# Solution to puzzle 154: Triangle in a trapezoid

In trapezoid ABCD, with sides AB and CD parallel, DAB = 6° and ABC = 42°.   Point X on side AB is such that AXD = 78° and CXB = 66°.  If AB and CD are 1 inch apart, prove that AD + DX − (BC + CX) = 8 inches.

Dropping a perpendicular (of length 1) from D to AX, and similarly from C to BX, we see that:

• DX = cosec 78°
• BC = cosec 42°
• CX = cosec 66°

Thus, we are asked to prove that cosec 6° + cosec 78° − cosec 42° − cosec 66° = 8.

Notice that, for x = 6°, 78°, −42°, −66°, and 30°, sin 5x = ½.  We now express sin 5x in terms of sin x.

De Moivre's theorem states that for any real number x and any integer n,

cos nx + i sin nx = (cos x + i sin x)n

Setting n = 5, expanding the right-hand side using the binomial theorem, and equating imaginary parts, we obtain

 sin 5x = sin5x − 10 sin3x cos2x + 5 sin x cos4x = sin5x − 10 sin3x (1 − sin2x) + 5 sin x (1 − sin2x)2,  since sin2x + cos2x = 1 = 16 sin5x − 20 sin3x + 5 sin x

This result can also be obtained by means of trigonometric identities.  Setting s = sin x, it follows that the five distinct real numbers,
sin 6°, sin 78°, −sin 42°, −sin 66°, and sin 30° = ½, (1)
are roots of the equation 16s5 − 20s3 + 5s = ½, or, equivalently, of 32s5 − 40s3 + 10s − 1 = 0. (2)
By the Fundamental Theorem of Algebra, (2) has exactly five roots, up to multiplicity, and hence these must be precisely the distinct roots identified in (1).

Since s = ½ is a root of (2), the equation factorizes:
(2s − 1)(16s4 + 8s3 − 16s2 − 8s + 1) = 0,
yielding the quartic equation whose roots are sin 6°, sin 78°, −sin 42°, and −sin 66°.

As s = 0 is not a root of this quartic equation, we may divide by s4, and, setting t = 1/s, obtain
t4 − 8t3 − 16t2 + 8t + 16 = 0,
an equation whose roots are cosec 6°, cosec 78°, −cosec 42°, and −cosec 66°.

By Viète's formulas, the sum of the roots of this equation is 8.

Thus, AD + DX − (BC + CX) = 8 inches.

Source: Original; inspired by a similar problem in M500 Magazine, Issue 210.