2024 AMC 12B Problems/Problem 24

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 1

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, \begin{align}\label{e1} \frac{1}{r}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \end{align}

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 $\fbox{\textbf{(B) }3}$

~tsun26

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=4x4EilBvIA8

2024 AMC 12B (Problems • Answer Key • Resources)
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2024 AMC 12B Problems/Problem 24

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Problem 24

Solution 1

Video Solution 1 by SpreadTheMathLove

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