Enigma 1299 - Pipe dreams

Enigma 1299 - Pipe dreams New Scientist magazine, 24 July 2004. by Ian Bell.

A pipe manufacturer ships pipes of 3 different radii in the same, square-section, box. The pipes just touch each other and the box as shown (see description below). If the middle -sized pipes have radii 4 centimetres what are the radii of the smallest pipes?

Description of the diagram: A square with a circle inscribed within, just touching the inside of each of the four sizes of the square. Four middle-sized circles, one in each of the four voids between the outside of the large circle and the inside of the square. Eight small circles, one in each of the eight voids between the outside of the large circle, the outside of the middle-sized circle, and the inside of the square.

Ciao,

Good old fashioned geometry! What use Excel now?

Let's have a go anyway...

Say the square is centred on the origin and aligned with the axes, so its vertices are (+- s, +- s). Then the biggest pipe A is centred on O and has radius s.

Let the top-right middle-sized pipe B have centre P. Since we know radius( = 4, and B is tight into the corner, we have P = (s-4, s-4).

Let K be the point of contact between pipes A and B, X be the point of contact between B and the right wall. Clearly X = (s, s-4). It is also obvious that angle KPX = 135 degrees. This along with the facts that PK = PX = 4 and sin 45deg = cos 45def = sqrt(2)/2 gives us K = (s-4-2sqrt(2), s-4-2sqrt(2))

But since A is a circle it has constant radius, so OK = s, so by Pythagoras 2(s-4-2sqrt(2))^2 = s^2. Let d = 4+2sqrt(2), then

2(s-d)^2 = s^2 s^2 -4ds + 2d^2 = 0 (s-2d)^2 = 2d^2 s-2d = sqrt(2)d s = (2+sqrt(2))d = (2+sqrt(2))(4+sqrt(2)) = 10 + 6sqrt(2) (~= 18.5)

This will surely prove useful soon. Also, K = (6+4sqrt(2), 6+4sqrt(2)).

Now let the smallest pipe C be centred on Q and have radius r. We have three constraints on Q:

1: OQ = s+r 2: PQ = 4+r 3: x coordinate of Q = s-r

Let Q = (x,y). Then these constraints become

1': x^2 + y^2 = (s+r)^2 2': (x-s+4)^2 + (y-s+4)^2 = (4+r)^2 3': x = s-r

Substitute 3' into 1' giving

(s-r)^2 + y^2 = (s+r)^2 s^2 - 2rs + r^2 + y^2 = s^2 + 2rs + r^2 4': y^2 = 4rs

Substitute 3' into 2' giving

(s-r-s+4)^2 + (y-s+4)^2 = (4+r)^2 (-r+4)^2 + (y-s+4)^2 = (4+r)^2 (y-s+4)^2 = (4+r)^2 - (4-r)^2 = 16r = 4y^2 / s (by 4' y^2 - 2ys + 8y + s^2 - 8s + 16 = 4y^2 / s (s-4)y^2 - 2ys^2 + 8ys + s^3 - 8s^2 + 16s = 0

Write q for sqrt(2) and expand out all these s expressions (s = 10+6q, s^2 = 172+120q, s^3 = 3160+2232q)

y^2(s-4) + y(-2s^2 + 8s) + (s^3 - 8s^2 + 16s) = 0 y^2(6+6q) + y(-344-240q+80+48q) + (3160+2232q-1376-960q+160+96q) = 0 y^2(6+6q) + y(-264-192q) + (1944+1368q) = 0 y^2(1+q) + y(-44-32q) + (324+228q) = 0

Discriminant D ('b^2-4ac' = 3984+2816q - 4.(1+q).(324+228q) = 3984 + 2816q - 4(780+552q) = 864 + 608q

D = 16(54+38q)

Solutions for y are therefore

(44+32q +- 4sqrt(54+38q)) / 2(1+q)

which are both positive but taking the + gives ~ 27 which is > s and hence too large. So we have

y = (44+32q-4sqrt(54+38q))/2(1+q)

y^2 = (1936 + 2816q - 352sqrt(54+38q) + 2048 - 256q.sqrt(54+38q) + 16(54+38q)) / 4(1+q)^2

By 4' we thus have

r = (4848 + 3424q - (352+256q)sqrt(54+38q)) / 4s.4(1+q)^2 = (4848 + 3424q - (352+256q)sqrt(54+38q)) / 16(10+6q)(1+q)^2 = 16 ( 303 + 214q - (22+16q)sqrt(54+38q) ) / 16 ( 54 + 38q ) = ( 303 + 214q - (22+16q)sqrt(54+38q)) / (54 + 38q)

~= 1.321

This satisfies 1' 2' and 3', and also looks in the right ballpark for a smallest pipe. Given the spectacularly ugly closed form I can't believe there's an 'elegant' solution waiting in the wings...

