Difference between revisions of "1983 AIME Problems/Problem 11"
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=== 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 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> | <math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math> |
Revision as of 20:57, 13 June 2012
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?
Solution
Solution 1
First, we find the height of the figure by drawing a perpendicular from the midpoint of to . The hypotenuse of the triangle is the median of equilateral triangle , and one of the legs is . We apply the Pythagorean Theorem to find that the height is equal to .
Next, we complete the figure into a triangular prism, and find the 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 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:
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 |