Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 8 Problems/Problem 25"
(→‎Solution 1)
Line 53: Line 53:
 
==Solution 1==
 
==Solution 1==
  
−
The tops of <math>A</math>, <math>B</math>, <math>C</math>,<math>D</math> , and <math>E</math> in the figure formed has sum <math> 1+1+1+1 = 4 </math> as do the bottoms. Thus, the total so far is <math>8</math>. Now, one of the sides has area one, since it combines all of the heights of <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math>, which is <math>1</math>. The other side also satisfies this. Thus the total area now is <math>10</math>. From the front, the surface are is half, because if you looked at it straight from the front it would look exactly like <math>A</math>, with surface area half. From the back it is the same thing. Thus, the total is <math> 10+\frac{1}{2}+\frac{1}{2}= 11 </math>, or <math> \boxed{\textbf{(E)}\:11 } </math>.
+
The tops of <math>A</math>, <math>B</math>, <math>C</math>,<math>D</math> , and <math>E</math> in the figure formed has sum <math> 1+1+1+1 = 4 </math> as do the bottoms. Thus, the total so far is <math>8</math>. Now, one of the sides has area one, since it combines all of the heights of <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math>, which is <math>1</math>. The other side also satisfies this. Thus the total area now is <math>10</math>. From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like <math>A</math>, with surface area half. From the back it is the same thing. Thus, the total is <math> 10+\frac{1}{2}+\frac{1}{2}= 11 </math>, or <math> \boxed{\textbf{(E)}\:11 } </math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:00, 24 November 2014

Problem

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cube is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? [asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]

[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]

$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$

Solution 1

The tops of $A$, $B$, $C$,$D$ , and $E$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$. Now, one of the sides has area one, since it combines all of the heights of $A$, $B$, $C$, $D$, and $E$, which is $1$. The other side also satisfies this. Thus the total area now is $10$. From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like $A$, with surface area half. From the back it is the same thing. Thus, the total is $10+\frac{1}{2}+\frac{1}{2}= 11$, or $\boxed{\textbf{(E)}\:11 }$.

Solution 2

The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be $\boxed{\textbf{(E)}\:11 }$ since it is the only answer greater than 10.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_25&oldid=66027"