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.
The Intermediate Value Theorem - The Location Principle (Bolzano Theorem)
Rolle's Theorem - The Mean Value Theorem
Existence of a tangent parallel to a chord - existence of a tangent parallel to the x-axis.
Binomial theorem for (1 + x)n and (x + y)n
The Maclaurin and Taylor series.
Two properties of Greatest Common Divisor
Pythagoras' Theorem and the Cosine Rule
Pythagoras' Theorem and its particular case of an isosceles right triangle
Pythagoras' Theorem and Larry Hoehn's generalization
Combining pieces of 2 and N squares into a single square
Measurement of inscribed and (more generally) secant angles
Probability of the union of disjoint events and any pair of events
Butterfly theorem in orthodiagonal and arbitrary quadrilaterals
Pascal's theorem in ellipse and in circle
A triangle is embeddable into a rectangle twice its area and so is any convex polygon
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.
Bottema's Theorem and McWorter's generalization
Butterfly theorem
Butterflies in a Pencil of Conics
Carnot's theorem and Wallace's theorem.
Concyclic Circumcenters: A Dynamic View.
Concyclic Circumcenters: A Sequel.
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
A Fine Feature of the Stern-Brocot tree
Four Construction Problems
Lucas' Theorem and its variant
Matrix Groups
Napoleon's Theorem and one theorem about similar triangles
Four Pegs That Form a Square
On the Difference of Areas
One Trigonometric Formula and Its Consequences
Pythagorean Theorem and the Parallelogram Law
Pythagorean Theorem - General Pythagorean Theorem
Asymmetric Propeller and Several of Its Generalizations
A construction problem that combines several apparently unrelated ones
Two-Parameter Solutions to Three Almost Fermat Equations
Three circles with centers on their pairwise radical axes
Square Roots and Triangle Inequality
Miguel Ochoa's van Schooten is a Slanted Viviani
Scalar Product Optimization
Barycenter of cevian trangle generalizes Marian Dinca's criterion.
Pairs of statements in which a more general is directly implied by a more specific one.
2D isoperimetric theorem - a similar theorem with a fixed line segment
Construct a line tangent to two circles - find a tangent from a point to a circle
Arithmetic mean of N numbers is never less than their geometric mean. N arbitrary and N=2n
The equation xxx3 = 3 is no more difficult than x3 = 3
Every convex polygon of area 1 is contained in a rectangle of area 2 because this is true for a triangle
x1x2 + x2x3 + ... + xn-1xn ≤ 1/4, for x1 + x2 + ... + xn = 1, privided it is true for n = 2.
Vietnamese Extension of a Japanese Theorem.
Property of Two Pencils of Parallel Lines
Problems that allow a meaningful generalization.
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?
Lines in a triangle intersecting at a common point. Ceva's theorem.
5109094x171709440000 = 21!, find x.
Given a 1x1 square. Is it possible to put into it not intersecting circles so that the sum of their radii will be 1996?
Criteria of divisibility by 9 and 11.
The game of Fifteen and Puzzles on Graphs
Weierstrass Product Inequality
Fermat's Little Theorem and Euler's Theorem
Problems that allow more than one generalization.
Pythagorean Theorem
Napoleon's Theorem
Fermat Point and Generalizations
A problem in extension fields
Butterfly theorem
A System of Equations Begging for Generalization
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.)
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:
Inequality
1/2·3/4·5/6· ... ·99/100 < 1/10. A Low Bound for Inequality 1/2·3/4·5/6· ... ·(2n - 1)/2n.
Inequality
(1 + 1-3)(1 + 2-3)(1 + 2-3)...(1 + n-3) < 3. Inequality
1 + 2-2 + 3-2 + 4-2 + ... + n-2 < 2. Inequality
1/(n+1) + ... + 1/2n < 25/36.
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