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Difference between revisions of "1988 AJHSME Problems/Problem 24"

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</asy>
 
</asy>
  
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The square in the first diagram "rolls" clockwise around the fixed regular hexagon until it reaches the bottom.  In which position will the solid triangle be in diagram <math>4</math>?
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The [[square]] in the first diagram "rolls" clockwise around the fixed regular [[hexagon]] until it reaches the bottom.  In which position will the solid [[triangle]] be in diagram <math>4</math>?
  
 
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<asy>
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=== Solution 1 ===
 
=== Solution 1 ===
  
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The inner angle of the hexagon is <math>120^\circ</math>, and the inner angle of the square is <math>90^\circ</math>. Hence during each rotation the square is rotated by <math>360^\circ - 120^\circ - 90^\circ = 150^\circ</math> clockwise. In the diagram <math>4</math> the square is therefore rotated by <math>3\cdot 150^\circ = 450^\circ</math> clockwise from its original state. Rotation by <math>450^\circ</math> is identical to rotation by <math>450^\circ - 360^\circ = 90^\circ</math>, hence the black triangle in the diagram <math>4</math> will be pointing to the right, and the answer is <math>\boxed{\text{(A)}}</math>.
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The inner [[angle]] of the hexagon is <math>120^\circ</math>, and the inner angle of the square is <math>90^\circ</math>. Hence during each [[rotation]] the square is rotated by <math>360^\circ - 120^\circ - 90^\circ = 150^\circ</math> clockwise. In the diagram <math>4</math> the square is therefore rotated by <math>3\cdot 150^\circ = 450^\circ</math> clockwise from its original state. Rotation by <math>450^\circ</math> is identical to rotation by <math>450^\circ - 360^\circ = 90^\circ</math>, hence the black triangle in the diagram <math>4</math> will be pointing to the right, and the answer is <math>\boxed{\text{(A)}}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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==See Also==
 
==See Also==
  
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[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=23|num-a=25}}
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[[Category:Introductory Geometry Problems]]

Revision as of 16:00, 4 June 2009

Problem

[asy] unitsize(15); for (int a=0; a<6; ++a) { draw(2*dir(60a)--2*dir(60a+60),linewidth(1)); } draw((1,1.7320508075688772935274463415059)--(1,3.7320508075688772935274463415059)--(-1,3.7320508075688772935274463415059)--(-1,1.7320508075688772935274463415059)--cycle,linewidth(1)); fill((.4,1.7320508075688772935274463415059)--(0,3.35)--(-.4,1.7320508075688772935274463415059)--cycle,black); label("1.",(0,-2),S); draw(arc((1,1.7320508075688772935274463415059),1,90,300,CW)); draw((1.5,0.86602540378443864676372317075294)--(1.75,1.7)); draw((1.5,0.86602540378443864676372317075294)--(2.2,1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(4,1.7320508075688772935274463415059)--(3,0)--(4,-1.7320508075688772935274463415059)--(6,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(7.7320508075688772935274463415059,2.7320508075688772935274463415059)--(8.7320508075688772935274463415059,1)--cycle,linewidth(1)); label("2.",(5,-2),S); draw(arc((7,0),1,30,240,CW)); draw((6.5,-0.86602540378443864676372317075294)--(7.1,-.7)); draw((6.5,-0.86602540378443864676372317075294)--(6.8,-1.5)); draw((14,0)--(13,1.7320508075688772935274463415059)--(11,1.7320508075688772935274463415059)--(10,0)--(11,-1.7320508075688772935274463415059)--(13,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((14,0)--(13,-1.7320508075688772935274463415059)--(14.7320508075688772935274463415059,-2.7320508075688772935274463415059)--(15.7320508075688772935274463415059,-1)--cycle,linewidth(1)); label("3.",(12,-2.5),S); draw((21,0)--(20,1.7320508075688772935274463415059)--(18,1.7320508075688772935274463415059)--(17,0)--(18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--(20,-3.7320508075688772935274463415059)--(18,-3.7320508075688772935274463415059)--cycle,linewidth(1)); label("4.",(19,-4),S); [/asy]

The square in the first diagram "rolls" clockwise around the fixed regular hexagon until it reaches the bottom. In which position will the solid triangle be in diagram $4$?

