Cool problem of the day! (1 Viewer)

[nightweaver066]

Algebra:

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



University students/Postgrads: Please refrain from answering, as tempting as it may be.

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

 

[RealiseNothing]

Geometry:

Consider a circular ring of radius R.

A stick of length k such that 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 . Hence the radius of the smaller circle is:



The area of the smaller circle is thus:

 

[Carrotsticks]

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 . Hence the radius of the smaller circle is:



The area of the smaller circle is thus:

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.

 

[RealiseNothing]

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.

 

[Carrotsticks]

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 =)

 

[Fus Ro Dah]

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.

 

[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 , find the length of the two unknown sides. All four corners of the quadrilateral are equidistant from some point inside the shape.

 

[Fus Ro Dah]

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?

 

[RealiseNothing]

Are you sure there is a unique solution?

When I did it I got a unique solution.

 

[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.

 

[AAEldar]

Hint: Use this http://en.wikipedia.org/wiki/Brahmagupta's_formula.

The proof for this is Q8 of the 2006 Independent Trial.

 

[barbernator]

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 )

 

[RealiseNothing]

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 .

This should work now.

 

[uniquee]

is this some 4u shit?

 

[RealiseNothing]

is this some 4u shit?

My question could be done by a 2U student if you use the formula I posted.

 

[barbernator]

ill do it tomorrow, i cbf to do it tonight

 

[seanieg89]

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.

 

[ZombieApocalypse]

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


brb, running back to nonschool

 

[IamBread]

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


brb, running back to nonschool

But it's maffs!

 

[Carrotsticks]

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