Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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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{triangle}[thick] | ||
+ | \draw(0,0) -- (90:2cm) node[midway,left]{<math>opposite leg</math>} -- (0:4cm) node[midway,above right]{<math>hypotenuse</math>} -- (0,0) node[midway,below]{<math>adjacent leg</math>}; | ||
+ | \draw[fill=lightgray, thick] (0,0) -- (0:0.8cm) arc (0:90:0.8cm) node at (45:0.5cm) {<math>\gamma</math>} -- cycle; | ||
+ | \end{triangle} | ||
==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 12:45, 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 .
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{triangle}[thick] \draw(0,0) -- (90:2cm) node[midway,left]{} -- (0:4cm) node[midway,above right]{} -- (0,0) node[midway,below]{}; \draw[fill=lightgray, thick] (0,0) -- (0:0.8cm) arc (0:90:0.8cm) node at (45:0.5cm) {} -- cycle; \end{triangle}
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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