Many famous problems in mathematics are phrased as a search for a specific construction. This often requires some creativity and understanding of the theory behind the problem. The typical examples are the three geometric problems of antiquity: squaring of the circle, doubling of the cube, and trisection of angles through a compass and straightedge only.

To solve these problems, we develop the concept of a constructible number in the complex plane. A complex number is constructible if its corresponding point in the Euclidean plane is constructible by using a unruled straightedge, compass and line segment of unit length. Using the language of field theory, it turns out that the constructible numbers are the "quadratic closure of the rational numbers". While this complete characterization is somewhat beyond us for now, we can show that a constructible number must be an algebraic number (which means that it is the root of a polynomial with integer coefficients). This follows by considering what the various compass and straightedge constructions actually allow. As such, since π \pi is transcendental, it shows that we are unable to square the circle.

More recently, Godel's Incompleteness Theorems proved in 1931 establishes that there is no complete and consistent set of axioms for mathematics, which negatively answers Hilbert's second problem. This is a somewhat surprising theorem in logic, which has far-reaching consequences.

Closer to home, we come across this idea often in geometry problems, in which construction of points crucial in the diagram helps us to establish the proof easily. Various algebraic identities like

(a2+b2)(c2+d2)=(ac+bd)2+(adbc)2 (a^2+b^2)(c^2+d^2 ) = (ac+bd)^2 + (ad-bc)^2

can help shed light on deeper implications in mathematics.

In a construction problem, we seek an explicit characterization or identification, which allows us to verify that the statement is true. This is different from existence problems, in which we merely know that the statement is true, but do not know for what values it is true.

Worked Examples

1. A longevity chain is a sequence of consecutive integers, whose digit sums are never a multiple of 9. What is the longest possible length of a longevity chain?

Solution: We know that any number which is a multiple of 9, has a digit sum which is a multiple of 9. Hence, the longevity chain can have at most 8 elements.

As an explicit example, the sequence 1,2,3,4,5,6,7,8 1, 2, 3, 4, 5, 6, 7, 8 is a sequence of 8 consecutive integers, whose digit sums are never a multiple of 9.

Hence, the answer is 8. _\square

In this problem, the understanding that we needed was how digit sums relate to multiples of 9. With that understanding, the construction of the solution is straightforward.


2. Does there exist a family of concentric circles, such that each circle contains exactly 1 lattice point, and each lattice point is on a circle?

Solution: Consider the center of these concentric circles. Since each lattice point is on a unique concentric circle, this means that the distance of each lattice point to the center must be distinct.

Claim: The point (13,2) ( \frac {1}{3}, \sqrt{2} ) works as the center of the concentric circles.

Suppose not, then we must have 2 lattice points (x1,y1),(x2,y2) (x_1, y_1), (x_2, y_2) whose distances are the same. This implies that

(x113)2+(y12)2=(x213)2+(y22)2. ( x_1 - \frac {1}{3} )^2 + ( y_1 - \sqrt{2} ) ^2 = ( x_2 - \frac{1}{3} )^2 + (y_2 - \sqrt{2}) ^2 .

Expanding the terms, we get that

x1223x1+y12x22+23x2y22=2(2y22y1). x_1^2 - \frac{2}{3} x_1 + y_1^2 - x_2 ^2 + \frac {2}{3} x_2 - y_2 ^2 = \sqrt{2} ( 2 y_2 - 2 y_1 ).

If y2y1 y_2 \neq y_1 , then the LHS is rational while the RHS is irrational, which is a contradiction. Thus y1=y2 y_1 = y_2 . Hence, LHS=RHS=0 LHS=RHS=0 , which implies that x1223x1x22+23x2=0 x_1^2 - \frac {2}{3} x_1 - x_2^2 + \frac {2}{3} x_2 = 0 , or that 0=(x1x2)(x1+x223) 0 = (x_1 - x_2) ( x_1 + x_2 - \frac {2}{3} ) . Since the second term is never 0 (because x1,x2 x_1, x_2 are integers), we must have x1=x2. x_1 = x_2. This contradicts the assumption that the points are distinct. _\square

In this problem, we played around with the conditions and rephrased it into a number-theoretic statement, which made it easier to approach.

Note: There is an existence solution which works by showing that the plane cannot be covered by countably many lines.


3. Is it possible to find 7 points in the plane, such that out of every subset of 3 points, there are 2 points that are a unit distance apart?

Solution: Let ABCD ABCD be 4 points such that AB=BC=CD=DA=BD=1 AB=BC=CD=DA=BD = 1 . Let's rotate this slightly about A A , to get ABCD AB^*C^*D^* such that CC=1 CC^* =1 .

image image

Claim: This set of 7 points satisfy the conditions.

Proof: If the 3 points lie in {A,B,C,D} \{A, B, C, D \} or {A,B,C,D} \{ A, B^*, C^*, D^*\} , then we are done. Otherwise, we must have 2 points that are in each of these sets. If these points are not {A,C} \{A, C\} or {A,C} \{A, C^*\} , then they will be a unit distance apart. Hence, the 3 points must thus be {A,C,C} \{A, C, C^* \} , but this gives us CC=1 CC^* = 1 , so we are done. _\square

Note by Calvin Lin
7 years, 4 months ago

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For the third question, if CC=1CC^*=1, then DD would be the same point as BB^*, which I do not know if it is allowed.

Kenny Lau - 6 years, 10 months ago

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No, DD will not be the same point as BB^* .

Note that CC is 3 \sqrt{3} away from AA, and so we are not rotating by 60 60 ^ \circ , but by a slightly smaller angle, to get CC=1 C C^* = 1 .

Calvin Lin Staff - 6 years, 10 months ago

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Oops, you're correct, and it's a pentagon!

Image here Image here

Kenny Lau - 6 years, 10 months ago

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@Kenny Lau Indeed!

I used markdown to display your image, by adding a "!" to the start of the hyperlink, since you already gave it in the image URL format.

I've also added it to the writeup. Thanks for the helpful image!

Calvin Lin Staff - 6 years, 10 months ago

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