Difference between revisions of "2011 AMC 10B Problems/Problem 18"
m (→Solution) |
(→Solution 1) |
||
(28 intermediate revisions by 11 users not shown) | |||
Line 6: | Line 6: | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
<center><asy> | <center><asy> | ||
Line 21: | Line 21: | ||
draw(anglemark(A,M,D)); | draw(anglemark(A,M,D)); | ||
draw(anglemark(D,M,C)); | draw(anglemark(D,M,C)); | ||
− | + | ||
pair[] ps={A,B,C,D,M}; | pair[] ps={A,B,C,D,M}; | ||
Line 31: | 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> | ||
Line 43: | Line 43: | ||
x &= \boxed{\textbf{(E)} 75} | x &= \boxed{\textbf{(E)} 75} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ===Easier Way to Continue=== | ||
+ | After finding <math>MC = 6,</math> we can continue using trigonometry as follows. | ||
+ | |||
+ | We know that <math>\angle{BMC} = 180-2x</math> and so <math>\sin (180-2x) = \frac{3}{6} = \frac{1}{2}</math> | ||
+ | |||
+ | It is obvious that <math>\sin (30) = \frac{1}{2}</math> and so <math>180-2x=30.</math> | ||
+ | |||
+ | Solving, we have <math>x = \boxed{75}</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | == Solution 4 (elimination) == | ||
+ | Let <math>\angle AMD=\angle DMC=\theta</math>. Thus, <math>\angle BMC=180-2\theta\implies\angle MCB=2\theta-90</math>. Since all angles should be positive, <math>\theta>45^\circ</math>, narrowing the options to D and E. Trying <math>60^\circ</math> (option D), <math>\Delta AMD</math> is a 30-60-90 triangle. <math>AD=3</math>, so it follows that <math>AM=\sqrt3</math>. | ||
+ | Since <math>\angle BMC=180-2\theta</math>, <math>\angle BMC=60^\circ</math>, too. However, that would imply that <math>\Delta MBC</math> is also a <math>30-60-90</math> triangle, which would, in turn, imply that <math>MB=3\sqrt3</math>, since <math>BC=3</math>. We know that <math>AM+MB=AB</math> and <math>AB=6,</math> but we know that <math>AM=\sqrt3</math> and <math>MB=3\sqrt3</math>. <math>AM+MB</math> is clearly not <math>6</math>, so this is not possible. | ||
+ | Thus, the answer must be <math>\boxed{\textbf{(E)}~75}</math>. | ||
+ | ~ Technodoggo | ||
== See Also== | == See Also== |
Latest revision as of 19:18, 2 September 2024
Contents
Problem
Rectangle has
and
. Point
is chosen on side
so that
. What is the degree measure of
?
Solution 1
![[asy] unitsize(10mm); defaultpen(linewidth(.5pt)+fontsize(10pt)); dotfactor=3; pair A=(0,3), B=(6,3), C=(6,0), D=(0,0); pair M=(0.80385,3); draw(A--B--C--D--cycle); draw(M--C); draw(M--D); draw(anglemark(A,M,D)); draw(anglemark(D,M,C)); pair[] ps={A,B,C,D,M}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,N); label("$6$",midpoint(C--M),SW); label("$3$",midpoint(B--C),E); label("$6$",midpoint(C--D),S); [/asy]](http://latex.artofproblemsolving.com/7/3/0/730ddc517474ace5c79ccbe56c72ca8c918276c6.png)
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
Easier Way to Continue
After finding we can continue using trigonometry as follows.
We know that and so
It is obvious that and so
Solving, we have
~mathboy282
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
Solution 4 (elimination)
Let . Thus,
. Since all angles should be positive,
, narrowing the options to D and E. Trying
(option D),
is a 30-60-90 triangle.
, so it follows that
.
Since
,
, too. However, that would imply that
is also a
triangle, which would, in turn, imply that
, since
. We know that
and
but we know that
and
.
is clearly not
, so this is not possible.
Thus, the answer must be
.
~ Technodoggo
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.