geometry
incenter
circumcircle
angle bisector
geometry solved

A triangle ABC is inscribed into a circle with a center O. A circle (K) lies inside the circle
(O) and touches it at a point T. The circle (K) touches AC at F and AB at E. Let I be the
midpoint of EF, and let M be the midpoint of the arc BC on the circle (O) which doesn't
contain the point A. Let N be a point on the circle (O) such that MN is a diameter of (O).
Prove that N, I, T are collinear.

[color=red][Moderator edit: Clarified the wording of the problem.][/color]

[quote="ttd"]A triangle ABC is inscribed into a circle with a center O. A circle (K) lies inside the circle
(O) and touches it at a point T. The circle (K) touches AC at F and AB at E. Let I be the
midpoint of EF, and let M be the midpoint of the arc BC on the circle (O) which doesn't
contain the point A. Let N be a point on the circle (O) such that MN is a diameter of (O).
Prove that N, I, T are collinear.[/quote]

Equivalent versions of this problem have been discussed a number of times on MathLinks.

First, note that a chord intercepts two arcs on a given circle, and the midpoints of these two arcs are diametrically opposite to each other on the circle. Hence, if we take the midpoint of one of these arcs, and construct the point diametrically opposite to it, then this new point is the midpoint of the other arc. Particularly, the chord BC intercepts two arcs on the circle (O): the one containing the point A and the one not containing the point A. Since the point M is the midpoint of the latter one, the point N, lying diametrically opposite to the point M on the circle (O), must therefore be the midpoint of the former one. In other words, the point N is the midpoint of the arc BAC on the circle (O).

The next thing is that the point I, being the midpoint of the segment EF, is the incenter of triangle ABC. This is a rather hard-to-prove, but classical fact; it was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=31739 , http://www.mathlinks.ro/Forum/viewtopic.php?t=42396 , http://www.mathlinks.ro/Forum/viewtopic.php?t=1314 and in some other places (by the way, in http://www.mathlinks.ro/Forum/viewtopic.php?t=15664 and http://www.mathlinks.ro/Forum/viewtopic.php?t=16171 , the corresponding fact was proven for the circle touching the sides AC and AB and the circumcircle of triangle ABC [i]externally[/i]; for this circle, the midpoint of the segment corresponding to EF will be not the incenter, but the A-excenter of triangle ABC).

Thus, the problem becomes equivalent to the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=5131 .

If we want, we can also reduce the problem to the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=22925 . In fact, that problem states that the line TI is the angle bisector of the angle BTC. But the angle bisector of an angle of a triangle bisects the arc intercepted on the circumcircle by the opposite side; the circumcircle of triangle BTC is the circle (O), and the arc intercepted by the side BC is the arc BAC. Hence, the line TI, being the angle bisector of the angle BTC, must bisect the arc BAC, i. e. it must pass through the midpoint N of the arc BAC. Thus, the points N, I, T are collinear.

  Darij


Find all the $ a_1,a_2,a_3,...,a_n$ such that :
$ \sqrt{a_1 }\plus{}\sqrt{a_2\minus{}1}\plus{}\sqrt{a_3\minus{}2}\plus{}...\plus{}\sqrt{a_n\minus{}(n\minus{}1)}\equal{}\frac{1}{2}(a_1\plus{}a_2\plus{}...\plus{}a_n)\minus{}\frac{n(n\minus{}3)}{4}$

[hide="Solution"]Lets make the substituion:

$ \sqrt{a_1}\equal{}b_1\\
\sqrt{a_2\minus{}1}\equal{}b_2\\
\sqrt{a_3\minus{}2}\equal{}b_3\\
\vdots\\
\sqrt{a_n\minus{}(n\minus{}1)}\equal{}b_n$

So we have:

$ \sum_{i\equal{}1}^{n}b_i\equal{}\frac{1}{2}\left(b_1^2\plus{}b_2^2\plus{}1\plus{}b_3^2\plus{}2\plus{}\ldots\plus{}b_n^2\plus{}n\minus{}1 \right)\minus{}\frac{n(n\minus{}3)}{4}$

$ 2\cdot \sum_{i\equal{}1}^{n}b_i\minus{}\sum_{i\equal{}1}^{n}b_i^2\equal{}\frac{n(n\minus{}1)\minus{}n(n\minus{}3)}{2}\equal{}n$

$ \sum_{i\equal{}1}^{n}b_i^2\minus{}2\cdot \sum_{i\equal{}1}^{n}b_i\plus{}\sum_{i\equal{}1}^{n}1\equal{}0$

$ \sum_{i\equal{}1}^{n}(b_i\minus{}1)^2\equal{}0$

Which implies:

$ b_1\minus{}1\equal{}\sqrt{a_1}\minus{}1\equal{}0\,\therefore\,a_1\equal{}1\\
b_2\minus{}1\equal{}\sqrt{a_2\minus{}1}\minus{}1\equal{}0\,\therefore\,a_2\equal{}2\\
b_3\minus{}1\equal{}\sqrt{a_3\minus{}2}\minus{}1\equal{}0\,\therefore\,a_3\equal{}3\\
\vdots$

So we have the sequence defined as $ a_n\equal{}n$[/hide]

Good !

Try this one now :

Find the positive numbers $ a_1,a_2,...,a_n$ that satisfies :

$ a_n\plus{}2a_{n\minus{}1}\plus{}3a_{n\minus{}2}\plus{}...\plus{}na_1\equal{}\frac{n(n\plus{}1)(n\plus{}2)}{6}\ ,\ (\forall)n\in \mathbb{N}^*$ .

[quote="alex2008"]Good !

Try this one now :

Find the positive numbers $ a_1,a_2,...,a_n$ that satisfies :

$ a_n \plus{} 2a_{n \minus{} 1} \plus{} 3a_{n \minus{} 2} \plus{} ... \plus{} na_1 \equal{} \frac {n(n \plus{} 1)(n \plus{} 2)}{6}\ ,\ (\forall)n\in \mathbb{N}^*$ .[/quote]
nice problem, Alex  :)
We have:
$ a_n \plus{} 2a_{n \minus{} 1} \plus{} 3a_{n \minus{} 2} \plus{} ... \plus{} na_1 \equal{} \frac {n(n \plus{} 1)(n \plus{} 2)}{6}$
$ \Leftrightarrow \sum_{k\equal{}1}^{n} a_k\plus{}\left(a_{n\minus{}1}\plus{}2a_{n\minus{}2}\plus{}....\plus{}(n\minus{}1)a_1\right)\equal{}\frac {n(n \plus{} 1)(n \plus{} 2)}{6}$
$ \Leftrightarrow \sum_{k\equal{}1}^{n} a_k\plus{}\frac {(n \minus{} 1)n(n \plus{} 1)}{6}\equal{}\frac {n(n \plus{} 1)(n \plus{} 2)}{6}$

$ \Leftrightarrow \sum_{k\equal{}1}^{n} a_k\equal{}\frac {n(n \plus{} 1)}{2}$

If $ n\equal{}i$ ,:$ \sum_{k\equal{}1}^{i} a_k\equal{}\frac {i(i \plus{} 1)}{2} \text{        (1)}$
If $ n\equal{}i\plus{}1$ ,:$ \sum_{k\equal{}1}^{i\plus{}1} a_k\equal{}\frac {(i \plus{} 1)(i\plus{}2)}{2}       \text{                 (2)}$
From (1) and (2): $ a_{i\plus{}1}\equal{}\sum_{k\equal{}1}^{i\plus{}1} a_k\minus{}\sum_{k\equal{}1}^{i} a_k\equal{}\frac {(i \plus{} 1)(i\plus{}2)}{2} \minus{}\frac {i(i \plus{} 1)}{2}\equal{}i\plus{}1$
So, $ a_n\equal{}n$


Find all polynomial $ P(x)$ with real coefficients satisfying $$ P^2(x)\plus{}P^2(x\minus{}1)\equal{}2(P(x)\minus{}x)^2.$$
Note that by $ P^2(x)$ we mean $P(x)\cdot P(x)$.

[quote="SUPERMAN2"]Find all polynomial $ P(x)$ with real coefficients satisfying $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$
Note:$ P^2(x) \equal{} P(x)P(x)$[/quote]

We can write this equality as $ P^2(x)\minus{}(P(x)\minus{}x)^2\equal{}(P(x)\minus{}x)^2\minus{}P^2(x\minus{}1)$ and so :

$ \boxed{x(2P(x)\minus{}x)\equal{}(P(x)\plus{}P(x\minus{}1)\minus{}x)(P(x)\minus{}P(x\minus{}1)\minus{}x)}$

If the degree of $ P(x)$ is $ n>1$, LHS has degree $ n\plus{}1$, but $ P(x)\plus{}P(x\minus{}1)\minus{}x$ has degree $ n$ and so $ P(x)\minus{}P(x\minus{}1)\minus{}x$ has degree $ 1$ and so $ P(x)$ has degree 2. So $ P(x)$ has degree $ 0,1$ or $ 2$.

Let then $ P(x)\equal{}ax^2\plus{}bx\plus{}c$ and so $ P(x\minus{}1)\equal{}ax^2\plus{}(b\minus{}2a)x\plus{}a\minus{}b\plus{}c$. The equation becomes :

$ x(2ax^2\plus{}(2b\minus{}1)x\plus{}2c)\equal{}(2ax^2\plus{}(2b\minus{}2a\minus{}1)x\plus{}a\minus{}b\plus{}2c)((2a\minus{}1)x\minus{}a\plus{}b)$

Equality between powers of $ x^3$ is $ 2a\equal{}2a(2a\minus{}1)$ and so $ a\equal{}0$ or $ a\equal{}1$

$ a\equal{}0$ $ \implies$ $ x((2b\minus{}1)x\plus{}2c)\equal{}((2b\minus{}1)x\minus{}b\plus{}2c)(b\minus{}x)$. Comparing $ x^2$ terms implies $ b\equal{}\frac{1}{2}$ and then $ 2cx\equal{}(\minus{}\frac{1}{2}\plus{}2c)(\frac{1}{2}\minus{}x)$ which is impossible.

So $ a\equal{}1$ and we have : $ x(2x^2\plus{}(2b\minus{}1)x\plus{}2c)\equal{}(2x^2\plus{}(2b\minus{}3)x\plus{}1\minus{}b\plus{}2c)(x\minus{}1\plus{}b)$ and so :

$ (2b\minus{}1)x^2\plus{}2cx\equal{}(4b\minus{}5)x^2\plus{}(4\plus{}2c\plus{}2b^2\minus{}6b)x  \plus{}(b\minus{}1) (1\minus{}b\plus{}2c)$ and so :

$ (2b\minus{}4)x^2\plus{}(4\plus{}2b^2\minus{}6b)x  \plus{}(b\minus{}1) (1\minus{}b\plus{}2c) \equal{} 0$ $ \implies$ $ b\equal{}2$ and $ c\equal{}\frac{1}{2}$

And so the only polynomial such that $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$ is $ \boxed{P(x)\equal{}x^2\plus{}x\plus{}\frac{1}{2}}$

[quote="pco"][quote="SUPERMAN2"]Find all polynomial $ P(x)$ with real coefficients satisfying $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$
Note:$ P^2(x) \equal{} P(x)P(x)$[/quote]

And so the only polynomial such that $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$ is $ \boxed{P(x) \equal{} x^2 \plus{} x \plus{} \frac {1}{2}}$[/quote]

[b]Solution:[/b]

The condition $ \Leftrightarrow P^2(x) \minus{} 4xP(x) \plus{} 2x^2 \equal{} P^2(x \minus{} 1)$
$ \Leftrightarrow (P(x) \plus{} P(x \minus{} 1) \minus{} 2x)(P(x) \minus{} P(x \minus{} 1) \minus{} 2x) \equal{} 2x^2$

So $ P(x)$ is of degree $ 0,1,2$

[b]Case 1:[/b] $ P(x)$ is of degree $ 0,1$ and it's easy to check that there are no solutions in this case.

[b]Case 2:[/b] $ P(x)$ is of degree $ 2$
Let $ P(x) \equal{} ax^2 \plus{} bx \plus{} c$, then

$ P(x) \minus{} P(x \minus{} 1) \minus{} 2x \equal{} a(2x \minus{} 1) \plus{} b \minus{} 2x$ must be of degree $ 0$, so $ a \equal{} 1$

Now $ L.H.S. \equal{} (b \minus{} 1)(P(x) \plus{} P(x \minus{} 1) \minus{} 2x) \equal{} (b \minus{} 1)(2x^2 \plus{} (2b \minus{} 4)x \plus{} (2c \minus{} b \plus{} 1)) \equal{} 2x^2$

Coefficient of $ x$ is $ 0$ $ \Rightarrow 2b \minus{} 4 \equal{} 0 \Rightarrow P(x) \equal{} x^2 \plus{} 2x \plus{} \frac {1}{2}$

Sorry, pco . Also the mistake is now corrected now.

[quote="pco"] ... $ \implies$ $ b \equal{} 2$ and $ c \equal{} \frac {1}{2}$

And so the only polynomial such that $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$ is $ \boxed{P(x) \equal{} x^2 \plus{} x \plus{} \frac {1}{2}}$[/quote]

Just a typo error. $ b \equal{} 2$ so :

the only polynomial such that $ P^2(x) \plus{} P^2(x \minus{} 1) \equal{} 2(P(x) \minus{} x)^2$ is $ \boxed{P(x) \equal{} x^2 \plus{} 2x \plus{} \frac {1}{2}}$

And I checked it ... twice :)

And your own error is here :
[quote="stephencheng"] ... Now $ L.H.S. \equal{} (b \minus{} 1)(P(x) \plus{} P(x \minus{} 1) \minus{} 2x) \equal{} (b \minus{} 1)(2x^2 \plus{} (2b \minus{} 2)x \plus{} (2c \minus{} b \plus{} 1)) \equal{} 2x^2$ [/quote]

Coefficient of $ x$ in LHS is $ 2b\minus{}4$ and not $ 2b\minus{}2$


trigonometry
inequalities
quadratics
function
logarithms
number theory
modular arithmetic

[quote="K81o7"][hide="methods"]
2) Bounding yields $x=2$ (you could even use modular arithmetic to get $x \equiv 2\mod{4}$ first)

4) It is equivalent to $\sin{x}+\cos{x}=0$, which holds for $x=\frac{3\pi}{4}+k\pi$ for the different values of $k$ which put $x$ in the given interval (assuming radians)

5)Solution only when $x=2k\pi, 2k\pi+\frac{\pi}{2}$, by the triangle inequality

6)Set $k=\sqrt{x^{2}+3x}$ an use the quadratic equation.

7) $=\frac{x^{10}-1}{x-1}$, so solutions are all 10th root of unity except 1.

[/hide][/quote]

For 2, you aren't allowed to suppose that $x\in\mathbb{N}$. However, considering the strictly decreasing function $f: \mathbb{R}\to\mathbb{R}: x\mapsto \left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$ should work. Also, number 5 is easily solved by squaring both sides.

