2007 AMC 12B Problems/Problem 23
Contents
Problem 23
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution
Let and
be the two legs of the triangle.
We have .
Then .
We can complete the square under the root, and we get, .
Let and
, we have
.
After rearranging, squaring both sides, and simplifying, we have .
Putting back and
, and after factoring using
, we've got
.
Factoring 72, we get 6 pairs of and
And this gives us solutions
.
Solution #2
We will proceed by using the fact that , where
is the radius of the incircle and
is the semiperimeter (
).
We are given .
\begin{tikzpicture}[thick] \coordinate (O) at (0,0); \coordinate (A) at (4,0); \coordinate (B) at (0,2); \draw (O)--(A)--(B)--cycle;
\tkzLabelSegment[below=2pt](O,A){\textit{adjacent leg}} \tkzLabelSegment[left=2pt](O,B){\textit{opposite leg}} \tkzLabelSegment[above right=2pt](A,B){\textit{hypotenuse}}
\tkzMarkAngle[fill= orange,size=0.65cm,%
opacity=.4](A,O,B)
\tkzLabelAngle[pos = 0.35](A,O,B){}
\tkzMarkAngle[fill= orange,size=0.8cm,%
opacity=.4](B,A,O)
\tkzLabelAngle[pos = 0.6](B,A,O){}
\tkzMarkAngle[fill= orange,size=0.7cm,%
opacity=.4](O,B,A)
\tkzLabelAngle[pos = 0.5](O,B,A){}
\end{tikzpicture}
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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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