Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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We can factor <math>72</math> to arrive with <math>6</math> pairs of solutions: <math>(7, 78), (8,42), (9, 30), (10, 24), (12, 18),</math> and <math>(14, 15) \Rightarrow \mathrm{(A)}</math>. | We can factor <math>72</math> to arrive with <math>6</math> pairs of solutions: <math>(7, 78), (8,42), (9, 30), (10, 24), (12, 18),</math> and <math>(14, 15) \Rightarrow \mathrm{(A)}</math>. | ||
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+ | == Solution 3 == | ||
+ | |||
+ | <math>[ABC] = r\cdot s</math>, <math>3(a+b+c) = r \cdot \frac{a+b+c}{2}</math>, <math>r = 6</math>, <math>r = \frac{a+b-c}{2}</math>, <math>a+b-c = 12</math>, <math>c = a + b - 12</math> | ||
+ | |||
+ | <math>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</math>, <math>ab = 6(a+b-c)</math>, <math>\frac{ab}{6} = a+b-c</math>, <math>c = a + b - \frac{ab}{6}</math> | ||
==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 09:54, 5 December 2022
Contents
[hide]Problem
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution 1
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 Simon's Favorite Factoring Trick, we've got .
Factoring 72, we get 6 pairs of and
And this gives us solutions .
Alternatively, note that . Then 72 has factors. However, half of these are repeats, so we have 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 .
The incircle of breaks the triangle's sides into segments such that , and . Since ABC is a right triangle, one of , and is equal to its radius, 6. Let's assume .
The side lengths then become , and . Plugging into Pythagorean's theorem:
We can factor to arrive with pairs of solutions: and .
Solution 3
, , , , ,
, , ,
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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