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Calculate \[ \lim_{n\to \infty}\left \{ \frac{1}{n^4}\left ( \sum_{k\equal{}1}^{n}k^2\int_{k}^{k\plus{}1}x\ln\left [ (x\minus{}k)(k\plus{}1\minus{}x) \right ] dx\right ) \right \}.\]

Let $ x \minus{} k \equal{} t$, we have $ \int_k^{k\plus{}1} \left [ (x \minus{} k)(k \plus{} 1 \minus{} x) \right ] dx \equal{} \int_0^1 (t \plus{} k)\ln t(1 \minus{} t)\ dt$

$ \equal{} \int_0^1 (t \plus{} k)\ln t\ dt \plus{} \int_0^1 (t \plus{} k)\ln (1 \minus{} t)\ dt$.

It is problem 859, vol.38, n.4 Sept.2007 of College Math.Jour., author Jose' Luis Diaz-Barrero. The answer is $ \minus{} 1/2.$

The integral is
\[ \begin{array}{ll} & \int_k^{k \plus{} 1} \bigl(x\ln (x \minus{} k) \plus{} x\ln (k \plus{} 1 \minus{} x)\bigr)dx \equal{} \\
& \lim_{a\to k^ \plus{} } \int_a^{k \plus{} 1} x\ln (x \minus{} k)dx \plus{} \lim_{b\to (k \plus{} 1)^ \minus{} } \int_k^b x\ln (k \plus{} 1 \minus{} x)dx \\
\end{array}\]

\[ \begin{array}{ll} & \int_a^{k \plus{} 1} x\ln (x \minus{} k)dx \equal{} {x^2\over 2}\ln(x \minus{} k)\Bigl\vert_a^{k \plus{} 1} \minus{} {1\over 2}\int_a^{k \plus{} 1} {x^2\over x \minus{} k} dx \equal{} \\
& {x^2\over 2}\ln(x \minus{} k)\Bigl\vert_a^{k \plus{} 1} \minus{} {1\over 2}\int_a^{k \plus{} 1} (x \plus{} k)dx \minus{} {1\over 2} \int_a^{k \plus{} 1} {k^2\over x \minus{} k} dx \equal{} \\
& \minus{} {a^2\over 2}\ln(a \minus{} k) \plus{} {1\over2} \bigl({a^2\over2} \plus{} ka) \minus{} {1\over2} ({(k \plus{} 1)^2\over2} \plus{} k(k \plus{} 1)\bigr) \plus{} {k^2\over 2}\ln(a \minus{} k) \\
\end{array}\]
whose limit $ a\to k^ \plus{}$ is $ \minus{} (4k \plus{} 1)/4$
\[ \begin{array}{ll} & \int_k^b x\ln (k \plus{} 1 \minus{} x)dx \equal{} {x^2\over 2}\ln(k \plus{} 1 \minus{} x)\Bigl\vert_k^b \plus{} {1\over 2}\int_k^b {x^2\over k \plus{} 1 \minus{} x} dx \equal{} \\
& {b^2\over 2}\ln(k \plus{} 1 \minus{} b) \minus{} {1\over 2}\int_k^b (x \plus{} k \plus{} 1)dx \plus{} {1\over 2} \int_k^b {(k \plus{} 1)^2\over k \plus{} 1 \minus{} x} dx \equal{} \\
& {b^2\over 2}\ln(k \plus{} 1 \minus{} b) \plus{} {1\over2} \bigl({k^2\over2} \plus{} k(k \plus{} 1) \minus{} {b^2\over2} \plus{} b(k \plus{} 1)\bigr) \minus{} {(k \plus{} 1)^2\over 2}\ln(k \plus{} 1 \minus{} x)\Bigl\vert_k^b \equal{}\\
& {b^2\over 2}\ln(k \plus{} 1 \minus{} b) \plus{} {1\over2} \bigl({k^2\over2} \plus{} k(k \plus{} 1) \minus{} {b^2\over2} \plus{} b(k \plus{} 1)\bigr) \minus{} {(k \plus{} 1)^2\over 2}\ln(k \plus{} 1 \minus{} b) \\
\end{array}\]
whose limit $ b\to (k \plus{} 1)^ \minus{}$ is $ \minus{} (4k \plus{} 3)/4.$ The sum of the integrals gives $ \minus{} 2k \minus{} 1.$
Now we can perform the sum
\[ \sum_{k \equal{} 1}^n k^2 ( \minus{} 2k \minus{} 1) \equal{} \minus{} 2(\sum_{k \equal{} 1}^n k)^2 \minus{} \sum_{k \equal{} 1}^n k^2 \equal{} \minus{} 2 \Bigl( {n(n \plus{} 1)\over2}\Bigr)^2 \minus{} {n(n \plus{} 1)(2n \plus{} 1)\over6}\]
and finally
\[ \lim_{n\to\infty} {1\over n^4} \left\{ \minus{} 2 \Bigl( {n(n \plus{} 1)\over2}\Bigr)^2 \minus{} {n(n \plus{} 1)(2n \plus{} 1)\over6}\right\} \equal{} \minus{} {1\over 2}\]

yes, i've heard this problem was originally taken from a journal..i just don't know what it is..but this also appeared as a contest problem in the last year's San Martin College's Search for the Math Wizard in Philippines...and fortunately, i have the same solution with the original proposer..thanks Arrigo-Sacchi for telling me where this problem from...it's not a hard problem though, it's just that - lengthy routine calculations...

A solution to a somewhat more general problem:

$ \lim_{n\to\infty}\left\{\frac {1}{n^{4}}\left (\sum_{k \equal{} 1}^{n}k^{2}\int_{k}^{k \plus{} 1}x f(x \minus{} k)\,dx\right )\right\}$

For the integral, use the substitution $ x \minus{} k \equal{} t$:

$ \lim_{n\to\infty}\left\{\frac {1}{n^{4}}\left (\sum_{k \equal{} 1}^{n}k^{2}\int_{0}^{1}(t \plus{} k) f(t)\,dt\right )\right\} \equal{}$

$ \lim_{n\to\infty}\left\{\frac {1}{n^{4}}\left (\sum_{k \equal{} 1}^{n}k^{3} \int_{0}^{1}f(t)\,dt \plus{} k^{2}\int_{0}^{1}t f(t)\,dt\right )\right\} \equal{}$

$ \lim_{n\to\infty}\left\{\frac {1}{n^{4}}\left (\alpha \sum_{k \equal{} 1}^{n}k^{3} \plus{} \beta \sum_{k \equal{} 1}^{n}k^{2}\right )\right\} \equal{}$

$ \lim_{n\to\infty}\left\{\frac {1}{n^{4}}\left (\alpha \sum_{k \equal{} 1}^{n}k^{3} \right )\right\} \equal{}$

$ \alpha \lim_{n\to\infty}\left\{\frac {1}{n}\left ( \sum_{k \equal{} 1}^{n}\left(\frac {k}{n}\right)^{3} \right )\right\} \equal{}$

$ \alpha \int ^1 _ 0 x^3 \, dx \equal{} \frac {\alpha}{4} \equal{} \frac {1}{4} \int_{0}^{1}f(t)\,dt$

In our example, we have $ f(t) \equal{} \ln\left(t(1 \minus{} t)\right) \equal{} \ln t \plus{} \ln(1 \minus{} t)$ so:

$ \frac {1}{4}\int_{0}^{1}f(t)\,dt \equal{} \frac {1}{4}\int_{0}^{1}\ln t(1 \minus{} t)\,dt \equal{} \frac {1}{2}\int_{0}^{1}\ln t\,dt \equal{} \frac {1}{2}\,t\, (\ln t \minus{} 1)\Big|^1_0 \equal{} \minus{} \frac {1}{2}$

[quote="milin"]
In our example, we have $ f(t) \equal{} \ln\left(t(1 \minus{} t)\right) \equal{} \ln t \plus{} \ln(1 \minus{} t)$ so:

$ \frac {1}{4}\int_{0}^{1}f(t)\,dt \equal{} \frac {1}{4}\int_{0}^{1}\ln t(1 \minus{} t)\,dt \equal{} \frac {1}{2}\int_{0}^{1}\ln t\,dt \equal{} \frac {1}{2}\,t\, (\ln t \minus{} 1)\Big|^1_0 \equal{} \minus{} \frac {1}{2}$[/quote]

At first sight, the integration smelt Riemman Integral.

I thought in such a way, but $ \frac {1}{4}\int_{0}^{1}\ln t(1 \minus{} t)\,dt$ was a wall to me.

That part is somewhat short in my previous post, so to clear all doubt:

$ \frac{1}{4}\int_{0}^{1}\ln t(1\minus{}t)\,dt \equal{}\frac{1}{4}\int_{0}^{1}\ln t\,dt \plus{} \frac{1}{4}\int_{0}^{1}\ln (1\minus{}t)\,dt$

Now for the second integral use the substitution $ 1\minus{}t\equal{}u$:

$ \frac{1}{4}\int_{0}^{1}\ln t\,dt \plus{} \frac{1}{4}\int_{1}^{0}\ln u \, (\minus{}du)\equal{}$

$ \frac{1}{4}\int_{0}^{1}\ln t\,dt \plus{} \frac{1}{4}\int_{0}^{1}\ln u \, du\equal{}$

$ \frac{1}{2}\int_{0}^{1}\ln t\,dt$

Use partial integration with $ dv \equal{} 1 \to v \equal{} t$ and $ u \equal{} \ln t \to du \equal{} \frac{1}{t}$:

$ \frac{1}{2}\int_{0}^{1}\ln t\,dt \equal{}$

$ \frac{1}{2} t \ln t \Big| ^1_0 \minus{} \frac{1}{2} \int^1_0 dt \equal{}$

$ \minus{}\frac{1}{2} \lim _{t \to 0} t \ln t\minus{} \frac{1}{2} t \Big|^1_0 \equal{}$

$ \frac{1}{2} \lim _{t \to \infty}\frac{\ln t}{t}\minus{} \frac{1}{2} \equal{} \minus{} \frac{1}{2}$

[quote="kunny"]Thank you for the explanation in detail, but I can understand it.
I feel strange that $ \int_0^1 \ln t\ dt$, how do we define $ \ln 0?$[/quote]

$ \minus{}\int_{t\equal{}0}^1 \log t \, dt \equal{} \int_{t\equal{}0}^\infty e^{\minus{}t} \, dt \equal{} \lim_{n \to \infty} \left[ \minus{}e^{\minus{}t} \right]_{t\equal{}0}^n \equal{} 1 \minus{} \lim_{n \to \infty} e^{\minus{}n} \equal{} 1 \minus{} 0 \equal{} 1$,

from which it follows that the given integral is equal to -1.

We can even use a double integration approach for that integral.

[quote="atomicwedgie"][quote="kunny"]Thank you for the explanation in detail, but I can understand it.
I feel strange that $ \int_0^1 \ln t\ dt$, how do we define $ \ln 0?$[/quote]

$ \minus{} \int_{t \equal{} 0}^1 \log t \, dt \equal{} \int_{t \equal{} 0}^\infty e^{ \minus{} t} \, dt \equal{} \lim_{n \to \infty} \left[ \minus{} e^{ \minus{} t} \right]_{t \equal{} 0}^n \equal{} 1 \minus{} \lim_{n \to \infty} e^{ \minus{} n} \equal{} 1 \minus{} 0 \equal{} 1$,

from which it follows that the given integral is equal to -1.[/quote]

That's an understandable solution to me including my high school students.

Thank you! atomicwedgie :lol:

than you guys for sharing your thoughts!  :lol:


geometry
analytic geometry
graphing lines
slope
vector
parallelogram
AMC

Triangle $ABC$ lies in the Cartesian Plane and has an area of 70. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20)$, respectively, and the coordinates of $A$ are $(p,q)$. The line containing the median to side $BC$ has slope $-5$. Find the largest possible value of $p+q$.

[hide="Solution"]$BC$ is the base of the altitude with length $\sqrt{122}$ by Pythagorean Theorem and slope $\frac{1}{11}$, and therefore equation $y=\frac{1}{11}x+\frac{197}{11}$. The median has slope $-5$ and goes through the midpoint of $BC$, which is at $\left(\frac{35}{2},\frac{39}{2}\right)$, so it has equation $y=-5x+107$. Finally, the altitude is perpendicular to the base, so it has slope $-11$. Since we know the area and the base, we have $70=\frac{1}{2}(\sqrt{122})h\Rightarrow h=\frac{140}{\sqrt{122}}$. Drawing the right triangle with the altitude as the hypotenuse, then connecting the longer leg to the base, we have two similar triangles, the larger with longer leg $11x$ and the shorter leg $x$, and therefore the smaller with longer leg $x$ and shorter leg $\frac{1}{11}x$. Next, we draw a line parallel to the base through $A(p,q)$, which the altitude is also perpendicular to. $x^2+(11x)^2=\frac{9800}{61}\Rightarrow 122x^2=\frac{9800}{61}\Rightarrow x=\frac{70}{61}$, so the vertical distance between the two parallel lines is $\frac{122}{11}\cdot \frac{70}{61}=\frac{140}{11}$, so we add that distance to the y-intercept of the base and see that the equation of the parallel line is $y=\frac{1}{11}x+\frac{337}{11}$.

Next, we solve the system of equations for where $\left(y=\frac{1}{11}x+\frac{337}{11}\right)=(y=-5x+107)$, so we have $y=32$, and substituting that back in for the equation of the median, we have $x=15$, so $32+15=\boxed{47}$.[/hide]

[quote="JesusFreak197"]Drawing the right triangle with the altitude as the hypotenuse, then connecting the longer leg to the base...[/quote] I don't really understand this part. Can you please explain?

Sorry, that was a little weird. So, you have a right triangle, with the longer leg being the shortest distance between the two lines you're looking at. Then, you draw the altitude to the hypotenuse of that right triangle, and this forms a new right triangle, where that altitude is the hypotenuse. This is the one I was talking about.

Does anyone know an area formula for a triangle using coordinates?  It would have helped here.

