Cool problem of the day! (2 Viewers)

asianese · May 28, 2012

I'm assuming that means sit snugly inside the parabola? OMG I JUST SOLVED IT!! SO HAPPY WTF OMG HAHHAHAHAHA. I feel...so smart...yet among some of these questions...

The general equation of a circle is $(x-h)^2+(y-k)^2=r^2$ . Here $h=0, r=1$ . Solving simultaneously the parabola and circle, $y+(y-k)^2=1\\y^2+y(1-2k)+k^2-1=0$ . As the solution is a double root, the discriminant equals 0. $(1-2k)^2-4(k^2-1)=0$ . $\therefore k=\frac{5}{4}$ . The circle has centre $(0,\frac{5}{4})$ .

By a similar process, the centre of the circle in the parabola $y=kx^2$ is $(0,\frac{1+4k}{4})$ (Different k to top one)

Carrotsticks · May 28, 2012

asianese said:
I'm assuming that means sit snugly inside the parabola?

More snuggled up than me on a cold Sunday morning under my warm blanket.

qwerty44 · Jun 16, 2012

This thread is practically dead, so thought I might post a question. But unlike many of the others posted so far, a 2U student can do it.

$\begin{align*}&A~and~B~are~positive~real~numbers~for~which~\log_{A}B=\log_{B}A.\\&If~neither~A~or~B~is~1~and~A\neq{B},~find~the~value~of~AB.\end{align*}$

ismeta · Jun 16, 2012

1?

I didn't use any particular mathematical method to do it, though.

qwerty44 · Jun 16, 2012

ismeta said:
1? I didn't use any particular mathematical method to do it, though.

haha yeh, but you have to show how to get the answer.

I guessed that the first time aswell

Sy123 · Jun 16, 2012

qwerty44 said:
This thread is practically dead, so thought I might post a question. But unlike many of the others posted so far, a 2U student can do it.

$\begin{align*}&A~and~B~are~positive~real~numbers~for~which~\log_{A}B=\log_{B}A.\\&If~neither~A~or~B~is~1~and~A\neq{B},~find~the~value~of~AB.\end{align*}$

$\log_b{a}=\log_a{b} \\ \frac{\log{a}}{\log{b}}=\frac{\log{b}}{\log{a}} \\ \\ \text{Case 1:}\log^2{a}=\log^2{b} \\ \\ \log{a}=+\log{b} \\ \\ \log{a}-\log{b}=0 \\ \\ \log{\frac{a}{b}}=0 \\ \\ \therefore \frac{a}{b}=1 \ \ \ \ a=b???? \\ \\ \\ \text{Case 2:}\log^2{a}=\log^2{b} \\ \\ \log{a}=-\log{b} \\ \\ \log{a}+\log{b}=0 \\ \\ \log{ab}=0 \\ \\ \therefore ab=1$

I think the conclusion is that a can indeed equal to b, but you gave us restricted boundaries, so yeah heh.

qwerty44 · Jun 16, 2012

Sy123 said:
$\log_b{a}=\log_a{b} \\ \frac{\log{a}}{\log{b}}=\frac{\log{b}}{\log{a}} \\ \\ \text{Case 1:}\log^2{a}=\log^2{b} \\ \\ \log{a}=+\log{b} \\ \\ \log{a}-\log{b}=0 \\ \\ \log{\frac{a}{b}}=0 \\ \\ \therefore \frac{a}{b}=1 \ \ \ \ a=b???? \\ \\ \\ \text{Case 2:}\log^2{a}=\log^2{b} \\ \\ \log{a}=-\log{b} \\ \\ \log{a}+\log{b}=0 \\ \\ \log{ab}=0 \\ \\ \therefore ab=1$

I think the conclusion is that a can indeed equal to b, but you gave us restricted boundaries, so yeah heh.

$\begin{align*}&Let~\log_{A}B=\log_{B}A=x\\&A^{x}=B~~and~~B^{x}=A\\&\therefore AB= A^{x} \cdot B^{x}=(AB)^{x}\\&(AB)^{x-1}=1\\&AB=1~~or~~x=1\\&but~it~is~given~A\neq B ,~so~x\neq 1\\\therefore &AB=1\end{align*}$

^I tried to do it without log laws, so that's how I did it.

qwerty44 · Jun 18, 2012

The curves are arcs of circles whose centers are vertices of the square, side 6cm. Find the area of the shaded region.

seanieg89 · Jun 18, 2012

$15\pi/2-9\sqrt{3}\, \, cm^2$

Hopefully I didn't make a calculation error.

qwerty44 · Jun 18, 2012

seanieg89 said:
$15\pi/2-9\sqrt{3}\, \, cm^2$

Hopefully I didn't make a calculation error.

Correct.

Carrotsticks · Jun 19, 2012

Here's a question I posted a while ago but I don't remember if anybody answered it (don't think so). Best done by the interested and well-read HSC student.

I am standing at the origin. I move to the right by one unit, then up by 1/2, then left by 1/3, then down by 1/4, then right again by 1/5 etc etc.

If I continue this infinitely, is there a coordinate I will (eventually) reach? If so, what is it?

---------------------------------------

Another:

I am a 1-Dimensional person (so 1 degree of freedom). This means I can only either move forwards or backwards.

I move forward 1 unit, then backward 1/2 units, then forward 1/3 units, then backwards 1/4 units etc etc.

If I continue this infinitely, is there a coordinate I will (eventually) reach? If so, what is it?

---------------------------------------

Another:

Consider a regular n-gon. What is the total number of m-gons (not necessarily regular) that I can make such that m < n ? (careful about rotational symmetry).

IamBread · Jun 19, 2012

Is the answer to the first one
$(\frac{\pi}{4},\frac{ln2}{2})$

Carrotsticks · Jun 19, 2012

Yep! Methinks that is a really cool result. No doubt if you could the first, then the second is trivial. Try going up 1 more dimension to 3D.

IamBread · Jun 19, 2012

Yeah, and it was a lot of fun to do too. I am doing the first one now, so far got the z direction coordinate to be $\frac{ln2}{3}$

Fus Ro Dah · Jun 19, 2012

Carrotsticks said:
Consider a regular n-gon. What is the total number of m-gons (not necessarily regular) that I can make such that m < n ? (careful about rotational symmetry).

I like this question. The hardest part is finding a closed expression for the sum that determines the number of m-gons. I think a good Extension 2 student should be able to do so if they play around a bit with Complex Numbers. I will post a solution later if nobody else does.

Fus Ro Dah · Jun 22, 2012

qwerty44 said:
The curves are arcs of circles whose centers are vertices of the square, side 6cm. Find the area of the shaded region.

I know this question can be done by elementary means by partitioning various areas, but is it possible to evaluate it using calculus?

$A= \iint_D f(x_1,x_2) \left | \frac{\partial (x_1,x_2)}{\partial (y_1,y_2)} \right | dy_1 dy_2$

Carrotsticks · Jun 22, 2012

Pretty sure it can be done using the transformation formula you have there (by perhaps mapping it out to a triangular domain) but then you would need to find the points of intersections etc, which would be quite tedious.

qwerty44 · Jun 29, 2012

Found a pretty good one that most 2U students should try.

Find the area of the rectangle.

Carrotsticks · Jun 29, 2012

Nice!

bleakarcher · Jun 29, 2012

qwerty44 said:
Found a pretty good one that most 2U students should try.

Find the area of the rectangle.

Is it sqrt(3)?

Cool problem of the day! (2 Viewers)

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