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How to Recognize a Parabola
Bettina Richmond and Tom Richmond
Parabolas have many interesting properties which were perhaps more well known in
centuries past. Many of these properties hold only for parabolas, providing a characteriza-
tion which can be used to recognize (theoretically, at least) a parabola. Here, we present
a dozen characterizations of parabolas, involving tangent lines, areas, and the well-known
reflective property. While some of these properties are widely known to hold for parabo-
las, the fact that they hold only for parabolas may be less well known. These remarkable
properties can be verified using only elementary techniques of calculus, geometry, and
differential equations.
A parabola is the set of points in the plane which are equidistant from a point F called
the focus and a line l called the directrix. If the directrix is horizontal, then the parabola
is the graph of a quadratic function p(x) = αx 2
- βx + ∞. We will not distinguish between
a function and its graph. A chord of a function f (x) is a line segment whose endpoints lie
on the graph of the function. A chord of f (x) is a segment of a secant line to f (x). By
the equation of a chord, we mean the equation of the corresponding secant line.
Our first characterization of parabolas involves the area between a function and a chord
having horizontal extent h.
Theorem (How to Recognize a Parabola) 1 Suppose f (x) is a differentiable func-
tion and for all real numbers a and h with h > 0 , l(a, h, x) is the secant line determined by
the two points (a, f (a)) and (a + h, f (a + h)) on the graph of f (x), separated horizontally
by h units. Then f (x) is a parabola if and only if the signed area
A(a, h) =
Z (^) a+h
a
l(a, h, x)dx −
Z (^) a+h
a
f (x)dx
between the line l(a, h, x) and the function f (x) over the interval [a, a + h] is a nonzero
function of h alone, not dependent on a.
Figure 1. (Thm. 1) The shaded area depends on h alone, independent of the location a.
Proof. Archimedes knew that parabolas satisfy this property, illustrated in Figure 1.
With techniques of calculus, it is easily verified that the area between q(x) = x 2 and
the chord from (a, a 2 ) to (a + h, (a + h) 2 ) is h 3 /6, independent of a. Since any parabola
p(x) = αx 2
- βx + ∞ is obtained from q(x) = x 2 by a translation, which does not change
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the area, and a vertical scaling/reflection by α, it follows that the signed area between
p(x) = αx 2
- βx + ∞ and the chord from (a, p(a)) to (a + h, p(a + h)) is αh 3 /6. This is
also proven without calculus in Swain and Dence [ 16 ].
Conversely, let f (x), h, l(a, h, x), and A(a, h) be as in the statement of the theorem.
The integral of l(a, h, x) occurring in the definition of A(a, h) is easily calculated and
represents the area of a trapezoid under l(a, h, x) and over [a, a + h] if f (a) and f (a + h)
are both positive, giving
A(a, h) =
f (a) + f (a + h)
2
h −
Z (^) a+h
a
f (x)dx.
Since this is constant relative to a, differentiating with respect to a and applying the
Fundamental Theorem of Calculus for the derivative of the indefinite integral gives
f 0 (a) + f 0 (a + h)
2
h − f (a + h) + f (a),
or
f 0 (a + h) + f 0 (a)
2
h = f (a + h) − f (a) for all a, for all h > 0. (1)
Fixing a = a 0 and letting x = a 0 + h, the differential equation above becomes
x − a 0
2
f 0 (x) − f (x) =
−(x − a 0 )
2
f 0 (a 0 ) − f (a 0 ), (2)
a linear first-order differential equation with nonconstant coefficients. In standard form
the differential equation becomes
f 0 (x) −
x − a 0
f (x) = −f 0 (a 0 ) −
x − a 0
f (a 0 ). (3)
Note that the coefficient of f (x) has a discontinuity at x = a 0 , as does the “forcing
function” on the right side of the equation. The standard existence and uniqueness theorem
guarantees a unique solution to any linear first-order differential equation through any
specified initial condition, valid on the largest interval over which the coefficient functions
and forcing functions are continuous. Thus, the differential equation (3) will have a unique
solution f 1 on (−1, a 0 ) through any specified initial values (x 0 , y 0 ) with x 0 < a 0 and a
unique solution f 2 on (a 0 , 1 ) through any specified initial values (x 1 , y 1 ) with a 0 < x 1.