I don't know if that would have helped to find the last point, although I suppose it's possible. It might have cut short some of the calculations, but you'd still have to do others. Also, I'm not sure that one even exists. :P

[quote="JesusFreak197"]Also, I'm not sure that one even exists. :P[/quote]

Of course it does :) Look [url=http://mathworld.wolfram.com/TriangleArea.html]here[/url]

the area of the triangle with vertices at $(a, b)$, $(c, d)$, and $(e, f)$ is given by the absolute value of:

 \[ \frac{1}{2} \left| \begin{array}{ccc} a & b & 1 \\ c & d & 1\\ e & f & 1 \end{array} \right| \]

Proof: use the fact that cross products of two vectors gives a vector perpendicular to both vectors with a magnitude equivalent to the parallelogram spamed by the two vectors.

I haven't actually tried this, and it may be taking the whole brute force approach to an extreme, but given that a compass, straightedge, and graph paper are all allowed on the test, wouldn't a euclidean construction of the triangle (based on the median, and the fact that the square root of two times area is the geometric means of the base and height) be faster for finding (p,q)?

knowing the final solution must be integer, I'm thinking the margin for error could be negligible...

Don't make a mistake. ;)

I actually heard that someone on AoPS once tried that to find the degree measure of an angle. He got $81$ degrees, and the answer was $82$ (or maybe the other way around, I can't remember).

Heh, it's funny how I had no idea how to solve this last year, but I saw the solution right after reading the problem today.  :D

[hide="Vectors"]Denote the midpoint of $\overline{BC}$ as $M,$ so $M=(17.5, 19.5).$ Since $A$ lies on the line through $M$ with slope $-5,$ the equation of the line $A$ must lie on is
\[y-19.5=-5(x-17.5)\implies y=-5x+107. \]Now let $A=(x,-5x+107).$

Note: I'm going to use the cross product for area in the 3D form, as it should come out more nicely. (Only one non-zero term in the cross product.)  :)

It follows that $\overrightarrow{AB}=<x-12,-5x+88,0>$ and $\overrightarrow{AC}=<x-23,-5x+87,0>.$ Hence,
\begin{eqnarray*}[ABC]=70&=& \frac{1}{2}\left| \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x-12 &-5x+88 & 0\\ x-23 &-5x+87 & 0 \end{array}\right| \\ &=&\frac{1}{2}\parallel <0,0,(x-12)(-5x+87)-(x-23)(-5x+88)>\parallel \end{eqnarray*}Some quick calculations show that\begin{eqnarray*}70&=& \frac{1}{2}(x-12)(-5x+87)-\frac{1}{2}(x-23)(-5x+88)\\ &=& 490-28x\implies x=15. \end{eqnarray*}Therefore, $y=107-5(15)=32$ and $(p,q)=(15,32)\implies p+q=47.$
[/hide]

The coordinate of the centroid is:
$ (\frac{p\plus{}35}{3},\frac{q\plus{}39}{3}),$ and the median to BC passes through $ (\frac{35}{2}, \frac{39}{2}),$ so we have that $ \frac{\frac{q\plus{}39}{3}\minus{}\frac{39}{2}}{\frac{p\plus{}35}{3}\minus{}\frac{35}{2}}\equal{}\minus{}5$, and we can use the determinant formula for area to get two linear equations in p and q, which we solve to get $ p \plus{} q \equal{} 47$

Could someone post a solution using the Shoelace Theorem?

[quote="gaussintraining"][hide="Another Solution"] We know that the area of the triangle is

$ \frac12\left|\begin{array}{ccc}12 & 19 & 1 \\
23 & 20 & 1 \\
p & q & 1\end{array}\right| \equal{} 70$.

Evaluating this, we find that

$ 12(20 \minus{} q) \minus{} 19(23 \minus{} p) \plus{} 23q \minus{} 20p \equal{} 140$

$ \implies 240 \minus{} 12q \minus{} 437 \plus{} 19p \plus{} 23q \minus{} 20p \equal{} 140$

$ \implies 11q \minus{} p \equal{} 337$.

Now, the midpoint of side $ BC$ is $ (\frac {35}{2},\frac {39}{2})$. Thus, from point-slope form of a line, the equation of the median to BC is

$ \frac {y \minus{} \frac {39}{2}}{x \minus{} \frac {35}{2}} \equal{} \minus{} 5$

$ \implies 5x \minus{} \frac {175}{2} \equal{} \minus{} y \plus{} \frac {39}{2}$

$ \implies 5x \plus{} y \equal{} 107$.

$ (p,q)$ is a point on this line, so we have that

$ 5p \plus{} q \equal{} 107$.

Thus, we now have a system of equations.

$ 11q \minus{} p \equal{} 337$

$ 5p \plus{} q \equal{} 107$.

Multiplying the first equation by $ 5$ and adding the resulting equation to the second yields

$ 56q \equal{} 1792 \implies q \equal{} 32$.

We can plug this back into the second equation to find that $ p \equal{} 15$.

Thus, $ p \plus{} q \equal{} 15 \plus{} 32 \equal{} \boxed{47}$.[/hide]

This is kind of what mathemonster did.[/quote]

That is exactly how I did it, but I accidently started with slope 5. :wallbash:

[url]http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem[/url]

A really cool theorem to use for this problem... The proof is a little dense but its easy to remember and an awesome go-to when encountering a geometry problem in the coordinate plane.

You can solve this problem with the distance from point to line formula, if you actually remember it.

[hide="Solution"]
Denote $M$ as the midpoint of $\overline{BC}$. Since $B = (12,19), C = (23,20)$, we have that $M = (17.5,19.5)$. Then since the slope of $\overline{AM}$ is $-5$, we have the equation \[\dfrac{q-19.5}{p-17.5} = -5 \implies q = -5p + 107\]
Now, we want another equation in terms of $p$ and $q$. We can use the other given fact, $[ABC] = 70$, to help us find this equation. Since $BC = \sqrt{(23-12)^2+(20-19)^2} = \sqrt{122}$, we can express $[ABC]$ in terms of the altitude from $A$ to $BC$. To find the altitude, we can use the distance from point to line formula, which states that the perpendicular distance from a point $(x_0,y_0)$ to a line $Ax + By + C = 0$ is just \[\dfrac{Ax_0 + By_0 + C}{\sqrt{A^2+B^2}}\]
Thus, we now want the equation of $\overline{BC}$. Standard methods (i.e. finding the slope and plugging in a point) give $-x+11y -197 = 0$. Now, we can apply the above formula to find the distance from $(p,q)$ to $\overline{BC}$, and then equate areas:
\begin{align*}\dfrac{\sqrt{122}\left(\dfrac{-p+11q-192}{\sqrt{122}}\right)}{2} &= 70 \\ -p+11q-192&= 140 \\ -p + 11q &= 337 \end{align*}
We found earlier that $q = -5p + 107$. Using this fact, we get
\begin{align*} -p + 11(-5p+107) &= 337 \\ -p - 55p + 1177 &= 337 \\ -56p &= -840 \\ p &= 15 \\ q &= -5(15) + 107 = 32 \\ \implies p + q &= 15 + 32 = \boxed{047} \end{align*}
[/hide]

This solution goes pretty fast if you have the formula memorized.

PS: When I solved this problem, I got that $-56p = 337 - 1177 \implies -56p = 840$. Darn. This gave me that $p=-15, q = 182$, so I got the incorrect answer of $167$. Careless mistakes will wreck you...

its just shoelace theorem and then slope.

It's really stupid how this problem was so trivial if you know Shoelace. This type of question should not have been an AIME #10 considering it was basically a small coordinate bash in conjunction with Shoelace.

AIME Grind: By Shoelace the area of this triangle is $|12 \cdot 20 + 23 \cdot q + p \cdot 19 - 12 \cdot q - 23 \cdot 19 - p \cdot 20| = \frac{1}{2}|11q-p-197| = 70$.

The midpoint of $BC$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$, so the slope of the median is $\frac{\frac{39}{2}-q}{\frac{35}{2}-p} = -5$. Taking cases based on the absolute value gives solutions of $(p,q)=(15,32), (20,7) \implies 15+32=\boxed{47}$.


Source: Austria MO 1985
5. A sequence $(a_{n})$ of positive integers satisfies $a_{n}=\sqrt{\frac{a_{n-1}^{2}+a_{n+1}^{2}}{2}}$ for all $n\geq 1$. Prove that this sequence is constant.

Can anyone check my approach, thanks
[hide="my approach"]
First, squaring both sides and multiply by 2 we get:
$2a_{n}^{2}=a_{n-1}^{2}+a_{n+1}^{2}$----$(1)$
Now replace $n$ with $n+1$ we get:
$2a_{n+1}^{2}=a_{n}^{2}+a_{n+2}^{2}$----$(2)$ or
$a_{n}^{2}=2a_{n+1}^{2}-a_{n+2}^{2}$.
Sub in this $a_{n}^{2}$ value into $(1)$ and rearranging
$3a_{n+1}^{2}=a_{n-1}^{2}+2a_{n+2}^{2}$------$(3)$.
Also from $(2)$, we get
$a_{n+1}^{2}=\frac{a_{n}^{2}+a_{n+2}^{2}}{2}$.
Sub in this into $(1)$, we get:
$3a_{n}^{2}=2a_{n-1}^{2}+a_{n+2}^{2}$------$(4)$.
Adding $(3)$ and $(4)$ and dividing by $3$ on both sides and rearranging:
$a_{n}^{2}-a_{n-1}^{2}=a_{n+2}^{2}-a_{n+1}^{2}$.
I don't know what to do now.... am i heading for a right direction? or is it totally wrong(hopefully not)??
[/hide]

wait, what does it mean by the constant sequence?
I thought that it meant that the sequence doesn't change...
(ex. $1,1,1,1,1,1,1,...$).

Yes...thats what my results imply too.  :wink:

[quote="boxedexe"]Let $a_{n}=\sqrt{x_{n}}$.  Thus we have $2x_{n}=x_{n-1}+x_{n+1}\implies u^{2}-2u+1=0$, which gives us a root of multiplicity 2, namely 1.  The equation in $u$ is the characteristic equation of the sequence. So $x_{n}=A(1)^{n}=A=a_{n}^{2}$, thus $a_{n}=\sqrt{A}$, where $A\in\mathbb{Z}^{+}$, a constant. Q.E.D.

Masoud Zargar[/quote]
Sorry, I don't really understand your solution... :blush:

[quote="FieryHydra"][quote="kimby_102"]Source: Austria MO 1985

Can anyone check my approach, thanks
[hide="my approach"]
First, squaring both sides and multiply by 2 we get:
$2a_{n}^{2}=a_{n-1}^{2}+a_{n+1}^{2}$----$(1)$
Now replace $n$ with $n+1$ we get:
$2a_{n+1}^{2}=a_{n}^{2}+a_{n+2}^{2}$----$(2)$ or
$a_{n}^{2}=2a_{n+1}^{2}-a_{n+2}^{2}$.
Sub in this $a_{n}^{2}$ value into $(1)$ and rearranging
$3a_{n+1}^{2}=a_{n-1}^{2}+2a_{n+2}^{2}$------$(3)$.
Also from $(2)$, we get
$a_{n+1}^{2}=\frac{a_{n}^{2}+a_{n+2}^{2}}{2}$.
Sub in this into $(1)$, we get:
$3a_{n}^{2}=2a_{n-1}^{2}+a_{n+2}^{2}$------$(4)$.
Adding $(3)$ and $(4)$ and dividing by $3$ on both sides and rearranging:
$a_{n}^{2}-a_{n-1}^{2}=a_{n+2}^{2}-a_{n+1}^{2}$.
I don't know what to do now.... am i heading for a right direction? or is it totally wrong(hopefully not)??
[/hide][/quote]
I think there is no need to manipulate so many terms.
$2{a_{n}}^{2}={a_{n-1}}^{2}+{a_{n+1}}^{2}$
$\Rightarrow{a_{n+1}}^{2}-{a_{n}}^{2}={a_{n}}^{2}-{a_{n-1}}^{2}$
$\Rightarrow a_{n}^{2}-a_{n-1}^{2}=a_{n+2}^{2}-a_{n+1}^{2}$, which is the equation you derived last.[/quote]
Wow , how stupid i am..... :D


Two real numbers x and y are randomly selected on a number line between 0 to 12. Find the probability that their sum is greater than 15.
using geometric probability i get 9/32 as the solution... is this right???

[quote="fizsle"]
using geometric probability i get 9/32 as the solution... is this right???[/quote]

I assume you used a triangle?

Well, here is another solution:

1+2+3+4+5+6+7+8+9 = 45 possibilities such that the sum is greater than 15.

Total sums are: 6x12 = 72

45/72 = 5/8 is my answer.

Neelnal, it says "REAL numbers" - you can't reduce the problem to integers.

Fiszle, your answer is correct.

Oops.

so, yes, using triangles and the fact that in triangle ABC,

AB + BC > AC
Yes, the answer is 9/32


function
algorithm
trigonometry
advanced fields
advanced fields theorems

Hi!
I have a bit weird request, I'm working with genetic algorithms and have written a program for parallel computation which finds max in function.
to my request, those functions I have are a bit "week" in sence that they are "easy" to solve.
I hope some of you can help me with function that on interval [0..MAX(2147483647)] have many local maximum, but only one real global

Thank you for you help!

Did you try $\prod_{k=1}^{10}(1+\sin(2\pi x/p_{k}))$ where $p_{k}$ is the $k-th$ odd  prime number ? The maximum of this one is easy to find in your head, but can your program come anywhere close to the true value?

thanx man! :D
could you by more precise and explaine what you mean by odd primes?
and maybe post what max is?

The odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, ..., i.e. all the prime numbers other than 2.

3,5,7,11,13,17,19,23,...

OK, 10 may be actually 8 here but what I meant is that as long as the product of your primes is less than that intMax number you mentioned, you have all sets of remainders modulo each $p_{k}$, so the maximum of the product is just the product of the maxima.

