Ruzzle/Riddle Thread

The lady was a rope-dancer at a theatre. She was practicing at an evening when the the girl put off the main electricity switch of the building (or caused a short-cut)..
 
Sorry no ProCop (are you a policeman?)

The lady dancer "dies" when the music stops..........hint
 
Thomo said:
Sorry no ProCop (are you a policeman?)
Click to expand...

No it's just a name..

As to the riddle it's tough one, the lady can be a dancer on a box (when the box is open the dancer dances, when it closes she dies (falls on the ground/cover/lid with her legs streatched...))
 
Yes, close enough
A little girl puts one of her treasures in her jewelery box, when the lid closes the little ballerina stops dancing and the music stops
 
OK, lets have some easy meat for a change,,,,

If three birds provide (together) 3 eggs in 3 days how many eggs will 9 birds produce in 9 days?
 
Easy, 27.
Okay...
781937Triangle1.jpg

What proportion of the area of the triangle is taken up by the circles, assuming there are an infinite number of them?
 
The proportion woudln't change very much from what it is now:

as an example I will describe only the space which the <b>b</b>iggest circle <b>B</b> at the botom-line of the triangle - creates: the space between the part of the curve of the big circle and the lines of the angle creating the "corner". You can put an another, <b>s</b>maller circe <b>S</b> so that it fits perfectly in that space and then imagine encreasing the size of S so that it equals the size of B (the rest of the triangle, B, and all will get bigger proportionally). If you than place a new circle nS in the corner proportion wouldn't change at all. So you can do this ad infinitum....
 
But there is an exact answer, which is relatively easy to find - doesn't involve infinite sums, just a bit of geometry.
 
I'd go for pi/4 parts of it. About 78.5%

oh.. wait.. no, thats not it.

right now i cant think of any ways to do this without knowing the relative sizes of the circles. I'll keep looking.
 
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Um ...area approaches (1/2 perimeter X apothem) - (pi X apothem^2)

if you consider the triangle to be a series of regular polygons with the perimeter approaching zero for the upper ones ........................I think
 
http://www.gomath.com/Questions/question.php?question=15501

I have found this on the web,


<i>

Question 15505
Submitted on 11/16/2001

I need help solving a calculs problem dealing with infinite seqences and series: There are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. IF the triangle has sides of lenght 1, find the total area occupied by the circles.

Solution 15505

If there are infinitely many circles in the triangle, then the infinitely many circles get infinitesimally small. The infinite, infinitesimally small circles completely fill the triangle and fill in any gaps. So the total area occupied by the circles equals the area of the triangle which is sqrt(3)/4

</i>

" <b>total area occupied by the circles equals the area of the triangle </b> " cannot be true, can it?
 
geodesic said:
Easy, 27.
Okay...
781937Triangle1.jpg

What proportion of the area of the triangle is taken up by the circles, assuming there are an infinite number of them?
Click to expand...

Okay - I have a solution - not sure if it's correct though...

The ratio of the area is (pi*r*b) / (4*r^2 + b^2) where r < b/2

where:
r = radius of the bottom circle
b = length of base of the triangle

I am not sure if you can relate r to b as they can both be arbitrary (and both together determine the height of the triangle).
You could use x as the ratio between b and r to simplify matters - letting x = 2r/b (i.e. x is between 0 and 1)...
This will then simplify the above ratio to pi*x / 2(x^2+1)
 
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ProCop said:
http://www.gomath.com/Questions/question.php?question=15501

I have found this on the web,


<i>

Question 15505
Submitted on 11/16/2001

I need help solving a calculs problem dealing with infinite seqences and series: There are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. IF the triangle has sides of lenght 1, find the total area occupied by the circles.

Solution 15505

If there are infinitely many circles in the triangle, then the infinitely many circles get infinitesimally small. The infinite, infinitesimally small circles completely fill the triangle and fill in any gaps. So the total area occupied by the circles equals the area of the triangle which is sqrt(3)/4

</i>

" <b>total area occupied by the circles equals the area of the triangle </b> " cannot be true, can it?
Click to expand...

This is true if you are continually filling in ALL the gaps in the triangle with the largest possible circle that will fill it (which is different to the posed question by geodesic).

So imagine the picture that geodesic submitted and then start filling in ALL the spaces with the largest possible circles, even if that circle will only touch other circles and not the sides of the triangle.

Then the answer given above holds true - the area of all the circles = the area of the triangle.
 
Using the triangles base angle, I got %= pi*tan(alpha) / (4tan(alpha)+4)
 
It has come to my attention that while it is hopefully obvious from the picture, I neglected to state that the triangle is isocelles.

Sarkus said:
I am not sure if you can relate r to b as they can both be arbitrary (and both together determine the height of the triangle).
Click to expand...
My solution gives the ratio of areas in terms of theta, the angle between the base and the side of the triangle. Your answer looks right though.

fo3:
Hmm, that's not the answer I've got, although it is very similar.
 
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Aha - then the answer is pi * SIN(theta) / 4 - where theta is the angle between the base and one side of the triangle.

Although this relies on me remembering correctly that tan(x)+1/tan(x) = 2/sin(2x) - so I may be wrong.
 
pi * SIN(theta) / 4
Click to expand...
Yep, that's what I got, though apparently not the same way as you. As for the identity, if you substitute 2sin(x)cos(x) for sin(2x) and multiply through by tan(x), you get tan^2(x)+1=sec^2(x) - so well remembered!
 
geodesic said:
Yep, that's what I got, though apparently not the same way as you. As for the identity, if you substitute 2sin(x)cos(x) for sin(2x) and multiply through by tan(x), you get tan^2(x)+1=sec^2(x) - so well remembered!
Click to expand...
Once you said that your answer was using the angle as the variable then it was relatively easy, whereas for some reason I started off trying to use the radius of the circle and the base of the triangle.
Ah well.
 
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