Volume of a Part of a Cube

Call the point where the plane cuts $BF$ , $P$ . Now extend $NP$ and $DM$ until they meet at a point $T$ . We can see that this forms a triangular pyramid. Through similar triangles we see that $BP = \frac{1}{4}$ . We find the volume of the solid $DNCBMP$ by finding the volume of the larger pyramid and subtracting the volume of the smaller pyramid.

The pyramids are triangular-right-pyramids, so finding the volumes is pretty straight forward.

$V_{\text{big pyramid}} = \frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}$ $V_{\text{small pyramid}} = \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{48}$ $V_{\text{desired solid}} = \frac{1}{6} - \frac{1}{48} = \frac{7}{48}$

Therefore, the bigger solid is $1 - \frac{7}{48} = \frac{41}{48} \implies 41 + 48 = \boxed{89}$

The problem is from AIME. However, this should not be a level 5 problem...

The plane passing through $D, M, N$ intersects $FB$ at point $X$ where $XB=\dfrac{1}{4}$ . The smaller solid can be divided into three pyramids. If the volume formula for a pyramid is

$V\left( a,b,h \right) =\dfrac { 1 }{ 3 } \dfrac { 1 }{ 2 } abh$

then

$V\left( \dfrac { 1 }{ 2 } ,1,\dfrac { 1 }{ 4 } \right) +V\left( 1,1,\dfrac { 1 }{ 4 } \right) +V\left( \dfrac { 1 }{ 2 } ,1,1 \right) =\dfrac { 7 }{ 48 }$

Reorient the cube such that $D$ is the origin and $AD$ is the $x$ axis. We can easily see that $DM$ is equivalent to the line $y=\frac{1}{2}x$ . Now note that the cross sections taken perpendicular to $AD$ (the $x$ -axis) are right triangles with $h=\frac{1}{2}b$ . Setting $b=1-\frac{1}{2}x$ , we get the volume as equivalent to $\int_0^1(\frac{1}{4}(1-\frac{1}{2}x)^2)dx$ , which evaluates to $\frac{7}{48}$ , so the volume is $1-\frac{7}{48}=\frac{41}{48}$ , giving the final answer as $41+48=\boxed{89}$ .