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 ![$[ABC] = r\cdot s$](http://latex.artofproblemsolving.com/2/8/9/289668c9ffccf64525a50cf707bd6a55f5edf196.png)
, where 
is the radius of the incircle and 
is the semiperimeter (
).
We are given ![$[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$](http://latex.artofproblemsolving.com/6/f/a/6fa0b1a3f620e4e16086645fa3e07b02c60c7449.png)
.
\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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