Difference between revisions of "1983 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
The solid shown has a [[square]] base of side length <math>s</math>. The upper edge is [[parallel]] to the base and has length <math>2s</math>. All edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?
+
The solid shown has a square base of side length <math>s</math>. The upper edge is parallel to the base and has length <math>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?
<!--<center><asy>
+
<center><asy>
 +
size(180);
 +
import three; pathpen = black+linewidth(0.65); pointpen = black;
 +
currentprojection = perspective(30,-20,10);
 +
real s = 6 * 2^.5;
 +
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);
 +
draw(A--B--C--D--A--E--D);
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draw(B--F--C);
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draw(E--F);
 +
label("A",A,W);
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label("B",B,S);
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label("C",C,SE);
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label("D",D,NE);
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label("E",E,N);
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label("F",F,N);
 +
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms -->
  
</asy></center> -->[[Category:Asymptote needed]]
+
== Solutions ==
  
[[Image:1983Number11.JPG]]
+
=== Solution 1 ===
== Solution ==
+
First, we find the height of the solid by dropping a perpendicular from the midpoint of <math>AD</math> to <math>EF</math>. The hypotenuse of the triangle formed is the [[median]] of equilateral triangle <math>ADE</math>, and one of the legs is <math>3\sqrt{2}</math>. We apply the Pythagorean Theorem to deduce that the height is <math>6</math>.
First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of <math>AD</math> to <math>EF</math>. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] <math>ADE</math> one of the legs is <math>3\sqrt{2}</math>. We apply the [[Pythagorean Theorem]] to find that the height is equal to <math>6</math>.
+
<center><asy>
 +
size(180);
 +
import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);
 +
currentprojection = perspective(30,-20,10);
 +
real s = 6 * 2^.5;
 +
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);
 +
triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);
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draw(A--B--C--D--A--E--D);
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draw(B--F--C);
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draw(E--F);
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draw(B--Ba--Ca--C,dashed+d);
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draw(A--Aa--Da--D,dashed+d);
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draw(E--(E.x,E.y,0),dashed+l);
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draw(F--(F.x,F.y,0),dashed+l);
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draw(Aa--E--Da,dashed+d);
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draw(Ba--F--Ca,dashed+d);
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label("A",A,S);
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label("B",B,S);
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label("C",C,S);
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label("D",D,NE);
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label("E",E,N);
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label("F",F,N);
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label("$12\sqrt{2}$",(E+F)/2,N);
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label("$6\sqrt{2}$",(A+B)/2,S);
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label("6",(3*s/2,s/2,3),ENE);
 +
</asy></center>
 +
Next, we complete the figure into a triangular prism, and find its volume, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>.
  
Next, we complete the figure into a triangular prism, and find the area, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>.
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Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is <math>2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>.
  
Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is <math>2\cdot \left( \frac{12\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>.
+
Thus, our answer is <math>432-144=\boxed{288}</math>.
 +
 
 +
=== Solution 2 ===
 +
<center><asy>
 +
size(180);
 +
import three; pathpen = black+linewidth(0.65); pointpen = black;
 +
currentprojection = perspective(30,-20,10);
 +
real s = 6 * 2^.5;
 +
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6);
 +
draw(A--B--C--D--A--E--D);
 +
draw(B--F--C);
 +
draw(E--F);
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draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed);
 +
label("A",A,(-1,-1,0));
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label("B",B,( 2,-1,0));
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label("C",C,( 1, 1,0));
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label("D",D,(-1, 1,0));
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label("E",E,(0,0,1));
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label("F",F,(0,0,1));
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label("G",G,(0,0,-1));
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label("H",H,(0,0,-1));
 +
</asy></center>
 +
Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. Now, we have a regular tetrahedron <math>EFGH</math>, which by symmetry has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is:
 +
 
 +
<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math>.
 +
 
 +
=== Solution 3 ===
 +
We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is <math>6</math>; thus, we will integrate with respect to height from <math>0</math> to <math>6</math>, noting that each cross section of height <math>dh</math> is a rectangle. The volume is then <math>\int_0^h(wl) \ \text{d}h</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. We can express <math>w</math> in terms of <math>h</math> as <math>w=6\sqrt{2}-\sqrt{2}h</math> since it decreases linearly with respect to <math>h</math>, and <math>l=6\sqrt{2}+\sqrt{2}h</math> since it similarly increases linearly with respect to <math>h</math>. Now we solve:<cmath>\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}</cmath>.
  
Thus, our answer is <math>432-144=\boxed{288}</math>.
+
==Solution 4==
 +
Draw an altitude from a vertex of the square base to the top edge. By using <math>30,60, 90</math> triangle ratios, we obtain that the altitude has a length of <math>3 \sqrt{6}</math>, and that little portion that hangs out has a length of <math>3\sqrt2</math>. This is a triangular pyramid with a base of <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math>, and a height of <math>3\sqrt{2}</math>. Since there are two of these, we can compute the sum of the volumes of these two to be <math>72</math>. Now we are left with a triangular prism with a base of dimensions <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math> and a height of <math>6\sqrt2</math>. We can compute the volume of this to be 216, and thus our answer is <math>\boxed{288}</math>.
 +
 
 +
pi_is_3.141
  
== See also ==
+
== See Also ==
 
{{AIME box|year=1983|num-b=10|num-a=12}}
 
{{AIME box|year=1983|num-b=10|num-a=12}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 10:30, 17 July 2020

Problem

The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid?

[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label("A",A,W); label("B",B,S); label("C",C,SE); label("D",D,NE); label("E",E,N); label("F",F,N); [/asy]

Solutions

Solution 1

First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$.

[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F);  draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label("A",A,S); label("B",B,S); label("C",C,S); label("D",D,NE); label("E",E,N); label("F",F,N); label("$12\sqrt{2}$",(E+F)/2,N); label("$6\sqrt{2}$",(A+B)/2,S); label("6",(3*s/2,s/2,3),ENE); [/asy]

Next, we complete the figure into a triangular prism, and find its volume, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$.

Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$.

Thus, our answer is $432-144=\boxed{288}$.

Solution 2

[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label("A",A,(-1,-1,0)); label("B",B,( 2,-1,0)); label("C",C,( 1, 1,0)); label("D",D,(-1, 1,0)); label("E",E,(0,0,1)); label("F",F,(0,0,1)); label("G",G,(0,0,-1)); label("H",H,(0,0,-1)); [/asy]

Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is:

$V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}$.

Solution 3

We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\int_0^h(wl) \ \text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\sqrt{2}-\sqrt{2}h$ since it decreases linearly with respect to $h$, and $l=6\sqrt{2}+\sqrt{2}h$ since it similarly increases linearly with respect to $h$. Now we solve:\[\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}\].

Solution 4

Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$, and that little portion that hangs out has a length of $3\sqrt2$. This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$, and a height of $3\sqrt{2}$. Since there are two of these, we can compute the sum of the volumes of these two to be $72$. Now we are left with a triangular prism with a base of dimensions $3\sqrt6, 3\sqrt6, 3\sqrt2$ and a height of $6\sqrt2$. We can compute the volume of this to be 216, and thus our answer is $\boxed{288}$.

pi_is_3.141

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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