The longest line you can draw in a circle is a diameter, so AB is a diameter of the circle. If X is the center of the circle, then AX is a radius. The question tells us AX is 3, so the radius of the circle is 3.

If you draw a diagram, and look at triangle ACX, two of the sides of that triangle, AX and CX, are radii, so have length 3. The third side is AC, which the question tells us also has length 3. So triangle ACX is equilateral, and all of its angles are 60 degrees.

If you now look at triangle ABC, it shares the angle at point A with triangle ACX, so triangle ABC has a 60 degree angle. We're also connecting a diameter AB to a point C on the circle, which always creates a 90 degree angle at that point. So triangle ABC has a 60 and a 90 degree angle, and must be a 30-60-90 triangle. We already know its hypotenuse is the diameter, so is 6, and its short side is AC which we know is 3, and then using either Pythagoras or 30-60-90 triangle ratios (the sides are in a 1 to √3 to 2 ratio), we find that the length of the missing side, BC, is 3√3. Adding the sides we get the perimeter of 6+3+3√3 = 9 + 3√3.

AAPL wrote:Princeton Review

\(A\) and \(B\) are the endpoints of the longest line that can be drawn in a circle with center \(X\). If \(C\) is a point of the circle such that \(AC=AX=3\), what is the perimeter of triangle \(ABC\)?

A. \(\frac{9}{2}\)
B. \(9\)
C. \(6+3\sqrt{3}\)
D. \(9+3\sqrt{3}\)
E. \(9\sqrt{3}\)

OA D

We see that AB is a diameter of the circle, and AX is a radius. Since AX = 3, AB = 6. Furthermore, since triangle ABC contains the diameter AB as a side, it must be a right triangle, and AB is its hypotenuse.

We know AC = 3 and AB = 6, and we need to determine BC. Since AB is twice AC, triangle ABC must be a 30-60-90 right triangle. Therefore, BC = 3√3, and the perimeter of triangle ABC is AC + AB + BC = 3 + 6 + 3√3 = 9 + 3√3.

Answer: D