I agree that it’s a weird problem but it can be solved without knowing trig or any geometry theorems.
Instead, simply assume the problem has an answer. Because point A is unconstrained (aside from being on the circle), the angle it forms must not depend on its position.
Therefore, I move it to be another 60 degrees clockwise from C. With the other points, it forms two equilateral triangles, and line AB bisects angle OBC. BAC must therefore be 30 degrees, and x=2.
I like that kind of reasoning. A math teacher once asked a class I was in “If there were a band of steel that went all the way around the Equator, and it were lifted one foot above the surface, how much would its length increase?”. Most of the class furrowed our brows and started working furiously, in our heads or on paper. But then the teacher added “It turns out, you don’t need to know the size of the Earth”, and at that point, my hand shot up. Because if you don’t need to know the size of the Earth, then you can use a zero-size Earth, and the answer to that one is obviously 2pi feet.
You could come up with one similar to the football problem. A gunner on a battleship needs to swivel his gun starting at the 20 degree mark. His gun swivels at 5 degrees per second. How long will it take to reach the 30 degree mark? Or various other scenarios involving movement through an angle.
I got at it a slightly different way than Tim did:
Because lines OB and OC are both radii of the same circle, we know that they’re equal to each other. We also know that a line drawn from B to C would be equal to OB and OC because it’s subtended by a 60-degree angle, and an equilateral triangle is composed of equal-length sides subtended by 60-degree angles. So BOC is an equilateral triangle.
Let’s draw a line along the diameter of the circle bisecting the 60-degree angle, thus passing through both A and O, making a 30-degree angle with line OB. It also bisects angle BAC.
Let’s use D to refer to the point where the diameter crosses line BC.
If we take the circle radius as a unit length, we know BD is 0.5 units long not only because BC has been bisected, but also because the sin(30) is 0.5.
The cosine of 30 gives us the length of line OD, which is cos(30). The bisection of angle BAC gives us a right triangle, BAD. We know that side BD is equal to 0.5 and AD must be equal to the radius plus line OD, or 1+cos(30), or about 1.866. The tangent of the angle BAD is equal to 0.5/1.866.
To get the angle of BAD, take the inverse tangent of 0.5/1.866, which works out to 15 degrees. BAD is half of BAC, so angle BAC is 30 degrees. After that, the algebraic portion is trivial.
Yep, the steel band puzzle is another good one. I’ve gotten a fair amount of mileage out of the technique. Another I remember off the top of my head was from a difficult Sudoku. I’d narrowed a box down to two numbers. Playing one number would have left the board in an ambiguous state–even after filling in everything else, there would have been two possible solutions. It couldn’t be that, since all Sudokus have one unique solution. Therefore it must have been the other number.
A small addition to the above: A must be on the obtuse side of BOC. That is, it must stay within the circular segment it starts in.
I’m with Chronos; I like this approach a lot.
But aren’t you implicitly assuming the existence of the central angle theorem? And if it’s self evident, why does anyone bother with a proof?
If there were no proof of the central angle theorem, it would not be at all clear that angle BAC doesn’t depend on A’s position aside from being on a circle. For instance, without knowing the central angle theorem, this problem may rest on A being diametrically opposed to a bisection of AOC. It turns out it’s not, but that’s only because proofs of the central angle theorem prove it.
Is your idea a bit tautological or am I being a bit dense? In all sincerity, it really could be either.
I once heard a story about a German kid who used a similar technique to quickly add all the numbers between 1 and 100. I imagine he grew up to be a pretty good mathematician.
Combining things helps illustrate that you actually understand the concepts thoroughly. Unlike the history problem, geometry uses algebra throughout, and so thorough knowledge of it is essential. You can’t do your proofs without understanding algebra.
If you know your math, then artificial problems should not trip you up.
It’s a weaker theorem that I’m assuming (I’m not sure if there’s a name for it): that for any two points B and C, the angle BAC is independent of A’s position (aside from being on one side or the other of the chord BC). But that’s an assumption the problem makes, by virtue of not constraining A.
You do have to be careful in not assuming more than the problem allows; despite BOC being double BAC, I can’t further generalize that it holds true for all angles, just 60 degree ones.
To be clear, if one was building up geometry from scratch, one would have to go through all the steps in proving this the right way. This technique is only legitimate because the problem doesn’t make sense unless the weakened CAT is true.
Which still isn’t the OP’s problem, because it still doesn’t involve the central angle theorem.
So write a problem that uses some of the algebra that geometry uses, instead of artificially introducing some that it doesn’t use.
