10 Favorite Theorems of Mystery

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Theorems of Mystery

2004 T3

New Orleans

March 2004

Michael Keyton

Illinois Mathematics and Science Academy (IMSA)

1500 W. Sullivan Road, Aurora, IL 60506

[email protected]



Theorems of Mystery

The conditions of collinearity, concurrence, equidistance, parallelism, and

perpendicularity are very special in the study of euclidean geometry. If objects are chosen

at random in the system, each of these conditions has a zero probability of existence.

Definition: A theorem of mystery is a result that has considerable structure with minimal

hypotheses.

I have assembled my favorite collection of theorems of mystery not necessarily in

order of significance and certainly subject to my unique, personal, capricious nature. The

theorems are not necessarily covered in high school geometry courses, but could easily beincluded with the ability now to make excellent drawings using computer programs such

as Cabri II.

The theorems shown today will be:

9) Varignon’s Theorem

8) Midpoints of midpoints

7) Orthopole

6) Carnot’s Theorem

5) Parallel, Circle, Isogonals

4) General Quadrilateral

3) The Third Pedal Triangle

2) Miquel’s Theorem

1) Brianchon’s Theorem

0) Pascal’s Mystic Hexagram Theorem

References:[1] Coxeter, H.S. and S.L. Greitzer, Geometry Revisited , MAA, 1967.

[2] Eves, Howard, A Survey of Geometry, Allyn and Bacon, 1963.

[3] F. G.-M., Exercices de Géométrie, Editions Jacques Gabay, 1991.[4] Altshiller-Court, N., College Geometry, Johnson Publ. Co., 1925.

[5] Honsberger, R., Episodes in 19th and 20th Century Euclidean Geometry, MAA, 1995.



Theorems of Mystery

Number 9

Varignon’s Theorem: The midpoints of the sides of a quadrilateral are the vertices of a

parallelogram.

The first mention of this theorem is in a book by the French mathematician Pierre

Varignon (1654-1722).

For many years this was my favorite theorem of mystery. It takes an unordered object andcreates a very structured one. In addition, since its discovery was relatively late for such a

simple discovery (1700), I ask what other theorems lie undetected? It is sometimes calledthe Varignon quadrilateral.

There are many corollaries of this theorem, which makes it even more interesting.

C1: If the diagonals of the quadrilateral are congruent, then the Varignon quadrilateral is

a rhombus.

C2: IF the diagonals of the quadrilateral are perpendicular, then the Varignonquadrilateral is a rectangle.

C3: Given a parallelogram, there are infinitely many quadrilaterals that have it as their

Varignon quadrilateral.

C4: The four triangles inside the quadrilateral but outside the Varignon quadrilateral can be rearranged to form the Varignon quadrilateral.

A related theorem to this, another construction of a parallelogram is:

Theorem: Construct equilateral triangles on the sides of a quadrilateral, alternating

internally and externally. The third vertices of the triangles form a parallelogram.



Number 8

These two are treated as one, for they are immediate extensions of 9).

Theorem: (a) The midpoints of the midpoints of the opposite sides of a quadrilateral are

the same. (b) The point in (8a) is the midpoint of the midpoints of the diagonals of thequadrilateral.

One of the questions always is “Why” is this true. Both (8) and (9) go back to a basic

theorem, The Midsegment Theorem.

Theorem: The line through midpoints of two sides of a triangle is parallel to the third

side and the distance between the midpoints is half the length of the third side.



Number 7: Orthopole

Theorem: From the vertices of a triangle, perpendiculars are drawn to a line. From theintersection points of the perpendiculars and the line, perpendiculars are drawn to the

sides opposite the respective vertices. These lines are concurrent.

The point of concurrence is called the orthopole.

I particularly like this theorem for it takes non-collinearity, uses perpendicularity goingthrough collinearity to concurrence.

There are many consequences of this theorem.

Corollary: If the line contains the circumcenter of the triangle, then the orthopole lies on

the nine-point circle of the triangle.



Number 6:Carnot’s Theorem:

Theorem: Give a circle that intersects a triangle in 6 points, two on each side. The product of the ratios of the distances around the triangle is the same, regardless of the

direction. Let circle with center P intersect ∆ ABC at the points M, N, P, Q, S and T

where the order is A-M-N-B, B-P-Q-C, and C-S-T-A.

Corollary: Take one line from each vertex through one of the intersection points of the

circle and the opposite side of the triangle. If these lines are concurrent, then the other

triple of lines are concurrent.



Number 5: Parallel, Circle, Isogonals

Theorem: If chord XY of circle O is parallel to BC of inscribed ∆ ABC in circle O,

then ∠ BAY ∠ XAC.

This implies that the bisector of ∠ BAC is the same as the bisector of ∠ YAX.

Definition: AY and AX are isogonals of each other if they make equal angles with the

bisector of an angle. (and consequently with the sides of the angle.)

Theorem: If AY bisects ∠ BAC where Y is on the circumcircle with center O of ∠ BAC,

then YO is the perpendicular bisector of BC

Such a simple problem, but such a staggering result when phrased as:

Theorem: The bisector of an angle of a triangle and the perpendicular bisector of the

opposite side of a triangle intersect on the circumcircle of the triangle.