[hide="Method for 3"]
Remark (by dividing by $x^{4}$) that $\zeta$ is a solution to the equation if and only if $\zeta^{-1}$ is a solution. Therefore, $2x^{4}+5x^{3}-12x^{2}+5x+2 = 2(x^{2}+ax+1)(x^{2}+bx+1)$ for some $a,b$. Expand and solve for $a$ and $b$. Then solve for $x$.[/hide]

[hide="Hint"] [quote="giannis"][i][color=red]Solve the equations[/color][/i]
$2x^{4}+5x^{3}-12x^{2}+5x+2=0$
[/quote]
$\Longleftrightarrow 2\left(x+\frac{1}{x}\right)^{2}+5\left(x+\frac{1}{x}\right)-16=0$

[quote="giannis"][i][color=red]Solve the equations[/color][/i]
$\left(\sqrt{25+2\sqrt 6}\right)^{x}+\left(\sqrt{25-2\sqrt 6}\right)^{x}=10$
[/quote]
Let $a=\left(\sqrt{25+2\sqrt 6}\right)^{x}\Longrightarrow \left(\sqrt{25-2\sqrt 6}\right)^{x}=\frac{1}{a}$
And we have to solve the equation: $a+\frac{1}{a}=10$[/hide]

[hide="8)"]
\[\iff 2^{(x-1)^{2}}3^{(y-2)^{2}}4^{(z-3)^{2}}=1 \]
Now $2^{(x-1)^{2}}\geq 1, 3^{(y-2)^{2}}\geq 1, 4^{(z-3)^{2}}\geq 1$ by the trivial inequality so that the above equality occurs only when there is equality in each of these three inequalities, i.e. $(x,y,z)=(1,2,3)$
[/hide]
edit
[hide="9)"]
Solutions observed at $x=0,1$. Consider $f(x)=2^{x}+5^{x}-3^{x}-4^{x}$. Note $f'(x)=\ln 2 \cdot 2^{x}+\ln 5 \cdot 5^{x}-\ln 3 \cdot 3^{x}-\ln 4 \cdot 4^{x}$
\[\iff \frac{5^{x}-4^{x}}{5-4}=\frac{3^{x}-2^{x}}{3-2}\]
Now holds $x$ constant with $x\not =0,1$. Then considering the function $f(y)=y^{x}$ we determine that $LHS\geq f'(4)>f'(3)\geq RHS$ or the exact opposite depending on $x$ so that there is no solution. Thus $x=0,1$ is the solution set.
[/hide]

Continue:
1)Solve the equation:
$\frac{1}{(2x-1)^{2}}+\frac{1}{(3x+1)^{2}}= \frac{5}{4(x+2)^{2}}$
2)Find all positive real roots of the following equation:
$5xyz=x+5y+7z+10$

i have an idea for 2)
case 1 : $x,y,z \geq 2$
we obtain $\frac{xyz}{4}\geq x ; \frac{5xyz}{4}\geq 5y ; \frac{7xyz}{4}\geq 7z ; \frac{7xyz}{4}>10$
by adding up those inequalities , we get a contradiction
i think we can easily solve the case 2 , if one of ${x;y;z}=1$

[hide]i dont think 1) can be done in some traditional way :D [/hide]

[quote="HTA"]i have an idea for 2)
case 1 : $x,y,z \geq 2$
we obtain $\frac{xyz}{4}\geq x ; \frac{5xyz}{4}\geq 5y ; \frac{7xyz}{4}\geq 7z ; \frac{7xyz}{4}>10$
by adding up those inequalities , we get a contracdition
i think we can easily solve the case 2 , if one of ${x;y;z}=1$[/quote]
Right :)
But I know there is another solution for this problem, try to find it :D

[hide]About problem 1, if you solve it it the traditional way, you will find it hard :P
Let's think of other ways to solve it (ex. substitution, ...) [/hide]

[hide="my ugly solution"]with some caculations , we get the final equation:
$128x^{4}-276x^{3}-303x^{2}-54x-27=0$
which can be rewrite $(x-3)(4x+3)(32x^{2}+3x+3)=0$
so $x=3;\frac{-3}{4}$ checks :D[/hide]

[hide="1 Partial Solution  :maybe: "]
Let $a=3x+1$ and $b=x+2$.
Getting $4a^{2}(b)^{2}+4b^{2}(a-b)^{2}= 5a^{2}(a-b)^{2}$ are some manipulation.

[/hide]

Can we post here our favourite equations . Thanks :roll:

[quote="bbbbbbbbbb"]$2^{x}+5^{x}=3^{x}+4^{x}$
$f(t)=t^{a}-(t-1)^{a}$
$f(5)=f(3)$
We use Larrange  :P
x=0 or x=1  :lol:[/quote] Can somebody explain to me what Larrange is?

[quote="tunganh"][quote="K81o7"]
7) $=\frac{x^{10}-1}{x-1}$, so solutions are all 10th root of unity except 1.
[/quote]
Or you can rewrite the equation:
$1+x+...+x^{9}= (x+1)(x^{8}+x^{6}+x^{4}+x^{2}+1)$
Notice that $x^{4}+x^{2}+1>0$ so that $x^{8}+x^{6}+x^{4}+x^{2}+1>0$
$\Rightarrow x=-1$[/quote]

True only if you restrict $x$ to reals.  You miss $8$ complex solutions.

[quote="t0rajir0u"]
True only if you restrict $x$ to reals.  You miss $8$ complex solutions.[/quote]

Oh, I think we should work with real numbers :P

[hide]About my first equation, we denote $a=3x+1, b=2x-1$ then $x+2=a-b$
After reducing to the same denominator, we have to solve the equation:
$4a^{4}+8a^{3}b+3a^{2}b^{2}+8ab^{3}+4b^{4}$
But this one is trivial (like the third equation of giannis)[/hide]

[hide]the following equation has real roots if and only if , $\bar{abc}\in(691;661;373;571;761;283;293;179;199;167;383;281;193;191;197;163;173;263;271;881;251;131;593;151;181;241)$
but from any values above , no values can give us the rational roots , so we get  QED[/hide]

[quote="tunganh"]Prove that the following equation have no rational roots:
$ax^{2}+bx+c$
for all 3-digits prime number $\bar{abc}$[/quote]
[hide]
If there were a rational root $m/n$ with $m,n$ relatively prime integers and $n>0$, then $m<0$ (the root can't be nonnegative) and $nx-m$ would divide $ax^{2}+bx+c$. In particular, $10n+m$ divides $100a+10b+c$. But we're given that this is a prime $p$. Therefore, one of the following must occur:
\[10n-m=\pm 1, 10n-m=\pm p \]
But also $m$ divides $c$ so $m\geq-9$, also $n$ divides $a$ so $n\leq 9$, by Rational Root Theorem. This gives $11\leq 10n-m\leq 99$. CONTRADICTION!
[/hide]


inequalities
function
algebra
IMO Shortlist

Prove the inequality:
\[\sum_{i < j}{\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}}}\leq \frac {n}{2(a_{1} \plus{} a_{2} \plus{}\cdots \plus{} a_{n})}\cdot \sum_{i < j}{a_{i}a_{j}}\]
for positive reals $ a_{1},a_{2},\ldots,a_{n}$.

[i]Proposed by Dusan Dukic, Serbia[/i]

Multiply both sides by $ 2(a_{1}+a_{2}+...+a_{n})=2S$ and see that:
$ \sum \frac{2a_{i}a_{j}S}{a_{i}+a_{j}}= \sum \frac{2a_{i}a_{j}(a_{i}+a_{j})}{a_{i}+a_{j}}+\sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}= 2 \sum a_{i}a_{j}+\sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}$.
So we want to prove that:
$ \sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}\leq (n-2) \sum a_{i}a_{j}$.
But for $ i \not = j \not = k \not = i$ we have $ \frac{2a_{i}a_{j}a_{k}}{a_{i}+a_{j}}\leq \frac{1}{2}(a_{i}a_{k}+a_{j}a_{k})$ since we can divide by $ a_{k}$ and transform it to $ (a_{i}-a_{j})^{2}\geq 0$. So after writing such inequalities and summing them up we get the desired inequality.

The inequality is equivalent to $ \sum_{i < j}\frac{(a_{i}-a_{j})^{2}}{a_{i}+a_{j}}\geq \frac{\sum_{i < j}(a_{i}-a_{j})^{2}}{\sum a_{i}}$ and the result follows...

RDeepMath91 > Could you show me more precisly how did you get this inequality's form please

"And the result follows" : please, explain more, I don't see how.

[hide="Solution"]
Note: For any function, $ f$, when we say "$ \sum f(a_i)$", we mean "$ \sum_{i \equal{} 1}^n f(a_i)$".  Furthermore, whenever we say "$ \sum f(a_i, a_j)$", we mean "$ \sum_{1\le i < j \le n} f(a_i, a_j)$".

Notice that $ \sum (a_i \plus{} a_j) \equal{} (n \minus{} 1)\sum a_i$ since $ a_i$ is counted once for each $ a_j$ with $ j\ne i$, implying that $ a_i$ is counted $ n \minus{} 1$ times for each $ i$.  Furthermore, $ (n \minus{} 1)\sum a_i^2 \plus{} 2\sum a_ia_j \equal{} \sum (a_i^2 \plus{} a_j^2 \minus{} 2a_i a_j) \equal{} \sum (a_i \minus{} a_j)^2$.  Hence,\[ \sum \frac {a_ia_j}{a_i \plus{} a_j} \le (\frac {1}{\sum a_i})(\frac {n}{2})\sum a_ia_j\iff\]\[ \frac {\sum (a_i \plus{} a_j)}{4} \minus{} \sum \frac {a_ia_j}{a_i \plus{} a_j} \ge \frac {n \minus{} 1}{4} \cdot \sum a_i \minus{} \frac {n}{2}\cdot \frac {1}{\sum a_i}\cdot \sum a_ia_j\iff\]\[ \sum \frac {(a_i \minus{} a_j)^2}{a_ia_j}\ge \frac {(n \minus{} 1)\cdot (\sum a_i)^2 \minus{} 2n\cdot \sum a_ia_j}{\sum a_i}\]\[ \equal{} \frac {(n \minus{} 1)(\sum a_i^2 \plus{} 2\sum a_ia_j) \minus{} 2n\sum a_ia_j}{\sum a_i}\]\[ \equal{} \frac {(n \minus{} 1)\sum a_i^2 \minus{} 2\sum a_ia_j}{\sum a_i}\]\[ \equal{} \frac {\sum (a_i \minus{} a_j)^2}{\sum a_i}\]which is true since $ \sum_{k \equal{} 1}^n a_k > a_i \plus{} a_j$ for all pairs $ (i, j)$. [/hide]

[quote="The QuattoMaster 6000"][hide="Solution"]

 Furthermore, $ (n \minus{} 1)\sum a_i^2- 2\sum a_ia_j \equal{} \sum (a_i^2 \plus{} a_j^2 \minus{} 2a_i a_j) \equal{} \sum (a_i \minus{} a_j)^2$.

****
Hence,\[ \sum \frac {a_ia_j}{a_i \plus{} a_j} \le (\frac {1}{\sum a_i})(\frac {n}{2})\sum a_ia_j\iff\] [/hide][/quote]

I think the first point has to be changed and how did you prove this second one?

Use the identity $\dfrac {2a_ia_j} {a_i + a_j} = a_i + a_j - \dfrac {a_i^2 + a_j^2} {a_i + a_j}$ to obtain

\begin{align*}
\sum a_ia_j &\ge \dfrac {\sum a_i} {n} \sum \dfrac {2a_ia_j} {a_i + a_j} \iff\\
\sum a_ia_j &\ge \dfrac {\sum a_i} {n} \left (\sum a_i + a_j - \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} \right) \iff\\
\sum a_ia_j + \dfrac {\sum a_i} {n}\sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} &\ge \dfrac {\sum a_i} {n} (n - 1) (\sum a_i) \iff \\
\dfrac {\sum a_i} {n} \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge \dfrac {n - 1} {n} (\sum a_i^2 + 2\sum a_i a_j) - \sum a_i a_j \iff \\
\sum a_i \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge (n - 1) \sum a_i^2 + (n - 2) \sum a_i a_j \iff \\
(n - 1)\sum a_i^2 + \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} &\ge (n - 1) \sum a_i^2 + (n - 2) \sum a_i a_j \iff \\
\sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge (n - 2) \sum a_i a_j
\end{align*}

Since AM-GM gives us $\dfrac {a_i^2 + a_j^2} {a_i + a_j} \ge \dfrac {a_i + a_j} {2}$, we have
\begin{align*}
\sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} \ge \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i + a_j} {2} = (n - 2) \sum a_i a_j
\end{align*}
as desired.

[hide="Solution"]Let $S_k=\sum_{i=1}^ka_i$ and $T_k=\sum_{i<j\le k}a_ia_j$.  We can show the following lemma:

Lemma: \[\frac{k-1}{2(k-2)}\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}\le \frac{k}{2}\cdot \frac{T_k}{S_k}\] for all $k\ge 3$.

[hide="Proof"] If \[\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}\ge \frac{T_k-S_k(a_j)+a_j^2}{S_k-a_j}\]then
\begin{align*}
(T_k-S_k(a_i)+a_i^2)(S_k-a_j)&\ge (T_k-S_k(a_j)+a_j^2)(S_k-a_i)\\
(a_j-a_i)(S_k^2-T_k+a_ia_j-S_ka_i-S_ka_j)&\ge 0
\end{align*}
Notice that
\[S_k^2-T_k+a_ia_j-S_ka_i-S_ka_j=\sum_{p\le q; p,q\neq i,j}a_pa_q>0\]
Thus, $a_j\ge a_i$; so $S_k-a_i\ge S_k-a_j$.  Therefore, by Chebyshev's inequality
\begin{align*}\frac{1}{k}\left(\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}\right)\frac{1}{k}\left(\sum_{i=1}^kS_k-a_i\right)&\le \frac{1}{k}\left(\sum_{i=1}^kT_k-S_k(a_i)+a_i^2\right)\\
\left(\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}\right)(k-1)S_k&\le k\left(kT_k-S_k^2+\sum_{i=1}^ka_i^2\right)\\
&=k(k-2)T_k\\
\left(\frac{k-1}{2(k-2)}\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}\right)&\le \frac{k}{2}\cdot \frac{T_k}{S_k}\end{align*}
as desired. $\boxed{}$
[/hide]
We will now prove the problem statement by induction of $n$.  If $n=2$, the statement is an equality.  If the inequality is true when $n\le k-1$.  Then notice that
\begin{align*}
\frac{k}{2(a_1+a_2+\cdots +a_k)}\sum_{i<j}a_ia_j&=\frac{k}{2}\cdot \frac{T_k}{S_k}
&\ge \frac{k-1}{2(k-2)}\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}
\end{align*}
Notice that $T_k-S_k(a_i)+a_i^2=\sum_{p<q; p,q\neq i}a_pa_q$.  Hence, we can use the induction hypothesis of $k-1$ to show
\begin{align*}
\frac{k-1}{2(k-2)}\sum_{i=1}^k\frac{T_k-S_k(a_i)+a_i^2}{S_k-a_i}&\ge \frac{1}{(k-2)}\sum_{i=1}^k\sum_{p<q;p,q\neq i}\frac{a_pa_q}{a_p+a_q}\\
&=\frac{1}{(k-2)}\sum_{p<q}\frac{(k-2)a_pa_q}{a_p+a_q}\\
&=\sum_{p<q}\frac{a_pa_q}{a_p+a_q}
\end{align*}
as desired.[/hide]

[quote="TheBernuli"]The inequality is equivalent with:
$ \sum_{i < j}{\frac{a_{i}a_{j}}{a_{i}+a_{j}}}((n-1)(a_{1}+a_{2}+\cdots+a_{n}))\leq\frac{n(n-1)}{2}\cdot\sum_{i < j}{a_{i}a_{j}}$. Let $x_{ij}=\frac{a_{i}a_{j}}{a_{i}+a_{j}}$ and $y_{ij}=a_i+a_j$.
Now we apply Chebyshev's inequality to numbers $x_{ij}$ and $y_{ij}$ and we will get inequality on the beginning.[/quote]

I had the same idea but can you please elaborate how you ordered the sequences ? I guess the ordering of x_{ij}'s and y_{ij} 's can not be uniquely ordered if we ONLY assume  a certain ordering of variables a_1,...a_n .
So , in that case , in case of some variables a_1,a_2,...,a_n , Can you please show that x_{ij} and y_{ij} are similarly sorted. ?  Plz elaborate it if possible :)

My solution:
Let $x_{ij}=a_ia_j$ and $y_{ij}=a_i+a_j$ and $S=\sum a_i$
\begin{align*}
RHS-LHS&=\sum_{i<j}\frac{nx_{ij}}{2S}- \frac{x_{ij}}{y_{ij}}
\\ &=\sum_{i<j} x_{ij} . \frac{\sum_{k=1}^n a_i-a_k+a_j-a_k}{2.S.y_{ij}}
\\ &=\sum_{i<k} (a_i-a_k).\sum_{j\not\in \lbrace i,k \rbrace }( \frac{x_{ij}}{2S.y_{ij}}-\frac{x_{jk}}{2Sy_{jk}})
\\ &= \sum_{i<k} \frac{a_i-a_k}{2.S}.\sum_{j\not\in \lbrace i,k \rbrace } a_j(\frac{a_i}{y_{ij}}-\frac{a_k}{y_{jk}})
\\ &= \sum_{i<k} \frac{(a_i-a_k)^2}{2.S}.\sum_{j\not\in \lbrace i,k \rbrace } a_j^2 \geq 0
\end{align*}
QED .$ \blacksquare$

Hello:

My solution:

We will use induction on $n$.For $n=2$ the inequality is true (as an equality).Suppose that for $n=k$ the inequality is also true.

