JOURNAL OF MATHEMATICAL BEHAVIOR 15, 151-154 (1996)

Solving the Pentagon Problem

BRIAN GREER

Queen’s University of Belfast

At a party at the house of Carolyn Maher of Rutgers University, I was introduced to this problem by Bob Davis:

Given a pentagon ABCDE in which triangles ABC, BCD, CDE, DEA, and EAB all have area = 1. is the area of the pentagon determined?

(1 do not know whether it was specified that the pentagon is convex, but I am going to assume that it is).

The first insight was that BE is parallel to CD. This follows because triangles BCD and ECD are equal in area and CD may be taken as a common base. Likewise, CA is parallel to DE, and so on. (I did not label points with letters when working on the problem, but have done so here to facilitate description). In fact, it is clear that having these pairs of lines parallel and having the triangles equal in area are equivalent conditions.

Jerry Goldin was also working on the problem at this time, and he investi- gated whether there are any nonregular pentagons with the given property. He was able to show that there are, as follows: Construct any isosceles triangle ACD. Construct DX parallel to CA and CY parallel to DA. Let BIHE be parallel to CD, with B on CY, I on AC, H on AD, and E on DX. Consider what happens

Correspondence and requests for reprints should be sent to Brian Greer, School of Psychology, The Queen’s University of Belfast, Belfast BT7 INN, Northern Ireland.

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152 GREER

C D

as the position of BIHE varies while remaining parallel to CD. As I approaches A, angle ABE approaches zero, whereas angle BEC increases; the converse holds as 1 approaches C. Because everything is varying continuously, there must be an intermediate point at which the two angles are equal and BA is parallel to CE. By symmetry, then, AE and BD will also be parallel and ABCDE will be a pentagon with the desired property. The question of whether there are nonsym- metrical pentagons with the property is taken up again later.

Within the pentagon, there are two types of triangle, labeled x and y in the first figure.

Area of pentagon = area ABC + area CDEI + area AEH + area AHI = 1 + 2 + area AEH + area AHI

because CDEI is a parallelogram of area twice that of CDE)

A different dissection shows that:

Area of pentagon = 1 + 2 + area ABI + area AHI

proving that triangles ABI and AEH are equal in area. Repetition of this argu- ment shows that all triangles of type x are equal in area, as are all triangles of type y. Henceforth, I use x and y to stand for the areas of these triangles.

Area of the pentagon ABCDE is 3 + x + y (shown already), which can be reexpressed as 4 - X, because Zr + y = 1. The area of the inner pentagon FGHIJ can also be expressed in terms of X:

area FGHIJ = area BFE - 2y = area ABE - 23’ = 4x - I

Having taken the problem to bed with me, the next key step was to conjecture that the inner and outer pentagons are similar. Because of the parallelisms, it is easy to show that corresponding angles are equal, but it is necessary also to show that the ratios of corresponding sides are equal.

SOLVINGTHEPENTAGONPROBLEM 153

IH/CD = IH/BH because BCDH is a parallelogram and IH/CD = y/(x + y) because triangles ABI and AIH with Bl and IH as bases have the same altitude, so the ratio of the bases is the same as the ratio of the areas. The same argument shows that the ratio of any pair of corresponding sides is likewise

Y/(X + y), or (1 - 2x)/(1 - x).

At this point, I had no clear plan of a proof in mind; I was just exploring. However, it occurred to me that 1 could express the ratio of the areas of the inner and outer pentagons in two ways, leading to the equation:

4x - I _ (1 - 2x12

4-x (I - x)2

(The expression on the right reflects the fact that the areas of similar figures are proportional to the squares of corresponding sides). This produces the equation:

8x' - 29x2 + 23x - 5 = 0

Because 2x + y = 1, we are looking for solutions between 0 and 112, and a quick check showed that this equation has one solution in this interval. This implies that x is fixed, and so, therefore, is the area of the original pentagon. I went to sleep.

In the morning, it occurred to me that I could check the solution by working out x for the special case of a regular pentagon. This is straightforward and leads to the solution:

x = l/2 - l/2 tan 18” tan 36”

which a calculator allowed me to compute as about 0.382. I plugged this value into the equation and was pleased to find that it works!

(It occurred to me that this could provide a rather difficult examination ques- tion, namely: “Show that l/2 - l/2 tan 18” tan 36” is a root of the equation 8x3 - 29x2 + 23x - 5 = 0.” Sawyer (1955) wrote about “reconstructing an exam- iner” (p. 34) from questions like this (i.e., inferring from a question how the examiner had hit on it in the first place).

On a later visit to Rutgers, Bob Davis asked me to write this proof up. This prompted me to return to the question of whether there are asymmetrical penta- gons satisfying the stated conditions. I was able to extend the argument devel- oped by Jerry Goldin and referred to earlier.

Start with a triangle ACD and construct DX parallel to CA and CY parallel to DA. BIHE is parallel to CD, as before, and the same argument as used earlier proves there is a position for BIHE that makes AB parallel to CE. It remains to show that then AE is also parallel to BD. After a little thought, I saw that:

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Area EAB = Area ABC (because AB and CE are parallel)

= Area BCD (because AD and BC are parallel) = Area CDE (because BE and CD are parallel) = Area DEA (because CA and DE are parallel)

Therefore AE and BD are also parallel. So there is a whole class of asymmetrical pentagons with the property under

investigation. A question I may well return to in the future is whether the invariance of area for such pentagons can be demonstrated through a construction using a software package such as Geometer’s Sketch Pad.

Postscript On seeing the alternative solution by Shteingold (1995), I was reminded of the link between the regular pentagon and the golden ratio (@), which I had previ- ously not remembered. In fact, what I call x equals 2 - o, which is a root of the equation x2 - 3x + 1 = 0. The cubic equation referred to in my proof factorizes to (8x - 5)(x* - 3x + 1).

Having found my original solution, I had not thought much more about the problem, nor had I looked for a simpler solution. However, looking again at the proof, I immediately saw that I had introduced unnecessary complications. Re- ferring to the first figure, I showed that IH/CD = y/(x + y) = (1 - 2x)/( 1 - x). But HI/CD also = AI/AC = xl 1 by a similar argument about areas and bases of triangles. So (1 - 2x)/( I - x) = x, whence x2 - 3x + 1 = 0 immediately. This exploitation of the tight interrelationship, echoing what you get from the golden ratio rectangle leading to the equation x2 - x - 1 = 0, feels much more right.

REFERENCES

Sawyer, William W. (1955). Prelude to mathematics. London: Penguin. Shteingold, Semion (1995). One solution of the pentagon problem. Journal of Muthematica/ Behav-

ior, /j(3), 299-302.