Theorem: The isogonals of concurrent cevians are concurrent.



Number 4: General Quadrilateral:

Theorem 1: Circumcircles of the four triangles that form a general quadrilateral areconcurrent.

Theorem 2: The orthocenters of the four triangles are collinear

These two theorems are consequences of Miquel’s theorem, which will be shown in (2).

These seem to be difficult; however, the proofs are quite simple by just looking at the

problems with the proper view.

Another basic property of the quadrilateral is the following surprising result:

Theorem: The diagonals of a quadrilateral intersect the line containing the intersectionsof the opposite sides of the quadrilateral harmonically. (for ABCD, P = AB ∩ CD and

Q = AD ∩ BC, S = BD ∩ PQ, T = AC ∩ PQ in order P, S, Q, T; PS/ST = PT/TQ.



Number 3:The Third Pedal Triangle

The pedal triangle of a point with respect to a triangle is the triangle formed by thefeet of the altitudes from the point to the lines containing the sides of the triangle.

Theorem: Given a point and a triangle, the pedal triangle of the point with respect to(wrt) the pedal triangle wrt the pedal triangle wrt the triangle is similar to the triangle.

This triangle is affectionately known as the third pedal triangle. (J. Neuberg, 1840-1926)

The fundamental theorem about pedal triangles is: The sums of the squares of the lengths

of every other subsegment of the sides of the triangle formed by the feet of the

perpendiculars that determine the pedal triangle of a point are equal.



Number 2: Miquel’s Theorem and Consequences:

Theorem: Given three points, one on each line containing a side of a triangle. The threecircles containing a vertex and the two points on the sides of its angle are concurrent. The

point is called Miquel’s point for the three points and the triangle.

This theorem has a large number of consequences. A few are:

Corollary 1: The angles formed clockwise using the Miquel point, a side point and the

vertex point are congruent.

Corollary 2: The triangle formed by the centers of the three circles is similar to the

original triangle.Corollary 3: The triangles formed by any set of three points that produce the sameMiquel point are similar.

Corollary 4: If the three points are the midpoints of the sides of the triangle, then the

three circles are congruent, and the Miquel point is the _________.

Corollary 5: The three points are collinear iff the Miquel point is on the circumcircle of

the triangle.



Number 1: Brianchon’s Theorem

Theorem: If a hexagon is circumscribed about a circle, then the main diagonals areconcurrent.

This theorem due to C.J. Brianchon (1760-1854) couples nicely with the next theorem,

Pascal’s Mystic Hexagram. Brianchon’s proof uses projective geometry and can begeneralized to an ellipse. He used the duality of points and lines. There is a nice euclidean

proof due to A. S. Smorgorzhevskii. [1]



Number 0: Pascal’s Mystic Hexagram Theorem

The primary theorem has to be Pascal’s theorem from which there are over 400corollaries.

Theorem: The intersection points of the opposite sides of a hexagon inscribed in a circleare collinear.

This is the dual of Brianchon’s theorem.

Blaise Pascal (1623-1662) proved this theorem by the time he was 16. The line is calledthe Pascal line. The converse of the theorem is not true; however, by extending this

theorem to a conic, both it and its converse are true (6 points lie on a conic iff the

intersections of the opposite sides of the hexagon formed by the 6 points are collinear.)

Since there are 60 ways to order the 6 points on the circle, there are 60 different Pascal

lines that result from this theorem. These 60 lines are concurrent in threes in 20 points,

called Steiner points. The Steiner points lie in quartets on 15 lines, called Plücker lines.The Pascal lines are also concurrent in trios in another set of 60 points, called Kirkman

points. Each Steiner point and 3 of the Kirkman points are collinear in the 20 Cayley

lines. They are concurrent in quartets in 15 Salmon points. [2]



Number –1: Keyton – Jia

I’m not really sure why I am so fond of this theorem, for it was just recently discovered

and proved so has not lasted long enough to attain the status of mystery. It was found by

Keyton, solved for special cases; then generalized and proved by Rouzhou Jia. (2003)

The question was under what circumstances will the corresponding sides of a triangle

inscribed in a triangle inscribed in a triangle intersect the corresponding sides of the

original triangle in collinear points. The first author discovered and proved that if the twoinscribed triangles were chosen from the medial, orthic, or intouch then the points were

collinear. His student, the second author, generalized and proved the following result.

Theorem: Let Ak (k=1, 2, 3) be three non-collinear points in the place. Let Bk ∈ Ak-1Ak .

Definek 1 k

k

k k 1

A B

B A

+

−

λ = (subscripts modulo 3), directed distances. Let Ck ∈ Bk-1Bk+1. Define

k 1 k

k

k k 1

B CC B

+

−

µ = (subscripts modulo 3), directed distances. Let k k 1 k 1 k 1 k 1D A A C C

− + − += I .

Dk (k = 1, 2, 3) are collinear iff 1 2 3 1 2 31λ λ λ µ µ µ =

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