Now,consider a $k+1$ positive reals $a_1,a_2,...,a_k$.Since the inequality is symmetric,we can suppose that $a_1\leq a_2\leq \ldots a_{k+1}$.

We have to show that $\left(\sum_{1\leq i<j\leq k+1} a_ia_j\right)\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \sum_{1\leq i<j\leq k+1} \frac{a_ia_j}{a_i+a_j} \ (\bigstar)$.

However,we have the following equalities:

$\bullet$ $\sum_{1\leq i<j\leq k+1} a_ia_j=\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i$.

$\bullet$ $\sum_{1\leq i<j\leq k+1} \frac{a_ia_j}{a_i+a_j}=\left(\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\right)+a_{k+1}\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}}$.

Thus,$(\bigstar)$ becomes $\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \left(\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\right)+a_{k+1}\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}} \ (1)$

We know (from our inductive hypothesis) that $\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\leq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)$,

thus,instead of $(1)$,it suffices to show that

$\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1} \frac{a_i}{a_i+a_{k+1}}$.

The function $f(x)=\frac{x}{x+a_{k+1}}$ is concave,thus,from Jensen we obtain $\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}}\leq \frac{k\sum_{i=1}^{k}a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}$.Thus,it suffices to show that

$\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\frac{k\sum_{i=1}^{k}a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}$

which is equivalent to $\frac{1}{2}\left(\sum_{1\leq i<j\leq k} a_ia_j\right)\left(\frac{k+1}{\sum_{i=1}^{k+1} a_i}-\frac{k}{\sum_{i=1}^{k} a_i}\right)\geq a_{k+1}\sum_{i=1}^{k} a_i\left(\frac{k}{ka_{k+1}+\sum_{i=1}^{k} a_i}-\frac{(k+1)}{2\sum_{i=1}^{k+1} a_i}\right)$

$\Leftrightarrow \frac{1}{2}\left(\sum_{1\leq i<j\leq k} a_ia_j\right)\frac{\sum_{i=1}^{k}a_i-ka_{k+1}}{\left(\sum_{i=1}^{k+1}a_i\right)\left(\sum_{i=1}^{k} a_i\right)}\geq a_{k+1}(k-1)\sum_{i=1}^{k}a_i\cdot \frac{\sum_{i=1}^{k} a_i-ka_{k+1}}{2\left(\sum_{i=1}^{k+1}a_i\right)\left(ka_{k+1}+\sum_{i=1}^{k} a_i\right)}$.

Since $a_{k+1}=\max \bigcup_{i=1}^{k+1} a_i$ we get $ka_{k+1}\geq \sum_{i=1}^{k} a_i$ thus,the latter inequality is equivalent to

$\frac{\left(\sum_{1\leq i<j\leq k} a_ia_j\right)}{\sum_{i=1}^{k}a_i}\leq \frac{a_{k+1}(k-1)\sum_{i=1}^{k} a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}\Leftrightarrow \sum_{1\leq i<j\leq k}a_ia_j\leq \frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{k+\frac{1}{a_{k+1}}\cdot \sum_{i=1}^{k} a_i}$.

Since $a_{k+1}=\max \bigcup_{i=1}^{k+1} a_i$ we get $\frac{\sum_{i=1}^{k}a_i}{a_{k+1}}\leq k$ thus

$\frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{k+\frac{1}{a_{k+1}}\cdot \sum_{i=1}^{k} a_i}\geq \frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{2k}$.

Hence,it suffices to show that $\left(\sum_{i=1}^{k}a_i\right)^2\geq \frac{2k}{k-1}\sum_{1\leq i<j\leq k} a_ia_j\Leftrightarrow \sum_{i=1}^k a_i^2\geq \frac{2}{k-1}\sum_{1\leq i<j\leq k} a_ia_j$.

However,from AM-GM we get $\sum_{1\leq i<j\leq k} a_i^2+a_j^2\geq 2\sum_{1\leq i<j\leq k} a_ia_j$.Each $a_i$ appears in $k-1$ pairs,thus

$LHS=(k-1)\sum_{i=1}^{k} a_i^2$ thus $(k-1)\sum_{i=1}^{k} a_i^2\geq 2\sum_{1\leq i<j\leq k}a_ia_j$ q.e.d.

[quote=Modesti]By Titu's inequality and Cauchy-Schwarz,

$$\sum_{i<j}\dfrac{a_ia_j}{a_i+a_j}\le \dfrac{(\sum_{i<j}\sqrt{a_ia_j})^2}{2(a_1+a_2+...+a_n)}\le\dfrac{n}{2(a_1+a_2+...+a_n)}\sum_{i<j}a_ia_j$$

and we are done. $\square$[/quote]
Check please your first step.

You are right, I reversed the sign. Thank you. Will delete the wrong solution.

We will prove the inequality with both sides multiplied by $\sum_{i=1}^n a_i:$ $$\sum_{i<j} \left( \frac{a_ia_j}{a_i+a_j} \right) \cdot \sum_{i=1}^n a_i=\sum_{i<j}  a_ia_j +\sum_{i<j} \frac{a_ia_j \sum_{k\neq i,j} a_k}{a_i+a_j} \stackrel{\text{AM-HM}}{\leq} \sum_{i<j} a_ia_j+\sum_{i<j} \frac{(a_i+a_j) \sum_{k\neq i,j} a_k}{4}=$$ $$=\sum_{i<j}  a_ia_j+\frac{(n-2)}{2}\sum_{i<j}a_ia_j=\frac n2 \sum_{i<j} a_ia_j\quad \blacksquare$$

Let $S=a_1+a_2+\dots + a_n$. For the sake of brevity, all summations will denote either the sum from $1$ to $n$, and with $i<j$ if there are two variables. Note that \[(a_i+a_j)-4\left(\frac{a_ia_j}{a_i+a_j}\right)=\frac{(a_i+a_j)^2-4a_ia_j}{a_i+a_j}=\frac{(a_i-a_j)^2}{a_i+a_j}\]
Thus, \[(n-1)S-4\sum{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}=\sum{\frac{(a_i-a_j)^2}{a_i+a_j}}\]
Meanwhile, \begin{align*}
(n-1)S-4\left(\frac {n}{2S}\cdot \sum{a_{i}a_{j}}\right) &= \frac{(n-1)S^2-2n\sum{a_ia_j}}{S} \\
&= \frac{(n-1)^2\sum{a_i^2}+2(n-2)\sum{a_ia_j}-2n\sum{a_ia_j}}{S} \\
&= \frac{(n-1)^2\sum{a_i^2}-2\sum{a_ia_j}}{S} \\
&= \frac{\sum{(a_i-a_j)^2}}{S}
\end{align*}
Therefore, it suffices to show the following inequality, which is obvious because $a_i+a_j \le S$.
\[\sum{\frac{(a_i-a_j)^2}{a_i+a_j}} \ge \frac{\sum{(a_i-a_j)^2}}{S}\]

Just an exercise in summations, but took me 2 hours :(
$$\sum_{i<j}\frac{a_ia_j}{a_i+a_j}\cdot\sum_{k=1}^{n}a_k\le\sum_{i<j}a_ia_j+\sum_{i<j} \frac{a_ia_j\sum_{k\neq i,j}a_k}{a_i+a_j}\stackrel{\text{AM-HM}}{\le}\sum_{i<j}a_ia_j+\sum_{i<j} \frac{(a_i+a_j)\sum_{k\neq i,j} a_k}{4}=\frac{n}{2}\sum_{i<j}a_ia_j$$ $$\iff\sum_{i<j}\frac{(a_i+a_j)\sum_{k\neq i,j}a_k}{2}=(n-2)\sum_{i<j}a_ia_j=(n-2)n.$$ But this is evident, since $a_i,a_k$ (and $a_j,a_k$) satisfy the same conditions as $i<j$, with any $a_la_m$ appearing n-2 times when $i=l, j\ne l,m$ or when $i=m, j\ne l,m$. (Note that even if $i=l>j\ne l,m$ or the same with m, we can just switch i and j since there's only one way they can be ordered increasing.)


How many odd five-digit counting numbers can be formed by
choosing digits from the set $ \{1, 2, 3, 4, 5, 6, 7\}$ if digits can be repeated?

there are 7 numbers in the set, which means that there are 7 possibilities for the first number, the second number, etc, because numbers can be repeated

however, the last number (ones place) has to be an odd number, so there are only 4 choices

therefore we have $ 7 \times 7 \times 7 \times 7 \times 4 \equal{} 9604$ possibilities

[quote="myyellowducky82"]Or just $ 7^4\cdot4 \equal{} 9,604$[/quote]
I think that everyone here knows that $ 7\cdot7\cdot7\cdot7\cdot4$ is the same thing as $ 7^4\cdot4$. So you don't need to post the same thing as vallon.


[quote="awesome08"]I am taking the AMC10 next week.I only scored a 70 on the past AMC10.I really want to get at least 120!!!!
I only have a week left...What can I do?any advices???I know how to get the answers.But I am really slow.I spend 5 mins for the every easiest questions on AMC10.How can I be suceessful on AMC10???HELP!!!please! :([/quote]
I don't know...going from 70 to 120 is a pretty big jump. You could do as much practice as possible this week, do as many problems as you can.

Well doing the past tests will help, but jumping 50 points, i dunno, around how many do you attempt and how many do you get right?

Well one thing to do is focus on getting at least 20 or 19 (the lowest possible raw scores that could potentially give you a score of 120 or higher).

Last year, i only did around 20 or 21 and i got 19 correct so that easily boosted me to AIME, besides the fact that the cutoff was lowered to 117.

If you scored 70 on that AMC10 because you made lots of mistakes on problems you solved, learn to avoid stupid mistakes. There are some resources on this site about that, but rechecking your work is one of the easiest fixes.

If you scored 70 on that AMC10 because you couldn't solve enough problems, here is my best suggestion given the time frame: Identify your worst subject (out of algebra, geometry, combinatorics, number theory). Study that subject like crazy, and do lots of problems with it.


trigonometry
limit
logarithms
ceiling function
factorial
floor function

In this question you must make all numbers of a clock, each with using 2, exactly 3 times and Mathematical symbols. You are not allowed to use English alphabets and words like $ \sin$ or $ \lim$ or $ a,b$ and no other digits.
[img]http://i2.tinypic.com/5x73dza.png[/img]

This is an [i]olympiad[/i] question?  :|
[hide="1"] $ 1 \equal{}\left(\frac{2}{2}\right)^{2}$ [/hide] [hide="2"] $ 2 \equal{} 2\plus{}2\minus{}2$ [/hide] [hide="3"] $ 3 \equal{}\frac{2}{2}\plus{}2$ [/hide] [hide="4"] $ 4 \equal{} (2\plus{}2)\cdot (!2)$ [/hide] [hide="5"] $ 5 \equal{} !2\plus{}2\plus{}2$ [/hide] [hide="6"] $ 6 \equal{} 2\cdot 2\plus{}2$ [/hide] [hide="7"] $ 7 \equal{} !(2\plus{}2)\minus{}2$ [/hide] [hide="8"] $ 8 \equal{} 2\cdot 2\cdot 2$ [/hide] [hide="9"] $ 9 \equal{} !( (2\plus{}2)\cdot (!2) )$ [/hide] [hide="10"] $ 10 \equal{} (2\plus{}2)!!\plus{}2$ [/hide] [hide="11"] $ 11 \equal{} !(2\plus{}2)\minus{}2$ [/hide] [hide="12"] $ 12 \equal{}\frac{(2\plus{}2)!}{2}$ [/hide]
Note that I have used the multifactorial and the subfactorial.  If these are not allowed (the problem does not specify), then it would be nice to have a list of what operations were allowed.

Yes, it really is  :D
And your answer is correct, I think(not checked carefully).

[quote="t0rajir0u"]This is an [i]olympiad[/i] question?  :|
Note that I have used the multifactorial and the subfactorial.  If these are not allowed (the problem does not specify), then it would be nice to have a list of what operations were allowed.[/quote]

Or also : [hide="11"]$ 11 \equal{}\frac{22}{2}$[/hide]

And, just for fun : $ \forall n > 0$ : $ n \equal{}\frac{\ln(\frac{\ln(2)}{\ln(\sqrt{\sqrt{\sqrt{\cdots\sqrt{2}}}})})}{\ln{2}}$ with exactly $ n$ $ \sqrt{.}$

I just realized I was neglecting some other common symbols.

[hide="Example"] $ 4 \equal{}\left\lceil\sqrt{\frac{22}{2}}\right\rceil$ [/hide]
I'm pretty sure the sub- and multi-factorial answers can be rewritten in this form as well.  I'll do it later.  :)

Official Solution:
Some possible answers are shown in the following:
\[ \begin{array}{ccc}
1\equal{}2^{2\minus{}2}&2\equal{}2\plus{}2\minus{}2&3\equal{}2\plus{}\frac 22\\
4\equal{}\sqrt{2\times2}^2&5\equal{}2\times2\plus{}\lfloor\sqrt{2}\rfloor&6\equal{}2\plus{}2\plus{}2\\
7\equal{}\lceil\sqrt{(2\plus{}2)!}\rceil\plus{}2&8\equal{}2\times2\times2&9\equal{}(2\plus{}\lfloor\sqrt{2}\rfloor)^2\\
10\equal{}2\times\lceil\sqrt{(2\plus{}2)!}\rceil&11\equal{}\frac{22}{2}&12\equal{}\frac{(2\plus{}2)!}2\\
\end{array}\]

[b]t0rajir0u[/b]: For 11, you wrote $ !(2 \plus{} 2) \minus{} 2$, when it's actually $ !(2 \plus{} 2)\ \textcolor{red}{ \plus{} }\ 2$.

What is ! sign in front of a number??

behind the number is factorial, but i havent seen in front of a number...

The [url=http://en.wikipedia.org/wiki/Subfactorial]subfactorial[/url]: $ !n \equal{} n! \sum^{n}_{k\equal{}0} \frac{(\minus{}1)^{k}}{k!}$


Find the integer solutions (x,y,z) such that x^2+y^2=2z^2.

 :)  :lol:

ML

It's easily seen that $x,y$ have the same parity so there exist some integers $a,b$ such that $x=a-b,y=a+b$ and thus $a^{2}+b^{2}=z^{2}.$ Therefore $a=k(m^{2}-n^{2}),b=2kmn$ and $z=k(m^{2}+n^{2})$ for some  $(m,n)=1.$


A car accelerates uniformly from rest. Initially, its door is slightly ajar. Calculate how
far the car travels before the door slams shut. Assume the door has a frictionless hinge, a uniform
mass distribution, and a length L from front to back.

I was thinking
[hide]Air pressure outside and inside the car, and Bernoulli's Law[/hide]

If anyone has any other ideas it will be helpful.

I'm thinking torque balance!!! - balance toques on the dorr in the critical case of door closing.

Right, I don't think you're supposed to take into account the air pressure, it would be difficult to calculate the pressure difference because of the geometry of the door and everything... and it is probably neglectible.

What Rushil says is that the car's acceleration alone creates a torque which makes the door close.

Can you explain that solution in more detail please, Rushil or bus?

Use the force  and torque created due to the pressure of wind on the door in equations of motion.

Even if car is in vacuum the door will close. The mention of uniform acc lead me to think that somehow pseudo-force(force arising due to observations conducted in a non-inertial reference frame) is involved.

Consider this:
the torque provided by pseudo force is maLsin(phi)/2 . This is easy to realize. Force is ma and it acts through the centre.
Now torque= Iw

so maLsin(phi)/2=Iw where I is moment of inertia of square plate of length L about axis along the "hinge" . That can be calculated easily. w is angular acc.

so maLsin(phi)(T^2)/4 = I(phi) after some integration. As door is slightly ajar phi=sin(phi)
so T=sqrt(4I/maL)
so distance covered is s=0.5aT^2= 2I/mL. Thus distance will be a function of L only as once I is found m will cancel out. I think my reasoning is correct, but you never know!!

That sounds correct. Or at least the first two equations :). By the way, you can use $ tags to write math formulas which are much easier to read...


In a regular 26-gon ABCDEFGHIJKLMNOPQRSTUVWXYZ what is measure of angle ACE?

 :lol:

[hide="Solution"]Draw out a part of it, namely ABCDE.  Draw in the anglle, and see that it is making two sides one.  So a 26-gon becomes a 13-gon, and the angle is just that of a 13-gon.  Thus, the angle is $180-\frac{360}{13}$, which is about equal to $152.308$[/hide]

wow thanks I get it. Now I will know what to do next time I see a similar one. ( a good bet I will )


Euler
function
inequalities
number theory open
number theory

Show that if the equation f(x)=n as one solution it always has a second solution,n being given and x being the unknown.
where f(x) represent the Euler Function

why,this inequality is seen in a book!!!!!!!!!!!!!!
I just forgot the source!

Can you put them in the Unsolved section,Zetax?

Inequality¿  :huh:

I think it's pretty good here in Open Problems, see: http://mathworld.wolfram.com/CarmichaelsTotientFunctionConjecture.html

I think I've also seen it as an exercice (!) in Leo Moser's "Elementary Number Theory". This is weird.

Sure that it wasn't for odd numbers only or something like that¿  :maybe:


these are the problems from AMOC senior contest 2004:
(1) We have 8 points in the plane: 4 points vertices of a square and also the midpoints of every edge. We colour the 8 points red, blue or green. Prove there is a more than 50% chance of finding a triangle with all vertices red. [This question is stupid, yeh.]
(2) Consider a1 >= a2 >= ... >= a(n)
      and a1 + a2 ... + a(n) <= 1.
Prove:
      a1^2 + 3*a2^2 + 5*a3^2 ... + (2n-1)* a(n)^2 <= 1.
(3) Let f(n) denote the integer closest to  :sqrt: n, where n is an integer.
Find 1/f(1) + 1/f(2) ... + 1/f(10^4).
(4) x,y and z are integers s.t. (1/x^2) + (1/y^2) = (1/z^2). Prove 20|xy.
(5) Let ABC be a triangle. Let B' and C' be the feet of the perpendicular from B and C to AC and AB respectively. H is the orthocentre of ABC. Let K1 be the midpoint of HB and K2 the midpoint of HC. If C'K1 is parellel to B'K2, prove those lines are also parellel to AT where T is the circumcentre of BHC.

In $Rt\triangle C'HB$ we have $\angle K_1HC'=\angle K_1C'H$ and in $Rt\triangle B'HC$ we have $\angle B'HK_2 = \angle HB'K_2$ and also $C'K_1//B'K_2 \Longrightarrow \angle C'K_1H = \angle HB'K_2$, we know that $\triangle C'HK_1$ and $\triangle B'HK_2$ are equilateral triangles.

thus $C'K_2 = K_1B'$ and $\angle A=60^\circ$ and $\angle BC'K_1 = 30^\circ$.

Suppose $TP$ and $TQ$ are perpendicular to $AB$ and $AC$ respectively where $P,Q$ lie on $AB,AC$.
it is clear to see that $C'K_2TP$ and $B'K_1TQ$ are rectangles.
thus $TQ=K_1B'=K_2C'=TP$, $T$ lies on the bisector of $\angle A$. thus $AT$ bisects $\angle A$ and $\angle BAT= 30^\circ=\angle BC'K_1$. therefore, $AT//CK_1$.

(3)
$f(n)$ has two 1s, four 2s, six 3s, ...., $2(n+1)$ $n$s....
sum over those parts are all $2$
there are 99 $2$s.
and $f(9901) = \cdots = f(10^4) = 100$, there are a hundred $100$s here and summing over those numbers get $1$
the sum is $99\cdot 2+1=199$.


$L$ is a line passing through $(1,5)$ and intersects the circle $(x-1)^{2}+(y-1)^{2}=4$ at two different points $A,B$ . Find the locus of the midpoint of $AB$ .

I wonder is there any nice ways to do it without messing around with ugly algebra ?

It can be easily seen that $(1,5)$ is outslide of given circle. If $M(1,5)$, center of circle is $O$ and midpoint of $AB$ is $N$, then $\angle ONM=90^{o}$, so locus is part of the cirle with diametar $OM$ which is inside the given cirlce.

Bye


Can anyone help me with this:

prove using strictly combinatorial argument:

$\sum_{k=1}^{n}\binom{2k-2}{k-1}\binom{2n-2k}{n-k}=2^{2n-2}$

[hide="Hint"]
Because of the 2 on the right side, we know that the left side must be related to the number of ways to make a certain type of binary scenario. I don't know what this scenario is, however. :( [/hide]

The sum can be written as \[\sum_{1\leq k \leq n}\binom{2(k-1)}{k-1}\binom{2(n-k)}{n-k}=4^{n-1}\]
or \[\sum_{0 \leq k-1 \leq n-1}\binom{2(k-1)}{k-1}\binom{2(n-k)}{n-k}=4^{n-1}\]
now let $i=k-1$ then \[\sum_{0\leq i \leq n-1}\binom{2i}{i}\binom{2(n-1-i)}{n-1-i}=4^{n-1}\]
but this already solved in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=bijection&t=40150[/url]


inequalities
linear algebra
linear algebra unsolved

Let $ A, B, C, D$ be positive definite matrices. Show

 $ Tr[(A \minus{} B)(A \plus{} C)^{ \minus{} 1}(C \minus{} D)(B \plus{} D)^{ \minus{} 1}]$

$ \ge Tr[(A \minus{} B)(B \plus{} C)^{ \minus{} 1}(C \minus{} D)(A \plus{} D)^{ \minus{} 1}].$

Hi;

This set of matrices make the counterexample easier to see, The eigenvalues are on the diagonal.

$ A=\left(
\begin{array}{ccc}
 11 & 0 & 0 \\
 0 & 8 & 0 \\
 0 & 0 & 5
\end{array}
\right),
B=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 11 & 0 \\
 0 & 0 & 4
\end{array}
\right),
C=\left(
\begin{array}{ccc}
 11 & 0 & 0 \\
 0 & 8 & 0 \\
 0 & 0 & 10
\end{array}
\right),
\\ \\ D=\left(
\begin{array}{ccc}
 7 & 0 & 0 \\
 0 & 8 & 0 \\
 0 & 0 & 6
\end{array}
\right)$

$ Tr[(A-B)(A+C)^{-1}(C-D)(B+D)^{-1}] = \frac{-2}{75}$

$ Tr[(A-B)(B+C)^{-1}(C-D)(A+D)^{-1}]. =\frac{-2}{77}$

[quote="bobbym"]Hi;

This set of matrices make the counterexample easier to see, The eigenvalues are on the diagonal.

$ A \equal{} \left( \begin{array}{ccc} 11 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 5 \end{array} \right), B \equal{} \left( \begin{array}{ccc} 1 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 4 \end{array} \right), C \equal{} \left( \begin{array}{ccc} 11 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 10 \end{array} \right), \\
 \\
D \equal{} \left( \begin{array}{ccc} 7 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 6 \end{array} \right)$

$ Tr[(A \minus{} B)(A \plus{} C)^{ \minus{} 1}(C \minus{} D)(B \plus{} D)^{ \minus{} 1}] \equal{} \frac { \minus{} 2}{75}$

$ Tr[(A \minus{} B)(B \plus{} C)^{ \minus{} 1}(C \minus{} D)(A \plus{} D)^{ \minus{} 1}]. \equal{} \frac { \minus{} 2}{77}$[/quote]
Your example cannot be true. Coz the trace ineq is true for diagonal matrices

Hi;

[quote]Your example cannot be true. Coz the trace ineq is true for diagonal matrices[/quote]

You are right; I rechecked the calculation and found an error.

$ Tr[(A \minus{} B)(A \plus{} C)^{ \minus{} 1}(C \minus{} D)(B \plus{} D)^{ \minus{} 1}] \equal{} \frac {419}{1650}$

$ Tr[(A \minus{} B)(B \plus{} C)^{ \minus{} 1}(C \minus{} D)(A \plus{} D)^{ \minus{} 1}] \equal{} \frac {439}{2079}$


integration
calculus
real analysis
real analysis solved

Prove that: $ \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{1}{a^{2}\cdot b+a\cdot b^{2}}=2\cdot\zeta\left(3\right)$.

The proof I know is quite difficult and long, so I would be interested in another one...
Moreover, it is well-known that dzeta(3) is irrational, but it is not difficult to prove that, for each integer n > 0, the number
S(n) =  \sum {a  \geq  1} \sum { b  \geq  1} 1/(a 2 b + ab 2  + nab)  is a rational number.

Pierre.

zeta(t)=\sum{i=1->00}1/it. It's well-known that this is finite iff t>1.

I don't know that much about it, but here's where you can find lots of info:

[url]http://mathworld.wolfram.com/RiemannZetaFunction.html[/url]

All I've managed to do is to "reduce" it to $\displaystyle \sum_{n\ge 1} (\frac 1{n^2}\sum_{1\le m\le n} \frac 1m)=2\zeta(3)$. Any idea if this is easier?

The idea to get a quick proof is $\frac{1}{a^2b+ab^2}=\frac{1}{ab}\int_{0}^{1}t^{a+b-1}dt$

$A=\sum_{a,b\geq 1}\frac{1}{a^2b+ab^2}=
\int_{0}^{1}\sum_{a\geq 1}\frac{t^a}{a}\sum_{b \geq 1}\frac{t^b}{b}\frac{1}{t}dt$
With expansion in power serie of $ln(1-t)$
$A=\int_{0}^{1}\frac{(ln(1-t))^2}{t}dt$
An integration by part gives
$A=[lnt.(ln(1-t))^2]_{0}^{1} + 2\int_{0}^{1}\frac{lnt.ln(1-t)}{1-t}dt$
the term in $[ ...]$ is $0$, do a change variable in last integrale $1-t=y$
$A=2\int_{0}^{1}\frac{ln(1-y).lny}{y}dy$
Use the expansion $ln(1-y)=-\sum_{k\geq1}\frac{y^k}{k}$
$A=-2\sum_{k\geq 1}\frac{1}{k}\int_{0}^{1}y^{k-1}lnydy$
To finish  the last integral is $-\frac{1}{k^2}$
$A=2\sum \frac{1}{k^3}=2\zeta(3)$

Remark:To intervating $\sum$ and $\int$ use the theorem wich allow to
permute integral and sum, to   make the proof rigourously for an oral examination
or written exam.

So Alekk it is your turn.
Can you post your proof ?


if min(x,y,z)=z>0 then prove that
sqrt((x+z)(y+z))+sqrt((x-z)(y-z))<=2sqrt(xy)

[i]unknown source[/i]

Square both sides.

Then we need to prove (x^2 - z^2)(y^2 - z^2) \leq x^2y^2 + z^4 - 2xyz^2

<=> 2xyz^2 \leq x^2z^2 + y^2z^2, which is true by AM-GM.

Equality iff z = 0 or x = y.


linear algebra
matrix
algebra
polynomial
linear algebra theorems

Something in Axler confused me as to the following point: If a linear map $ T$ has distinct eigenvalues $ \lambda_1, \ldots, \lambda_n$, can dim null $ (T\minus{}\lambda_i I)$ be anything other than $ 1$?

If $ T$ is an $ n\times n$ matrix and there are $ n$ distinct eigenvalues, then $ \text{dim}\,\text{nul}\,(T\minus{}\lambda_iI),$ also known as the geometric multiplicity of the eigenvalue, must be $ 1.$

The sum of the algebraic multiplicities of the eigenvalues must be $ n,$ (that's just basic polynomial algebra, nothing to do with matrices) and each geometric multiplicity is less than or equal to the corresponding algebraic multiplicities. So the sum of the geometric multiplicities is less than or equal to $ n.$ But for each distinct eigenvalue, the geometric multiplicity is at least $ 1.$ The rest is pigeonhole.

For infinite -dimensional case the statetment is,of course wrong.


prove that every natural number can be expresed as the finite sum of distint fibonacci numbers.

F: a_1=1,a_2=1,a_n=a_n-1 + a_n-2

Just repeatedly subtract the largest fibonacci number less than a given number, using the fact that each successive fibonacci number is less than (edit: or equal to) double the previous.  The summation is analogous to that for powers of two.

Now, how does one go about proving that this summation is unique?

[quote="t0rajir0u"]Now, how does one go about proving that this summation is unique?[/quote]

What exactly are you asking? As I understood it, I don't think it's true.

Clearly, the statement is true for $F_{1}$ and $F_{2}$, so suppose $n\geq 3$. Suppose all number up to $F_{n}$ can be written as the sum of distinct Fibonacci numbers. Let $S_{n}$ be the set of all the Fibonacci numbers that are used to sum up to $n$. Clearly, $F_{n}\not\in S_{i}$, for all $i<F_{n}$. Therefore, $F_{n}+i$ will be the sum of distinct Fibonacci numbers for all $i<F_{n}$, i.e. all numbers up to $F_{n}+(F_{n}-1) = 2F_{n}-1$ are the sum of distinct Fibonacci numbers. But \[F_{n+1}\leq 2F_{n}-1\ \Leftrightarrow\ F_{n-1}= F_{n+1}-F_{n}\leq F_{n}-1\ \Leftrightarrow\ 1\leq F_{n}-F_{n-1}= F_{n-2},\] which is obvious for $n\geq 3$. Finish it off with induction.

[quote="t0rajir0u"]Ah, right, it isn't.  My apologies.[/quote]

You did have the correct idea. You were referring to [b]Zeckendorf's Theorem[/b] which states that any positive integer $N$ can be expressed [i]uniquely[/i] as a sum of distinct Fibonacci numbers containing [i]no neighbours[/i].  You can prove this using induction.

Masoud Zargar


[quote="spx2"]$x^{k-1}(1+x^{k^2-1})=2$
how can this equation be solved in k so that for every x it is true.
$x\in R$,$k\in R$

i'm not sure where to put this equation so i'll put it here
it can be moved by one of the moderators but if someone already posted something
please let it here[/quote]
Obviously $k=1$ is a solution.

Suppose $k\ne 1$ is another real solution. For $x > 0$, set $z = f(x) = x^{k-1} > 0$. Then $f: (0,\infty) \to (0,\infty)$ is a diffeomorphism. Since $x^{k^2-1} = \left(x^{k-1}\right)^{k+1} = z^{k+1}$, we therefore require $z (1+ z^{k+1}) = 2$ for all $z > 0$. But that is impossible, since the expression on the left hand side is never constant.

So $k = 1$ is the only solution.

ok i understood...
now how can we find all pairs $\( k,x\)$ so that
they are solutions to the equation.
hence ,from what you say $x=\varphi(k)$
and what does diffeomorphism mean ?

[quote="spx2"]and what does diffeomorphism mean ?[/quote]In this context it just means a change of variables that is differentiable in both directions.


Let x, y, z be positive reals such that xyz(x + y + z) = 1.
Find the least possible value of (x + y)(y + z).

This isn?t as hard as it seems. From the first condition we get xzy^2+xz(x+z)y=1. If we divide this by xz we get y^2+(x+z)y=1/xz. (x+y)(y+z)=y^2+(x+z)y+xz=1/xz+xz >= 2. We have equality iff xz=1.


number theory
prime numbers
relatively prime
number theory unsolved

let $ a,b,c$ positive integers such that:

$ \frac {a^2}{a \plus{} b}$,  $ \frac {b^2}{b \plus{} c}$,  $ \frac {c^2}{c \plus{} a}$ are all prime numbers then prove that  $ a \equal{} b \equal{} c$

Suppose $ p$ is a prime dividing $ a$.  Then since $ a^2 / (a \plus{} b)$ is a prime, $ p$ must also divide $ b$.  And vice versa.  And do the same for $ a$ and $ c$.  This means that the prime factors of $ a$, $ b$, and $ c$ are exactly the same.  So to show $ a \equal{} b \equal{} c$, i just need to show that the largest power of any prime $ p$ dividing $ a$,$ b$,$ c$ are the same.