Fedja, you have made a small mistake: you need all the $p_{k}$ to be congruent mod 4 or something:

$2\pi x/3 = (2m+1/2)\pi$ implies $x = 3m+\frac{3}{4}$ for some integer $m$.  Then $2\pi x/5 \neq (2n+1/2)\pi$ for any integer $n$.

Yes, you are right :blush:. For some strange reason I was thinking of integers only :roll:. The idea is the same, nevertheless: take some particular $x$ you want to be your maximum; design a few ($1+\sin$)-like periodic functions with different periods  each of which takes the maximal value at $x$ and multiply them out.


For $ a,b,c$ positive reals

$ \sum \frac{1}{a\plus{}b}\plus{}\frac{3abc}{2(ab\plus{}bc\plus{}ca)^2}\geq \frac{5}{a\plus{}b\plus{}c}$

The inequality can be transformed into

$ (a\plus{}b\plus{}b\plus{}c\plus{}c\plus{}a)\left(\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\right) \plus{} \frac{3abc(a\plus{}b\plus{}c)}{(ab\plus{}bc\plus{}ca)^2} \ge 10$

which is equivalent to

$ \sum_{cyc}\frac{(a\minus{}b)^2}{(c\plus{}a)(c\plus{}b)} \ge \frac{1}{2}\cdot \frac{\begin{displaystyle}\sum_{cyc}{a^2(b\minus{}c)^2}\end{displaystyle}}{(ab\plus{}bc\plus{}ca)^2}$

and I am stuck in this mess.  :blush:

I think this inequality has a nice solution although I have not found it so far. :huh:

By Am-Gm, easily to see that: $ \frac {3abc(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
So, it's equivalent to: $ \sum\ \frac{a}{b\plus{}c} \plus{} \frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 2$ (obviously trues)  :)


When an excess of magnesium was added to 20cm^3 of 0.2M copper sulphate solution the temp. of the solution increased from 18.3 degrees C to 43.0 degrees C.

How many moles of copper sulphate were used? (moles = concentration (in M) X Volume (in dm^3)

Calculate the enthalpy change for the reaction in KJ per mole.

The answer may well be spelled out for me there, I'm just not sure how to go about doing the calculation. The number of moles is actually stated in the question..damn confusing.
Please help, before I lose any more sleep over this piece of work which has dragged on for too long.  ;)

I managed to do the question, i think. I forgot to mention one small detail, the energy change, which was 24.7 degrees C. From this, and the other bits of data I was able to work out the number of copper sulphate moles used, which was 0.004.

For the enthalpy change I got 6175 KJ per mole (approx).

Please let me know if this is correct.

(Look, I actually attempted the question and so deserve a little help surely!)  :ninja:

Your number for the moles of copper sulfate is correct, but I don't believe that your enthalpy change (per mole) is correct.

Answer:

We have:

$\Delta q=mc\Delta t$

[hide]$\Delta q=\left(20\text{cm}^3\cdot \frac{1\text{g}}{\text{cm}^3}\right)\cdot\left(\frac{4.186\ \text{J}}{\text{g}\ ^\circ\text{C}}\right)\cdot\left(43.0^\circ\text{C}-18.3^\circ\text{C}\right)=2068\ \text{J}=2.068\ \text{kJ}$[/hide]
Now we calculate the molar enthalpy change:

[hide]$\Delta H=\frac{q}{n}=\frac{2.068\ \text{kJ}}{0.004\ \text{mol}}=517\ \text{kJ/mol}$[/hide]

It works now, thanks a lot for ur help.  :D

$\frac{4.186\text{ J}}{\text{g}\ ^\circ\text{C}}$ is the specific heat of water.  It is equivalent to $\frac{\text{1 cal}}{\text{g}\ ^\circ\text{C}}$

g degrees C? For the purposes of this question the specific heat capacity is 4.2  :P  decimal places are far too complicated for us students apparently. I don't see which terms you multiplied and divided etc. Could you rewrite it without latex please? Sorry to be such a pain in the ass.

Are you learning factor labeling?  It's a common technique, and worth some investigation.

The first pair of parentheses simplifies to $20\text{g}$, which is $m$ in the equaiton $\Delta q=mc\Delta t$.
The second pair of parentheses is $c$ (sometimes $c_p$).  Note that the grams from $m$ cancels the $\text{g}$ in the denominator of $c$.
The third pair is $\Delta t$, is in units $^\circ\text{C}$.  This cancels the $^\circ\text{C}$ in $c$, leaving us with units of $\text{J}$, which is what we want ($q$ is in (kilo)joules).


Find the $n$ th term of the sequence $\{a_{n}\}$ which is defined by $a_{1}=0,\ a_{n}=\left(1-\frac{1}{n}\right)^{3}a_{n-1}+\frac{n-1}{n^{2}}\ (n=2,\ 3,\ \cdots).$

Multiply the expression by $n^{3}$ and then substitute $b_{n}=n^{3}a_{n}$:
$b_{n}=b_{n-1}+2\binom{n}{2}=b_{1}+2\sum_{i=2}^{n}\binom{i}{2}=0+0+2\binom{n+1}{3}$ (using the hockey-stick pattern in Pascal's triangle).
So $a_{n}=\frac{2(n+1)n(n-1)}{6n^{3}}=\frac{n^{2}-1}{3n^{2}}$ for $n>1$.

Yes, my solution is exactly same as yours. :wink:


ABC  is  triangle. Show that

[tex] r+4R-\sqrt{3}p\leq   \Sigma(r_a-h_a) \leq  4(R-2r) [/tex]

it is easy to verify that $\sum r_a=r+4R$, so the inequality reduces to proof
$9r\leq \sum h_a\leq \sqrt{3}p$.
the left is equivalent to
$\frac{9}{a+b+c}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$,
which is just AM-HM. the right is (after some algebra) equivalent to(substitute $a=y+z,b=x+z,c=x+y$):
$2\sqrt{xyz}(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z})\leq \sqrt{3(x+y+z)}$
which immediately follows since $2\frac{\sqrt{xyz}}{x+y}\leq \sqrt{z}$(by AM-GM), so the left side is at most $\sqrt{x}+\sqrt{y}+\sqrt{z}$ which is at most the right side by AM-QM.

Peter


Fie $\ ABCD$ un paralelogram.Pe latura $\ BC$ se considera un punct variabil $\ M$, astfel incat $\ BM>MC$ si se noteaza cu $\ N$ simetricul punctului $\ B$ fata de $\ M$.Pe prelungirea laturii $\ AD$ se considera punctul $\ P$ astfel incat $\ CN=AP$.Sa se arate ca dreapta variabila $\ MP$ trece printr-un punct fix.
$\text{ Mihaela Sterpu, Craiova}$

Trece prin punctul O, unde {O} = MP intersectat cu BD si...BO=1/3 BD

Demonstratzia este imediata: PD =2y, BM = y si PD || BM ...si aplicam TFA...:)

Inca ceva.
b)   Demonstrati ca suma proiectiilor punctelor $\ A,B,C$ pe dreapta variabila $\ MP$ este egala cu $\ DD'$, unde $\ D'$ este proiectia punctului $\ D$ pe $\ MP$.

Pai iese misto problema :)

Fie AA` =a , BB`= b, CC`= c , DD` = d , unde AA`, BB`,CC`, DD` sunt proiectile:)

Aplicam TFA de 3 ori si avem:

* a/d = (x-y)/2x

* b/d= 1/2 (s-a demonstrat mai sus)

* c/d = y/x

Prin insumare obtinem ca a+b+c/d =1 deci a+b+c =d

Am notat cu x = BM, iar cu y=CM, x-y fiind CN si AP, iar x+y = AD

Eu m-am cam complicat pentru punctul b.Am facut cu arii....e frumoasa solutia ta.

Iesea si asa, mai.Dar faceai infarct, ca era mult de calculat. :blush:


In square $ ABCD$ with sides of length 4 cm, $ N$ is the midpoint of side $ BC$ and $ M$ is the midpoint of side $ CD$. What is the area of triangle $ AMN$, in cm$ ^2$?

[asy]pair a,b,c,d,m,n;
a=(0,10);
b=(10,10);
c=(10,0);
d=(0,0);
m=(5,0);
n=(10,5);
draw(a--b--c--d--cycle);
draw(a--m--n--cycle);
label("4 cm",(5,10),N);
label("$A$",a,N);
label("$B$",b,N);
label("$C$",c,S);
label("$D$",d,S);
label("$M$",m,S);
label("$N$",n,E);[/asy]

$ [ADM]\plus{}[ABN]\equal{}8$ and $ [MCN]\equal{}2$
$ 16\minus{}8\minus{}2\equal{}\boxed6$

Oh so [ADM] is area of triangle ADM. I got it now. Thanks!


inequalities
geometry
inequalities solved

If $ a$, $ b$, $ c$ are the sides of a triangle, then prove that

$ 3\left(\frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\right)\geq \left(a^2\plus{}b^2\plus{}c^2\right)\cdot\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)$.

As far as I know, even the following problem is hard:
  If a,b,c are the sides of a triangle, then
  3(a/b+b/c+c/a)>=(a+b+c)(1/a+1/b+1/c) (it appears in Kvant, author Kurliandcik, I don't know the year or issue).
   So, it may happen that this problem stayed here for a while.

Just an idea, not a complete solution:
   Put f(a,b,c)=LHS-RHS. Put m=sqrt(b^2+c^2)/2). Try to prove that f(a,b,c)-f(a,m,m) is positive if a is maximum or minimum between a,b,c. Anyway, the computaions are not at all easy and I cannot see where I can use the fact that a,b,c are sides of a triangle. Very difficult inequality!

This is an extraordinary problem. The idea is so simple, but so hard to find. Why didn't I know how to use the fact that a,b,c are sides of a triangle? Because I tried an algebraic proof.
   But there is a simple geometric solution:
  Let us denote T the first Brocard point of the triabgle ABC, whose sides are a,b,c. It is well known that (AT)^2=(a^4c^2)/(a^2b^2+b^2c^2+c^2a^2). Thus the problem states that 3sum (AT)^2>=a^2+b^2+c^2. But from Leibnitz we have 3 sum (AT)^2=9MT^2+3sum AG^2>=3 sum AG^2=sum a^2.
    Absolutely extraordinary problem. Thank you very much, Manlio, for giving me the chance of thinking about it.
   Still, a question bothers me: what about an algebraic proof?

Your solution is simply GENIAL.
Congratulations, Harazi.

Well, I've the 'official' solution of this problem. Are you interested in? I have to tell you before you answer that this is a very long and boring algebraic proof.

Pierre.

If it isn't very bad for you to write lots of computations, please, post it.
  Did anyone think at the problem from Kvant, related to this one? It is also very nice.

You are right. It is very nice but also very hard.
Have you found a solution?

Yes, but surprise: the solution is this time easy with some algebra. This doesn't mean of course that the problem is simple:
   Take a=y+z, b=z+x and c=x+y, with x,y,z>=0. The inequality can be written 3(a^2c+b^2a+c^2b)>=(a+b+c)(ab+bc+ca) and after some boring computations it becomes x^3+y^3+z^3+xz^2+yx^2+zy^2>=2(x^2z+y^2x+z^2y). Luckily, this can be written as x(x-z)^2+y(y-x)^2+z(z-y)^2>=0, obvious.

Ok, since your solution is amazing, I give here the published one, just to make people see how boring it is. Just note that the correct reference is Math Magazine.

Clearing the denominators, the stated inequality is equivalent to :
3(a 4 c 2  + b 4 a 2  + c 4 b 2 )  \geq  (a 2  + b 2  + c 2 )(b 2 c 2  + c 2 a 2  + a 2 b 2 )
This can be written as :
3(a 4 c 2  + b 4 a 2  + c 4 b 2 - (a 2 c 4 + b 2 a 4  + c 2 b 4 ) ) + (a 4 c 2  + b 4 a 2  + c 4 b 2 + a 2 c 4 + b 2 a 4  + c 2 b 4 )  \geq 6a 2 b 2 c 2 .

Since, from AM/GM, we have :
(a 4 c 2  + b 4 a 2  + c 4 b 2 + a 2 c 4 + b 2 a 4  + c 2 b 4 )  \geq 6a 2 b 2 c 2 , it suffices to prove that  :
 (\sum a 4 b 2  - 6a 2 b 2 c 2 ) 2  \geq 9(a 2  - b 2 ) 2 (b 2  - c 2 ) 2 (c 2  - a 2 ) 2 .   (1)

Now put p =  \sum a 2  q =  \sum a 2 b 2  and r = a 2 b 2 c 2 .
By the formula of the discriminant of a cubic, we have :
(a 2  - b 2 ) 2 (b 2  - c 2 ) 2 (c 2  - a 2 ) 2 = -27r 2  + 18pqr - 4p 3 r - 4q 3  + p 2 q 2
thus (1) becomes :
81r 2  + 9q 3 + 9p 3 r  \geq  45pqr + 2p 2 q 2 ,
that is :
4[81r 2  + 9p(p 2 - 5q)r + 9q 3  - p 2q 2 ]  \geq  0.  (2)

But the LHS of '2) is
[18r + p(p 2  - 5q)] 2  + (4q - p 2 )(p 2  - 3q) 2 .
Since 4q - p 2  = 4 \sum a 2 b 2  - ( \sum a 2 ) 2 = 16K 2   \geq  0  (where K denotes the area of the given triangle), the conclusion follows.

Pierre.

Oh, my God, this remembers me about that inequality published in Crux with 9/4. Another example for which all the solutions are with lots and lots of computations. But this one is even more! But anyone should remark the ingeniozity of the solution. It is long, but we all can learn from it.

Sorry for my mistake in reference's name. I have confused on my notesbook AMM (American Mathematical Monthly) with MM(Mathematics Magazine).

Thank you, Pierre for your correction.

If x,y,z  \in R, a,b,c the sides of a triangle then:

(ya2 + zb2 + xc2)(za2 + xb2 + yc2)  \geq (xy + yz + zx)(a2b2 +  b2c2 + c2a2).

To prove the inequality, after some manipulations we obtain:
(x/a - y/b cosC - z/c cosB)2 + (y/b sinC - z/c sinB)2  \geq 0.

