Difference between revisions of "2017 AMC 12B Problems/Problem 15"
m (→Solution 2: Inspection) |
m (→Solution 4: Computing the Areas) |
||
Line 31: | Line 31: | ||
==Solution 4: Computing the Areas== | ==Solution 4: Computing the Areas== | ||
− | Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}ab\sin\theta</math> and letting side <math>AB = 1</math> for ease, we get: <math>4\cdot3\cdot\frac{sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}</math> | + | Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}ab\sin\theta</math> and letting side <math>AB = 1</math> for ease, we get: <math>4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}</math> |
Solution by Aadileo | Solution by Aadileo |
Revision as of 23:54, 20 June 2020
Contents
Problem 15
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 1: Law of Cosines
Solution by HydroQuantum
Let .
Recall The Law of Cosines. Letting , Since both and are both equilateral triangles, they must be similar due to similarity. This means that .
Therefore, our answer is .
Solution 2: Inspection(easiest solution)
Note that the height and base of are respectively 4 times and 3 times that of . Therefore the area of is 12 times that of .
By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of , or .
Solution 3: Coordinates
First we note that due to symmetry. WLOG, let and Therefore, . Using the condition that , we get and . It is easy to check that . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is
Solution by mathwiz0803
Solution 4: Computing the Areas
Note that angle is °, as it is supplementary to the equilateral triangle. Then, using area and letting side for ease, we get: as the area of . Then, the area of is , so the ratio is
Solution by Aadileo
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.