2012 AIME I Problems/Problem 8

Problem

Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

$[asy]import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465); pair[] dotted = {A,B,C,D,E,F,G,H,M,N}; D(A--B--C--G--H--E--A); D(E--F--B); D(F--G); pathpen=dashed; D(A--D--H); D(D--C); dot(dotted); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NW); label("$E$",E,W); label("$F$",F,SE); label("$G$",G,NE); label("$H$",H,NW); label("$M$",M,S); label("$N$",N,NE); [/asy]$

Solution: think outside the box (pun intended)

Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$ , $y$ , and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.$ So $[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}$

Another way to finish after using coordinates: Take the volume of DMBCNQ as the sum of the volumes of DMBQ and DBCNQ. Then, we have the sum of the volumes of a tetrahedron and a pyramid with a trapezoidal base. Their volumes, respectively, are $\tfrac{1}{48}$ and $\tfrac{1}{8}$ , giving the same answer as above. $\blacksquare$ ~mathtiger6

Solution 2

Define a coordinate system with $D = (0,0,0)$ , $M = (1, \frac{1}{2}, 0)$ , $N = (0,1,\frac{1}{2})$ . The plane formed by $D$ , $M$ , and $N$ is $z = \frac{y}{2} - \frac{x}{4}$ . It intersects the base of the unit cube at $y = \frac{x}{2}$ . The z-coordinate of the plane never exceeds the height of the unit cube for $0 \leq x \leq 1, 0 \leq y \leq 1$ . Therefore, the volume of one of the two regions formed by the plane is $\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}$ Since $\frac{7}{48} < \frac{1}{2}$ , our answer is $1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$ .

Alternative Solution (using calculus) : think inside the box

Let $Q$ be the intersection of the plane with edge $FB,$ then $\triangle MQB$ is similar to $\triangle DNC$ and the volume $[DNCMQB]$ is a sum of areas of cross sections of similar triangles running parallel to face $ABFE.$ Let $x$ be the distance from face $ABFE,$ let $h$ be the height parallel to $AB$ of the cross-sectional triangle at that distance, and $a$ be the area of the cross-sectional triangle at that distance. $A(x)=\frac{h^2}{4},$ and $h=\frac{x+1}{2},$ then $A=\frac{(x+1)^2}{16}$ , and the volume $[DNCMQB]$ is $\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.$ Thus the volume of the larger solid is $1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}$

Alternative Solution : think inside the box

The volume of a frustum is $\frac{h_2b_2 -h_1b_1}3$ where $b_i$ is the area of the base and $h_i$ is the height from the chopped off apex to the base.

We can easily see that from symmetry, the area of the smaller front base is $\frac{1}{16}$ and the area of the larger back base is $\frac{1}4$

Now to find the height of the apex.

Extend the $DM$ and (call the intersection of the plane with $FB$ G) $NG$ to meet at $x$ . Now from similar triangles $XMG$ and $XDN$ we can easily find the total height of the triangle $XDN$ to be $2$

Now straight from our formula, the volume is $\frac{7}{48}$ Thus the answer is:

$1-\text{Volume} \Longrightarrow \boxed{089}$

Alternative Solution : think inside the box with pyramids

We will solve for the area of the smaller region, and then subtract it from 1.

Let $X$ be the point where plane $DMN$ intersects $FB$ . Then $DMBCNX$ can be split into triangular pyramid $DMBX$ and quadrilateral pyramid $BCNXD$ .

Pyramid $DMBX$ has base $DMB$ with area $\frac{1}2 \cdot 1 \div 2 = \frac{1}4$ . The height is $BX = \frac{1}4$ , so the volume of $DMBX$ is $\frac{1}4 \cdot \frac{1}4 \div 3 = \frac{1}{48}$ .

Similarly, pyramid $BCNXD$ has base $BCNX$ with area $(\frac{1}4 + \frac{1}2) \cdot 1 = \frac{3}8$ . The height is $CD = 1$ , so the volume of $BCNXD$ is $\frac{3}8 \cdot 1 \div 3 = \frac{1}8$ .

Adding up the volumes of $DMBX$ and $BCNXD$ , we find that the volume of $DMBCNX$ is $\frac{1}{48} + \frac{6}{48} = \frac{7}{48}$ . Therefore the volume of the larger solid is $1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$

This is basically mathtiger6's solution, but you don't need coordinates or thinking outside the box.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/344

~ dolphin7

Video Solution

https://www.youtube.com/watch?v=LWUN_ZymNnw ~Shreyas S

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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