Angular Angst

 

What is the angle denoted by α in the figure below?

 

Documents30

 

It isn't too difficult to determine with simple trigonometry that α is very nearly equal to 30 degrees, but to rigorously prove that it is exactly 30 degrees is not trivial. The problem seems to have been first posed by E. M. Langley in 1922, and has become known as the problem of "adventitious angles", because only for certain special combinations of angles is it possible for all the angles in the figure to be rational multiples of π. In the particular case illustrated above there is a clever geometrical proof, devised by Mercer in 1923. Label the vertices of the quadrilateral A,B,C,D, and construct an additional line AE from the lower left vertex making an angle of 20 degrees with the horizontal, as shown below.

tri30-3

The angles ABD and ADB are both 50 degrees, so |AB| = |AD|. Also, since ADE is 80 degrees and EAD is 20 degrees, it follows that AED is 80 degrees and therefore |AD| = |AE|, and hence |AB| = |AE|. Now, since BAE is isosceles and the angle at A is 60 degrees, we know that ABE is an equilateral triangle, which implies that |AE| = |BE|. In addition, since the angles EAC and ECA are both 40 degrees, we know that |AE| = |CE|, and therefore |BE| = |CE|, so the triangle BEC is isosceles. We also know the angle BEC equals 40 degrees (because BEA equals 60 degrees and AED equals 80 degrees), so it follows that the base angles of BEC are both equal to 70 degrees, and hence the angle BCA equals 30 degrees, which was to be proven.

 

Another solution can be deduced from drawing a line AF through A at 20 degrees from the line AC, and another line EB, and extending the line AB through the point G, as shown in the figure below.

 

 

If a circle is inscribed so that it is tangent to the three lines GB, BE, and EF, the perpendiculars from that center to those three lines will have equal lengths, so the center lies on the bisectors of the angles GAF, BEF, and GBE. (In other words, the angular bisectors of any two exterior angles and the opposite interior angle of a triangle pass through a single point.) Now, by construction and simple angular relations, point C in our figure is the unique point that lies on the bisectors of GAF and BEF, so it must be the center of the inscribed circle, and hence it also lies along the bisector of GBE. Consequently we have μ = ν, and since 50+30+μ+ν = 180, it follows that μ = ν = 50°, and therefore α = 30°, which was to be proven.

 

For an algebraic solution, note that the diagonal going from lower left to upper right makes an angle of 60° with the horizontal, so α + β also equals 60°. Therefore, in order to show that α = 30° it suffices to show that β = 30°, i.e., that the upper side of the quadrilateral makes an angle of 30° with the horizontal. To show this, we begin by assigning variables to the important lengths as shown below. (Since the scale is unimportant, we have normalized so that the horizontal distance between the two altitudes is unity.)

 

 

We have tan(β) = (v – u) tan(80°) where u and v are related by

 

 

For convenience, we will hereafter denote tan(k°) by tk. From the above relations we get

 

 

Therefore we have

 

 

where we have made use of the facts that t60 = 1/t30, t50 = 1/t40, and t80 = 1/t10. Now, substituting for t40 with the trigonometric identity t40 = (t30 + t10)/(1 – t30 t10), we get

 

 

Subtracting t30 from both sides, we have

 

 

Making use of the fact that 1/t302 = 3, this can be written as

 

 

The numerator of the fraction can also be written as

 

 

The quantities in parentheses are simply t20, so the numerator of the fraction in the previous expression is

 

 

where the first term on the left side is simply the tangent of 10° + 20°. Therefore, tan(β) = tan(30°), so we have β = 30° and hence α = 30°, which was to be proven. (Incidentally, this leads to several interesting identities, such as t10 = t20 t30 t40.)

 

In general we can consider an arbitrary quadrilateral and its diagonals, as shown below.

 

 

The law of sines implies that the eight angles around the perimeter satisfy the relation

 

 

Given the values of four consecutive interior angles around the perimeter, such as a, b, c, and d, the quadrilateral is completely determined, so we can solve for the remaining angles. The values of h and e are immediate, as is the sum g+f, but to determine the individual values of g and f requires some trigonometry. Since sin(π−x) = sin(x), we can substitute for e and h in the preceding expression to give

 

 

We can also replace g with b + c − f and expand the resulting sine to give

 

 

Dividing through by cos(f) and solving for tan(f), we have

 

 

Equivalently, by applying the previous analysis to the general quadrangle, we find that the angle β which the upper edge (from the vertex of angle f) makes with the horizontal is

 

 

where ta denotes tan(a) and so on. Expanding the tangents of the summed angles, this can be written in terms of the tangents of the individual angles as

 

 

