AMC12 2024 B

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, i.e., $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$. With this in mind, it remains to find all positive integer solutions $(r, a, b, c)$ to the above such that $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$, and $a\le b\le c\le 9$. We do this by doing casework on the value of $r$. Since $r$ is a positive integer, $r\ge 1$. Since $a\le b\le c\le 9$, $\frac{1}{r}\ge \frac{1}{3}$, so $r\le3$. The only possible values for $r$ are 1, 2, and 3. Case 1: $r=1$ For this case, we can't have $a\ge 4$, since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ would be too small. When $a=3$, we must have $b=c=3$. When $a\le2$, we would have $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$, which doesn't work. Hence this case only yields one valid solution $(1, 3, 3, 3)$ Case 2: $r=2$ For this case, we can't have $a\ge 7$, for the same reason as in Case 1. When $a=6$, we must have $b=c=6$. When $a=5$, we must have $b=5, c=10$ or $b=10, c=5$. Regardless, $10$ appears, so it is not a valid solution. When $a\le4$, $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$. Hence, this case also only yields one valid solution $(2, 6, 6, 6)$ Case 3: $r=3$ The only possible solution is $(3, 9, 9, 9)$, and clearly it is a valid solution. Hence the only valid solutions are $(1, 3, 3, 3), (2, 6, 6, 6), (3, 9, 9, 9)$, and our answer is $\boxed{\textbf{(B) }3}$