Following standard procedures, (3) is solved by multiplying through by the integrating
factor μ = e−^
R 2 /(x−a 0 ) dx (^) = (x − a 0 )
− (^2). This gives
(x − a 0 ) − 2 f 0 (x) − 2(x − a 0 ) − 3 f (x) = −(x − a 0 ) − 2 f 0 (a 0 ) − 2(x − a 0 ) − 3 f (a 0 ),
or d
dx
[(x − a 0 ) − 2 f (x)] = −(x − a 0 ) − 2 f 0 (a 0 ) − 2(x − a 0 ) − 3 f (a 0 ).
Antidifferentiating both sides of the equation above gives
(x − a 0 ) − 2 f (x) = (x − a 0 ) − 1 f 0 (a 0 ) + (x − a 0 ) − 2 f (a 0 ) + c,
December 2009] HOW TO RECOGNIZE A PARABOLA 911
Again, this property characterizes parabolas, but we present some other preliminary results
before presenting this as Theorem 4.
Theorem (How to Recognize a Parabola) 3 Suppose f is a nonlinear function with
a continuous derivative. Then f (x) is a parabola if and only if the average of the slopes
of the tangent lines at the endpoints of any interval equals the slope of the tangent line at
the midpoint of the interval. That is, f (x) is a parabola if and only if
f 0 (a − h) + f 0 (a + h)
2
= f 0 (a) for all a, for all h > 0. (6)
See Figure 3.
Figure 3. (Thm. 3) The slope at a is the average of the slopes at a + h and a − h.
Proof. It is easy to verify that a quadratic function f (x) satisfies (6), so suppose f (x) has
a continuous derivative and satisfies (6). Rewriting (6) as
f 0 (a) − f 0 (a − h)
h
f 0 (a + h) − f 0 (a)
h
we see that over any interval [a − h, a + h] the point (a, f 0 (a)) on f 0 (x) at the midpoint
lies on the line L(x) determined by the two points (a − h, f 0 (a − h)) and (a + h, f 0 (a + h)).
Thus, the three points (a + nh, f 0 (a + nh)) (n = 0 , ±1) all lie on L(x). Iterating this
over the two halves of [a − h, a + h], we find that the five points (a + nh/ 2 , f 0 (a + nh/2))
(n = 0 , ± 1 , ±2) lie on L(x). Further iterations show that f 0 (x) coincides with L(x) at all
points x = a + nh/ 2 k for k ∈ N, n ∈ { 0 , ± 1 ,... , ± 2 k }. Thus, f 0 (x) coincides with L(x) on
a dense subset of [a − h, a + h], so by the continuity of f 0 (x), it follows that f 0 (x) = L(x)
on [a − h, a + h]. Since f 0 (x) is linear on any interval [a − h, a + h] and R can be written
as a union of overlapping intervals, it follows that f 0 (x) is linear on R, and thus f (x) is a
quadratic function.
This characterization of parabolas provides a nice way to recognize a linear function:
Lemma (How to Recognize a Line) 1 Suppose f is a continuous function. Then f (x)
is a linear function if and only if the average value of f (x) over every interval is the value
of f at the midpoint of the interval. That is, f (x) is a linear function if and only if
R (^) a+h a−h f^ (x)dx 2 h
= f (a) for all a, for all h > 0. (7)
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Proof. If f (x) = mx + b is a linear function, it is geometrically clear and easy to verify
using calculus that (7) holds. Thus, suppose f (x) satisfies (7). Since the left-hand side
of (7) is a differentiable function of a, f is differentiable, and we have
f 0 (a) =
f (a + h) − f (a − h)
2 h
f (a + h) − f (a)
h
f (a) − f (a − h)
h
f 0 (a + h/2) + f 0 (a − h/2)
2
by two applications of (8). Thus, the average of the slopes f 0 (a − h/2) and f 0 (a + h/2) at
the endpoints of any interval [a − h/ 2 , a + h/2] is simply the value f 0 (a) of the derivative
at the midpoint of that interval. Since the right-hand side of (8) is a continuous function
of a, f has a continuous derivative. Now by the proof of Theorem 3, it follows that f (x)
has the form f (x) = αx^2 + βx + ∞. We will show that α = 0, so that f (x) is in fact a
linear function.
