markr1963

The big circle has a diameter of 1m and fits exactly into a right angled corner. A second circle fits exactly into the space between the larger circle and the corner. What is the diameter of the smaller circle?

Reffro

20.7 cm but that's only a guess.

dnb

0.207m

I'll put my proof up later.

SWRTWannabe

Radius of circle = 50cm, so distance from centre of circle to 0,0 point on axis (using pythagoras theorom)is sqrt(50^2 + 50^2) i.e. 70.7cm

Therefore, the diameter of the small circle is 70.7 - 50cm i.e. 20.7cm

I think

dnb

Same answer - simple - solve the quadratic x^2 + x + (1/2)^2 = 0 for x

(where ^ is to the power)

[Edited by dnb - 4/28/2003 2:53:54 PM]

markr1963

If my maths is up to it then the distance from the edge of the larger circle to the corner(0,0) is 0.207m. This includes the diameter of the smaller circle and the gap between it and the corner.

555-Shaggy

Yep, you are right Mark.. everyone else is WRONG

I'll try to post the correct answer as soon as I've managed to work it out..

TonyG

Think you could do it with ratios and similar triangles, but it may take me a while to work out as I am at...er.....work

Dr Steve

How about 14.645cm?

Wrong. Now I get 17.1572cm.

[Edited by Dr Steve - 4/28/2003 4:51:17 PM]

ajm

removed as it was a lame effort!

[Edited by ajm - 4/28/2003 4:13:11 PM]

Apparition

Diameter is 25cm .

nkh

I make it 17.16cm

Working along the diagonal from centre of large circle to 0,0 there are three areas, the radius of the large circle (50cm), the diameter of the small circle which I let equal 2x, and the little bit in the corner which is sqrt(2x^2)-x. That lot equals 70.7 as mentioned before so re-arrange and solve the resulting equation.

Maybe I'm wrong though...

dnb

Yep - missed the little bit in the corner! Shouldn't try to do 2 things at once while at work

If x is the bit in the corner and n is the length I first calculated and d is the diameter, x = (n-d) then solve 2d^2 = 4x^2 + d^2 + 4xd to get the correct answer... I can't be bothered with maths when I'm home... (don't have mathcad to do numbers for me)

(Or like all good engineers, assume the bit in the corner is neglegable! )

[Edited by dnb - 4/28/2003 4:42:31 PM]

markr1963

Here's what I get





The things you do when you're bored

Hope it's right

[Edited by markr1963 - 4/28/2003 4:42:46 PM]

Tino

20.7cm is the long bit of a triangle whose other two sides are the same as diameter of circle so pythagoras thingy again.
Admits-I initially made the same mistake pointed out by mark1963.Well spotted mate.

I'll get me coat!!

[Edited by Tino - 4/28/2003 9:06:47 PM]

[Edited by Tino - 4/29/2003 4:20:01 PM]

SWRTWannabe

Ah yes - nice one

Gordo

Help me out here - I still think 0.146m is the answer.

I agree that the ratios will be equal (diameter of circle/hypotenuse of square = constant) but think the rearrangement of the formula where one of the lengths is replaced by a 1 is flawed. You have to use the absolute ratio to get the answer.

Coming at it from two different methods:

1) ratios

diameter of large circle (1) / hypotenuse of large square (1.412 or sq rt of 2) = 0.7071

I agree the length of AE is 0.2071 ((1.412-1)/2). The diameter of the small circle / AE (0.2071) = 0.7071

The small circle diameter is, therefore, 0.146m

2) Good old pythagoras

Length of hypotenuse of small 'square' = AE = 0.2071

Cos 45degress = adjacent (side of square, diameter of small circle)/ hypotenuse (AE, 0.2071)

Cos 45degrees x 0.2701 = 0.146m


Is my logic flawed?

[Edited by Gordo - 4/29/2003 3:02:35 PM]

dnb

Gordo, yes your logic is flawed in this case - there is only one length AF in the drawing, and you are allowing for the length AF twice.

Gordo

doh! got there in the end - 0.171 it is !

Pete Croney

The tangent point is 353.6mm down from the centre of the large circle. The height up to this point is 146.4mm. This point as a ratio of the diameter is 1000/853.6

So the diameter of the small circle is 1000/853.6*146.4 = 171.5mm

Clarebabes

Why don't you just use a ruler?

dnb

Where's the fun in using a ruler, when you can have no end of fun writing a computer simulation of the problem?

some people have no idea!