So take any prime $ p$ dividing $ a$ (so it must divide $ b$ and $ c$).  Let's say
$ p^n$ is the largest power of $ p$ dividing $ a$.  Suppose, on the contrary, $ p^m$ is the largest power of $ p$ dividing $ b$ or $ c$, where $ m \neq n$.  Without a loss of generality, let's say $ p^m$ is the largest power of $ p$ dividing $ b$ and $ n > m$ (since the condition of the problem is symmetric).  Then $ a^2$ has $ p^{2n}$.  But $ a \plus{} b$ has only $ p^m$.  So $ a^2 / (a \plus{} b)$ is divisible by $ p^{2n \minus{} m}$, so it cannot be a prime, contradiction.

I proved it using the substitution a= da'

I ll post my solution later ...

[quote="FOURRIER"][quote="jinjohn"]Suppose $ p$ is a prime dividing $ a$.  Then since $ a^2 / (a \plus{} b)$ is a prime, $ p$ must also divide $ b$.[/quote]

why?......[/quote]

because otherwise $ a^2 / (a\plus{}b)$ would have at least $ p^2$ factors.

Supose $ (a,b)\equal{}d_1$ and $ a\equal{}d_1p_1$ and $ b\equal{}d_1q_1$ (with $ p_1$ and $ q_1$ relatively prime) then substituting: $ \frac{d_1^2p_1^2}{d_1p_1\plus{}d_1q_1}$ is a prime integer. Reducing terms: $ \longrightarrow$ $ \frac{d_1p_1^2}{p_1\plus{}q_1}$ is a prime. But $ ([p_1\plus{}q_1],p_1^2)\equal{}1$ this means $ [p_1\plus{}q_1]|d_1$ but then, $ p_1^2$ must divide the prime number. This implies $ p_1\equal{}1$. Then, $ a\equal{}d_1$ $ \longrightarrow$ $ a|b$ analogously: $ b|c$ and $ c|a$ then $ a<\equal{}b<\equal{}c<\equal{}a$ $ \longrightarrow$ $ a\equal{}b\equal{}c$


Determine and locate the resultant $ R$ of the two forces and one couple acting on the I-beam.

[img]http://i33.tinypic.com/24ln32v.jpg[/img]

Unfortunately, I can't quite understand the drawing.  It looks to me like there are torques involved, and it looks like the rod has mass, but I can't quite tell.  Please explain this better.

I think it would be more easy to simply scan the picture from the source and post it here.


calculus integration function trigonometry

Say you drop an object straight down onto a spring standing upright (and the object is centered exactly over the top of the spring... it doesn't jostle around). Assume that all forces are conservative (no friction or air resistance). What is an equation that gives the amount the spring has been compressed from its original position (x) in terms of the elapsed time since the object comes into contact with the spring (t)? Express your answer in terms of k (the spring constant), g, the mass of the object (m), the length of the spring (L) and the velocity at the time the object hits the spring (v).

I started by finding the force acting on the object as it pushes on the spring, which is mg-kx because gravity is pushing down and the spring is pushing up with force kx (by Hooke's law). Using Newton's second law, the acceleration at each point is g-kx/m.

At this point I tried to use calculus to integrate both sides with respect to time (because the displacement (total compression of the spring) equals the second integral of acceleration), but that didn't work because g-kx/m isn't a known function of t. Presumably, x is a function of t, but that function is what the problem is trying to solve for... so it seems like a dead end to me.

Any hints?

What you've got there is a second-order homogeneous linear ordinary [url=http://en.wikipedia.org/wiki/Differential_equation]differential equation[/url].  These equations have well-known solutions, but it looks like you aren't familiar with them; was this problem assigned to you in class?

Sweet, thanks :)

No, it wasn't assigned--we are, however, studying Hooke's law currently and this particular problem occurred to me during one of our labs.

You can ask a simpler version of this question:  consider a spring attached to a wall with spring constant $ k$ and a mass $ m$ attached to that spring sliding on a frictionless surface.  You stretch the spring a distance of $ A$ past the equilibrium and let go at time $ t \equal{} 0$.  Describe the displacement $ x(t)$ from the equilibrium.

The relevant equation is $ F \equal{} ma \equal{} m \frac{dx^2}{dt^2} \equal{} \minus{}kx$ and the relevant solution is

$ x(t) \equal{} A \cos \omega t$

where $ \omega^2 \equal{} \frac{k}{m}$.  This is simple harmonic motion.  When gravity is thrown in, the situation gets a bit more complicated.


geometry
incenter
power of a point
radical axis
geometry unsolved

4-Point $ P$ is taken on the segment $ BC$ of the scalene triangle $ ABC$ such that $ AP \neq AB,AP \neq AC$.$ l_1,l_2$ are the incenters of triangles $ ABP,ACP$ respectively. circles $ W_1,W_2$ are drawn centered at $ l_1,l_2$ and with radius equal to $ l_1P,l_2P$,respectively. $ W_1,W_2$ intersects at $ P$ and $ Q$. $ W_1$ intersects $ AB$ and $ BC$ at $ Y_1( \mbox{the intersection closer to B})$ and $ X_1$,respectively. $ W_2$ intersects $ AC$ and $ BC$ at $ Y_2(\mbox{the intersection closer to C})$ and $ X_2$,respectively.PROVE THE CONCURRENCY OF $ PQ,X_1Y_1,X_2Y_2$.

Let $ R \equiv X_1Y_2 \cap X_2Y_2.$ $ BX_1 \equal{} BY_1, CX_2 \equal{} CY_2$ and $ AY_1 \equal{} AP \equal{} AY_2$  $ \Longrightarrow$

$ \angle Y_2Y_1X_1 \equal{} 180^\circ \minus{} (\angle AY_1Y_2 \plus{} \angle X_1Y_1B) \equal{}$
$ \equal{} \frac {_1}{^2}\angle A \plus{} \frac{_1}{^2}\angle B \equal{} 90^\circ \minus{} \frac {_1}{^2} \angle C \equal{} \angle Y_2X_2C$

$ \Longrightarrow$ $ X_1X_2Y_2Y_1$ is cyclic $ \Longrightarrow$ $ \overline {RX_1} \cdot \overline{RY_1} \equal{} \overline {RX_2} \cdot \overline{RY_2}$ $ \Longrightarrow$ $ R$ is on radical axis $ PQ$ of the circles $ \mathcal W_1, \mathcal W_2.$


Please state problem sources. This is 2002 AMC 10A I believe, might be 10B.

[hide]Factoring gives
$(4y+1)(2x-3)=0$
This is true iff one of the two factors is 0. For $y$, it is always true when $4y+1=0$, so $y=\boxed{-\frac{1}{4}}$.[/hide]

I suggest you read the problem more carefully.

[hide]Factors to $(2x-3)(4y+1)=0$, which is true for all y iff $x=3/2$.[/hide]

[quote="now a ranger"]Find the value(s) of $x$ such that $8xy-12y+2x-3=0$ is true for all values of $y$.[/quote]
[hide]
$x(8y+2)=12y+3$
$2x(4y+1)=3(4y+1)$
$2x=3$
$x=1.5$
Plugging this back in, we get
$12y-12y+3-3=0$
which is clearly always true.[/hide]


integration
trigonometry
logarithms
calculus
calculus computations

Let $ f: \mathbb{R} \to \mathbb{R}$, $ f(x)\equal{}\arcsin \left( \frac{2x}{1\plus{}x^2} \right)$ .

Find $ F$, such that $ F^{\prime} (x)\equal{}f(x)$ and $ F(0)\equal{}2\ln(2)$.

$ \int_{0}^{x}F^{\prime }(u)du \equal{} \int_{0}^{x}\sin ^{ \minus{} 1}\left( \frac {2u}{1 \plus{} u^{2}}\right) du$

$ F(x) \minus{} F(0) \equal{} \allowbreak x\sin ^{ \minus{} 1}\left( \frac {2x}{1 \plus{} x^{2}}\right) \plus{} \ln \left( 1 \plus{} x^{2}\right)$

Hence $ F(x) \equal{} \allowbreak x\sin ^{ \minus{} 1}\left( \frac {2x}{1 \plus{} x^{2}}\right) \plus{} \ln \left( 1 \plus{} x^{2}\right) \plus{} 2\ln 2$

Also, you should be aware of the fact that

$ f$ is not differentiable at $ \minus{}1$ and $ 1$, so you have to consider the following cases:

$ x \in I\equal{} (\minus{}1;1)$ , $ x \in (\minus{}\infty;\minus{}1)$ and $ x \in (1; \infty)$.

Now, for each case you can integrate by parts :)


quadratics
algebra
polynomial
probability
number theory unsolved
number theory

Find all the natural number $n$ such that : $(2^{n}-1)|(3^{n}-1)$

Yes, but there is nothing there. Clearly $n$ is odd. Suppose that $n$ is at least $3$, then $2^{n}=8(mod 12)$. Thus there exists a prime number $p$ of the form $12k\pm 5$ that divides $2^{n}-1$. Using the quadratic reciprocity for $p$ and $3$ we immediately find a contradiction. By the way, this is a very beautiful problem.

I still wonder if there exists an elementary solution for the fact that if $b$ is not a power of $a$ then the density of the set of those $n$ for which $a^{n}-1|b^{n}-1$ is zero.  :idea:

I think if $\frac{a^{n}-1}{b^{n}-1}$ is integer for infinetely many n, then exist k: $a=b^{k}$.

Yes, this is true, it follows from a very recent result of Corvaja and Zannier, but the proof is more than nonelementary and difficult.

What is your elementary solution about density 0?

I said I still wonder, so I don't have a solution.

Let $m=lcm(a,b)$ and P is subset of odd primes, suth that $(m,p)=1, (\frac{p}{a})=-1, (\frac{p}{b})=1$. If \[\sum_{p\in P}\frac{1}{p}=\infty\] , then density is 0. Because exist k, suth that all primes p $p=k(mod \ 4m)$ belongs P, it is true for case when $b\not |a$.

Can you elaborate a little bit your argument, please? If $a$ is a multiple of $b$, then it's very easy to iterate, because if $a=bc$ then $(a^{n}-1)/(b^{n}-1)$ is an integer if and only if $(c^{n}-1)/(b^{n}-1)$ is an integer. But I'm confused about your argument for the case when $a$ is not a multiple of $b$. Please, give some details. Thanks. PS: note that this is a much more general problem than the masterpiece problem that was so much discussed.

Let a is not square and $d=(a,b),a_{1}=a/d, b_{1}=b/d$, then $a_{1}>1$ and exist $k_{0}$, suth that $(\frac{a}{p})=-1$ if $p=k_{0}(mod \ 4a)$. If $b\not |a$ then $b_{1}>1$, therefore we can find $l$, suth that $k=k_{0}+4la$ and Jakobi symbol $(\frac{b}{k})=1$. It give that all prime $p=k(mod \ 4m)$ belongs P. Let $c_{b}(p)$ is minimal T, suth that $p|(b^{T}-1)$ and $c_{a}(p)$ is minimal T, suth that $p|(a^{T}-1)$. Then $c_{a}(p)\not |c_{b}(p)$, because $ord_{2}(c_{a}(p))>ord_{2}(c_{b}(p))$. Let $e(p)=\frac{c_{a}(p)}{(c_{a}(p),c_{b}(p))}>1$. If $(b^{n}-1)|(a^{n}-1)$ then for all $p\in P$ or $p\not |n$ or $e(p)|n$. Therefore density of n is \[\prod_{p\in P}(\frac{1}{pe(p)}+1-\frac{1}{p})\le \prod_{p}(1-\frac{1}{2p})=0.\]

Thanks, Rust, your proof seems very interesting, I will study it. By the way, could you post a more detailed proof for the problem with special polynomials and probability? I have another solution, that does not give the exact value of the probability.

[quote="harazi"]Ok, I read your proof, I tried to understand it, but I must confess I did not understand practically anything from your arguments. I may be extremely stupid, but can you clearly explain the following:
  1) If I correctly understood, then $c_{b}(p)$ and $c_{a}(p)$ are the orders of $b$ and $a$ modulo $p$. Now, what is $ord_{2}(c_{a}(p))$? Next, why do you say that $c_{a}(p)$ does not divide $c_{b}(p)$? I just cannot see why: for instance, if $n=c_{b}(p)$ is one of those numbers for which $a^{n}-1$ is a multiple of $b^{n}-1$, it follows that $c_{b}(p)$ is a multiple of $c_{a}(p)$.
 2)  Why if $p\in P$ verifies $p|n$ then $e(p)|n$?
     Sorry for all these questions, but your proof seems very unclear for me (probably I am the only one who finds it extremely cryptic).[/quote]
1.$ord_{2}(n)$ is maximal m, suth that $2^{m}|n$.
$c_{a}(p)\not |c_{b}(p)$, because $ord_{2}(c_{a}(p))>ord_{2}(c_{b}(p))$. If $m|n$ then $ord_{p}(n)\ge ord_{p}(m) \ \forall \ prime \ p$.
2. If $p\in P$ and $p|n, \ (b^{n}-1)|a^{n}-1)$ then $e(p)|n$.


tangent
trigonometry
geometry
similar triangles
IMO
IMO 1990

Chords $ AB$ and $ CD$ of a circle intersect at a point $ E$ inside the circle.  Let $ M$ be an interior point of the segment $ EB$.  The tangent line at $ E$ to the circle through $ D$, $ E$, and $ M$ intersects the lines $ BC$ and $ AC$ at $ F$ and $ G$, respectively.  If
\[ \frac {AM}{AB} \equal{} t,
\]
find $\frac {EG}{EF}$ in terms of $ t$.

Very Nice Problem.  :)

with simple angle chasing, we find that triangles $CEG$ and $MDB$ are similar.

so, $GE/EC = MD/MB$. (*)

again with simple angle chasing, we find that triangles $CEF$ and $AMD$ are similar.

so, $EF/DM = CE/AM$. (**)

so, by (*) and (**), we have $GE/EF = MA/MB = t/(t-1)$.

E.L

here is my 'quick & dirty' solution:

    Out of projective considerations the ratio in question could be one of only

     6 Salmon ratios:

    t  ,  t -1  ,  1 / t ,   1/ (t - 1)  ,    t / (t - 1)  ,    (t - 1) / t

    A simple checking of limiting positions  leads  to the ratio   t / (t - 1)

    Cute  problem!

    Thank you.

    M.T.

I got $\frac{GE}{EF}=\frac{t}{1-t}$. I think this is right since $\frac{BM}{AB}=1-t$ and using the ratios from the similar triangles previously mentioned one obtains:\[ (1) \frac{BM}{MD}=\frac{CE}{EG}  \]\[ (2) \frac{AM}{MD}=\frac{CE}{EF}  \]Dividing $(1)$ by $(2)$ we obtain\[ \frac{EF}{EG}=\frac{BM}{AM}  \]\[ \Leftrightarrow \frac{EG}{EF}=\frac{AM}{BM}  \]\[ \Leftrightarrow \frac{EG}{EF}=\frac{AM}{AB} \cdot \frac{AB}{BM}  \]Substituting from **\[ \Leftrightarrow \frac{EG}{EF}=\frac{t}{1-t}  \]

Assume that $ M\in AE$ since the other case is analogous. Let $ N$ and $ N'$ be on $ EC$ and $ AC$, repectively, such that $ BNN'//GF$. Then $ \angle MDE\equal{}\angle GEA\equal{}\angle BEF\equal{}\angle NBE$ so $ MDBN$ is cyclic. Hence $ \angle MNE\equal{}\angle EBD\equal{}\angle ACD$, so $ MN//AC$. Using the parallel lines, $ GE: EF\equal{}N'N: NB\equal{}AM: MB\equal{}t: 1\minus{}t$.

I will present a proof using trigonometry.

