Difference between revisions of "2003 AMC 10A Problems/Problem 10"

(added pic)
(Redirected page to 2003 AMC 12A Problems/Problem 13)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
== Problem ==
+
#REDIRECT[[2003 AMC 12A Problems/Problem 13]]
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
 
 
 
[[Image:2003amc10a10.gif]]
 
 
 
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
 
 
 
== Solution ==
 
[[Image:2003amc10a10solution.gif]]
 
 
 
Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>.
 
 
 
When the polygon is folded, the "right" edge of square <math>A</math> becomes adjacent to the "bottom edge" of square <math>C</math>, and the "bottom" edge of square <math>A</math> becomes adjacent to the "bottom" edge of square <math>D</math>.
 
 
 
So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.
 
 
 
Therefore, squares <math>1</math>, <math>2</math>, and <math>3</math> will prevent the polygon from becoming a cube with one face missing.
 
 
 
Squares  <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math> will allow the polygon to become a cube with one face missing when folded.
 
 
 
Thus the answer is <math>6 \Rightarrow E</math>.
 
 
 
== See Also ==
 
*[[2003 AMC 10A Problems]]
 
*[[2003 AMC 10A Problems/Problem 9|Previous Problem]]
 
*[[2003 AMC 10A Problems/Problem 11|Next Problem]]
 
 
 
[[Category:Introductory Geometry Problems]]
 

Latest revision as of 17:21, 31 July 2011