2013 AMC 10A Problem 23

The point $C$ is outside the circle (since $AC = 97 > 86$), and line $CB$ is a secant meeting the circle at $X$ and $B$. By power of a point,
$$CX \cdot CB = AC^2 - AB^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 11 \cdot 183 = 2013.$$

Since $BX$ and $CX$ are integers, so is $CB = CX + BX$, and $CX < CB$. Factor $2013 = 3 \cdot 11 \cdot 61$; the divisor pairs with the smaller factor first are
$$(1, 2013), \quad (3, 671), \quad (11, 183), \quad (33, 61).$$

The triangle inequality on $ABC$ requires $97 - 86 < BC < 97 + 86$, i.e. $11 < BC < 183$ (strictly, since $A$, $B$, $C$ form a genuine triangle). Only the pair $(33, 61)$ has its larger member in this range, so $CB = 61$ and $CX = 33$, giving $BX = 28$.

The answer is $\boxed{\textbf{(D)}\ 61}$.

A circle centered at a vertex with radius equal to a side is a strong hint for power of a point from the other vertex, and the difference of squares makes the product easy to factor. Integer-length conditions then convert geometry into a divisor search, and the triangle inequality is the tool that discards the extraneous divisor pairs. Notice that $BX = 28$ and $CX = 33$ are both answer choices, planted as distractors.

Drop the altitude from $A$ to $BC$; it bisects chord $BX$ at its midpoint $M$. Then $AC^2 - AB^2 = CM^2 - BM^2 = (CM - BM)(CM + BM) = CX \cdot CB$, which re-derives the power of a point without quoting the theorem.

• Choosing the divisor pair (11, 183) and answering BC = 183 (not a choice) or 11, without checking the strict triangle inequality.
• Reporting $BX = 28$ or $CX = 33$ instead of the requested $BC$.