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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==
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Let <math>a</math> and <math>b</math> be the two legs of the triangle.
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We have <math>\frac{1}{2}ab = 3(a+b+c)</math>.
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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>.
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After squaring both sides, then simplifying and combining, we have <math>p=12s-72</math>.
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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>.
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Factoring 72, we get 6 pairs of <math>a</math> and <math>b</math>
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<math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math>
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 +
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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

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

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

$ab = 6(a+b+c)$

Using Euclid's formula for generating primitive triples: $a = m^2-n^2$, $b=2mn$, $c=m^2+n^2$ where $m$ and $n$ 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. $a = k(m^2-n^2)$, $b=2kmn$, $c=k(m^2+n^2)$

$(m^2-n^2)\cdot 2mn \cdot k^2 = 6(2m^2 + 2mn)k$

$mn(m-n)(m+n)k = 6m(m+n)$

$n(m-n)k = 6$

Now we do some casework.

For $k=1$

$n(m-n) = 6$ which has solutions $(7,1)$, $(5,2)$, $(5,3)$, $(7,6)$

Removing the solutions that do not satisfy the conditions of Euclid's formula, the only solutions are $(5,2)$ and $(7,6)$

For $k=2$

$n(m-n)=3$ has solutions $(4,1)$, $(4,3)$, both of which are valid.

For $k=3$

$n(m-n)=2$ has solutions $(3,1)$, $(3,2)$ of which only $(3,2)$ is valid.

For $k=6$

$n(m-n)=1$ has solution $(1,2)$, which is valid.

This means that the solutions for $(m,n,k)$ are

$(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)$

$6$ 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 squaring both sides, then simplifying and combining, 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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