Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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<math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math> | <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math> | ||
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+ | ==Solution 2== | ||
+ | Let <math>a</math> and <math>b</math> be the two legs of the triangle. | ||
+ | |||
+ | We have <math>\frac{1}{2}ab = 3(a+b+c)</math>. | ||
+ | |||
+ | Then <math>ab=6\cdot (a+b+\sqrt {a^2 + b^2})</math>. | ||
+ | |||
+ | We can complete the square under the root, and we get, <math>ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})</math>. | ||
+ | |||
+ | Let <math>ab=p</math> and <math>a+b=s</math>, we have <math>p=6\cdot (s+ \sqrt {s^2 - 2p})</math>. | ||
+ | |||
+ | After squaring both sides, then simplifying and combining, we have <math>p=12s-72</math>. | ||
+ | |||
+ | |||
+ | Putting back <math>a</math> and <math>b</math>, and after factoring using <math>SFFT</math>, we've got <math>(a-12)\cdot (b-12)=72</math>. | ||
+ | |||
+ | |||
+ | Factoring 72, we get 6 pairs of <math>a</math> and <math>b</math> | ||
+ | |||
+ | |||
+ | <math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math> | ||
+ | |||
+ | |||
+ | And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</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 12:28, 14 September 2013
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
Using Euclid's formula for generating primitive triples: , , where and are relatively prime positive integers, exactly one of which being even.
Since we do not want to restrict ourselves to only primitives, we will add a factor of k. , ,
Now we do some casework.
For
which has solutions , , ,
Removing the solutions that do not satisfy the conditions of Euclid's formula, the only solutions are and
For
has solutions , , both of which are valid.
For
has solutions , of which only is valid.
For
has solution , which is valid.
This means that the solutions for are
solutions
Solution 2
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 squaring both sides, then simplifying and combining, 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 .
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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