CONCURRENCY THEOREMS FOR PENTAGRAMS

Larry Hoehn

Although concurrency theorems for cevians of triangles are very well known, there appear to be few comparable results for either simple polygons or for star polygons. A short history of regular star polygons, or regular polygrams, is given in [1, pp. 44–45]. Most of this history is prior to 1678 when the Italian mathematician, Giovanni Ceva (1647–1736), published his famous Ceva’s theorem. In this note we give what may be a new proof of Ceva’s theorem and use the same techniques to prove several results about pentagrams.

By definition a cevian of a triangle is a line segment from one vertex to a point on the opposite side which is not an endpoint. In the same manner we define a cevian of a pentagon as a line segment from one vertex to a point on the opposite side which is not an endpoint. The most ubiquitous examples of cevians of triangles are medians, angle–bisectors, and altitudes; there does not, however, appear to be any special named cevians for pentagons. We make frequent use of the following.

\underline{Lemma 1}. If \overline{CD} is a cevian of △ABC (see Figure 1), then

\frac{AD}{DB} = \frac{AC sin ACD}{BC sin DCB} .

Figure 1.

\underline{Proof}. By the law of sines in △ADC and △BDC, respectively, we have AD/AC = sin ACD/ sin ADC and DB/BC = sin DCB/ sin BDC. Dividing the first equation by the second, noting that sin ADC = sin BDC, and simplifying yields the result.

As an application of this lemma, we prove the following:

\underline{Theorem 1}. (Ceva’s Theorem): If cevians \overline{AD}, \overline{BE}, and \overline{CF} of △ABC are concurrent at a point P, then

\frac{AF}{F B} · \frac{BD}{DC} · \frac{CE}{EA} = 1,

and conversely.

Figure 2.

\underline{Proof}. With notation as in Figure 2 and by Lemma 1, we have

\frac{AF}{F B} = \frac{P A sin^{}1̸}{P B sin^{}2̸},

\frac{BD}{DC} = \frac{P B sin^{}3̸}{P C sin^{}4̸},

and \frac{CE}{EA} = \frac{P C sin^{}5̸}{P A sin^{}6̸}.

By multiplying these three equations, noting that vertical angles are congruent, and simplifying we obtain the result. With minor variations this proof also holds when P is outside △ABC. The converse, which is sometimes considered a part of Ceva’s theorem, can be proved in the usual way.

From an examination of approximately thirty available “college geometry” textbooks, it was found that the methods of proof of Ceva’s theorem involved either areas of triangles, similar triangles, or else Menelaus’ theorem. None used the above approach which appears (to this author) to be a much simpler method.

Next we use this approach to prove a generalization of Ceva’s theorem for pentagons. We note that this result is also given as a problem in [1, p. 67].

\underline{Theorem 2}. Let ABCDE be a convex pentagon with a point P inside the pentagon such that there exist cevians from each vertex through P, to points A′, B′, C′, D′, and E′ on the opposite sides (see Figure 3), then

\frac{AD′}{D′B} · \frac{BE′}{E′C} · \frac{CA′}{A′D} · \frac{DB′}{B′E} · \frac{EC′}{C′A} = 1,

and conversely, if four cevians are concurrent at P and the formula holds, then the fifth cevian is concurrent at P.

Figure 3.

\underline{Proof}. Proceeding as in the proof of Ceva’s theorem and using Lemma 1 five times we have

\frac{AD′}{D′B} · \frac{BE′}{E′C} · \frac{CA′}{A′D} · \frac{DB′}{B′E} · \frac{EC′}{C′A}

= \frac{P A sin^{}1̸}{P B sin^{}2̸} · \frac{P B sin^{}3̸}{P C sin^{}4̸} · \frac{P C sin^{}5̸}{P D sin^{}6̸} · \frac{P D sin^{}7̸}{P E sin^{}8̸} · \frac{P E sin^{}9̸}{P A sin^{}1̸0}

= 1.

Conversely, suppose that cevians \overline{AA′}, \overline{BB′}, \overline{CC′}, and \overline{DD′} are concurrent at P. We must show that \overline{EE′} also passes through P. Let \overline{EP} be extended through P to meet \overline{BC} at E^{∗}. Then by the first part of this theorem we have

\frac{AD′}{D′B} · \frac{BE^{∗}}{E^{∗}C} · \frac{CA′}{A′D} · \frac{DB′}{B′E} · \frac{EC′}{C′A} = 1.

