As a companion to the equiangular point (or the Fermat point, F) we have just discussed, there are two significant points in a triangle made popular by the French mathematician Henri Brocard (1845–1922). Before we can locate these points we must first establish how to construct a circle (of course, using only an unmarked straightedge and a compass) containing a given point and tangent to a given line (not containing that point), since we will be using this construction to locate this new point.

A NECESSARY CONSTRUCTION

Consider the point A and the line l, as shown in figure 4-1. We will show how to construct a circle through point A and tangent to line l. First, we construct a perpendicular line to l at the planned point of tangency, B. Then, we construct the perpendicular bisector of AB to meet the previous perpendicular at point O. We can then draw the circle with center O and radius OA, which gives us our required circle. This construction will now be used several times to locate some truly amazing points in a triangle, known as the Brocard points.

BROCARD POINTS

We are now ready to locate a triangle's Brocard points. Consider the following:

In figure 4-2 you will find the three circles constructed as we did above. Each is tangent to a side of the triangle ABC and contains a vertex of the triangle. That is,

 

circle P is tangent to side AC at point C and contains vertex B,

circle Q is tangent to side AB at point A and contains vertex C, and

circle R is tangent to side BC at point B and contains vertex A.

 

Two amazing things occur in this construction: First, the three circles meet at a common point, the first Brocard point, B₁, and, second, the angles made from this point of concurrency have the following relationship: ∠B₁AB = ∠B₁BC = ∠B₁CA = ω.

A second Brocard point, B₂, can also be found in this triangle ABC by taking the point of concurrency of these circles (see figure 4-3) :

 

circle S is tangent to side AB at point B and contains vertex C,

circle T is tangent to side BC at point C and contains vertex A, and

circle U is tangent to side AC at point A and contains vertex B.

 

The Brocard points lead to lots of further advanced relationships that would be a bit beyond the scope of this book, yet they can be appreciated, nevertheless, for the unusual properties they present.

There is also a time when one concurrency generates another concurrency, as we have seen earlier. In figure 4-4, the three lines AP, BP, and CP are concurrent at point P. If we now take three line segments that make equal angles with the adjacent sides of the angle of the triangle so that

 

∠QAC = ∠PAB (= ρ), ∠QBC = ∠PBA (= σ), and ∠QCA = ∠PCB (= ),

 

then QA, QB, and QC are also concurrent. This quite-surprising result is one that would not be easily anticipated yet continues to exhibit the secrets of the beauties that are embedded in triangles.

THE MIQUEL POINT

The Brocard points were established from concurrent circles. Circles and triangles can also determine other interesting concurrencies. Consider a randomly selected triangle ABC and three randomly selected points, D, E, and F, one on each side of the triangle, as shown in figure 4-5. We then draw three circles, where each contains two of the three side points and the included vertex. Lo and behold, the three circles share a common point M—the point of concurrency. This point is known as the Miquel point, named after the nineteenth-century French mathematician Auguste Miquel, who published this theorem in Liouville's Journal in 1838.¹ Yet, as is so often the case in mathematics, there is strong evidence that others had already known about this fantastic relationship—such as the Scottish mathematician William Wallace (1768–1843) and the Swiss mathematician Jakob Steiner (1797–1863).

This concurrency can be easily justified if we recall that the opposite angles of a quadrilateral inscribed in a circle (called a cyclic quadrilateral) are supplementary. Consider the case when M is inside triangle ABC, as shown in figure 4-6. (Bear in mind that a similar argument can be made for a point M outside the triangle.) Points D, E, and F are any points on sides AC, BC, and AB, respectively, of triangle ABC. Let circles Q and R, determined by points F, B, and E; and D, C, and E, respectively, meet at M.

To justify this concurrency we draw in figure 4-6 FM, EM, and DM. In cyclic quadrilateral BFME, we have ∠FME = 180° – ∠B, since opposite angles of a cyclic quadrilateral are supplementary—that is, they have a sum of 180°. Similarly, in cyclic quadrilateral CDME, ∠DME = 180° – ∠C. By addition, ∠FME + ∠DME = 360° – (∠B + ∠C). Therefore, ∠FMD = ∠B + ∠C. However, in triangle ABC, ∠B + ∠C = 180° – ∠A. Thus, ∠FMD = 180° – ∠A and quadrilateral AFMD is cyclic. Thus, point M lies on all three circles.

In figure 4-7, we see that Miquel's theorem also holds true when the Miquel point, M, is outside the original triangle ABC.

THE MIQUEL TRIANGLE

Were that all there was to say about the concurrency of these three circles, we would already have a beautiful phenomenon. However, as you might have expected, there are more gems to be found in this configuration. In figure 4-8, triangle DEF is called the Miquel triangle. Consider the following: The segments joining the Miquel point of a triangle to the vertices of the Miquel triangle form equal angles with the respective sides of the original triangle. This means that in figure 4-8, ∠AFM = ∠CDM = ∠BEM (= µ), as well as ∠ADM = ∠CEM = ∠BFM (= 180° – µ).

This can be easily justified. Because quadrilateral AFMD is cyclic (see figure 4-8), ∠AFM is supplementary to ∠ADM. But ∠ADM is supplementary to ∠CDM. Therefore ∠AFM = ∠CDM, whereupon it follows that ∠BFM = ∠ADM. The remaining equalities are arrived at by applying the same argument to the other cyclic quadrilaterals.