[u][b]Notations[/b][/u]:$\angle{DCB}=\theta,\angle{DME}=x$

[u][b]Proof[/b][/u]:Applying sine rule in $\triangle{AMD}$ we get
$\frac{AM}{sin(x+\theta)}=\frac{AD}{sinx}$
We also have $\frac{AB}{sinC}=\frac{AD}{sin(C-\theta)}$

Dividing the two relations we get
$t=\frac{AM}{AB}=\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}$

Applying sine rule in $\triangle{EFC}$ we get
$\frac{EF}{sin\theta}=\frac{EC}{sin(x+\theta)}$

Applying sine rule in $\triangle{EGC}$ we get
$\frac{EG}{sin(C-\theta)}=\frac{EC}{sin(x-C+\theta)}$

Dividing the two relations we get
$\frac{EG}{EF}=\frac{sin(C-\theta)sin(x+\theta)}{sin(x-C+\theta)sin\theta}$.

Again
 $\frac{t}{1-t}=\frac{\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}}{1-\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}}
=\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC-sin(x+\theta)sin(C-\theta)}
=\frac{2sin(x+\theta)sin(C-\theta)}{cos(C-x)-cos(C+x)-cos(C-x+2\theta)+cos(C+x)}
=\frac{sin(x+\theta)sin(C-\theta)}{sin\theta sin(x-C+\theta)}$

From these relations we get $\frac{EG}{EF}=\frac{t}{1-t}$.We are done!!!

Dear Mathlinkers,
this problem appears again at
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=498513
Sincerely
Jean-Louis

Just Notice, $\Delta EGC \sim \Delta MDB$ and, $\Delta ECF \sim \Delta MAD \implies EC \cdot MD= EF \cdot AM=EG \cdot MB \implies \frac{EG}{EF}=\frac{AM}{MB}=\boxed{\frac{t}{1-t}}$

[hide=Solution]This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets $\frac{MA}{MB}=\frac{t}{1-t}$.
By Ratio Lemma, $\frac{FB}{FC}=\frac{BE}{CE} \cdot \frac{\sin{\angle{FEB}}}{\sin{\angle{FEC}}}=\frac{DE}{AE} \cdot \frac{\sin{\angle{EDM}}}{\sin{\angle{DME}}}=\frac{DE}{AE} \cdot \frac{EM}{DE}=\frac{ME}{EA}$. Similarly, $\frac{GA}{GC}=\frac{ME}{EB}$. We can rewrite these equalities to get $\frac{AM}{EM}=\frac{BC}{FB}$ and $\frac{BM}{EM}=\frac{AC}{GC}$.
Using Ratio Lemma, $\frac{GE}{\sin{\angle{ACD}}}=\frac{GC}{\sin{\angle{GED}}}$ and $\frac{EF}{\sin{\angle{BCD}}}=\frac{FC}{\sin{\angle{FEC}}}$. Since $\angle{GED}=\angle{FEC}$, we have $\frac{FE}{GE}=\frac{FC}{GC} \cdot \frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}$ (eq 1). Note that by Ratio Lemma, $\frac{\sin{\angle{BCD}}}{\sin{\angle{ACD}}}=\frac{CA}{CB} \cdot \frac{EB}{EA}$. Plugging this into (eq 1), we get $\frac{EF}{GE}=\frac{FC}{GC} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{\frac{EA}{EM} \cdot FB}{\frac{EB}{EM} \cdot GA} \cdot \frac{CA}{CB} \cdot \frac{EB}{EA}=\frac{FB}{GA} \cdot \frac{CA}{CB}=\frac{EM}{AM} \cdot \frac{MB}{EM}=\frac{MB}{MA}=\frac{1-t}{t}$. So $\frac{EG}{EF}=\boxed{\frac{t}{1-t}}$.[/hide]

A sweet problem.

My solution is as follows:

Let's first recognize some similarities among triangles.

 Since $ABCD$ is cyclic, $\angle{ECF}=\angle{DCB}=\angle{DAB}$. Furthermore, due to the tangent, $\angle{CEF}=\angle{GED}=\angle{EMD}$. Thus, $\triangle{CEF} \sim \triangle{AMD}$, leading to the following equality of ratios:

$\frac{EF}{CE} = \frac{MD}{AM}    (1)$

Once again since $ABCD$ is cyclic, $\angle{GCE}=\angle{ACD}=\angle{ABD}=\angle{MBD}$. In addition, because of the supplementary angles, $\angle{CEG}=\angle{BMD}$. Therefore $\triangle{CGE} \sim \triangle{BDM}$, leading to the following equality of ratios:

$\frac{EG}{CE} = \frac{DM}{BM}    (2)$

Dividing $(1)$ by $(2)$ gives us the following equality:

$\frac{EF}{EG}=\frac{BM}{AM}=\frac{AB-AM}{AM}=\frac{AB}{MA}-\frac{AM}{AM}=\frac{1}{t}-1=\frac{1-t}{t}$

Therefore,
$\frac{EG}{EF}=\frac{t}{1-t}$ $\blacksquare$

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/A%20variant%20of%20IMO%201990.pdf

Sincerely
Jean-Louis

[quote=jayme]Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/A%20variant%20of%20IMO%201990.pdf] see here [/url]

Sincerely
Jean-Louis[/quote]

[b]Claim 1:[/b] $\triangle ECG\sim \triangle MBD$.
[b]Proof:[/b] Angle Chasing. $\angle ECG = \angle DCA = \angle DBA = \angle DBM$ and $\angle GEC = \angle GED = \angle EMD = \angle BMD$. $\square$.

[b]Claim 2: [/b] $\triangle CEF\sim \triangle AMD$
[b]Proof: [/b] $\angle CEF = \angle GED = \angle DMA$ and $\angle ECF = \angle DCB = \angle DAB = \angle DAM$. $\square$.

Thus,
\[\frac{EG}{EF} = \frac{\frac{EG}{EC}}{\frac{EF}{EC}} = \frac{\frac{MD}{MB}}{\frac{MD}{AM}}=\frac{AM}{MB} = \frac{t}{1-t} \]
$\blacksquare$.

The answer is $\frac{EG}{EF}=\frac{t}{t-1}$.
------
From two triangle similarities $\triangle{CEG} \sim \triangle{BMD}$ and $\triangle{CEF}\sim \triangle{AMD}$ we get $$\frac{EG}{MD}=\frac{CE}{BM}, \frac{EF}{MD}=\frac{CE}{AM}\Rightarrow \frac{\frac{EG}{MD}}{\frac{EF}{MD}}=\frac{\frac{CE}{BM}}{\frac{CE}{AM}}\Rightarrow \frac{EG}{EF}=\frac{AM}{BM}.$$which is easy to derive once we know $\frac{AM}{AB}=t$.


trigonometry
geometry
circumcircle
trig identities
Law of Sines
Law of Cosines

The triangle $\triangle ABC$ has $AB = c$, $BC = a$, $CA = b$ as usual. Find $\cot C/(\cot A + \cot B)$ if $a^2 + b^2 = 1989c^2$.

[hide]By the law of sines and cossines:

$cos\ A=\frac{b^2+c^2-a^2}{2bc}$ and $sin\ A=\frac{a}{2R}$, $R$ the circumradius then:

$cot\ A=\frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{a}{2r}}=\frac{b^2+c^2-a^2}{4.\Delta ABC}$ doing the same for the other angles:

$cot\ B=\frac{a^2+c^2-b^2}{4.\Delta ABC}$ and $cot\ C=\frac{a^2+b^2-c^2}{4.\Delta ABC}$ in consequence:

${\frac{cot\ C}{cot\ A+cot\ B}=\frac{a^2+b^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2}=\frac{1989c^2-c^2}{2c^2}=994}$[/hide]

Wow, I love this problem.

[hide]
By plugging in $\cot = \frac{ \cos }{ \sin }$, we get the desired expression to be $\frac{ \sin A \sin B \cos C }{ \sin C \sin (A + B) }$.  Since $A+B = 180-C$, the denominator simplifies to $\sin ^2 C$.  By law of sines, $\sin C = \frac{c \sin A }{a} = \frac{ c \sin B}{b}$.  Plug each of those in the denominator and cancel to get our new expression to be $\frac{ ab \cos C}{c^2}$.  This is just begging for law of cosines, and good thing it works so elegantly.  We get $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{1988c^2}{2ab} = \frac{994c^2}{ab}$.  Frantically plugging this in, all the variables cancel, and we get $\boxed{994}$.
Lovely.
[/hide]

What I did when I did this problem was brute force it.
a=1989
b=1989
c= \frac{\sqrt{2}*1989}{2}

Two right triangles can be formed, and then some ugly brute forcing....... not fun. But I got the same answer.
Moral: If you can assume something, do it!


AoPS=the BIGGEST forum I've seen(in font size)
1, Choose AoPS style on your 'Profile' page.
2, In Internet Explorer, select the Greatest Font Size.
Then you get the BIGGEST forum in font size, as the picture below shows. There's no forum bigger than that, all over the Internet!
The mystery is:
[hide]
Usually, no matter what software a forum uses: phpBB, vBulletin or anything else, the font size can't be changed(vBulletin 2 is an exception, but its font size is not as big as this one :lol: ) . This is also correct for MathLinks forum. But AoPS has a special design, that's why everything becomes so big.
[/hide]

With the usually notice that AoPS forum and MathLinks forum are the one and only (the same), but the skins are different, I do that see what your [b]questions or suggestion[/b] is :)

it's not ques or suggestion but I didn't find any other forum to put in ;)
just ignore it :)

paganinio,

Understand that the AoPS skin is meant to be kid-friendly.  I also like the big font because it's dyslexic-friendly.

You can change the font size in two different ways: both in the upper right hand corner of this screen and in your own browser.  That should allow for plenty of flexibility.


Anyone got an answer for these questions?

Given any three natural numbers, show that there are two of them, say a and b,
such that a^3b – b^3a is divisible by 30.
2. In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is
on AB. If angle B is 60 degrees, show that AC = CD + AE.
3. Let A, B, C and D be points of the plane. Let P be the disk for which AB is a
diameter, and Q the disk for which CD is a diameter. Assume that A, B do not
belong to Q and C, D do not belong to P. Prove that the segments AB and CD
are disjoint.
4. A tennis tournament has every participant playing every other participant. No
ties are allowed. A player A is superior if for every other player B, either A
beats B or there is a third player C such that A beats C and C beats B. Prove
that if there is only one superior player then he or she beats every other player.

NOOOOO DO NOT POST HERE

THIS IS AN ONGOING MATH COMPETITION FROM THE IUPUI MATH CONTEST
Link here says it does not end until APril 11, 2008.
http://www.math.iupui.edu/news/contest/contest08/instructions08.pdf

CAN A MOD PLEASE LOCK THIS?

OR JUST DELETE THE QUESTIONS/WHOLE TOPIC?


[quote="joejia"]Running at a speed of 8 miles per hour, how many minutes will it take Edwina to run 14 miles?[/quote]

[hide]It takes Edwina 2 hours to run 16 miles. So it should take her $\frac{6}{8}$ or $\frac{3}{4}$ of an hour to run 6 miles. That is 45 minutes.

$45+60=105 \text{minutes}$[/hide]

[hide]divide miles by miles.$\frac{14}{8}=\frac{7}{4}$which is $\frac{7}{4}$hrs or 1 hr and 45 minutes which is $60+45=105 minutes$

The method is a bit longer but it still works! :D[/hide]

[hide]

$\frac{8}{14} = \frac{60}{x}$ You cross multiply: $8x = 840$ Simplify: $x = \framebox{105 minutes}$[/hide]

[hide]i set up a proportion:
${\frac{x}{60}=\frac{14}{8}\\\\\text{cross multiply}\\\\x=\boxed{\boxed{\boxed{\boxed{105}}}}}$[/hide]


The lengths, in order, of four consecutive sides of an equiangular hexagon are 1, 7, 2 and 4 units, respectively. What is the sum of the lengths of the two remaining sides?

Since this is an equiangular hexagon, we can extend the sides to form an equiangular triangle, which must be equilateral. The rest follows.

[geogebra]8ea25a0deea4ffda07ca799af09625b99e026dda[/geogebra]

From this, we see that the answer is $ 4\plus{}5\equal{}\boxed9$.

Note that the sum of the lengths any two consecutive sides of an equiangular hexagon is equal to the sum of the lengths of the two opposite sides.


For an isosceles triangle ABC with AB = BC. AH is an altitude and D is a point on AC. The incircle of triangle ABD and the excircle of triangle BDC, relative to CD, are equal circles with radius r, Prove that AH = 4r

[quote]Let $\triangle ABC$ be an isosceles triangle with $ AB \equal{} AC \equal{} L.$ $ D$ is  a point on  $ BC,$ such that the radii of the incircle of $ \triangle ABD$ and the A-exincircle of $ \triangle ADC$ are equal to  $r.$ Show that the altitude $ h$ on  the leg $ L$ is four times $ r.$[/quote]
Drop perpendiculars $ DP$ and $ DQ$ from $ D$ to $ AB$ and $ AC,$  respectively. Then, using the well-known formulae of the inradii and exinradii in terms of altitudes, we get

$ DP \equal{} \frac {r(L \plus{} AD \plus{} BD)}{L} \ , \ DQ \equal{} \frac {r(AD \plus{} L \minus{} DC)}{L}$

$ DP \plus{} DQ \equal{} \frac {r(2L \plus{} 2AD \plus{} BD \minus{} DC)}{L}$

On the other hand, $ DP \plus{} DQ \equal{} h  \Longrightarrow h \equal{} \frac {r(2L \plus{} 2AD \plus{} BD \minus{} DC)}{L} \  (\star)$

Since these two circles are congruent, the tangent segments from D to both are equal.

$ L \plus{} DC \minus{} AD \equal{} AD \plus{} BD \minus{} L \ \Longrightarrow \ 2L \equal{} 2AD \plus{} BD \minus{} DC$

Combining with  $ (\star)$ yields  $ h \equal{} \frac {r(2L \plus{} 2L)}{L} \equal{} 4r.$


geometry
trapezoid
incenter
angle bisector
congruent triangles
geometry unsolved

Let $ ABCD$ be a circumscribed trapezoid such that the lines $ AD$ and $ BC$ intersect at point $ R$. Denote by $ I$ the incenter of the trapezoid, and let the incircle touch the sides $ AB$ and $ CD$ at points $ P$ and $ Q$, respectively. Let the line through $ P$, which is perpendicular to $ PR$, meet the lines angle bisector $ AI$ and $ BI$ at points $ A_1$ and $ B_1$, respectively. Similarly, let the line through $ Q$, perpendicular to $ QR$, meet $ CI$ and $ DI$ at $ C_1$ and $ D_1$, respectively. Show that $ A_1D_1\equal{}B_1C_1$.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given, with incircle $ (I),$ which tangents to $ BC$ at point so be it $ D.$ We draw the line through the point $ D$ and perpendicular to $ AD,$ which intersects the line segments $ BI,\ CI,$ at points $ B',\ C',$ respectively. Prove that $ AB' \equal{} AC'$ and $ \angle B'AC' \equal{} \angle BAC.$[/color][/size][/b]

$ \bullet$ [b][size=100]PROOF OF THE LEMMA.[/size][/b] – [ [b][size=100]t=265731[/size][/b] ] - Because of $ ID\perp BC$ and $ AD\perp B'C',$ we have that $ \angle BDB' \equal{} \angle ADI$ $ ,(1)$

So, in the configuration of the triangle $ \bigtriangleup BAD,$ we have the points $ B',\ I,$ as the intersection points of its angle bisector of the angle $ \angle ABD,$ from two lines through its vertex $ D,$ isogonal conjugates with rspect to the angle $ \angle ADB.$

Hence, we conclude that the lines $ AB',\ AI,$ are isogonal conjugates with respect to the angle $ \angle BAD$ $ ($ the points $ B',\ I,$ are isogonal conjugates, with respect to the triangle $ \bigtriangleup BAD$ $ ).$

That is, we have $ \angle B'AB \equal{} \angle DAI$ $ ,(2)$

By the same way, we can prove that $ \angle C'AC \equal{} \angle DAI$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \angle B'AB \equal{} \angle C'AC$ $ \Longrightarrow$ $ \angle B'AC' \equal{} \angle BAC$ $ ,(4)$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

[b][size=100]REMARK.[/size][/b] - [ [b][size=100]t=265731(a)[/size][/b] ] - We can use the same argument also, in the alternative configuration of the triangle $ \bigtriangleup ABC,$ with $ A$-excircle $ (I_{a})$ which tangents to $ BC,$ at point so be it $ D'.$

Then, the line through the point $ D'$ and perpendicular to $ AD',$ intersects the lines segments $ BI_{a},\ CI_{a},$ at points $ B'',\ C''$ respectively, such that $ AB'' \equal{} AC''$ and $ \angle B''AC'' \equal{} \angle BAC.$

$ \bullet$ [b][size=100]PROOF OF THE PROPOSED PROBLEM.[/size][/b] – [ [b][size=100]t=265731(b)[/size][/b] ] - Based on the above [b][size=100]Lemma[/size][/b], we have that the triangles $ \bigtriangleup RA_{1}B_{1},\ \bigtriangleup RC_{1}D_{1},$ are isosceles with $ RA_{1} \equal{} RB_{1}$ and $ RC_{1} \equal{} RD_{1}$ and $ \angle A_{1}RB_{1} \equal{} \angle C_{1}RD_{1} \equal{} \angle ARB$ $ ,(5)$

From $ (5)$ $ \Longrightarrow$ $ \angle A_{1}RD_{1} \equal{} \angle B_{1}RC_{1}$ $ ,(6)$

Hence, from the congruent triangles $ \bigtriangleup RA_{1}D_{1},\ \bigtriangleup RB_{1}C_{1},$ we conclude that $ A_{1}D_{1} \equal{} B_{1}C_{1}$ and the proof of the proposed problem is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Alexandros Sigelakis[/size][/b].

Kostas Vittas.


Circle $O$ is the incircle of the square $ABCD$. $O$ is tangent to $AB$ and $AD$ at $E$ and $F$, respectively. Let $K$ be a point on the minor arc $EF$, and let the tangent of $O$ at $K$ intersect $AB$, $AC$, $AD$ at $X$, $Y$, $Z$, respectively. Show that $\displaystyle \frac{AX}{XB} + \frac{AY}{YC} + \frac{AZ}{ZD} =1$.

Denote $x = \frac{AX}{AB},\ y = \frac{AY}{AC},\ z = \frac{AZ}{AD}$. AY bisects the right angle $\angle XAZ$, hence $AY = \frac{AX \cdot AZ\ \sqrt 2}{AX + AZ}$ and $y = \frac{AY}{AB \sqrt 2} = \frac{xz}{x + z}.$ (O) is the excircle of the right angle triangle $\triangle XAZ.$ Therefore, $AX + XY + AZ = 2AE = AB$ and $XY = \sqrt{AX^2 + AZ^2}$, which implies

$x + \sqrt{x^2 + z^2} + z = 1$

$x^2 + z^2 = (1 - x - z)^2 = 1 + x^2 + z^2 - 2z - 2x + 2xz$

$x + z - xz = \frac 1 2,\ \ \ (1 - x)(1 - z) = 1 - x - z + xz = \frac 1 2$

Substituting the above into the sum $\frac{AX}{XB} + \frac{AY}{YC} + \frac{AZ}{ZD},$

$\frac{x}{1 - x} + \frac{y}{1 - y} + \frac{z}{1 - z} = \frac{x}{1 - x} + \frac{xz}{x + z - xz} + \frac{z}{1 - z} =$

$= \frac{x}{1 - x} + 2xz + \frac{z}{1 - z} = \frac{x(1 - z) + 2xz(1 - x)(1 - z) + z(1 - x)}{(1 - x)(1 - z)} =$

$= \frac{x - xz + xz + z - xz}{\frac 1 2} = 2(x + z - xz) = 1$


Let $ x,\ y,\ z$ be integers such that $ 2x\plus{}3y\equal{}5x\minus{}2z\equal{}6$. Find the minimum value of $ |x\plus{}y\plus{}z|,\ |xy\plus{}yz\plus{}zx|.$

[hide="part 1"]First, solve for $ y$ in terms of $ x$ and $ z$:
$ 2x\plus{}3y\equal{}5x\minus{}2z$

$ 3y\equal{}3x\minus{}2z$

$ y\equal{}x\minus{}\frac{2}{3}z$

Plug this into $ \left|x\plus{}y\plus{}z\right|$ to get $ \left|x\plus{}x\minus{}\frac{2}{3}z\plus{}z\right|\equal{}\left|2x\plus{}\frac{1}{3}z\right|\equal{}\left|\frac{12x\plus{}2z}{6}\right|$

$ \equal{}\left|\frac{17x}{6}\minus{}\frac{5x\minus{}2z}{6}\right|\equal{}\left|\frac{17x}{6}\minus{}\frac{6}{6}\right|\equal{}\left|\frac{17x}{6}\minus{}1\right|$

This is an integer, since $ x, y, z$ are all integers.  So that means $ x$ must be divisible by 6.  Then the function we're trying to minimize is either 1 or -1 mod 17.  The positive least of these is obviously 1, and for this $ x\equal{}0$.  If this is true, then $ y\equal{}2$ and $ z\equal{}\minus{}3$.  Thus the minimum value is $ \boxed{1}$.[/hide]

[hide="part 2"]We know already that $ y\equal{}x\minus{}\frac{2}{3}z$.  We try to find the minimum value of

$ \left|\left(x\plus{}z\right)y\plus{}xz\right|\equal{}\left|\left(x\plus{}z\right)\left(x\minus{}\frac{2}{3}z\right)\plus{}xz\right|\equal{}\left|x^2\plus{}xz\minus{}\frac{2}{3}xz\minus{}\frac{2}{3}z^2\plus{}xz\right|$
$ \equal{}\left|x^2\plus{}\frac{4}{3}xz\minus{}\frac{2}{3}z^2\right|\equal{}\left|x^2\plus{}2\left(\frac{2}{3}xz\right)\plus{}\frac{4}{9}z^2\minus{}\frac{10}{9}z^2\right|\equal{}\left|\left(x\plus{}\frac{2}{3}z\right)^2\minus{}\frac{10}{9}z^2\right|$
$ \equal{}\frac{1}{9}\left|\left(3x\plus{}2z\right)^2\minus{}10z^2\right|\equal{}\frac{1}{9}\left|\left(\frac{3}{5}\left(5x\minus{}2z\right)\plus{}\frac{16}{5}z\right)^2\minus{}10z^2\right|$
$ \equal{}\frac19\left|\left(\frac{3}{5}\cdot6\plus{}\frac{16}{5}z\right)^2\minus{}10z^2\right|\equal{}\frac1{225}\left|\left(18\plus{}16z\right)\minus{}250z^2\right|$
$ \equal{}\frac{1}{225}\left|6z^2\plus{}576z\plus{}324\right|\equal{}\frac2{75}\left|z^2\plus{}96z\plus{}54\right|$

Now $ z$ must be divisible by 3, so let $ z\equal{}3k$.

$ \equal{}\frac{2}{75}\left|9k^2\plus{}3\cdot96k\plus{}54\right|\equal{}\frac2{25}\left|3k^2\plus{}96k\plus{}18\right|$
$ \equal{}\frac6{25}\left|k^2\plus{}2\cdot16k\plus{}256\minus{}250\right|\equal{}6\left|\left(\frac{k\plus{}16}{5}\right)^2\minus{}10\right|$.

To minimize this, we look for the closest square to 10, which is obviously 9.  Solving for $ k$, we get:

$ \left(\frac{k\plus{}16}{5}\right)^2\equal{}9\Rightarrow k\plus{}16\equal{}\pm 15\Rightarrow k\equal{}\minus{}1, \minus{}31$

We plug these in for $ z$: $ 3\left(\minus{}1\right)\equal{}\minus{}3\equal{}z$ or $ 3\left(\minus{}31\right)\equal{}\minus{}93\equal{}z$.

Now solve for $ x$: $ 5x\equal{}6\plus{}2\left(\minus{}3\right)\equal{}0 \Rightarrow x\equal{}0$ or $ 5x\equal{}6\plus{}2\left(\minus{}93\right)\equal{}\minus{}180\Rightarrow x\equal{}\minus{}36$

Finally, we find $ y$: $ y\equal{}x\minus{}\frac{2}{3}z\equal{}0\minus{}\frac{2}{3}\left(\minus{}3\right)\equal{}2$ or $ y\equal{}\minus{}36\minus{}\frac{2}{3}\left(\minus{}93\right)\equal{}26$.

We see that for both triples $ \left(x, y, z\right)\equal{}\left(0, \minus{}3, 2\right), \left(\minus{}36, 26, \minus{}93\right)$, $ \left|xy\plus{}yz\plus{}zx\right|\equal{}\boxed{6}$ and we've shown that this is the minimum.[/hide]

[quote="kunny"]Let $ x,\ y,\ z$ be integers such that $ 2x \plus{} 3y \equal{} 5x \minus{} 2z \equal{} 6$. Find the minimum value of $ |x \plus{} y \plus{} z|,\ |xy \plus{} yz \plus{} zx|.$[/quote]

Solution:

From the given equations and conditions we easily get $ \ x\equal{}6\lambda,\ y\equal{}\minus{}4\lambda \plus{}2,\ z\equal{}15\lambda \minus{}3,\ \lambda\in\mathbb I$

For part one

$ |x \plus{} y \plus{} z|\equal{}|17\lambda \minus{}1|\ge 0$

Note $ \ 17\lambda \minus{}1$ can't be zero for any integral value of $ \ \lambda .$

$ \implies|17\lambda \minus{}1|\ge 1$ and equality for $ \ \lambda\equal{}0$

$ \implies\min |x \plus{} y \plus{} z|\equal{}1$ for $ \ x\equal{}0,\ y\equal{}2,\ z\equal{}\minus{}3\ .$

For part two

$ |xy \plus{} yz \plus{} zx|\equal{}|6\lambda^2 \plus{}36\lambda\minus{}6|\equal{}6|\lambda^2 \plus{}6\lambda\minus{}1|\equal{}6|(\lambda \plus{} 3)^2\minus{}10|$

Note $ \ (\lambda \plus{} 3)^2\minus{}10$ can't be zero for any integral value of $ \ \lambda .$

$ \implies|(\lambda \plus{} 3)^2\minus{}10|\ge 1$ and equality for $ \ \lambda\equal{}0,\ \minus{}6$

$ \implies\min |xy \plus{} yz \plus{} zx|\equal{}6$ for $ \ x\equal{}0,\ y\equal{}2,\ z\equal{}\minus{}3$ or for $ \ x\equal{}\minus{}36,\ y\equal{}26,\ z\equal{}\minus{}93\ .$

That's correct. :lol:

@tenniskidperson3

[quote="tenniskidperson3"]

[hide="part 2"]We know already that $ y \equal{} x \minus{} \frac {2}{3}z$.  We try to find the minimum value of

$ \left|\left(x \plus{} z\right)y \plus{} xz\right| \equal{} \left|\left(x \plus{} z\right)\left(x \minus{} \frac {2}{3}z\right) \plus{} xz\right| \equal{} \left|x^2 \plus{} xz \minus{} \frac {2}{3}xz \minus{} \frac {2}{3}z^2 \plus{} xz\right|$
$ \equal{} \left|x^2 \plus{} \frac {4}{3}xz \minus{} \frac {2}{3}z^2\right| \equal{} \left|x^2 \plus{} 2\left(\frac {2}{3}xz\right) \plus{} \frac {4}{9}z^2 \minus{} \frac {10}{9}z^2\right| \equal{} \left|\left(x \plus{} \frac {2}{3}z\right)^2 \minus{} \frac {10}{9}z^2\right|$
$ \equal{} \frac {1}{9}\left|\left(3x \plus{} 2z\right)^2 \minus{} 10z^2\right| \equal{} \frac {1}{9}\left|\left(\frac {3}{5}\left(5x \minus{} 2z\right) \plus{} \frac {16}{5}z\right)^2 \minus{} 10z^2\right|$
$ \equal{} \frac19\left|\left(\frac {3}{5}\cdot6 \plus{} \frac {16}{5}z\right)^2 \minus{} 10z^2\right| \equal{} \frac1{225}\left|\left(18 \plus{} 16z\right) \minus{} 250z^2\right|$
$ \equal{} \frac {1}{225}\left|6z^2 \plus{} 576z \plus{} 324\right| \equal{} \frac2{75}\left|z^2 \plus{} 96z \plus{} 54\right|$

Now $ z$ must be divisible by 3, so let $ z \equal{} 3k$.

$ \equal{} \frac {2}{75}\left|9k^2 \plus{} 3\cdot96k \plus{} 54\right| \equal{} \frac2{25}\left|3k^2 \plus{} 96k \plus{} 18\right|$
$ \equal{} \frac6{25}\left|k^2 \plus{} 2\cdot16k \plus{} 256 \minus{} 250\right| \equal{} 6\left|\left(\frac {k \plus{} 16}{5}\right)^2 \minus{} 10\right|$.

To minimize this, we look for the closest square to 10, which is obviously 9.  Solving for $ k$, we get:

$ \left(\frac {k \plus{} 16}{5}\right)^2 \equal{} 9\Rightarrow k \plus{} 16 \equal{} \pm 15\Rightarrow k \equal{} \minus{} 1, \minus{} 31$

We plug these in for $ z$: $ 3\left( \minus{} 1\right) \equal{} \minus{} 3 \equal{} z$ or $ 3\left( \minus{} 31\right) \equal{} \minus{} 93 \equal{} z$.

Now solve for $ x$: $ 5x \equal{} 6 \plus{} 2\left( \minus{} 3\right) \equal{} 0 \Rightarrow x \equal{} 0$ or $ 5x \equal{} 6 \plus{} 2\left( \minus{} 93\right) \equal{} \minus{} 180\Rightarrow x \equal{} \minus{} 36$

Finally, we find $ y$: $ y \equal{} x \minus{} \frac {2}{3}z \equal{} 0 \minus{} \frac {2}{3}\left( \minus{} 3\right) \equal{} 2$ or $ y \equal{} \minus{} 36 \minus{} \frac {2}{3}\left( \minus{} 93\right) \equal{} 26$.

We see that for both triples $ \left(x, y, z\right) \equal{} \left(0, \minus{} 3, 2\right), \left( \minus{} 36, 26, \minus{} 93\right)$, $ \left|xy \plus{} yz \plus{} zx\right| \equal{} \boxed{6}$ and we've shown that this is the minimum.[/hide][/quote]

The pair of $ (x,\ y,\ z)$ giving the equality is wrong. Please correct this as I informed it by private mail.

I was going to as soon as I read the PM, but the 1-day grace period in which I could edit it had already expired.  Here's the original post corrected:

[hide="part 1"]First, solve for $ y$ in terms of $ x$ and $ z$:
$ 2x \plus{} 3y \equal{} 5x \minus{} 2z$

$ 3y \equal{} 3x \minus{} 2z$

$ y \equal{} x \minus{} \frac {2}{3}z$

Plug this into $ \left|x \plus{} y \plus{} z\right|$ to get $ \left|x \plus{} x \minus{} \frac {2}{3}z \plus{} z\right| \equal{} \left|2x \plus{} \frac {1}{3}z\right| \equal{} \left|\frac {12x \plus{} 2z}{6}\right|$

$ \equal{} \left|\frac {17x}{6} \minus{} \frac {5x \minus{} 2z}{6}\right| \equal{} \left|\frac {17x}{6} \minus{} \frac {6}{6}\right| \equal{} \left|\frac {17x}{6} \minus{} 1\right|$

This is an integer, since $ x, y, z$ are all integers.  So that means $ x$ must be divisible by 6.  Then the function we're trying to minimize is either 1 or -1 mod 17.  The positive least of these is obviously 1, and for this $ x \equal{} 0$.  If this is true, then $ y \equal{} 2$ and $ z \equal{} \minus{} 3$.  Thus the minimum value is $ \boxed{1}$.[/hide]

[hide="part 2"]We know already that $ y \equal{} x \minus{} \frac {2}{3}z$.  We try to find the minimum value of

$ \left|\left(x \plus{} z\right)y \plus{} xz\right| \equal{} \left|\left(x \plus{} z\right)\left(x \minus{} \frac {2}{3}z\right) \plus{} xz\right| \equal{} \left|x^2 \plus{} xz \minus{} \frac {2}{3}xz \minus{} \frac {2}{3}z^2 \plus{} xz\right|$
$ \equal{} \left|x^2 \plus{} \frac {4}{3}xz \minus{} \frac {2}{3}z^2\right| \equal{} \left|x^2 \plus{} 2\left(\frac {2}{3}xz\right) \plus{} \frac {4}{9}z^2 \minus{} \frac {10}{9}z^2\right| \equal{} \left|\left(x \plus{} \frac {2}{3}z\right)^2 \minus{} \frac {10}{9}z^2\right|$
$ \equal{} \frac {1}{9}\left|\left(3x \plus{} 2z\right)^2 \minus{} 10z^2\right| \equal{} \frac {1}{9}\left|\left(\frac {3}{5}\left(5x \minus{} 2z\right) \plus{} \frac {16}{5}z\right)^2 \minus{} 10z^2\right|$
$ \equal{} \frac19\left|\left(\frac {3}{5}\cdot6 \plus{} \frac {16}{5}z\right)^2 \minus{} 10z^2\right| \equal{} \frac1{225}\left|\left(18 \plus{} 16z\right) \minus{} 250z^2\right|$
$ \equal{} \frac {1}{225}\left|6z^2 \plus{} 576z \plus{} 324\right| \equal{} \frac2{75}\left|z^2 \plus{} 96z \plus{} 54\right|$

Now $ z$ must be divisible by 3, so let $ z \equal{} 3k$.

