AIME 1983 · 第 11 题

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 $6$.

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

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

Solution 5

From solution 1, the height of the solid is $6$. Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this triangular prism is $( 6\sqrt{2} )^2 \cdot 6 \cdot 1/2 = 216$ 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 $6\sqrt{2}$ and since the formula for the volume of a regular tetrahedron is $\frac{s^3}{6\sqrt{2}}$ the volume of this tetrahedron is $72$. $216 + 72 = \boxed{288}$.

~skibidi solver