Who Invented Euler's Circle?

Actually, we are not concerned with a particular circle (like the Arctic Circle), but one circle for every conceivable triangle. As shown in Figure 1, any triangle ABC spawns (at least) three other triangles: A*B*C*, A'B'C', and A''B''C'', the first of these being formed by the midpoints of the sides, the second by the feet of the altitudes, and the third by the points midway between the vertices and the so-called orthocentre H of ABC. As every triangle has a unique circle containing its three vertices (its circumcircle), we would have three derived circumcircles K*, K', and K''. The subject of the present article is the amazing fact that these all turn out to be one and the same! Since "orthocentre" is not exactly a household word, an explanation is in order. The diagram hints that H is the intersection of the altitudes (perpendiculars dropped from each vertex to the opposite side), but why should these three lines, unlike the sides of ABC itself, intersect in a single point?

Figure 2 shows why. If you focus on the thin red line sprouting from midpoint C* of AB, you will readily agree that any point on this perpendicular bisector of AB must be equidistant from A and B (why?). By the same token, all points of the perpendicular bisector of BC are equidistant from B and C. The point of intersection of these two bisectors, traditionally labelled O, is therefore equidistant from of A, B, and C. In other words, the circle of radius OA (=OB=OC) is the circumcircle K of ABC. Now comes a surprise: the three perpendicular bisectors of ABC are exactly the altitudes of A*B*C* (because AB is parallel to A*B*, etc. -- see?). Since A*B*C* is similar to ABC (why?) the altitudes of ABC must behave like those of A*B*C* and intersect in a single point. Thus the orthocentre H is born -- or discovered. Since the triangle A'B'C' formed by the feet of the altitudes of ABC (blue outline in Figure 1) will play a very important role, it deserves a special name. To bring together the notions of orthogonal (Greek for "right-angled") and feet, it would seem resonable to call it "orthopaedic" -- but tradition obliges us to shorten this to orthic.

The blue dot which shows up between O and H in Figure 2 is called the centroid of ABC and usually denoted by G for "gravity". It is where you would stick a pin through a cardboard triangle to make it spin freely. We shall not call on Archimedes to explain this (though you may do so), but instead define G wihout reference to it as the common meeting point of the three medians AA*, BB*, CC*. A careful study of Figure 3 will convince you that they do indeed meet in a single point -- which cuts each median in the ratio 2:1, and is also the centroid of the smaller triangle A*B*C*. We shall presently see that G lies on OH, which -- though only a segment -- is known as the Euler line of ABC. In fact, the collinearity of OGH is one of the by-products of an article written by Euler in 1765. With that study, Euler launched what might be called the "Modern geometry" of the triangle, as opposed to its "Ancient geometry". The latter was an old hat, amply described in Euclid's Elements, further extended by Hipparchus with his trigonometry, visited by the likes of Heron of Alexandria, and refined by numerous Islamic scientists. It was by no means dead-wood, but of daily beneficicial use in navigation, construction, artillery, etc. -- and even in the exploration of fancier geometric and kinematic phenomena by mathematicians and physicists. But to worry about the humble triangle itself must have seemed like a waste of time: all that could be known about it appeared to be out in the open.

Euler's paper was published by the "Petropolitan" (from St. Petersburg) Academy of Science, under the title "An easy solution to some very difficult geometrical problems" (in Latin). It set out to explore the relations, in terms of distances and angles, between the following four special points associated to any non-equilateral triangle:

(1) the orthocentre, (2) the centroid, (3) the circumcentre, (4) the incentre.

The arguments to be given here are, however, quite unlike Euler's. He was, after all, first and foremost a master of algebraic computation, perhaps the greatest virtuoso of all times on this difficult instrument, and he derived those relations with consummate artistry by means of cleverly concocted formulas -- whereas this column of Pi in the Sky has consistently aimed at side-stepping algebra to bring up the geometric rear.

