n}n$, with equality when $ x\equal{}\frac{\sqrt[n\plus{}1]{n^n}}n$.
[/hide]

You could also do it using derivatives.

[hide="1"]
$ f(x) \equal{} 2x \plus{} \frac {1}{x^2} \implies f'(x) \equal{} 2 \minus{} \frac {2}{x^3}$

Extremes can be found if we let $ f'(x) \equal{} 0 \implies 2 \minus{} \frac {2}{x^3} \equal{} 0 \implies x \equal{} 1$

$ f''(x) \equal{} (f'(x))' \equal{} \frac {6}{x^5} \implies f''(x) > 0 \;\; \forall x > 0$ so $ f(x)$ is convex $ \implies$ for $ \boxed{x \equal{} 1}$ minimum of the function is achieved.[/hide]

[hide="2"]
$ f(x) \equal{} x^n \plus{} \frac {1}{x} \implies f'(x) \equal{} nx^{n \minus{} 1} \minus{} \frac {1}{x^2}$ letting $ f'(x) \equal{} 0 \implies x \equal{} \sqrt [n \plus{} 1]{\;\frac {1}{n}\;}$

$ f''(x) \equal{} n(n \minus{} 1)x^{n \minus{} 2} \plus{} \frac {2}{x^3} \implies f''(x) > 0 \;\; \forall x > 0, n \in \mathbb{N} \implies$ for $ \boxed{x \equal{} \sqrt [n \plus{} 1]{\;\frac {1}{n}\;}}$ minimum of the function is achieved[/hide]


LaTeX
geometry
inradius
inequalities
incenter
invariant
radical axis

Consider three circles with radiuses $r_1, r_2, r_3$ such that
every two are externally tangent to each other and there exist circles
with radiuses $R$ and $r$ that are tangent to all our given circles.
($r$ - small one, $R$ - big one)

Prove $R*r \geq  \frac {r_1r_2r_3}{ r_1+r_2+r_3}$

P.S I think I have two different solutions,
     second one I think impossible to find in olimpiad time. :)

I think I have seen this before, but I can't remember where :? Anyway a small $\LaTeX$ tip. Use \cdot instead of *, like $R \cdot r \geq \frac {r_1r_2r_3}{ r_1+r_2+r_3}$ ;)

I believe there are two topics on the forum where $R,r$ are actually computed.

But in olimpiad time you won't find them fast.

First I rewrite the problem to change the notations:

[color=blue][b]Problem.[/b] Let $k_a$, $k_b$, $k_c$ be three circles with centers A, B, C, respectively, such that each two of these circles $k_a$, $k_b$, $k_c$ are externally tangent to each other. Also, let the circle internally tangent to these three circles $k_a$, $k_b$, $k_c$ have radius $\Sigma$, and let the circle externally tangent to these three circles $k_a$, $k_b$, $k_c$ have radius $\sigma$. Prove that $\Sigma\cdot\sigma\geq\frac{r_1r_2r_3}{r_1+r_2+r_3}$, where $r_1$, $r_2$, $r_3$ are the radii of the circles $k_a$, $k_b$, $k_c$, respectively.[/color]

[i]Solution.[/i] Let a = BC, b = CA, c = AB be the sides of the triangle ABC. Since the circles $k_a$ and $k_b$, with their respective centers A and B and their respective radii $r_1$ and $r_2$, are externally tangent to each other, we have $AB=r_1+r_2$. Similarly, $BC=r_2+r_3$ and $CA=r_3+r_1$. Thus, $BC+CA-AB=\left(r_2+r_3\right)+\left(r_3+r_1\right)-\left(r_1+r_2\right)=2r_3$, so that $r_3=\frac{BC+CA-AB}{2}=\frac{a+b-c}{2}=\frac{a+b+c}{2}-c$. In other words, $r_3=s-c$, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Similarly, $r_1=s-a$ and $r_2=s-b$. Hence, $r_1+r_2+r_3=\left(s-a\right)+\left(s-b\right)+\left(s-c\right)=3s-\left(a+b+c\right)=3s-2s=s$, and thus

$\frac{r_1r_2r_3}{r_1+r_2+r_3}=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$.

On the other hand, the Heron formula for the inradius of a triangle yields

$r^2=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$,

where r is the inradius of triangle ABC. Thus, $\frac{r_1r_2r_3}{r_1+r_2+r_3}=r^2$. Consequently, the inequality in question, $\Sigma\cdot\sigma\geq\frac{r_1r_2r_3}{r_1+r_2+r_3}$, becomes $\Sigma\cdot\sigma\geq r^2$.

Now, after [1], formula (2), we have

$\frac{1}{\sigma}=\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}+\frac{2}{r}$;

in other words,

$\frac{1}{\sigma}=\frac{2}{r}+\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)$.

Similarly,

$\frac{1}{\Sigma}=\frac{2}{r}-\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)$.

Thus,

$\frac{1}{\Sigma\cdot\sigma}=\frac{1}{\Sigma}\cdot\frac{1}{\sigma}$
$=\left(\frac{2}{r}-\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\right)\cdot\left(\frac{2}{r}+\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\right)$
$=\left(\frac{2}{r}\right)^2-\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)^2$.

But, again by [1], we have

$\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\geq\frac{\sqrt3}{r}$;

thus,

$\frac{1}{\Sigma\cdot\sigma}=\left(\frac{2}{r}\right)^2-\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)^2$
$\leq\left(\frac{2}{r}\right)^2-\left(\frac{\sqrt3}{r}\right)^2=\frac{4}{r^2}-\frac{3}{r^2}=\frac{1}{r^2}$.

Thus, $\Sigma\cdot\sigma\geq r^2$, and the problem is solved.

[b]References[/b]

[b][1][/b] My solution of problem 2 from the 3rd MathLinks Contest, Round 2. See http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pdf&t=4796 post #3.

  Darij

Problem:
Let three circles with centers $A, B, C$ are externally tangent to each other in points
$U, V, W$ and there exist two circles with centers $X$ and $Y$ tangent to all these
three circles. (circle with center $X$ is the bigger one)

Solution:
Let $I$ be the radical center of our three circles:
 $IU = IV = IW$ and $IU$ is perpendicular to BC(others similarly).
Hence we can consider $I$ as incenter of triangle $ABC$ with radius $t$.

Consider inversion $H$ from pole $I$ and power $H = t^2$.
$H(U) = U,  H(V) = V,  H(W) = W$. (three circles remain invariant)

Big circle being tangent to all three our circles will transform in small one.
From basic properties of inversion we have $I, X, Y$ are collinear.

Let $IX$ intersects big circle in $M_1, N_1$ and small circle in $M, N$.
We have $IM*IM_1 = t^2, IN*IN_1= t^2$.

Using that $I$ is inside the big circle, after inversion it follows that also inside the small one,
as a result  $IM_1*IN_1 = R^2 - IX^2, IM*IN = r^2 - IY^2$.

Hence$R^2r^2\geq (R^2 - IX^2)(r^2- IY^2) = t^4$ or $Rr\geq t^2$

But $r_1 = p-a, r_2= p-b, r_3 = p-c$ and $t ^2 = \frac {r_1r_2r_3}{r_1+r_2+r_3}$

@ Darij Grinberg
Your solution is very nice.
My second solution is similar to your's - the main idea is taken from
ISL 2003 G7 (very hard problem from South Africa) :)

[quote="prowler"]
My second solution is similar to your's - the main idea is taken from
ISL 2003 G7 (very hard problem from South Africa) :)[/quote]

Well, it can be seen from Darij's references :D .


quadratics
function
analytic geometry
geometry
geometric transformation
rotation
algebra

Functions $ f$ and $ g$ are quadratic, $ g(x) \equal{} \minus{} f(100 \minus{} x)$, and the graph of $ g$ contains the vertex of the graph of $ f$. The four $ x$-intercepts on the two graphs have $ x$-coordinates $ x_1$, $ x_2$, $ x_3$, and $ x_4$, in increasing order, and $ x_3 \minus{} x_2 \equal{} 150$. The value of $ x_4 \minus{} x_1$ is $ m \plus{} n\sqrt p$, where $ m$, $ n$, and $ p$ are positive integers, and $ p$ is not divisible by the square of any prime. What is $ m \plus{} n \plus{} p$?

$ \textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752\qquad \textbf{(E)}\ 802$

[hide="Solution (without ugly computation!) edit: thanks azjps"]The two quadratics are $ 180^{\circ}$ rotations of each other about $ (50,0)$. Since we are only dealing with differences of roots, we can translate them to be symmetric about $ (0,0)$. Now $ x_3 = - x_2 = 75$ and $ x_4 = - x_1$. Say our translated versions of $ f$ and $ g$ are $ p$ and $ q$, respectively, so that $ p(x) = - q( - x)$. Let $ x_3 = 75$ be a root of $ p$ and $ x_2 = - 75$ a root of $ q$ by symmetry. Note that since they each contain each other's vertex, $ x_1$, $ x_2$, $ x_3$, and $ x_4$ must be roots of alternating polynomials, so $ x_1$ is a root of $ p$ and $ x_4$ a root of $ q$
\[ p(x) = a(x - 75)(x - x_1) \\
q(x) = - a(x + 75)(x + x_1)
\]
The vertex of $ p(x)$ is half the sum of its roots, or $ \frac {75 + x_1}{2}$.
\begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{2}\right) \\
& = & - \frac {a}{4}(x_1 - 75)^2 \\
- \frac {a}{4}(x_1 - 75)^2 & = & q\left(\frac {75 + x_1}{2}\right) \\
& = & - a\left(\frac {x_1 + 225}{2}\right)\left(\frac {3x_1 + 75}{2}\right) \\
& = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}
Let $ x_1 = 75u$ and divide through by $ 75^2$, since this is a timed competition and it will drastically simplify computations. We know $ u < - 1$ and that $ (u - 1)^2 = (3u + 1)(u + 3)$, or
\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\
& = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\
& = & 2u^2 + 12u + 2 \\
& = & u^2 + 6u + 1 \end{eqnarray*}
So $ u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2$. Since $ u < - 1$, $ u = - 3 - 2\sqrt2$ and $ x_1 = - 225 - 150\sqrt2$. $ x_4 - x_1 = - 2x_1 = 450 + 300\sqrt2$, so our answer is $ 752\rightarrow\boxed{\textbf{(D)}}$.[/hide]

[hide="Darn, I should learn the quadratic formula; edit: wawb that should be D"]
[asy]import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;
real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);}
xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6));

draw(graph(f,-450,300),red+linewidth(0.8),Arrows(6));draw(graph(g,-200,550),blue+linewidth(0.8),Arrows(6));
draw((50,ymax)--(50,ymin),linetype("4 4"),Arrows(4));dot((-rt,0));dot((100+rt,0));dot((-25,0));dot((125,0));[/asy]

We note that the graphs of $ g(x)$ and $ \minus{} f(x) \equal{} g(100 \minus{} x)$ are symmetric about the line $ x \equal{} 50$. Without loss of generality, $ r_1,r_3$ are roots of $ f(x)$, with $ r_1 < r_2 < 50$. Then $ g(r_2) \equal{} \minus{} f(100 \minus{} r_2) \equal{} 0$, so $ 100 \minus{} r_2 > 50$ is a root of $ f(x)$; hence, $ r_3 \equal{} 100 \minus{} r_2$. It follows that $ r_2 \equal{} \minus{} 25, r_3 \equal{} 125$.

Now,
\[ f(x) \equal{} a(x \minus{} 125)(x \minus{} r_1)\ \mathrm{and}\ g(x) \equal{} \minus{} a(x \plus{} 25)(x \minus{} 100 \plus{} r_1).
\]
Since the vertex of $ f$, $ \left(\frac {125 \plus{} r_1}{2}, f\left[\frac {125 \plus{} r_1}{2}\right] \equal{} a\left[\frac {r_1 \minus{} 125}{2}\right]\left[\frac {125 \minus{} r_1}{2}\right]\right)$, lies on the graph of $ g(x)$, it follows that
\[ g\left(\frac {125 \plus{} r_1}{2}\right) \equal{} \frac { \minus{} a}{4}\left(r_1 \plus{} 175\right)( \minus{} 75 \plus{} 3r_1) \equal{} \frac { \minus{} a}{4}(r_1 \minus{} 125)^2.
\]
Expanding, this is just a quadratic:
\[ 3r_1^2 \plus{} 450r_1 \minus{} 175 \times 75 \equal{} r_1^2 \minus{} 250r_1 \plus{} 125^2 \Longrightarrow 2r_1^2 \plus{} 700r_1 \minus{} 175 \times 75 \minus{} 125^2
\]
Hence $ r_1 \equal{} \frac { \minus{} 700 \pm \sqrt {700^2 \plus{} 4 \times 2 \times (175 \times 75 \plus{} 125^2)}}{4}$ $ \equal{} \minus{} 175 \pm \frac {\sqrt {25^2(28^2 \plus{} 8(3 \times 7 \plus{} 5^2))}}{4}$ $ \equal{} \minus{} 175 \pm 150\sqrt {2}$.

