Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 12B Problems/Problem 24"
 
(9 intermediate revisions by 3 users not shown)
Line 10: Line 10:
 
</math>
 
</math>
  
−
==Solution==
+
==Solution 1==
  
−
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>.
 
Using an area argument, we can get the following well known result
 
Using an area argument, we can get the following well known result
−
<cmath>\left(\frac{AB+BC+AC}{2}\right)R=A</cmath>
+
<cmath>\left(\frac{AB+BC+AC}{2}\right)r=A</cmath>
 
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
 
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>
+
<cmath>\frac{r}{c}=\frac{AB}{AB+BC+AC}</cmath>
 
Similarly, we can get
 
Similarly, we can get
−
<cmath>\frac{R}{b}=\frac{AC}{AB+BC+AC}</cmath>
+
<cmath>\frac{r}{b}=\frac{AC}{AB+BC+AC}</cmath>
−
<cmath>\frac{R}{a}=\frac{BC}{AB+BC+AC}</cmath>
+
<cmath>\frac{r}{a}=\frac{BC}{AB+BC+AC}</cmath>
 
Hence,
 
Hence,
 
\begin{align}\label{e1}
 
\begin{align}\label{e1}
−
\frac{1}{R}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
+
\frac{1}{r}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
 
\end{align}
 
\end{align}
  
−
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. With this in mind, it remains to find all positive integer solutions <math>(R, a, b, c)</math> to the above such that <math>\frac{1}{a}, \frac{1}{b}, \frac{1}{c}</math> from a non-degenerate triangle, and <math>a\le b\le c\le 9</math>. We do this by doing casework on the value of <math>R</math>. Since <math>R</math> is a positive integer <math>R\ge 1</math>. Since <math>a\le b\le c\le 9</math>, <math>\frac{1}{R}\ge \frac{1}{3}</math>, so <math>R\le3</math>. The only possible values for <math>R</math> are 1, 2, 3.
+
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, i.e., <math>\frac{1}{b}+\frac{1}{c}>\frac{1}{a}</math>.  
  
−
<math>\textbf{Case 1: R=1}</math>
+
With this in mind, it remains to find all positive integer solutions <math>(r, a, b, c)</math> to the above such that <math>\frac{1}{b}+\frac{1}{c}>\frac{1}{a}</math>, and <math>a\le b\le c\le 9</math>. We do this by doing casework on the value of <math>r</math>.
 +
 
 +
Since <math>r</math> is a positive integer, <math>r\ge 1</math>. Since <math>a\le b\le c\le 9</math>, <math>\frac{1}{r}\ge \frac{1}{3}</math>, so <math>r\le3</math>. The only possible values for <math>r</math> are 1, 2, and 3.
 +
 
 +
Case <math>1</math>: <math>r=1</math>
 +
 
 +
For this case, we can't have <math>a\ge 4</math>, since <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}</math> would be too small. When <math>a=3</math>, we must have <math>b=c=3</math>. When <math>a\le2</math>, we would have <math>\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}</math>, which doesn't work. Hence this case only yields one valid solution <math>(1, 3, 3, 3)</math>
 +
 
 +
Case <math>2</math>: <math>r=2</math>
 +
 
 +
For this case, we can't have <math>a\ge 7</math>, for the same reason as in Case 1. When <math>a=6</math>, we must have <math>b=c=6</math>. When <math>a=5</math>, we must have <math>b=5, c=10</math> or <math>b=10, c=5</math>. Regardless, <math>10</math> appears, so it is not a valid solution. When <math>a\le4</math>, <math>\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}</math>. Hence, this case also only yields one valid solution <math>(2, 6, 6, 6)</math>
 +
 
 +
Case <math>3</math>: <math>r=3</math>
 +
 
 +
The only possible solution is <math>(3, 9, 9, 9)</math>, and clearly it is a valid solution.
 +
 
 +
Hence the only valid solutions are <math>(1, 3, 3, 3), (2, 6, 6, 6), (3, 9, 9, 9)</math>, and our answer is <math>\fbox{\textbf{(B) }3}</math>
 +
 
 +
~tsun26
 +
 
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=4x4EilBvIA8
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Latest revision as of 14:37, 25 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, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_24&oldid=235575"