By equating this formula with the formula of our hypothesis, and simplifying we obtain BE^{∗}/E^{∗}C = BE′/E′C. By adding one to each side of this equation we obtain BC/E^{∗}C = BC/E′C which implies E^{∗} = E′. Hence, \overline{EE′} is concurrent with the other cevians.

As an aside we note that we can obtain a trigonometric form of Ceva’s theorem for pentagons by using the law of sines in △AP B, △BP C, △CP D, △DP E, and △EP A, respectively (Figure 3) to obtain

\frac{sin P AB}{sin P BA} = \frac{P B}{P A} , \frac{sin P BC}{sin P CB} = \frac{P C}{P B} , \frac{sin P CD}{sin P DC} = \frac{P D}{P C} ,

\frac{sin P DE}{sin P ED} = \frac{P E}{P D} , and \frac{sin P EA}{sin P AE} = \frac{P A}{P E} .

By multiplying these five equations and simplifying, we have

\frac{sin P AB}{sin P BA} · \frac{sin P BC}{sin P CB} · \frac{sin P CD}{sin P DC} · \frac{sin P DE}{sin P ED} · \frac{sin P EA}{sin P AE} = 1.

Next we give a new proof of a result that appeared in [2]. Not only is the new proof shorter and easier to follow, but it also shows that the result can be easily generalized.

\underline{Theorem 3}. Let ABCDEF GHIJ be a pentagram (Figure 4), then

\frac{AB}{BC} · \frac{CD}{DE} · \frac{EF}{F G} · \frac{GH}{HI} · \frac{IJ}{JA} = 1.

Figure 4. \underline{Proof}. By using the law of sines in △ABC, △CDE, △EF G, △GHI, and

△IJA, respectively, we have

\frac{AB}{BC} = \frac{sin ACB}{sin BAC} , \frac{CD}{DE} = \frac{sin CED}{sin DCE} , \frac{EF}{F G} = \frac{sin EGF}{sin F EG},

\frac{GH}{HI} = \frac{sin GIH}{sin HGI} , and \frac{IJ}{JA} = \frac{sin IAJ}{sin JIA}.

By multiplying these five equations, noting that vertical angles are congruent, and simplifying we obtain the result.

Finally, by combining the above theorems we have the following theorem. \underline{Theorem 4}. Let ABCDEF GHIJ be a pentagram (Figure 5) in which the

“diagonals” \overline{BG}, \overline{DI}, \overline{F A}, \overline{HC}, and \overline{JE} are concurrent at some point P inside pentagon ACEGI, then

\frac{sin^{}1̸}{sin^{}2̸} · \frac{sin^{}3̸}{sin^{}4̸} · \frac{sin^{}5̸}{sin^{}6̸} · \frac{sin^{}7̸}{sin^{}8̸} · \frac{sin^{}9̸}{sin^{}1̸0} = 1,

and conversely, if four diagonals are concurrent and this formula holds, then the fifth diagonal is also concurrent at this point.

Figure 5.

\underline{Proof}. Let B′, D′, F′, H′, and J′ be the respective intersections of the diagonals of the pentagram with the corresponding sides of pentagon ACEGI as shown in Figure 5. By Lemma 1 for △ABC, △CDE, △EF G, △GHI, and △IJA, respectively, we have

\frac{AB′}{B′C} = \frac{AB sin^{}1̸}{BC sin^{}2̸}, \frac{CD′}{D′E} = \frac{CD sin^{}3̸}{DE sin^{}4̸}, \frac{EF′}{F′G} = \frac{EF sin^{}5̸}{F G sin^{}6̸},

\frac{GH′}{H′I} = \frac{GH sin^{}7̸}{HI sin^{}8̸}, and \frac{IJ′}{J′A} = \frac{IJ sin^{}9̸}{JA sin^{}1̸0}.