To see the collinearity OGH, forget triangles for a moment, and just think of a single point G stranded in the middle of the plane. For any other point P in this plane, one could define the "Euler image through G" (a term ad hoc just invented a minute ago) of P to be the point P* which lies on the line PG beyond G but only half as far away from G as P. Putting it differently, P* is obtained from P by turning the segment GP around G by 180 degrees, and then shrinking it by 1/2 to get P*G. If you keep this firmly in mind, you will see that every segment PP* is cut by G in the ratio 2:1. It will also be clear that A* is the Euler image (through G) of A, and so on; in fact all, of A*B*C* can be thought of as the Euler image of ABC.

Since the two stages of creating P* from P -- namely the 180 degree turn as well as the shrinkage by 1/2 -- both preserve angles (why?), the whole transition from some planar struture to its Euler image is conformal (i.e., angle-preserving). In particular, the altitudes of A*B*C* are the Euler images of the altitudes of ABC, and their intersection O must be H* . Thus O and H lie on either side of G on a single line, the latter twice as far from G as the former. Since the wondrous circle K* is the Euler image of the circumcircle K, its centre O* is the Euler image of O. Just like O itself, O* also gets a special label, namely N. Now we know four points of the Euler line: O, G, N, H in that order. As the segment OG, which is 1/3 of OH, must be 2/3 of OO*=ON, the latter must be 1/2 of OH. In other words,

Everything said so far would apply verbatim to triangles with an obtuse angle. Indeed, if you can look past the the confusing lines and letters of Figure 4, you will see a triangle ABC with an obtuse angle at B. Now the feet of the altitudes from A and C lie on the extensions of the sides BC and CA, respectively, and the orthocentre H lies outside the bluish halo of the circumcircle. This makes for a nice and long Euler line OH, on which G and N are clearly visible. For a first time through, you should probably skip Figure 4 and the text that goes with it. All it does is justify what is stated under "Upshot" a couple of paragraphs further down, namely that obtuse angles can be circumvented. It looks messy but it is only a question of sorting things out -- no tricky thinking -- you can can come back to it later.

Transferring your attention from ABC to the triangle AHC, you will notice that B is the orthocentre of the latter (see?), in other words H and B have switched places. In fact, the vertices of the triangles A*B*C*, A'B'C', and A''B''C'' associated with ABC play an interesting game of musical chairs with the corresponding points associated with AHC. For example, A' is the foot of the perpendicular from C onto AH, while C' is the foot of the perpendicular from A to CH. Hence A'B'C' is the orthic triangle of both ABC and AHC (though A' and C' switch positions). For the triangles A*B*C* and A''B''C'', things are not as simple: their vertices are permuted within the sextuple A*B*C*A''B''C'', the corresponding one for AHC being C''B*A''C*B''A*. The reader is urged to identify, at least partially, what goes where -- it's fun. But please don't expect the Euler lines of ABC and AHC to coincide: the only point they have in common is N. Upshot: to show that A*B*C*, A'B'C', A''B''C'' lie on a single circle, we never need to consider obtuse triangles like ABC, because we can transfer the drama (with the same cast of characters) to the acute triangle AHC. Why is that acute? Because of the right angles AC'C and AA'C, the angles HCA and CAH are acute; so is AHC, since it is caught in the quadrilateral A'HC'B with two right angles and one obtuse angle. Why do we want it acute? Because our arguments will rest on the principle that any vertex is as good as any other.

To prove the promised identity K*=K'=K'', we have to reach into geometric prehistory for a theorem which seems to have been known forever -- almost certainly to Thales, who was (with Pythagoras) one of the founders of Greek geometry. In school texts, it goes by the handle of "Angles Subtended by a Chord" or "Circle Theorems", and it says:

You may look this up in a book or an electronic resource to get an idea of what it is saying. But in many of these places, the reasoning is horribly convoluted, and it is much more fun to figure it out by counting up angles in the isosceles triangles PQO, PON, and NOQ. For now, we require only two corollaries of this useful old-timer:

Corollary (1) follows from the theorem by taking N to be the apex of an isosceles right triangle with hypotenuse PQ. For Corollary (2), you could let PQ and NR be the two parallel sides of the trapezoid. Then there are are several ways of concluding the congruence of the triangles PRQ and QNP from that of the segments PR and QN. It is worthwhile to reflect on these and to fill in all details of the proof.

