Difference between revisions of "2011 AMC 10B Problems/Problem 18"
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− | == Problem | + | == Problem== |
Rectangle <math>ABCD</math> has <math>AB = 6</math> and <math>BC = 3</math>. Point <math>M</math> is chosen on side <math>AB</math> so that <math>\angle AMD = \angle CMD</math>. What is the degree measure of <math>\angle AMD</math>? | Rectangle <math>ABCD</math> has <math>AB = 6</math> and <math>BC = 3</math>. Point <math>M</math> is chosen on side <math>AB</math> so that <math>\angle AMD = \angle CMD</math>. What is the degree measure of <math>\angle AMD</math>? | ||
<math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math> | <math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
<center><asy> | <center><asy> | ||
Line 30: | Line 31: | ||
label("$M$",M,N); | label("$M$",M,N); | ||
label("$6$",midpoint(C--M),SW); | label("$6$",midpoint(C--M),SW); | ||
− | |||
label("$3$",midpoint(B--C),E); | label("$3$",midpoint(B--C),E); | ||
+ | label("$6$",midpoint(C--D),S); | ||
</asy> | </asy> | ||
</center> | </center> | ||
− | It is given that <math>\angle AMD \ | + | It is given that <math>\angle AMD \sim \angle CMD</math>. Since <math>\angle AMD</math> and <math>\angle CDM</math> are alternate interior angles and <math>\overline{AB} \parallel \overline{DC}</math>, <math>\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM</math>. Use the [[Base Angle Theorem]] to show <math>\overline{DC} \cong \overline{MC}</math>. We know that <math>ABCD</math> is a [[rectangle]], so it follows that <math>\overline{MC} = 6</math>. We notice that <math>\triangle BMC</math> is a <math>30-60-90</math> triangle, and <math>\angle BMC = 30^{\circ}</math>. If we let <math>x</math> be the measure of <math>\angle AMD,</math> then |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2x + 30 &= 180\\ | 2x + 30 &= 180\\ | ||
Line 42: | Line 43: | ||
x &= \boxed{\textbf{(E)} 75} | x &= \boxed{\textbf{(E)} 75} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ==Solution 2 (with trig, not recommended)== | ||
+ | |||
+ | 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 the law of cosine, 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 relatively difficult to deal with. 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 | ||
+ | |||
+ | ==Solution 3(Easier Trig)== | ||
+ | We have <math>DC=CM=6</math>. By the Pythagorean Theorem, <math>BM=\sqrt{6^2-3^2}=3\sqrt{3}</math>, and thus <math>AM=6-3\sqrt{3}</math>, We have <math>\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}</math>, or <math>\angle AMD=\boxed{75}</math> | ||
+ | ~awsomek | ||
+ | |||
+ | == See Also== | ||
+ | |||
+ | {{AMC10 box|year=2011|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Revision as of 11:13, 13 July 2021
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 (with trig, not recommended)
Let . If we let , we have that , by the Pythagorean Theorem, and similarily, . Applying the law of cosine, we see that and YAY!!! We have two equations for two variables... that are relatively difficult to deal with. 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
Solution 3(Easier Trig)
We have . By the Pythagorean Theorem, , and thus , We have , or ~awsomek
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 |
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