Difference between revisions of "2024 AMC 12B Problems/Problem 24"
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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> | ||
− | <math>\textbf{Case | + | <math>\textbf{Case 2: 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, 10 appears, so it is not a valid solution. When <math>ale4</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, 10 appears, so it is not a valid solution. When <math>ale4</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> | ||
− | <math>\textbf{Case | + | <math>\textbf{Case 3: 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. | ||
− | Hence our answer is <math>\fbox{\textbf{(B) }3}</math> | + | Hence the only 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 | ~tsun26 |
Revision as of 02:06, 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
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, 10 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 solutions are , and our answer is
~tsun26