Point P lies inside ABC, and is such that PAC = 18°, PCA = 57°, PAB = 24°, and PBA = 27°. Show that ABC is isosceles.

Let PBC = x°, so that BCP = 54° − x.

Since PCA alone is greater than CAB, if ABC is to be isosceles, we must have CAB = ABC. We will prove this below.

Applying the law of sines (also known as the sine rule) to triangles PAB, PBC, and PCA, respectively, we obtain:

Hence sin 27° · sin(54° − x) · sin 18° = sin 24° · sin x° · sin 57°. (1)

Clearly, ABC is isosceles PBC = 15°. We first show that x = 15° is a solution to the above equation; that is,

sin 27° · sin 39° · sin 18° = sin 24° · sin 15° · sin 57°. (2)

Firstly, applying product-to-sum trigonometric identities, we obtain

4 sin a · sin b · sin c | = 2 sin a [cos(b − c) - cos(b + c)]. |

= sin(a − b + c) + sin(a + b − c) − sin(a − b − c) − sin(a + b + c). |

Hence, from (2) we obtain

sin 6° + sin 48° − sin(−30°) − sin 84° = sin 66° + sin(−18°) − sin(−48°) − sin 96°.

Since sin 96° = sin(180° − 96°) = sin 84°, this simplifies to

½ + sin 18° = sin 66° − sin 6°.

Then, using a sum-to-product trigonometric identity, we have

sin 66° − sin 6° = 2 cos 36° sin 30° = cos 36°.

Writing sin 18° = cos 72°, we hence obtain cos 36° − cos 72° = ½. (3)

Now consider regular pentagon ABCDE, with unit length sides. Draw diagonal EC, and drop a perpendicular from D to AB, meeting at X. Extend BA to O.

By symmetry, OAE = 360°/5 = 72°, EC is parallel to AB, DX bisects AB, and DEC = 180° − 2×72° = 36°.

Hence AX = −cos 72° + cos 36° = ½, which establishes (3).

Since all of the above steps are reversible, we have proved (2).

Having found one solution, we must show that the solution is unique.

Equation (1) is of the form sin x° / sin (54° − x) = k, for 0° < x < 54°, where k is a positive constant.

Since sin x° is a strictly increasing and continuous, and sin (54° − x) a strictly decreasing and continuous, function over that range, sin x° / sin (54° − x) is a strictly increasing and continuous function, and therefore takes the value k exactly once. Hence x = 15° is the only solution in the range.

Therefore ABC is isosceles.

Lest the above solution seem somewhat ad hoc, as with many trigonometric problems there is a more uniform approach that uses complex arithmetic. For example, to establish sin 18° + sin 30° − sin 54° = 0, equivalent to (3), above, we would take the following approach.

We use Euler's formula, which states that, for any real number t, e^{it} = cos t + *i* sin t.

Hence sin 2t = [e(t) − e(−t)] / 2*i*, where for convenience e(t) denotes e^{2it}.

Applying this formula to sin 18° + sin 30° − sin 54° = 0, after cancelling the (2*i*) terms, we get

- e(3/60) − e(−3/60) + e(5/60) − e(−5/60) − e(9/60) + e(−9/60) = 0.

We now invoke the periodicity rule

e(t + n) = e(t), for all t and all integers n,

and the cyclotomic rule

e(t) + e(t + (1/n)) + e(t + (2/n)) + ... + e(t + ((n−1)/n)) = 0, for all t and all integers n > 1.

Applying these rules yields

- e(3/60) + e(27/60) + e(5/60) + e(25/60) + e(39/60) + e(51/60) = 0.

Now note that by the cyclotomic rule, e(5/60) + e(25/60) = −e(45/60) = e(15/60).

Hence we have e(3/60) + e(27/60) + e(15/60) + e(39/60) + e(51/60) = 0.

This equation is true by the cyclotomic rule. Since all of the above steps are reversible, we have proved (3).

We could also apply this approach directly to equation (2), expanding both sides to a product of eight terms using the identity e(x) · e(y) = e(x + y). The difference is a linear combination of terms that we wish to show is equal to zero. This would obviate the need to use trigonometric identities.

Source: Traditional