Difference between revisions of "2011 AMC 10B Problems/Problem 18"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
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x &= \boxed{\textbf{(E)} 75} | x &= \boxed{\textbf{(E)} 75} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>\angle{DMC} = \angle{AMD} = \theta</math>. If we let <math>AM = x</math>, we have that <math>MD = \sqrt{x^2 + 9}</math>, by the Pythagorean Theorem, and similarily, <math>MC = \sqrt{x^2 - 12x + 45}</math>. Applying LOC, we see that | ||
+ | <cmath>2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot cos (\theta) = 36</cmath> and <cmath>tan (\theta) = \frac{3}{x}</cmath>. YAY!!! We have two equations for two variables... that are terribly ugly. Well, we'll try to solve it. First of all, note that <math>sin (\theta) = \frac{3}{\sqrt{x^2+9}}</math>, so solving for <math>cos (\theta)</math> in terms of <math>x</math>, we get that <math>cos (\theta) = \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9}</math>. The equation now becomes | ||
+ | |||
+ | <cmath>2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9} = 36</cmath> | ||
+ | Simplifying, we get | ||
+ | |||
+ | <cmath>4x^4 - 48x^3 + 216x^2 - 432x + 324</cmath> | ||
+ | |||
+ | Now, we apply the quartic formula to get | ||
+ | |||
+ | <cmath>x = 6 \pm 3 \sqrt{3}</cmath>. | ||
+ | |||
+ | We can easily see that <math>x = 6 + 3 \sqrt{3}</math> is an invalid solution. Thus, <math>x = 6 - 3 \sqrt{3}</math>. | ||
+ | |||
+ | Finally, since <math>\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}</math>, <math>\theta = \frac{5 + 12n}{12} \pi</math>, where <math>n</math> is any integer. Converting to degrees, we have that <math>\theta = 75 + 180n</math>. Since <math>0 < \theta < 90</math>, we have that <math>\theta = \boxed{75}</math>. <math>\square</math> | ||
+ | |||
+ | ~ilovepi3.14 | ||
== See Also== | == See Also== |
Revision as of 20:41, 24 November 2018
Contents
Problem
Rectangle has and . Point is chosen on side so that . What is the degree measure of ?
Solution 1
It is given that . Since and are alternate interior angles and , . Use the Base Angle Theorem to show . We know that is a rectangle, so it follows that . We notice that is a triangle, and . If we let be the measure of then
Solution 2
Let . If we let , we have that , by the Pythagorean Theorem, and similarily, . Applying LOC, we see that and . YAY!!! We have two equations for two variables... that are terribly ugly. Well, we'll try to solve it. First of all, note that , so solving for in terms of , we get that . The equation now becomes
Simplifying, we get
Now, we apply the quartic formula to get
.
We can easily see that is an invalid solution. Thus, .
Finally, since , , where is any integer. Converting to degrees, we have that . Since , we have that .
~ilovepi3.14
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.