Pretty Area Puzzle

ray giraffe

Construct 3 circles such that each centre lies on the edge of the other 2 circles.

What is the total area enclosed by the circles?

This problem would be considered too long to put on a Leaving Cert paper.

However there is a fast easy solution using only maths taught in first year of secondary school! :pac:

(Hint: Easiest is to let each radius be 1. Or call it r if you prefer)

Area is sqrt(3) + 3*pi/2 if radii are 1, or (sqrt(3) + 3*pi/2)*r^2 in general

Solution in post number 10 kindly provided by cnote2!

(I made up this problem while teaching a Leaving Cert student. However I'm sure the problem and solution are old as the hills...)

sponsoredwalk

πr²




Are you allowed to put the three circles on top of each other? Each center lies on the other two circles, (they are even on top of each other), fulfilling the requirement.

I get the feeling I've missed the trickery in the question

ray giraffe

sponsoredwalk wrote: »

πr²




Are you allowed to put the three circles on top of each other? Each center lies on the other two circles, (they are even on top of each other), fulfilling the requirement.

I get the feeling I've missed the trickery in the question

I meant that the centre lies on the edge of the other two circles!

c-note

This is not the answer!!

(4*r^2)+((5*pi*r^2)/6)

= about 6.61799*r^2

??

i drew it on mspaint and divided the shape into different areas.
http://img707.imageshack.us/img707/8305/3circles.jpg

RolandIRL

i think this was on a LC paper in the 90's. remember doing it in exam papers. can't remember the solution though

equivariant

ray giraffe wrote: »

Construct 3 circles such that each centre lies on the edge of the other 2 circles.

What is the total area enclosed by the circles?

This problem would be considered too long to put on a Leaving Cert paper.

However there is a fast easy solution using only maths taught in first year of secondary school! :pac:

(I made up this problem while teaching a Leaving Cert student. However I'm sure the problem and solution are old as the hills...)

The given information does not determine the total area. It is easy to contruct an example where the total area is 1 and to construct another example where the total area is 2 (or indeed any other positive real number)

c-note

:rolleyes:

i'll
help

give the anser in terms of r, the radius of the circles!

equivariant

cnote2 wrote: »

:rolleyes:

i'll
help

give the anser in terms of r, the radius of the circles!

Then I think that your answer is not right. I think it is

r^2( (root of 3) + 3pi/2)

* If the problem is as stated, then it should include a phrase like "in terms of the radius of the circles", or it should just fix a specific radius. Otherwise a student would be logically correct to say "it can be any positive real number"

c-note

yes
that would be protocoligorically correct,
Luckily this is just a casual forum:p

you're right about my ans being wrong...
the real answer should be less that what i've calculated!
(which your ans is),

c-note

equivariant was on the money

i'd forgotten about little corner that were cut off!

i did it again in paint!
http://img686.imageshack.us/img686/3280/3circ2.jpg

i wonder did john bonham ever wonder about this!?

ray giraffe

equivariant wrote: »

Then I think that your answer is not right. I think it is

r^2( (root of 3) + 3pi/2)

* If the problem is as stated, then it should include a phrase like "in terms of the radius of the circles", or it should just fix a specific radius. Otherwise a student would be logically correct to say "it can be any positive real number"

While we're being precise, the question didn't say "non-degenerate circles", so the answer could be zero too. :eek:

Your formula would be right except that nowhere in the question did I mention a space with zero curvature Good try though :rolleyes:

In seriousness, thanks to equivariant for the answer, and to cnote for the nice drawing and solution

MathsManiac

Here's a related but more challenging puzzle. Given two circles, each of radius 1, how far apart must their centres be in order that half of the area of each circle is in the overlapping region?

(Ans. to 3 dec. places!)

P.S. pushes the boundaries of LC maths a bit!

c-note

i made a mistake, heres my wrong answer

.398

(0.3977303951)

???

i didnt use any lc maths, (that i know of)

equivariant

MathsManiac wrote: »

Here's a related but more challenging puzzle. Given two circles, each of radius 1, how far apart must their centres be in order that half of the area of each circle is in the overlapping region?

(Ans. to 3 dec. places!)

P.S. pushes the boundaries of LC maths a bit!

I won't give a decimal approximation, but the answer can be described as follows

Let x be the unique number between 0 and \pi/2 such that

x-1/2sin(2x) = pi/4

Then the required distance is 2cos(x).
If you want a decimal approximation, then use your favourite numerical to estimate the value of x (e.g. Newton's method should do it)

equivariant

Here's another one for you.

Construct three circles of radius 1, each of which is tangent to the other two. Now construct a fourth circle that is tangent to the first 3. What is the radius of this fourth circle? (There are two possible answers)

Please do not give any decimal approximations (I hate 'em):p

c-note

(ans to question on two overlapping circles)

i got the answer wrong on my first guess,

heres my final answer

.808

(.807945)

http://img31.imageshack.us/img31/6336/solnlarge.jpg

c-note

ans to eqiuvariants question


is it

(2+sqrt(3)) / (sqrt(3))
(2-sqrt(3)) / (sqrt(3))






2.1547
0.1547
(sorry i couldnt resist:D)

equivariant

OK that was too easy . One more. This is one that I really like, that I was reminded of recently.

An annulus is the region in between two concentric circles (i.e. circles with the same centre). Suppose that you have an annulus and that you construct a tangent to the inner circle. This tangent line will intersect the outer circle at two points. Suppose that the distance between these two points is 2010. What is the area of the annulus?

Fremen

Answer:

Pi (2010)^2 / 4.
Very nice problem. It's quite counterintuitive that the area should be completely determined by that condition. You would expect the answer to be in terms of the outer or inner radius.