The answer is $ r_4 \minus{} r_1 \equal{} 100 \minus{} 2r_1 \equal{} 450 \pm 300\sqrt {2}$. Discarding the negative root, our answer is $ 450 \plus{} 300 \plus{} 2 \equal{} 752\ \mathbf{(D)}$.
[/hide]

[hide="Algebraic solution (not too bad)"]
Use vertex form. Let
$ f(x) \equal{} a(x \minus{} h)^2 \plus{} k$, $ g(x) \equal{} \minus{} a(100 \minus{} x \minus{} h)^2 \minus{} k$
Then g(h)=k since they have the same vertex, so we have the relation
$ \minus{} a(100 \minus{} 2h)^2 \equal{} 2k$
Let $ \sqrt { \minus{} \frac {k}{a}} \equal{} n$. From the above, we have
$ \minus{} \sqrt {2} \cdot n \plus{} 2h \equal{} 100$.
The roots of the quadratics are n+h, n-h+100, -n+h and -n-h+100. We will assume the roots are in that order.
Then we have $ n \minus{} h \equal{} 25$ and $ \minus{} \sqrt {2} \cdot n \plus{} 2h \equal{} 100$. Solving, we get $ n \equal{} \frac {150}{2 \minus{} \sqrt {2}}$. The answer is $ 2(n \plus{} h) \plus{} 100 \equal{} 2(2n \minus{} 25) \plus{} 100 \equal{} 450 \plus{} 300\sqrt {2}$.
[/hide]

It makes all of these solutions easier to translate.

[quote="Complex_Ninja"]Why is there a box around the right answer?? lol[/quote]Fixed.

I know this thread is way out of date, but I was wondering how you know f(x) and g(x) are symmetric around (50,0).

Here we have g(x) = -f(100-x), now let us substitute x = 50+y to get g(50+y) = -f(50-y). Ignoring the - sign for a moment, we see that there is a common value at (50,0), and moving y to the right of 50 for g is the same as moving y to the left of 50 for f. So this is a reflection around x=50. Now the minus sign just means we reflect the graph over the y axis, hence it is a 180 rotation around (50,0).


geometry
3D geometry
pyramid
tetrahedron

[quote="t0rajir0u"]Why exactly is this in Intermediate?[/quote]

Another case of overestimating the difficulty of the problem, and failing to recognize that Intermediate, in fact, does not mean any given high school-level problem.

[quote="xxxyyyy"]Take a solid pyramid with a square base. All edges are one inch long.

Take a solid tetrahedron. Again, all edges are one inch long.

Glue these two solids together by matching two triangular faces.

How many faces does the resulting solid have?![/quote]

Using the terrible drawing skills I get:

 7 faces


How many square units are in the area of the parallelogram with
vertices at (0, 0), (6, 0), (2, 8) and (8, 8)?

The parallelogram has base length $ 8$ and height $ 8$, and thus has area $ \boxed{32}$

you could do what jxl28 said or shoestring it

uh I don't know how I got 32 either :|

jxl is correct :|

using the shoelace, (this is just supplying a solution to those who are curious)

$ \frac {1}{2}|(2\cdot8 \plus{} 8\cdot0 \plus{} 6\cdot0 \plus{} 0\cdot8) \minus{} (8\cdot8 \plus{} 8\cdot6 \plus{} 0\cdot0 \plus{} 0\cdot2)| \equal{} \frac {1}{2}| \minus{} 96| \equal{} \frac {1}{2}96 \equal{} \boxed{48}$

gj jxl

[quote=ab2024][quote=jxl28]Actually, we have a base length of $ 6$ and a height of $ 8$, so we have $ 6 \times 8\equal{}\boxed {48}$  I believe.  Unless I totally fail.  I think you may have found the length of both bases, which are supposed to be congruent.  I don't really understand how you got $ 32$ though...[/quote]

He thought it was a triangle probably. He did BH/2.[/quote]

wait so is the correct answer 32 or 48. BTW i got 32


If $ f_n\rightarrow f$ in $ L^1$ and $ f_n\rightarrow g$ point wise a.e. , can we conclude that$ f\equal{}g$ a.e. ?

there's a subsequence of $ f_n$ which converges pointwise to $ g$, which must converge pointwise to $ f$ as well, so $ f\equal{}g$ a.e.


quadratics
geometry
trapezoid
analytic geometry
USAMTS
algebra
system of equations

How many problems did you guys get correct? I'm not asking for the points, just the questions correct.

I got 4 of them right. Missed number 3. ;)

I missed 3 and 2. Misread number 2...

Now 3 is so obvious. I tried to find the roots then factor into quadratics when it was better to reverse that and I wouldn't arrived at the correct conclusion.

I got 1,4, and 5 correct. :(

all correct!!!!! IM so happy :lol:  :lol:  :lol:  :lol:  :lol:

now I'll just see if i'll get deducted from all my dumb work.....

Good job pkerichang: I was off a little on #2, didn't try #3, and got stuck on #4: I tried to make a pretty solution :( But I got #1 and #5 just fine. Ah, well, next round I won't procrastinate: I got #5 at midnight EST Monday night...

I got three of them (I was particularly satisfied with my solution for No. 5), and am farily confident that I would have gotten the other two if I had sat down and slogged through them earlier.  Next time, I won't procrastinate either.

I got them all right, so I can't complain  :)

I got them all right, thankfully!  I'm interesting in seeing other peoples' solutions, since I missed almost all of the Math Jam.

[quote="4everwise"]I'll have to admit. I felt really dumb when Mr. Crawford hinted at factoring into quadratics for number 3.  :fool:[/quote]
I saw that right away - I thought "the solutions must be imaginary, so the next best way to factor it is into quadratics, by the FTA."  Then I expanded and assigned trial solutions to b,d.

Repre a zentin' CANADA...on Georgia soil.  I got them all right.  I particularly liked MCrawford's solution to #2 in the Math Jam.  Hopefully Round 2 won't be too hot to handle...

For number three, I figured out how to find the area of the trapezoid about 10 minutes after I saw the problem, and I tried to eliminate r after that by using newton sums, partial sums, etc...

Then, with two days left, I saw the 9r^8 + 10r^4 + 1 term... and noticed that it was a quadratic.  'Twas smooth sailing from there.

I also got 5/5 correct.  The question is, however, will my brevity on solutions be too much.  I tried to be "grader-friendly," and had one-page solutions for all but #4, which was 2 pages.  We'll see how that goes...

I got #2 right. I think I made an arithmetic error on #1, though I haven't looked into it, and I thought #4 said b>a>0 instead of b>a>1 for some reason, which pretty much makes my solution worthless. Didn't solve #3 and #5. Eeeaargh.

[quote="jackb1115"]I got them all rigt ... mine solution legnths were 3,6,3,6,3 all handwritten ... I tend to fall on the longer side to make sure i dont leave out information

Do they take off for too long solutions?

JB[/quote]
Not usually - they might remark about how long it is, but unless it's REALLY long, so much so that the grader basically can't follow your logic through the entire course of your argument, they won't take off.

i only got two right (1 and 5) :(

good job to all those who got all right though  :P

and shoutout to j "the binder" b.  :P

I wasn't there and can't find the transcript, but from what I hear I think I got all five. It seems as though even Mr. Crawford had a not-so-pleasant solution for 3. Mine was quite simple, really. Anyway, I'd like to see solutions.

[quote="4everwise"]I'll have to admit. I felt really dumb when Mr. Crawford hinted at factoring into quadratics for number 3.  :fool:[/quote]

Same here.

[quote="Shogia"]I ended up getting them all right so i guess its just up to graders how clear and concise my solutions are on whether to take off points.[/quote]

ditto

Well....I actually got answers for 2 of them...(probs 1&2)

Problem 4 I mostly figured it out, just got tired of checking. I stopped after just 11 thinking a+b could be in the 30s. Well, a+b turned out to be the next possible one, 17  :dry:

Problem 5, I managed to get it down to a messy system of equations whose solution was the answers...too bad I used coordinate geometry instead of just geometry (bad idea).

Problem 3, I just avoided, then realized it wasn't as compilcated as I had thought after I sent in my problems.

Well, pretty good for my first ever USAMTS :)

I got all five, but my solutions weren't that elegant.  Hopefully, I'll receive full credit for each, even if they are a bit messy.


inequalities
geometry
inequalities proposed

prove that for any triangle of area 1/2 we have:
1/a +1/b+1/c <= r1+r2+r3
where ri are the radiuses of the excircles of ABC and a,b,c are the sides of triangle.

As it's in fact algebra and not geometry, I can try it ....

With classical notation with have (r1=rA, ...)
K = rA (s-a) = rB (s-b) = rC(s-c) = 1

So the inequality to prove is (with cyclic sums) :

\sum 2/a+b-c >= \sum 1/a

Now b-c <= a so a+b-c <= 2a and so on.

I think Belenos did misteake somewhere (rA(p-a)=1/2!). The inequality is equivalent to Sum 1/(a+b-c) >= Sum 1/a. We can use 1/x + 1/y >= 4/(x+y) to prove this inequality.

Namdung

The inequality is trivial.It reduces to ${{{{ frac\{1}{b+c-a}+frac\{1}{c+a-b}+frac\{1}{a+b-c}\geq frac\{1}{a}+frac{1}{b}+frac{1}{c}$ which follows from $AM-GM$


Has this ever been talked about? I heard about drink but not food...

I don't think this has been talked about but my favorite food is chocolate, brownies, cookies, ice cream.  This should be a poll

pizza, ice cream, brownies, fried chicken, burgers, actually i like lots of foods so i'll stop here before i add too much.

[quote="math92"]This should be a poll.[/quote]

It would be nice, but there are too many types of foods to list. Too many "others"

[quote="Iversonfan2005"]Sushi..It's so good, I don't know why people get turned off because it's raw, but it tastes so good..[/quote]

SUSHI is awesome!!! Pizza is good. Ice Cream is good. Pasta, OMG, is SO good.

1. Chinese
2. Italian
3. Japanese

I don't have favorite "foods" but I do have favorite food groups :) ^^ And of course all of those are Americanized versions of the real stuff (except Italian which my mom cooks - really good b/c of her italian heritage).

[quote="Iversonfan2005"]Sushi..It's so good, I don't know why people get turned off because it's raw, but it tastes so good..[/quote]
Yeah, sushi is the best :D

[quote="WindSlicer"]1. Chinese
2. Italian
3. Japanese [/quote]
Mine would be...
[hide]1. Japanese
2. Chinese
3. Korean
I love Asian foods :) [/hide]

how can you have a favorite food? There's so many good foods in the world!

[quote="Derek"]Chinese
Japanese
Korean

Azn f00d  :D[/quote]

asian food is yum ;D I like non-korean sushi with spam!

[quote="left_unattended"]asian food is yum ;D I like non-korean sushi with spam![/quote]
Sushi with spam? I've never tasted it. (At least, there's nothing like that in Japan.) But I wanna try it anyway :)

Nutella, yummy..
But not with toast.

I like pizza and fried chicken the best  :lol:
There are too many ties for second and third, can't list them all...

I like very much pancakes with bananas and finetti :).  I eat first time them like this in a camp. They are awesome :D!

[quote="236factorial"]I like pancakes plain. With bananas? No.[/quote]
Too bad for you :harhar:

[quote="noneoftheabove"]i like peanutbutter and banana sandwich[/quote]

You're joking, right?

I like peanut butter, but not banana, and not both!

And.... I like salsa  :D

[quote="math92"]i like salsa with nachos.[/quote]

I can eat that non-stop forever.

I hate bananas to begin with. They make me sick.

[quote="math92"][quote="236factorial"][quote="math92"]i like salsa with nachos.[/quote]

I can eat that non-stop forever.

[/quote]
i once had a nacho eating contest with my sister....i choked on the 3rd chip....[/quote]

your sister beat you?  :o

I don't really care what chips, just the salsa.  :lol:

[quote="236factorial"][/quote][quote="math92"][quote="236factorial"][quote="math92"]i like salsa with nachos.[/quote]

I can eat that non-stop forever.

[/quote]
i once had a nacho eating contest with my sister....i choked on the 3rd chip....[/quote][quote="236factorial"]

your sister beat you?  :o
[/quote]yeah, because i choked and i didn't want to choke again.  So i guess she won by default.

i like bannannas, and i like pancakes, but i do ot like bananna pankakes

I like a pancake with raisins in it. (It's my mom's original recipe :) )

i hate salsa but bananas are so good with peanut butter its even better and i hate most spicy foods i like chips with cheese tho although u get sick of them really quickly i like ben and jerrys ice cream

I don't like bananas theyre to mushy. they're ok in a banana split though...


[quote="jagdish"][b](1) If f:[1,infinity)-->R, and f(x) is monotonic and Increasing function, Then find the no. of solution of
f(f(x)) = 1 /(x^2-2x+2)[/b][/quote]

None, since $ f(f(x))$ is increasing and  $ \frac 1{x^2\minus{}2x\plus{}2}$ decreasing.


x - y = 5

x^2 + y^2 = 15

solve for y

I solved for x in x - y = 5 and got x = y + 5

I then plugged the value of x into x^2 + y^2 = 15 but was unable to find the value of y.

Can you help me?

