Question

What is the number of ordered triples $(a, b, c)$ of positive integers, with $a \le b \le c \le 9,$ such that there exists a (non-degenerate) triangle $\triangle ABC$ with an integer inradius for which $a,$ $b,$ and $c$ are the lengths of the altitudes from $A$ to $\overline{BC},$ $B$ to $\overline{AC},$ and $C$ to $\overline{AB},$ respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)

$2$

$3$

$4$

$5$

$6$

Solution:

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 $\tfrac1a \lt \tfrac1b + \tfrac1c.$

Searching $a \le b \le c \le 9,$ the triples with $\tfrac1a + \tfrac1b + \tfrac1c = \tfrac1r$ that also satisfy the triangle inequality are exactly the equilateral ones: $(3,3,3)$ with $r = 1,$ $(6,6,6)$ with $r = 2,$ and $(9,9,9)$ with $r = 3.$ Other solutions such as $(2,3,6)$ or $(4,8,8)$ give degenerate triangles. So there are $3$ triples.

Thus, the correct answer is B.

2024 AMC 12B Problem 24

Solution:

Problem 24 in Other Years