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 15:28, 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,
\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 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.