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Difference between revisions of "2024 AMC 12B Problems/Problem 24"

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First we derive the relationship between the inradius of a triangle <math>R</math>, and its three altitudes <math>a, b, c</math>.
 
First we derive the relationship between the inradius of a triangle <math>R</math>, and its three altitudes <math>a, b, c</math>.
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~tsun26
 
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{{AMC12 box|year=2024|ab=B|num-b=23|num-a=25}}
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Revision as of 14:28, 14 November 2024

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, 3.

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

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 $(1, 3, 3, 3)$

$\textbf{Case 2: R=2}$

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 $ale4$, $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$. Hence, this case also only yields one valid solution $(2, 6, 6, 6)$

$\textbf{Case 3: R=3}$

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


See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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