Angular Angst
What is the angle denoted by a in the figure below?
It isn't too difficult to determine with simple trigonometry that a 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 p. 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.
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.
It's interesting to explore some of the trigonometric identities that are responsible for this seemingly simple construction. Note that the diagonal going from lower left to upper right makes an angle of 60 degrees with the horizontal, so it is sufficient to show that the upper side of the quadrilateral makes an angle of 30 degrees with the horizontal. To show this, we begin by assigning variables to the important lengths as shown below.
We wish to prove that
Now, by similar triangles, we have h/A = H/C. In addition, we have
Substituting for H and h in equation (1), we see that our objective is to prove that
To evaluate this, we can make use of the relations
to give
Similarly we have
and so
Solving (3) and (4) for the ratios A/B and A/C gives
Inserting these expressions into the putative equation (2), and making use of the facts that
we find that the equality (2) is satisfied if and only if tan(40) is a root of the equation
Using the trigonometric identity
we can substitute for the left hand side of (5) to give
Dividing both sides by 4tan(40), and noting that tan(160) = -tan(20), this becomes
Using the double-angle formula to express the tangent of 40 degrees in terms of the tangent of 20 degrees, we find that this equation is satisfied if and only if
Notice that this polynomial has the same form as (5) except for the sign of the cube term. Substituting for the left hand side again, we get
Dividing both sides by 4tan(20), and noting that tan(80) = 1/tan(10), this becomes
Again using the double-angle formula, this reduces to the condition
Making use of the identity (6) again, and simplifying, we get
Since the numerator of the left hand side is 2tan(10)/tan(20), this is equivalent to
Making the substitution tan(80) = 1/tan(10), this can be written as
Hence in order to prove that a equals 30 degrees it is sufficient to prove this identity. For convenience, let tn denote the tangent of 10n degrees. Then we note the special identities
Now, since 20 = 30 - 10 and 40 = 30 + 10, we can use the addition formula for tangents to re-write the putative equation (8) as
Of course, there is some value of t1 that satisfies the above equation, namely, a real root of the cubic polynomial
Our task is to show that t1 is actually the tangent of 10 degrees, in which case the chain of implication back to the original problem is valid. Let us first define t2 such that arg(t1) + arg(t2) = 30 , which implies the relation
Substituting this into the preceding equation for t1, we find that t2 is a root of the cubic
Let us now define t4 such that arg(t2) + arg(t4) = 60, which implies the relation
Substituting this for t2 in f20, we find that t4 is a root of the cubic
Repeating this procedure once more, we define t8 such that arg(t4) + arg(t8) = 120, which implies the relation
Substituting for t4 in f40, we find that t8 must be a root of the cubic
The coefficients of this polynomial are the reverse of the coefficients of f10, confirming that t8 = 1/t1, consistent with the fact that arg(t1) + arg(t8) = 90.
Now, in order to prove that t1 is the tangent of 10, it would be sufficient to show that the argument of t2 is twice the argument of t1, because t2 is defined such that arg(t1) + arg(t2) = 30. If arg(t2) happens to be twice arg(t1), we have the relation
Substituting for t2 in f20, we find that the resulting equation is satisfied if and only if t1 is a root of the polynomial
This shows that if f10(t1) = f20(t2) = 0, then arg(t2) = 2arg(t1). Together with the fact that arg(t1) + arg(t2) = 30, this implies that arg(t1) = 10, and hence the angle a in the original drawing above equals 30.
Langley's original construction is far from unique. 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+h, but to determine the individual values of g and h requires some trigonometry. Since sin(p-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
One of the interesting categories of solutions consists of quadrilaterals with perpendicular diagonals. If we focus on those whose angles are unequal multiples 10 degrees (i.e., p/18), it's clear that 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.
Langley's original example of adventitious angles involves the upper left case, as illustrated in the figure below:
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.