Since f (x) satisfies (7), we have R (^) a+h a−h (αx 2
2 h
= αa 2
Evaluating the integral above, we get
αa 2
αh 2
- βa + ∞ = αa 2
- βa + ∞ for all a, for all h > 0.
Thus, αh 2 / 3 = 0 for all h > 0, and consequently α = 0, so f (x) = βx + ∞ is a linear
function, as needed.
Now we will show that the mean-value-at-the-midpoint property (5) satisfied by parabo-
las is in fact a characterization of parabolas: f (x) is a parabola if and only if “centered
chords are parallel,” that is, chords horizontally centered at x = a are parallel to the
tangent line at x = a. See Figure 4.
Figure 4. (Thm. 4) Chords horizontally centered at x = a are parallel to the tangent at x = a.
Theorem (How to Recognize a Parabola) 4 A nonlinear differentiable function f (x)
is a parabola if and only if over every interval [a − h, a + h], the secant line l(a, x) deter-
mined by (a − h, f (a − h)) and (a + h, f (a + h)) is parallel to the line tangent to f at the
midpoint a of the interval. That is, f is a parabola if and only if
f 0 (a) =
f (a + h) − f (a − h)
2 h
for all a, for all h > 0. (9)
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intersect f (x) in at most two points: If a line intersected f (x) at three points x 1 < x 2 < x 3 ,
then the chords of f over the intervals [x 1 , x 2 ] and [x 2 , x 3 ] each have the same slope as the
line, but do not have vertically aligned midpoints, contrary to the hypothesis.
Suppose a and h > 0 are given and let l(x) = mx + b be the chord of f over the interval
[a − h, a + h]. Let c ∈ (a − h, a + h) be the point guaranteed by the Mean Value Theorem
with f 0 (c) = l 0 (c) = m, and let t(x) = mx + b 0 be the line tangent to f (x) at x = c. Now
l(x) and t(x) are distinct parallel lines, and we assume that l(x) lies above t(x). The case
of l(x) below t(x) is similar. Let (bn)^1 n=1 be a decreasing sequence in (b 0 , b) converging to
b 0 and consider the parallel lines ln(x) = mx + bn. By the Intermediate Value Theorem
(applied to f (x) − t(x)), ln(x) must intersect f (x) at a point vn ∈ (a − h, c) and a point
wn ∈ (c, a + h), and as noted above, at no other points. Now the chord of f over [vn, wn]
determined by ln(x) is parallel to the chord over [a − h, a + h] and therefore has the same
center a. Thus, [vn, wn] = [a − hn, a + hn] for some hn > 0, and the sequence (hn)^1 n=1 is
a decreasing sequence in [0, h]. As a decreasing sequence bounded below, (hn)^1 n=1 must
converge to a limit h 0 , and since c ∈ (a − hn, a + hn) for every natural number n, we have
c ∈ [a − h 0 , a + h 0 ]. It follows that a = c, for otherwise a − h 0 , c, and a + h 0 would be
three points on the intersection of f (x) and the line limn→1 ln(x) = t(x). Now the chord
l(x) is parallel to the line t(x) tangent to f at c = a, which is the midpoint of the interval
[a − h, a + h], so Theorem 4 implies f is a parabola.
In his Quadrature of the Parabola, Archimedes states that parabolas satisfy the prop-
erties of Theorems 4 and 5, and cites the proofs from a Greek work Elements of Conics
attributed to Aristaeus and Euclid ([ 6 , pp. 50,78]). Again, the converses were not men-
tioned.
Before characterizing parabolas by the maximal vertical distance from the function to
a chord, we present a lemma.
Lemma 2 Let g be a continuous function on [a, b] with g 00 defined on (a, b) and with
g(a) = g(b) = 0. If g 00 (x) < 0 for x ∈ (a, b), then g(x) > 0 for x ∈ (a, b). If g 00 (x) > 0 for
x ∈ (a, b), then g(x) < 0 for x ∈ (a, b).
