Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>. | Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>. | ||
− | Now we see that <math>\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - | + | Now we see that <math>\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}</math>. Thus <math>\triangle CDE</math> is a right triangle, with side lengths of <math>3x</math>, <math>4x</math>, and <math>5x</math> (by the Pythagorean Theorem, or simply the Pythagorean triple <math>3-4-5</math>). Therefore <math>AC=4x</math> (by definition), <math>BC=5x+3x = 8x</math>, and <math>AB=4\sqrt{5}x</math>. Hence <math>\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1</math> (by the double angle formula), giving <math>2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}</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>. | 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>. |
Revision as of 22:59, 24 March 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 | |
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