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

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(→‎Solution #2)
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We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>.
 
We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>.
  
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\begin{tikzpicture}[thick]
+
The incircle of ABC breaks the triangle's sides into segments such that <math>AB = x + y</math>, <math>BC = x + z</math> and <math>AC = y + z</math>. Since ABC is a triangle, one of <math>x</math>, <math>y</math> and <math>z</math> is equal to its radius, 6. Let's assume <math>z = 6</math>.
−
\coordinate (O) at (0,0);
 
−
\coordinate (A) at (4,0);
 
−
\coordinate (B) at (0,2);
 
−
\draw (O)--(A)--(B)--cycle;
 
  
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\tkzLabelSegment[below=2pt](O,A){\textit{adjacent leg}}
+
The side lengths then become <math>AB = x + y</math>, <math>BC = x + 6</math> and <math>AC = y + 6</math>. Plugging into Pythagorean's theorem:
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\tkzLabelSegment[left=2pt](O,B){\textit{opposite leg}}
 
−
\tkzLabelSegment[above right=2pt](A,B){\textit{hypotenuse}}
 
  
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\tkzMarkAngle[fill= orange,size=0.65cm,%
+
(x + y)^2 = (x+6)^2 + (y + 6)^2
−
opacity=.4](A,O,B)
 
−
\tkzLabelAngle[pos = 0.35](A,O,B){<math>\gamma</math>}
 
  
−
\tkzMarkAngle[fill= orange,size=0.8cm,%
+
x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36
−
opacity=.4](B,A,O)
 
−
\tkzLabelAngle[pos = 0.6](B,A,O){<math>\alpha</math>}
 
  
−
\tkzMarkAngle[fill= orange,size=0.7cm,%
+
2xy - 12x - 12y = 72
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opacity=.4](O,B,A)
 
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\tkzLabelAngle[pos = 0.5](O,B,A){<math>\beta</math>}
 
  
 +
xy - 6x - 6y = 36
  
−
\end{tikzpicture}
+
(x - 6)(y - 6) - 36 = 36
 +
 
 +
(x - 6)(y - 6) = 72
 +
 
 +
<math>72 = 2^3 \cdot 3^2</math> so it has <math>(3 + 1)\cdot (2 + 1) = 4\cdot 3 = 12</math> factors, meaning <math>x - 6</math> and <math>y - 6</math> can take on 12 values. But for each pair of factors that multiply to 72, they produce one distinct triangle. Thus, the number of right triangles ABC that satisfy the given condition is <math>\frac{12}{2} = 6 \Rightarrow</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 12:57, 29 November 2014

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

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

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 ($s = \frac{p}{2}$).

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 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

$72 = 2^3 \cdot 3^2$ so it has $(3 + 1)\cdot (2 + 1) = 4\cdot 3 = 12$ factors, meaning $x - 6$ and $y - 6$ can take on 12 values. But for each pair of factors that multiply to 72, they produce one distinct triangle. Thus, the number of right triangles ABC that satisfy the given condition is $\frac{12}{2} = 6 \Rightarrow$

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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