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Difference between revisions of "2024 AMC 12B Problems/Problem 24"
(→‎Problem 24)
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\textbf{(E) }6\qquad
 
\textbf{(E) }6\qquad
 
</math>
 
</math>
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==Solution==
 +
 +
First we derive the relationship between the inradius of a triangle <math>R</math>, and its three altitudes <math>a, b, c</math>.
 +
Using an area argument, we can get the following well known result
 +
<cmath>\left(\frac{AB+BC+AC}{2}\right)R=A</cmath>
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where <math>AB, BC, AC</math> are the side lengths of <math>\triangle ABC</math>, and <math>A</math> is the triangle's area. Substituting <math>A=\frac{1}{2}\cdot AB\cdot c</math> into the above we get
 +
<cmath>\frac{R}{c}=\frac{AB}{AB+BC+AC}</cmath>
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Similarly, we can get
 +
<cmath>\frac{R}{b}=\frac{AC}{AB+BC+AC}</cmath>
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<cmath>\frac{R}{a}=\frac{BC}{AB+BC+AC}</cmath>
 +
Hence,
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<cmath>\frac{1}{R}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}</cmath>
 +
 +
Note that there exists a unique, non-degenerate triangle with altitudes <math>a, b, c</math> if and only if <math>\frac{1}{a}, \frac{1}{b}, \frac{1}{c}</math> are the side lengths of a non-degenerate triangle

Revision as of 01:38, 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

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, \[\frac{1}{R}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\]

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

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