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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==
+
==Solution 1==
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<math>\frac{1}{2}ab = 3(a+b+c)</math>
 
  
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<math>ab = 6(a+b+c)</math>
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Let <math>a</math> and <math>b</math> be the two legs of the triangle.
 +
 
 +
We have <math>\frac{1}{2}ab = 3(a+b+c)</math>.
 +
 
 +
Then <math>ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)</math>.
 +
 
 +
We can complete the square under the root, and we get, <math>ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)</math>.
 +
 
 +
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>.
 +
 
 +
 
 +
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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Using Euclid's formula for generating primitive triples:
 
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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>
+
Factoring 72, we get 6 pairs of <math>a</math> and <math>b</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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<math>mn(m-n)(m+n)k = 6m(m+n)</math>
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<math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math>
  
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<math>n(m-n)k = 6</math>
 
  
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Now we do some casework.
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And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
  
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For <math>k=1</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>
+
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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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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==Solution 2==
 +
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>.
  
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For <math>k=2</math>
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We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 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.
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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>.
  
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For <math>k=3</math>
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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>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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<math>(x + y)^2 = (x+6)^2 + (y + 6)^2</math>
  
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For <math>k=6</math>
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<math>x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36</math>
  
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<math>n(m-n)=1</math> has solution <math>(1,2)</math>, which is valid.
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<math>2xy - 12x - 12y = 72</math>
  
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This means that the solutions for <math>(m,n,k)</math> are
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<math>xy - 6x - 6y = 36</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>(x - 6)(y - 6) - 36 = 36</math>
  
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<math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>
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<math>(x - 6)(y - 6) = 72</math>
  
 +
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 2==
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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.
+
 
 +
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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We have <math>\frac{1}{2}ab = 3(a+b+c)</math>.
+
By using <math>Area = \frac{r}{2} (a+b+c)</math>, where <math>r</math> is the in-radius, we get:
  
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Then <math>ab=6\cdot (a+b+\sqrt {a^2 + b^2})</math>.
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<cmath>3(a+b+c) = \frac{r}{2} (a+b+c)</cmath>
 +
<cmath>r=6</cmath>
  
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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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In right triangle, <math>r = \frac{a+b-c}{2}</math>
 +
<cmath>a+b-c = 12</cmath>
 +
<cmath>c = a + b - 12</cmath>
  
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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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After rearranging, squaring both sides, and simplifying, we have <math>p=12s-72</math>.
+
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>
  
−
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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By substituting <math>c</math> in:
  
 +
<cmath>ab = 6(a+b+a + b - 12)</cmath>
 +
<cmath>ab - 12a - 12b + 72 = 0</cmath>
 +
<cmath>(a - 12)(b - 12) = 72</cmath>
  
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Factoring 72, we get 6 pairs of <math>a</math> and <math>b</math>
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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>.
  
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
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<math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math>
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== 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.
  
  
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And this gives us <math>8</math> solutions <math>\Rightarrow \mathrm{(C)}</math>.
+
==Solution 5 (very cheesy)==
 +
Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is <math>\boxed{\textbf{(A) } 6}</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}}

Latest revision as of 18:59, 4 January 2024

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.


Solution 5 (very cheesy)

Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is $\boxed{\textbf{(A) } 6}$.

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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