2013 AMC 10A Problem 23

1. What the circle tells us. $B$ and $X$ are both on the circle centered at $A$, so $AX = AB = 86$. Triangle $ABX$ is isosceles.
2. Drop a perpendicular. Let $M$ be the foot of the perpendicular from $A$ to line $BC$. In an isosceles triangle this perpendicular cuts the base in half, so $M$ is the middle of $BX$: $BM = MX$. Along the line the order is $B, M, X, C$.
3. Pythagoras twice. Let $AM = h$.
$$AC^2 = h^2 + CM^2, \qquad AB^2 = h^2 + BM^2.$$
Subtract to get rid of $h$:
$$AC^2 - AB^2 = CM^2 - BM^2.$$
4. Factor both sides. On the right, $CM^2 - BM^2 = (CM - BM)(CM + BM)$.
- $CM - BM = CM - MX = CX$.
- $CM + BM = CM + MB$ is the whole length $CB$.

On the left, $97^2 - 86^2 = (97 - 86)(97 + 86) = 11 \times 183 = 2013$. So
$$CX \times CB = 2013.$$
5. Use whole numbers. $CX$ and $BX$ are whole numbers, so $CB = CX + BX$ is too, and $CX$ is the smaller one. Since $2013 = 3 \times 11 \times 61$, the choices are:

1. Use the triangle rule. In triangle $ABC$, side $BC$ must be longer than $97 - 86 = 11$ and shorter than $97 + 86 = 183$. Only $CB = 61$ fits (and $183$ is not allowed, since it would make the triangle flat).
2. Check. $CX = 33$, $CB = 61$, so $BX = 28$, a whole number.

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

A circle centered at one corner of a triangle, passing through another corner, sets up the step "subtract two Pythagoras equations". The difference of two squares then factors into a product of lengths, and whole-number conditions turn the geometry into a list of factor pairs. The triangle rule throws out the wrong pairs. Notice that $28$ and $33$ are also answer choices, waiting to catch anyone who answers $BX$ or $CX$.

The "power of a point" theorem says that for a point $C$ outside a circle, and a line through $C$ that meets the circle at $X$ and $B$, the product $CX \times CB$ equals (distance from $C$ to the center)$^2$ minus (radius)$^2$. Here that is $97^2 - 86^2 = 2013$ at once, and the rest is the same as Steps 5 to 7.