Difference between revisions of "1995 AJHSME Problems/Problem 21"

Revision as of 14:10, 5 July 2012

Problem

A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides as shown. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?

$[asy] draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw(circle((2,2),1)); draw((4,0)--(6,1)--(6,5)--(4,4)); draw((6,5)--(2,5)--(0,4)); draw(ellipse((5,2.5),0.5,1)); fill(ellipse((3,4.5),1,0.25),black); fill((2,4.5)--(2,5.25)--(4,5.25)--(4,4.5)--cycle,black); fill(ellipse((3,5.25),1,0.25),black); [/asy]$

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

It is rather easy to see that the only way that you can possibly have three cubes without protruding snaps is to have them in a triangle formation. However, in this case, the exterior angles are $120^\circ$ , whereas the outside angle of the cubes are $90^\circ$ each. A square like formation will work in the case of 4 cubes, so the answer is $\text{(B)}$

@@ Line 15: / Line 15: @@
 <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
+==Solution==
+It is rather easy to see that the only way that you can possibly have three cubes without protruding snaps is to have them in a triangle formation. However, in this case, the exterior angles are <math>120^\circ</math>, whereas the outside angle of the cubes are <math>90^\circ</math> each. A square like formation will work in the case of 4 cubes, so the answer is <math>\text{(B)}</math>

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Difference between revisions of "1995 AJHSME Problems/Problem 21"

Revision as of 14:10, 5 July 2012

Problem

Solution