Difference between revisions of "1983 AIME Problems/Problem 11"
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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); | ||
+ | label("B",B); | ||
+ | label("C",C); | ||
+ | label("D",D); | ||
+ | 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 == | == Solution 1 == | ||
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>. | 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>. |
Revision as of 15:59, 26 March 2011
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?
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 one of the legs is . We apply the Pythagorean Theorem to find that the height is equal to .
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); D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); D(B--Ba--Ca--C,dashed+d);D(A--Aa--Da--D,dashed+d);D(E--(E.x,E.y,0),dashed+l);D(F--(F.x,F.y,0),dashed+l); D(Aa--E--Da,dashed+d); D(Ba--F--Ca,dashed+d); MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);MP("12\sqrt{2}",(E+F)/2,N);MP("6\sqrt{2}",(A+B)/2);MP("6",(3*s/2,s/2,3),ENE); (Error compiling LaTeX. 4236047701a23aee0aec75276861479a62d4d1f1.asy: 9.2: no matching function 'D(void(flatguide3))')
Next, we complete the figure into a triangular prism, and find the area, which is .
Now, we subtract off the two extra pyramids that we included, whose combined area 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 |