In the figure above, showing circle with center O and points

Guys I did this question with this approach, can you kindly evaluate the approach
\(AB= \frac{1}{3} Circumference\)
So,\(AB = \frac{2}{3}Pi* R\)
Also \(AB=12\sqrt{3}\)
Therefore, \(12\sqrt{3} = \frac{2}{3}Pi* R\)
Manipulating\(R = 12\)

How do you that
since \(\angle AOB = 120^{\circ}\)
so \(\angle\) ACB = \(60^{\circ}\) ?

I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: https://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem

Please, follow this link: math-circles-87957.html

Do we just assume that AC is the diameter?

Ok, assuming we know this is a right triangle because we know that the picture is accurate...

If the triangle is a right triangle, we know that angle B is the right angle. Furthermore, we are given that arc AB = 1/3 of the circle's circumference (i.e it = 120 degrees) This means that AOB, the central angle = 120 degrees and because AO and BO are radii, we know that triangle AOB is an isosceles triangle. We also know the angle measure of angle BOC which = 180 - 120 = 60. If angle AOB is the interior angle, it is twice the measure of the "exterior" angle ACB, so ACB = 60. Now we know that triangle BOC is equilateral. More importantly, we know that ABC is a 30:60:90 triangle. With this, we can find the measure of any given side knowing just the measure of one. The ratio of side lengths in a 30:60:90 is (x/2):(3/2*x):(x) in this case, x (the radius) = 12.

Please read here: in-the-figure-above-showing-circle-with-center-o-and-points-147885.html#p1187477

We did not assume that. AC passes through the center of the circle thus it's the diameter.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

I understand the rule, but in other words, how do we know that the drawing isn't misleading and that AC passes above the center?

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hi Bunuel

Particularly in this question with the given options , do we even need this much detail.

We know that the angle AOB = 120 degrees as it is 1/3 the circumference. AO=OB as they are the radius. So the two angles OAB and OBA will be equal in the triangle AOB. Thus Angle OBA=OBA=30 ( 180-120)/2. In a triangle the greatest side is opposite the largest angle . Here AB is opposite the largest angle and is 12sqrt2 so the other sides ( the radii) will have to be less that this value and only A satisfies the condition over here

hi please help, got stuck. why dont i get a clean answer of 12?

12\(\sqrt{3}\) = \(\frac{1}{3}\) * 2 * pie * R

does the wording simply states that the length of the arc AB is one third WITHOUT INDICATING THE EXACT VALUE, and the value of 12 root 3 is given for the side AB (rather than the arc AB)? i think the probles is poorly worded. look how it looks in the original. veritas should not have placed that bar above AB if they meant it was not the arc but the length

AB Arc and not AB line segment is equal to 2/3 of circumference. Moreover AB arc is not 12root3, AB Line segment is.

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