CB/NM or 2
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03 Oct 2025, 01:10
We are given that line $C A B$ is the diameter of the circle, as $A$ is the center. Also, we know that angle NAM is a right angle (even though it does not look like a 90-degree angle in the diagram), and, since sides $\(A M\)$ and $\(A N\)$ are both radii of the circle and therefore the same length, triangle MAN is a $\(45-45-90\)$ triangle.
Let's use $x$ for the radius of this circle.
Now let's find the ratio of $\(C B\)$ to $\(N M\)$. Even though we are not given exact lengths, we can solve for this. We know that because the triangle is a $\(45-45-90\)$ triangle, NM must be equal to $\(x \sqrt{2}\)$ (because it is opposite the 90-degree angle) and both $\(N A\)$ and $\(M A\)$ are of size $x$ in length.
We also know that $\(C B\)$ (the diameter) is equal to $\(2 x\)$ (because a diameter is equal to 2 times the radius and $N A$ and $M A$ are radii). Therefore, the ratio is equal to $\(\frac{2 x}{x \sqrt{2}}\)$.
We can cancel out the $x$ 's to get $\(\frac{2}{\sqrt{2}}\)$, which equals $\(\sqrt{2}\)$.
Since the root of any integer greater than 1 must be less than that integer, $\(\sqrt{2}\)$ must be less than 2. Thus, the correct answer is (B).