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

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==Solution==
 
==Solution==
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<math>\frac{1}{2}ab = 3(a+b+c)</math>
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<math>\fracqwb = 3(a+bet)</math>
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qwetwqet
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qwet
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Using Euclid's formula for generating primitive triples:wqe
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<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>
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wettqwwqetqwet
  
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<math>ab = 6(a+b+c)</math>
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</math>n(m-n)k = 6qwe<math>
  
−
Using Euclid's formula for generating primitive triples:
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Now we dqweto some casework.
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<math>a = m^2-n^2</math>, <math>b=2mn</math>, <math>c=m^2+n^2</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, exactly one of which being even.
 
  
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Since we do not want to restrict ourselves to only primitives, we will add a factor of k. <math>a = k(m^2-n^2)</math>, <math>b=2kmn</math>, <math>c=k(m^2+n^2)</math>
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For </math>k=1<math>et
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</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>
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ethe conditions of Euclid's formula, the only solutions are </math>(5,2)<math> and </math>(7,6et<math>
  
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<math>(m^2-n^2)\cdot 2mn \cdot k^2 = 6(2m^2 + 2mn)k</math>
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For </math>k=2<math>
  
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<math>mn(m-n)(m+n)k = 6m(m+n)</math>
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</math>n(m-n)=3<math> has solutions </math>(4,qwet1)<math>, </math>(4,3)<math>, both of which are valid.
  
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<math>n(m-n)k = 6</math>
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For </math>k=3<math>
  
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Now we do some casework.
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</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.
  
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For <math>k=1</math>
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For </math>k=6<math>
  
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<math>n(m-n) = 6</math> which has solutions <math>(7,1)</math>, <math>(5,2)</math>, <math>(5,3)</math>, <math>(7,6)</math>
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</math>n(m-n)=1<math> has solution </math>(1,2)<math>, which is valid.
  
−
Removing the solutions that do not satisfy the conditions of Euclid's formula, the only solutions are <math>(5,2)</math> and <math>(7,6)</math>
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This means that the solutions for </math>(m,n,k)<math> are
  
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For <math>k=2</math>
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</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>n(m-n)=3</math> has solutions <math>(4,1)</math>, <math>(4,3)</math>, both of which are valid.
 
−
 
 
−
For <math>k=3</math>
 
−
 
 
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<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.
 
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For <math>k=6</math>
 
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<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
 
−
 
 
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<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)}</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

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 $8$ solutions $\Rightarrow \mathrm{(C)}$.

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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