mathematics - A triangular Quidditch field

A Quidditch field is usually in the shape of an oval, a hundred and eighty feet wide and five hundred feet long. But today the Gryffindor house team is training on a smaller field in the shape of a triangle, with small towers in its corners $A,B,C$. The angles at $A$, $B$, $C$ are respectively $14^{\circ}$, $62^{\circ}$, $104^{\circ}$.

Harry is standing and resting in a point $H$ on the sideline $AC$ of the field, as he suddenly sees the Golden Snitch popping up at point $S$ on the sideline $AB$. Harry notes in a flash that $\angle HBC=50^{\circ}$ and that $\angle SCB=94^{\circ}$.

Question: How large is the angle $\angle CSH$?

(The answer to this question will be an integer. A good solution will clearly explain the reason why an integer number shows up here.)

Answer

Here is our situation. We are given the following setup:

I have omitted the "top" of the triangle (vertex $A$). We know the following angles:

$$\begin{align} \angle ABC = \angle SBC &= u + v = 62^{\circ} \\ \angle ACB = \angle HCB &= s + t = 104^{\circ} \\ \angle HBC &= u = 50^{\circ} \\ \angle SCB &= t = 94^{\circ} \\ \end{align}$$

We wish to find:

$$\angle HSC = x = {?}^{\circ}$$

We can solve for the unknown angles $w$, $y$, and $z$ by noticing two facts: the angles in a triangle add up to a constant ($\pi$) and the diagonals of the quadrilateral $HSBC$ create two pairs of triangles that share an angle. Therefore:

$$\begin{align} t + u &= x + y \\ s + z &= v + w \\ \end{align}$$

To get a third equation, note that the total of the interior angles of any quadrilateral add up to $2\pi$:

$$s + t + u + v + w + x + y + z = 2\pi$$

Solving this system of three equations gives us:

$$\begin{align} w &= \pi - t - u - v \\ y &= t + u - x \\ z &= \pi - s - t - u \end{align}$$

Now, we can use the law of sines to write a second set of relations:

$$\begin{align} \frac{\overline{HS}}{\sin s} &= \frac{\overline{CH}}{\sin x} & \frac{\overline{CH}}{\sin u} &= \frac{\overline{BC}}{\sin z} & \frac{\overline{BC}}{\sin w} &= \frac{\overline{SB}}{\sin t} & \frac{\overline{SB}}{\sin y} &= \frac{\overline{HS}}{\sin v} \end{align}$$

Multiplying them together, the lengths cancel:

$$\sin s \sin u \sin w \sin y = \sin t \sin v \sin x \sin z$$

Adding in our substitutions from before (and remembering $\sin(\pi-x)=\sin x$):

$$\sin s \sin u \sin (t+u+v) \sin (t+u-x) = \sin t \sin v \sin x \sin (s+t+u)$$

(This approach is adapted from the webpage Angular Angst that Fimpellizieri posted in this comment on the question.)

Now proving that the answer is $x=34^{\circ}$ (oops, did I say that out loud?) reduces to proving the following trigonometric identity:

$$\sin 10^{\circ} \sin 50^{\circ} \sin 110^{\circ} \sin 156^{\circ} = \sin 12^{\circ} \sin 34^{\circ} \sin 94^{\circ} \sin 154^{\circ}$$

or equivalently:

$$\begin{multline} \sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{11\pi}{18}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{17\pi}{90}\right) \sin\left(\frac{47\pi}{90}\right) \sin\left(\frac{77\pi}{90}\right) \end{multline}$$

We can expand some of the fractions into some interesting sums:

$$\begin{multline} \sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right) \sin\left(\frac{5\pi}{6}+\frac{\pi}{45}\right) \end{multline}$$

Remember that $\sin x=\sin(\pi-x)$; this allows us to put our identity into an even more intriguing form:

$$\begin{multline} \sin\left(\frac{2\pi}{15}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right) \end{multline}$$

Let's see if we can't make a useful identity with the pattern that we see:

$$\sin\left(\frac{\pi}{6}-\alpha\right) \sin\left(\frac{\pi}{6}+\alpha\right) \sin\left(\frac{\pi}{2}+\alpha\right) \\ = \left(\frac{1}{2}\cos\alpha - \frac{\sqrt{3}}{2}\sin\alpha \right)\left(\frac{1}{2}\cos\alpha + \frac{\sqrt{3}}{2}\sin\alpha \right)\cos\alpha \\ = \frac{1}{4}\left(\cos^2\alpha - 3\sin^2\alpha\right)\cos\alpha \\ = \frac{1}{4}\left(4\cos^3\alpha - 3\cos\alpha\right) \\ = \frac{\cos(3\alpha)}{4}$$

With this in mind, we can collapse our identity to:

$$\sin\left(\frac{2\pi}{15}\right)\cos\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)$$

This is beginning to look a lot more convenient. Using the double angle formula on $\sin(2\cdot\pi/15)$ and evaluating $\cos(\pi/3)$ we obtain:

$$2\sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)\frac{1}{2} = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)$$

Which is obviously true.

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Since we know that the solution is unique, this proves that the solution is $34^{\circ}$. As to why, the only answer I can offer is "because it makes the beautiful cancellations above possible."