At one vertex, connect the midpoints of the three edges that meet there. These three midpoints and the vertex form a small pyramid (a tetrahedron). Identify its base and its height in terms of $p$.

Instead of using the slanted triangular face through the three midpoints as the base, use the triangle on a flat face of the cube made by the vertex and two midpoints. What is the area of that right triangle, and how far is the third midpoint from that face? Ѕоurсe​: ​а‌n​iko.аi

After you find the volume of one pyramid, multiply by 8 to get the total volume removed. Subtract from the cube’s volume, then form the ratio (remaining volume)/(original volume) and simplify.

In Desmos, type p = 1 (or any positive number). Then enter:

• Vcube = 8*p^3
• Vpyr = p^3/6
• Vremain = Vcube - 8*Vpyr These expressions give the cube’s volume, one pyramid’s volume, and the remaining volume after all 8 pyramids are removed.

In Desmos, type Vremain / Vcube. Note the decimal value shown, and compare it to the decimal forms of the answer choices (for example, $2/3$, $3/4$, $5/6$, $7/8$) to see which one matches. SAТ prеp⁠ bу Аn⁠ікο.a‍i

The cube has side length $2p$.

So its volume is

All later volumes will be compared to this.

Focus on one corner of the cube.

Label the corner (vertex) as $A$, and the midpoints of the three edges from $A$ as $B$, $C$, and $D$.

• Each edge has length $2p$, so each midpoint is $p$ units from $A$ along its edge.
• The removed solid is the pyramid (a tetrahedron) with vertices $A$, $B$, $C$, and $D$.

Now choose a convenient face of this pyramid to treat as the base:

• Let $A$, $B$, and $C$ lie on one face of the cube (for example, the bottom face), forming a right triangle.
• Point $D$ is on the vertical edge above $A$.

We will use triangle $\triangle ABC$ as the base and $D$ as the top vertex (apex). The volume of the pyramid does not depend on which face we call the base. Рο‌wered bу‌ ⁠А‌n⁠іko

First find the area of the base $\triangle ABC$:

• $AB = AC = p$ (each is half of a cube edge).
• $\angle BAC$ is a right angle (they are perpendicular edges of the cube).

So the base area is

Next, find the height of the pyramid relative to this base:

• The base triangle $\triangle ABC$ lies in a flat face of the cube.
• Point $D$ is directly above $A$ along an edge perpendicular to that face and is $p$ units away.

So the height is

Use the pyramid volume formula $V = \tfrac{1}{3}Bh$:

There are 8 identical pyramids (one at each vertex), so the total removed volume is

The volume that remains is

The fraction of the original volume that remains is