Thirty degrees without a circle

In 1922 Edward Langley, the founding editor of The Mathematical Gazette, published a problem that has since been passed around as a test of ingenuity. Take a triangle $ABC$ with angles $20°$ at $A$ and $80°$ at $B$ and $C$. Draw a line from $B$ making $60°$ with the base $BC$, meeting $AC$ at $D$, and a line from $C$ making $50°$ with the base, meeting $AB$ at $E$. Find the angle $BDE$.

The answer is $30°$, exactly, and the computation in the first figure confirms it to nine decimal places from coordinates alone. What makes the problem famous is that the obvious method fails. Chasing angles around the figure, using only that the angles of a triangle add to $180°$, leaves one equation short, and every attempt to close the gap by guessing ends in a circle — and there is no circle.

Why a whole number is surprising #

Angle-chasing problems usually have whole-number answers for a reason, and the reason is the inscribed-angle theorem: two points on a circle subtend the same angle from everywhere on the far arc. If four points $B$, $C$, $D$, $E$ lie on a circle, then the angle $BDE$ equals the angle $BCE$, because both stand on the chord $BE$. When the given angles are whole degrees and the four points are concyclic, the unknown angle is one of the given ones, and its being a whole number is no surprise at all.

Here that explanation is unavailable. The angle $BCE$ is $50°$ and the angle $BDE$ is $30°$; if the four points were on a circle the two would be equal. So the 30 is not inherited from any given angle. It is what the geometry calls an adventitious angle: a whole-number answer that arrives with no circle to deliver it.

To see how special that is, fix the cevian from $C$ at $50°$ and turn the one from $B$ through every angle from $1°$ to $79°$, computing the angle $BDE$ each time.

The angle varies smoothly, and at only five of the 79 whole-degree positions of $BD$ is it a whole number of degrees. One of those, $BD$ at $50°$, gives $50°$, and is explained by a circle: the cevians are then symmetric, the four points are the corners of an isosceles trapezoid, and an isosceles trapezoid is cyclic. The other four are adventitious. Langley’s $60°$ giving $30°$ is one of them, and $20°$, $40°$ and $70°$ give $60°$, $60°$ and $10°$ for equally little apparent reason.

How to tell whether four points are on a circle #

Since everything turns on whether $B$, $C$, $D$, $E$ are concyclic, it is worth having the tests in view. Four points form a cyclic quadrilateral exactly when opposite angles add to $180°$, or equivalently when a side subtends equal angles at the two vertices across from it — the inscribed-angle theorem read backwards. A third test is metric: two chords $BD$ and $CE$ meeting at a point $X$ come from a circle exactly when $XB \cdot XD = XC \cdot XE$, which is the power of a point, the one number shared by every chord through it. In Langley’s figure the two products differ, by a factor that the coordinates give as about 1.7, so there is no circle and no hope of inheriting the answer from a given angle.

Cyclic quadrilaterals are special in other ways that make the contrast sharper. Among all quadrilaterals with four given side lengths, the cyclic one encloses the most area, Brahmagupta’s formula gives that area from the sides alone, and the angles of a cyclic quadrilateral are determined by three of them. A cyclic quadrilateral is in that sense rigid in its angles, and the inscribed-angle theorem is the statement of that rigidity. An adventitious quadrangle has the same rigidity — all its angles whole when four of them are — without the circle that usually supplies it.

The proof by one extra point #

The classical proof adds a single point and then sees only isosceles triangles. Draw a line from $B$ making $20°$ with the base, meeting $AC$ at $F$.

Triangle $BCF$ has angles $20°$ at $B$, $80°$ at $C$, and so $80°$ at $F$; it is isosceles with $BF = BC$. Triangle $BCE$ has $80°$ at $B$, $50°$ at $C$, and so $50°$ at $E$; it is isosceles with $BE = BC$. So $BE = BF$, and the angle between them, $EBF$, is $80° - 20° = 60°$: triangle $BEF$ is equilateral, and $EF = BF$. Triangle $BFD$ has $40°$ at $B$ (that is $60° - 20°$), and at $F$ the angle $BFD = 180° - 80° = 100°$, so the angle at $D$ is $40°$; it is isosceles with $FD = BF$.

Now $FE = FD$, and the angle between them at $F$ is $180° - 60° - 80° = 40°$, so in the isosceles triangle $EFD$ the base angles are $70°$. The angle $BDE$ is the angle $FDE$ minus the angle $FDB$, which is $70° - 40° = 30°$. Every step is an isosceles triangle; the whole difficulty is in finding $F$, which nothing in the problem suggests. The figure checks all five equal lengths in coordinates — they agree to twelve decimal places — and that check is the honest form of the picture, since equal-looking segments in a drawing prove nothing.

