Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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− | <math>\ | + | <math>\fracqwb = 3(a+bet)</math> |
+ | qwetwqet | ||
+ | qwet | ||
+ | Using Euclid's formula for generating primitive triples:wqe | ||
+ | <math>a = m^2-n^2</math>, <math>b=2we</math>, <math>c=m^2+n^2</math> whwqtere <math>m</math> and <math>n</math> are relatively prime positive integers, exactly one of which being eveet=2kmn<math>, </math>c=k(mqw^2+n^2)<math> | ||
+ | wettqwwqetqwet | ||
− | <math> | + | </math>n(m-n)k = 6qwe<math> |
− | + | Now we dqweto some casework. | |
− | |||
− | + | For </math>k=1<math>et | |
+ | </math>n(m-n) = 6<math> wethich has solutions </math>(7,1)<math>, </math>(5,2)<math>, </math>(5,3)<math>, </math>(7,6)<math> | ||
+ | ethe conditions of Euclid's formula, the only solutions are </math>(5,2)<math> and </math>(7,6et<math> | ||
− | <math> | + | For </math>k=2<math> |
− | <math> | + | </math>n(m-n)=3<math> has solutions </math>(4,qwet1)<math>, </math>(4,3)<math>, both of which are valid. |
− | <math> | + | For </math>k=3<math> |
− | + | </math>n(m-n)=2<math> has solutions </math>(3,1)<math>, </math>(3,2)<math> of which only </math>(3,2)<math> is valid. | |
− | For <math>k= | + | For </math>k=6<math> |
− | <math>n(m-n) = | + | </math>n(m-n)=1<math> has solution </math>(1,2)<math>, which is valid. |
− | + | This means that the solutions for </math>(m,n,k)<math> are | |
− | + | </math>(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)<math> | |
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+ | </math>6<math> solutions </math>\Rightarrow \mathrm{(A)}$ | ||
==Solution 2== | ==Solution 2== |
Revision as of 20:20, 17 December 2013
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
$\fracqwb = 3(a+bet)$ (Error compiling LaTeX. Unknown error_msg)
qwetwqet
qwet
Using Euclid's formula for generating primitive triples:wqe
,
,
whwqtere
and
are relatively prime positive integers, exactly one of which being eveet=2kmn
c=k(mqw^2+n^2)
n(m-n)k = 6qwe$Now we dqweto some casework.
For$ (Error compiling LaTeX. Unknown error_msg)k=1n(m-n) = 6
(7,1)
(5,2)
(5,3)
(7,6)
(5,2)
(7,6et
k=2$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=3
(4,qwet1)
(4,3)$, both of which are valid.
For$ (Error compiling LaTeX. Unknown error_msg)k=3$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=2(3,1)
(3,2)
(3,2)$is valid.
For$ (Error compiling LaTeX. Unknown error_msg)k=6$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=1(1,2)$, which is valid.
This means that the solutions for$ (Error compiling LaTeX. Unknown error_msg)(m,n,k)(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)$$ (Error compiling LaTeX. Unknown error_msg)6
\Rightarrow \mathrm{(A)}$
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 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
.
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.