2024 AMC 12B Problems/Problem 24

Revision as of 01:47, 14 November 2024 by Tsun26 (talk | contribs) (Solution)

Problem 24

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.)

$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }4\qquad \textbf{(D) }5\qquad \textbf{(E) }6\qquad$

Solution

First we derive the relationship between the inradius of a triangle $R$, and its three altitudes $a, b, c$. Using an area argument, we can get the following well known result \[\left(\frac{AB+BC+AC}{2}\right)R=A\] where $AB, BC, AC$ are the side lengths of $\triangle ABC$, and $A$ is the triangle's area. Substituting $A=\frac{1}{2}\cdot AB\cdot c$ into the above we get \[\frac{R}{c}=\frac{AB}{AB+BC+AC}\] Similarly, we can get \[\frac{R}{b}=\frac{AC}{AB+BC+AC}\] \[\frac{R}{a}=\frac{BC}{AB+BC+AC}\] Hence, 1R=1a+1b+1c

Note that there exists a unique, non-degenerate triangle with altitudes $a, b, c$ if and only if $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are the side lengths of a non-degenerate triangle. With this in mind, it remains to find all positive integer solutions $(R, a, b, c)$ to the above such that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ from a non-degenerate triangle, and $a\le b\le c\le 9$. We do this by doing casework on the value of $R$.

$\textbf{Case 1: R=1}$