[hide]$ (x\minus{}y)^{2}\plus{}2xy\equal{}15 \implies 5^{2}\plus{}2xy\equal{}15 \implies 2xy\equal{}\minus{}10 \implies xy\equal{}\minus{}5 \implies x\equal{}\frac{\minus{}5}{y}$ Now, substituting THIS back in gives us the equation $ (y\plus{}3)(y\plus{}2)\equal{}0 \implies \boxed{y\equal{}\minus{}3,\minus{}2}$[/hide]

[hide]
Substituting $ x \equal{} \frac { \minus{} 5}{y}$ back in to $ x\minus{}y\equal{}5$ gives $ y^{2} \plus{} 5y \plus{} 5 \equal{} 0$, not $ (y \plus{} 3)(y \plus{} 2) \equal{} 0$
Use the quadratic formula to get that $ y\equal{}\frac { \minus{} 5 \pm \sqrt {5}}{2}$[/hide]

$ (y\plus{}5)^2\plus{}y^2\equal{}2y^2\plus{}10y\plus{}25\equal{}15\implies 2y^2\plus{}10y\plus{}10\equal{}y^2\plus{}5y\plus{}5\equal{}0$
using the quadratic formula $ y\equal{}\frac{\minus{}\sqrt{5}\plus{}5}{2}$ or $ \frac{\sqrt{5}\minus{}5}{2}$
substitue back the $ y$ values back in one the first equations and you will get the value of $ x$


This is the set of all complex numbers $ c$ such that the sequence $ z_0 \equal{} 0, z_{n \plus{} 1} \equal{} z_n^2 \plus{} c$ is bounded.

This is the set of all complex numbers $ c$ such that the sequence $ z_0\equal{}0, z_{n\plus{}1}\equal{}z_n^3\plus{}c$ is bounded.

This is the set of all complex numbers $ c$ such that the sequence $ z_0 \equal{} 0, z_{n \plus{} 1} \equal{} z_n^5 \plus{} c$ is bounded.

This is the set of all complex numbers $ z_0$ such that the sequence $ z_{n \plus{} 1} \equal{} z_n^2 \plus{} (1\minus{}\phi)$ is bounded.

This is the set of all complex numbers $ z_0$ such that the sequence $ z_{n \plus{} 1} \equal{} z_n^2 \plus{} (\phi\minus{}2)\plus{}(\phi\minus{}1)i$ is bounded.

This uses the sequence $ z_0\equal{}0, z_{n\plus{}1}\equal{}\overline{z_n}^2\plus{}c$.

I have more of these, but they're so complex that they go over the file size limit... I'll try to post them somehow!

http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1387164#1387164


A rhombus has sides of length 10 cm and one diagonal of length 12 cm. How long is the other diagonal, in cm?

By 3-4-5 right triangles, half the length of the other diagonal is 8 since half the length of the 12 diagonal is 6 (and the hypotenuse is 10, which is a side of the rhombus). So the other diagonal is $ \boxed{16}$ centimeters long.


geometry
trigonometry
incenter
parallelogram
geometric transformation
reflection
projective geometry

OK, I suggested starting a geometry marathon on this forum, so here is the first question:
The distance between two well-known points in a triangle ABC is:
$\frac{bc}{a+b+c}\sqrt{2(\cos A+1)}$

What are the points?
The solver should post a new question, and give a solution that is sufficiently detailed that others who have not solved it will be able to understand.
Dominic

$\frac{bc}{2p}\sqrt{2(\cos A+1)}= d \Leftrightarrow bc\sqrt{\frac{\cos A+1}{2}}= dp \Leftrightarrow bc\cos \frac{A}{2}= dp$, where we've used the half-angle formula of the cosine.

By the Brigss formulas we get $bc\sqrt{\frac{p(p-a)}{bc}}= dp$ and so $\frac{bc(p-a)}{p}= d^{2}$.

On the other hand $bc\cos \frac{A}{2}= dp \Leftrightarrow \frac{bc \cdot \overline{AX}}{\overline{AI}}= \frac{bc(p-a)}{\overline{AI}}= dp$, where $I$ is the incenter and $X$ a touching point of the incircle with $b$ or $c$.

From these two relations we get $AI = d$.

Ok, here's the second problem from Crux.

Let $ABCD$ be a convex quadrilateral, and let $P, Q, R, S$ be the mid-points of $AB, BC, CD, DA$, respectively. Suppose that four distinct lines each passing through one of $P, Q, R, S$ concur at a point $O$. Draw lines parallel to these four lines but passing through the mid-points of the opposite sides. Prove that these four lines are also concurrent.

Given a hexagon ABCDEF such that AB,CD,EF are concurrent and BC,DE,FA are concurrent, show that AD,BE,CF are concurrent.

I am not sure how difficult this is without using a certain trick on a certain well known theorem, which would be worth discussing afterwards.

Is this not just a straight statement of Pappus' Theorem.
Look at the diagram on this page of Wikipedia [url]http://en.wikipedia.org/wiki/Image:Pappus_hexagon.svg[/url]
Then substitute (X,b,c,Y,C,B) for (A,B,C,D,E,F) and the result follows directly. Sorry to give such a blunt solution, but I am guessing that this is the solution you were after.
I was hoping that we might be able to use Brianchon's Theorem, but alas, no...
Dominic

I had always learnt Pappus' theorem as the reciprocal of this, so my solution was going to be "reciprocate the Pappus theorem with respect to an artibrary convenient circle". Sorry for setting something well known.

However, as a moral point, in a colinearity/concurrence question, reciprocation can be nice. :D

Do you want to post something? If I find something nice I will.

Question Four:
Given triangle ABC and D, E on AB, AC such that DE is parallel to BC and tangent to the incircle: Prove that
\[DE\leq \frac{1}{8}(AB+BC+CA)\]
A comment on the equality case would be nice too.
Incidentally, if you know which publication this problem is from, please do not just look it up: the point of this thread is to help improve everyone's geometry by attempting the problems themselves.
Dominic

I let a,b,c,s take their conventional meanings.

Let the incircle touch $AB$ at $L$, and the excircle opposite $A$ touch $AB$ at $L'$. By an obvious similarity argument, $DE/BC = AL/AL'$ and it is well known that $AL=s-a, AL'=s$.

Therefore $DE=\frac{a(s-a)}{s}\leq \frac{s^{2}}{4s}= \frac{1}{8}(AB+BC+CA)$ by AM-GM

Equality holds when $a=s-a$, so $3a=b+c$. In other words, this is when the incircle touches AB and AC at points L,M such that AL+AM=BL+CM.

Question 5:
Given a pair of circles $\Gamma_{1}, \Gamma_{2}$, with centres $O_{1}, O_{2}$, we let $O_{1}A_{2}$, $O_{1}B_{2}$ be tangent to $\Gamma_{2}$ and define $A_{1},B_{1}$ similarly. Let $O_{1}A_{2}$ cut $\Gamma_{1}$ at $P_{1}$ and let $O_{1}B_{2}$ cut it at $Q_{1}$. We define $P_{2},Q_{2}$ on $\Gamma_{2}$ similarly.
The [b]Eyeball Theorem[/b] states that $P_{1}Q_{1}$=$P_{2}Q_{2}$. Prove this by elementary means.

I hope it's ok for me to post not-that-well-known theorems as problems to solve. My intention is for this thread to end up containing lots of quite useful facts.

If anyone is interested, here is a diagram for the latest question:

Apologies for awful notation:
Drop perpendicular from P1 onto O1O2 at X and from P2 at Y. Let O1P1=R; O2P2=r and O1O2=d. Therefore P_1XO_1 is similar to O_2A_2O_1. Therefore P_1X=Rr/d. Exactly the same can be shown for P_2Y. Therefore P1Q1=2Rr/d=P2Q2.
I hope this is valid.
Tom, perhaps you could enlighten us as to why this is called the "eyeball theorem"...

I'd guess because it looks sort of like an eyeball and a couple of light rays focusing in, although you would be incredibly short sighted. I can't find the answer on the internet. I'd imagine Geoff might know.

Btw, Dom, I like your NRICH post. "my school (wherever that may be) "

Question Six:
This is at present part of a problem in the Unsolved Geometry Problems section of this forum, but I think this part is relatively simple and appropriate.
Given a triangle ABC, let X be a point on AB such that AX.BX=CX^2. Devise a ruler and compass construction to find all possible positions of X, if any exist.

Let us define K to be the point on CX such that CX = XK and K is not C.  by the converse of the intersecting chords theorem, K is on the circumcircle of ABC.  We can use this fact to construct X.
First, we construct D, the reflection of C in AB by dropping a perpendicular to AB, and then drawing a circle centre on the intersection of AB and the perpendicular.
We then construct a perpendicular to CD through D.  This line is the locus of P where AB bisects CP.  Thus K is on this line.  Call this line [i]l[/i]
We can also construct the circumcircle of ABC by taking perpendicular bisectors of AB, BC, etc.
Let the crossings of [i]l[/i] and the circumcircle be E and F (possibly the same).  Since K is on circle ABC, it is clear that E and F are the only possible locations for K.  Hence we bisect CE and CF to find our possible locations of X.

Sam siebert

Nice to see you. I will be impressed if you are able to remember your user name the next time you log in...
Anyway, it is now up to you to pose the next question.

OK Q7 coming later today.  ZVPCSPLFUIFDPEF is easy for me to remember as I have used it for a long time, but I am logged in automatically whenever I visit the site.

Question Seven:
Let ABC be an acute triangle with altitudes AL, BM and CN.
Let P be the centre of circle AMN.
Prove that PM.LN+PN.LM=PL.MN

Sam Siebert

OK, so it can be shown relatively easily that $\angle NLM=180-2\alpha$. Since $\triangle AMN \sim \triangle ABC$, it follows that $\angle MPN=2\alpha$. Clearly this means that PMLN is cyclic, so then we can apply Ptolemy's theorem on the given quadrilateral, which directly gives the required result!
I shall post a new question tomorrow.

Question Eight:
In triangle ABC, the sides satisfy AB+AC=2BC. Prove that the bisector of <BAC is perpendicular to IO.

Let the incircle be tangent to $BC$ AT $X$.

Then, $\angle IAX=\measuredangle \frac12 A$, $IX=r$, $x=s-a$.
Hence $AI=\frac{r}{\sin \measuredangle \frac12 A}$
and $AI=\frac{s-a}{\cos \measuredangle \frac12 A}=\frac{a}{2 \cos \measuredangle \frac12 A}$ by the condition of the question and considering the right-angled triangle $IAX$.

Hence $AI^{2}=r\frac{a}{2 \cos \measuredangle \frac12 A \sin \measuredangle \frac12 A}=r\frac{a}{\sin \measuredangle A}=2Rr$.

But $OI^{2}=R^{2}-2Rr$ and $OA^{2}=R^{2}$. Hence $OI^{2}+AI^{2}=OA^{2}$ and we are done by Pythagoras.

Next question soon.

I'll do another not-too-well-known theorem.

The butterfly theorem states that, given a chord $PQ$ of a circle, with midpoint $M$, and chords $AB$ and $CD$ through $M$, if we let chords $AD$ and $BC$ meet $PQ$ at $X$ and $Y$ then $M$ is the midpoint of $XY$.

Prove this.

Tim, since this forum appears to have been afflicted with a bad case of creative inertia recently, perhaps you would provide a hint towards the proof of the theorem, or give a whole proof and pose another problem. I think this will help to keep the thread moving...

Yes; apologies for the hold up.

I'll link to a [url=http://www.cut-the-knot.org/pythagoras/Butterfly.shtml]page of proofs[/url]rather than post one here (because I'm lazy).

The next problem:

Let $\triangle ABC$ be a triangle and $M_{A}, M_{B}, M_{C}$ the midpoints of the sides $BC, CA, AB$ respectively. Show that the triangle with side lengths $AM_{A}, BM_{B}, CM_{C}$ has area $\frac34$ that of $\triangle ABC$.

I think this sort of counts as a proof without words. ABC is congruent to CDA. I shall fill in some more rigour later

OK, I can now elaborate a bit further:
Construct D as described so ABCD is a parallelogram. We can construct two medians from A as shown. We need to prove that the line joining the feet of these two (purple & light green) is equal in length to the blue line BX. Well, it is well known that the diagonals of a parallelogram meet at their midpoints, therefore BX=1/2 BD. By a SAS similarity argument, the second blue line is 1/2 BD, so the two blue lines are equal. Thus the big triangle in the middle is the one given by the question. The area of this is given by [ABCD]/2 - [XCY] where X and Y are the midpoints of BC & CD. Clearly [XCY]=[ABCD]/8 by the same similarity argument so the area of the critical triangle is 3[ABCD]/8 =3[ABC]/4.

I shall post a new question this evening.
I am sure you are all numb with anticipation...

Dominic, we are feeling geometrically starved... :(.

How many people are following Geoff's recent e-mail suggestion, out of interest?

I'm thinking of some sort of more structured version of geometrical study.  e.g. This is how you prove somethings tangent to a circle, [i]this[/i] is how you prove that points are collinear, etc. etc.  It seems more efficient and useful.  I hope I am not too numb with anticipation to do BMO2 tomorrow.
Sam Siebert

I worked through PEG chapters 3-5 (about that far- maybe someway into chapter 6), and I got bored. I know I'm being incredibly picky but I dislike working from little A5 books which don't stay open on their own, and highly prefer A4 sheets. Hence I'm working through (parts of) Geometry Unbound instead. After that, I think I'll do Geometry revisited.

I also haven't yet followed Geoff's most recent suggestion on nRich (that if you are reading this post on 29th January, and are taking BMO2 tomorrow, then you should go to bed).

