Can You Solve “Unsolvable” Mathematical Problems? (Nov, 1931)

Can You Solve “Unsolvable” Mathematical Problems?

PROBLEMS of mathematics, long considered unsolvable, may soon give up their secrets to inquiring mathematicians. The problem of trisecting an angle by plane geometry, for example, is one which has puzzled mathematicians for 2500 years, and which has come to be regarded as insolvable as the squaring of a circle. Rev. Joseph J. Callahan, president of Duquesne University, recently announced that he had solved the trisection problem by the Euclidean method of geometry.

The diagram herewith was not made by Father Callahan, who has not made public his solution, but it shows a well-known way of trisecting an angle. Angle DOA is trisected and angles BOC and BCO each is one-third of the original angle DOA. Angles OAB and OBA each is twice angles BOC and BCO. The unsolvable problem is the angle’s trisection by ruler and compass, which is the method by which mathematicians assume Father Callahan has solved the puzzle.

“Squaring the circle”, or obtaining the exact area of a circle by mathematical formulae, is a familiar geometric problem which has defied solution for years, and which has probably caused as many headaches among mathematicians as the perpetual motion problem has been responsible for among inventors. With the development of mechanical methods of solving such problems as angle trisection and circle squaring, however, the geometric solution of the problems is mostly of academic interest. Surveyors and engineers have long been accustomed to trisecting angles by rule of thumb methods which prove entirely adequate for their purposes.

  1. Buddy says: December 5, 20087:08 am

    It is possible to bisect an angle, and by recursive bisections, 4-sect, 8-sect, 16-sect, etc. an angle.

    Using the limit:

    1/3 = 1/2 – 1/4 + 1/8 – 1/16 + 1/32 – 1/64 + …

    Multiplying by x:

    x/3 = x/2 – x/4 + x/8 – x/16 + x/32 – x/64 + …

    It is evident that x/2, x/4, x/8, x/16, x/32, x/64, etc. are equivalent to recursive bisections of x.

    Hence is possible to trisect an angle x to any finite precision using recursive bisections in a finite number of steps.

    It doesn’t count as a proof because in theory, an infinite number of steps would be required to trisect exactly.

    The limit can also be expressed 1/3 = 1/4 + 1/16 + 1/64 + … + 1/(n**4).

    It’s the same for all n not equal to -1, 0, 1, even non-integer.

    1/2 = 1/3 + 1/9 + 1/27 + …
    1/3 = 1/4 + 1/16 + 1/64 + … = 1/2 – 1/4 + 1/8 – 1/16 + 1/32 – 1/64 + …
    1/4 = 1/5 + 1/25 + 1/125 + … = 1/3 – 1/9 + 1/27 – 1/71 + …

    Any fractions which can be expressed in sums of powers of two are amenable to this technique:

    1/3 = 1/4 + 1/16 + 1/64 + …
    1/5 = 1/4 – 1/16 + 1/64 – …
    1/7 = 1/8 + 1/64 + 1/512 + …
    1/9 = 1/8 – 1/64 + 1/512 – …
    1/15 = 1/16 + 1/256 + 1/4096 + …
    1/17 = 1/16 – 1/256 + 1/4096 – …

    Meaning it’s possible to trisect, or 5-, 7-, 9-, 15-, 17-sect, etc. any angle to any finite precision in finitely many steps.

  2. fluffy says: December 5, 20084:55 pm

    I believe that trisecting an angle via plane geometry has been proven impossible by contradiction (i.e. presupposing that it’s possible, showing that the math falls apart). However, trisecting in plane geometry is possible via folding, which is a rather neat modern branch of geometry which has grown out of origami.

    The article also mis-states what squaring the circle is about. It’s not about finding the precise area of a circle mathematically, but geometrically constructing a square with the same area as any given circle. Given that the area of a circle is πr2 and the area of a square is r2, that means that you’d have to somehow construct a square whose side’s length is precisely √pi of the circle’s radius — not a feasible construction! (And, again, probably disprovable by contradiction.)

  3. fluffy says: December 5, 20084:56 pm

    Obviously that should be πr^2. This comment system eats <sup> tags.

  4. Toronto says: December 6, 200812:22 am

    Fluffy – I’m impressed you got a pi-like character to show up, never mind the superscript.

    I have to roll the platen back half a click manually via the right hand roller handle to do that on this typewriter…

  5. Matt says: December 6, 20083:53 pm

    For superscript 2, hold down the alt key, type 0178 on your numeric keypad (num lock must be on), and release the alt key. Presto ² shows up. Works for all the standard ASCII characters on PCs. You can look up the codes with the Character Map in Windows. I don’t know if Macs have a similar shortcut.

  6. fluffy says: December 6, 20084:37 pm

    Yeah, I’m on a Mac, which has plenty of mnemonic shortcuts for actual symbols, but a superscript is not semantically a symbol. I could just use the Unicode character entry palette, but writing proper HTML is usually a lot faster.

    π is just &pi;. It’s a standard HTML entity.

  7. George Trudeau says: December 8, 20083:27 pm

    I’ve discovered a remarkable solution to this problem, unfortunately this text box is too small to hold it. P. de Fermat

  8. Maaz says: October 7, 20091:40 pm

    There are no unsolvable problems…Maybe the problem is non existant in nature.

  9. Firebrand38 says: October 7, 20091:42 pm

    Maaz: Don’t just make an unsubstantiated statement. Solve the problems.

  10. Zyzzyva says: December 15, 200912:23 am

    It’s insoluble. It had been proved insoluble by then – IIRC, more than fifty years before. Callahan was (provably) a crank.

    That said, the article does get one thing right: squaring the circle has probably caused exactly “as many headaches among mathematicians as the perpetual motion problem has among inventors”, eg, none, if you don’t count the headache of fending off the cranks.

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