Donate to Remove ads

Got a credit card? use our Credit Card & Finance Calculators

Thanks to johnstevens77,Bhoddhisatva,scotia,Anonymous,Cornytiv34, for Donating to support the site

Circles

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
Has thanked: 231 times
Been thanked: 160 times

Circles

#395191

Postby cinelli » March 13th, 2021, 12:26 pm

.            .......
.... ....
.. ..
. . .....
. . .. ..
. . . .
. . . .
. . .
. A . .
. . B .
. .. .
. .. .
. . . .
. . . . .
.. .. . C. . .
..........................................................

My figure attempts to show three circles, centres A, B and C, each of which touches the other two. In addition they all touch a straight line. If the radii of the two large circles are 36 and 25 inches, what is the radius of the small circle?

Cinelli

9873210
Lemon Slice
Posts: 984
Joined: December 9th, 2016, 6:44 am
Has thanked: 226 times
Been thanked: 296 times

Re: Circles

#395288

Postby 9873210 » March 13th, 2021, 6:25 pm

let r be the radius of the small circle.

Let x be the horizontal distance between the center of A and B.
Let y be the horizontal distance between the center of A and C.
Let z be the horizontal distance between the center of A and C.

Then
x = y + z

By Pythagoras

(36+25)^2 = x^2 + (36-25)^2
x^2 = 61^2 - 11^2
x^2 = 3600
x = sqrt(3600)

(36+r)^2 = y^2 + (36-r)^2
y^2 = (36+r)^2 - (36-r)^2
y^2 = 4.36r
y = 12 sqrt(r)

(25+r)^2 = z^2 + (25-r)^2
z^2 = 4.25r
z = 10 sqrt(r)

so
22 sqrt(r) = sqrt(3600)
r = 3600/(22^2)
r = 900/121 = 7.4380...

Gengulphus
Lemon Quarter
Posts: 4255
Joined: November 4th, 2016, 1:17 am
Been thanked: 2628 times

Re: Circles

#395302

Postby Gengulphus » March 13th, 2021, 8:03 pm

cinelli wrote:
.            .......
.... ....
.. ..
. . .....
. . .. ..
. . . .
. . . .
. . .
. A . .
. . B .
. .. .
. .. .
. . . .
. . . . .
.. .. . C. . .
..........................................................

My figure attempts to show three circles, centres A, B and C, each of which touches the other two. In addition they all touch a straight line. If the radii of the two large circles are 36 and 25 inches, what is the radius of the small circle?

Spoiler 1...

If we assume that the straight line is the X axis, and the centre of the radius 36 circle is on the Y axis, then A's co-ordinates are (0, 36) and B's co-ordinates are (X,25) for some positive X (assuming the usual orientations of the axes and the radius 25 circle to the right of the radius 36 circle, as drawn). The distance between A and B must be 36+25 = 61 for the two circles to touch, so sqrt((X-0)^2 + (36-25)^2) = 61, from which it follows that X^2 + 11^2 = 61^2, or X = sqrt(61^2-11^2) = sqrt(3721-121) = sqrt(3600) = 60 since X is positive.

So A is at (0,36) and B is at (60,25). Now suppose that C is at (x,y) is the centre of the small circle, which has radius r. Since the small circle touches the straight line, y=r, and since it touches the two large circles:

sqrt((x-0)^2 + (r-36)^2) = 36+r
sqrt((x-60)^2 + (r-25)^2) = 25+r

From those, it follows that:

x^2 + r^2 - 72r + 36^2 = 36^2 + 72r + r^2
x^2 - 120x + 60^2 + r^2 - 50r + 25^2 = 25^2 + 50r + r^2

Collecting and cancelling similar terms gives:

x^2 = 144r
x^2 - 120x + 60^2 = 100r

Substituting the first into the second gives 144r - 120x + 60^2 = 100r, from which it follows that 120x = 44r + 3600, or x = (11r+900)/30. Substituting that in x^2 = 144r gives (11r+900)^2/900 = 144r, or 121r^2 + 19800r + 810000 = 129600r, i.e. 121r^2 - 109800r + 810000 = 0. That's a quadratic in r, with solutions r = (109800 +/- sqrt(109800^2-4*121*810000)) / 2*121 = (109800 +/- 108000) /242 = 1800/242 = 900/121 or 217800/242 = 900.

