Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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===Solution 2 === | ===Solution 2 === | ||
As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a 30-60-90 with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}</math>. | As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a 30-60-90 with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}</math>. | ||
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+ | ===Solution 3 === | ||
+ | Let <math>AB=BC=CA=x</math>. We start by noting that we can just write <math>AB'</math> as just <math>AB+BB'=4AB</math>. | ||
+ | Similarly <math>BC'=4BC</math>, and <math>CA'=4CA</math>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula, | ||
+ | <math>[ABC]=\sqrt{\frac{3x}{2}\cdot {\Bigg(\frac{3x}{2}-x\Bigg)}^3}=\frac{x^2\sqrt{3}}{4}</math>. | ||
+ | Next in order to evaluate <math>A'B'C'</math> we need to evaluate the area of the larger triangles <math>AA'B',BB'C', \text{ and } CC'A'</math>. | ||
+ | In this solution we shall just compute <math>1</math> of these as the others are trivially equivalent. | ||
+ | In order to compute the area of <math>\Delta{AA'B'}</math> we can use the formula <math>[XYZ]=\frac{1}{2}xy\cdot\sin{z}</math>. | ||
+ | Since <math>ABC</math> is equilateral and <math>A</math>, <math>B</math>, <math>B'</math> are collinear, we already know <math>\angle{A'AB'}=180-60=120</math> | ||
+ | Similarly from above we know <math>AB'</math> and <math>A'A</math> to be <math>4x</math>, and <math>3x</math> respectively. Thus the area of <math>\Delta{AA'B'}</math> is <math>\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}</math>. Likewise we can find <math>BB'C', \text{ and } CC'A'</math> to also be <math>3x^2\cdot\sqrt{3}</math>. | ||
+ | <math>[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x\sqrt{3}}{2}=\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)</math>. | ||
+ | Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37 : 1}</math> | ||
==See Also== | ==See Also== |
Revision as of 01:05, 25 February 2017
Problem
Let be an equilateral triangle. Extend side beyond to a point so that . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?
Solution
Solution 1
Note that by symmetry, is also equilateral. Therefore, we only need to find one of the sides of to determine the area ratio. WLOG, let . Therefore, and . Also, , so by the Law of Cosines, . Therefore, the answer is
Solution 2
As mentioned in the first solution, is equilateral. WLOG, let . Let be on the line passing through such that is perpendicular to . Note that is a 30-60-90 with right angle at . Since , and . So we know that . Note that is a right triangle with right angle at . So by the Pythagorean theorem, we find Therefore, the answer is .
Solution 3
Let . We start by noting that we can just write as just . Similarly , and . We can evaluate the area of triangle by simply using Heron's formula, . Next in order to evaluate we need to evaluate the area of the larger triangles . In this solution we shall just compute of these as the others are trivially equivalent. In order to compute the area of we can use the formula . Since is equilateral and , , are collinear, we already know Similarly from above we know and to be , and respectively. Thus the area of is . Likewise we can find to also be . . Therefore the ratio of to is
See Also
2017 AMC 10B (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 10 Problems and Solutions |
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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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