Apologies: I have been ludicrously busy for the last three days.
This should be fairly easy:
Find two isosceles triangles, each with two sides of length 106, with the other side having an integer length, which have the same area.
Dominic

Revisiting PEG is actually so much fun. You find lots of interesting facts which relate to advanced theory. I've just done a question from the tangency chapter just before Ptolemy, and it implies that if the cevian AK through the Lemoine point meets the circumcircle at X, and we call the midpoint of BC L, then the circles through L tangent to AB at B and AC at C have X as their other common point :). It's so cool. I made the Lemoine point actually do something :D:D:D

I see you cannot resist the temptation to call it Lemoine's point, rather than Symmedian point, which might actually tell people what it is!
Anyway, I feel I should apologise for posting such a dull question. If anyone has a better one, please feel free to post instead...

I always use confusing nomenclature. In Birmingham I just had to say "quadratic residues" and everyone was convinced that I was somehow clever, despite my infamous failure to actually apply knowledge of them. :)

To the question!

If we find a single right angled triangle with hypotenuse 106 then we are done, because we can just flip it round and glue two together.

I observe that $5^{2}+9^{2}=106$, so $2*5*9=90$, $9^{2}-5^{2}=56$ and we get (56,90,106) as such a right angled triangle, so taking isosceles triangles with base 112 height 90 and base 180 height 56 gives the desired result.

I have just been reading Christopher Bradley's "Challenges in Geometry", so the whole "integer-sided triangles" theme is becoming somewhat familiar.

This problem may be too well known, but it is very very nice, so I'll post it anyway, along with some interesting extensions I haven't yet considered in full detail myself.

Let P be a point inside a triangle ABC. I define p(ABC) as the triangle A'B'C' formed by the feet A',B',C' of the perpendiculars from P onto BC,CA,AB respectively. Show that p(p(p(ABC))) is directly similar to ABC. Is this true for a point P outside the triangle ABC? Does this generalise to n-gons?

I have seen this question before, but as it is rather nice and the forum is quietening down severely, I'll post a solution.

Let $A'$ be the foot from $P$ to $BC$, similarly $B'$, $C'$. Also let $A''$ be the foot from $P$ to $B'C'$ etc.

Clearly $AB'PC'$ is cyclic. Hence $\angle PAC' = \angle PB'C'$. Also $B'A''PC''$ is cyclic so $\angle PB'C'=\angle PB'A''=\angle PC''A''$. Also $C''B'''PA'''$ is cyclic and so $\angle PC''A''=\angle PC''B'''=\angle PA'''B'''$.
Hence $\angle PAC'=\angle PA'''B'''$.
Similarly $\angle PAB'=\angle PC'B'=\angle PB''A''=\angle PA'''C'''$.

Adding these together, $\angle B'''A'''C'''=\angle BAC$, and similarly on the other two vertices, proving the result.

I think the same kind of idea works with $P$ outside the triangle (although there would be problems with some special cases) but I haven't checked this through thoroughly.

As for other n-gons, I think that in the case of a quadrilateral, $P(P(P(P(ABCD))))$ is similar to $ABCD$, though the diagram quickly becomes to fiddly for me to want to attempt a formal proof, particularly at this time of day. I suppose this does rather suggest that for an n-gon, $P^{n}(AB...D)$ is similar to $AB...D$.

Anyway, another problem:

A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $\overline{AD}, \overline{BE}, \overline{CF}$ are concurrent. Let $P$ be the intersection of $\overline{AD}$ and $\overline{CE}$. Prove that $\frac{CP}{PE}=\left(\frac{AC}{CE}\right)^{2}$.

Replacement question:

$ABCD$ is a convex quadrilateral. $K,L,M,N$ are the midpoints of sides $AB,BC,CD,DA$. $BD$ bisects $KM$ at $Q$. $QA=QB=QC=QD$, and $\frac{LK}{LM}=\frac{CD}{CB}$. Prove that $ABCD$ is a square.

As BD bisects KM at Q and QB=QD, in effect, we have KM and BD bisecting each other. Thus, we have parallelogram KBMD, especially noting that AB is parallel to DC.
Now, as QA=QB=QC=QD, then ABCD is a cyclic quadrilateral too, which means that ABCD must be a rectangle.
As K, L, M are the midpoints of AB, BC, CD respectively, it is obvious that KL=LM. But as LK/LM=CD/CB, then CD=CB.
This tells us that ABCD must be a square.

Tim, your earlier question - does it involve areal coordinates? - It looks pretty interesting.

Alison

The hexagon one? Not as far as I know...

Alison, what makes you think areals will work? Which triangle are you thinking of using?

I hope you are right. It sounds interesting.

Having seen Brocard points, was attempting to use areals on triangle AEC, failed, but here is a solution (I hope):

Let the point where AD, BE and CF are concurrent be X.
As AB=CD=EF, we know that the angle subtended by these chords at the circumference will always be the same. We label these angles as having size 1. Then, we have angles subtended by arc BC having size 2, angles subtended by arc DE having size 3 and angles subtended by arc AF having size 4.
It is easy to see from the diagram by equal alternate angles that BC is parallel to AD, DE is parallel to CF and FA is parallel to BE. Thus we can also mark on the angles around X.

AC/CE=sin(1+2)/sin(1+3)
CP=AC* sin1/ sinAPC
PE=AE* sin3/ sinAPE

CP/PE=(AC/AE)* (sin1/sin3)

Note that there are lots of similar triangles in the diagram with angles (1+2), (1+3), (1+4). We now use these.

The ratio required can be turned into: [sin(1+2)/sin(1+4)]*[sin1/sin3]=sin2(1+2)/ sin2(1+3)
Which is that: 	sin1/sin3=[sin(1+2)/sin(1+3)]*[sin(1+4)/ sin(1+3)]
		BX/DX=(BX/AB)*(CD/DX)
As AB=CD, the LHS=RHS, so we are done.
Very sorry about the lack of LaTeXing, I will learn soon.

Alison

My question: O, I are the circumcentre and incentre of triangle ABC. AD is the height of the triangle with base BC (D is on the line BC). O, I and D are collinear. Prove that the radii of the circumcircle and the corresponding excircle are equal in length.

[quote="Ilthigore"]PQRS is the well known Varignon parallelogram of ABCD.

The lines PO and OR form a parallelogram with PM and MR (where M is the point of intersection of the two new lines), and the centre of this paralleogram is the midpoint of PR which is the centre of the Varignon parallelogram. Similarly, the centre of OQNS (N being the other point of intersection) is also the centre of the Varignon parallelogram, so M and N are both obtained by reflection O in the centre of PQRS, and so they are the same point.

I'll just think up another problem.[/quote]

also it can be soved easier...
$PS=QR,PS\parallel QR$($i$)
$SM=QR,SM\parallel QR$($ii$)

so we get:
$PM=RO,PM \parallel RO$ and that's what we wished for.


Similar to a 2000 problem:

It is currently September. What month will it be 150 months from now?

[hide]150/12=12 remainder 6

6 months after Sep= March[/hide]

[hide]150/12=12 r 6. Start counting. September will be 144 months, so March will be 150 months from September. [/hide]

[hide="solution"]
$\frac{150}{12}=12$r$6$
So, $6$ months from septembre is March
[/hide]

[hide]150/12=12 remainder 6

so six months after september is: MARCH[/hide]

[quote="solafidefarms"]Similar to a 2000 problem: [/quote]

It is currently September. What month will it be 150 months from now?
[hide="hide?"]150/12=12 r6. So MarCh[/hide]

How many months are in a year? How many years in 150 months? Figure it out. :D

[hide]150/12= 12 r 6

Sep, Oct, Nov, ... Mar
ans is March[/hide][/hide]

[quote="pianoforte"][quote="236factorial"][quote="bryantacademy"]How many months are in a year? [/quote]

I hope everyone knows this  :rotfl: 12.[/quote]

WHAT!?! THERE ARE 13 MONTHS IN A YEAR!!! What about UNDECEMBER! My little sister sayed so!!!
* :huh:  :?  :noo:  :fool: *   ;)[/quote]
is that supposed to be a joke, or do you really think there are 13 months in a year :?

[quote="math92"][/quote][quote="pianoforte"][quote="236factorial"][quote="bryantacademy"]How many months are in a year? [/quote]

I hope everyone knows this  :rotfl: 12.[/quote]

WHAT!?! THERE ARE 13 MONTHS IN A YEAR!!! What about UNDECEMBER! My little sister sayed so!!!
* :huh:  :?  :noo:  :fool: *   ;)[/quote][quote="math92"]
is that supposed to be a joke, or do you really think there are 13 months in a year :?[/quote]

think abotu it  :roll:

[quote="236factorial"][/quote][quote="math92"][/quote][quote="pianoforte"][quote="236factorial"][quote="bryantacademy"]How many months are in a year? [/quote]

I hope everyone knows this  :rotfl: 12.[/quote]

WHAT!?! THERE ARE 13 MONTHS IN A YEAR!!! What about UNDECEMBER! My little sister sayed so!!!
* :huh:  :?  :noo:  :fool: *   ;)[/quote][quote="math92"]
is that supposed to be a joke, or do you really think there are 13 months in a year :?[/quote][quote="236factorial"]

think abotu it  :roll:[/quote]
i am thinking and i don't get it.....i'm very slow at jokes....

[quote="math92"][/quote][quote="236factorial"][/quote][quote="math92"][/quote][quote="pianoforte"][quote="236factorial"][quote="bryantacademy"]How many months are in a year? [/quote]

I hope everyone knows this  :rotfl: 12.[/quote]

WHAT!?! THERE ARE 13 MONTHS IN A YEAR!!! What about UNDECEMBER! My little sister sayed so!!!
* :huh:  :?  :noo:  :fool: *   ;)[/quote][quote="math92"]
is that supposed to be a joke, or do you really think there are 13 months in a year :?[/quote][quote="236factorial"]

think abotu it  :roll:[/quote][quote="math92"]
i am thinking and i don't get it.....i'm very slow at jokes....[/quote]
It's a joke, math92, and there is a wink at the end of pianoforte's post.
[hide]
150/12=12r6, and March is 6 months after September.
$March$[/hide]


The stronger is also true
\[\sum\sqrt[5]{\frac{x}{y+k}}\ge \frac{3}{\sqrt[5]{k+1}}\]
:)

[quote="toanhocmuonmau"]For any $x,y,z,k>0$ such that $xyz=1$, we have
\[\sqrt[4]{\frac{x}{y+k}}+\sqrt[4]{\frac{y}{z+k}}+\sqrt[4]{\frac{z}{x+k}}\ge \frac{3}{\sqrt[4]{k+1}}\]
:)[/quote]
Could you please to give me some hint ?  :blush:
Holder ?  :D

hoder?
I don't think that
MV ?
What do your think ?
:D

Also look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111873

Very skilful and ingenious your solution, Toanhocmuonmau!  :lol:

Unthinkable proof, toanhocmuonmau! Congras!


group theory
abstract algebra
induction
modular arithmetic
superior algebra
superior algebra unsolved

1. Let $ G$ be a finite $ p \minus{}$group of order $ p^n$. Prove that for every $ 0\leq k\leq n$, $ G$ has a normal subgroup $ H$ of order $ p^k$.
2. Let $ G$ be a finite $ p \minus{}$group and $ H$ be a non-trivial subgroup of $ G$. Prove that $ H\cap Z(G)\ne 1$.
3.  Let $ G$ be a finite $ p \minus{}$group. Prove that every sungroup of order $ p^s$  is contained in a subgroup  of order $ p^{s \plus{} 1}$.
4. Let $ G$ be a group of order $ p^3$. Prove that : $ |Z(G)| \equal{} p$

2. holds only for normal $ H$. $ D_8$ has non-central involutions.
3. holds only for proper subgroups.
4. holds only for non-abelian groups.

Also they all follow from $ Z(G)\not\equal{}1$ for all $ p$-groups $ G\not\equal{}1$.

1. Take $ x\in Z(G)$ of order $ p$ and use induction on $ G/\langle x\rangle$.

2. $ 1\not\equal{} Z(H)\unlhd G$ and $ G/C_G(Z(H))$ operates faithfully on $ Z(H)$ by conjugation.
    So take $ K: \equal{}Z(H)\rtimes (G/C_G(Z(H)))\not\equal{}1$. Then $ 1\not\equal{}(x,y)\in Z(K)$ gives $ y\equal{}1$ and $ 1\not\equal{}x\in Z(G)\cap Z(H)$.

3. Take $ H\lneq G$.
For $ 1\not\equal{}Z(G)\le H$ use induction on $ H/Z(G)\lneq G/Z(G)$.
For $ Z(G)\not\le H$ take $ x\in HZ(G)\setminus H$ of minimal order. Then $ x^p\in H$ and $ |\langle H,x\rangle : H|\equal{}p$.

4. $ \langle Z(G),x\rangle$ is abelian for any $ x\in G$.
So $ 1\lneq Z(G)\lneq \langle Z(G),x\rangle\lneq G$ for some $ x\in G\setminus Z(G)$ and $ |G|\equal{}p^3$ gives $ |Z(G)|\equal{}p$.

@kidmath: it will be helpful [url]http://www.mathlinks.ro/viewtopic.php?t=217995[/url]

@olorin: What does $ \rtimes$ mean ?? I've seen it since before  :maybe:

$ \rtimes$ means [url=http://en.wikipedia.org/wiki/Semidirect_product]semidirect product[/url].

thanks very much, But in problem 2, incase having no knowledge about semidirect product, how can we solve this?

