Difference between revisions of "1983 AIME Problems/Problem 11"
(→Solution) |
Firebolt360 (talk | contribs) (→Solution 4) |
||
(16 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | The solid shown has a | + | 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); | size(180); | ||
Line 7: | Line 7: | ||
real s = 6 * 2^.5; | 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 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 --> | </asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --> | ||
− | == Solution 1 == | + | |
− | First, we find the height of the | + | == Solutions == |
+ | |||
+ | === Solution 1 === | ||
+ | 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> | <center><asy> | ||
size(180); | size(180); | ||
Line 19: | Line 29: | ||
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 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); | 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> | </asy></center> | ||
− | Next, we complete | + | 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 | + | 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>. | Thus, our answer is <math>432-144=\boxed{288}</math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
− | 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>. | + | <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 | |
− | == See | + | == 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 14:28, 20 June 2022
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?
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 .
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
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
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 |