2024 AMC 12B Problem 24

Writing each side as $\dfrac{2[\triangle]}{h},$ the semiperimeter is $[\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr),$ so the inradius $r = \dfrac{[\triangle]}{s}$ satisfies $\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c.$ We need this to be $\dfrac1r$ for a positive integer $r,$ with the sides (proportional to $\tfrac1a, \tfrac1b, \tfrac1c$) forming a non-degenerate triangle, requiring $\tfrac1a \lt \tfrac1b + \tfrac1c.$ Because $a\le b\le c\le9,$ the reciprocal sum is at least $\frac{3}{c}\ge\frac{1}{3},$ so the integer $r$ is one of $1,2,3.$ Also $\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{a},$ so $r\lt a\le3r.$ For each of these few values of $r,a,$ substitute $b=a,a+1,\ldots,9$ into $$c=\frac{abr}{ab-ar-br}.$$ Keeping only integral $c$ with $b\le c\le9$ gives the complete list $$\begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9). \end{array}$$ The triples $(2,3,6),(2,4,4),$ and $(4,8,8)$ have $\tfrac1a=\tfrac1b+\tfrac1c$ and therefore give degenerate triangles. The remaining triples are $(3,3,3),(6,6,6),$ and $(9,9,9),$ so the answer is $3.$ Thus, the correct answer is B.