# Tough math puzzler
**URL:** https://boards.straightdope.com/t/tough-math-puzzler/28642
**Category:** Factual Questions
**Created:** [August 9, 2000, 12:14am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642 "2000-08-09T00:14:15Z")
**Posts on this page:** 18
**Page:** 1
### Author:  [@mrblue92](https://boards.straightdope.com/u/mrblue92)
#### Post date: [August 9, 2000, 12:14am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/1 "2000-08-09T00:14:15Z")
There was this puzzle my dad gave me once that’s pretty difficult… Goes like this:
You have a pipe with a 3 inch inner diameter. Jammed inside this pipe are two solid rods, one with a 2 inch diameter, the other with a 1 inch diameter. What is the diameter of the largest possible rod that can fit inside the pipe along with the two rods?
AFAIK, the question is purely a math puzzle, not a logic puzzle, where you can remove the rods or cut the pipe…
Some calculus may be required. No graphing or iterative computer programs are allowed. You must show your work (at minimum, throughly explain the method of calculation). This is worth 50 percent of your term grade.
Ready? Begin!
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [August 9, 2000, 1:05am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/2 "2000-08-09T01:05:06Z")
This is going to be tough without a diagram, but I can reduce it to a system of equations. Call the center of the pipe A, the center of the large rod B, the center of the small rod C, and the center of the mystery rod D. The angle formed by DAB is theta, and the angle formed by DAC is phi. The radius of the unknown rod is x. We now have three unknowns, theta, phi, and x, and three equations, as follows:
By the Law of Cosines, .5[sup]2[/sup] + (1.5 - x)[sup]2[/sup] = (1 + x)[sup]2[/sup] + 2\*.5\*(1.5-x)_cos(theta) . Similarly,
1[sup]2[/sup] + (1.5 - x)[sup]2[/sup] = (.5 + x)[sup]2[/sup] + 2_1\*(1.5-x)\*cos(phi) . Finally, because theta and phi are complements, theta + phi = 180[sup]o[/sup] . Somebody take it from here.
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### Author:  [@scr4](https://boards.straightdope.com/u/scr4)
#### Post date: [August 9, 2000, 1:05am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/3 "2000-08-09T01:05:33Z")
It depends on the thickness of the rod. Is it zero?
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### Author:  [@scr4](https://boards.straightdope.com/u/scr4)
#### Post date: [August 9, 2000, 1:07am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/4 "2000-08-09T01:07:03Z")
Um, never mind, I misunderstood the question.
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### Author:  [@KJ](https://boards.straightdope.com/u/KJ)
#### Post date: [August 9, 2000, 1:10am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/5 "2000-08-09T01:10:18Z")
I haven’t been through Geometry yet, but I’m sure I can figure this one out.
BTW, it’s interesting to note that you can place 2 smaller rods into the pipe, and then more smaller rods in between the corners of that one, again and again and again, forever, without ever reaching a solid state (i.e. a fractal)
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [August 9, 2000, 2:53am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/6 "2000-08-09T02:53:09Z")
> [@](#):
>
> _Originally posted by Chronos \*
> By the Law of Cosines, .5[sup]2[/sup] + (1.5 - x)[sup]2[/sup] = (1 + x)[sup]2[/sup] + 2.5(1.5-x)cos(theta) . Similarly,
> 1[sup]2[/sup] + (1.5 - x)[sup]2[/sup] = (.5 + x)[sup]2[/sup] + 21_(1.5-x)\*cos(phi) . Finally, because theta and phi are complements, theta + phi = 180[sup]o[/sup] . Somebody take it from here. \*\*
Cool. Same ones I got. The equations become linear, not quadratic. I found x to be 3/7, or the diameter to be 6/7 inch.
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### Author:  [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)
#### Post date: [August 9, 2000, 2:55am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/7 "2000-08-09T02:55:25Z")
I’m too slow in my old age.
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [August 9, 2000, 5:42am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/8 "2000-08-09T05:42:30Z")
Two circles that touch in one point are said to kiss. The following poem describes kissing circles, spheres and hyper-spheres:
```
The Kiss Precise
by Frederick Soddy
For pairs of lips to kiss maybe
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance form the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.
To spy out spherical affairs
An oscular surveyor
Might find the task laborious,
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.
In *Nature*, June 20, 1936
Later another verse was written by Thorold Gosset to describe the even more general case in N dimensions for N+2
hyperspheres of the Nth dimension.
The Kiss Precise (Generalized) by Thorold Gosset
And let us not confine our cares
To simple circles, planes and spheres,
But rise to hyper flats and bends
Where kissing multiple appears,
In n-ic space the kissing pairs
Are hyperspheres, and Truth declares -
As n + 2 such osculate
Each with an n + 1 fold mate
The square of the sum of all the bends
Is n times the sum of their squares.
In *Nature*, January 9, 1937.
```
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [August 9, 2000, 6:33am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/9 "2000-08-09T06:33:27Z")
Remember, that the outside circle has negative curvature.
