Solution to my little trig puzzle

I've posted a trig puzzle in my previous blog post "My current plans, and a little (or not) puzzle".  And then, I proceded to post it in my future high school's Mu Alpha Theta organization's Discord.

And the responses were... surprising, yet somewhat anticipated.  I've sufficiently hidden the underlying diagram, and so the puzzle worked!!!!

The problem

Recall the simple algebraic statement of the puzzle:

Find $\tan\theta$ if

$$\sin^2\theta + x^2 + x\sqrt3 \sin\theta = 3,$$

$$\cos^2\theta + x^2 + x \cos \theta = 4{,}$$

where $0<\theta<\frac\pi2$ and $x>0$.

The attempts

Now, some of y'all may start to recognize how these two equations look like Law of Cosines.  And some of the people in that Discord did correctly piece the entire geometric configuration together, which looks roughly like this:

Now, another otter was further solving this problem... and suggested coordinate bashing.  When I told them (I haven't asked for pronouns...) that this would not be optimal, they suggested using $\sqrt{bc}$ inversion at $B$ in the above diagram..

[proceeds to post another complicated diagram]

And of course, they were very orz people for even knowing these technique which I have absolutely no idea of, and I was glad to see my little problem attract so many creative solutions.  But this was not the original solution I intended; depending on what you know, either one of the alternate solutions or my original solution may be simpler.

My solution

So let me drop my writeup.

Note that both equations look like the result of using the Law of Cosines on some triangles.  Using $\cos120^{\circ}=-\frac12$ and $\cos150^\circ=-\frac{\sqrt3}2$, we can rewrite the equations as

$$\sin^2\theta-2(\sin\theta)x\cos150^{\circ}=(\sqrt3)^2,$$

$$\cos^2\theta+x^2-2(\cos\theta)x\cos120^{\circ}=2^2.$$

which is the result of the Law of Cosines on these two triangles:

The variable $x$ is present in both figures as an edge, and the angles $150^\circ$ and $120^\circ$ sum to $270^\circ$, exactly a right angle away from $360^\circ$.  Since right angles are useful, we assemble the two right triangles together in this way, so that the right angle appears:

Since $\angle ADC$ in the figure above is right, and $AD=\cos\theta$ and $DC=\sin\theta$, $\angle DAC=\theta$ and $AC=1$.  Also, because the lengths of $AB=2$, $BC=\sqrt3$ and $AC=1$ form a $1$-$\sqrt 3$-$2$ ratio, $\triangle ABC$ is a 30-60-90 triangle with $\angle BAC=60^\circ$, $\angle ACB=90^\circ$, and $\angle ABC=30^\circ$.

Now, since $ABCD$ is a quadrilateral, we know that its interior angles sum to $360^\circ$. Since $\angle ABC=30^\circ$, and the reflex angle $\angle ADC=270^\circ$, we have

$$\angle DAB+\angle DCB=360-(120+150)-30=60^\circ=\angle BAC=\angle DAB+\angle DAC.$$

Subtracting $\angle DAB$ from the leftmost and rightmost parts of the equation, we get $\angle DCB=\angle DAC=\theta$.  We also know $\angle BAD=60^\circ-\theta$ from the same quadrilateral angle sum.

Marking everything into our diagram, we get this:

Since we now know two sides and two angles in each of them, we use the Law of Sines to reformulate the two triangles.  As they share a side, some algebra can be simplified later; we have

$$\frac x{\sin\theta}=\frac{\sqrt3}{\sin150^\circ}$$

for $\triangle BDC$ and

$$\frac x{\sin(60^\circ-\theta)}=\frac{2}{\sin120^\circ}$$

for $\triangle BDA$.

By plugging in the known $\sin$ values and solving for $x$, we get

$$x=2\sqrt3\sin\theta$$

and

$$x=\frac{4\sqrt3}3\sin(60^\circ-\theta).$$

Equating the two expressions for $x$,

$$\frac{4\sqrt3}3\sin(60^\circ-\theta)= 2\sqrt3\sin\theta.$$

Using the $\sin$ subtraction formula on the left,

$$\frac23(\sin60^\circ\cos \theta-\cos60^\circ\sin\theta)=\sin\theta.$$

Plugging in our known trig values,

$$\frac23\left(\frac{\sqrt3}2\cos\theta-\frac12\sin\theta\right)=\sin\theta$$

From here, we rearrange to get $\frac43\sin\theta=\frac{\sqrt3}3\cos\theta$ and divide both sides by $\frac43\cos\theta$ to get $\boxed{\tan\theta=\frac{\sqrt3}4}$.

What now?

Well, note each of my essays usually starts with a personal section.  For this personal section... well I was procrastinating on Chinese math homework and spent 40 minutes late night trying to solve the geometric formulation I had which was from the problem using the Law of Cosines.  In other words, I reformulated another problem into the prompt of my puzzle with some angle chasing.  It was also the night before a TMSCA competition.

It turns out that some of my angle chasing was spurious and made the problem harder to solve with my reformulated system.  Oh well.  40 minutes I'll never get back.  But then I figured out how to solve the problem properly, wrote it down and was gonna go to sleep.

I figured however the law of cosines is a much better puzzle, so I wrote it down and brought it to the TMSCA event the next morning, and challenged some people with it before the event started.  During lunch I met Dr. Alan Haynes from U of H by way of him being a parent of someone I knew at school, and we checked the problem, and I posted it on Math.StackExchange to find alternate solutions.  After seeing all the alternate solutions are not that easy and I needed a blog puzzle the other day I posted it, and reposted it to the discord as well.  After some days of discussion, I decided to put the writeup on my blog...

Such distant yet beautiful memories of the past, where marching band doesn't consume 3 hours of your day.  Not that I don't like marching band, though...

But sometimes, the best math puzzles come from the management missteps...

Math.stackexchange link: https://math.stackexchange.com/questions/5124583/