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

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==Problem 23==
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==Problem==
 
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to <math>3</math> times their perimeters?
 
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to <math>3</math> times their perimeters?
  
 
<math>\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12</math>
 
<math>\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12</math>
  
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==Solution==
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==Solution 1==
  
 
Let <math>a</math> and <math>b</math> be the two legs of the triangle.
 
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>.
 
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>.
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Then <math>ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)</math>.
  
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We can complete the square under the root, and we get, <math>ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})</math>.
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We can complete the square under the root, and we get, <math>ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)</math>.
  
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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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Let <math>ab=p</math> and <math>a+b=s</math>, we have <math>p=6 \left(s+ \sqrt {s^2 - 2p}\right)</math>.
  
 
After rearranging, squaring both sides, and simplifying, we have <math>p=12s-72</math>.
 
After rearranging, squaring both sides, and simplifying, 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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Putting back <math>a</math> and <math>b</math>, and after factoring using Simon's Favorite Factoring Trick, we've got <math>(a-12)(b-12)=72</math>.
  
  
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And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
 
And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
  
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==Solution #2==
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We will proceed by using the fact that <math>[ABC] = r\cdot s</math>, where <math>r</math> is the radius of the incircle and <math>s</math> is the semiperimeter (<math>s = \frac{p}{2}</math>).
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Alternatively, note that <math>72 = 2^3 \cdot 3^2</math>. Then 72 has <math>(3+1)(2+1) = (4)(3) = 12</math> factors. However, half of these are repeats, so we have <math>\frac{12}{2} = 6</math> solutions.
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 +
==Solution 2==
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We will proceed by using the fact that <math>[ABC] = r\cdot s</math>, where <math>r</math> is the radius of the incircle and <math>s</math> is the semiperimeter <math>\left(s = \frac{p}{2}\right)</math>.
  
 
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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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>.
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The incircle of <math>ABC</math> 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 right 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>.
  
 
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:
 
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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<math>(x - 6)(y - 6) = 72</math>
 
<math>(x - 6)(y - 6) = 72</math>
  
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<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> A.
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We can factor <math>72</math> to arrive with <math>6</math> pairs of solutions: <math>(7, 78), (8,42), (9, 30), (10, 24), (12, 18),</math> and <math>(14, 15) \Rightarrow \mathrm{(A)}</math>.
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== Solution 3 ==
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 +
Let <math>a</math> and <math>b</math> be the two legs of the triangle, and <math>c</math> be the hypotenuse.
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 +
By using <math>Area = \frac{r}{2} (a+b+c)</math>, where <math>r</math> is the in-radius, we get:
 +
 
 +
<cmath>3(a+b+c) = \frac{r}{2} (a+b+c)</cmath>
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<cmath>r=6</cmath>
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 +
In right triangle, <math>r = \frac{a+b-c}{2}</math>
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<cmath>a+b-c = 12</cmath>
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<cmath>c = a + b - 12</cmath>
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 +
 
 +
By the triangle's area we get:
 +
 
 +
<cmath>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</cmath>
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<cmath>ab = 6(a+b+c)</cmath>
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 +
By substituting <math>c</math> in:
 +
 
 +
<cmath>ab = 6(a+b+a + b - 12)</cmath>
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<cmath>ab - 12a - 12b + 72 = 0</cmath>
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<cmath>(a - 12)(b - 12) = 72</cmath>
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 +
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>.
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 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
== Solution 4 ==
 +
All pythagorean triples can be parametrized in the form <math>(a, b, c) = k(r^2 - s^2), k(2rs), k(r^2 + s^2)</math> for positive integers <math>k, r, s</math>. The area being triple the perimeter implies that <cmath>k^2(r^2 - s^2)rs = 3(k(r^2 - s^2) + k(2rs) + k(r^2 + s^2)).</cmath> This can be simplified to get <cmath>ks(r - s) = 6.</cmath> Now, we get the triples <cmath>(k, r, s) = (1, 7, 1), (1, 5, 2), (1, 5, 3), (1, 7, 6), (2, 4, 1), (2, 4, 3), (3, 3, 1), (3, 3, 2), (6, 2, 1).</cmath> However, the ones where <math>r</math> and <math>s</math> are not different signs and relatively prime are redundant, so we get <math>6</math> triples total.
  
 
==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 20:00, 22 May 2023

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

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

By using $Area = \frac{r}{2} (a+b+c)$, where $r$ is the in-radius, we get:

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

In right triangle, $r = \frac{a+b-c}{2}$ \[a+b-c = 12\] \[c = a + b - 12\]


By the triangle's area we get:

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

By substituting $c$ in:

\[ab = 6(a+b+a + b - 12)\] \[ab - 12a - 12b + 72 = 0\] \[(a - 12)(b - 12) = 72\]

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

~isabelchen

Solution 4

All pythagorean triples can be parametrized in the form $(a, b, c) = k(r^2 - s^2), k(2rs), k(r^2 + s^2)$ for positive integers $k, r, s$. The area being triple the perimeter implies that \[k^2(r^2 - s^2)rs = 3(k(r^2 - s^2) + k(2rs) + k(r^2 + s^2)).\] This can be simplified to get \[ks(r - s) = 6.\] Now, we get the triples \[(k, r, s) = (1, 7, 1), (1, 5, 2), (1, 5, 3), (1, 7, 6), (2, 4, 1), (2, 4, 3), (3, 3, 1), (3, 3, 2), (6, 2, 1).\] However, the ones where $r$ and $s$ are not different signs and relatively prime are redundant, so we get $6$ triples total.

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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