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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Revision as of 14:28, 14 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, 3.
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
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
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 |
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