Difference between revisions of "1989 AJHSME Problems/Problem 23"
5849206328x (talk | contribs) (New page: ==Problem== An artist has <math>14</math> cubes, each with an edge of <math>1</math> meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surfa...) |
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The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a <math>3\times 3</math> square, hence this contributes <math>9</math> for the surface area. | The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a <math>3\times 3</math> square, hence this contributes <math>9</math> for the surface area. | ||
| + | |||
| + | You do not have to count the side underneath, since it is not exposed. | ||
Thus, the artist paints <math>24+9=33 \rightarrow \boxed{\text{C}}</math> square meters. | Thus, the artist paints <math>24+9=33 \rightarrow \boxed{\text{C}}</math> square meters. | ||
| − | == | + | ==See Also== |
{{AJHSME box|year=1989|num-b=22|num-a=24}} | {{AJHSME box|year=1989|num-b=22|num-a=24}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 10:08, 30 July 2023
Problem
An artist has 
cubes, each with an edge of 
meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint?
![[asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy]](http://latex.artofproblemsolving.com/c/9/0/c907b9d8e0a7e2f4eb5dc0d61c0fe69439512738.png)
Solution
We can consider the contributions of the sides of the three layers and the tops of the layers separately.
Layer 
(counting from the top starting at ) has 
side faces each with unit squares, so the sides of the pyramid contribute 
for the surface area.
The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a 
square, hence this contributes 
for the surface area.
You do not have to count the side underneath, since it is not exposed.
Thus, the artist paints 
square meters.
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