The fact that A', B', and C' lie on K* does not seem to be explicitly mentioned in Euler's paper (unless my Latin is too weak to detect it), but it was noted decades later by the distinguished French geometers Brianchon and Poncelet. To see that C' lies on K*, imagine the circumcircle (not shown in Figure 5) of the right triangle AC'C. Corollary (1) implies that AC is one of its diameters, whence B* is its centre and B*C' one of its radii. This makes B*C' congruent to B*A and hence to A*C* (see?). Corollary 2 now says that all vertices of the trapezoid A*B*C*C' lie on a single circle -- which can only be the circumcircle K* of A*B*C*. Analogous arguments, using different right triangles (which ones?), will pull B' and A' onto K*.

For the initially promised result, it remains to bring K'' into the story. This was done by a contemporary of Brianchon and Poncelet, a French mathematician with the remarkable name of Olry Terquem. People who know how to pronounce that name therefore tend to refer to the circle K*=K'=K'' as Terquem's Circle, while Terquem himself had named it the circle of the nine points (in French). In English this has become the nine-point circle, a puzzling label, since everyone knows that a circle has more than nine points.

Figure 6 is meant to show why A'' lies on K*=K (this equation having been proved already). If you draw a parallel to BH through C*, you will create a triangle AC*A'' (gray) which is half the size, by linear measure, of the triangle ABH (since C* is the midpoint of AB). Being parallel to BH and hence to BB', the segment C*A'' is perpendicular to AC as well as to C*A*. In other words, A''C*A* is a right triangle with hypotenuse A''A* -- but so is the triangle A''A'A*, for even more obvious reasons. Corollary (1) says that A"A* is a diameter of the circumcircles of both A''C*A* and A''A'A*, which must therefore coincide. Now we have A'', C*, A', and A* on a single circle which can only be the circumcircle of C*A'A*, i.e., the circle K*=K'. Again, analogous arguments also put B'' and C'' on this circle K*, which has O* (the midpoint of the Euler line) as its centre and half the radius of K.

This seems to be sufficient reason to call it "Euler's circle", especially since it was Euler who began to study triangles in these terms. Other names like "Terquem" or "Feuerbach" are used only by those who know how to pronounce them, and "nine-point" is somewhat misleading -- though it explains why the midpoint of the Euler line is usually labelled N.

And where did Feuerbach -- whoever that was -- come in? Young Karl Feuerbach put the icing on this cake in 1822, by showing that K* was tangent to the incircle as well as to the three "exocircles" of ABC. But thereby hangs another tale.

The sides of the quadrilateral are the bases of four isosceles traingles, each with apex at O. At each of its vertices, the quadrilateral has an angle formed by the base angles of two neighbouring isosceles triangles. In the upper diagram, four different colours are used for those base angles, and it is plain as day that any two vertices belonging to the same diagonal have exactly one angle of each colour. Therefore, < M + < N = < P + < Q, and since all four angles must add up to 360 degrees, each pair of opposite angles (such as < M + < N) must yield 180 degrees. In otherwords, Opposite angles in a cyclic quadrilateral are supplementary. The equation < M + < N = < P + < Q (with its conclusion) also holds in the lower diagram, provided the angles are counted correctly. The problem is that the traiangle NOP (with red base angles in the upper diagram) now lies outside the quadrilateral, and has its base angles subtracted instead of added at P and N. Thus < P consists of a full yellow angle, like the one at M, minus the base angle OPN, and < N consists of a full blue angle, like the one at Q, minus the base angle PNO. (In fact, we could have avoided this split into two cases, because strictly speaking < MPN = < MPO + < OPN in both diagrams, if you take into account that the angles on the right of the equation "turn" in the same direction for one and opposite directions for the other.) What we really need, however, is a criterion for deciding whether some given point R does or does not lie on a certain circle. Here is what we can prove: Consider a segment PQ and points R and M on opposite sides of the line PQ. Then R lies on the circumcircle of the triangle PMQ if and only if the angles PMQ and QRP are supplementary. Indeed, if R does lie on that circle, we can let it play the role of N in our previous considerations, and supplementarity is assured. Suppose, then, that we know only that the angles PMQ and QRP are supplementary, but nothing about the position of R. Let N be the point of intersection of the line QR with the circle in question. If N were not equal to R, the angles QRP (given) and QNP (from previous result) could not both be supplementary to angle PMQ.