A companion problem in the same triangle, with the cevians at $70°$ and $60°$, gives $20°$, and it is often set as the harder of the two because its auxiliary construction is longer.

Angle chasing is linear algebra, and it runs out #

Why does ordinary angle chasing fail here? Label every angle in Langley’s figure as an unknown. Each triangle in the figure gives a linear equation — its three angles add to $180°$ — each straight line gives one, and each point where several angles fill a full turn gives one. All of these are linear equations with whole-number coefficients, and if they determined the answer, the answer would be a rational combination of the given angles, and whole whenever the given angles were. In Langley’s figure they do not determine it: the system has one degree of freedom left over, the same freedom that lets the curve in the earlier figure vary smoothly as the cevian turns. Angle chasing, which is solving those linear equations, cannot see the answer, because the answer is fixed by something non-linear — the lengths.

That is the general shape of the subject. Linear relations among angles are what sums that a triangulation cannot change and the old temple problems of a total hung in a temple exploit: quantities whose invariance follows from adding angles up in two ways. The inscribed-angle theorem adds exactly one non-linear fact to the linear system — that angles on the same chord of a circle are equal — and with it, every cyclic case closes. The adventitious cases need more: an identity among sines, which is a statement about lengths in disguise, and the auxiliary point $F$ is the device that converts the length information into isosceles triangles, where it becomes angle information again.

Every pair of cevians in the triangle #

How many such coincidences does the $20°$–$80°$–$80°$ triangle hold? With whole-degree cevians from $B$ and $C$ there are $79 \times 79 = 6{,}241$ pairs, and each gives an angle $BDE$ that can be computed and tested.

Ninety-five pairs give a whole-degree answer. Seventy-nine of them lie on a line in the grid and are explained by a circle — for those, $D$ and $E$ are placed so that $B$, $C$, $D$, $E$ are concyclic, and the answer equals the cevian angle at $C$. The other sixteen are adventitious, and they are scattered: Langley’s $(60°, 50°) \to 30°$, the companion $(70°, 60°) \to 20°$, and fourteen more, such as $(20°, 50°) \to 60°$, $(30°, 60°) \to 80°$ and $(65°, 25°) \to 5°$. The pale line is the inscribed-angle theorem drawn as a picture; the dark cells are what it does not cover.

Every quadrilateral #

The triangle is a special case of a more symmetric problem. A convex quadrilateral $PQRS$ with its two diagonals is fixed, up to size, by four angles at the base $PQ$: the angle $a$ between the base and the diagonal $PR$, the angle $b$ between that diagonal and the side $PS$, and similarly $c$ and $d$ at $Q$. Every other angle between sides and diagonals follows, and one of them, the angle $x$ between the diagonal $PR$ and the side $RS$, is the interesting one. If the quadrilateral is cyclic, then $x = c$ by the inscribed-angle theorem. The question is when $x$ is a whole number of degrees for any other reason.

The computation tests every choice of four whole-degree angles for which the quadrilateral exists: 83,653,681 of them. In 1,002,733 the angle $x$ is a whole number of degrees to within $10^{-7}$. In 939,929 of those it is because the quadrilateral is cyclic, $x = c$. The remaining 62,804 are adventitious, and the figure plots them by the two base angles. Once all four angles may vary, they are not rare at all: they fill whole regions of the plot and run along lines, the trace of infinite families of solutions in which the angles vary together and the coincidence persists. Langley’s triangle is a single cell of this picture, and its sixteen coincidences are the adventitious quadrilaterals that fall there. Each quadrilateral is counted several times, once for each way of choosing which side is the base and which way round to read it, so the number of geometrically different quadrilaterals is smaller.

Colin Rigby classified these quadrangles geometrically in 1978, into infinite families and a finite list of sporadic exceptions. The census here does not attempt the classification; it shows the size of the problem and how much of it the families occupy, and it confirms that the circle explains almost every case, nearly fifteen out of sixteen, and leaves a large and structured remainder.

The lines and regions in the plot have a simple meaning. A family is a set of quadrilaterals in which the base angles can be varied together — one up, another down, keeping some combination fixed — while the coincidence survives at every whole-degree step. Langley’s triangle is the special case in which $a + b = 80°$ and $c + d = 80°$, a single cell of the plot, and the sixteen adventitious pairs of the earlier grid are the adventitious quadrilaterals in that cell. Which of them belong to families passing through that cell and which are sporadic is exactly what Rigby’s classification decides; the census shows only that both kinds are present, and that a uniform argument can prove a whole family at once while each sporadic case needs a construction of its own, like the point $F$.

