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Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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== Solution 3 ==
 
== Solution 3 ==
  
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<math>[ABC] = r\cdot s</math>, <math>3(a+b+c) = r \cdot \frac{a+b+c}{2}</math>, <math>r = 6</math>, <math>r = \frac{a+b-c}{2}</math>, <math>a+b-c = 12</math>, <math>c = a + b - 12</math>
+
By using <math>[ABC] = r\cdot s</math> we get:
  
−
<math>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</math>, <math>ab = 6(a+b-c)</math>, <math>\frac{ab}{6} = a+b-c</math>, <math>c = a + b - \frac{ab}{6}</math>
+
<cmath>3(a+b+c) = r \cdot \frac{a+b+c}{2}</cmath>
 +
<cmath>r=6</cmath>
 +
<cmath>r = \frac{a+b-c}{2}</cmath>
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<cmath>a+b-c = 12</cmath>
 +
<cmath>c = a + b - 12</cmath>
 +
 
 +
 
 +
By the triangle's area we get:
 +
 
 +
<cmath>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</cmath>
 +
<cmath>ab = 6(a+b+c)</cmath>
 +
<cmath>\frac{ab}{6} = a+b+c</cmath>
 +
<cmath>c = \frac{ab}{6} - a - b</cmath>
 +
 
 +
<math>a + b - 12 = \frac{ab}{6} - a - b</math>, <math>ab - 12a - 12b + 72 = 0</math>, <math>(a - 12)(b - 12) = 72</math>
 +
 
 +
As <math>72 = 2^3 \cdot 3^2</math>, there are <math>\frac{(3+1)(2+1)}{2} = 6</math> solutions, <math>\boxed{\textbf{(A) } 6}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==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 10:09, 5 December 2022

Problem

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 1

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

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

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

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

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

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


Putting back $a$ and $b$, and after factoring using Simon's Favorite Factoring Trick, we've got $(a-12)(b-12)=72$.


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 $\Rightarrow \mathrm{(A)}$.


Alternatively, note that $72 = 2^3 \cdot 3^2$. Then 72 has $(3+1)(2+1) = (4)(3) = 12$ factors. However, half of these are repeats, so we have $\frac{12}{2} = 6$ solutions.

Solution 2

We will proceed by using the fact that $[ABC] = r\cdot s$, where $r$ is the radius of the incircle and $s$ is the semiperimeter $\left(s = \frac{p}{2}\right)$.

We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$.

The incircle of $ABC$ breaks the triangle's sides into segments such that $AB = x + y$, $BC = x + z$ and $AC = y + z$. Since ABC is a right triangle, one of $x$, $y$ and $z$ is equal to its radius, 6. Let's assume $z = 6$.

The side lengths then become $AB = x + y$, $BC = x + 6$ and $AC = y + 6$. Plugging into Pythagorean's theorem:

$(x + y)^2 = (x+6)^2 + (y + 6)^2$

$x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36$

$2xy - 12x - 12y = 72$

$xy - 6x - 6y = 36$

$(x - 6)(y - 6) - 36 = 36$

$(x - 6)(y - 6) = 72$

We can factor $72$ to arrive with $6$ pairs of solutions: $(7, 78), (8,42), (9, 30), (10, 24), (12, 18),$ and $(14, 15) \Rightarrow \mathrm{(A)}$.

Solution 3

By using $[ABC] = r\cdot s$ we get:

\[3(a+b+c) = r \cdot \frac{a+b+c}{2}\] \[r=6\] \[r = \frac{a+b-c}{2}\] \[a+b-c = 12\] \[c = a + b - 12\]


By the triangle's area we get:

\[\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}\] \[ab = 6(a+b+c)\] \[\frac{ab}{6} = a+b+c\] \[c = \frac{ab}{6} - a - b\]

$a + b - 12 = \frac{ab}{6} - a - b$, $ab - 12a - 12b + 72 = 0$, $(a - 12)(b - 12) = 72$

As $72 = 2^3 \cdot 3^2$, there are $\frac{(3+1)(2+1)}{2} = 6$ solutions, $\boxed{\textbf{(A) } 6}$

~isabelchen

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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