# 1988 AJHSME Problems/Problem 24

## 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]$