Particular case:
1. For x=1/b, y=1/c, z=1/a we obtain:
a2b(a - b) + b2c(b - c) + c2a(c -a) \geq 0. (XXIV OIM)
2. For x=1/c2, y=1/a2, z=1/b2 we obtain:
3(a2/b2 + b2/c2 + c2/a2) \geq (a2 + b2 + c2)(1/a2 + 1/b2 + 1/c2).

Indeed, this is an extraordinary solution, but very difficult to find. Those sum of squares are not an easy job to find.

Hi!
About Gil's proof of AMM 1970 INEQUALITY I have checked the equivalence of

(ya^2 + zb^2 + xc^2)(za^2 + xb^2 + yc^2) \geq   (xy + yz + zx)(a^2b^2 + b^2c^2 + c^2a^2). (1)

with

(x/a - y/b cosC - z/c cosB)^2 + (y/b sinC - z/c sinB)^2 \geq   0. (2)

Expandig (1) I have obtained

a^2y^2c^2 +x^2b^2c^2+z^2b^2a^2 \geq   2axbyc^2cosC +2a^2byczcosA+2axb^2czcosB

while expanding (2) I have obtained

a^2y^2c^2((cosB)^2(senB)^2) +x^2b^2c^2((senC)^2(senB)^2(cosB)^2(cosC)^2)+z^2a^2b^2(cosC)^2(senC)^2)  2axbyc^2((cosC)(cosB)^2(senC)^2(senB)^2) +2a^2bycz((senB)(senC)(cosB)^2(cosC)^2)+2axb^2cz((cosC)^2(cosB)(senC)^2(senB)^2)

Now I have some problems on going on.

Please, a hint.

Thank you very much.

In fact, computing (x/a-y/bcosC-z/ccosB)^2+(y/bsinC-z/csinB)^2 yields
x^2/a^2+y^2/b^2(cos^2C+sin^2C)+z^2/c^2(cos^2B+sin^2B)-2xy/abcosC-2xz/accosB-2yz/bc( sinBsinC-cosBcosC)=sum x^2/a^2-2 sum xy/abcosC, and now the equivalence is immediate.


After five students take a quiz, the teacher asks them to exchange papers so that no one has thier own quiz.  How many arrangements of this type are possible?

[hide]This is just dearrangements size 5.

$\sum_{n=0}^{5}\frac{1}{n!}(-1)^{n}= 44$ [/hide]

of course you have to say $n!$ in front to be right.

not taht good at this stuff so i probably took a very long nonalgebraic way.

[hide="My Not-so-good answer"]

[1] [2] [3] [4] [5].

^5 students.

(1)(2)(3)(4)(5)

^5 test papers.

We have several combinations.

1-1
1-2
1-3
1-4
1-5
2-1
2-2
2-3
2-4
2-5
3-1
3-2
3-3
3-4
3-5
4-1
4-2
4-3
4-4
4-5
5-1
5-2
5-3
5-4
5-5

Now, that is 25 combinations, but if we subtract the same-numbered ones, we get 20 combinations in all.

The (shorter, nonalgebraic still) way to do it:

5 students, 5 tests. 5*5=25. 25-5(sametests)=20.

(----20----)[/hide]


linear algebra
matrix
linear algebra unsolved

Let $ M \in \mathcal{M}_{n}\left(\mathbb{Z}\right)$ defined by $ m_{i,0}\equal{}m_{0,i}\equal{}1$ for $ 0 \leq i < n$ and $ m_{i,j}\plus{}m_{i\minus{}1,j\minus{}1} \equal{} m_{i\minus{}1,j} \plus{} m_{i,j\minus{}1}$ for $ 0 \leq i, j < n$. Find an upper bound for $ M$'s rank.

[b]Claim[/b]: $ \forall i,j: m_{ij}\equal{}1$, in particular, $ rank(M)\equal{}1$

[i]Proof[/i]: Assume the existence of a minimal counterexample $ m_{ij}\not\equal{} 1$ with $ i\plus{}j$ minimal. Clearly, $ i,j>0$, therefore $ m_{ij}\equal{}m_{i,j\minus{}1}\plus{}m_{i\minus{}1,j}\minus{}m_{i\minus{}1,j\minus{}1}\equal{}1$, contradicting the existence of a minimal counterexample.

This is a discretized version of the partial differential equation $ f_{xy}\equal{}0$, which has solutions $ f(x,y)\equal{}g(x)\plus{}h(y)$. The boundary conditions force $ g$ and $ h$ to be constant.

On the other hand, any boundary conditions would still have solutions of that form, and tensor rank at most 2. The matrix version should thus have rank at most 2 even without specifying a row and a column.


The sum of two numbers is 22. Their difference is 4. What is the greater of the two numbers?

Let's say that the greater number is x.
Therefore, the smaller number is x-4.
The sum is x+x-4=22.
2x-4=22, so 2x=26 and x=13

Another way, similar:

Call the first number $ x$ and the second number $ y$.

$ x \plus{} y \equal{} 22$

$ x \minus{} y \equal{} 4$

We can add these two equations together.

$ 2x \equal{} 26$

As you can see, the $ y$ terms dropped out, leaving us with a simple one-step equation.

$ x \equal{} 13$

$ x \minus{} y \equal{} 4$, $ x$ is the greater number, so our answer is $ \boxed{13}$.


[hide]There are two of them: 46 and 96.
I got this result using a computer.[/hide]

[quote="aspiringchemlete"]Can someone confirm that there is only one positive number k less than 100 such that k^3 ends in 36?

[hide="The number I got using some medium NT"]96[/hide][/quote]
The solutions are $ 46,96$
[anyway it is not a good idea to check using computer :huh: ]
[hide="solution"]
It is easy to check using $ mod 10$ that only the numbers $ (10x+6)^{3}$ ends with $ 6$
$ (10x+6)^{3}\equiv 80x+16 \equiv 36 mod 100$.
So we get a equation $ 4x-1=5t$
solving we get $ x=4,9$.

$ QED$
[/hide]

[quote="buzzer11"][quote="Moonmathpi496"]
[anyway it is not a good idea to check using computer :huh: ]
[/quote]

Well if noone did, then we wouldn't have the proof to the four color theorem.  :)

or rather: the CONFIRMATION of the four color theorem.[/quote]
Ok look... (this is not to debate) God has given u one of the best brains among the creatures , so why not u use it when u can?  :)

well in some cases, the computer can help greatly in solving problems. in a conversative's mind, of course, this type of proof is rather "ugly", which why a non-computer generated proof to the four color theorem still is searched for today...

Here was my original solution:

We solve $ k^{3}\equiv 36\pmod{25}$ first:

First, $ \gcd{(36,25)}=1$. Since $ \phi(25)=20$, $ \gcd{(3,\phi(25))}=1$. Thus there will be integers $ u$, $ v$ that satisfy $ 3u-20v=1$. By inspection, $ u=7$, $ v=1$ will work. So we find $ 36^{u}=36^{7}\pmod{25}$, which can be done by the method of successive squaring. It turns out $ 36^{7}\equiv 21\pmod{25}$. We get $ k\equiv21$ as the desired solution.

Now we solve $ k^{3}\equiv 36\pmod 4$, and the obvious solutions are $ k\equiv 0,2$.

Aha and I just realized my mistake: I forgot to do the $ k\equiv 2$ case when finishing the problem with the Chinese Remainder Theorem (the $ k\equiv 0$ case yields 96). Well I guess that worked out after all.


It's a common aspect of every-day life.  People wake up idiots and go to sleep bearing 200+ IQ's.  What else is there to say?

[quote="jimhu"]I just woke up last night and somehow became more stupid. I tried solving some problems I've previously thought were easy, but now they're just not coming to me relally well. Do you think its only a psychological factor or is it really true?[/quote]
I don't think we become stupid overnight.

You are a Chinese, so I suggest you to read Chapter 48 and 49 of the Dream of the Red Mansion (or Chamber).

You were probably just having an off day..it happens to everyone, when it happens to me, I just intake a lot of sugar (coke, chocolate) and I'm back to normal :lol:

i always say: oh this is an wasy problem, then i go to solve it and i can't do it ;)

What may well have happened is that you lost a perspective that you had when you went to bed.  It may be useful to think about the approaches and concepts and not pushing the details.  Yes this has happened to many of us, including myself.

The same thing happens if you keep a dream log and write something significant down in the middle of the night.  By morning it seems trivial or meaningless.  Once again the only difference is loss of context.

Take care.

happens to me all the time  :roll:
jk

but seriously, there are days when i cant even solve simple math problems for some reasons...i think its because my brain refuses to think creatively

Sometimes i have this feeling, it happens to me to think that i was smarter yesterday, last month,last year...but i worry more when i see that my future life will be largely shaped in some short moments...what if i am only with 10 percent of my abilities at the "last fight"?...so the winners are always the opportunistic people who know how to give their best at the right moment..just a simple example :how would someone feel if he "is" Euler at home and cannot focus one day of a final exam? it also depends on how much we care about the others..if we are obsessed with showing our accomplishments, of course it disorients and makes us some illusions...
we shouldn't expect our favourite conditions,we should challenge all secondary things and arbitrary details, and whenever such ideas come to our mind, we try to disprove it by doing some great work and by the time getting free from these superstitions(i am not sure if it is a suitable word) :)
Scientifically, as we take care of our body, we should learn how to take care of our brain...

Thanks guys for your eideas. What happened to me was my dad brought me to a gym for the first time and I tried lifting some weights and then when I got home I slept and woke up next morning and can't think creatively anymore... I don't know whats the problem...


Hello I'm studying for the SATs and I hope you guys don't mind if I post problems I dont understand on here.

If a, b, and c are different positive integers and $2^a * 2^b * 2^c = 64$ then $2^a + 2^b +2^c = ?$

a. $14$
b. $17$
c. $21$
d. $28$
e. $34$

A regulatio for riding a certain amusement park ride requires that a child between 30 inches and 50 inches tall. Which of the following inequalities can be used to determine whether or not a child's height hsatisfies the regulation for this ride?

a. $|h - 10| < 50$
b. $|h - 20| < 40$
c. $|h - 30| < 20$
d. $|h - 40| < 10$
e. $|h - 45| < 5$

More problems later.

Can you guys explain how you got the answer thank you.

for 1:
$2^{a+b+c}=64=2^6$
$a+b+c=6$ with $a,b,c$ all different positive integers (nonzero I assume?)
only solution: $(a,b,c)=(1,2,3)$.
$2^1+2^2+2^3=\boxed{14}$

thanks K.W.M.A.N

what do you guys think of the second problem?

[hide="2nd problem"]
Consider a number line. Note that $40$ is the midpoint of $30$ and $50$. The distance between $40$ and any number within the interval ($30<h<50$) is less than $10$.

Therefore, the answer is $|h-40|<10$.[/hide]


quadratics trigonometry AMC AIME USA(J)MO USAMO algebra

ABCD is a square with side length of one unit, and equilateral triangle AED is constructed on top of ABCD. Find the length of EB. Express your answer in simplest radical form.

[hide]
Just so you know, $\sqrt{\sqrt{3}+2}= \frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}$. At least according to my calculator - I'm not entirely sure how it managed to get there though....[/hide]

[quote="bpms"][quote="13375P34K43V312"]Hint for simplifying that: set it equal to $\sqrt{a}+\sqrt{b}$.[/quote]
I tried something similar if you read my work...
I'm pretty sure that my ugly equation would've simplified it... Maybe not...
$\sin(15)$?[/quote]

that new hint should clean it up a little more than yours bpms...although i liked your thinking...close

Is it mere coincedence that the answer = $\sin(15)$?
I think not...

[hide] OK, so set it up like: $\sqrt{\sqrt{3}+2}= \sqrt{a}+\sqrt{b}$
Squaring both sides:
$\sqrt{3}+2 = a+2\sqrt{ab}+b$
Thus,
$a+b=2$,
and $2\sqrt{ab}= \sqrt{3}$
Squaring both sides in the second equation:
$4ab=3$
Substituting $b=2-a$:
$4a(2-a)=3$
Solving as a quadratic equation, you get the two solutions for a and b are 1/2 and 3/2. Putting these under square roots and rationalizing the denominators gives you: ${\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}}$
[/hide]

No, it's not a coincidence. You can do some angle chasing and find that the answer is not $\sin{15}$, but $2\cos{15}$.

[quote="Ignite168"]There was a problem pretty similar to this on the 2005 Nationals Target but it wanted it to the nearest tenth.

I think these nationals team questions were too hard and you won't see problems of this difficulty at Nationals.[/quote]
Um, yes you will. The question you are talking about was actually easier than this. It also had a clean answer, no radical. If it asked for to the nearest hundred the hundredths place would've been 0...

[hide]Here's how I did it.

Draw the midpoint of BC and call it F. Draw a line from A to F. So now we have  a right triangle. One leg is the height of the triangle plus the side length of the square, and the other leg is 1/2 the side length of the square

$(\frac{\sqrt{3}}{2}+1)^{2}+(\frac{1}{2})^{2}=EB^{2}$
$EB=\sqrt{2+\sqrt{3}}$
That simplifies to $\frac{\sqrt{2}+\sqrt{6}}{2}$[/hide]

[quote="13375P34K43V312"][quote="bpms"][quote="13375P34K43V312"]Yes, I agree that this was a very difficult set of team problems. But hey, if a school team can get an 8 on it...[/quote]
I'd say it is an average-hard state round difficulty. The hardest tean round questions at nats should be about AIME 6-10 difficulty...[/quote]

What?! In AIME 6-10, you usually see 1-2 Trig problems, 1-2 Complex number problems, and 1-2 Log problems.[/quote]
I wish there were 1-2 log problems and 1-2 trig problems. Nuh uh. This past years national  team round number 10 could've been AIME number 10. Same for 05. Note that none of the many USAMO qualifiers either of those years solved it.