[asy] unitsize(12); label("(A)",(0,0),W); fill((1,-1)--(1,1)--(5,0)--cycle,black); label("(B)",(6,0),E); fill((9,-2)--(11,-2)--(10,1)--cycle,black); label("(C)",(14,0),E); fill((17,1)--(19,1)--(18,-1.8)--cycle,black); label("(D)",(22,0),E); fill((25,-1)--(27,-2)--(28,1)--cycle,black); label("(E)",(31,0),E); fill((33,0)--(37,1)--(37,-1)--cycle,black); [/asy]

Solution

Solution 1

The inner angle of the hexagon is $120^\circ$, and the inner angle of the square is $90^\circ$. Hence during each rotation the square is rotated by $360^\circ - 120^\circ - 90^\circ = 150^\circ$ clockwise. In the diagram $4$ the square is therefore rotated by $3\cdot 150^\circ = 450^\circ$ clockwise from its original state. Rotation by $450^\circ$ is identical to rotation by $450^\circ - 360^\circ = 90^\circ$, hence the black triangle in the diagram $4$ will be pointing to the right, and the answer is $\boxed{\text{(A)}}$.

Solution 2

Alternately, we can simply keep track of the "bottom" side of the square. In the diagrams below, this bottom side is shown in red.

[asy] unitsize(15); for (int a=0; a<6; ++a) { draw(2*dir(60a)--2*dir(60a+60),linewidth(1)); } draw((1,1.7320508075688772935274463415059)--(1,3.7320508075688772935274463415059)--(-1,3.7320508075688772935274463415059)--(-1,1.7320508075688772935274463415059)--cycle,linewidth(1)); fill((.4,1.7320508075688772935274463415059)--(0,3.35)--(-.4,1.7320508075688772935274463415059)--cycle,black); label("1.",(0,-2),S); draw(arc((1,1.7320508075688772935274463415059),1,90,300,CW)); draw((1.5,0.86602540378443864676372317075294)--(1.75,1.7)); draw((1.5,0.86602540378443864676372317075294)--(2.2,1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(4,1.7320508075688772935274463415059)--(3,0)--(4,-1.7320508075688772935274463415059)--(6,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(7.7320508075688772935274463415059,2.7320508075688772935274463415059)--(8.7320508075688772935274463415059,1)--cycle,linewidth(1)); label("2.",(5,-2),S); draw(arc((7,0),1,30,240,CW)); draw((6.5,-0.86602540378443864676372317075294)--(7.1,-.7)); draw((6.5,-0.86602540378443864676372317075294)--(6.8,-1.5)); draw((14,0)--(13,1.7320508075688772935274463415059)--(11,1.7320508075688772935274463415059)--(10,0)--(11,-1.7320508075688772935274463415059)--(13,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((14,0)--(13,-1.7320508075688772935274463415059)--(14.7320508075688772935274463415059,-2.7320508075688772935274463415059)--(15.7320508075688772935274463415059,-1)--cycle,linewidth(1)); label("3.",(12,-2.5),S); draw((21,0)--(20,1.7320508075688772935274463415059)--(18,1.7320508075688772935274463415059)--(17,0)--(18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--(20,-3.7320508075688772935274463415059)--(18,-3.7320508075688772935274463415059)--cycle,linewidth(1)); label("4.",(19,-4),S); draw((1,1.7320508075688772935274463415059)--(-1,1.7320508075688772935274463415059),linewidth(1)+red); draw((6,1.7320508075688772935274463415059)--(7.7320508075688772935274463415059,2.7320508075688772935274463415059),linewidth(1)+red); draw((14.7320508075688772935274463415059,-2.7320508075688772935274463415059)--(15.7320508075688772935274463415059,-1),linewidth(1)+red); draw((18,-1.7320508075688772935274463415059)--(18,-3.7320508075688772935274463415059),linewidth(1)+red); [/asy]

See Also

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 Problems and Solutions
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