Note that all points N on the circular arc on one side of the chord PQ must yield the same angle QNP, namely the one which is supplementary to PMQ (imagine M fixed). Hence the criterion just derived can be reformulated by substituting "equal" for "supplementary" as the last word, and placing R and M on the same side of PQ instead of opposite sides. This leads directly to results quoted in the text following Figure 4.

At the Greater Vancouver Regional Science Fair of 2007, Bill Pang, a Grade 11 student from Sir Winston Churchill High School in Vancouver, presented three explorations involving orthic triangles. One of them is a proof of Euler's formula

relating the radii R and r of the circumcircle and incircle, respectively, to the distance d between their centres. This does follow directly from Euler's 1765 paper, and hence would be most suitable for inclusion in the present article. Unfortunately, the length of the latter does not allow a leisurely presentation. What follows is therefore a compression of Bill's reasoning, which might not at all be to his liking, and might even be a misrepresentation. In that case, the author of these lines will just have to eat crow.

Our first task is to establish two facts: (a) the orthocentre H of a triangle ABC is the incentre of its orthic triangle, and (b) every triangle is orthic for a suitable acute triangle ABC. To begin with, let us consider some acute triangle, a point P inside it, and perpendiculars PA', PB', PC' dropped to the sides. This divides ABC into three quadrilaterals, which we call "skew-kites", each with two right angles opposite each other. If we draw line segments (purple) from P to the vertices, they divide each skew-kite into two right triangles, making its four vertices lie on a circle. In the case of the skew-kite PA'CB', this makes the two red-dot angles equal (see?), and ditto for the red-square angles in PC'BA'. Note that the red-dot at C is complementary to the angle CPB', as the red-square at B is to B'PC. Hence A'P bisects the angle C'A'B' if and only if the angles BPC' and CPB' are equal. Together with its analogues for B'P and C'P, this implies that A'P, B'P, and C'P are angle bisectors in A'B'C' if and only if A'PA, B'PB, and C'PC are collinear triples.

(a) If P is the orthocentre of ABC, those triples will be collinear, and P will be the intersection of the angle bisectors (hence the incentre) of its orthic triangle. (b) Given any triangle A'B'C', draw lines through its vertices which are perpendicular to the respective angle bisectors. These lines will intersect in pairs and form an acute (why?) triangle ABC. Because of the bisections, the lines AA', BB', and CC' will all go through P, which makes them altitudes in ABC.

Because of (b), it suffices to prove Euler's formula for the orthic triangle A'B'C'. Because of (a), H is the incentre of the latter. The radius r of its incircle is the side opposite A' in the (green) right triangle on the hypotenuse of HA'. Since its angle at A' is equal to the angle B'A*A'' (why?), and since the triangle with those same vertices has a right angle at B' (as A''A* is a diameter of Euler's circle), the right triangles on the hypotenuses HA' and A''A* are similar. Therefore r is to HA' as A''B' is to 2R. But A''B' is a radius of the circumcircle of AHB' (see?), hence congruent to A''H. Therefore r : HA' = A''H : 2R, or 2rR = (A''H)(HA'). We now consider the narrow triangle A''HN which is divided into two right triangles A''PN (black) and NPH (green). Of course HN = d, and its square is the sum of those on PH and PN, the latter again being the difference of the square on A''N and that on A''P. Since NP is parallel to A*A', and N halves A''A*, the segments A''P and PA' are congruent. Abbreviating PA' = A''P = u, and PH = v, as well as remembering that A''N represents R as NP does d, you'll easily boil all this down to d2 = R2 - (u+v)(u-v) = R2 - (A''H)(HA') = R2 - 2rR.