Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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(diagram by djmathman) | (diagram by djmathman) | ||
− | ==Solution== | + | ==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>. | 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>. | ||
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Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | ||
− | ==Solution | + | ==Solution 3== |
First, we want to angle chase. Set <math><YXC</math> equal to <math>a</math> degrees. | First, we want to angle chase. Set <math><YXC</math> equal to <math>a</math> degrees. | ||
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Solution by IronicNinja | 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== | ==Video Solution by Richard Rusczyk== |
Latest revision as of 10:01, 27 December 2022
Contents
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 to
Since
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.