Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>. | By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>. | ||
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==Solution 3== | ==Solution 3== |
Revision as of 12:41, 23 December 2019
Problem
In with a right angle at , point lies in the interior of and point lies in the interior of so that and the ratio . What is the ratio
Solution 1
Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a triangle with .
Then , and is a triangle.
In isosceles triangles and , drop altitudes from and onto ; denote the feet of these altitudes by and respectively. Then by AAA similarity, so we get that , and . Similarly we get , and .
Solution 2
Let , and . (For this solution, is above , and is to the right of ). Also let , so , which implies . Similarly, , which implies . This further implies that .
Now we see that . Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ). Therefore (by definition), , and . Hence (by the double angle formula), giving .
By the Law of Cosines in , if , we have Now . Thus the answer is .
Solution 3
Draw a nice big diagram and measure. The answers to this problem are not very close, so it is quite easy to get to the correct answer by simply drawing a diagram. (Note: this strategy should only be used as a last resort!)
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
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