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> | |
+ | 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></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --> | ||
− | + | == Solutions == | |
− | == Solution == | + | === Solution 1 === |
− | First, we find the height of the | + | 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>. |
+ | <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); | ||
+ | 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></center> | ||
+ | Next, we complete t he 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>. | ||
− | + | 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>. | |
− | + | 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); | ||
+ | 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></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>. | |
− | |||
− | == | + | ===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 | |
+ | |||
+ | === Solution 5 === | ||
+ | From solution 1, the height of the solid is <math>6</math>. Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this triangular prism is <math>( 6\sqrt{2} )^2 \cdot 6 \cdot 1/2 = 216</math> Notice that the solid is symetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a regular tetrahedron. This tetrahedron has sidelength of <math>6\sqrt{2}</math> and since the formula for the volume of a regular tetrahedron is <math>\frac{s^3}{6\sqrt{2}}</math> the volume of this tetrahedron is <math>72</math>. <math>216 + 72 = \boxed{288}</math>. | ||
+ | |||
+ | ~skibidi solver | ||
+ | |||
+ | == See Also == | ||
+ | {{AIME box|year=1983|num-b=10|num-a=12}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 15:35, 27 May 2024
Contents
Problem
The solid shown has a square base of side length . The upper edge is parallel to the base and has length
. All other edges have length
. Given that
, 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]](http://latex.artofproblemsolving.com/5/6/8/5680a5bb8ba2d309d945d72d71950880a221cd5f.png)
Solutions
Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of to
. The hypotenuse of the triangle formed is the median of equilateral triangle
, and one of the legs is
. We apply the Pythagorean Theorem to deduce that the height is
.
![[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]](http://latex.artofproblemsolving.com/a/0/4/a0406c7f42ae451f7ee423c91bd5b18d4cd6b556.png)
Next, we complete t he figure into a triangular prism, and find its volume, which is .
Now, we subtract off the two extra pyramids that we included, whose combined volume is .
Thus, our answer is .
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]](http://latex.artofproblemsolving.com/c/0/d/c0daaf96bd5d22d9e77dbca583857221ea1bdb77.png)
Extend and
to meet at
, and
and
to meet at
. Now, we have a regular tetrahedron
, which by symmetry has twice the volume of our original solid. This tetrahedron has side length
. Using the formula for the volume of a regular tetrahedron, which is
, where S is the side length of the tetrahedron, the volume of our original solid is:
.
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 ; thus, we will integrate with respect to height from
to
, noting that each cross section of height
is a rectangle. The volume is then
, where
is the width of the rectangle and
is the length. We can express
in terms of
as
since it decreases linearly with respect to
, and
since it similarly increases linearly with respect to
. Now we solve:
.
Solution 4
Draw an altitude from a vertex of the square base to the top edge. By using triangle ratios, we obtain that the altitude has a length of
, and that little portion that hangs out has a length of
. This is a triangular pyramid with a base of
, and a height of
. Since there are two of these, we can compute the sum of the volumes of these two to be
. Now we are left with a triangular prism with a base of dimensions
and a height of
. We can compute the volume of this to be 216, and thus our answer is
.
pi_is_3.141
Solution 5
From solution 1, the height of the solid is . Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this triangular prism is
Notice that the solid is symetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a regular tetrahedron. This tetrahedron has sidelength of
and since the formula for the volume of a regular tetrahedron is
the volume of this tetrahedron is
.
.
~skibidi solver
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
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 |