By substituting these five equations into the formula,

\frac{AB′}{B′C} · \frac{CD′}{D′E} · \frac{EF′}{F′G} · \frac{GH′}{H′I} · \frac{IJ′}{J′A} = 1

of Theorem 2, and rearranging factors we have

(\frac{AB}{BC} · \frac{CD}{DE} · \frac{EF}{F G} · \frac{GH}{HI} · \frac{IJ}{JA})

· (\frac{sin^{}1̸}{sin^{}2̸} · \frac{sin^{}3̸}{sin^{}4̸} · \frac{sin^{}5̸}{sin^{}6̸} · \frac{sin^{}7̸}{sin^{}8̸} · \frac{sin^{}9̸}{sin^{}1̸0}) = 1.

Since the first set of factors is equal to one by Theorem 3, the result follows.

For the proof of the converse let diagonals \overline{BG}, \overline{DI}, \overline{F A}, \overline{HC}, and \overline{JE} meet the sides of pentagon ACEGI at B′, D′, F′, H′, and J′, respectively. Suppose that the first four diagonals are concurrent at P and let J^{∗} be the intersection of \overline{EP} and \overline{AI}. Additionally, let ^{}1̸1 = ^{}I̸JJ^{∗} and ^{}1̸2 = ^{}J̸^{∗}JA.

Since \overline{AF′}, \overline{CH′}, \overline{EJ^{∗}}, \overline{GB′}, and \overline{ID′} are concurrent at P, then by Theorem 2,

\frac{AB′}{B′C} · \frac{CD′}{D′E} · \frac{EF′}{F′G} · \frac{GH′}{H′I} · \frac{IJ^{∗}}{J^{∗}A} = 1.

But

\frac{AB′}{B′C} = \frac{AB sin^{}1̸}{BC sin^{}2̸}, \frac{CD′}{D′E} = \frac{CD sin^{}3̸}{DE sin^{}4̸}, \frac{EF′}{F′G} = \frac{EF sin^{}5̸}{F G sin^{}6̸},

\frac{GH′}{H′I} = \frac{GH sin^{}7̸}{HI sin^{}8̸}, and \frac{IJ^{∗}}{J^{∗}A} = \frac{IJ sin^{}1̸1}{JA sin^{}1̸2}

by Lemma 1. By substituting these five formulas into the formula above and rearranging terms we have

(\frac{AB}{BC} · \frac{CD}{DE} · \frac{EF}{F G} · \frac{GH}{HI} · \frac{IJ}{JA})

· (\frac{sin^{}1̸}{sin^{}2̸} · \frac{sin^{}3̸}{sin^{}4̸} · \frac{sin^{}5̸}{sin^{}6̸} · \frac{sin^{}7̸}{sin^{}8̸} · \frac{sin^{}1̸1}{sin^{}1̸2}) = 1.

But the first set of factors equals 1 by Theorem 3. By equating

\frac{sin^{}1̸}{sin^{}2̸} · \frac{sin^{}3̸}{sin^{}4̸} · \frac{sin^{}5̸}{sin^{}6̸} · \frac{sin^{}7̸}{sin^{}8̸} · \frac{sin^{}1̸1}{sin^{}1̸2} = 1

with the formula in the hypothesis and simplifying we have

\frac{sin^{}9̸}{sin^{}1̸0} = \frac{sin^{}1̸1}{sin^{}1̸2}.

Hence,

\frac{IJ sin^{}9̸}{JA sin^{}1̸0} = \frac{IJ sin^{}1̸1}{JA sin^{}1̸2}.

Substituting via Lemma 1 yields

\frac{IJ′}{J′A} = \frac{IJ^{∗}}{J^{∗}A}.

Hence, J′ = J^{∗}. Therefore diagonal \overline{EJ} passes through P so that all five diagonals are concurrent.

By the nature of the above proofs it should be clear that all the theorems can be generalized to polygrams with an odd number of sides and that some can be generalized to arbitrary polygrams. This is left to the interested reader.

\underline{References}

1. H. Eves, A Survey of Geometry (rev. ed.), Allyn and Bacon, Inc., Boston, 1972.

2. L. Hoehn, “A Menelaus–type Theorem for the Pentagram,” Mathematics Magazine, 66 (1993), 121–123.

Larry Hoehn Mathematics and Computer Science Department Austin Peay State University Clarksville, TN 37044–0001 email: hoehnl@apsu01.apsu.edu

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