Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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<math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | <math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | ||
− | == | + | <asy> |
+ | size(200); | ||
+ | defaultpen(linewidth(1)); | ||
+ | pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); | ||
+ | pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; | ||
+ | draw(A--B--C--D--E--F--cycle); | ||
+ | draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); | ||
+ | dot("$A$",A,W,linewidth(4)); | ||
+ | dot("$B$",B,dir(0),linewidth(4)); | ||
+ | dot("$C$",C,dir(0),linewidth(4)); | ||
+ | dot("$D$",D,dir(20),linewidth(4)); | ||
+ | dot("$E$",E,dir(100),linewidth(4)); | ||
+ | dot("$F$",F,W,linewidth(4)); | ||
+ | dot("$X$",X,dir(0),linewidth(4)); | ||
+ | dot("$Y$",Y,N,linewidth(4)); | ||
+ | dot("$Z$",Z,W,linewidth(4)); | ||
+ | </asy> | ||
− | Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and | + | (diagram by djmathman) |
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | We can, <math>\textsc{wlog}</math>, assume <math>Y</math> coincides with <math>D</math> and <math>CD\parallel AF</math> as before. In which case, we will have <math>BC=EF=41(\sqrt{3}-1)</math>. So we have square <math>AXDZ</math> inscribed in equiangular hexagon <math>ABCDEF</math> with <math>X</math> on <math>\overline{BC}</math> and <math>Z</math> on <math>\overline{EF}</math>. | ||
+ | <asy> | ||
+ | size(200); defaultpen(fontsize(10)+linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; pair Cp=extension(B,C,Y,Y+dir(-60)); draw(A--B--Cp--Y--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); dot("$A$",A,W,linewidth(4)); dot("$B$",B,dir(0),linewidth(4)); dot("$C$",Cp,dir(0),linewidth(4)); dot("$E$",E,dir(100),linewidth(4)); dot("$F$",F,W,linewidth(4)); dot("$X$",X,dir(0),linewidth(4)); dot("$D$",Y,N,linewidth(4)); dot("$Z$",Z,W,linewidth(4)); label("$u$", B--X, SE);label("$v$", X--Cp, SE); label("$40$", A--B, S); label("$s$", A--X, NW); label("$s$", Y--X, SW); </asy> | ||
+ | Let <math>\angle BXA = \theta</math>; then <math>\angle BAX=60^\circ -\theta</math>. Let <math>BX=u</math>. In <math>\triangle ABX</math> we have | ||
+ | <cmath>\begin{align} | ||
+ | \frac{2s}{\sqrt{3}}=\frac{u}{\sin(60^\circ-\theta)}=\frac {40}{\sin\theta} | ||
+ | \end{align}</cmath> | ||
+ | We also have <math>\angle CXD=90^\circ - \theta</math> and <math>\angle CDX = \theta-30^\circ</math>. Let <math>CX=v</math>. In <math>\triangle CDX</math> we have | ||
+ | <cmath>\begin{align}\tag{2} | ||
+ | \frac{2s}{\sqrt{3}}=\frac{v}{\sin(\theta-30^\circ)}=\frac {CD}{\cos\theta} | ||
+ | \end{align}</cmath> | ||
+ | Now <math>BC=u+v=41(\sqrt{3}-1)</math>. From <math>(1)</math> and <math>(2)</math> we get<cmath>\begin{align*} | ||
+ | 41(\sqrt{3}-1) &= \frac{2s}{\sqrt{3}}\left(\sin(60^\circ-\theta)+\sin(\theta-30^\circ)\right) \ | ||
+ | &= \frac{2s}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}2\cdot (\sin\theta + \cos\theta) | ||
+ | \end{align*}</cmath> | ||
+ | From <math>(1)</math> we get <math>s\sin\theta = 20\sqrt{3}</math> and therefore <math>s\cos\theta = \sqrt{s^2-3\cdot 20^2}</math>. Thus | ||
+ | <cmath>41(\sqrt{3}-1) = \frac{\sqrt{3}-1}{\sqrt{3}}(20\sqrt{3}+\sqrt{s^2-3\cdot 20^2})</cmath>which simplifies to<cmath>3\cdot 21^2 = s^2-3\cdot 20^2.</cmath>Since <math>(20, 21, 29)</math> is a Pythagorean triple, we get <math>s=29\sqrt{3}</math>, i.e. <math>\framebox{A}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and because <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle YEZ</math>. We now get <math>YE=AB=40</math>. | ||
Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | ||
Line 32: | Line 72: | ||
Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | First, we want to angle chase. Set <math><YXC</math> equal to <math>a</math> degrees. | ||
+ | |||
+ | Now the key idea is that you want to relate the numbers that you have. You know <math>\overline{AB} = 40</math> and that <math>\overline{EZ} + \overline{ZF} = 41(\sqrt{3}-1)</math>. We proceed with the Law of Sines. | ||
+ | |||
+ | Call the side length of the square x. Then we are going to set a constant k equal to <math>\frac{\sin 120^{\circ}}{x}</math>, and this is consistent for every triangle in the diagram because all the angles of the hexagon are equiangular (and so they are all <math>120^{\circ}</math>). | ||
+ | |||
+ | Then we get the following process: | ||
+ | <cmath>\frac{\sin(90-a)}{40} = k</cmath> | ||
+ | <cmath>\cos a = 40k</cmath> | ||
+ | |||
+ | <cmath>\frac{\sin(a-30)}{\overline{EZ}} = k</cmath> | ||
+ | <cmath>\sin(a-30) = \overline{EZ}\cdot k</cmath> | ||
+ | <cmath>\frac{\sin(60-a)}{\overline{ZF}} = k</cmath> | ||
+ | <cmath>\sin(60-a) = \overline{ZF}\cdot k</cmath> | ||
+ | <cmath>\sin(a-30) + \sin(60-a) = k\cdot 41(\sqrt{3}-1)</cmath> | ||
+ | |||
+ | And now expanding using our trig formulas, we get: | ||
+ | <cmath>(\sin a + \cos a)(\frac{\sqrt{3}-1}{2} = k\cdot 41(\sqrt{3}-1)</cmath> | ||
+ | <cmath>\sin a + \cos a = 82k</cmath> | ||
+ | <cmath>\sin a = 42k</cmath> | ||
+ | |||
+ | And so now we have a triangle where <math>\cos a = 40k</math> and <math>\sin a = 42k</math>. Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get: | ||
+ | <cmath>\sqrt{(40k)^2 + (42k)^2} = 1</cmath> | ||
+ | <cmath>3364k^2 = 1</cmath> | ||
+ | <cmath>k = \frac{1}{58}</cmath> | ||
+ | |||
+ | And since <math>k = \frac{\sin(120^{\circ})}{x}</math>, then: | ||
+ | <cmath>x = \frac{\sqrt{3}}{2}\cdot58</cmath> | ||
+ | <cmath>x = \boxed{29\sqrt{3}}</cmath> | ||
+ | |||
+ | Solution by IronicNinja | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>EZ = x</math>, <math>\angle XAB = \alpha</math> | ||
+ | |||
+ | <math>\angle BAX = \angle EYZ</math>, <math>AX = YZ</math>, <math>\angle ZEY = \angle XBA</math>, <math>\triangle BAX \cong \triangle EYZ</math> by <math>AAS</math>, <math>BX = EZ = x</math> | ||
+ | |||
+ | <math>\angle AXB = 180^\circ - 120^\circ - \alpha = 60^\circ - \alpha</math> | ||
+ | |||
+ | <math>\frac{XB}{\sin \angle XAB} = \frac{AX}{\sin \angle ABX} = \frac{AB}{\angle AXB}</math>, <math>\frac{x}{\sin \alpha} = \frac{AX}{\sin 120^\circ} = \frac{40}{\sin (60^\circ - \alpha)}</math> | ||
+ | |||
+ | <math>\angle ZAF = 120^\circ - 90^\circ - \alpha = 30^\circ - \alpha</math> | ||
+ | |||
+ | <math>ZF = 41(\sqrt{3} - 1) - x</math>, <math>\frac{ZF}{\sin \angle ZAF} = \frac{AZ}{ \sin \angle ZFA}</math>, <math>\frac{41(\sqrt{3} - 1) - x}{\sin (30^\circ - \alpha)} = \frac{AZ}{ \sin 120^\circ}</math> | ||