The same way you show $ Z(G)\not \equal{} 1$.
$ G$ operates by conjugation on $ H$ with fixed points $ H\cap Z(G)$.
The length of each [url=http://en.wikipedia.org/wiki/Group_orbit]orbit[/url] divides $ |G| \equal{} p^n$,
so the non-fixed points $ H\setminus(H\cap Z(G))$ split into orbits of lengths divisible by $ p$.
This gives $ 0\equiv |H|\equiv |H\cap Z(G)|\pmod{p}$ and $ 1\in H\cap Z(G)$ gives $ |H\cap Z(G)|\ge p$.


A cylinder of radius $18.0$ cm is free to rotate about a smooth horizontal axle. A mass of $0.380$kg hangs from a string wound round the cylinder. When the system is released from rest, the mass takes $2.0$ s to fall through a height of $5.00$ m. What is the moment of inertia of the cyliner?

I cant seem to find the answer , can anyone help ?

 :)

[color=blue]$\large mgR=I\gamma$

$\large \gamma=\frac{a}{R}$

$\large h=\frac{1}{2}at^2$

From these three equations, you get

$I=\frac{mgt^2R^2}{2h}$

where $\gamma$ is the acceleration of the mass, and $a$ is the angular acceleration of the cylinder[/color]


function
number theory unsolved
number theory

I guess many of you have seen/considered the following problem:

Prove that there exist infinite set of natural numbers $ n$ such that $ P(n) < P(n \plus{} 1) < P(n \plus{} 2)$ where $ P(n)$ is the largest prime number dividing $ n$.

Link to this problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=2005

And what about $ P(n) > P(n \plus{} 1) > P(n \plus{} 2)$ ?

//

[quote="Rust"]1. $ P(x)$ is not decrease, therefore exist $ x$, suth that $ P(x-1)<P(x)$ (for example x is any odd prime).
If $ P(x+1)>P(x)$ it give solution for one. If $ P(x+1)<P(x)$ ($ P(x+1)\not = P(x)$), then $ P(x^2-1)<P(x^2)$. If $ P(x^2+1)>P(x)$ we have solutione for one $ n=x^2-1$. If $ P(x^{2^l}+1)<P(x) \ \forall l<k$ and ${ P(x^{2^k}+1}>P(x)$ work $ n=x^{2^k}$. Because odd prime divisor $ P(x^{2^k}+1)>2^{k}>x$ if $ k>log_2 x$, then we find suth k for all x.
2. If $ S_k=P(2^k-1)$ is not monotonik, exist k, suth that $ P(2^{k-1}-1)=P(2^k-2)>P(2^k-1)>P(2^k)=2$.[/quote]

It is not obviously that:"$ S_k=P(2^k-1)$ is not monotonik"

My solution is this. Take $n=p^2-1$ where $p$ is a prime Now $P(p^2-1)<P(p^2)$. If $P(p^2)<P(p^2+1)$ then we are done.  If no then $P(p^4-1)<P(p^4)$. If $P(p^4)<P(p^4+1)$ then we are done again. So let $P(p)=P(p^4)>P(p^4+1)$. Proceeding this way if for some $k$ $P(p)<P(p^{2^k}+1)$ then we are done. So let for all $k$ $P(p)>P(p^{2^k}+1)$. Hence the set of all prime factors of $p^{2^k}+1$ for all $k$ is finite. But you can easily show that the function $f(p^{2^k}+1)$ is strictly increasing. Where $f(x)$ is the number prime divisors of $x$.


Do colleges look at the score such that the best of each section is taken into account?
for example if one got 700 for reading 800 maths 650 writing(total 2150) and then he took it again and got 750 reading, 780 maths, and 720 writing(total:2250)

is your best score 750, 800, 720
or 750, 780, 720?

Colleges have differing policies about how they treat scores if you have taken the SAT I more than once. Most top private colleges give you the benefit of your best score on each section. Some colleges, notably the University of California colleges, consider your best overall score from ONE test date. Look up the policies of each college you are interested in. It's never a disadvantage to take the SAT I more than once, although it would be silly to take it more than three times, in my opinion, because there are better things to do with your weekends.


Prove that the number $a_1a_2\cdots a_na_n\cdots a_2a_1$ is divisible by 11

What's the matter?
A number is divisible by 11 if the difference between the sum of the odd digits with the even digits of the number is $0$ or a multiple of $11$. Prove.
In this case we are in the first possibility.


geometry inequalities AMC AIME MATHCOUNTS USA(J)MO USAMO \/closed

Monday Classes:
Introductory Geometry   Oct 4  Dec 20
Olympiad Inequalities   Sept 20  Dec 6

Tuesday Classes:
Intermediate Algebra  Sept 14  Nov 30
AIME Problem Series  Dec 21  Mar 1
Olympiad Problem Solving - Sept 21  Dec 7

Wednesday Classes:
AMC-10 Problem Series - Nov 10  Jan 26
Olympiad Geometry  Sep 29  Dec 15

Thursday Classes:
AMC-12 Problem Series  Oct 21  Jan 27 (no class Nov 25, Dec 23, or Dec 30)
Intermediate Counting/Prob  Sept 16  Dec 9 (no class Nov 25)

Friday Classes:
Intermediate Complex Numbers/Trigonometry -- Sep 24  Dec 10
Introductory Number Theory -- Sep 17  Dec 3

Sunday Class:
MATHCOUNTS Problem Series  Nov 14  Feb 13 (no class Dec 26 or Jan 2)

What level is the Olympiad Problem Solving class? I know that it is called "Olmpiad," but is it for intermediate students who want to make the step towards olympiad type problems or advanced students who are already doing those types of questions? I doubt I'll even make USAMO as a senior though... : (

[quote="joml88"]You mean Fall, right?[/quote]

Thanks, fixed.

[quote="mathnovice"]What level is the Olympiad Problem Solving class?[/quote]

It's going to be pretty challenging.  I expect it will be only slightly less challenging than the geometry course.  Probably 1/2 of it will be very good for the student who is taking his/her first steps beyond AIME-style find the answer questions into Olympiad-style proofs.  The other half will be more challenging than that.  The goal of the course is to introduce some standard problem solving tools, examine how to tell when these tools might be useful, and then apply them to very challenging problems.

As for making the USAMO as a senior - work hard & challenge yourself and good things will happen, even if 'making the USAMO' isn't one of those things. . .

[quote="Rep123max"]Is the MATHCOUNTS/AMC Geometry going to be challenging to help on AIME?[/quote]

The MATHCOUNTS/AMC Geometry course will serve as a very solid foundation.  Since the AIME covers a wide variety of Geometry concepts, there is no doubt that the course will be helpful in providing a base.  However, the course will not get into much material as difficult as the AIME.  The difference will be a few theorems that might be tested for on the AIME, a broader array of techniques involving introducing algebra to geometry problems is required on the AIME, and AIME problems tend to be 3-4 step problems whereas the MATHCOUNTS/AMC class will typically including 1-3 step problems -- good for establishing the leap to AIME problems, but not yet AIME problems.

Your confidence in Geometry should be the deciding factor as to whether or not you want to take that course.  If you typically knock down all AMC geometry problems without looking at the solutions, then you would be better off focusing on something else.

[quote="Rep123max"]Will it kind of be the difficulty of the MATHCOUNTS/AMC Number Theory? I think I will probably take the geometry course since its my worst subject.[/quote]

I believe the difficulty levels of those two classes will be at least comparable.

[quote="Rep123max"]Will it kind of be the difficulty of the MATHCOUNTS/AMC Number Theory? I think I will probably take the geometry course since its my worst subject.[/quote]

Exactly how hard is MC/AMC Number Theory? I haven't taken the class, but is it suitable for someone who has already done most of the number theory in AoPS? How much more in depth does the class go?

[quote="mathnovice"]
Exactly how hard is MC/AMC Number Theory? I haven't taken the class, but is it suitable for someone who has already done most of the number theory in AoPS? How much more in depth does the class go?[/quote]

One of the nice things about the course is that we prove virtually everything related to AMC number theory.  Once you have proved something, you rarely miss problems about it because you understand it better and use the facts more creatively.  Many people who take the class say they get virtually every AMC number theory problem now.  We even discuss some AIME and USAMO problems toward the end.

If you're already getting all the AMC number theory problems, you should wait until we have an intermediate course.

[quote="MCrawford"] If you're already getting all the AMC number theory problems, you should wait until we have an intermediate course.[/quote]

Even if someone was getting all the AMC number theory questions right, it's still worth taking. Number theory gets real hard real fast. A good foundation is very important for understanding higher-level material (especially USAMO and up).

[quote="SingTheSorrow"]Will you be offering the Trig/Complex numbers in the winter also? Because I would really want to take the class, however, I am taking MC Numbers Theory instead.[/quote]

It is highly likely that the Complex numbers/Trig course will again be offered starting in January.

[quote="MCrawford"]It is highly likely that the Complex numbers/Trig course will again be offered starting in January.[/quote]

That would be cool. We'd take it here right away, except I think that course is slated to be on Thursday in the fall, and Thursday is our day with a schedule conflict with a curricular math course.

whoaaaaa OLYMPIAD PROBLEM SOLVING that looks tasty

The AMC 12 Problem Series is aimed mostly at the last few questions on the AMC.  Can I assume that the AIME Problem Series is similar in that it is aimed at the last few problems on the AIME?

The AIME Problem Series targets roughly the whole last half of the test, as much of the first half (difficulty-wise) is largely covered by the AMC-12 Problem Series.

Hmm...I'm looking forward to taking the AMC 10 problem series... There, will we be learning how to solve the hardest problems on the AMC 10 AND the easier (first half, I guess) problems on the AMC 12?

It will be almost all old AIME problems, though they will be selected to highlight special problem solving strategies (which we'll point out and emphasize as we go along).

There will be a heavier emphasis on geometry in the AMC-10 Problem Series class.  If the only thing you're looking for is geometry practice, you might opt for the Intro Geometry class would probably be a better choice (though about half of it might well be too easy for you, the rest should give you a good grounding in geometry).

[quote="aznphatso"]what comes after Mathcounts/AMC Counting/probability?
Does it just jump to AIME/Olympiad Counting/probablility?[/quote]

Yes, though we are changing the names of those classes to Intro and Intermediate.  It is likely that we will eventually offer an Advanced class in that area as well.

If you have taken the Intro class and absorbed it pretty well then you should be ready for the Intermediate class.

[quote="h_s_potter2002"]I'm taking regular high-school Geometry in the fall.  Would that be about the same as "Introduction to Geometry"?[/quote]

I can't say that I know what your high school course will be like, but judging from the curriculum I had in school the AoPS course will include a few pieces of curriculum not covered in high school, but the biggest difference will be that the AoPS course will be focused on problem solving as opposed to solving simpler textbook problems.

[quote="gadler"]What level are these classes and for whom are they geared (amateur or expert problem solver, or in-between, what high-school grades, etc.)?  And how does one enroll? I assume they are online, correct?

I'm new to this forum...

Thanks!

Gabi[/quote]

Gabi,

Welcome to the forum.  You can find more information about the level of the online classes in the Online Classes and Enroll tabs on the left sidebar.  The classes have defined levels based on their content and we have several levels of many of our offered courses.  The range will only become broader as time goes on.

We have pre and post-tests available to help guide students into the right courses.  You might also find out the best fit by offering up your own "problem solving resume" that might include the subjects you have studied, strong points, weak points, and AMC/AIME scores if you have them.  You could do that in an email to one of the instructors or on the site where other students who have taken classes could give their own opinions of which classes they think would be best for you from their own perspectives.

Let me know if you have any other questions.

[quote="NightFlarer"]I heard that the AMC 10 problem series course is focused mainly on geometry.  I'm not sure whether I should take Intro geometry or the amc 10 problem series. I just finished Geometry in highschool, and I'm pretty good at it. I can't solve all amc 10 geometry problems though.  Like the last 8 or so I sometimes have trouble with.  Also, I've looked at most of the amc 10 problems and their solutions so would it be a waste to enroll in the amc 10 class?[/quote]

I would suggest taking the Geometry class.  It will focus on developing the problem solving techniques more deeply while the AMC 10 class should be viewed more as practice and reinforcement.

Also, where did you hear that the AMC 10 class focuses on geometry?  About a third of it will be geometry, but it will cover many areas of problem solving mathematics.

Somebody on the forums, I can't remember who, said the AMC 10 class focuses a lot on Geometry.  I think I'll take the geometry class, especially since I'm a bit rusty on certain topics.  How many weeks will the class be?  And are the lessons 1.5 hours each?

Both the AMC-10 class and the geometry class will be twelve 90-minute classes.

[quote="gadler"]Any new information on the Intro to Geometry class - i.e., pre and post tests? Is it recommended for entering 9th graders (i.e., very little previous exposure to geometry)?[/quote]

Very recommended for entering 9th graders with little previous geometry.

We'll have pre- and post-tests up next week.

Also, about the AMC 10 problem series, do you think a person who got a 68, 74.5, and a 79.5 on the Mock AMC's on this board would do okay in that class?  I don't know how hard the AMC 10 will be in relation to the Mock AMC's, but I don't know if my scores are too low either.

Thanks.

I'm guessing from what I've heard that the mock AMC's are plenty hard.  I think you'd be fine in the AMC-10 class.

[quote="joml88"]I'm just curious but how hard is the AMC 10 PS in comparison to the AMC 12 PS?[/quote]

The AMC12 is significantly harder than the AMC10.  The AMC12 problem series is designed for students who are already quite confident they'll make at least a 110 on the AMC12.  We focus on only the hardest AMC problems in that course.

The AMC10 class on the other had would be good for students taking either the AMC10 or AMC12 but are borderline as to making the AIME.


