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Difference between revisions of "1988 AJHSME Problems/Problem 24"
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==Solution==
 
==Solution==
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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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=== 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.
 
Alternately, we can simply keep track of the "bottom" side of the square. In the diagrams below, this bottom side is shown in red.

Latest revision as of 10:05, 11 June 2024

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

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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