Fremen

Thinking further about that problem, I have a feeling there's a much more beautiful proof lurking around than the junior-cert level one I used first.

Spoilers here, don't read if you still want to work on the problem.

If you could show that the condition we're assuming on the annulus (relevant distance = 2010) implies that area is conserved as we change the radius of the inner circle, then you could just let that radius tend to zero. The answer could easily be found by passing to the limit, giving the area of a circle with diameter 2010.

seandoiler

Fremen wrote: »

Thinking further about that problem, I have a feeling there's a much more beautiful proof lurking around than the junior-cert level one I used first.

Spoilers here, don't read if you still want to work on the problem.

If you could show that the condition we're assuming on the annulus (relevant distance = 2010) implies that area is conserved as we change the radius of the inner circle, then you could just let that radius tend to zero. The answer could easily be found by passing to the limit, giving the area of a circle with diameter 2010.

not sure i follow you here, surely the fact that the tangent to the inner circle is a chord to the outer circle is the reason why the area is thus determined by the condition on the chord length....sorry if i've missed something easy but i just don't follow your chain of thought

equivariant

seandoiler wrote: »

not sure i follow you here, surely the fact that the tangent to the inner circle is a chord to the outer circle is the reason why the area is thus determined by the condition on the chord length....sorry if i've missed something easy but i just don't follow your chain of thought

What Fremen is saying is that if we can somehow prove that the area is independent of the inner radius (without actually calculating the area) then we can calculate it by passing to the limiting case, where it is easy to calculate.

It is a good strategy that works for many different types of problems. However in this case, I can see no way (yet!) of proving that the area is fixed without essentially having to work out the relationship between the two radii first. Once you have done that, the solution is clear without the limiting argument.

Fremen

Yeah, something like

A = Pi(R^2 - r^2)
where R and r are the radii.

Now, find dA/dr and apply the assumption to show dA/dr = 0.

equivariant

Fremen wrote: »

Thinking further about that problem, I have a feeling there's a much more beautiful proof lurking around than the junior-cert level one I used first.

Spoilers here, don't read if you still want to work on the problem.

If you could show that the condition we're assuming on the annulus (relevant distance = 2010) implies that area is conserved as we change the radius of the inner circle, then you could just let that radius tend to zero. The answer could easily be found by passing to the limit, giving the area of a circle with diameter 2010.

I would tend to think that it is the "junior cert level" proofs that are the beautiful ones. Calculus is an ugly SOB in my view. I prefer to find solutions that are as elementary as possible rather than to smash a problem over the head with a sledgehammer such as differential calculus. Just my opinion however.

Fremen

Funny, my views are pretty much diametrically opposed. I go for ideas more than details. The JC-level proof is more or less all details. It gives the answer, but it doesn't tell you why the answer is true (at least not in an intuitive way that I can see).

Once you get past all the calculus in the other approach (assuming there is another approach), the intuition is clear, and you can more or less just say "ah, I see how that works".

seandoiler

is it not simply the case that Area=pi(R^2-(R^2-(d/2)^2))=pi(d/2)^2
where R is the radius of the outer circle and d the length of the tangent chord?

can't see how we can use a limiting argument

Fremen

Well, yes. If the big radius is R, the little one r and the chord d, then

R^2 = r^2 + (d^2)/4

which is equivalent to what you wrote.
Since d is a constant, this gives a relationship between R and r. If you set r=0, the set-up describes a circle with radius d/2. I was saying that the proof would be conceptually much nicer if you could bypass some of the geometry and justify setting r=0 from the outset.

equivariant

Fremen wrote: »

Funny, my views are pretty much diametrically opposed. I go for ideas more than details. The JC-level proof is more or less all details. It gives the answer, but it doesn't tell you why the answer is true (at least not in an intuitive way that I can see).

Once you get past all the calculus in the other approach (assuming there is another approach), the intuition is clear, and you can more or less just say "ah, I see how that works".

You should be an algebraic geometer/topologist (if you are not already)

If you like using powerful and general machinery rather than clever but ad hoc arguments to solve problems. I spent years doing algebraic topology before I realised that I don't like it all that much for that reason. Of course many great mathematicians like it for precisely that reason - its a personality thing I think.

ray giraffe

Fremen wrote: »

Thinking further about that problem, I have a feeling there's a much more beautiful proof lurking around than the junior-cert level one I used first.

Spoilers here, don't read if you still want to work on the problem.

If you could show that the condition we're assuming on the annulus (relevant distance = 2010) implies that area is conserved as we change the radius of the inner circle, then you could just let that radius tend to zero. The answer could easily be found by passing to the limit, giving the area of a circle with diameter 2010.

That was my first thought also, Fremen. Unless you're thinking about why the statement is true and trying to generalize, you don't learn much from the puzzle.

I can say something on the annulus problem:

Pythagoras's theorem says that the square on the hypotenuse R^2 is equal to the sum of the squares on the other two sides, r^2 + d^2, say.

Areas of circles and squares both grow as the square of their sizes, so therefore we can say "the circle on the hypotenuse is equal to the sum of the circles on the other two sides". (meaning will become clear)

For a right angled triangle, if we constuct a square on the hypotenuse and a square on another side, the difference of areas is just the area on the third side, which is constant if the third side is fixed.

The annulus problem effectively constucts a circle on the hypotenuse (large Radius R) and a circle on another side (small radius r), and asks for the difference in areas. Answer is "the circle on the third side d", which is constant since d is just half the chord and the chord is fixed!

So the problem is a version of Pythagoras's Theorem. :pac: