Difference between revisions of "2008 AMC 12A Problems/Problem 22"
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<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math> | <math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math> | ||
− | + | == Solution 1 (Trigonometry) == | |
− | |||
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: | Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: | ||
Line 45: | Line 44: | ||
draw((0,0)--(-0.5,3.9686));</asy> | draw((0,0)--(-0.5,3.9686));</asy> | ||
− | Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. | + | Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]] (the hexagon with side length <math>x</math> is regular). So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. |
By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. | By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. | ||
− | Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C</math>. | + | Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)</math>. |
− | + | == Solution 2 (without trigonometry) == | |
− | Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math> | + | Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math>. |
Line 87: | Line 86: | ||
That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math> | That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math> | ||
− | Since <math> \Delta | + | Since <math> \Delta ODE</math> is a right triangle, |
<math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math> | <math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math> | ||
− | Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math> | + | Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow (C) </math> |
+ | |||
+ | == Solution 3 (simply Pythagorean Theorem) == | ||
+ | <asy>unitsize(8mm); | ||
+ | defaultpen(linewidth(.8)+fontsize(8)); | ||
+ | draw(Circle((0,0),4)); | ||
+ | path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; | ||
+ | draw(mat); | ||
+ | draw(rotate(60)*mat); | ||
+ | draw(rotate(120)*mat); | ||
+ | draw(rotate(180)*mat); | ||
+ | draw(rotate(240)*mat); | ||
+ | draw(rotate(300)*mat); | ||
+ | label("\(x\)",(-1.55,2.1),E); | ||
+ | label("\(x\)",(0.03,1.5),E); | ||
+ | label("\(A\)",(-3.6,2.5513),E); | ||
+ | label("\(B\)",(-3.15,1.35),E); | ||
+ | label("\(C\)",(0.05,3.20),E); | ||
+ | label("\(D\)",(-0.75,4.15),E); | ||
+ | label("\(E\)",(0,4.17)); | ||
+ | label("\(F\)",(0.75,4.15),W); | ||
+ | label("\(O\)",(0.00,-0.10),E); | ||
+ | label("\(1\)",(-0.1,3.8),S); | ||
+ | label("\(4\)",(-0.4,2.2),S); | ||
+ | draw((0,0)--(0,3.103)); | ||
+ | draw((0,0)--(-2.687,1.5513)); | ||
+ | draw((0,0)--(-0.5,3.9686)); | ||
+ | draw((0,0)--(-0.5,3.9686));</asy> | ||
+ | |||
+ | By symmetry, <math>E</math> is the midpoint of <math>DF</math> and <math>OE</math> is an extension of <math>OC</math>. Thus <math>\angle OED = 90^\circ</math>. Since <math>OD = 4</math> and <math>DE = \frac{1}{2}</math>, <math>OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}</math>. Since <math>\triangle CED</math> is <math>30-60-90</math>, <math>CE = \frac{\sqrt{3}}{2}</math> (this can also be deduced from Pythagoras on <math>\triangle CED</math>). | ||
+ | |||
+ | Thus <math>OC = \frac{3\sqrt{7}-\sqrt{3}}{2}</math>. As previous solutions noted, <math>\triangle BOC</math> is equilateral, and thus the desired length is <math>x = OC \implies (C)</math>. | ||
==Solution 3== | ==Solution 3== | ||
Line 121: | Line 151: | ||
draw((0.5,-3.9686)--(-0.5,3.9686));</asy> | draw((0.5,-3.9686)--(-0.5,3.9686));</asy> | ||
− | Looking at the diagram above, we know that <math>BE</math> is a diameter of circle <math>O</math> due to symmetry. Due to Thales' theorem, triangle <math>ABE</math> is a right triangle with <math>A = 90 ^\circ</math>. <math>AE</math> lies on <math>AD</math> and <math>GE</math> because <math>BAD</math> is also a right angle. To find the length of <math>DG</math>, notice that if we draw a line from <math>F</math> to <math>M</math>, the midpoint of line <math>DG</math>, it creates two <math>30</math> - <math>60</math> - <math>90</math> triangles. Therefore, <math>MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x</math>. <math>AE = 2 + \sqrt{3}x</math> | + | Looking at the diagram above, we know that <math>BE</math> is a diameter of circle <math>O</math> due to symmetry. Due to [[Thales' theorem]], triangle <math>ABE</math> is a right triangle with <math>A = 90 ^\circ</math>. <math>AE</math> lies on <math>AD</math> and <math>GE</math> because <math>BAD</math> is also a right angle. To find the length of <math>DG</math>, notice that if we draw a line from <math>F</math> to <math>M</math>, the midpoint of line <math>DG</math>, it creates two <math>30</math> - <math>60</math> - <math>90</math> triangles. Therefore, <math>MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x</math>. <math>AE = 2 + \sqrt{3}x</math> |
− | + | Applying the Pythagorean theorem to triangle <math>ABE</math>, we get <cmath>(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0</cmath> Using the quadratic formula to solve, we get <cmath>x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}</cmath> <math>x</math> must be positive, therefore <cmath>x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow (C)</cmath> | |
~Zeric Hang | ~Zeric Hang | ||
+ | |||
+ | ==Solution 4 (coordinate bashing)== | ||
+ | <asy>unitsize(8mm); | ||
+ | defaultpen(linewidth(.8)+fontsize(8)); | ||
+ | draw(Circle((0,0),4)); | ||
+ | path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; | ||
+ | draw(mat); | ||
+ | draw(rotate(60)*mat); | ||
+ | draw(rotate(120)*mat); | ||
+ | draw(rotate(180)*mat); | ||
+ | draw(rotate(240)*mat); | ||
+ | draw(rotate(300)*mat); | ||
+ | label("\(x\)",(-1.55,2.1),E); | ||
+ | label("\(1\)",(-0.5,3.8),S);</asy> | ||
+ | |||
+ | We will let <math>O(0,0)</math> be the origin. This way the coordinates of <math>C</math> will be <math>(0,y)</math>. By <math>30-60-90</math>, the coordinates of <math>D</math> will be <math>\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)</math>. The distance any point with coordinates <math>(x, y)</math> is from the origin is <math>\sqrt{x^2 + y^2}</math>. Therefore, the distance <math>D</math> is from the origin is <math>4</math> and <math>\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16</math>. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)</math> | ||
+ | |||
+ | Note: Since <math>C</math> and <math>D</math> are not labeled in the diagram, refer to solution 1 for the location of points <math>C</math> and <math>D</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | <asy>unitsize(8mm); | ||
+ | defaultpen(linewidth(.8)+fontsize(8)); | ||
+ | draw(Circle((0,0),4)); | ||
+ | path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; | ||
+ | draw(mat); | ||
+ | draw(rotate(60)*mat); | ||
+ | draw(rotate(120)*mat); | ||
+ | draw(rotate(180)*mat); | ||
+ | draw(rotate(240)*mat); | ||
+ | draw(rotate(300)*mat); | ||
+ | label("\(x\)",(-1.55,2.1),E); | ||
+ | label("\(A\)",(-3.6,2.5513),E); | ||
+ | label("\(B\)",(-3.15,1.35),E); | ||
+ | label("\(C\)",(0.05,3.20),E); | ||
+ | label("\(D\)",(-0.75,4.15),E); | ||
+ | label("\(E\)",(0.3,4.15),E); | ||
+ | label("\(F\)",(-3.4,1.89),E); | ||
+ | draw((0.5,3.9686)--(-3.13,2.45)); | ||
+ | draw((0.5,3.9686)--(-2.95,2));</asy> | ||
+ | |||
+ | Notice that <math>\overarc{AE}</math> is <math>\frac16</math> the circumference of the circle. Therefore, <math>\overline{AE}</math> is the side length of an inscribed hexagon with side length <math>4</math>. <math>\triangle AFE</math> is a right triangle with <math>\overline{AF}=\frac12</math>. The length of <math>\overline{EF}=x+\frac{\sqrt{3}}{2}</math>. Using the [[Pythagorean Theorem]], we get | ||
+ | |||
+ | |||
+ | <math>4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2</math> | ||
+ | |||
+ | |||
+ | Solving for <math>x</math>, we get <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}</math> | ||
+ | |||
+ | ==Solution 6 (Answer Choices & Estimation)== | ||
+ | The smallest distance between the intersection between two adjacent placemats and the circle is a bit less than <math>1</math>. Thus, the answer will be a bit more than <math>4-1=3.</math> Going up, we guess that the next one up is the answer, <math>\frac{3\sqrt{7}-\sqrt{3}}{2},</math> which is about <math>3.1</math>. The next one up after that is <math>2\sqrt{3},</math> which is about <math>3.5</math> and seems too big. So we circle <math>\boxed{\text{(C) }\frac{3\sqrt{7}-\sqrt{3}}{2}}</math> and are happy we just cheesed an <math>AMC</math> <math>10</math> problem <math>25.</math> | ||
+ | |||
==See Also== | ==See Also== | ||
+ | {{AMC10 box|year=2008|ab=A|num-b=24|after=Last question}} | ||
{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}} | ||
− | |||
− | |||
− | |||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:00, 16 May 2023
- The following problem is from both the 2008 AMC 12A #22 and 2008 AMC 10A #25, so both problems redirect to this page.
Contents
Problem
A round table has radius . Six rectangular place mats are placed on the table. Each place mat has width
and length
as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length
. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is
?
Solution 1 (Trigonometry)
Let one of the mats be , and the center be
as shown:
Since there are mats,
is equilateral (the hexagon with side length
is regular). So,
. Also,
.
By the Law of Cosines: .
Since must be positive,
.
Solution 2 (without trigonometry)
Draw and
as in the diagram. Draw the altitude from
to
and call the intersection
.
As proved in the first solution, .
That makes
a
triangle, so
and
Since is a right triangle,
Solving for gives
Solution 3 (simply Pythagorean Theorem)
By symmetry, is the midpoint of
and
is an extension of
. Thus
. Since
and
,
. Since
is
,
(this can also be deduced from Pythagoras on
).
Thus . As previous solutions noted,
is equilateral, and thus the desired length is
.
Solution 3
Looking at the diagram above, we know that is a diameter of circle
due to symmetry. Due to Thales' theorem, triangle
is a right triangle with
.
lies on
and
because
is also a right angle. To find the length of
, notice that if we draw a line from
to
, the midpoint of line
, it creates two
-
-
triangles. Therefore,
.
Applying the Pythagorean theorem to triangle , we get
Using the quadratic formula to solve, we get
must be positive, therefore
~Zeric Hang
Solution 4 (coordinate bashing)
We will let be the origin. This way the coordinates of
will be
. By
, the coordinates of
will be
. The distance any point with coordinates
is from the origin is
. Therefore, the distance
is from the origin is
and
. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that
Note: Since and
are not labeled in the diagram, refer to solution 1 for the location of points
and
.
Solution 5
Notice that is
the circumference of the circle. Therefore,
is the side length of an inscribed hexagon with side length
.
is a right triangle with
. The length of
. Using the Pythagorean Theorem, we get
Solving for , we get
Solution 6 (Answer Choices & Estimation)
The smallest distance between the intersection between two adjacent placemats and the circle is a bit less than . Thus, the answer will be a bit more than
Going up, we guess that the next one up is the answer,
which is about
. The next one up after that is
which is about
and seems too big. So we circle
and are happy we just cheesed an
problem
See Also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 |
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.