[quote="jhredsox"]is the answer

[hide]
$\sqrt{4.25}$[/hide]
-jorian[/quote]

1. No
2. That is not simplified
3. Please give a solution.

[quote="bpms"][hide]Build a right triangle. The height is the sum of 1 and the height of the triangle $1+.5 \sqrt{3}$. The width is half the width of the triangle, or .5. Using pythagorean, we get $\sqrt{2+\sqrt{3}}$. Lets see if this can be simplified...
$\sqrt{2+\sqrt{3}}=a+b\sqrt{3}$
$2+\sqrt{3}=a^{2}+3b^{2}+2ab\sqrt{3}$
$a^{2}+3b^{2}=2$
$ab=\frac{1}{2}$
$b=\frac{1}{2a}$
$a^{2}+\frac{3}{4a^{2}}=2$
$4a^{4}-3=8a^{2}$
That is too ugly. So, I'm going to go with my answer $\sqrt{2+\sqrt{3}}$ [/hide][/quote]
That's not too ugly...

[hide]$4a^{4}-8a^{2}-3=0$

Let $z=a^{2}$

$4z^{2}-8z-3=0$

$z=\frac{8\pm\sqrt{64+48}}{8}$

$z=\frac{8\pm4\sqrt{7}}{8}$

$z=\frac{2\pm\sqrt{7}}{2}$

$a=\sqrt{\frac{2\pm\sqrt{7}}{2}}$

$a=\frac{2\sqrt{2}+\sqrt{14}}{2}$[/hide]

wait, i thought this is a really easy problem, considering the units of 1....what am i doing wrong?
-jorian

[quote="jhredsox"]k, fine[hide]
$\sqrt{\frac{17}{4}}$

i first drew a picture in my head

and i drew a right triangle from the top of the equilateral triangle (E) to B. i then used pythagoream to find the length. well, i knew the height was 2, since both the square and the eq. triangle had a height of 1, and i knew the width was .5, because the top of the equilateral trianle was halfway throught the square. then, i squared both to get 4 + .25 = x ^ 2. this is 4.25 = x ^ 2. since the problem asked for radical in simplest form, i got the answer above[/hide]
-jorian[/quote]
First: Assuming you did everything else correctly, that's not simplified radical form.

Second: You didn't do everything correctly. An equilateral triangle in a square doesn't touch the top side.

[quote="i_like_pie"][quote="bpms"][hide]Build a right triangle. The height is the sum of 1 and the height of the triangle $1+.5 \sqrt{3}$. The width is half the width of the triangle, or .5. Using pythagorean, we get $\sqrt{2+\sqrt{3}}$. Lets see if this can be simplified...
$\sqrt{2+\sqrt{3}}=a+b\sqrt{3}$
$2+\sqrt{3}=a^{2}+3b^{2}+2ab\sqrt{3}$
$a^{2}+3b^{2}=2$
$ab=\frac{1}{2}$
$b=\frac{1}{2a}$
$a^{2}+\frac{3}{4a^{2}}=2$
$4a^{4}-3=8a^{2}$
That is too ugly. So, I'm going to go with my answer $\sqrt{2+\sqrt{3}}$ [/hide][/quote]
That's not too ugly...
Okay, but generally, when you want to [b]simplify[/b] something, you want to go with something that isn't complicated...

[hide]$4a^{4}-8a^{2}-3=0$

Let $z=a^{2}$

$4z^{2}-8z-3=0$

$z=\frac{8\pm\sqrt{64+48}}{8}$

$z=\frac{8\pm4\sqrt{7}}{8}$

$z=\frac{2\pm\sqrt{7}}{2}$

$a=\sqrt{\frac{2\pm\sqrt{7}}{2}}$

$a=\frac{2\sqrt{2}+\sqrt{14}}{2}$[/hide][/quote]

[quote="jhredsox"]OUCH!

do you know what i did wrong?

i had the isosceles triangle laying on top of the square, making this incredibly easy

wait, if its inscribed inside, how hard can it be

can someone attatch a file from paint or something because i just dont get this

-jorian[/quote]

Umm it is laying on top of the square..

[quote="jhredsox"]can someone attatch a file from paint or something because i just dont get this[/quote]
Everything has side length 1.

[img]http://img152.imageshack.us/img152/8237/squaretriangletd3.gif[/img]

nooooooo! The triangle is on TOP of the square.

I don't get it. :?:

Two of the points are A and D, and the triangle is equilateral and inside the square, so there's only one point to place E.

It's not inside the square. It's outside the square.

[quote="i_like_pie"]Oh!

[hide]The equilateral triangle can be divided into two 30-60-90 triangles with their common side being perpendicular to $AD.$

The base is $\frac{1}{2},$ and the height is $\frac{\sqrt{3}}{2}.$

Using Pathagoras' Theorem, $\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=x^{2}.$

$\frac{1}{4}+\frac{7+4\sqrt{3}}{4}=x^{2}$

$x=\sqrt{\frac{8+4\sqrt{3}}{4}}=\frac{\sqrt{8+4\sqrt{3}}}{2}$[/hide][/quote]

Radicals are not allowed inside radicals and neither is a common factor of 4.

[quote="i_like_pie"]Oh!

[hide]The equilateral triangle can be divided into two 30-60-90 triangles with their common side being perpendicular to $AD.$

The base is $\frac{1}{2},$ and the height is $\frac{\sqrt{3}}{2}.$

Using Pathagoras' Theorem, $\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=x^{2}.$

$\frac{1}{4}+\frac{7+4\sqrt{3}}{4}=x^{2}$

$x=\sqrt{\frac{8+4\sqrt{3}}{4}}=\sqrt{2+\sqrt{3}}$[/hide][/quote]

A square root inside a square root looks nasty and might not be accepted. Try to fix that.

Also, it's spelled Pythagorean Theorem.

[hide="Hint"]$\sqrt{2+\sqrt{3}}=\sqrt{x}+\sqrt{y}$
Square both sides and solve for x and y.[/hide]

I know it's Pythagorean, I just wanted to do something... different. :ninja:

How do you simplify radicals in radicals? :?:


A rectangle with vertices at (2, 5), (7, 5), (7, 8) and (2, 8) is revolved around the line
y = 5 to obtain a cylinder. What is the volume of the cylinder? Express your answer in terms of π.

This problem basically requires a lot of imagination.

The height of this cylinder is 5, and the radius is 3, as shown by the vertices.
Now we rotate this across the line y=5 to create a cylinder. The volume of this cylinder is just $ 3^2 * 5 * \pi$ = $ \boxed {45 \pi}$

I think (emphasis on think) that this can be done with transformation matrices... (it wouldn't be MC level though...)


induction
algebra
polynomial
strong induction

[hide="Induction"]
Note that $ (x\plus{}y)(x^2\plus{}y^2) \equal{} x^3\plus{}y^3 \plus{} xy(x\plus{}y)$ so $ xy$ is an integer. This is an important information.

Now, for $ n \equal{} 1,$ the problem is held true. Assume it is true for $ n \equal{} k$ for some positive integer $ k$. Then let's prove for $ n \equal{} k\plus{}1$.

$ x^{k\plus{}1} \plus{} y^{k\plus{}1} \equal{} (x^k \plus{} y^k)(x\plus{}y) \minus{} xy(x^{k\minus{}1} \plus{} y^{k\minus{}1})$

So, if we prove $ x^{k\minus{}1} \plus{} y^{k\minus{}1}$ is an integer, then $ x^{k\plus{}1} \plus{} y^{k\plus{}1}$ is an integer and the induction's done.

$ x^{k\minus{}1} \plus{} y^{k\minus{}1} \equal{} \frac{x^k \plus{} y^k \plus{} xy(x^{k\minus{}2} \plus{} y^{k\minus{}2})}{x\plus{}y}$ so now consider $ x^{k\minus{}2} \plus{} y^{k\minus{}2}$ since everything else is an integer.

$ x^{k\minus{}2} \plus{} y^{k\minus{}2} \equal{} \frac{x^{k\minus{}1} \plus{} y^{k\minus{}1} \plus{} xy(x^{k\minus{}3} \plus{} y^{k\minus{}3})}{x\plus{}y}$

....

If this continues, $ x^2 \plus{} y^2 \equal{} \frac{x^3 \plus{} y^3 \plus{} xy(x\plus{}y)}{x\plus{}y}$ which obviously is true. Going up, $ x^3\plus{}y^3 \equal{} \frac{x^4 \plus{} y^4 \plus{} xy(x^2\plus{}y^2)}{x\plus{}y}$ and $ x^{4} \plus{} y^{4} \equal{} \frac{x^{5} \plus{} y^{5} \plus{} xy(x^3 \plus{} y^3)}{x\plus{}y}$. Note that on the last equation, we just proved that $ x^5 \plus{} y^5$ is an integer.

So continuing this pattern, we can see that $ x^{k\minus{}1} \plus{} y^{k\minus{}1}$ is an integer and $ x^{k\plus{}1} \plus{} y^{k\plus{}1}$ is an integer as well. Thus, induction's done.

$ \therefore \ x^n \plus{} y^n$ is an integer for all positive values $ n$. [/hide]

A better way to show that $ xy$ is an integer is to note that $ x^{2} \plus{} 2xy \plus{} y^{2}$ and $ x^{2} \plus{} y^{2}$ are integers, so $ 2xy$ is an integer. So if $ xy$ isn't an integer, it's of the form $ k/2$, where $ k$ is an odd integer. But in a similar way we find that $ 2x^{2}y^{2}$ is an integer, so $ k^{2}/2$ is an integer. But $ k$ is odd, contradiction. So $ xy$ is an integer.

Silverfalcon's proof can be shortened substantially by just using strong induction, making the inductive hypothesis that $ x^{n}\plus{}y^{n}$ is an integer for [i]all[/i] integers $ n\le k$. Then you can just do the same thing, but you don't have to deal with $ x^{k\minus{}1}\plus{}y^{k\minus{}1}$.

[quote="matt276eagles"][quote="Silverfalcon"]Note that $ (x \plus{} y)(x^2 \plus{} y^2) \equal{} x^3 \plus{} y^3 \plus{} xy(x \plus{} y)$ so $ xy$ is an integer.[/quote]

How does this imply that $ xy$ is an integer?[/quote]

LHS is product of two integers so it is an integer. This means that RHS must be an integer as well. $ x^3\plus{}y^3$ is an integer so all we now need is to see if $ xy(x\plus{}y)$ is an integer. Since $ x\plus{}y$ is an integer, $ xy$ must be an integer (if it's not then, RHS will not be an integer.

[quote="CatalystOfNostalgia"]A better way to show that $ xy$ is an integer is to note that $ x^{2} \plus{} 2xy \plus{} y^{2}$ and $ x^{2} \plus{} y^{2}$ are integers, so $ 2xy$ is an integer. So if $ xy$ isn't an integer, it's of the form $ k/2$, where $ k$ is an odd integer. But in a similar way we find that $ 2x^{2}y^{2}$ is an integer, so $ k^{2}/2$ is an integer. But $ k$ is odd, contradiction. So $ xy$ is an integer.

Silverfalcon's proof can be shortened substantially by just using strong induction, making the inductive hypothesis that $ x^{n} \plus{} y^{n}$ is an integer for [i]all[/i] integers $ n\le k$. Then you can just do the same thing, but you don't have to deal with $ x^{k \minus{} 1} \plus{} y^{k \minus{} 1}$.[/quote]

I'm not sure if I know the difference between the strong induction and weak induction. I've heard of it but I'm not familiar with its difference. Thanks  :)

[quote="Silverfalcon"]...is an integer so all we now need is to see if $ xy(x \plus{} y)$ is an integer. Since $ x \plus{} y$ is an integer, $ xy$ must be an integer (if it's not then, RHS will not be an integer.[/quote]

You claim that if $ AB$ (here $ A \equal{} xy,B \equal{} x \plus{} y$ to make it more clear) and $ B$ are both integers, $ A$ must be an integer. But just take $ A \equal{} 5/2,B \equal{} 4$, making $ AB \equal{} 10$.

[quote="Silverfalcon"]I'm not sure if I know the difference between the strong induction and weak induction. I've heard of it but I'm not familiar with its difference. Thanks[/quote]

So say we wanted to prove some statement S(n) for all positive integers n, for example. We can show the base case, S(1). In weak induction, our inductive hypothesis would be that S(k) is true for [i]some[/i] positive integer k, then we would prove that S(k+1) follows. This is a "weak" hypothesis, since we're only assuming the truth of one statement.

However, in strong induction, we assume that S(1), S(2), ..., S(k) are all true, then prove that S(k+1) follows. This is a "strong" hypothesis, as we are assuming much more, but it makes the proof no less valid.

Pwnage way:

Let:
$ x \plus{} y \equal{} p$
$ xy \equal{} q$

Given:
$ x \plus{} y \equal{} p$ therefore p is an integer.

$ x^2 \plus{} y^2 \equal{} p^2 \minus{} 2q$ therefore 2q is an integer since $ p^2$ is an integer. Let 2q=k <--> q=k/2 w/ k integer.

$ x^3 \plus{} y^3 \equal{} p^3 \minus{} 3pq \equal{} p(p^2 \minus{} q) \minus{} pq$ therefore pq is an integer, since $ p$ and$ p^2 \minus{} 2q$ are given to be integers.

$ x^4 \plus{} y^4 \equal{} p^4 \minus{} 2(2q)(p^2 \minus{} 2q) \minus{} 6q^2$ therefore $ 6q^2 \equal{} 3k^2/2$ is an integer. so $ 2|3k^2$ so $ |k^2$ so $ 2|k$

therefore k/2 is an integer, so q is an integer.

Now, $ x^{n \plus{} 1} \plus{} y^{n \plus{} 1} \equal{} p(x^n \plus{} y^n) \minus{} q(x^{n \minus{} 1} \plus{} y^{n \minus{} 1})$ so strong induction pwns.

It seems that the fact that $ x^3\plus{}y^3$ is an integer is extraneous...

I guess given that x+y, x^2+y^2, x^4+y^4 are integers only sort of points to the fact that you want to use squares in some way, as I did. Another thing to note is that once you get that xy is an integer, you can just finish by being all x^n+y^n is a symmetric polynomial in x and y so it can be written in terms of the elementary symmetric sums, x+y and xy. But those are both integers.