+ | |||
+ | <math>\frac{x}{\sin \alpha} = \frac{41(\sqrt{3} - 1) - x}{\sin (30^\circ - \alpha)}= \frac{40}{\sin(60^\circ - \alpha)}</math> | ||
+ | |||
+ | <math>40 \cdot \sin \alpha = x(\sin 60^\circ \cos \alpha - \cos 60^\circ \sin \alpha)</math> <math>(1)</math> | ||
+ | |||
+ | <math>x(\sin 30^\circ \cos \alpha - \cos 30^\circ \sin \alpha) = [41(\sqrt{3} - 1) - x] \sin \alpha</math> <math>(2)</math> | ||
+ | |||
+ | By simplifying <math>(1)</math> we get, <math>\frac{\sqrt{3}}{2} \cdot x \cdot \cos \alpha - \frac{\sin \alpha}{2} \cdot x = 40 \cdot \sin \alpha</math> <math>(3)</math> | ||
+ | |||
+ | By simplifying <math>(2)</math> we get, <math>\frac{\cos \alpha}{2} \cdot x + \frac{2 - \sqrt{3}}{2} \cdot \sin \alpha \cdot x = 41(\sqrt{3} - 1)\sin \alpha </math> <math>(4)</math> | ||
+ | |||
+ | By <math>\sqrt{3} \cdot (4) - (3)</math> we get, <math>\frac{2\sqrt{3} - 3 +1}{2} \cdot \sin \alpha \cdot x = [41(3-\sqrt{3}) - 40] \sin \alpha</math> | ||
+ | |||
+ | <math>(\sqrt{3} - 1)x = 83 - 41\sqrt{3}</math>, <math>x = 21 \sqrt{3} - 20</math> | ||
+ | |||
+ | By the law of cosine <math>AX = \sqrt{(21 \sqrt{3} - 20)^2 + 40^2 - 2 \cdot (21 \sqrt{3} - 20) 40 \cdot \cos 120^\circ} = \boxed{\textbf{(A) } 29 \sqrt{3}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc12b/273 | ||
+ | |||
+ | ~dolphin7 | ||
== See Also == | == See Also == |
Latest revision as of 09:01, 27 December 2022
Contents
[hide]Problem 21
Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?
(diagram by djmathman)
Solution 1
We can, , assume coincides with and as before. In which case, we will have . So we have square inscribed in equiangular hexagon with on and on . Let ; then . Let . In we have We also have and . Let . In we have Now . From and we get From we get and therefore . Thus which simplifies toSince is a Pythagorean triple, we get , i.e. .
Solution 2
Extend and so that they meet at . Then , so and because is parallel to . Also, since is parallel and equal to , we get , hence is congruent to . We now get .
Let , , and .
Drop a perpendicular line from to the line of that meets line at , and a perpendicular line from to the line of that meets at , then is congruent to since is complementary to . Then we have the following equations:
The sum of these two yields that
So, we can now use the law of cosines in :
Therefore
Solution 3
First, we want to angle chase. Set equal to degrees.
Now the key idea is that you want to relate the numbers that you have. You know and that . We proceed with the Law of Sines.
Call the side length of the square x. Then we are going to set a constant k equal to , and this is consistent for every triangle in the diagram because all the angles of the hexagon are equiangular (and so they are all ).
Then we get the following process:
And now expanding using our trig formulas, we get:
And so now we have a triangle where and . Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get:
And since , then:
Solution by IronicNinja
Solution 4
Let ,
, , , by ,
,
, ,
By simplifying we get,
By simplifying we get,
By we get,
,
By the law of cosine
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12b/273
~dolphin7
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.