But if geometry uses algebra throughout, then why add a little extra (and trivial) algebra onto the just-fine-by-itself geometry problem?
I think the complaint is not that the arbitrary algebra makes the problem unsolvable for students, but rather that it makes the problem throb with arbitrariness. It’s patronizing.
You’re right; students who know their math won’t be tripped up by artificial problems. But they will be annoyed, and rightly so.
OK; works for me. Thanks for explaining.
I like your answer, but I’m wondering what your teacher’s original intent was. Was it simply that you could present your answer with the earth’s size as a variable? Or was your teacher just mistaken? If the former, I’d call your teacher’s statement a bit misleading. Which is, I guess, what you did at the time.
Hm. I thought everyone knew the sketches in geometry problems were just rough ways to organize the information you’ve actually been given, and so shouldn’t be taken very seriously, such that eyeballing angles and saying “well, that looks like it’s bigger” is a good way to get a wrong answer, and not just because you failed to show your work.
Maybe that kind of thinking is too close to actual (that is, axiomatic) mathematics for some to grasp.
My 5th grader has been testing up in the 98th percentile for math this past year so I gave him a shot at it.
After examining it a bit he drew a line from A to the center of the circle and out to the other circle edge. Filled in the angles he knew and came up with 30 for the angle.
The rest he did with algebra and gave me “2” as the answer.
Smart ass wants to know if we have any more for him.
You have NO idea how often my Geometry students would answer a question on a test by working out the theorem that they were being asked (covertly) to apply to solve the problem. Which always gave me a laugh, since I was quite well-known in my own Geometry class way back when for doing the same thing. 
If that’s directed at me, I only pulled the image up in CAD to see if Okrahoma was right about the center being…off center. It didn’t look noticeably so to me and frankly I assumed that I would be telling him that he was wrong. While he was right, it’s so close that I don’t know that it could reliably be eyeballed as opposed to being a lucky guess.
Anyhow, once I was looking at the correctly scaled image is when I realized how out of whack the rest of the linework was. I hadn’t even noticed until then how far off the 60 degree/right angle was for exactly the reasons you state - I solved the problem using the information given without consideration for the accuracy of the sketch.
My little exercise doesn’t say anything about the level of my math education or my approach to problem solving but it does speak volumes to my boredom this afternoon and my ready access to CAD software.
Eh, it’s not an attack, more like an observation about two world systems: One, the more “mathematical” system, is willing to ignore the diagram and simply work with the numbers, even if that thing marked as a right angle is sixty degrees if it’s a day, and if that “midpoint” is centered so’s yer mamma… And the other world system, the more concrete thinkers, who may well bring a protractor or ruler in order to make a “point”, unaware that to the other world system, dissecting the diagram is getting hung up on a mostly-irrelevant sketch and finding flaws in it says more about the failings of some textbook company than anything else.
This comes to the fore again, when the concrete thinkers attempt to square a circle and get all discombobulated over the difference between a ruler and a straightedge, or attempt to prove the difference between 0.999… and 1 by rattling on about verniers and just get vernier and vernier at infinite length.
Smartasses with CAD software don’t really fit into either of those categories, granted. 
As Derleth also noted, such pictures are just rough sketches to show how the problem is arranged.
If the picture for this problem were to be completely accurate with the angles and such, then a student would be able to get the answer by just physically measuring the BAC angle and then solving the algebra problem. That does not lead to learning/understanding geometry at all.
It also sets up the student to struggle later on by teaching him/her to make assumptions based on not-intended-to-be-accurate-nor-give-any-additional-info pictures that just happen to be nearby problems. By making the pictures slightly off, it forces the student to have to figure out which parts of the problem are required/factual information and what parts are extraneous/variable/unknowable.
I’m actually not sure if you’re disagreeing with my point.
What I’m saying is that the answer is obvious, whether stated or not, and we’d like the student to demonstrate they know the method to find the answer.
When I used to sit maths exams, the rule was: if you write the correct answer you get full marks, with or without showing your working (but you should show your working so you get partial credit in the event your answer is wrong).
So for the geometry questions it doesn’t matter how complex of a bird’s nest the diagram is: just find an angle that looks visually the same (or 2x, like in the OP, or looks like would sum to 180 degrees) and does have a literal value, and you’re done. Or, to be less of a bad example to the OP’s son: write that answer in as a placeholder and come back to that question later if you have time. But when I used this strategy with past papers I always got full marks, even for questions where I couldn’t remember the actual principle for calculating the angle.