Proof. The second conclusion follows from the first by replacing g by −g, so we will prove
only the first statement. Now g 00 (x) < 0 on (a, b) implies that g 0 is strictly decreasing on
[a, b]. Since g(a) = g(b) = 0, it follows from the Mean Value Theorem that there exists
w ∈ (a, b) with g^0 (w) = 0. So, g^0 (x) > 0 for x ∈ (a, w) and g^0 (x) < 0 for x ∈ (w, b), and
hence g is strictly increasing on [a, w] and strictly decreasing on [w, b]. Since g(a) = g(b) =
0, it follows that g(x) > 0 for x ∈ (a, b).
Applied to the difference g(x) of a parabola p(x) and a twice differentiable function
f (x) of the same concavity, this lemma simply tells us that if f (x) is flatter than a parabola
between two of its points of intersection with the parabola then, between these points, f (x)
is closer than the parabola to the secant line determined by the intersection points.
Theorem (How to Recognize a Parabola) 6 Suppose f is a function for which f 00 is
continuous. For any a ∈ R and h > 0 , let l(a, h, x) be the secant line connecting (a, f (a))
and (a + h, f (a + h)), and let
v(a, h) = max x∈[a,a+h]
|l(a, h, x) − f (x)|
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Figure 6. (Thm. 6) The maximum vertical distance from function to chord is a function of h alone, independent of a.
be the maximum vertical distance between f (x) and l(a, h, x). Then f (x) is a parabola if
and only if v(a, h) is a nonzero function of h alone, independent of a. (See Figure 6.)
Proof. If f (x) = p(x) = αx 2 +βx+∞ is a parabola, it is easy to verify that v(a, h) = |α|h 2 /4,
independent of a.
Conversely, suppose f (x) satisfies the hypotheses above but is not a parabola. Then
f 00 (x) is not constant, and by the continuity of f 00 , we can find disjoint intervals on
which the values of f 00 have the same sign but are separated by some constant k. That
is, there exist a 1 , a 2 , k ∈ R and h 0 > 0 such that either f 00 (x) > k > f 00 (y) > 0 or
0 > f 00 (x) > k > f 00 (y) for all x ∈ [a 1 , a 1 + h 0 ] and all y ∈ [a 2 , a 2 + h 0 ]. The cases are
analogous, so we assume the former condition, in which f is concave up on the intervals in
question. For i = 1 , 2, let pi(x) be the parabola determined by p^00 i (x) = k, pi(ai) = f (ai),
and pi(ai + h 0 ) = f (ai + h 0 ). The condition on the second derivatives implies that
α = k/ 2 is the leading coefficients of both p 1 (x) and p 2 (x). Now applying Lemma 2 to
g 1 (x) = f (x)−p 1 (x), we have g 100 (x) = f 00 (x)−p^001 (x) > 0 over (a 1 , a 1 +h 0 ) so f (x) lies below
the parabola p 1 (x) on (a 1 , a 1 + h 0 ). Thus, the maximum vertical distance between f and
l(a 1 , h 0 , x) over [a 1 , a 1 +h 0 ] is greater than the maximum vertical distance |α|h^20 / 4 between
p 1 and l(a 1 , h 0 , x) over the same interval. Consequently, if v(a, h) = v(h) is a function of
h alone, we have v(h 0 ) > |α|h^20 /4. But applying Lemma 2 to g 2 (x) = f (x) − p 2 (x), we
have g 200 (x) = f 00 (x) − p^002 (x) < 0 over (a 2 , a 2 + h 0 ) so f lies above p 2 on (a 2 , a 2 + h 0 ),
and v(h 0 ) < |α|h 2 0 /4,^ a^ contradiction.^ Thus,^ if^ f^ is^ not^ a^ parabola,^ then^ v(a,^ h)^ is^ not^ a function of h alone.
We note that if f 00 is continuous and v(a, h) = v(h) is a nonzero function of h alone,
then v(h) = ch^2 for some constant c, and v(h) occurs at the midpoint a + h 2 of the interval
[a, a + h] for any a and any h > 0.
In Theorem 2 we saw a characterization of parabolas based on tangent lines at the ends
of a chord. We now present another characterization of parabolas based on pairs of tangent
lines. It has long been known (see [ 14 ]) that a parabola p(x) has the property that the
lines tangent to p(x) at any points x = a and x = b intersect at the point x = (a + b)/2,
as illustrated in Figure 7. The converse gives another way to recognize parabolas.