SPOILER

Let the radius of the large pipe be R, the radius of the medium pipes be r, and the radius of the small pipes be x.

For the line joining the centres of the large and medium pipes we have sqrt(2)*(R+r)/2+r = R, and so R = (1+sqrt(2))^2*r.

The distance between the perpendiculars from the centres of the medium and small pipes to the bounding box side they both touch is found from the Pythagoras theorem to be d = sqrt((r+x)^2-(r-x)^2) = 2 * sqrt(r*x). Therefore the distance between the perpendiculars from the centres of the large and small pipes to the bounding box side they both touch is R-r-d.

For the line joining the centres of the large and small pipes we have (R+x)^2 = (R-x)^2+(R-r-d)^2. Substituting for d, rearranging terms, and factoring out R-r yields 4*x+4*sqrt(r*x)-(R-r) = 0. This is a quadratic equation in sqrt(x), and the only positive solution is sqrt(x) = (sqrt(R)-sqrt(r))/2 = sqrt(2)*r/2, and so x = r/2.

The radius of the small pipes is thus 2 centimetres.

Please reply to drgmayer at hotmail dot com

Umm, nope. Your value for s is too low.

[puzzle as spoiler space]

Let (0,0) be the lower left corner of the square.

The center of the lower left middle-sized circle is at (4,4).

The lower left middle-sized circle touches the large circle at [4 + 2*sqrt(2), 4 + 2*sqrt(2)].

The center of the large circle is at (x,x).

Consider the diagonal of the square from lower left to upper right: 4*sqrt(2) + 4 + x = x*sqrt(2) 4*sqrt(2) + 4 = x * [sqrt(2) - 1] x = [4*sqrt(2) + 4] / [sqrt(2) - 1]

We can now apply Descartes' theorem, treating the straight line as a degenerate circle with curvature 0:

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

where k4 is the curvature (1 / radius) of the small circle, k1 and k2 the curvatures of the medium and large circles.

1/4 + [sqrt(2)-1]/[4*sqrt(2)+4] +- 2*sqrt([sqrt(2)-1]/[16*sqrt(2)+16]) ~= 0.5 or 0.0857864378

The radius of the small circles is 2 cm. The other value gives 11.6568542259 cm; this is the radius of a circle tangent to the large circle, one of the medium circles, and one side of the square, but intersected by a perpendicular side of the square.

Oops, d seems to have changed definition. Tsk tsk. No wonder the rest is wrong. Corrected version follows, using q=sqrt(2) as a notational convenience...

s = (2+q)d = (2+q)(4+2q) = 12 + 8q. Also, P = (8+8q, 8+8q)

(s=12+8q, s^2=272+192q, s^3=6336+4480q

y^2(8+8q) + y(-448-320q) + (4352+3072q) = 0 y^2(1+q) + y(-56-40q) + (544+384q) = 0

Discriminant D ('b^2-4ac' = 1088+768q = 64(17+12q)

(56+40q +- 8sqrt(17+12q)) / 2(1+q)

By an amazing coincidence, sqrt(17+12q) = 3+2q so we have

y = (56+40q +- (24+16q))/2(1+q)

so

y = 40+28q / 1+q or y = 16+12q / 1+q = 16+12q or y = 8+4q

Since the problem dictates that y<s we must have y = 8+4q

so y^2 = 96+64q = 4rs = r(48+32q)

Giving

r=2

which simple result means there was a much easier way to do this. Which Ilan has already posted

Lesson: Enigmas usually have neat answers. If you get a stupid answer, it's late, go home

Nicely done Ilan, and correct of course (as intuition must have told you when it dropped out to a round number). Is this a surprise, that the small inscribed circle is half the radius of the medium-sized one?

I have to say that I find these geometry Enigmas more satisfying than those that can be solve with Excel or a computer program. Some more (harder I think) geometry to come with Enigma 1302 in a couple of weeks.

Ciao,