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### Author:  [@mrblue92](https://boards.straightdope.com/u/mrblue92)
#### Post date: [August 9, 2000, 6:19pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/10 "2000-08-09T18:19:58Z")
Excellent job guys–that was quick. The only thing you forgot to mention is that you have to use the identity:
cos 180-A = -cos A (in degrees, of course…)
I haven’t looked at that particular problem in a long while, and I think I never got it because I either didn’t spend enough time or wasn’t looking at the second triangle.
The other problem he gave me was one involving cutting a small section of a large inner diameter with an end mill by angling the mill head so the cut was an elliptical section. Frankly that one is probably too difficult to explain without a diagram, and I solved it anyway, so I’ll digress…
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### Author:  [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)
#### Post date: [August 9, 2000, 10:29pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/11 "2000-08-09T22:29:55Z")
> [@](#):
>
> \*Originally posted by mrblue92 \*
> \*\*Excellent job guys–that was quick. The only thing you forgot to mention is that you have to use the identity:
>
> cos 180-A = -cos A (in degrees, of course…)
> \*\*
You don’t need that. I’ll explain later if anyone cares.
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### Author:  [@jcgmoi](https://boards.straightdope.com/u/jcgmoi)
#### Post date: [August 9, 2000, 11:23pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/12 "2000-08-09T23:23:39Z")
> [@](#):
>
> \*Originally posted by Lance Turbo \*
> \*\*
>
> > [@](#):
> >
> > \*Originally posted by mrblue92 \*
> > \*\*Excellent job guys–that was quick. The only thing you forgot to mention is that you have to use the identity:
> >
> > cos 180-A = -cos A (in degrees, of course…)
> > \*\*
>
> You don’t need that. I’ll explain later if anyone cares. \*\*
**Lance** , I’m curious about your approach. FWIW, here’s how I reached the general solution:
If 2 pipes of radius **a** and **b** plug a pipe of radius **c = a+b** , then the largest pipe that can be inserted has radius **x = abc/[c[sup]2[/sup] - ab]**.
Using **Chronos’** nomenclature you have triangle BCD and line DA intersecting BC, with AB = a, AC = b, AD = a+b-x ,BD = b+x, CD = a+x. Triangles ABD and ACD share the same height so their areas are in the ratio of their bases, namely, a:b. This fact and Heron’s formula for the area of a triangle in terms of its sides, plus a little algebra, yield the above result.
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### Author:  [@Manlob](https://boards.straightdope.com/u/Manlob)
#### Post date: [August 10, 2000, 12:50am UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/13 "2000-08-10T00:50:16Z")
I think the following is easiest (no need to remember law of cosines, Heron’s formula, or long poems):
Call radius of the unknown circle “a”, and the coordinates of its center x and y (center of big circle is the origin of coordinates). There are 3 right triangles that all have a leg of length y (hypoteneus of the three, using Chronos’s notation are AD, BD, and CD). So by Pythagorean Theorem:
y[sup]2[/sup]=(1.5-a)[sup]2[/sup]-x[sup]2[/sup]=(1+a)[sup]2[/sup]-(x+0.5)[sup]2[/sup]=(0.5+a)[sup]2[/sup]-(1-x)[sup]2[/sup]
Expanding simplifies to:
2.25-3a=0.75+2a-x=-0.75+a+2x
x is easily eliminated from those equations to give a=3/7
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### Author:  [@mrblue92](https://boards.straightdope.com/u/mrblue92)
#### Post date: [August 10, 2000, 12:54pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/14 "2000-08-10T12:54:16Z")
Well, obviously you do need that cosine identity if you use law of cosines…
**jcgmoi** : The _diameter_ is given by your formula, not the radius. -5 points…
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### Author:  [@jcgmoi](https://boards.straightdope.com/u/jcgmoi)
#### Post date: [August 10, 2000, 1:51pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/15 "2000-08-10T13:51:58Z")
> [@](#):
>
> \*Originally posted by mrblue92 \*
>
> **jcgmoi** : The _diameter_ is given by your formula, not the radius. -5 points…
Oh? The formula gives **x** as a diameter if you substitute **a, b, c** as diameters. If **a, b, c** are radii, as I specified, **x** is radius too.
Do I get my 5 points back?
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### Author:  [@mrblue92](https://boards.straightdope.com/u/mrblue92)
#### Post date: [August 10, 2000, 2:14pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/16 "2000-08-10T14:14:04Z")
Actually, I’ll give you 10 because I’m such a complete idiot for posting that too quickly. I just steal these five from Dr. Matrix, because we all know that poetry is NEVER acceptable as a mathematical proof.
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### Author:  [@mrblue92](https://boards.straightdope.com/u/mrblue92)
#### Post date: [August 10, 2000, 2:15pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/17 "2000-08-10T14:15:41Z")
Oh, and it was WAY too early in the morning. Yea, that’s the ticket…
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### Author:  [@jcgmoi](https://boards.straightdope.com/u/jcgmoi)
#### Post date: [August 10, 2000, 6:21pm UTC](https://boards.straightdope.com/t/tough-math-puzzler/28642/18 "2000-08-10T18:21:00Z")
**mrblue92** :
Thanks for the reinstatement. And I’ll take the 5 points from **DrMatrix** too; he has plenty to spare.