Theorem (How to Recognize a Parabola) 7 A differentiable function f (x) is a
parabola if and only if the lines tangent to f (x) at x = a and x = b intersect only at
x = (a + b)/ 2 for all a, b ∈ R.
Proof. A direct argument shows that a parabola has the indicated property (see [ 14 ]).
Assume f (x) satisfies the hypotheses of the theorem. The lines la(x) = f 0 (a)(x − a) + f (a)
December 2009] HOW TO RECOGNIZE A PARABOLA 917
rays parallel to the axis of symmetry into the focus are easily found, proofs that this
property characterizes parabolas are less common. Drucker and Locke [ 8 ] give a nice proof
of this using concepts of geometry and calculus without differential equations. See also [ 5 ]
and [ 12 ]. Both directions of this characterization were addressed by Diocles circa 200 B.C.
in his treatise On Burning Mirrors [ 17 ].
Theorem (How to Recognize a Parabola) 9 Suppose f (x) is a differentiable func-
tion. Then f (x) is a concave up parabola if and only if it has the property that all vertical
rays coming downward from above f (x) are reflected to a single point F.
(a) (b)
Figure 9. (Thm. 9) (a) Vertical rays are reflected to a single point F.
Proof. Suppose f (x) is a concave up parabola and, without loss of generality, has vertex
at the origin. Referring to Figure 9b, if a vertical ray intersects the function f (x) at
A(a) = (a, f (a)) and intersects the x-axis at B(a) = (a, 0), then that vertical ray is
reflected by the function f (x) with an angle of reflection equal to the angle of incidence.
Let O be the origin and let F = (0, c) be the focus, so that f (x) = x^2 /(4c). Let D(a) be
the point on ray AB such that AF = AD, and let E(a) be the intersection of F D and
the line tangent to f (x) at x = a. Since vertical angles are equal, the angle of incidence
is ∠EAB. Then by the focus-directrix definition, D(a) lies on the directrix, and hence
D(a) = (a, −c). Since F D has slope − 2 c/a and EA has slope f 0 (a) = a/(2c), we see that
F D ⊥ EA. It follows that 4 EAF is congruent to 4 EAD, so ∠EAF = ∠EAD = the
angle of incidence. Thus, if f (x) is a parabola, any vertical ray is reflected at an angle
which will pass through the focus F. This is the proof of [ 14 ] and [ 18 ].
Conversely, suppose vertical rays coming from above f (x) are reflected to a point F ,
and without loss of generality, assume F is at the origin. Consider the right half of
the curve—that is, the curve f (x) for x > 0. Let A = (x, f (x)) be any point on this
part of the curve. Let D be the point directly below A such that AF = AD. Then
D has coordinates (x, f (x) −
p x^2 + f (x)^2 ). Let E be the midpoint of F D, which has
coordinates (x/ 2 , (f (x) −
p x^2 + f (x)^2 )/2). Then 4 AF E is congruent to 4 ADE, so AE
bisects ∠F AD. Since the downward vertical ray through A reflects to F , AE must coincide
with the tangent line at A, so we have
f 0 (x) =
f (x) − (f (x) −
p x^2 + f (x)^2 )/ 2
x − x/ 2
f (x) +
p x^2 + f (x)^2
x
December 2009] HOW TO RECOGNIZE A PARABOLA 919
Mueller and Thompson [ 12 ] indicate how to solve this differential equation. We proceed,
avoiding the techniques of differential equations. We will show that the points D lie on
a horizontal line y = c. Then from the focus-directrix definition, for x > 0, f would
bpe a parabola with focus F and directrix y = c. The second coordinate of D is f (x) −
x^2 + f (x)^2. To show that this is constant, consider its derivative:
d
dx
f (x) −
p x^2 + f (x)^2
= f 0 (x) −
x + f (x)f 0 (x) p x^2 + f (x)^2
f 0 (x)(
p x^2 + f (x)^2 − f (x)) − x p x^2 + f (x)^2
Now substituting the value of f 0 (x) from equation (11) shows that the value of the expres-
sion above is zero. Thus, there is a constant c such that f (x) −
p x^2 + f (x)^2 = c. Noting
that c < 0 and solving for f (x), we get f (x) = −x^2 /(2c) + c/2. Similar reasoning applies
to the left half of the curve, and since the function is continuous at x = 0, the values of c
for the two halves must agree, so f (x) is a single parabola on (−1, 1 ).
