Difference between revisions of "2019 AMC 12A Problems/Problem 19"
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math> | <math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44</math> | ||
+ | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (<math>\sin{x}</math> is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there). | Notice that by the Law of Sines, <math>a:b:c = \sin{A}:\sin{B}:\sin{C}</math>, so let's flip all the cosines using <math>\sin^{2}{x} + \cos^{2}{x} = 1</math> (<math>\sin{x}</math> is positive for <math>0^{\circ} < x < 180^{\circ}</math>, so we're good there). | ||
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=== Solution 2 === | === Solution 2 === | ||
<math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer. | <math>\angle ACB</math> is obtuse since its cosine is negative, so we let the foot of the altitude from <math>C</math> to <math>AB</math> be <math>H</math>. Let <math>AH=11x</math>, <math>AC=16x</math>, <math>BH=7y</math>, and <math>BC=8y</math>. By the Pythagorean Theorem, <math>CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}</math> and <math>CH=\sqrt{64y^2-49y^2}=y\sqrt{15}</math>. Thus, <math>y=3x</math>. The sides of the triangle are then <math>16x</math>, <math>11x+7(3x)=32x</math>, and <math>24x</math>, so for some integers <math>a,b</math>, <math>16x=a</math> and <math>24x=b</math>, where <math>a</math> and <math>b</math> are minimal. Hence, <math>\frac{a}{16}=\frac{b}{24}</math>, or <math>3a=2b</math>. Thus the smallest possible positive integers <math>a</math> and <math>b</math> that satisfy this are <math>a=2</math> and <math>b=3</math>, so <math>x=\frac{1}{8}</math>. The sides of the triangle are <math>2</math>, <math>3</math>, and <math>4</math>, so <math>\boxed{\textbf{(A) } 9}</math> is our answer. | ||
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=== Solution 3 === | === Solution 3 === |
Revision as of 13:24, 19 January 2021
Problem
In with integer side lengths, , , and. What is the least possible perimeter for ?
Solutions
Solution 1
Notice that by the Law of Sines, , so let's flip all the cosines using ( is positive for , so we're good there).
These are in the ratio , so our minimal triangle has side lengths , , and . is our answer.
Solution 2
is obtuse since its cosine is negative, so we let the foot of the altitude from to be . Let , , , and . By the Pythagorean Theorem, and . Thus, . The sides of the triangle are then , , and , so for some integers , and , where and are minimal. Hence, , or . Thus the smallest possible positive integers and that satisfy this are and , so . The sides of the triangle are , , and , so is our answer.
Solution 3
Using the law of cosines, we get the following equations:
Substituting for in and simplifying, we get the following:
Note that since are integers, we can solve this for integers. By some trial and error, we get that . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is .
~hiker
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.