Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>. | And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>. | ||
+ | |||
+ | \usepackage{tkz-euclide} | ||
+ | \usetkzobj{all} | ||
==Solution #2== | ==Solution #2== | ||
Line 35: | Line 38: | ||
We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>. | We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>. | ||
− | \begin{ | + | \begin{tikzpicture}[thick] |
− | \ | + | \coordinate (O) at (0,0); |
− | \ | + | \coordinate (A) at (4,0); |
− | \end{ | + | \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){<math>\gamma</math>} | ||
+ | |||
+ | \tkzMarkAngle[fill= orange,size=0.8cm,% | ||
+ | opacity=.4](B,A,O) | ||
+ | \tkzLabelAngle[pos = 0.6](B,A,O){<math>\alpha</math>} | ||
+ | |||
+ | \tkzMarkAngle[fill= orange,size=0.7cm,% | ||
+ | opacity=.4](O,B,A) | ||
+ | \tkzLabelAngle[pos = 0.5](O,B,A){<math>\beta</math>} | ||
+ | |||
+ | |||
+ | \end{tikzpicture} | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:46, 29 November 2014
Contents
[hide]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 .
\usepackage{tkz-euclide} \usetkzobj{all}
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.