I don't understand your problem. Can you please restate it a little more clearly? What exactly are you asking for?

[quote="Pirkuliyev Rovsen"]Let us assume  conditions of positive numbers of a, b.c.d and he pays a≤b≤c≤d,ab+ac+ad+bc+bd+cd=1
Can buy  ad sum   what  prices ?[/quote]

please don't spam :spam:


geometry
circumcircle
trigonometry
geometric transformation
reflection
incenter
inequalities

It was posted in mathlink, I can't find the link therefore I will post my solution
By law of sine
$\frac{OD}{\sin OBD}=\frac{a}{\sin BOC}$
it isn't hard to show $\angle OBD=\frac{\pi}{2}-(\angle B-\angle C),\angle BOC=2\angle A$ when $O$ inside triangle(because triangle is acute)
thus $OD=\frac{a\cos(B-C)}{\sin 2A}=\frac{2R\sin A\cos(B-C)}{2sin A\cos A}=R\frac{\cos(B-C)}{cos A}$ similarly for $B,C$ thus we need to prove
$\prod\cos(B-C)\ge 8\prod\cos A$ multiply with $\prod\sin A$ we have
$\prod\sin A\cos(B-C)\ge 8\prod\cos A\sin A=\prod\sin 2A$
But $\prod\sin A\cos(B-C)=\prod\sin(B+C)\cos(B-C)=\prod\frac{1}{2}(sin 2B+sin 2C)$ thus we need to prove
$\frac{1}{8}\prod(\sin 2B+\sin 2C)\ge\prod\sin 2A$ but it is quite obvious by $AM-GM$
 :D  :D

[quote="N.T.TUAN"]Let $O$ be the circumcenter and $R$ be the circumradius of an acute triangle $ABC$. Let $AO$ meet the circumcircle of $OBC$ again at $D$, $BO$ meet the circumcircle of $OCA$ again at $E$, and $CO$ meet the circumcircle of $OAB$ again at $F$. Show that $OD.OE.OF\geq 8R^{3}$.[/quote]
[color=darkblue][b][u]Proof.[/u][/b] Note $I\in BC\cap AD$ and

the reflection $A'$ of the point $A$ w.r.t. the point $O$. Then

$\boxed{\mathrm{\ the\ points\ A\ ,\ A'\ are\ harmonical\ conjugate\ w.r.t.\ the\ points\ I\ ,\ D\ }}$.

Let $\rho$ be the circumradius of the triangle $BOC$. Suppose $AB<AB$.

$BA\perp BA'$ $\Longrightarrow$ $\widehat{CBD}\equiv 2\cdot \widehat{CBA'}$ $\Longrightarrow$ $\widehat{OBD}\equiv 90^{\circ}+C-B$.

$\left\{\begin{array}{c}BC=2\rho\sin\widehat{BOC}\\\\ m(\widehat{BOC})=2A\end{array}\right\|\Longrightarrow\left\|\begin{array}{c}\rho =\frac{a}{2\sin 2A}\\\\ a=2R\sin A\\\\ \sin 2A=2\sin A\cos A\end{array}\right\|\Longrightarrow\rho =\frac{R}{2\cos A}\ .$

Therefore, $\left\{\begin{array}{c}OD=2\rho \sin\widehat{OBD}\\\\ \rho=\frac{R}{2\cos A}\\\\ m(\widehat{OBD})=90^{\circ}+C-B\end{array}\right\|\Longrightarrow\boxed{\ OD=\frac{R\cos (B-C)}{\cos A}\ }$

a.s.o.[/color]

If we change circumcenter $O$ by incenter $I$ we will get the inequalities
$8r^{3}\le ID\cdot OE\cdot IF\le R^{3}$ and we will make a question ourself
Find the locus of $P$ in triangle such that $PD\cdot PE\cdot PF\le R^{3}$(or $\ge R^{3}$)


linear algebra
matrix
vector
superior algebra
superior algebra theorems

This is basically a general question:Why is the concept of trace of a matrix so important?Moreover any you suggest any book where I can find any hair-raising problems where the concept of trace is needed in a very non-traditional way.

I will try to give you an answer because I am afraid otherwise the topic is pushed back too much for others to reply.

Usually matrices (n*n) are associated with linear maps from an n dimensial space towards that same space

if A is a matrix of a map using a certain basis, the matrix of that same map using another basis would be $D^(-1)AD$ with D invertible

One easily proves that A and $D^(-1)AD$ have the same trace
So trace of a map 'exists', meaning that you can define the trace of a map independently of the basis you choose

I have a very concrete applicatio :
if you solve the system of differential equations :
$\frac{dv}{dt}=A v$ with v a twodimensial vector
you diagonalise (if possible) A to calculate the exact solutions
if you want to know how they behave, that depends of the eigenvalues in that diagonalised form, more precisely of their sign
by using determinant and trace you can see this without the sometimes tedious way of explicitly calculating them

well , perhaps other people have far better ideas


Let be $ \alpha,\beta \in \mathbb{R}$ such that $ 3\sin\ (3\alpha\plus{}2\beta)\equal{}\sin \ \alpha$ . Prove that :

$ \mathrm{tg} (2\alpha \plus{}\beta)\plus{}\mathrm{tg}(\alpha\plus{}\beta)\equal{}0$

Do you mean tan(2a+b)+2tan(a+b)=0?
[hide="If so"]3sin(x+y)=sin(x-y)
3sin(x)cos(y)+3cos(x)sin(y)=sin(x)cos(y)-cos(x)sin(y)
2sin(x)cos(y)+4cos(x)sin(y)=0
tan(x)+2tan(y)=0

Substituting x=2a+b, y=a+b, we get tan(2a+b)+2tan(a+b)=0[/hide]


The positive difference between two consecutive perfect squares is 35. What is the greater of the two squares?

With some guess and check, we find that $ 18^2 \minus{} 17^2 \equal{} 35$. So the greater square is $ 18^2$, or $ \boxed{324}$.

[quote="FantasyLover"]$ (x \plus{} 1)^2 \minus{} x^2 \equal{} 35$
$ x^2 \plus{} 2x \plus{} 1 \minus{} x^2 \equal{} 35$
$ 2x \plus{} 1 \equal{} 35, x \equal{} 17$
therefore, x=18
$ 18^2 \equal{} 324$[/quote]

easier:

use diff of squares

(x+1+x)(x+1-x)=35
2nd term=1
2x+1=35
x=17, (x+1)^2=324


Hello, I've had a broken thumb for the 3 weeks (ruining my scores and stoping me from posting) so I was wondering how many of you have had injuries. And just for your information I_ broke it playing american football, hint: do [b]NOT[/b] look down to make sure your don't step in dog poop after you catch an interception.

I am relatively injury-free.  I was once knocked rather badly on the head when I was slide-tackled from behind playing soccer, but it was luckily not a concussion.  Other than that, nothing much.

Mmm, I got hit by cars on several occasions but was never really hurt. For example my family fell off a ridge on a highway in Canada and everyone thought we were knocked out but we weren't hurt at all, just some bruises, we were just sleeping :lol: Although our rent-a-car was trashed...no wonder people were worried :lol:
I think I already mentioned the time I was hit with a BMW while on a bicycle and the other car-crash last year was just a minor one (a drunk guy crashed into us from the behind...in a parking garage :lol: ) The only serious injury I got was when I broke my pinky on a door (not an automated one) in preschool.

How could someone vote for Died and be truthful? :lol:

[quote="Tare"]Mmm, I got hit by cars on several occasions but was never really hurt. [/quote]

...... :shock:  care to elaborate?

i overstretched my knee during tennis and got osgood-schlatter's disease. its not an actual disease, a doctor just said that at the time it was discovered, it was all the rage to call everything a disease.

I rode my bike off a pavement down a steep hill, when i tripped over something  and lost control. fortunately i did not break my leg or had any head injuries (not sure if i wore a helmut or not) but you can probably guess what happened in the end.

[quote="Tare"]How could someone vote for Died and be truthful? :lol:[/quote]

It is actually possible. ``Death'' is a rather technical term. (Unfortunately, I will not tell you the meaning of it because I do not know; I think it has to do with the heart stopping for a certain amount of time.) Thus it is possible for a person to have died but still be alive. Naturally, it's a rather uncommon occurrence.

I've sprained/twisted my ankle several times playing soccer. And I brok my right pinky toe somehow and it healed wrong, and now it's crooked and I have to go into the doctor and have him rebreak it so it heals right. That was a really long sentence, but oh well. I've also broken some fingers several times playing goalie/getting stepped on after being slide tackled in soccer. Soccer can be dangerous ya kno??? :)

i get rann over alot in soccer. i got a hyperextended knee in wrestling, but i haven't broken anything.

ive had an ankle problem for a year due to basketball season and practicing etc. but nothing more than that.

really surprising... i have none... except blisters (if they're injuries at all...)

But it seems that most of the asian people i know had no serious injuries...

hehehe... sorry for double-posting... i think the site was down.

:lol:, maybe we're more flexible in general...I know that that reduces a lot of joint injuries like elbows knees and ankles.

ya well besides sprains (mostly ankle from running and bball) ive been pretty injury free... *sighs* so happy!  :lol:

I have had stiches twice, staples once, and I almost broke one of my toes.

Never injured myself, unless you count the time in 6th grade when I was playing basketball and fell and slid into a volleyball-net-pole-thingy, busticating one of my teeth.  Yup.

It seems like a lot of people play soccer.  Interesting.  So do I.

the top of the inside of my mouth was cut off. a football hit me while i had a sharp candy cane in my mouth. i'm glad i'm still alive.
wow 2 ppl have voted for died.

ouchy.. iv never been in a emergency room, cept when my big sis had to go... my... i wonder who voted 4 died?


probability
combinatorics proposed
combinatorics

Hi
I have a probability question.
X is sending information to Y.  There are 2 wires between X and Y.  X send the same packets to Y on both wires. Wire 1 has probability of failure = 0.1 and wire 2 has probability of failure = 0.2. Y successfully recieving the packet if either wire successfully sent the packet to Y.

a) What is the probability of Y successfully receiving the packets? if X sends the packets thru both wires.

b) Suppose X doesn't not send packets to Y on both links, but randomly picks one of the 2 links with probability 50% each, what is the probability of Y successfully receiving the packets?

my answer:
a) P (wire 1 successfully sent) = 1 - 0.1 = 0.9
P (wire 1 successfully sent) = 1 - 0.2 = 0.8
P (Y successfully receiving the packets) = (0.9 + 0.8) / 2 = 0.85

b) P (Y successfully receiving the packets) = 0.5 * 0.9 + 0.8 * 0.5 = 0.85

It doesn't make sense, I don't think I did it right! any help plz?

The answer to question (a) depends on whether the failure on one wire is independent of the failure on the other wire.  Supposing it is, why do you think that averaging the probabilities of failures on the two wires gives the probability of failure on both?

Also, this question really belongs in one of the high school forums (say, Intermediate Topics).


I just wanted to know how to redirect the LaTeX files produced from TeXnicCentre to a folder I want? I probably did not alter it during the configuration, but is there any way to shift all documents henceforth from My Documents to somewhere else? Thanks.

pdflatex saves files in the same directory as the tex file being compiled. So the simplest method is to put the tex file wherever you want the output.

You can tell pdflatex to save files in another folder in Build, Define Output Profiles. The command line options include:
[quote]Usage: pdflatex [OPTION...] [COMMAND...]
  -aux-directory=DIR              Use DIR as the directory to write auxiliary files to.
  -output-directory=DIR           Use DIR as the directory to write output files to. [/quote]but of course this may mean you need to change the command lines for all the other programs that are being used.


I have heard about MOEMS, and I know it stands for Math Olympiads for Elementary and Middle Schools, but that is all I know. Could someone give me some information, like how to register for it (my school isn't in it)? Is it possible to take the test without getting all the other materials? (just the test, not the extra sample questions)? Also, if you do good in the competition, do you advance to somewhere? I can't find these on their website [url=http://www.moems.org/]here[/url].

There is no advancing to the next round in MOEMS. I dont know if you can take it as an individual, but you could contact them to find out (Im sure there is an email address on their website somewhere). It would be a really great competition to get your school involved in (if they have the money to do it).
As for what it is, MOEMS is a 5 round competition with 5 questions in each round. Each round occurs once a month. The difficulty of MOEMS in my opinion ranges from easy chapter mathcounts level to pretty hard chapter mathcounts level. I would say its fairly easy because I aced it the past 3 years (hoping to make it 4 :P ), but there are usually one or two questions that really trip people up and prevent LOTS of people from getting a perfect.

To sum it up, its a really great competition to get your school involved in, but Im not sure it would be worthwhile for just you to spend the money to take the test as an individual.

[hide=" Sample Problem From This Year"]
Here is a sample problem from this years MOEMS that I thought was the hardest question be far for this year:
A pentagon has all of its diagonals drawn. How many triangles of all sizes result from the new figure?

[hide="Answer"]35[/hide]
[/hide]

You could also check out this: [url]http://www.moems.org/enroll_files/0708Enrollment%20Form.pdf[/url]

Go check out their website.

The format is 5 rounds of 5 questions taken under timed conditions. If you can find a proctor, as few as a single student can take the test. I took it with 1 other person in something like 2nd or 3rd grade (and then didn't take it again till 7th; I don't really advise taking at that young age.)

[quote="RoFlLoLcOpT"]The sample problem was basically taken from MATHCOUNTS.[/quote]
Wait, my sample problem? That was an actual question from this years MOEMS.

