# Thread: Cool problem of the day!

ahhh i see

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

I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.

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

Originally Posted by Sy123
I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.
I've got an idea for one, just have to go about typing it up >.< I'll try and have it up by tomorrow - Carrot will hopefully have his bunch up as well, but he is a busy man!

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

Originally Posted by Sy123
I would love more questions to come along, some of them albeit hard are fun to do, and it excercise obscure math techniques.
Sorry, been a bit busy lately. I might be able to do something tonight.

Originally Posted by AAEldar
I've got an idea for one, just have to go about typing it up >.< I'll try and have it up by tomorrow - Carrot will hopefully have his bunch up as well, but he is a busy man!
That would be great! Please PM me =)

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

Geometry:

Consider a circular ring of radius R.

A stick of length k such that $k \leq 2R$ is placed within this circle so it becomes a 'chord'.

The stick is then 'spun' around inside the circle such that the two ends of the stick are always touching the circumference of the circle.

Find the area of the shape created by the midpoint of the stick once it's made a full rotation.

Algebra:

Determine all real values of X and Y satisfying the following system of equations:

$\\ x + xy + xy^2 = 26 \\\\ x^2y + x^2 y^2 + x^2 y^3 = 156$

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

Originally Posted by Carrotsticks
Algebra:

Determine all real values of X and Y satisfying the following system of equations:

$\\ x + xy + xy^2 = 26 \\\\ x^2y + x^2 y^2 + x^2 y^3 = 156$

Algebra: (18, 1/3) or (2, 3)

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

Originally Posted by Carrotsticks
Geometry:

Consider a circular ring of radius R.

A stick of length k such that $k \leq 2R$ is placed within this circle so it becomes a 'chord'.

The stick is then 'spun' around inside the circle such that the two ends of the stick are always touching the circumference of the circle.

Find the area of the shape created by the midpoint of the stick once it's made a full rotation.

Here's my attempt:

The midpoint will make a smaller circle concentric with the original circle in the diagram. To find the radius of this circle, we use pythagoras.

Let's join the centre of the circle to the midpoint of the line and join the centre of the circle to the end of the line to form a right angled triangle. The hypotenuse is R (the radius of the original circle) and the base is half the length of the line, which is $\frac{k}{2}$. Hence the radius of the smaller circle is:

$\sqrt{R^2 - (\frac{k}{2})^ 2$

The area of the smaller circle is thus:

$\pi (R^2 - \frac{k^2}{4})$

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

Originally Posted by RealiseNothing
Here's my attempt:

The midpoint will make a smaller circle concentric with the original circle in the diagram. To find the radius of this circle, we use pythagoras.

Let's join the centre of the circle to the midpoint of the line and join the centre of the circle to the end of the line to form a right angled triangle. The hypotenuse is R (the radius of the original circle) and the base is half the length of the line, which is $\frac{k}{2}$. Hence the radius of the smaller circle is:

$\sqrt{R^2 - (\frac{k}{2})^ 2$

The area of the smaller circle is thus:

$\pi (R^2 - \frac{k^2}{4})$
That is correct, and you can confirm it by observing that as k --> 2R, Area --> 0.

Here's the hard(er) part... prove that the locus of the midpoint is actually a circle. You've assumed it here in your answer.

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

Originally Posted by Carrotsticks
That is correct, and you can confirm it by observing that as k --> 2R, Area --> 0.

Here's the hard(er) part... prove that the locus of the midpoint is actually a circle. You've assumed it here in your answer.
The reason I knew it was a circle was because instead of spinning the chord around the circle, I spun the circle around the chord.

So I let the chord be fixed, and rotated the circle, and obviously from this the midpoint will make a circle since the centre of the circle is fixed, and the midpoint is fixed, hence they are always the same distance apart and thus the locus is a circle.

But I'll try a more rigorous proof if necessary.

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

Originally Posted by RealiseNothing
The reason I knew it was a circle was because instead of spinning the chord around the circle, I spun the circle around the chord.

So I let the chord be fixed, and rotated the circle, and obviously from this the midpoint will make a circle since the centre of the circle is fixed, and the midpoint is fixed, hence they are always the same distance apart and thus the locus is a circle.

But I'll try a more rigorous proof if necessary.
Very clever =)

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

Originally Posted by RealiseNothing
The reason I knew it was a circle was because instead of spinning the chord around the circle, I spun the circle around the chord.

So I let the chord be fixed, and rotated the circle, and obviously from this the midpoint will make a circle since the centre of the circle is fixed, and the midpoint is fixed, hence they are always the same distance apart and thus the locus is a circle.

But I'll try a more rigorous proof if necessary.
Argument by rotational symmetry of the circle as a limiting case of an N-gon.

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

This may be easy, or it may be hard. I just came up with it then, so not sure how difficult it is.

Given a quadrilateral with two of it's sides of length 4 and 4, perimeter of 24, and area of $\frac{40}{\sqrt5}$, find the length of the two unknown sides. All four corners of the quadrilateral are equidistant from some point inside the shape.

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

Originally Posted by RealiseNothing
This may be easy, or it may be hard. I just came up with it then, so not sure how difficult it is.

Given a quadrilateral with two of it's sides of length 4 and 4, perimeter of 24, and area of 40, find the length of the two unknown sides. All four corners of the quadrilateral are equidistant from some point inside the shape.
Are you sure there is a unique solution?

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

Originally Posted by Fus Ro Dah
Are you sure there is a unique solution?
When I did it I got a unique solution.

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

Originally Posted by RealiseNothing
This may be easy, or it may be hard. I just came up with it then, so not sure how difficult it is.

Given a quadrilateral with two of it's sides of length 4 and 4, perimeter of 24, and area of 40, find the length of the two unknown sides. All four corners of the quadrilateral are equidistant from some point inside the shape.
Hint: Use this http://en.wikipedia.org/wiki/Brahmagupta's_formula.

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

Originally Posted by RealiseNothing
The proof for this is Q8 of the 2006 Independent Trial.

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

Originally Posted by RealiseNothing
This may be easy, or it may be hard. I just came up with it then, so not sure how difficult it is.

Given a quadrilateral with two of it's sides of length 4 and 4, perimeter of 24, and area of 40, find the length of the two unknown sides. All four corners of the quadrilateral are equidistant from some point inside the shape.
no quadrilateral of perimeter 24 can have an area 40.

If there were no restrictions upon side length, the maximum area of a 24 unit quadrilateral would be 6x6=36 units squared ( when it is a square )

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

Originally Posted by barbernator
no quadrilateral of perimeter 24 can have an area 40.

If there were no restrictions upon side length, the maximum area of a 24 unit quadrilateral would be 6x6=36 units squared ( when it is a square )
Woops when I did the question I forgot to square root something.

The area should be $\frac{40}{\sqrt5}$.

This should work now.

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

is this some 4u shit?

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

Originally Posted by uniquee
is this some 4u shit?
My question could be done by a 2U student if you use the formula I posted.

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

ill do it tomorrow, i cbf to do it tonight

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

A hint for the first two number theory questions I posted on the previous page. Consider the prime factors of squares and try to use properties of prime numbers.

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

oh god, how did I end up on this side of bos?

brb, running back to nonschool

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

Originally Posted by ZombieApocalypse
oh god, how did I end up on this side of bos?

brb, running back to nonschool
But it's maffs!

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

Sorry been a bit busy lately. Here's a quick problem for the mean time.

A ball of radius 1 is 'dropped' into the parabola y=x^2. Find the centre of the circle.

EXTENSION: Same as above, but for the parabola $y=k x^2, where k \geq 1$

Page 4 of 6 First ... 23456 Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•