We round out our dozen characterizations of parabolas with three which are proved
elsewhere. The characterization below, given in [ 13 ], requires working in R 3 .
Theorem (How to Recognize a Parabola) 10 A twice differentiable increasing func-
tion f (x) on [0, 1 ) is a parabola on [0, 1 ) with vertex at the origin if and only if revolving
that portion of f over the interval [0, r] about the y-axis gives, for every r > 0 , a bowl with
exactly as much volume under it as inside it. (See Figure 10.)
Figure 10. (Thm. 10) The volume under the bowl equals the volume in the bowl.
Another property of parabolas known to Archimedes is that the area between the
secant line l(a, h, x) and the parabola p(x) as in Theorem 1 occupies two-thirds of the
area of the circumscribing parallelogram having two vertical sides. See Figure 11. From
Theorem 1, the area between the parabola p(x) = αx 2
- βx + ∞ and the chord over
[a, a + h] is |α|h 3 /6. The result then follows since the height (by Theorem 6) and width of
the circumscribing parallelogram are |α|h 2 / 4 and h, respectively. Archimedes’ method of
proof from his Quadrature of the Parabola is discussed in [ 7 ], [ 15 ], and [ 16 ], and a geometric
proof from the 19th century is given in [ 11 ]. The proof of the converse is nontrivial and is
found in [ 2 ]. We note that it requires the assumption that the function f be three times
differentiable.
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[6] E. J. Dijksterhuis, Archimedes, Princeton University Press, Princeton, NJ, 1987.
[7] H. D¨orrie, 100 Great Problems of Elementary Mathematics: Their History and Solution (trans. D. Antin), Dover, New York, 1965.
[8] D. Drucker and P. Locke, A natural classification of curves and surfaces with reflection properties, Math. Mag. 69 (1996) 249–256.
[9] J. Gallego-Diaz, M. Goldberg, and D. C. B. Marsh, Problem E1659, this Monthly 71 (1964) 1136–1137. doi:10.2307/
[10] S. Haruki, A property of quadratic polynomials, this Monthly 86 (1979) 577–579.
doi:10.2307/
[11] O. L. Mathiot, Geometric determination of the area of the parabola, The Analyst 9 (1882)
106–107. doi:10.2307/
[12] W. Mueller and R. Thompson, Discovering differential equations in optics, College Math. J.
28 (1997) 217–223. doi:10.2307/
[13] M. B. Richmond and T. A. Richmond, Characterizing power functions by volumes of revo-
lution, College Math. J. 29 (1998) 40–41. doi:10.2307/
[14] C. Smith, An Elementary Treatise on Conic Sections, Macmillan, London, 1885. 1904 edition
available at http://www.archive.org/details/anelementconicsec00smitrich.
[15] S. Stein, Archimedes: What Did He Do Besides Cry Eureka? Mathematical Association of
America, Washington, DC, 1999.
[16] G. Swain and T. Dence, Archimedes’ quadrature of the parabola revisited. Math. Mag. 71
(1998) 123–130.
[17] G. J. Toomer, Diocles on Burning Mirrors: The Arabic Translation of the Lost Greek Orig-
inal, Springer-Verlag, New York, 1976.
[18] R. C. Williams, A proof of the reflective property of the parabola, this Monthly 94 (1987)
667–668. doi:10.2307/
BETTINA RICHMOND received her Ph.D. from Florida State University under the direction
of Warren Nichols in 1985. Her primary research interests are Hopf algebras and semigroups. This
is her third article in this Monthly.
Department of Mathematics, Western Kentucky University, 1906 College Heights Blvd.,
Bowling Green, KY 42104.
bettina.richmond@wku.edu
TOM RICHMOND received his Ph.D. from Washington State University under the direction
of Darrell Kent in 1986. His research interests lie at the intersection of topology and order. He
recently completed a 4-year stint as Chair-Elect and Chair of the Kentucky Section of the MAA.
Department of Mathematics, Western Kentucky University, 1906 College Heights Blvd.,
Bowling Green, KY 42104.
tom.richmond@wku.edu
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