1997 AHSME Problems/Problem 23

Problem

$[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); draw((-1,1)--(2,1)); draw((-1,0)--(1,0)); draw((-1,1)--(-1,0)); draw((0,-1)--(0,3)); draw((1,2)--(1,0)); draw((-1,1)--(1,1)); draw((0,2)--(1,2)); draw((0,3)--(1,2)); draw((0,-1)--(2,1)); draw((0,-1)--((0,-1) + sqrt(2)*dir(-15))); draw(((0,-1) + sqrt(2)*dir(-15))--(1,0)); label("$\textbf{A}$",foot((0,2),(0,3),(1,2)),SW); label("$\textbf{B}$",midpoint((0,1)--(1,2))); label("$\textbf{C}$",midpoint((-1,0)--(0,1))); label("$\textbf{D}$",midpoint((0,0)--(1,1))); label("$\textbf{E}$",midpoint((1,0)--(2,1)),NW); label("$\textbf{F}$",midpoint((0,-1)--(1,0)),NW); label("$\textbf{G}$",midpoint((0,-1)--(1,0)),2SE);[/asy]$

In the figure, polygons $A$ , $E$ , and $F$ are isosceles right triangles; $B$ , $C$ , and $D$ are squares with sides of length $1$ ; and $G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is

$\textbf{(A)}\ 1/2\qquad\textbf{(B)}\ 2/3\qquad\textbf{(C)}\ 3/4\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 4/3$

Solution

Since there are three squares, the figure may appear to be based upon a unit cube.

Since there is an equilateral triangle of side length $\sqrt{2}$ , this triangle has the length of a facial (non-spacial) diagonal of a unit cube.

Since there are three isoceles right triangles that are half the area of a face, they might represent faces of the cube that have been cut in half along the diagonal.

Putting these three clues together, we can mentally fold the solid back together to form a unit cube that has had one vertex removed. Along with this vertex, the triangular half of each of the three adjoining faces has been removed. When the unit cube is sliced like this, it will cut three of the faces into isoceles triangles like $A$ , $E$ , and $F$ . Additionally, it will create an equilateral triangle that has as its sides three facial diagonals, just like $G$ .

So the volume is a unit cube withh a tetrahedron removed. The tetrahedron has three right triangular faces like $A$ , $E$ , and $F$ , and one equilateral triangular face like $G$ . The volume of this removed tetrahedron can be calculated in two ways:

1) Place one of the three right triangles on the ground as the base. It has an area of $\frac{1}{2}$ . The height is the height of one of the other two right triangles - the equilateral triangle is slanted and is not a height. Thus, the volume of the tetrahedron is $\frac{1}{3}\cdot \frac{1}{2}\cdot 1 = \frac{1}{6}$ , and when this volume is removed, the remaining volume is $\frac{5}{6}$ , which is option $\boxed{D}$ .

2) Place the equilateral triangle on the ground as the base. Its side is $\sqrt{2}$ , so its area is $s^2\cdot\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ . The highest point of this tetrahedron is the vertex of the original cube that was removed, which lies directly over the center of the equilateral triangle. Thus, the distance from the center of the triangle to the vertex of the cube is the height of the tetrahedron. Since the center of the equilateral triangle is also the center of the cube, the distance from the center of the cube to the vertex will be the distance from $(\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$ to $(1,1,1)$ . This distance is $\frac{1}{3}\cdot \sqrt{3}$ . The volume of the tetrahedron is $\frac{1}{3}Bh = \frac{1}{3}\cdot \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{3} = \frac{1}{6}$ , which is the removed volume from the unit cube. Again, the volume remaining is $\boxed{\frac{5}{6}}$ .

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1997 AHSME Problems/Problem 23

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