Why the answers are whole numbers #

Every step in the quadrilateral can be written with the law of sines, which measures every side by one diameter, and eliminating the lengths leaves a single equation among sines of the angles, in the form of the trigonometric version of Ceva’s theorem:

$$\frac{\sin a}{\sin b}\cdot\frac{\sin c}{\sin d}\cdot\frac{\sin x}{\sin y} = 1,$$

with $y$ determined by the others. The question “when is $x$ a whole number of degrees, given that $a$, $b$, $c$, $d$ are?” is therefore a question about products of sines of rational multiples of $\pi$, and writing each sine as a difference of two roots of unity turns it into a question about sums of roots of unity that vanish.

That is the surprising connection. The arithmetic that decides Langley’s puzzle is the arithmetic of the sums of roots of unity that add to nothing, whose short cases John Conway and Antonia Jones classified in 1976. Most vanishing sums are unions of regular polygons, rotated, and those give identities that hold along whole families; a few are not, the minimal exceptional sums, and those are where isolated coincidences come from. Bjorn Poonen and Michael Rubinstein used the same arithmetic in 1998 to count the points where three or more diagonals of a regular polygon meet, a problem that is the same equation seen from the polygon’s side: an adventitious angle is a coincidence of three lines in a regular polygon, with angles that are multiples of $180°/n$.

Morley’s trisectors, which always meet in an equilateral triangle, are the other famous surprise of angle chasing, and the comparison is instructive: Morley’s theorem holds for every triangle and is an identity, not a coincidence of particular angles. The difference is precisely the difference between the families and the sporadic cases: an identity holds along a whole continuum, a coincidence only at isolated points.

Measured to a tolerance, not proved #

The whole-degree tests are numerical. An angle is counted as a whole number when it is within $10^{-7}$ of one, which is far below the size of the gap between neighbouring candidate answers at these scales but is not a proof that any particular angle is exactly whole; that comes from the trigonometric identity or from a synthetic proof like the one with the point $F$. Conversely, an angle could in principle be whole only in the limit, outside the tolerance, and be missed. Neither is likely at whole degrees, but the census is a measurement.

The census counts labelled quadrilaterals, and it does not identify which ones are the same figure seen from different bases, nor which lines in the plot are which families. The figure suggests the structure; Rigby’s classification states it. And everything here is in whole degrees: allowing angles that are rational multiples of a degree, or of $\pi$ more generally, gives more solutions and the same structure.

Eighty-three million quadrilaterals, and Rigby’s classification #

Computed here: the angles of Langley’s figure and its companion from coordinates; the five equal segments of the classical proof to twelve places; the angle $BDE$ at every whole-degree position of one cevian; the 6,241 pairs of cevians in the $20°$–$80°$–$80°$ triangle; and the 83,653,681 quadrilaterals with whole-degree base angles, with the cyclic and adventitious cases separated and plotted.

Quoted: Langley’s publication of 1922; the classical proof, which is reconstructed and verified here; Rigby’s classification of adventitious quadrangles; the Conway–Jones classification of vanishing sums of roots of unity; and Poonen and Rubinstein’s count of concurrent diagonals.

Still open: a proof that sees the answer #

The classification is complete: the adventitious quadrangles are known, as families and sporadic cases, and the census is consistent with that description. What is not settled is a question about proofs rather than answers. Each sporadic case can be proved by trigonometry, mechanically, and many by an auxiliary construction like the point $F$; but there is no known general method that produces, for each adventitious quadrangle, a synthetic proof using only isosceles triangles and the inscribed-angle theorem, and for some of the sporadic cases no short synthetic proof is known at all. Whether every such coincidence has one, and how long it must be, is a question about the power of elementary geometry that has been explored by enthusiasts and by automated provers and is not answered.

A related question asks for the analogue on the sphere or in the hyperbolic plane, where the angles of a triangle no longer add to $180°$ and the inscribed-angle theorem fails. Whole-number coincidences there are a different equation, and their classification has not been carried out.

A circle’s worth, and a remainder #

In the $20°$–$80°$–$80°$ triangle, cevians at $60°$ and $50°$ make an angle of exactly $30°$, though the four points are not on a circle; one added point and five equal segments prove it. Among all 83,653,681 quadrilaterals with whole-degree base angles, 1,002,733 have a whole-degree fifth angle: 939,929 because the quadrilateral is cyclic and the inscribed-angle theorem hands the answer over, and 62,804 adventitious, most of them belonging to infinite families and a few to a finite list of exceptions.

The inscribed-angle theorem explains almost every whole-degree answer in angle chasing, and Langley’s puzzle is a member of the small remainder it does not. What explains the remainder is not geometry but arithmetic — the ways roots of unity can add to nothing — and the puzzle that has tested schoolchildren for a century turns out to be a question about cyclotomic fields.