Thanks CatalystOfNostalgia, hmm, I didn't actually [i]try[/i] the values so I just assumed the $ xy$ to be an integer. Blah, I got it right luckily. I get what you mean by strong vs. weak induction now.

You can write any symmetric polynomial as a polynomial in the elementary symmetric polynomials...but that is not strong enough obviously...you need something about integer coefficients in there.

Is it true that any symmetric polynomial with integer coefficients can be written as a polynomial with integer coefficients in the elementary symmetric sums?

I believe it is true, but it is not that obvious...and certainly, it should be stated clearly.

Yes.  The proof of the fundamental theorem of symmetric polynomials works over an arbitrary ring.

[quote="Silverfalcon"]Let $ x,y$ be real numbers such that the numbers $ x\plus{}y, x^2\plus{}y^2, x^3\plus{}y^3$ and $ x^4\plus{}y^4$ are integers. Prove that for all positive integers $ n$, the number $ x^n \plus{} y^n$ is an integer.[/quote]
We use the identity $x^n+y^n=(x+y)(x^{n-1}+y^{n-1})-xy(x^{n-2}+y^{n-2})$.

First we prove that $xy$ is an integer. Note that $2xy=(x+y)^2-(x^2+y^2)\in \mathbb{Z}$ and $2x^2y^2=(x^2+y^2)^2-(x^4+y^4)\in \mathbb{Z}$. From the first equation we see that $2xy=m$ for some integer $m$, so $xy=\frac{m}{2}$. From the second equation we have $2x^2y^2=n$ for some integer $n$. Plugging $xy=\frac{m}{2}$ yields $2\left(\frac{m^2}{4}\right)=n\Leftrightarrow \frac{m^2}{2}=n\in \mathbb{Z}$. So $2\mid m$. Hence $xy=\frac{m}{2}\in \mathbb{Z}$, as desired. //

Now, let $p(n)=\text{true}\Rightarrow x^n+y^n\in \mathbb{Z}$. By the identity, $p(1), p(n-1), xy, p(n-2)\Rightarrow p(n)$. So $p(1), p(4), xy, p(3)\Rightarrow p(5)$; $p(1), p(5), xy, p(4)\Rightarrow p(6)$, and by straight-forward strong induction on $n$, $p(n)=\text{true}\ \forall n\in \mathbb{N}$. $\Box$

Assumption $x^3+y^3\in Z$ is not necessary


rotation
trigonometry
calculus
calculus computations

I am having trouble with the following related rates problem:
A beacon of light 200m offshore rotates 2 times in a minute forming a spot of light that moves along a straight wall along the shore. How fast is the spot of light moving when the spot is at point P directly opposite the beam? When it is 200m from point P?

How do I set this up?

I would put my rotating source at the origin. Then I'd place a vertical line $ x\equal{}200$. Now draw an arbitrary light line, say angle $ \theta$ above the $ x$-axis. You should be able to simply relate these quantities now. The light line meets the "wall" a distance $ y$ up and $ x$ out, so $ \tan\theta\equal{}\frac{y}{x}$. Eh, but remember $ x\equal{}200$ is unchanging as $ \theta$ rotates around. You have $ \frac{d\theta}{dt}$ and you want $ \frac{dy}{dt}$. So differentiate and plug in.


Support function integration calculus trigonometry symmetry

[color=darkred]HC VERMA
EXERCISES PROBLEM NO.50
Two blocks A and B,each having a mass of 320g connected by a light string passing over a smooth light pulley.The horizontal surface on which the block A can slide is smooth.The block A is attached to a spring os spring constant $ 40 N(m^{ \minus{} 1})$whose other end is fixed to a support 40cm above the horizontal surface. Initially,the spring is vertical and unstretched when the system is released to move.Find the velocity of the block A at the instant it breaks off the surface below it. Take $ g \equal{} 10 \frac {m}{s}$[/color]

i am really sry cuz i dont kno how to post the diag whoever has the book can help me by posting the figure!
i cant even interpret the question properly
what does "breaks off the surface" mean exactly?

Let $ x$ be the distance of mass A on the horizontal from the starting position. Let $ h$ be the length of the loose spring. i.e. the length of the spring when $ x\equal{}0$. So then the stretching of the spring will be $ \sqrt{x^2\plus{}h^2}\minus{}h$.
Let the spring constant be $ D$. So the our equations of motion:

\[ mg\minus{}T\equal{}m\ddot x\]
\[ m\ddot x \equal{} T\minus{}D(\sqrt{x^2\plus{}h^2}\minus{}h)\]

eliminating $ T$ we have:

\[ 2m\ddot x \equal{} mg\plus{}Dh\minus{}D\sqrt{x^2\plus{}h^2}\]

Now we dont the $ x(t)$ function, just the velocity as a function of $ x$ i.e. the $ \dot x(x)$ function, hence we should do the following:

\[ \ddot x\equal{}\frac{d\dot x}{dt}\equal{}\frac{d\dot x}{dx}\frac{dx}{dt}\equal{}\dot x \frac{d\dot x}{dx}\equal{}\frac12\frac{d\dot x^2}{dx}\]

With this we can separate the variables, and integrate (using $ v(x)\equal{}\dot x$):

\[ v^2(x)\plus{}C\equal{}\left(g\plus{}\frac{Dh}{m}\right)x\minus{}\frac{D}{m}\int\sqrt{x^2\plus{}h^2}dx\]

Performing the integration and using that at $ x\equal{}0$ we have $ v\equal{}0$ i.e. C=0, we obtain for the velocity:

\[ v(x)\equal{}\left(g\plus{}\frac{Dh}{m}\right)x\minus{}\frac{D}{2m}x\sqrt{x^2\plus{}h^2}\minus{}\frac{Dh^2}{2m}\text{arsh}\left(\frac{x}{h}\right)\]

now we just need the $ x_0$ where the block will rise from the floor. This will be when the vertical component of the spring force will be greater than the gravitational force on the block. That is the critical condition will be:

\[ \frac{mg}{Dh}\equal{}1\minus{}\frac{h}{\sqrt{x_0^2\plus{}h^2}}\]

Now express $ x_0$ from here and plug it into the velocity equation to get the required velocity.

sorry for spamming, but yeahnigga can u plz use, $ v$ for velocity,$ a$ for acceleration etc...

From Mechanical Energy conservation we can get

\[ v^2 \equal{} x_0 g \minus{} {k \over{2m}}({\sqrt{x_0^2\plus{}h^2} \minus{} h)}^2\]

$ v$ following  geniusbliss wishes, and $ k$ for the spring's elastic constant.

That could also be get from  the movement equation, but taking only the horizontal component of the elastic force in $ \ddot x$'s (sorry, geniusbliss) equation.

Hey Immanuel, thats my line  :rotfl:

Yes In my hurry i forgot to write up the horizontal component of the spring force. So the second equation correctly:

\[ m\ddot x\equal{} T\minus{}D(\sqrt{x^2\plus{}h^2}\minus{}h)\dfrac{x}{\sqrt{x^2\plus{}h^2}}\]

I was surprized that inverse-hyperbolic functions joined the game anyway..

Unnecessary. :D
If Conservation of Energy gives the answer so fast and quick, why do you need all this scientific mumbo-jumbo? :wink:

I did the "scientific mumbo-jumbo" since 99% of the time, energy conservation cannot be used. So i just wanted to show you guys, how to do a problem like this when you are "out of the box of high school physics", and to stop thinking of conservation laws as universal. It is only in very special cases that conservation laws can be applied..., when the system has a symmetry. For example energy conservation can be used if the Hamiltonian, or Lagrangian of the system doesnt depend explicitely on time etc.

So try to go a bit more inside, and study the dynamics of the system, and if you see some symmetry only then apply some conservation law...

Relief, at last Yeahnigga! You won't recieve any arguments from me here :wink:

BTW, is my answer correct?

[quote="Immanuel"]
How about increasing both masses to 800g?

[/quote]

He deleted that post!

Correction:
How about increasing the mass of each block to 1.1 kg ?

[quote="Rajiv"]Thank you.
 :D
I suppose yeahnigga is too busy to read private messages! So I'd make this an open question: Can anyone tell me where and when (and how?) Conservation of Energy fails? So far I have found it a very useful tool in solving problems but yeahnigga blasted a big hole in my confidence in the principle! I'd expect an example in support, a problem will also do! I am of the opinion that the principle is never violated in the universe. Apparently, yeahnigga has some old rivalry with CoE...[/quote]

I think it never fails in classical mechanics but see he is talking about time dependence and all; this might be related to relativity and/or quantum. There is no obvious reason why conservation of energy should always remain valid.

[quote=Rajiv]Conservation of Energy:

$ kx\sin\theta \equal{} mg \\
 \\
k\sin\theta(l\csc\theta \minus{} l) \equal{} mg \\
 \\
 \\
\sin\theta \equal{} 1 \minus{} \frac {mg}{kl} \equal{} \frac {4}{5} \\
 \\
 \\
\therefore \boxed{\theta \equal{} 53^{^0}} \\
 \\
mgs \minus{} \frac {1}{2}kx^2 \equal{} mv^2 \\
 \\
 \\
mg\frac {3l}{4} \minus{} \frac {1}{2}k \left(\frac {l}{4}\right)^2 \equal{} mv^2 \\
 \\
 \\
\therefore\boxed{v \equal{} \sqrt {\frac {3gl}{4} \minus{} \frac {kl^2}{32m}}}$

Subtituting the values, my calculator says $ v \equal{} 1.5411 \approx 1.5$[/quote]

Can you explain why did you use kx sinθ instead of kx cosθ?

[quote=ciao_potter]Because sine of 90 degrees is 1 and cosine of 90 degrees is 0.

So the first equation means when the block is perpendicular to the surface (90 degrees), then kx = mg.[/quote]

Dude are you serious? ?

[quote=TheOreticalGuy][quote=ciao_potter]If he used $kx \cos(\theta)$, then in the initial condition, $0=mg$ which is not true.[/quote]

Cos θ is not zero. It is not perpendicular
 -_-[/quote]

Sorry if I wasn't being specific enough. I was giving an example why using cosine would not work. Rajiv used $\theta$ as the angle between the spring and the horizontal surface. $kx \cos(\theta)$ is the horizontal component of the spring force so it would not equal $mg$ which is a vertical force.

You could use $\cos(\theta)$ but you would have to make $\theta$ the angle between the spring and the vertical.

[b][u]Explanation of the solution:[/u][/b]
Let $l$ be the length of the spring. Angle $\theta$ is the angle between the spring and the horizontal.

We know that the vertical component of the spring force will cancel out the gravitational force because the block does not have an acceleration in the vertical direction.
[asy]
size(5cm);
draw((3,5)--(8,2));
draw((3,5)--(3,2));
draw((3,2)--(8,2));
label("$\theta$", (7.2,2.3), W);
label("$kx$",(5.5,3.5), NE);
label("$kx \sin(\theta)$",(3,3.5), W);
label("$kx \cos(\theta)$",(5.5,2), S);
[/asy]
Therefore, $kx \sin(\theta) = mg$
($k$ is the spring constant and the $x$ is the length that the spring gained.)

[asy]
size(10cm);
draw((0,0)--(11,0));
draw((2,0)--(4,0)--(4,2)--(2,2)--cycle, blue);
draw((7,0)--(9,0)--(9,2)--(7,2)--cycle, red);
draw((3,5)--(8,2));
draw((3,5)--(3,2));
draw((3,2)--(8,2), dashed);
label("$l$", (3,3.5), W);
label("$\theta$", (7.2,2.3), W);
label("$l \cdot \csc(\theta)$", (5.5,3.5), NE);
[/asy]

The length of the new spring will be $l \csc(\theta)$. Therefore, the spring will have gained $l \csc(\theta) - l$.

The first equation can be rewritten as,
\[k\sin(\theta) (l\csc(\theta)-l)= mg\]

Rest is just calculation.

I hope that this clears your confusion!


What is the constant term in the Laurent polynomial
\[ \prod_{1\le i\neq j\le n}\left(1 \minus{} \frac {x_i}{x_j}\right)^{k_i}\]
where $ k_i$'s are given naturals.

Please continue the discussion in [url=http://www.mathlinks.ro/viewtopic.php?t=76800]Constant term of an expression[/url]


calculus
integration
induction
inequalities
algebra
functional equation
algebra solved

Let {$ f(n) $} be an intergral sequence satisfying :
 $ f(0) = 0 $ , $ f(n) = n -f(f(n-1)) $
 Prove that :
 a) $ f(k) \geq f(k-1) $ for all $ k \geq 1 $
 b) $ f(n) = [nw] $ , where w is the root of the equation : $ x^2+x-1 = 0 $

Did I miss something or $0<w<1$ so that $f(1)=1$ but $[nw] = 0$ ?

Pierre.

Hmm.. I'd rather say $f(n) = [(n+1)w]$... ;)

Pierre.

I think, I posted it some time ago in another thread.
The right equality is $f(n)=[(n+1)w]$.

[quote="Myth"]I think, I posted it some time ago in another thread.
[/quote]

What do you mean? That you already solved this problem here? ;)
Anyway, here is my proof :

Direct checking shows that $f(0)=0,f(1)=f(2)=1$. And since $f$ is defined on the non-negative integers, we deduce from the functional equation that $f(n-1) $ is a non-negative integer for all $n$. Therefore, $f(n) \leq n$. (1)

Now assume that $f$ is non-decreasing on $\{1, 2, \cdots , n-1 \}$.
For all $a$ in this set, we have $f(a-1) \leq f(a) \leq a.$ From the induction hypothesis, it follows that $f(f(a-1)) \leq f(f(a))$ so that $f(a+1)-f(a) = 1 + f(f(a-1)) - f(f(a)) \leq 1$, that is $f(a+1) \leq f(a)+1.$
Thus $f(f(n-1)) \leq 1 + f(f(n-1)-1)$ and $f(n-1) - 1 \leq f(n-2) \leq n-2$. Thus $f(f(n-1) - 1) \leq f(f(n-2))$ and then $f(f(n-1)) \leq 1 + f(f(n-2))$.
Therefore $f(n)-f(n-1) = 1 + f(f(n-2)) - f(f(n-1)) \geq 0$, which means that $f$ is non-decreasing on $\{1,2, \cdots, n \}$.
Thus a) follows by induction.

