Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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+ | ==Solution 2== | ||
+ | <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as "t". Then <math><ACD</math> is <math>180-2t</math> degrees, which implies that <math><DCB</math> is <math>2t - 90</math> degrees. Similarly, the angle of point B is <math>90 - t</math> degrees, which implies that <math><BED</math> is<math>2t</math> degrees. This further implies that <math><DEC</math> is <math>180 - 2t</math> degrees. | ||
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+ | This may seem strange, but if you draw the diagram, the solution will work itself out like this. | ||
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+ | Now we see that <math><CDE = 180 - <ECD - <CED \Rightarrow 180 - 2x + 90 - 180 + 2x \Rightarrow 90</math>. Thus triangle CDE is a right triangle, with side lengths of 3x, 4x, and by the pythaogrean theorem, 5x. Now we see that AC is 4x (by definition), BC is 5x+3x = 8x, and AB is <math>4\sqrt{5}</math>x. Now, we find the cosine of 2y - this is <math>2cos^2x - 1</math>. which is <math>2*(\frac{1}{\sqrt{5}})^2 - 1 \Rightarrow \frac{-3}{5}</math> Using law of cosines on triangle BED, and denoting the length of BD as "d", we get <cmath>d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x)</cmath> <cmath>d^2 = 18x^2 + \frac{54x^2}{5} \Rightarrow {144x^2}{5}</cmath> <cmath>d = \frac{12x}{\sqrt{5}}</cmath> Since this is DB, and we know AB, to find the ratio we find AD, which is <math>\frac{4x}{\sqrt{5}} - \frac{12x}{\sqrt{5}}</math>, which is <math>\frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)\frac{2}{3}}</math> | ||
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+ | iron (note from me, if anyone wants to edit it to make it more clear/look better, that's fine with me - you can give yourself credit if you wish) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:14, 15 February 2019
Contents
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
Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a 3-4-5 triangle with .
Then , and is a 1-2- triangle.
On 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 .
-scrabbler94
Solution 2
, and . (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as "t". Then is degrees, which implies that is degrees. Similarly, the angle of point B is degrees, which implies that is degrees. This further implies that is degrees.
This may seem strange, but if you draw the diagram, the solution will work itself out like this.
Now we see that . Thus triangle CDE is a right triangle, with side lengths of 3x, 4x, and by the pythaogrean theorem, 5x. Now we see that AC is 4x (by definition), BC is 5x+3x = 8x, and AB is x. Now, we find the cosine of 2y - this is . which is Using law of cosines on triangle BED, and denoting the length of BD as "d", we get Since this is DB, and we know AB, to find the ratio we find AD, which is , which is . Thus the answer is
iron (note from me, if anyone wants to edit it to make it more clear/look better, that's fine with me - you can give yourself credit if you wish)
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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