And @ PoP:
Like I said before, I definitely wouldnt recommend taking the test by yourself (or with 1 or 2 other people for that matter). I would say that it would be better to take it with your school, partly so that you dont have to pay the money, and partly because its just fun taking it with your entire school.

I thought everyone knew about MOEMS....lol
MOEMS is an easy contest. You dont even have to practice for it.
Even I got  perfect in it and I'm in 4th grade..so you probably can easily get a perfect. You get a half hour to do 5 easy problems, so dont rush through it.

[color=indigo]That sample problem was easy .:surf: Probably a about a very easy MATHCOUNTS chapter problem.
But to tell the truth I should'nt be taking since I kinda scored bad in the Nationals this year.[/color] :blush:

@ supernova

It was an easy MOEMS question, but compared to all the other easy questions this year, it was the hardest (in my opinion).

@PoP

I forgot to mention, there are 2 levels of MOEMS, division E and division M. Division M is for 7-8 and E is for 6th and below.

Thanks for all the help. I will try to convince my school to register.

I agree with Thunder365 and it is hard in my opinion (though I can still solve it).

@supernova: If you made it to Nats, then you did pretty good.

Some more sample problems can be found here in case any body was intrested http://www.moems.org/sample_files/SampleE.pdf

you don't always get 30 minutes. Sometimes you get 20 minutes, 26 minutes, 24 etc. It depends on how long the people who write the test think it would take. It is still pretty easy and a lot of people finis in the first 10 minutes or so.

yes, i gots a perfect in E and 23/25 in M tis year.
Me friend got two perfets.

One piece of info: The MOEMS is NOT a really high olympiad although it says olympiad in its name.

And two: The creator of MOEMS died on my birthday :( :noo: :(

Anyhow, MOEMS is much easier than mathcounts, AMC 8 and other contests.

[hide=" Sample Problem From This Year"]
Here is a sample problem from this years MOEMS that I thought was the hardest question be far for this year:
A pentagon has all of its diagonals drawn. How many triangles of all sizes result from the new figure?

[hide="Answer"]35[/hide]
[/hide][/quote]

Hey....whats funny is that...that sample problem came up in a REAL math olympiad contest for middle schoolers.
I think it was this year,(oh yeah BTW..i got a perfect). THE SAME PROBLEM WAS IN A MATHCOUNTS BOOK. Specifically the All time greatest MC Problems Book. The same answer too...lol...so that kinda helped me out..but it was easy so I woulda gotten it anyway.


The incircle of triangle $ABC$ touches its sides $BC$, $AC$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$ respectively. A segment $AA_{1}$ intersects the incircle at the point $Q\ne A_{1}$. A line $\ell$ through $A$ is parallel to $BC$. Lines $A_{1}C_{1}$ and $A_{1}B_{1}$ intersect $\ell$ at the points $P$ and $R$ respectively. Prove that $\angle PQR=\angle B_{1}QC_{1}$.
[i]A. Polyansky[/i]

[b][u]Solution[/u][/b]

We have: $\angle{C_{1}QA_{1}}=\angle{BA_{1}P}=\angle{APA_{1}}$ so the quadrilateral $PAQC_{1}$ is cyclic and therefore we get:

$\angle{C_{1}QA_{1}}=\angle{BA_{1}P}=\angle{PC_{1}A}=\angle{PQA}$

Similarly we get $\angle{A_{1}QB_{1}}=\angle{AQR}$ Q.E.D.

[u][i]Nick[/i][/u]

Let $T$ be the antipode of $A_{1}$ (in the incircle).
It is well-known that each of the triplets $B_{1}, T, P$ and $C_{1}, T, R$ are collinear.
Furthermore, if $\{V\}=C_{1}B_{1}\cap{PR}$ then $Q,T,V$ are collinear (by considering the radical axes of the following circles : incircle, Euler's circle of $\triangle{A_{1}RP}$ and the circumcircle $\triangle{AQTM}$ where $\{M\}=TA_{1}\cap{PR}$ ).
Therefore $PQTR$ is cyclic and thus: \[\angle{PQR}=\angle{PTR}=\angle{B_{1}QC_{1}}.\]

This initial problem is easy, but the configuration hides many projective results, which are not specified, but I won't do that either.

We have that $\angle{B'QC'}=\angle{C'QA'}+\angle{B'QA'}=\angle{C'B'A'}+\angle{B'C'A'}$ because $A'B'QC'$ is cyclic.

Thus $\angle{B'QC'}=\angle{BA'C'}+\angle{CA'B'}$, but from the paralelism we have that:

1) $\angle{BA'C'}=\angle{BC'A'}=\angle{AC'P}=\angle{APC'}$

2) $\angle{CA'B'}=\angle{CB'A'}=\angle{AB'R}=\angle{ARB'}$

Therefore $\angle{B'QC'}=\angle{AC'P}+\angle{AB'R}$ $(\star)$

But we have that $\angle{APC'}=\angle{C'QA'}$, so the quadrilateral $APC'Q$ is cyclic, and similar the quadrilateral $ARB'Q$ is cyclic, so:

1) $\angle{AC'P}=\angle{PQA}$

2) $\angle{AB'R}=\angle{RQA}$

And now by simply replacing this into $(\star)$ we obtain that $\angle{B'QC'}=\angle{PQA}+\angle{RQA}= \angle{PQR}$, thus the problem is solved.

[quote="pohoatza"]Another super nice problem on this same configuration is the following:

Denote $E, F$ the midpoints of $(A'P), (A'R)$ respectively. Prove that $A,B',C',E,F$ lie on the same circle.[/quote]
it is really easy to proof $B',C',E,F$ lie on circle but how can we proof $A$ lie on this circle?

[quote="pohoatza"]Another super nice problem on this same configuration is the following:

Denote $E, F$ the midpoints of $(A'P), (A'R)$ respectively. Prove that $A,B',C',E,F$ lie on the same circle.

PS: I remember I solved it some time ago, and I think it was from AMM, or from a FG note.[/quote]
Very simple, the points $ A, B', C', E, F $ all lie on the nine point circle of $ \Delta A'PR $

[quote]The incircle of triangle $ABC$ touches its sides $BC$, $AC$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$ respectively. A segment $AA_{1}$ intersects the incircle at the point $Q\ne A_{1}$. A line $\ell$ through $A$ is parallel to $BC$. Lines $A_{1}C_{1}$ and $A_{1}B_{1}$ intersect $\ell$ at the points $P$ and $R$ respectively. Prove that $\angle PQR=\angle B_{1}QC_{1}$.
[/quote]
[b]Solution:[/b] Let $K$ be the reflection of $A_1$ over $I$
$$-1=( A_1, ~ Q ; ~ B_1 , ~ C_1 ) \overset{A_1}{=} ( \infty_{BC}, ~ A ; ~ R, ~P) \implies AP=AR$$
$\angle A_1C_1B_1=\angle B_1A_1C=\angle PRB_1 \implies PRB_1C_1$ is cyclic. Also, $\angle C_1KA_1$ $=$ $\angle BA_1P$ $=$ $\angle C_1PA$ and, $\angle C_1QA_1$ $=$ $\angle BA_1P$ $=$ $\angle C_1PA$ $\implies$ $K$ is orthocenter, $AA_1$ is median and $Q$ is the $A_1-$humpty point WRT $\Delta PQA_1$ $\implies$ $PQKR$ is cyclic $\implies$ $\angle PQR$ $=$ $\angle PKR$ $=$ $\angle B_1QC_1$

Dear Mathlinkers,

https://artofproblemsolving.com/community/c6t48f6h1589300_geometry

Sincerely
Jean-Louis

Nice configuration
[hide=Solution]Note that it suffices to show $\angle PQR=90^{\circ} + \dfrac{A}{2}$.Its easy to see $ARQB_1,AQC_1P$ concylic.Hence $\angle PQR=\angle PQA + \angle RQA= \angle PC_1A + \angle AB_1R=\angle A_1C_1B+\angle A_1B_1C=90^{\circ} + \dfrac{A}{2}$.So we are done.[/hide]

∠C1QA1 = ∠C1A1B = ∠C1PA so C1PAQ is cyclic and ∠B1QA1 = ∠B1A1C = ∠B1RA so B1RAQ is cyclic. ∠PQR = ∠PQA + ∠RQA = ∠AC1P + ∠AB1R = ∠C1QA1 + ∠B1QA1 = ∠C1QB1. we're Done.

By Reim's $AQB_1R,AQC_1P$ are cyclic and therefore $$\angle PQR=\angle PQA+\angle AQR=\angle PC_1A+\angle AB_1R=\angle BA_1C_1+\angle CA_1B_1=\angle B_1QC_1.$$

By projecting $(A_1,Q;B_1,C_1) =-1$ onto $\ell$ we get $A$ is the midpoint of $PR$. Also since $CA_1=CA_2$, we get that $A$ is the circumcentre of $PRB_1 C_1$, hence $B_1$ is the foot of the altitude from $P$ onto $A_1 R$.

This means that $Q$ is the $A_1$-Humpty point in $A_1 PR$ and so $\angle PQR = \pi - \angle B_1 A_1 C_1 = \angle B_1 Q C_1$.

Rename $A_1,B_1,C_1$ to $D,E,F$ respectively.

Since $\measuredangle QAP=\measuredangle QDB=\measuredangle QED=\measuredangle QFD=\measuredangle QFP$, $AQFP$ is cyclic. Likewise, $AQER$ is cyclic. Then
$$\measuredangle PQR=\measuredangle PQA+\measuredangle AQR=\measuredangle PFA+\measuredangle AER=\measuredangle DFB+\measuredangle CED=\measuredangle DEF+\measuredangle EFD=\measuredangle EDF=\measuredangle EQF,$$
which finishes.


Let $ a,b,c>0$. Prove that

$ \sum[\frac{a^4+b^}{(a+b)^4}]\geq\frac{15}{8}-\sum\frac{a^2+\sqrt{ab}-bc}{a+b}$

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \sum[\frac {a^4 \plus{} b^4}{(a \plus{} b)^4}]\geq\frac {15}{8} \minus{} \sum\frac {a^2 \plus{} \sqrt {ab} \minus{} bc}{a \plus{} b}$[/quote]

I edited it as above

by cauchy shwartz we can easily show that :
$ \sum{\frac {a^2 \minus{} bc}{a \plus{} b}} \geq 0$

hence it sufficies to show that :
$ \sum{\frac {a^4 \plus{} b^4}{(a \plus{} b)^4}} \plus{} \sum{\frac {\sqrt {ab}}{a \plus{} b}} \geq \frac {15}{8}$

by AM-GM : $ \sqrt {ab} \geq \frac {2ab}{a \plus{} b}$

hence it's enough to show that :  $ \ sum{\frac { a^4 \plus{} b^4}{(a \plus{} b)^4}} \plus{} 2\sum{\frac { ab}{(a \plus{} b)^2}} \geq \frac {15}{8}$

but : $ \frac {a^4 \plus{} b^4}{(a \plus{} b)^4} \plus{} \frac {2ab}{(a \plus{} b)^2} \minus{} \frac {5}{8} \equal{} \frac { (a \minus{} b)^2(3(a^2 \plus{} b^2) \plus{} 2ab)}{8(a \plus{} b)^4}$   and we are done :wink:

by the way the following holds with same conditions:
   $ \frac {9}{8} \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} ac \plus{} bc)} \geq \sum{\frac {2a^4}{(a \plus{} b)^4}} \plus{} \sum{\frac {2ab}{(a \plus{} b)^2}} \geq \frac {15}{8}$  :)


What is the difference between $ 32\%$ of 24 and $ 24\%$ of 32?

Both values are equivalent to $ \frac{(32)(24)}{100}$. So their difference is $ \boxed{0}$.


inequalities
vector
inequalities unsolved

prove that for an arbitrary pair of vectors $ f$ and $ g$ in the plane the inequality
$ af^2\plus{}bfg\plus{}cg^2\ge0$
holds if and only if the following conditions are fulfilled:
$ a\ge0$,$ c\ge0$, $ 4ac\ge b^2$

I believe $ \boxed{a\equal{}b\equal{}0\ and\  c\geq 0}$ or $ \boxed{a>0\ and\ b^2\leq 4ca}$.

[quote="puuhikki"]How do you define a product of two vectors or a square of a vector?[/quote]

I think he meant that the former is the scalar product and the latter is the square of the norm of the vector.

i wrote the problem just like the one in the IMO compendium book
i think its scalar product

[quote="ishfaq420haque"]prove that for an arbitrary pair of vectors $ f$ and $ g$ in the plane the inequality
$ af^2 \plus{} bfg \plus{} cg^2\ge0$
holds if and only if the following conditions are fulfilled:
$ a\ge0$,$ c\ge0$, $ 4ac\ge b^2$[/quote]
but isn't this obvious? Put $ f\equal{}(x,0,...,0)$ and $ g\equal{}(y,0,...,0)$ to see by discriminant that $ a\ge0$,$ c\ge0$, $ 4ac\ge b^2$ are necessary (& sufficient for this case to hold for all $ x,y\in\mathbb R$).

So , if the three conditions are fulfilled, $ af^2 \plus{} bfg \plus{} cg^2\equal{}\sum_i ( ax_i^2\plus{}bx_iy_i\plus{}cy_i^2)\ge0$.
Or do I miss something?