Let $w$ be defined as in the statement. We prove by induction that $f(n) = [(n+1)w].$
Note that $w = \frac 1 {\varphi}$ where $\varphi$ is the golden mean.
According to (1) and the induction hypothesis (here we assume the result holds for all $k \leq n-1$), we want to prove that
$[ \frac {n+1} {\varphi} ] = n - [ \frac { [\frac n {\varphi}] + 1 } {\varphi} ]$

Let $L = [ \frac {n+1} {\varphi} ]$ and $R = n - [ \frac { [\frac n {\varphi}] + 1 } {\varphi} ]$.
Note that, since $\frac 1 {\varphi} = \varphi - 1$ we have $\frac { [\frac n {\varphi}] + 1 } {\varphi} = \varphi [ \frac n {\varphi}+ 1]- [\frac n {\varphi} + 1]$ so that $R = n - [\frac n {\varphi} + 1] + \varphi [ \frac n {\varphi}+ 1].$

- If $ \frac n {\varphi} < [ \frac n {\varphi} ] + 1 < \frac n {\varphi} + \frac 1 {\varphi}$ (2) :
Then $[\varphi [ \frac n {\varphi}+ 1] ] = n$ and $[ \frac n {\varphi} + \frac 1 {\varphi} ] =  [ \frac n {\varphi} ] + 1$ (since the mid-term in (2) is an integer and the two extremal terms differ by less than 1).
It easily follows that $L =  [ \frac n {\varphi} ] + 1 = R$ in that case.

- $\frac n {\varphi} + \frac 1 {\varphi} <  [ \frac n {\varphi} ] + 1 <  \frac n {\varphi}  + 1$ (the first inequality is strict because $\varphi $ is irrational), then $[\varphi [ \frac n {\varphi}+ 1] ] = n+1$ and $[ \frac n {\varphi} + \frac 1 {\varphi} ] =  [ \frac n {\varphi} ] $ .
It easily follows that $L = [ \frac n {\varphi} ]  = R$ in that case, and we are done.

Pierre.

There are $0 < w < 1 , w^2 + w - 1 = 0 $ . From the definition of the integer part we have :
 $n-(n+1)w \leq  n-[(n+1)w] \leq  n-(n+1)w+1$ and $  nw^2< ([nw]+1)w \leq  (nw+1)w $
So :
$n-(n+1)w -  (nw+1)w < n-[(n+1)w] - ([nw]+1)w  <  n-(n+1)w+1 - nw^2  \Longrightarrow$ $ -w  < n-[(n+1)w] - ([nw]+1)w < 1-w $
Now it's trivial :
$ -1 < - w < 0 < 1-w < 1 \Longrightarrow [ n-[(n+1)w] - ([nw]+1)w]=0 $ $  \Longrightarrow n - [([nw]+1)w] = [(n+1)w] $
 :cool:  Could you conclude by induction ? :cool:

:oops:  Sorry, I made a mistake. Forget my previous message !  :cool:

[quote="pbornsztein"][quote="Myth"]I think, I posted it some time ago in another thread.
[/quote]What do you mean? That you already solved this problem here? ;) [/quote]
I solved it many years ago. ;) I propose it on forum, and grobber found an answer, but, as I remember, he didn't write his proof.
My solution was similar to yours. But I think thre is another nice approach.

My solution was pretty much equally long, so I decided I didn't want to post it :). I'd like to see a shorter one though.

[quote="grobber"]I'd like to see a shorter one though.[/quote]

Ask for someone else. You know I can't do such thing  :D

Pierre.


A circle$\Gamma$ with radius $R$ and center $\omega$, and a line $d$ are drawn on a plane,
such that the distance of $\omega$ from $d$ is greater than $R$. Two points $M$ and
$N$ vary on $d$ so that the circle with diameter $MN$ is tangent to $\Gamma$. Prove
that there is a point $P$ in the plane from which all the segments $MN$ are
visible at a constant angle.

a solution of this problem follows straigthforward from a solution of this

http://www.mathlinks.ro/Forum/viewtopic.php?t=69620


Given a $n/times n$ checker board (n is even and n>2 ) which is colored by $\frac{n^2}{2}$ colors ( each color is used for 2 squares )
Prove that there exist $n$ squares that lie in different row, different column, and have different  color

I suppose you forgot to add that $n$ is even and $>2$?

I posted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=colors&t=6164]here[/url], and I even remember seeing a correct solution, but that post got deleted for some reason :?.

Given a certain pair of squares of the board which have the same color, there are at most $(n-2)!$ combinations of squares, one on each row and column of the board, that contain those two squares. This gives at most $(n-2)!\cdot\frac{n^2}2$ combinations of $n$ squares, one on each row and column of the board, which contain two squares with the same color. Since there are $n!>(n-2)!\cdot\frac{n^2}2$ combinations of $n$ squares, one on each row and column, there must be one which has no pair of squares with the same color.

Oops  :blush:  I forgot that condition, I'll correct it.
Thanks for your hint grobber  :)


arithmetic sequence
geometric sequence
algebra unsolved
algebra

Prove: if an infinte arithmetic sequence ($ a_n\equal{}a_0\plus{}nd$) of positive real numbers contains two different powers of an integer $ a>1$, then the sequence contains an infinite geometric sequence ($ b_n\equal{}b_0q^n$) of real numbers.

Let $ x^m,x^n$ be the two different powers of $ a$, where $ x^m \ge x^n$,
Then $ x^m\minus{}x^n\equal{}kd$ for some $ k$.
So $ x^{2m\minus{}n}\equal{}x^{m\minus{}n}(x^m\minus{}x^n)\plus{}x^m\equal{}[x^{m\minus{}n}kd\plus{}m]$ which also belongs to the sequence.

[quote="stephencheng"]${}x^{m\minus{}n}(x^m\minus{}x^n)\plus{}x^m\equal{}[x^{m\minus{}n}kd\plus{}m]$ [/quote]

I found typo in this statement. It must be $x^{m-n}. kd + x^m$, since $x^m = a_0 + cd$. So, $x^{m-n}. kd + a_0 + cd = a_0 + d(k.x^{m-n} + c)$.  :maybe:


The diameter of a circle is divided into $n$ equal parts.  On each part a semicircle is constructed.  As $n$ becomes very large, the sum of the lengths of the arcs of the semi-circles approaches a length:

(A) equal to the semi-circumference of the original circle
(B) equal to the diameter of the original circle
(C) greater than the diameter but less than the semi-circumference of the original circle
(D) that is infinite
(E) greater than the semi-circumference but finite

[hide]We'll say that the original circle has a radius of 1. The arc length of a semicircle is equal to $\pi r$. When $n=1$, $\pi r=\pi$, which is equal to the semiperimeter of the original circle. When $n=2$, $\pi r=\frac{\pi}{2}$ of each semicircle, but there are two, meaning that the total is $\pi$, or again equal to the semiperimeter of the original circle. The formula for the sum of the arc lengths of the semicircles is $\frac{n \pi}{n}=\pi$, which is the semiperimeter of the original circle, so the correct answer is $A$.[/hide]

yeah it's always A for any n.  If the diamter of the circle is x, then half the circumference is (pi/2)x.  The length of each arc is (pi/2)(x/n).  And there are n of them.  Go figure.


Let $a, b, c$ be such real numbers that $a+b+c=2$ and $ab+bc+ca=1$. Prove that

$\max \left\{a, b, c \right\} -\min \left\{a, b, c \right\} \leq \sqrt{\frac{4}{3}}$.

some things that can be obtained by the condition are

$a^2+c^2=2-b^2\ \ \ \ (1) \\ac=1-b(a+c)=1-b(2-b)=(b-1)^2 \ \ \ \ (2)$

wlog, suppose $a\geq b \geq c$, we must prove that
$a-c\leq \sqrt{\frac{4}{3}} \Leftrightarrow (a^2+c^2)-2(ac)\leq \frac{4}{3} \\\Leftrightarrow (2-b^2)-2(b-1)^2\leq \frac{4}{3} \\\Leftrightarrow 0\leq \frac{9b^2+12b+4}{3}=\frac{(3b+2)^2}{3}$
and we're done!

...but this result seems kind of weird to me, since equality is obtained when $b=\frac{-2}{3}$, but (let me prove something else)...

by viête's formula we have $a,c$ are the roots of the equation
$x^2-x(2-b)+(b-1)^2=0$, and since $a,c$ are real we must have the discriminant to be greater or equal than 0, i.e.
$(b-2)^2-4(b-1)^2\geq 0 \Rightarrow 0\geq 3b^2-4b \Rightarrow \frac{4}{3}\geq b\geq 0$, hence $b\neq \frac{-2}{3}$, and that's what seems weird to me!


$ ABC$ and $ PQR$ are both equilateral triangles of area 1. The centre $ M$ of $ PQR$ lies on the perimeter of $ ABC$. Determine the minimal area of the intersection of the two triangles.

Well you could start off by finding the side length of PQR, and then the height, and then with that knowledge, you could draw a line from the center of PQR that is perpendicular to the closest side of PQR, and then that distance would be the height of the intersection of PQR and ABC. Hope that helped!

I am still confuse
the intersection area can be triangle or quadrilateral. Also the question didnt specify the location of the center of $ PQR$ in the perimeter of $ ABC$

Well, here's what I did. I drew a triangle that was equilateral and ressembled PQR, and then I picked a point that I though was the center, and then I drew a perpendicular line to the closest side to it. Then, lets say the height of PQR is x, and the length of the perpendicular line is y. What I did after that was I extended the perpendicular line by (x-y) units. This would be the height of ABC. Then I finished drawing ABC with the line that I had. Then the small little triangle that was the intersection would be the smallest possible intersection. You don't seem to understand it.

Does anybody else understand my solution, if you do please explain it to mingming.

Do you have the answer ming ming? Or is this just a problem that you got without an answer?

If you had the answer, then I would solve it, but because you don't understand, that might mean that you don't believe my answer.


algebra polynomial Vieta quadratics function

$let a, b, c and d be a reels numbers wish satisfy$\\
 $a=\sqrt{{4}-\sqrt{5-a}}$\qqua$b=\sqrt{{4}+\sqrt{5-b}}$\\
$c=\sqrt{{4}-\sqrt{5-c}}$\qquad$d=\sqrt{{4}+\sqrt{5-d}}$find the value of $abcd$

Maybe I'm missing something totally obvious, but $b$ and $d$ have a plus sign under the outer radical...

I am sorry I missed that completely. But on second thought, $a,b,c,d$ would still be the four roots of the biquadratic equation obtained by removing the square-root signs from $x=\sqrt{4+\sqrt{5-x}}$.

Anyone has any comment on that?

$its true but this polinomial accept just a;c and-b;-d$ :lol:

[quote="FieryHydra"][hide="Hint"]$a,b,c,d$ are the roots of the equation $x=\sqrt{4-\sqrt{5-x}}.$ Transform this into a biquadratic equation and use Viete's relation. [/hide][/quote]

when you say biquadratic equation, do you just mean 4th degree polynomial (quartic)?

also this solution is not complete.... take something like \[a^{3}=2b+2\qquad b^{3}=2a+2\] Let $s_{k}=a^{k}+b^{k}$

Those equations added imply: \[s_{3}-2s_{1}-4=0\] but you cannot directly state that $a^{3}-2a-2=0$

(it takes more argument)

[quote="Peter VDD"]Please stop writing every english word in latex. It's ugly, difficult to read and causes needless server costs for extra image storing.[/quote]
Ok sorry. :blush:  I will not do it again. In fact this is my first time I did this...
However, the first sentence of your signature is Latexed, and this sentence is added to nearly every one of your 2000+ posts, and this uses more needless server costs for image sorting. :wink:

That image is stored only once, obviously. ;)

And I just said it to prevent the whole topic from becoming like that...

[quote="Farenhajt"][quote="Altheman"]when you say biquadratic equation, do you just mean 4th degree polynomial (quartic)?[/quote]

Let me answer to that. "Biquadratic equation" is the term for the special kind of quartic equation, namely a quadratic in $x^{2}$ i.e. $ax^{4}+bx^{2}+c=0$ with $a\neq 0$. But this problem seems to yield a regular quartic and not a biquadratic.[/quote]
I think a general biquadratic equation is just a general $4^{th}$ degree poynomial equation, also called a quartic equation. A "biquadratic" (or "quartic") isn't always a special case of a $4^{th}$ degree polynomial. The terms can be used interchangeably.
http://mathworld.wolfram.com/QuarticEquation.html
http://en.wikipedia.org/wiki/Biquadratic_equation
http://en.wikipedia.org/wiki/Quartic_function

On the first Wikipedia page - namely [url]http://en.wikipedia.org/wiki/Biquadratic_equation[/url] - read the section called "Biquadratic equations" under "Special cases".

And keep in mind that "biquadratic [b]equation[/b]" isn't the same as "biquadratic [b]function[/b]".

I would still stick to the definition offered in http://www.mathworld.com.

Obviously that's the same kind of problem as what does $\mathbb{N}$ mean, i.e. does it include zero or not. The best is to avoid the ambiguity, since as far as I can tell, both definitions are widely accepted and used.


AMC
AIME
MATHCOUNTS
geometry
trigonometry
ratio
number theory

First of all, "not really three digits" implies that this is an AIME problem... besides, you could modify the problem to be like "... in the form $ \frac {a}{b}$, where $ a$ and $ b$ are relatively prime positive integers. Find $ a \plus{} b$," but that's off topic.

The answer is 7/24, but the solution is a bit advanced for MATHCOUNTS level. But since this is supposedly AIME level, I suppose I am allowed to post the solution.

Anyway...

[hide]
The shape or side lengths of the outer triangle is not given, so it doesn't change the answer. That said, let's make a shape and a set of side lengths up! I say that the triangle is equilateral and that the side lengths are all 12. Then, the area of the outer triangle is $ 36\sqrt {3}$.

