Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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− | ==Problem | + | ==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> | ||
− | ==Solution== | + | ==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. | ||
Line 19: | Line 19: | ||
− | Putting back <math>a</math> and <math>b</math>, and after factoring using | + | 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>. |
Line 30: | Line 30: | ||
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>. | ||
− | ==Solution | + | |
+ | 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. | ||
+ | |||
+ | ==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>. | 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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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>. | 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>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let <math>a</math> and <math>b</math> be the two legs of the triangle, and <math>c</math> be the hypotenuse. | ||
+ | |||
+ | 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> | ||
+ | <cmath>r=6</cmath> | ||
+ | |||
+ | In right triangle, <math>r = \frac{a+b-c}{2}</math> | ||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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] | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | |||
+ | ==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 19:59, 4 January 2024
Contents
Problem
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution 1
Let and
be the two legs of the triangle.
We have .
Then .
We can complete the square under the root, and we get, .
Let and
, we have
.
After rearranging, squaring both sides, and simplifying, we have .
Putting back and
, and after factoring using Simon's Favorite Factoring Trick, we've got
.
Factoring 72, we get 6 pairs of and
And this gives us solutions
.
Alternatively, note that . Then 72 has
factors. However, half of these are repeats, so we have
solutions.
Solution 2
We will proceed by using the fact that , where
is the radius of the incircle and
is the semiperimeter
.
We are given .
The incircle of breaks the triangle's sides into segments such that
,
and
. Since ABC is a right triangle, one of
,
and
is equal to its radius, 6. Let's assume
.
The side lengths then become ,
and
. Plugging into Pythagorean's theorem:
We can factor to arrive with
pairs of solutions:
and
.
Solution 3
Let and
be the two legs of the triangle, and
be the hypotenuse.
By using , where
is the in-radius, we get:
In right triangle,
By the triangle's area we get:
By substituting in:
As , there are
solutions,
.
Solution 4
All pythagorean triples can be parametrized in the form for positive integers
. The area being triple the perimeter implies that
This can be simplified to get
Now, we get the triples
However, the ones where
and
are not different signs and relatively prime are redundant, so we get
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.