We say that a triangle is inscribed in a second triangle if each of the vertices of the first triangle lies on the sides of the second triangle. We can then state the following further application of the Miquel point M: Two triangles inscribed in the same triangle and having a common Miquel point are similar. We can justify this in the following manner.

In figure 4-9, we show triangle DEF and triangle D′E′F′, which have the same Miquel point M. We have just established that ∠MFB = ∠MDA, and ∠MF′A = ∠MD′C. Therefore, we have ΔMF′F ~ ΔMD′D, and ΔMD′D ~ ΔME′E. It follows that ∠FMF′ = ∠DMD′ = ∠EME′. By addition, we get ∠F′MD′ ∠FMD, ∠F′ME′ = ∠FME, and ∠E′MD′ = ∠EMD.

Also as a result of the above similar triangles, we get

We can establish two triangles similar if two pairs of corresponding sides are proportional and the included angles congruent. This gives us the following pairs of similar triangles:

 

ΔF′MD ~ ΔFMD,

ΔF′ME′ ~ ΔFME, and

ΔE′MD′ ~ ΔEMD.

 

Therefore, and . Thus, . Similarly, .

This proves that ΔDEF ~ ΔD′E′F′, since the pairs of corresponding sides are proportional.

We can still show another nifty relationship emanating from the Miquel triangles. That is, the centers of Miquel circles of a given triangle determine a triangle similar to the given triangle. Again, the justification is rather straightforward.

We can begin to justify this relationship by drawing common chords FM, EM, and DM , as shown in figure 4-10. Let PQ meet circle Q at point N and RQ meet circle Q at point L. Since the line of centers of two circles is the perpendicular bisector of their common chord, PQ is the perpendicular bisector of FM, so that . Similarly, QR bisects so that .

Now, the central angle and the inscribed angle . Therefore, ∠NQL = ∠FBE.

In a similar fashion it may be proved that ∠QPR = ∠BAC. Thus, ΔPQR ~ ΔABC, since their corresponding angles are equal.

You might wish to investigate the Miquel triangle of an equilateral triangle, or the Miquel triangle of a right triangle, as they yield some interesting properties.

CONCYCLIC POINTS

Auguste Miquel is also responsible for another truly amazing relationship that also involves intersections of circles. We begin with a randomly drawn (irregular) pentagram (a five-cornered star) as shown in figure 4-11. We will then draw the circumscribed circle for each of the five outside triangles (shaded). Amazingly, the intersection points (A, B, C, D, and E) of the five circles always lie on the same circle (dashed line).² Such points are said to be concyclic.

There is a somewhat similar “five-circle theorem” that states that if five consecutively intersecting circles, whose centers lie on a circle, are drawn so that one of their two intersection points with their adjacent circle also lies on the circle of centers, then by joining the remaining intersection points a (not necessarily regular) pentagram PQRST is formed with its vertices each lying on one of the circles. This is shown in figure 4-12.

MORE CONCURRENT CIRCLES

Before we leave the topic of concurrency of circles, we should admire another configuration that opens the door to lots of further such investigations. In figure 4-13, we have a randomly drawn triangle ABC and on each of its sides we have constructed the reflection of triangle ABC. Consequently, all four triangles are congruent (ΔABC ΔA′BC ΔAB′C ΔABC′). First, we notice that the three circumscribed circles of the reflected triangles are concurrent at the point Y. Then when we draw lines joining each of the centers of these circles with the remote vertex of the original triangle ABC, we find that these lines are also concurrent at a point X.

The beauty here is that the configurations that generate concurrency are limited only by one's imagination. Suppose we take the original triangle ABC and rotate it 180° about the midpoint (M_a, M_b, M_c) of each of its sides. We will then get the darker-shaded triangles: ΔA″BC, ΔAB″C, and ΔABC″, shown in figure 4-14. Once again, the circumscribed circles of each of the rotated triangles (shaded) are concurrent at point Y.

And furthermore, also the lines joining each of the centers, P, Q, and R, of these circles with the vertices A″, B″, and C″, respectively, of the rotated triangle ABC, are also concurrent at this point Y.

In addition, as in the previous example, the lines joining each of the centers of these circles with the remote vertex of the original triangle ABC are also concurrent at a point X. We leave other such concurrency findings to the reader.

To close out this chapter on concurrent circles there is a cute series of concurrent circles and points first published by the American mathematician Roger A. Johnson (1890–1954),3 where for a given triangle three concurrent congruent circles are drawn (these are often called Johnson circles), with each containing two vertices of the triangle. In figure 4-15, we have three concurrent congruent circles with centers J_a, J_b, and J_c, with each containing two vertices of the triangle ABC. The following surprising properties hold true:

• The three concurrent circles are unique and are congruent.
• The three concurrent Johnson circles have the same radius as the circumscribed circle of the original triangle ABC, and their centers, J_a, J_b, and J_c, lie on a circle with the same radius as that of the three concurrent circles.
• The point of intersection of the three concurrent Johnson circles is the orthocenter H (point of intersection of the altitudes) of the original triangle ABC.
• The triangle formed by the centers of the concurrent circles, triangle J_aJ_bJ_c (Johnson triangle) is congruent to the original triangle ABC.

With this amazing relationship, we have a fine way to close our chapter on concurrent circles.