$ \equal{} \frac {2}{75}\left|9k^2 \plus{} 3\cdot96k \plus{} 54\right| \equal{} \frac2{25}\left|3k^2 \plus{} 96k \plus{} 18\right|$
$ \equal{} \frac6{25}\left|k^2 \plus{} 2\cdot16k \plus{} 256 \minus{} 250\right| \equal{} 6\left|\left(\frac {k \plus{} 16}{5}\right)^2 \minus{} 10\right|$.

To minimize this, we look for the closest square to 10, which is obviously 9.  Solving for $ k$, we get:

$ \left(\frac {k \plus{} 16}{5}\right)^2 \equal{} 9\Rightarrow k \plus{} 16 \equal{} \pm 15\Rightarrow k \equal{} \minus{} 1, \minus{} 31$

We plug these in for $ z$: $ 3\left( \minus{} 1\right) \equal{} \minus{} 3 \equal{} z$ or $ 3\left( \minus{} 31\right) \equal{} \minus{} 93 \equal{} z$.

Now solve for $ x$: $ 5x \equal{} 6 \plus{} 2\left( \minus{} 3\right) \equal{} 0 \Rightarrow x \equal{} 0$ or $ 5x \equal{} 6 \plus{} 2\left( \minus{} 93\right) \equal{} \minus{} 180\Rightarrow x \equal{} \minus{} 36$

Finally, we find $ y$: $ y \equal{} x \minus{} \frac {2}{3}z \equal{} 0 \minus{} \frac {2}{3}\left( \minus{} 3\right) \equal{} 2$ or $ y \equal{} \minus{} 36 \minus{} \frac {2}{3}\left( \minus{} 93\right) \equal{} 26$.

We see that for both triples $ \left(x, y, z\right) \equal{} \left(0, 2, \minus{} 3\right), \left( \minus{} 36, 26, \minus{} 93\right)$, $ \left|xy \plus{} yz \plus{} zx\right| \equal{} \boxed{6}$ and we've shown that this is the minimum.[/hide]


If $ a,b,c \ge 0$
Prove that $ \frac 1 {a^2} \plus{} \frac 1 {b^2} \plus{} \frac 1 {c^2} \minus{} 27{\left( \frac {ab}c \plus{} \frac {bc}a \plus{} \frac {ca}b \right)}^{\minus{}2}$ $ \ge \frac 1 3 \left[ {\left(\frac 1 a \minus{} \frac 1 b \right)}^2 \plus{} {\left(\frac 1 b \minus{} \frac 1 c \right)}^2 \plus{} {\left(\frac 1 c \minus{} \frac 1 a \right)}^2 \right]$

[hide="solution"]
We have
$ \sum_{cyc}\frac{1}{a^2}\minus{}27\left(\sum_{cyc}\frac{ab}{c}\right)^{\minus{}2}\ge \frac{1}{3}\left[\sum_{cyc}\left(\frac{1}{a}\minus{}\frac{1}{b}\right)^2\right]\Longleftrightarrow$
$ \sum_{cyc}\frac{1}{a^2}\minus{}27\left(\sum_{cyc}\frac{a^2b^2}{c^2}\plus{}2\sum_{cyc}a^2\right)^{\minus{}1}\ge \frac{1}{3}\left[\sum_{cyc}\left(\frac{1}{a^2}\minus{}\frac{2}{ab}\plus{}\frac{1}{b^2}\right)\right]\Longleftrightarrow$
$ 3\sum_{cyc}\frac{1}{a^2}\minus{}81\left(\sum_{cyc}\frac{a^2b^2}{c^2}\plus{}2\sum_{cyc}a^2\right)^{\minus{}1}\ge \left(2\sum_{cyc}\frac{1}{a^2}\minus{}2\sum_{cyc}\frac{1}{ab}\right)\Longleftrightarrow$
$ \sum_{cyc}\frac{1}{a^2}\plus{}2\sum_{cyc}\frac{1}{ab}\ge 81\left(\sum_{cyc}\frac{a^2b^2}{c^2}\plus{}2\sum_{cyc}a^2\right)^{\minus{}1}\Longleftrightarrow$
$ \left(\sum_{cyc}\frac{1}{a^2}\plus{}2\sum_{cyc}\frac{1}{ab}\right)\left(\sum_{cyc}\frac{a^2b^2}{c^2}\plus{}2\sum_{cyc}a^2\right)\ge 81$
By Cauchy,
$ \left(\sum_{cyc}\frac{1}{a^2}\plus{}2\sum_{cyc}\frac{1}{ab}\right)\left(\sum_{cyc}\frac{a^2b^2}{c^2}\plus{}2\sum_{cyc}a^2\right)\ge \left(\sum_{cyc}\frac{b^2}{c^2}\plus{}2\sum_{cyc}\frac{a}{b}\right)^2\ge (3\plus{}2(3))^2\equal{}81$
where the last inequality comes from AM-GM.
[/hide]

[hide="Solution"]The inequality is homogeneous, so we may assume $ abc \equal{} 1$. Denote $ x \equal{} a^2b^2\plus{}b^2c^2\plus{}c^2a^2$. The LHS becomes $ x \minus{} 27/x^2$ and the RHS becomes $ \frac{1}{3}\sum (a\minus{}b)^2c^2 \equal{} \frac{1}{3}(2x \minus{} 2(a\plus{}b\plus{}c))$. The inequality is therefore equivalent to $ x^3 \plus{} 2(a\plus{}b\plus{}c)x^2 \geq 81$. But $ x \equal{} (ab)^2\plus{}(bc)^2\plus{}(ca)^2 \geq (ab)(bc)\plus{}(bc)(ca)\plus{}(ca)(ab) \equal{} a\plus{}b\plus{}c$, so $ x^3\plus{}2(a\plus{}b\plus{}c)x^2 \geq 3(a\plus{}b\plus{}c)^3$. AM-GM tells us $ a\plus{}b\plus{}c \geq 3\sqrt[3]{abc} \equal{} 3$, and that proves our inequality.[/hide]


induction
function
limit
continued fraction
calculus
calculus computations

Prove that if $ a_1\ne 2$ and $ a_n\equal{}a_{n\minus{}1}^2\minus{}2$ then
\[ \frac{1}{a_1}\plus{}\frac{1}{a_1\cdot a_2}\plus{}\frac{1}{a_1\cdot a_2\cdot a_3}\plus{}\dots\equal{}\frac{1}{2}\cdot\left(a_1\minus{}\sqrt{a_{1}^2\minus{}4}\right)\]
 :)

Substitute $ a_1\equal{}x\plus{}\frac{1}{x}$ (we need $ a_1>2$ for convergence). The identity becomes:
$ \sum_{n\equal{}1}^{\infty} \frac{1}{\prod_{i\equal{}1}^{n} (x^{2^{i\minus{}1}}\plus{}x^{\minus{}2^{i\minus{}1}})} \equal{} \frac{1}{x}$
or:
$ \sum_{n\equal{}1}^{\infty} \frac{x^{2^n}}{\prod_{i\equal{}1}^{n} (x^{2^i}\plus{}1)} \equal{}1$
And by induction we get:
$ S(m)\equal{}\sum_{n\equal{}1}^{m} \frac{x^{2^n}}{\prod_{i\equal{}1}^{n} (x^{2^i}\plus{}1)}\equal{}1\minus{}\frac{1}{\prod_{i\equal{}1}^{m} (x^{2^i}\plus{}1)}$.
Sending $ m$ to infinity gives us the limit $ 1$.

Question - can someone use the method of continued fractions here? It reminded me of the following topic (which I am not very familiar with): [url]http://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula[/url]
[BTW, what is the source? Or is it yours?]

PS. It reminded me of $ \sum_{n\equal{}1}^{\infty} \frac{(n\minus{}1)!}{\prod_{i\equal{}1}^{n} (x\plus{}i) }\equal{}\frac{1}{x}$ for $ x>0$ - a problem by Titu.

Sorry for double-posting, but I shared this nice problem with a friend and he came up with a beautiful solution (did I mention that he's an IMO gold medalist?  :lol: ).
Here it is-
Put $ a_1 \equal{} y \plus{} \frac {1}{y}, y \ge 1$. Denote the sum by $ S(y)$.
Because $ a_2\equal{}y^2\plus{}y^{\minus{}2}$, we easily get the following relation:
$ S(y) \equal{} \frac {1}{a_1}(1 \plus{} S(y^2)) \equal{} \frac {1}{y \plus{} \frac {1}{y}}(1 \plus{} S(y^2))$
In addition, $ S(y) \to 0$ when $ y \to \infty$, because $ a_i > y^i \ge y \implies S(y) \le \sum_{k \equal{} 1}^{\infty} \frac {1}{y^k} \equal{} \frac {1}{y \minus{} 1}$. So $ S$ converges for $ a_1 \ge 2$ and it is continuous as a function of $ y$, $ y \ge 1$ (this part is not so rigorous...).
Put $ h(y) \equal{} S(y) \minus{} \frac {1}{y}$. The relation we found is now equivalent to $ h(y)(y \plus{} \frac {1}{y}) \equal{} h(y^2), y \ge 1$.
Notice that $ h(1) \equal{} S(1) \minus{} 1 \equal{} \sum_{i \equal{} 1}^{\infty} \frac {1}{2^i} \minus{} 1 \equal{} 0, \lim_{y \to \infty} h(y) \equal{} 0$.
The relation for $ h$ implies the bound $ |h(y^2)| \ge 2|h(y)|$, or: $ |h(y)|\le \frac {|h(y^{2^k})|}{2^k}$. Because $ y^{2^k} \to \infty$ when $ k \to \infty$, and $ \lim_{y \to \infty} h(y) \equal{} 0$, we get that $ h(y) \equiv 0$, or: $ S(y) \equal{} \frac {1}{y}$.
Solving $ a_1 \equal{} y \plus{} \frac {1}{y}, y \ge 1$ gives us the claim.


could u clarify what is your question finally?
$ bc\equal{}\sqrt{a^2\plus{}(2b\plus{}1)^2\plus{}1}\plus{}1$
or
$ bc\equal{}\sqrt{a^2\plus{}(2b\plus{}1)^2\plus{}3}\plus{}1$.
thx.

[quote="SUPERMAN2"]Find all positive integers $ a,b,c$ satisfying $ bc \equal{} \sqrt {a^2 \plus{} (2b \plus{} 1)^2 \plus{} 3} \plus{} 1$[/quote]
The number $ 3$ is correct.There are no changes.


The base of an isosceles triangle is 6 cm long. If the area of the triangle cannot exceed 12 square centimeters, what is the maximum number of centimeters in the perimeter of the triangle?

Since the area can't exceed 12, $ \frac{6\times h}{2}\leq 12\Rightarrow h\leq 4$.  In this scenario, we can break down the original triangle into 2 right triangles, with base 3, height 4, and hence diagonal 5.  The the other two sides of the triangle have length 5.  Hence the perimeter can't exceed $ 6\plus{}5\plus{}5\equal{}16$.


Out of the Michigan Mathematics League problem set from today:

"There are no non-negative integers (x,y) that satisfy 9x+16y = 78. What is the largest integer n for which no non-negative integers (x,y) satisfy 9x+16y=n?"

None of the kids taking it, nor the head of the math department knew how to solve it. Ideas? (if this belongs in advanced, feel free to move it)

This is currently being discussed on another thread on this forum as PIE.

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1932

Isn't it well-known that if a,b are positive integers with gcd(a,b) = 1, then

ab - a - b

is the greatest positive integer which can be written as

xa + yb

for positive integers x, y ?

A proof of this useful fact can be found at

http://www.kalva.demon.co.uk/imo/isoln/isoln833.html

yup!
unfortunately, the mathleague test makers seem to not know this. their official solution relies on making a kind of sieve where you check each number to see if it has solutions. it is a very stupid solution.

it also says something like note: in the equation ax+by=c where (a,b)=1, ab-(a+b) is the largest value of c such that there are no solutions.  So they do know.  I dont think those questions were so great though.... i dont know.  all of the recent contest questions that i've done havent been too interesting.

Besides, a solution with a method is better than a solution that relies on a fact that some people might not know.  Because it shows that the problem wasn't just a "I don'tknow the formula, so I better just give up now" - type problem.

i must have missed that note. im just mentioning that they could have done the problem in a better way, like a few of the solutions in the other post. instead, the only solution posted is the list.


function
limit
logarithms
calculus
calculus computations

Hello
try find all functions f:[1,+00)->R
if they are continious and  $ 2xf\left( {{x^2}} \right) \minus{} \left( {x \plus{} 1} \right)f\left( x \right) \equal{} 0,x \ge 1$

$ f(x)\equal{}f(1)\cdot\prod_{n\equal{}1}^{\infty}\frac{1\plus{}x^{2^{\minus{}n}}}{2x^{2^{\minus{}n}}}$; that product always converges, and $ \lim_{t\to 1^\plus{}}f(t)\equal{}f(1)$.

We can evaluate that product: $ 1\plus{}x^{2^{\minus{}n}}\equal{}\frac{x^{2^{1\minus{}n}}\minus{}1}{x^{2^{\minus{}n}}\minus{}1}$, so $ \prod_{n\equal{}1}^{N}\frac{2x^{2^{\minus{}n}}}{1\plus{}x^{2^{\minus{}n}}}\equal{}\frac{x\minus{}1}{x^{2^{\minus{}N}}\minus{}1}\cdot 2^{\minus{}N}x^{\minus{}1\plus{}2^{\minus{}N}}$
Using that $ \lim_{t\to 0}\frac{x^t\minus{}1}{t}\equal{}\ln x$, the infinite product is $ \frac{x\minus{}1}{x\ln x}$, and $ f(x)\equal{}\begin{cases}c&x\equal{}1\\c\frac{x\minus{}1}{x\ln x}&x>1\end{cases}$.

jmerry thx for explanation and solution :lol:


The direct distance between city A and city B is 200 miles. The direct distance from B to city C is 300 miles. What could be the distance from A to C?

A. 50
B. 550
C. 350
D. 650

I got C as an answer, however I did it completely randomly... I drew a diagram as a triangle, and used the Triangle inequality theorem to solve it and got 350. Am I right or wrong? If wrong, can you explain how to do it?

1) if BC is the longest

AB+AC>300, 200+AC>300, AC>100.

2) if AC is the longest

AB+BC>CA, 500>AC.

so 500>AC>100.

theres only one letter that satisfies above interval.

answer : C(350)

yes; ur right.

Or just think of it this way. The lower limit is the difference of BC and AB. The upper limit is the sum of BC and AB.

These can be achieved by having A, B, and C be collinear, such that:

(1) A is between B and C.
(2) B is between A and C.

So your interval is $ 300\minus{}200<d<300\plus{}200$
$ \Longrightarrow100<d<500$ (in miles)