If every point of the plane is painted one of three colours, do there necessarily exist two points of the same color exactly one inch apart? What if "three" is replaced by "nine"?

The first part. Let's have a look at two equilaterar triangles ABC and ABD that have side one inch long. Let point A be green, point B red and C yellow. D cant' be neither green nor red, so it must be yellow. Points C and D must be the same colour.  Notice that such triangles can be drawn for each two points C, D, which are in the distance $\sqrt 3$. So each two points in distance  $\sqrt 3$ must be the same colour. Now draw a triangle XYZ that has sides $\sqrt 3$, $\sqrt 3$ and $1$ inch long. Points X, Y, and Z must be the same color. Contradiction.
And now the second part.  Divide the plane into squares of side 0.6 and color each of them by one color using patern
1 2 3
4 5 6
7 8 9
Two points of one color are then in distance lower or equal to $0.6\sqrt 2$ or greater then 1.2.
That was easy. Try it with six colours :)

what a coincidence. i just solved hits problem this morning. it is in the engel book

[quote="Kondr"]
Two points of one color are then in distance lower or equal to $0.6\sqrt 2$ or greater then 1.2.
[/quote]

How does the result follow?

And is it true for 6 colours?

[quote="indybar"]
How does the result follow?
[/quote]
If the two points of the same color are in the same square, their distance is lower then diagonal of this square, which is $0.6\sqrt 2$. If thez are in different squares of the same color, the smallest distance is obviously 2x0.6=1.2. If u don|t see it, just draw it.
[quote="indybar"]
And is it true for 6 colours?[/quote]
I'm sorry abot this case, I don't know how to solve it. But if you divide the plane into regular hexagons and color them with 7 colors, it's easy to show that for a suitible side lenght of the hexagons there are no two points in the distance 1inch of the same color.

So 4,5 and 6 colors are still a challange :)

[quote="Kondr"]So 4,5 and 6 colors are still a challange :)[/quote]

Indeed, not just a challenge, but an undecidable question, since the answer depends on the axiom of choice (and maybe on even more - the problem is not completely solved!). See http://www.math.princeton.edu/~bsudakov/soifer2003.pdf .

  Darij


hi , it's me again  :D

a,b,c are positif real numbers ,prove that :

$ \frac{a}{{\sqrt {b^2  \plus{} \frac{{bc}}{4} \plus{} c^2 } }} \plus{} \frac{b}{{\sqrt {a^2\plus{}\frac{{ac}}{4}\plus{}c^2 } }} \plus{} \frac{c}{{\sqrt {a^2  \plus{} \frac{{ab}}{4} \plus{} b^2 } }} \geq 2$

an elementary solution would be nice :D

[quote="anas"]hi , it's me again  :D

a,b,c are positif real numbers ,prove that :

$ \frac {a}{{\sqrt {b^2 \plus{} \frac {{bc}}{4} \plus{} c^2 } }} \plus{} \frac {b}{{\sqrt {a^2 \plus{} \frac {{ac}}{4} \plus{} c^2 } }} \plus{} \frac {c}{{\sqrt {a^2 \plus{} \frac {{ab}}{4} \plus{} b^2 } }} \geq 2$

an elementary solution would be nice :D[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183367
I think, by Holder it's elementary enough.  :wink:

I mean by elementary : (a-b)² \geq 0 ...  :D


trigonometry
geometry
incenter
geometric transformation
homothety
power of a point
geometry unsolved

Triangle $ABC$ is inscribed in circle $\omega$. Circle $\gamma$ is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively. Also circle $\gamma$ is tangent to circle $\omega$ at point $S$. Let the intesection of $AS$ and $PQ$ be $T$. Prove that $\angle{BTP}=\angle{CTQ}$.

Since $S$ is tangent point, so $\angle{RPS}=\angle{PBS}$. Also since $AB$ is tangent to the circle, so $\angle{PRS}=\angle{BPS}$. Therefore $\triangle{BPS} \sim \triangle{PRS}$. Similarly, we have $\angle{BSP}=\angle{ASP}$. Similarly we have: $\angle{CSQ}=\angle{ASQ}$. Thus:
\[ \frac{BP}{BS}=\frac{AP}{AS}=\frac{AQ}{AS}=\frac{CQ}{CS} \]
which implies $\frac{BP}{CQ}=\frac{BS}{CS}$.

Also, we have:
\[ \frac{PT}{QT}=\frac{\sin{\angle{BAS}}}{\sin{\angle{CAS}}}=\frac{BS}{CS} \]
togather with $\angle{BPT}=\angle{CQT}$, we conclude that $\triangle{BPT} \sim \triangle{CQT}$. Thus $\angle{BTP}=\angle{CTQ}$.

We can construct the common targent $ Sx$ of two circle to proof
We know that PQ pass through $ I$( $ I$ is the incenter of incircle $ ABC$).
Hence, $ SBPI$ and $ SCQI$ are cyclic quadrilater, we get the result

Dear Mathlinkers,
Circle  gamma is the A-mixtilinear incircle of ABC.
What can we say with the A-mixtilinear excircle of ABC?
Sincerely
Jean-Louis

Some other solutions: http://www.mathlinks.ro/viewtopic.php?t=140464  :D

[quote="jayme"]Dear Mathlinkers,
you can see my synthetic proof on my website
http://perso.orange.fr/jl.ayme   vol. 4  A new mixtilinear incircle adventure p. 56-58
Sincerely
Jean-Louis[/quote]
Few people understand the language. Can you post them here in English?

[hide="Solution"]Clearly, it suffices to prove that $\frac{TP}{TQ}=\frac{PB}{QC}$. Let $SP\cap \omega = X, SQ\cap \omega = Y, AS\cap \gamma = U$.

By Power of a Point, we have $AP\cdot PB = PX\cdot PS$ and $AQ\cdot QC=QY\cdot QS$, from which we obtain $\frac{PB}{QC}=\frac{PX\cdot PS}{QY\cdot QS}$.

On the other hand, since a homothety centered at $S$ takes $P, Q$ to $X, Y$ respectively, $\frac{PX\cdot PS}{QY\cdot QS}=\left(\frac{PS}{QS}\right)^2$.

Furthermore, since quadrilateral $UPSQ$ is harmonic, $\frac{PS}{QS}=\frac{UP}{UQ}$. Hence, it suffices to prove that $\frac{TP}{TQ}=\frac{PS\cdot PU}{QS\cdot QU}=\frac{[PSU]}{[QSU]}$, which is indeed true. We are done. $\blacksquare$[/hide]

Trivial from $S,I,P,B$ and $S,T,Q,C$ cyclic and $ST$ is symmedian in $SPQ$


linear algebra
matrix
limit
absolute value
linear algebra unsolved

related to a recent problem on the forum:
Find all complex matrix $B$ such that there exists a matrix $A \in M_n(\mathbb{C})$ satisfying $B= \lim \frac{A+A^2+A^3+..+A^n}{n}$

[quote="Diogene"]The solutions of $B^2=B$
 :cool:[/quote]

How do you prove that these are the only solutions ?

Can we say that:

$\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^{\infty}A^{i}}{n} = \lim_{n\rightarrow\infty}\frac{A*I}{n}+\lim_{n\rightarrow\infty}\frac{A*(\sum_{i=1}^{\infty}A^{i})}{n} = A*B$

so we need to solve : $A*B = B$

You're almost there: you have $AB=B$, so $\forall n\ge 1,\ A^nB=B$, which means that $\frac{A+A^2+\ldots+A^n}nB=B$. Taking limits on both sides, you get $B^2=B$, so this is indeed a necessary condition. Conversely, if $B^2=B$, you can take $A=B$.

Additional questions: what are those matrices for which the Cesaro average of $A^n$ is bounded, respectively convergent?

My previous "proof" was wrong. I think I have the answers now, in terms of the Jordan cells of $A$, but I'm not going to prove them (no time right now) :).

In order for the sequence of Cesaro means to be bounded, we need $A$ to have spectral radius $\le 1$, with the additional conditions that the Jordan cells corresponding to the eigenvalue $1$ (if there re any, of course) have dimension $1$, and those corresponding to other eigenvalues of absolute value $1$ have dimension $\le 2$. For convergence, we need all Jordan cells corresponding to eigenvalues of absolute value $1$ to have dimension $1$.

If it's still wrong, sorry :).

The spectral radius of a matrix is its greatest eigenvalue. (in absolute values)


2 playing cards, the 2 of spades and the 3 of clubs are put in a box. a card is drawn, value recorded, put back and then another card is drawn and recorded. what is the probabiility that the sum is greater than 4?
[Is it...] 2/3 [/hide]

a donut store has chocolate, plain, jelly, and powdered donuts. how many different ways can the bakery package 3 donuts to a pack?

[hide="1"]The sum won't be greater than 4 only if two 2 of spades are drawn. The probability that this happens is $ 1/4$, therefore probability that the sum is greater than 4 is $ 3/4$.[/hide]
[hide="2"]If only one of kind goes in package than it is $ \binom 4 3$ but if it can be more than one of a kind in a pack then let's make simple casework:

case 1: all 3 donuts are of the same kind - $ 4$ ways
case 2: two donuts are of one kind and third is of different kind - $ 4\cdot 3\equal{}12$ ways (4 to pick a kind of which two go in pack, and 3 to pick third kind)
case 3: all are different (we did this) - $ 4$ ways.

Total of $ 20$ ways.[/hide]

[quote="Flame"][hide="1"]The sum won't be greater than 4 only if two 2 of spades are drawn. The probability that this happens is $ 1/4$, therefore probability that the sum is greater than 4 is $ 3/4$.[/hide]
[hide="2"]If only one of kind goes in package than it is $ \binom 4 3$ but if it can be more than one of a kind in a pack then let's make simple casework:

case 1: all 3 donuts are of the same kind - $ 4$ ways
case 2: two donuts are of one kind and third is of different kind - $ 4\cdot 3 \equal{} 12$ ways (4 to pick a kind of which two go in pack, and 3 to pick third kind)
case 3: all are different (we did this) - $ 4$ ways.

Total of $ 20$ ways.[/hide][/quote]

for 1, how are there 4 total cases? i only have:
2 spades and 2 spades
2spades and 3 clubs
3 clubs and 3 clubs
Order doesnt matter, right?

for2, cant you just do 4*4*4? because there are no limits on donuts, and each spot can have any of the 4 donuts?

Order matters in 1. Easily proved by contradiction: there is $ 1/2$ chance to draw 2 of spades, therefore there is $ 1/2 \cdot 1/2 \equal{}1/4$ chance to draw two consecutive 2's of spades, but if order didn't matter then the probability would be $ 1/3$. Contradiction.

But order doesn't matter in 2, because for example pack of chocolate, plain and jelly is same as plain, jelly and chocolate. Therefore you are over-counting.


function
calculus
derivative
real analysis
real analysis unsolved

Suppose $ f$ is a function continous on $ [a,b]$ and derivable on $ (a,b)$ such that $ f(a) \equal{} f(b) \equal{} 0$
Prove that there exist $ x_{1},x_{2}\in (a,b)$ such that $ f(x_{1})\plus{}f(x_{2})\plus{}f'(x_{1})f'(x_{2}) \equal{} 0$

Any help appreciated!!

 :)

This appears to be false. Choose $ f$ so that $ f(x)>0$ for $ x\in(a,b).$ This can be done easily enough; for one concrete example, let $ f(x)\equal{}(x\minus{}a)(b\minus{}x).$

Then, for $ (x_{1},x_{2})\in(a,b)\times(a,b),$ the quantity $ f(x_{1})f(x_{2})\plus{}f(x_{1})\plus{}f(x_{2})$ is clearly positive.

Are you sure this is the question you meant to ask?

oh sorry...I ve just edited..I omitted the derivative symbol...

Sorry for the trouble..

A counterexample:

Let $ a \equal{}\minus{}3,b \equal{} 3$, and set $ f(x) \equal{}\begin{cases}\frac{1}{4}(x\plus{}3)^{2}&\minus{}3\le x\le\minus{}1\\ \frac{1}{2}(3\minus{}x^{2}) &\minus{}1 < x\le 1\\ \frac{1}{4}(3\minus{}x)^{2}& 1 < x\le 3\end{cases}$

This $ f$ has the property that $ |f'(x)|\le\sqrt{f(x)}$ everywhere, so $ |f'(x_{1})f'(x_{2})|\le\max(f(x_{1}),f(x_{2}))$.

[Edit- fixed a mistake from modifying my original example]

Now it is, and it's $ C^{1}$. The original version was on a broader interval, and I made a mistake modifying it.


[quote="white_horse_king88"]I don't know about others, but I find it difficult loading the site because it is so full of graphics. Weird - Latex is taking it's toll.[/quote]What do you mean?

Please paste your browser, system version and the debug line present in the footer of every page.

[quote="white_horse_king88"]Whoa. I'm not a computer expert like you man. I don't know what all that stuff is.[/quote]If you don't know what your browser and OS is, then it's 99% Windows + IE 6.0 :D

Good, now go down below anypage (especially the first page of the forums) and you will see a line with some numbers, like these:

[ Time: 0.2198s ][ Queries: 12 (0.0237s) ][ GZIP enabled - Debug on ]

Please paste it here (hit F5 - Refresh a couple of times before doing that).

[ Time: 0.7724s ][ Queries: 11 (0.1505s) ][ GZIP enabled - Debug on ]

That's not too bad. I guess. It just takes a long time to actually load pages with lots and lots of latex. This is the time for the main page.