The corresponding values of x are x = (11r+900)/30 = 360/11 and 360 respectively, and as a check:

(360/11,900/121) is at distance sqrt((360/11-0)^2 + (900/121-36)^2) = sqrt((360/11)^2 + (-3456/121)^2) = sqrt(129600/121 + 11943936/14641) = sqrt(27625536/14641) = 5256/121 = 36 + 900/121 from (0,36)
(360/11,900/121) is at distance sqrt((360/11-60)^2 + (900/121-25)^2) = sqrt((-300/11)^2 + (-2125/121)^2) = sqrt(90000/121 + 4515625/14641) = sqrt(15405625/14641) = 3925/121 = 25 + 900/121 from (60,25)

(360,900) is at distance sqrt((360-0)^2 + (900-36)^2) = sqrt(129600 + 746496) = sqrt(876096) = 936 = 36 + 900 from (0,36)
(360,900) is at distance sqrt((360-60)^2 + (900-25)^2) = sqrt(90000 + 765625) = sqrt(855625) = 925 = 25 + 900 from (60,25)

Since we're told the circles with radii 36 and 25 are "large" and the third circle is "small", the radius of the third circle is 900/121 = 7 53/121.

Spoiler 2...

Make use of Descartes' theorem (see its Wikipedia entry), which says that if four circles (or three circles and a straight line) mutually touch each other, then (k1 + k2 + k3 + k4)^2 = 2 * (k1^2 + k2^2 + k3^2 + k4^2), in which:

* k1, k2, k3 and k4 are the curvatures of the four circles, the curvature of a circle being defined to be the reciprocal of its radius if contacted from outside and minus the reciprocal of its radius if contacted from inside;

* a straight line is considered to be a circle of infinite radius / zero curvature.

And if we're given three of the curvatures (say k1, k2 and k3), we can solve the resulting quadratic equation in k4 to get:

k4 = k1 + k2 + k3 +/- 2sqrt(k1*k2 + k2*k3 + k3*k1)

In this case, k1 = 1/36, k2 = 1/25 and k3 = 0, so:

k4 = 1/36 + 1/25 +/- 2sqrt(1/36 * 1/25)

= 61/900 +/- 2*1/30

= 61/900 +/- 60/900

= 121/900 or 1/900

from which we deduce that the fourth radius is 900/121 or 900, and 900/121 is the desired answer as in spoiler 1.

Gengulphus

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
Has thanked: 231 times
Been thanked: 160 times

Re: Circles

#395711

Postby cinelli » March 15th, 2021, 11:45 am

Well solved, both, including two solutions for the price of one from Gengulphus. I admit I didn't know of Descartes' theorem so am very grateful for this extra knowledge. I had the solution just in terms of the radii of three circles. If the radii are a, b and c then the formula for c boils down to the pleasing

1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)

This can be proved with several applications of Pythagoras.

Cinelli

UncleEbenezer
The full Lemon
Posts: 10690
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1459 times
Been thanked: 2965 times

Re: Circles

#395721

Postby UncleEbenezer » March 15th, 2021, 12:12 pm

cinelli wrote:Well solved, both, including two solutions for the price of one from Gengulphus. I admit I didn't know of Descartes' theorem so am very grateful for this extra knowledge. I had the solution just in terms of the radii of three circles. If the radii are a, b and c then the formula for c boils down to the pleasing

1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)

This can be proved with several applications of Pythagoras.

Cinelli

Funny you should mention that.

I misread the dimensions as the two bigger radii being 6 and 5 - guess I was already assuming Pythagoras and read your squares as their roots. Only realised my misreading when I looked at the spoilers and got confused about why they were using the squares!

Solving for 6 and 5 factors out to that formula, which is kind-of obscured by 36,25 when 3600 turns out to be a perfect square.


Return to “Games, Puzzles and Riddles”

Who is online

Users browsing this forum: No registered users and 8 guests