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Difference between revisions of "2007 AMC 12B Problems/Problem 23"

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==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 14:47, 23 December 2013

Problem 23

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution

$\fracqwb = 3(a+bet)$ (Error compiling LaTeX. Unknown error_msg) qwetwqet qwet Using Euclid's formula for generating primitive triples:wqe $a = m^2-n^2$, $b=2we$, $c=m^2+n^2$ whwqtere $m$ and $n$ are relatively prime positive integers, exactly one of which being eveet=2kmn$,$c=k(mqw^2+n^2)$wettqwwqetqwet$n(m-n)k = 6qwe$Now we dqweto some casework.

For$ (Error compiling LaTeX. Unknown error_msg)k=1$et$n(m-n) = 6$wethich has solutions$(7,1)$,$(5,2)$,$(5,3)$,$(7,6)$ethe conditions of Euclid's formula, the only solutions are$(5,2)$and$(7,6et$For$k=2$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=3$has solutions$(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$has solutions$(3,1)$,$(3,2)$of which only$(3,2)$is valid.

For$ (Error compiling LaTeX. Unknown error_msg)k=6$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=1$has solution$(1,2)$, which is valid.

This means that the solutions for$ (Error compiling LaTeX. Unknown error_msg)(m,n,k)$are$(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$solutions$\Rightarrow \mathrm{(A)}$

Solution 2

Let $a$ and $b$ be the two legs of the triangle.

We have $\frac{1}{2}ab = 3(a+b+c)$.

Then $ab=6\cdot (a+b+\sqrt {a^2 + b^2})$.

We can complete the square under the root, and we get, $ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})$.

Let $ab=p$ and $a+b=s$, we have $p=6\cdot (s+ \sqrt {s^2 - 2p})$.

After rearranging, squaring both sides, and simplifying, we have $p=12s-72$.


Putting back $a$ and $b$, and after factoring using $SFFT$, we've got $(a-12)\cdot (b-12)=72$.


Factoring 72, we get 6 pairs of $a$ and $b$


$(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$


And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$.

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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