Furthermore, knowing that f + β = b, we can use the tangent addition formula to give the explicit expression

 

 

With these formulas it’s easy to show that Langley's original construction is far from unique. One of the interesting categories of solutions consists of quadrilaterals with perpendicular diagonals. If we focus on those whose angles are unequal multiples of 10 degrees (i.e., p/18), any such quadrilateral must be comprised of a 10:80, a 20:70, a 30:60, and a 40:50 triangle. It isn't self-evident that four triangles with these shapes can be combined into a single quadrilateral, but in fact they can, in the six distinct ways shown below.

advent rights

 

Langley's original example of adventitious angles involves the upper left case, as illustrated in the figure below:

tri30 right

 

This shows Langley's original quadrilateral ABCD, but the line AB has been extended to the point F, and the horizontal line FC has been drawn. The quadrilateral DBFC is a "10:20:30:40 " quadrilateral with perpendicular diagonals BC and DF.

 

Another example of this kind that is often mentioned along with Langley’s original puzzle is the isosceles triangle shown in the figure below.

 

 

To determine the angle α geometrically, we draw a line from A to the point on BC that is directly below D. By symmetry this line must make an angle of 60 degrees from AB, as shown in the figure below, where we have also indicated several of the angles that can be immediately deduced.

 

 

 

By construction, triangles ABG and DFG are equilateral, and triangle AFC is isosceles. It follows that AF = CF. Also, since the line bisecting angle CAF strikes CF at E, and the line (not drawn) bisecting ACF strikes AF at G, we have AG = CE, and therefore GF = FE. Now, since DF = GF, we see that DFE is isosceles. Combined with the fact that the angle DFE is 80°, it follows that angle DEF is 50°. We know that AEB is 30°, so AED is 20°.

 

Incidentally, if we draw a line bisecting the angle DFE and striking DC at the point H, the point E is an “excenter” of the triangle AFH.  In other words, E lies on the bisectors of two external angles and the opposite internal angle of AFH, as shown below.

 

 

From this we can infer that the angles at H are 60°. This is similar to the second proof of Langley’s original problem described previously, except in this case the relevant angle is of the radius of the circle to the tangent point on the line FH. However, to prove that the line DE is perpendicular to FH, we still need to show that DFE is isosceles, and once we know this, we can already determine the desired angle, without needing to invoke the excenter property of point E.

 

To solve the problem algebraically, we can proceed just as we did with the original problem, which in this case amounts to finding the angle β between vertical and the line DE. We know that AE makes an angle of 70° with vertical, so we expect to find β = 50°, which will imply that the desired angle AED is 20°. Proceeding just as in the previous case, we have tan(β) = (v – u) tan(80°) where u and v are related by

 

 

As before, we will denote tan(k°) by tk. From the above relations we get

 

 

Therefore we have

 

 

where we have made use of the facts that t60 = 1/t30, t70 = 1/t20, and t80 = 1/t10. Now, making the substitutions t30 = (t20 + t10)/(1 – t20 t10) and t20 = 2t10/(1−t102), we get

 

 

By use of the tangent addition formula, we have the trigonometric identity

 

 

Subtracting this from tan(β) and simplifying, we have

 

 

At this point it’s convenient to recall from the article on Linear Fractional Transformations that the quantities tan(kπ/(2n))2 are the roots of the polynomial whose coefficients are alternating terms (with alternating signs) of the nth row of binomial coefficients. For example, we have the trigonometric identities

 

 

Now, the first of these has the factorization

 

 

The second factor with x = t10 is identical to the numerator of tan(β) – tan(50°), which implies tan(β) = tan(50°), so we have β = 50° and hence α = 20°, which was to be proven.

 

In our algebraic solution of Langley’s original problem, we exploited the fact that the resulting angle was 30°, but we could just as well have applied the more general method, working entirely in terms of t10, noting that we have

 

 

and the trigonometric identity

 

 

Taking the difference between these two expressions gives

 

 

Again the numerator vanishes because it contains the same factor as discussed above. This is the minimal polynomial with integer coefficients having the value t10 as a root, so any expression in terms of t10 that vanishes must contain this factor.

 

For one last example, consider the quadrangle shown in the figure below.

 

 

Using the general formula given previously, we can write the tangent of the angle β that the edge CD makes with the horizontal (i.e. parallel to AB) as

 

 

Then we can prove that β = 10° by noting that the difference

 

 

has the very same polynomial in the numerator as the previous examples, which vanishes identically, proving that β = 10° and therefore α = 40°.

 

All the above was based on quadrangles whose angles were multiples of 10°, i.e., integer multiples of π/18, but we can obviously perform the analogous analysis for any quadrangle whose angles are any rational multiples of π.

 

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