$ \overline{AD} \equal{} 9, \overline{BE} \equal{} 8, \overline{AF} \equal{} \overline{FC} \equal{} 6, \overline{CE} \equal{} 4, \overline{BD} \equal{} 3$.

Triangle areas:
$ \triangle{ADF} \equal{} \frac {1}{2} \cdot 9 \cdot 6 \cdot \sin{60^\circ} \equal{} \frac {27\sqrt{3}}{2}$
$ \triangle{CEF} \equal{} \frac {1}{2} \cdot 6 \cdot 4 \cdot \sin{60^\circ} \equal{} 6\sqrt {3}$
$ \triangle{BDE} \equal{} \frac {1}{2} \cdot 8 \cdot 3 \cdot \sin{60^\circ} \equal{} 6\sqrt {3}$

The area of $ \triangle{DEF}$ is the area of the outer triangle minus the areas of the three triangles surrounding the inner triangle. First, we factor out $ \frac {\sqrt {3}}{2}$ to make the problem simpler.

$ \frac {\sqrt {3}}{2}(72 \minus{} 27 \minus{} 12 \minus{} 12) \equal{} \frac {\sqrt {3}}{2}(21)$

And the requested answer is

$ \frac {\frac {\sqrt {3}}{2}(21)}{\frac {\sqrt {3}}{2}(72)} \equal{} \boxed{\frac {7}{24}}$.
[/hide]

Edit: Fixed

[quote="mathcrazed"][hide="Not Aime"]$ \frac {3}{4}*\frac {2}{3}*\frac {1}{2} \plus{} \frac {1}{4}*\frac {1}{3}*\frac {1}{2} \equal{} \frac {1}{4} \plus{} \frac {1}{24} \equal{} \frac {7}{24}$
[/hide]

By your set up... $ AD \equal{} \frac {9}{2}$[/quote]

Oh! :blush:

I read "$ \overline{BD}$ is one-third of $ \overline{AB}$" instead of $ \overline{AD}$... so then, $ \overline{BD}$ is one-fourth of $ \overline{AB}$, right? Then, it's 7/24. :D

[quote="ZhangPeijin"]You ever try doing the same thing with another triangle?
Maybe Isoceles right?
I'm pretty sure the ratio isn't constant.[/quote]

I'm pretty sure the ratio is constant. If the side lengths or angles are unspecified, the side lengths and angles don't matter.

Any triangle with those specifications on the ratio of side lengths to segments will have that area ratio.

Except
1) This isn't a AIME problem.
Therefore, the problem could be wrong.
And. I'm pretty sure the ration isn't constant because
It doesn't specify what type of triangle it is.
It doesn't specify which sides are split into halfs, thirds, or fourths.
Therefore, in a triangle that is not equilateral, two or more of these triangles can be made, and I'm pretty sure they won't be the same area.
And to prove this to you. I will do the same thing with an isosceles right with leg lengths of 12.
Ratio is $ \frac {1}{8}$
Unless I did it wrong...

You did it wrong.  Any arbitrary triangle with [u]those specifications[/u] has that ratio.

Thank you guys for explaining this to Peijin. I tried to tell him that it does not have an infinite amount of solutions. This problem is a 5th grade Olympiad Problem from China. @Mathcrazed your solution was exactly the same as the way as the answer key.

Edit: the not really three digits was because I gave it to alan as a practice AIME problem that was good for like #1, but it was a fraction so...

[i]Fifth grade[/i]? Wow... I was pretty stupid in fifth grade.

Hao, alan agreed with me when I told him about this problem.

oh shoot, I used cosine instead of sine... dangit :oops:

You just have to draw lines of similar heights.

[quote="mathcrazed"][hide="Not Aime"]$ \frac {3}{4}*\frac {2}{3}*\frac {1}{2} \plus{} \frac {1}{4}*\frac {1}{3}*\frac {1}{2} \equal{} \frac {1}{4} \plus{} \frac {1}{24} \equal{} \frac {7}{24}$
[/hide]
[/quote]

@mathcrazed....I feel really stupid asking this, but why do you multiply those numbers like that and where did the 3/4ths come from?


geometry circumcircle AMC USA(J)MO USAMO

$ AB$ is a fixed chord of a circle that is not a diameter.  $ CD$ is a variable diameter.  Find the locus of the intersection of $ AC$ and $ BD$.

I'm going to be a bit lazy with the formal locus proof, but I'll give the main details.

Take some chord, $ XY$ so $ ABYX$ is convex (or a degenerate triangle). Then $ AX\cap BY \equal{} M$ and $ AY\cap BX \equal{} N$. (If $ AXBY$ is convex, then our triangle is $ MNY$

Now if we look at triangle $ MXY$, then $ N$ is the orthocenter of that triangle (remember the diameter...). Let $ O'$ be the midpoint of $ MN$. It is well known that $ O'AOB$ lie on the ninepoint circle of $ \triangle MXY$. This nine-point circle is fixed as $ A$, $ B$, and $ O$ are fixed! It is also well-known that $ OO'$ is a diameter of this circle (the midpoint of that segment is the nine-point center for that matter).

Next we have to cover the case when the two chords intersect, but that is pretty much the same.

Now we have to show that every point on the circle center $ O'$, diameter $ AO'$ is satisfactory. (We can easily find $ O'$ as the diametric opposite point of $ O$ in the circumcircle of $ \triangle ABO$). Chose $ M'$ on the circle, then draw $ AM$  intersecting the circle at $ C'$. The we note that the circles are orthogonal, and so $ M'AO' \plus{} 90 \plus{} C'AO \equal{} 180$. Then we angle chase to show that $ MB\perp BC'$. That pretty much finishes the problem...

Doing old USAMOs too I see...

I like these old USAMOs.  They're good confidence boosters :P

[hide="With cyclic quads"] Let $ P$ be an arbitrary point outside the circle.  Let $ O$ be the center of the circle, and let $ RS$ be the diameter perpendicular to $ PO$.  Let $ A, B$ be the intersections of the circle and $ RP, SP$ respectively.  $ AB$ then describes a generic chord that is not a diameter.

We claim that the desired locus is the circumcircle of $ \triangle ABP$.

Let $ CD$ be an arbitrary diameter (so that $ ABCD$ is a cyclic quadrilateral).  Let $ M$ be the intersection of $ AC, BD$ and let $ N$ be the intersection of $ AD, BC$.  Then $ \triangle DBN$ is a right triangle (because $ \angle CBD$ is a right angle), and since $ \angle ADB \equal{} \angle ARB$ and $ \triangle RBP$ is also a right triangle it follows that the two are similar and so $ \angle APB \equal{} \angle ANB$, so $ ABNP$ is cyclic.

Similarly, $ \triangle DAM$ is right.  Let $ \angle APB \equal{} \theta$; then $ \angle ADB \equal{} \angle ARB \equal{} 90 \minus{} \theta$, so $ \angle DMA \equal{} \theta$ and $ \angle AMB \equal{} 180 \minus{} \theta$, so $ AMBP$ is cyclic.

$ M, N$ are therefore both on the circumcircle of $ \triangle ABP$ as desired. [/hide]

Haha yea I've solved a combo problem...

Anyway, nice, I didn't notice that cyclic quad sol. I tend to just bash things with a lot of lemmas and not look for better solutions. I think that you need to start with the chord first, and not $ P$ because you don't know that $ P$ exists otherwise. And of course in locus proofs, one must show that every point on the circle is in the locus but maybe you were being concise.

[quote="Altheman"]And of course in locus proofs, one must show that every point on the circle is in the locus but maybe you were being concise.[/quote]

Good point.

[hide="Addendum"] The above construction works in reverse; given a point $ M$ on the circumcircle of $ \triangle ABP$, construct points $ C$ and $ D$ with the appropriate right triangles, and then $ CD$ is a diameter.  It's a little late at night to work out the details, but my main point is that the map from diameters $ CD$ to diameters $ MN$ is one-to-one and continuous, hence invertible :) [/hide]


Does anybody have a proof (without trig) from both of them¿?

What is Brahmagupta and Bretschneider formula ?

look at here :
[url]http://mathworld.wolfram.com/BrahmaguptasFormula.html[/url]

so if, this is prove http://en.wikipedia.org/wiki/Brahmagupta's_formula


function
number theory unsolved
number theory

Find all $ n$ such that

$ \phi(n) \equal{} 12$

Obviosly $ n\equal{}2^k3^m5^l7^a13^b$, were $ k\le 3,m\le 2,l\le 1,a\le 1,b\le 1.$.
If $ l\equal{}1$, then $ n\equal{}5*2^k$ - contradition. Therefore $ l\equal{}0$.
It give $ n\equal{}13,26,28,21,36$.


2. This L-shape piece of land is made of 43 unit squares. By constructing a fence, only along drawn segments in the figure, the one large piece of land can be divided into two congruent smaller pieces of land, each with area 16 square units. What is the number of units in the perimeter of one of these smaller pieces of land?(#21)*Picture attached below
I guessed it right, but how would you do this logically(for lack of better word)?
[hide="Answer"]20[/hide]

3. Jeff will pick a card at random from ten cards numbered 1 through 10. The number on this card will indicate his starting point on the number line shown below. He will then spin the fair spinner shown below(which has three congruent sectors) and follow the instruction indicated by his spin. From this new point he will spin the spinner again and follow the resulting instruction. What is the probability that he ends up at a multiple of 3 on the number line? Express your answer as a common fraction.(#28)*Picture attached below
[hide="Answer"]$ \frac {31}{90}$[/hide]

[quote="xpmath"][hide="First One"]I think you can divide it into 2 smaller L-shape things.
Like the left 2 by 6+top right 2 by 2 or something.[/hide][/quote]
So in order to achieve the smallest perimeter, the shapes have to be same?

Nooo... They have to be the same because that is stated in the problem. :D

[quote="tinytim"]just use casework for #2. Its not that bad.[/quote]

Do casework on the probability that one will reach a multiple of three when the first number is picked as a:
0 mod 3
1 mod 3
2 mod 3
And then multiply by the probabilities of each.

[hide="L shaped land"]
It's hard to teach how to solve a problem like this.  The biggest factor is your visualization skills.  You can exclude many possibilities by looking for a unique and distinctive feature of the uncut area that would be uncommon in random cuttings of the region.  (Perhaps there's a distinctive angle.)  In this case, it's the long 2x4 tail.  [It's got to be attached to something, so you have a 2x5 tail, or bigger.]  You'll have to find another 2x4 tail somewhere in the square-ish region to match it.
[/hide]

[hide="Cards and spinner casework"]
1,4,7,10:  We must add 1 and add 1 to get a multiple of 3:
$ p_1 \equal{} \frac {4}{10}\cdot\frac23\cdot\frac23 \equal{} \frac {16}{90}$

2,5,8:  We must subtract 1 and subtract 1 to get a multiple of 3:
$ p_2 \equal{} \frac3{10}\cdot\frac13\cdot\frac13 \equal{} \frac {3}{90}$

3,6,9:  We must add 1 and subtract 1 (or subtract 1 and add 1)
$ p_3 \equal{} \frac3{10}\cdot\frac23\cdot\frac13\times 2 \equal{} \frac {12}{90}$

Add the above:
$ \frac {16}{90} \plus{} \frac {3}{90} \plus{} \frac {12}{90} \equal{} \frac {31}{90}$[/hide]

Can anybody please draw the line into the diagram? I can't see it. I understand what rcv says above about the tail but what does he mean by finding something in the square region to match it? There's a lot of 2x4 rectangles.

Let the lower left corner of the shape be the origin, (0,0)
Draw one segment from (2,0 to (2,4).  Draw a second segment from (2,4) to (4,4).
The shape is divided into two identical L-shapes.


The legs of a right triangle have lengths 10 cm and 24 cm. What is the number of centimeters in the length of the hypotenuse?

This is simply a multiple of the triple $ 5\minus{}12\minus{}13$, so therefore the hypotenuse is just $ 13 \cdot 2 \equal{} \boxed{26}$.

Yes, or you can use Pythagorean to find the hypotenuse.

a^2+b^2=c^2 where c is the longest side (the hypotenuse)

Plug in the numbers and you get

10^2+24^2=hypotenuse

100+576=676

The 676 is c^2. So the hypotenuse is the square root of 676 which is 26.

The answer is 26.

██████████████████████████████████████████████████████████████████████████████████████

~R@d!0Act!vÉ

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Duke college USAMTS Princeton AMC AIME USA(J)MO

I shamelessy stole this from dts, and I will add more SC-specific contest dates soon.  Feel free to post any I left out.

Below are the dates for upcoming mathematical contests and programs.  [b]Please note that dates marked with asterisks (*) are tentative dates only![/b]

[u]November 2006[/u]
Duke Math Meet, November 11th, 2006 (Saturday)
AMC 8 on Tuesday, November 14th, 2006 (Tuesday)
USAMTS Postmark Deadline, November 20th, 2006 (Monday)

[u]December 2006[/u]
USC High School Math Contest on December 9th, 2006 (Saturday)
Princeton Mathematics Competition on December 16th, 2006 (Saturday)

[u]January 2007[/u]

[u]February 2007[/u]
AMC-10A/12A on Tuesday, February 6, 2007
AMC-10B/12B on Wednesday, February 21, 2007

[u]March 2007[/u]
Furman Math Contest on March 10, 2007 (Saturday)
AIME on Tuesday, March 13, 2007
Alternate AIME on Wednesday, March 28, 2007

[u]April 2007[/u]
USAMO on April 24-25, 2007

[u]May 2007[/u]

[u]June 2007[/u]
ARML - Dates Unknown

USAMTS first round deadline- October 10th

What was it...2002 or 2003 that was the best nats mathcounts year ever?  2003, right?

[quote="knexpert"]Some USAMTS dates and the Putnam date were not listed.[/quote]

I'll add anything you want, just post it here.  However, I don't think many SCers will be taking the Putnam...Joe?

Yeah, I'm taking it.  I'm going to get pwned badly, but w/e.