Difference between revisions of "2007 AMC 12B Problems/Problem 23"
LOTRFan123 (talk | contribs) (→Solution 2) |
|||
(22 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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== |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | </math> | + | 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>. |
− | |||
− | </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>. |
− | |||
− | < | + | Factoring 72, we get 6 pairs of <math>a</math> and <math>b</math> |
− | |||
− | < | + | <math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math> |
− | |||
− | + | And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</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. | |
− | We | + | ==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 are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>. | |
− | + | 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: | ||
− | + | <math>(x + y)^2 = (x+6)^2 + (y + 6)^2</math> | |
+ | <math>x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36</math> | ||
− | + | <math>2xy - 12x - 12y = 72</math> | |
+ | <math>xy - 6x - 6y = 36</math> | ||
− | <math>( | + | <math>(x - 6)(y - 6) - 36 = 36</math> |
+ | <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>. | |
==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 16:36, 15 February 2021
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 .
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.