Inventor's Paradox

The more ambitious plan may have more chances of success
G. Polya, How To Solve It
Princeton University Press, 1973

Polya continues: ... provided it is not based on a mere pretension but on some vision of the things beyond those immediately present.

Mathematical discovery is seldom a single step process. Often it's indeed the case where answering a more general question is easier than finding an answer to a specific one. Later, on this page, we'll see a collection of examples that I'll be updating from time to time. I would appreciate your mailing me additional examples.

Oftentimes, however, solving a specific problem first may suggest, in a hindsight, a formulation of and a solution to, a more general one. I'll call such a situation

Inventor's Paradise

Both circumstances underscore investigative part of doing Mathematics or actually of any kind of problem solving:

  • before solving a problem, look around for other formulations, dig your memory for similar known facts
  • after a problem has been solved, make this experience an edifying one. Try to stash away as much information as possible. If, in the process, you run into a problem you can't solve right away, go for it - you won't regret it.

I'll call the latter situation

Inventor's Paradigm

Always look for new problems; especially after successfully solving one - you may get much more than you expected to start with.

Pairs of statements in which one is a clear generalization of another whereas in fact the two are equivalent.
  1. The Intermediate Value Theorem - The Location Principle (Bolzano Theorem)
  2. Rolle's Theorem - The Mean Value Theorem
  3. Existence of a tangent parallel to a chord - existence of a tangent parallel to the x-axis.
  4. Binomial theorem for (1 + x)n and (x + y)n
  5. The Maclaurin and Taylor series.
  6. Two properties of Greatest Common Divisor
  7. Pythagoras' Theorem and the Cosine Rule
  8. Pythagoras' Theorem and its particular case of an isosceles right triangle
  9. Pythagoras' Theorem and Larry Hoehn's generalization
  10. Combining pieces of 2 and N squares into a single square
  11. Measurement of inscribed and (more generally) secant angles
  12. Probability of the union of disjoint events and any pair of events
  13. Butterfly theorem in orthodiagonal and arbitrary quadrilaterals
  14. Pascal's theorem in ellipse and in circle
  15. A triangle is embeddable into a rectangle twice its area and so is any convex polygon
  16. Brahmagupta's Theorem and Heron's Theorem

There are cases where a more general statement highlights the features of the original problem not otherwise obvious and by doing so spells a solution that works in both cases.

Pairs of statements in which one is a clear generalization of another and the more general statement is not more difficult to prove.
  1. Bottema's Theorem and McWorter's generalization
  2. Butterfly theorem
  3. Carnot's theorem and Wallace's theorem.
  4. Concyclic Circumcenters: A Dynamic View.
  5. Concyclic Circumcenters: A Sequel.
  6. Find a plane through a point outside of an octahedron such that the plane bisects the volume of the octahedron - same statement but replace octahedron with a solid with a center of symmetry
  7. A Fine Feature of the Stern-Brocot tree
  8. Four Construction Problems
  9. Lucas' Theorem and its variant
  10. Matrix Groups
  11. Napoleon's Theorem and one theorem about similar triangles
  12. Four Pegs That Form a Square
  13. On the Difference of Areas
  14. One Trigonometric Formula and Its Consequences
  15. Pythagorean Theorem - General Pythagorean Theorem
  16. Asymmetric Propeller and Several of Its Generalizations
  17. A construction problem that combines several apparently unrelated ones
  18. Two-Parameter Solutions to Three Almost Fermat Equations
  19. Three circles with centers on their pairwise radical axes
Pairs of statements in which a more general is directly implied by a more specific one.
  1. 2D isoperimetric theorem - a similar theorem with a fixed line segment
  2. Construct a line tangent to two circles - find a tangent from a point to a circle
  3. Arithmetic mean of N numbers is never less than their geometric mean. N arbitrary and N=2n
  4. The equation xxx3 = 3 is no more difficult than x3 = 3
  5. Every convex polygon of area 1 is contained in a rectangle of area 2 because this is true for a triangle
  6. x1x2 + x2x3 + ... + xn-1xn ≤ 1/4, for x1 + x2 + ... + xn = 1, privided it is true for n = 2.
Problems that allow a meaningful generalization.
  1. In a plane, given 3n points. Is it possible to draw n triangles with vertices at these points so that no two of them have points in common?
  2. Lines in a triangle intersecting at a common point. Ceva's theorem.
  3. 5109094x171709440000 = 21!, find x.
    • A number is written with 81 ones. Prove it's a multiple of 81.
    • Prove that the number 1010101...0101, 81 ones and 80 alternating zeros, is a multiple of 81.
  4. Given a 1x1 square. Is it possible to put into it not intersecting circles so that the sum of their radii will be 1996?
  5. Criteria of divisibility by 9 and 11.
  6. The game of Fifteen and Puzzles on Graphs
  7. Weierstrass Product Inequality
  8. Fermat's Little Theorem and Euler's Theorem
Problems that allow more than one generalization.
  1. Pythagorean Theorem
  2. Napoleon's Theorem
  3. Fermat Point and Generalizations
  4. A problem in extension fields
  5. Butterfly theorem

Thus we see that generalization is quite useful and often enjoyable. It's a great vehicle for discovering new facts. However, if unchecked, generalization may lead to erroneous results. I'd call such situations

Inventor's Paranoia

Following are a few examples where attempts to generalize lead one astray. (There eventually will be a few.)

  1. Formula for prime numbers.
  2. Partitioning a circle.

I have allowed myself to swerve from Pólya's maxim: The more ambitious plan may have more chances of success. "The more ambitious" may not necessarily mean "the more general". Following are a few examples where a "more powerful, stronger" statement comes out easier to prove:

  1. Inequality 1/2·3/4·5/6· ... ·99/100 < 1/10.
  2. A Low Bound for Inequality /2·3/4·5/6· ... ·(2n - 1)/2n.
  3. Inequality (1 + 1-3)(1 + 2-3)(1 + 2-3)...(1 + n-3) < 3.
  4. Inequality 1 + 2-2 + 3-2 + 4-2 + ... + n-2 < 2.

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