2024 AMC 12B Problem 24

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

Answer: B

Concepts:altitude incircle, incenter, and inradius triangle inequality

Difficulty rating: 2410

Small hint:

Since each side equals

\dfrac{2[\triangle]}{\text{altitude}},

the inradius satisfies

\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c

Big hint:

The sides are proportional to

\tfrac1a, \tfrac1b, \tfrac1c,

so non-degeneracy needs

\tfrac1a \lt \tfrac1b + \tfrac1c;

seek triples with

\tfrac1a + \tfrac1b + \tfrac1c

a unit fraction

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.

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.