A CHALLENGING AREA PROBLEM REVISITED

Robert L. Doucette

McNeese State University

Some years ago a problem was proposed in the American Mathematical Monthly [1] for which the editors received no correct solutions before the deadline. Although eventually a solution was published (under the title “One Tough Area Problem” [2]), it is relatively involved. I would like to present a quite different, simpler solution.

The problem is to find the area of the convex planar region

R = {P : P A + P B + P C ≤ 2a},

where ABC is an equilateral triangle of perimeter 3a.

For convenience we take a = 1. We start by imposing a rectangular coordinate system in which the coordinates of A, B, C are (−1/2, 0), (1/2, 0), and (0, \sqrt{3}/2) respectively. As mentioned in [2], the convexity of R is relatively easy to show using the triangle inequality. Let ∂R denote the boundary of R. Clearly A, B and C are on ∂R. We may deduce that the portion of ∂R in quadrant I is a convex curve connecting C and B. A parameterization of this curve may be obtained by constructing a circle of radius r, 0 ≤ r ≤ 1, with center C; and an ellipse with foci A and B, and major axis of length 2 − r. If P is the point of intersection in quadrant I, then P C = r and P A + P B = 2 − r, so that P A + P B + P C = 2. See the figure below. As r goes from 0 to 1, P travels along ∂R from C to B. The coordinates (x, y) of P can be found by solving the system

(1)  x^{2} + (y − \sqrt{3}/2)^{2} = r^{2}

(2)  \frac{x^{2}}{(\frac{2−r}{2} )2} + \frac{y^{2}}{(\frac{2−r}{2} )^{2} − (\frac{1}{2})^{2}} = 1.

Multiplying (1) by (\frac{2−r}{2} )^{−2} and subtracting the result from (2) we eliminate x^{2}, and find (after a bit of algebra) that y = (1 − r)\sqrt{3 − r}(2 − \frac{\sqrt{3}}{2} \sqrt{3 − r}). Setting

(3)  s = \sqrt{3 − r}

we obtain

(4)  y = s(s^{2} − 2)(2 − \frac{\sqrt{3}}{2}s).

From (1) we get x^{2} = r^{2} − (y − \sqrt{3}/2)^{2} = (r − y + \sqrt{3}/2)(r + y − \sqrt{3}/2). Therefore by (3) and (4) x^{2} equals

(5) [(s^{2} − 3) − s(s^{2} − 2)(2 − \frac{\sqrt{3}}{2}s) + \frac{\sqrt{3}}{2}][(s^{2} − 3) + s(s^{2} − 2)(2 − \frac{\sqrt{3}}{2}s) − \frac{\sqrt{3}}{2}].

We may make use of the system (1)–(2) to factor (5) completely. From equation (1) we see that r = 0 (and hence s = \sqrt{3}) implies that x = 0. We find that \sqrt{3} is a root of both factors in (5). In our search for other roots we note that by (2) r = 2 (and hence s = 1) also implies that x = 0. We verify that 1 is a double root of the first factor in (5). Next we observe that in (4) s = 1 gives y = −2 + \sqrt{3}/2. But, when r = 2, from (1) y = 2 + \sqrt{3}/2 is also a possibility. This value may be obtained in (4) by setting s = −1. We verify that s = −1 is a double root of the second factor in (5). From this we may factor (5):

x^{2} = \frac{3}{4}(s − \sqrt{3})^{2}(s − 1)^{2}(s + 1)^{2}[4 − (s − \frac{\sqrt{3}}{3})^{2}].

Setting 2t = s − \sqrt{3}/3, and simplifying we arrive at the following parameterization for the portion of ∂R in the first quadrant:

(6)  x(t) = (−8\sqrt{3}t^{3} + 4\sqrt{3}t − \frac{4}{3})\sqrt{1 − t^{2}}

(7)  y(t) = −8\sqrt{3}t^{4} + 8\sqrt{3}t^{2} − \frac{4}{3}t − \frac{5\sqrt{3}}{6}

where r ∈ [0, 1] ⟹ s ∈ [\sqrt{2}, \sqrt{3}] ⟹ t ∈ [(\sqrt{2} − \sqrt{3}/3)/2, \sqrt{3}/3].

From calculus,

α := area of R in quadrant I = ∫_{(\sqrt{2}−\sqrt{3}/3)/2}^{\sqrt{3}/3} x(t)y′(t) dt.

Once we determine the value of α the computation of area(R) follows. We may decompose R into the equilateral triangle ABC, and three regions each congruent to the region of R in quadrant I outside of the segment CB. This latter region has area α − \sqrt{3}/8 and hence

(8)  area(R) = area(△ABC) + 3(α − \sqrt{3}/8) = 3α − \sqrt{3}/8.

From (6) and (7) we obtain

(9)  x(t)y′(t) = 16(48t^{6} − 48t^{4} + \frac{10\sqrt{3}}{3}t^{3} + 12t^{2} − \frac{5\sqrt{3}}{3}t + \frac{1}{9})\sqrt{1 − t^{2}}.

The integrals I_{n} := ∫ t^{n}\sqrt{1 − t^{2}} dt; n = 0, 1, 2, 3, 4, 6; are standard and are given by

I_{0} = \frac{1}{2}t\sqrt{1 − t^{2}} + \frac{1}{2} sin^{−1} t

I_{1} = \frac{1}{3}(−1 + t^{2})\sqrt{1 − t^{2}}

I_{2} = \frac{1}{4}(−\frac{1}{2}t + t^{3})\sqrt{1 − t^{2}} + \frac{1}{8} sin^{−1} t

I_{3} = \frac{1}{5}(−\frac{2}{3} − \frac{1}{3}t^{2} + t^{4})\sqrt{1 − t^{2}}

I_{4} = \frac{1}{6}(−\frac{3}{8}t − \frac{1}{4}t^{3} + t^{5})\sqrt{1 − t^{2}} + \frac{1}{16} sin^{−1} t

I_{6} = \frac{1}{8}(−\frac{5}{16}t − \frac{5}{24}t^{3} − \frac{1}{6}t^{5} + t^{7})\sqrt{1 − t^{2}} + \frac{5}{128} sin^{−1} t.

Multiplying each of the I_{n} by the appropriate coefficient from (9) and combining terms we obtain

α = [16\{(\frac{\sqrt{3}}{9} − \frac{23}{72}t− \frac{7\sqrt{3}}{9}t^{2}+ \frac{15}{4}t^{3}+ \frac{2\sqrt{3}}{3}t^{4}−9t^{5}+6t^{7})\sqrt{1 − t^{2}}+ \frac{31}{72} sin^{−1} t\}]_{(\sqrt{2}−\sqrt{3}/3)/2}^{\sqrt{3}/3}.

By some tedious but straightforward computations we may evaluate this last expression, and using (8), we find that the desired area is

−\frac{10\sqrt{2}}{9} + (\frac{187}{72} − \frac{89\sqrt{6}}{72})\sqrt{5 + \sqrt{24}} + \frac{62}{3}[sin^{−1}(\frac{\sqrt{3}}{3}) − sin^{−1}(\frac{\sqrt{2} − \sqrt{3}/3}{2})] − \frac{\sqrt{3}}{8}.

\underline{References}

1. E 2983, American Mathematical Monthly, 90 (1983), 54.

2. American Mathematical Monthly, 96 (1989), 642–645.

Figure.