Difference between revisions of "2024 AMC 12B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
− | First we derive the relationship between the inradius of a triangle <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) | + | <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{ | + | <cmath>\frac{r}{c}=\frac{AB}{AB+BC+AC}</cmath> |
Similarly, we can get | Similarly, we can get | ||
− | <cmath>\frac{ | + | <cmath>\frac{r}{b}=\frac{AC}{AB+BC+AC}</cmath> |
− | <cmath>\frac{ | + | <cmath>\frac{r}{a}=\frac{BC}{AB+BC+AC}</cmath> |
Hence, | Hence, | ||
\begin{align}\label{e1} | \begin{align}\label{e1} | ||
− | \frac{1}{ | + | \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, i.e., <math>\frac{1}{b}+\frac{1}{c}>\frac{1}{a}</math>. | 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>. | ||
− | With this in mind, it remains to find all positive integer solutions <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> | + | 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> | + | 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> | 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> | + | 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> | 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> | + | 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. | The only possible solution is <math>(3, 9, 9, 9)</math>, and clearly it is a valid solution. |
Revision as of 10:41, 16 November 2024
Problem 24
What is the number of ordered triples of positive integers, with , such that there exists a (non-degenerate) triangle with an integer inradius for which , , and are the lengths of the altitudes from to , to , and to , respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)
Solution 1
First we derive the relationship between the inradius of a triangle , and its three altitudes .
Using an area argument, we can get the following well known result
where are the side lengths of , and is the triangle's area. Substituting into the above we get
Similarly, we can get
Hence,
Note that there exists a unique, non-degenerate triangle with altitudes if and only if are the side lengths of a non-degenerate triangle, i.e., .
With this in mind, it remains to find all positive integer solutions to the above such that , and . We do this by doing casework on the value of .
Since is a positive integer, . Since , , so . The only possible values for are 1, 2, and 3.
Case :
For this case, we can't have , since would be too small. When , we must have . When , we would have , which doesn't work. Hence this case only yields one valid solution
Case :
For this case, we can't have , for the same reason as in Case 1. When , we must have . When , we must have or . Regardless, appears, so it is not a valid solution. When , . Hence, this case also only yields one valid solution
Case :
The only possible solution is , and clearly it is a valid solution.
Hence the only valid solutions are , and our answer is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.