# Thread: Cool problem of the day!

1. ## Re: Cool problem of the day!

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)

2. ## Re: Cool problem of the day!

Originally Posted by asianese
I'm assuming that means sit snugly inside the parabola?
More snuggled up than me on a cold Sunday morning under my warm blanket.

3. ## Re: Cool problem of the day!

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~valu e~of~AB.\end{align*}

4. ## Re: Cool problem of the day!

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

5. ## Re: Cool problem of the day!

Originally Posted by ismeta
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 :P

6. ## Re: Cool problem of the day!

Originally Posted by qwerty44
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~valu e~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.

7. ## Re: Cool problem of the day!

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

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.

8. ## Re: Cool problem of the day!

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

9. ## Re: Cool problem of the day!

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

Hopefully I didn't make a calculation error.

10. ## Re: Cool problem of the day!

Originally Posted by seanieg89
$15\pi/2-9\sqrt{3}\, \, cm^2$

Hopefully I didn't make a calculation error.
Correct.

11. ## Re: Cool problem of the day!

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).

12. ## Re: Cool problem of the day!

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

13. ## Re: Cool problem of the day!

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.

14. ## Re: Cool problem of the day!

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 $\frac{ln2}{3}$

15. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks
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.

16. ## Re: Cool problem of the day!

Originally Posted by qwerty44

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?

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

17. ## Re: Cool problem of the day!

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.

18. ## Re: Cool problem of the day!

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

Find the area of the rectangle.

Nice!

20. ## Re: Cool problem of the day!

Originally Posted by qwerty44
Found a pretty good one that most 2U students should try.

Find the area of the rectangle.
Is it sqrt(3)?

22. ## Re: Cool problem of the day!

Originally Posted by bleakarcher
Is it sqrt(3)?
That's what i got, but I don't have answer lol

23. ## Re: Cool problem of the day!

I remember doing this a while ago now.

k=2/3

24. ## Re: Cool problem of the day!

Here is another good one:

The lengths of the sides of the octagon are 1, 2, 3, 4, 5, 6, 7 and 8 units in some
order. Find the maximum area of the hexagon (square units).

25. ## Re: Cool problem of the day!

Originally Posted by bleakarcher
Is it sqrt(3)?
Yeh sqrt(3) confirmed.

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