Difference between revisions of "2012 AIME I Problems/Problem 8"

(Alternative Solution (using calculus): think inside the box)
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==Problem 8==
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==Problem==
 
Cube <math>ABCDEFGH,</math> labeled as shown below, has edge length <math>1</math> and is cut by a plane passing through vertex <math>D</math> and the midpoints <math>M</math> and <math>N</math> of <math>\overline{AB}</math> and <math>\overline{CG}</math> respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form <math>\tfrac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
 
Cube <math>ABCDEFGH,</math> labeled as shown below, has edge length <math>1</math> and is cut by a plane passing through vertex <math>D</math> and the midpoints <math>M</math> and <math>N</math> of <math>\overline{AB}</math> and <math>\overline{CG}</math> respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form <math>\tfrac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
  
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</asy></center>
 
</asy></center>
  
== Solution: think outside the box ==
+
== Solution: think outside the box (pun intended) ==
Define a coordinate system with <math>D</math> at the origin and <math>C,</math> <math>A,</math> and <math>H</math> on the <math>x</math>, <math>y</math>, and <math>z</math> axes respectively. The <math>D=(0,0,0),</math> <math>M=(.5,1,0),</math> and <math>N=(1,0,.5).</math> It follows that the plane going through <math>D,</math> <math>M,</math> and <math>N</math> has equation <math>2x-y-4z=0.</math> Let <math>Q = (1,1,.25)</math> be the intersection of this plane and edge <math>BF</math> and let <math>P = (1,2,0).</math> Now since <math>2(1) - 1(2) - 4(0) = 0,</math> <math>P</math> is on the plane. Also, <math>P</math> lies on the extensions of segments <math>DM,</math> <math>NQ,</math> and <math>CB</math> so the solid <math>DPCN = DMBCQN + MPBQ</math> is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume of solid <math>DMBCQN,</math> which is one of the solids bounded by the cube and the plane, is <math>[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].</math> But the volume of <math>DPCN</math> is simply the volume of a pyramid with base <math>1</math> and height <math>.5</math> which is <math>\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.</math> So <math>[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.</math> Note, however, that this volume is less than <math>.5</math> and thus this solid is the smaller of the two solids. The desired volume is then <math>[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}</math>
+
Define a coordinate system with <math>D</math> at the origin and <math>C,</math> <math>A,</math> and <math>H</math> on the <math>x</math>, <math>y</math>, and <math>z</math> axes respectively. Then <math>D=(0,0,0),</math> <math>M=(.5,1,0),</math> and <math>N=(1,0,.5).</math> It follows that the plane going through <math>D,</math> <math>M,</math> and <math>N</math> has equation <math>2x-y-4z=0.</math> Let <math>Q = (1,1,.25)</math> be the intersection of this plane and edge <math>BF</math> and let <math>P = (1,2,0).</math> Now since <math>2(1) - 1(2) - 4(0) = 0,</math> <math>P</math> is on the plane. Also, <math>P</math> lies on the extensions of segments <math>DM,</math> <math>NQ,</math> and <math>CB</math> so the solid <math>DPCN = DMBCQN + MPBQ</math> is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume of solid <math>DMBCQN,</math> which is one of the solids bounded by the cube and the plane, is <math>[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].</math> But the volume of <math>DPCN</math> is simply the volume of a pyramid with base <math>1</math> and height <math>.5</math> which is <math>\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.</math> So <math>[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.</math> Note, however, that this volume is less than <math>.5</math> and thus this solid is the smaller of the two solids. The desired volume is then <math>[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}</math>
  
== Alternative Solution (using calculus): think inside the box ==
+
*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 <math>\tfrac{1}{48}</math> and <math>\tfrac{1}{8}</math>, giving the same answer as above. <math>\blacksquare</math> ~mathtiger6
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then triangle <math>MQB</math> is similar to triangle <math>DNC</math> and the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be the distance from face <math>ABFE,</math> let <math>h</math> be the height parallel to <math>AB</math> of the cross-sectional triangle at that distance, and <math>a</math> be the area of the cross-sectional triangle at that distance. <math>a=\frac{h^2}{4},</math> and <math>h=\frac{x+1}{2},</math> then <math>a=\frac{(x+1)^2}{16}</math>, and the volume <math>[DNCMQB]</math> is <math>\int^1_0{a}{dx}=\int^1_0{\frac{(x+1)^2}{16}}{dx}=\frac{7}{48}.</math> Thus the volume of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math>
+
 
 +
== Solution 2 ==
 +
Define a coordinate system with <math>D = (0,0,0)</math>, <math>M = (1, \frac{1}{2}, 0)</math>, <math>N = (0,1,\frac{1}{2})</math>. The plane formed by <math>D</math>, <math>M</math>, and <math>N</math> is <math>z = \frac{y}{2} - \frac{x}{4}</math>. It intersects the base of the unit cube at <math>y = \frac{x}{2}</math>. The z-coordinate of the plane never exceeds the height of the unit cube for <math>0 \leq x \leq 1, 0 \leq y \leq 1</math>. Therefore, the volume of one of the two regions formed by the plane is
 +
<cmath>\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}</cmath>
 +
Since <math>\frac{7}{48} < \frac{1}{2}</math>, our answer is <math>1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}</math>.
 +
 
 +
== Alternative Solution (using calculus) : think inside the box ==
 +
Let <math>Q</math> be the intersection of the plane with edge <math>FB,</math> then <math>\triangle MQB</math> is similar to <math>\triangle DNC</math> and the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be the distance from face <math>ABFE,</math> let <math>h</math> be the height parallel to <math>AB</math> of the cross-sectional triangle at that distance, and <math>a</math> be the area of the cross-sectional triangle at that distance. <math>A(x)=\frac{h^2}{4},</math> and <math>h=\frac{x+1}{2},</math> then <math>A=\frac{(x+1)^2}{16}</math>, and the volume <math>[DNCMQB]</math> is <math>\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.</math> Thus the volume of the larger solid is <math>1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}</math>
 +
 
 +
== Alternative Solution : think inside the box with formula==
 +
 
 +
If you memorized the formula for a frustum, then this problem is very trivial.
 +
 
 +
The formula for a frustum is:
 +
 
 +
<math>\frac{h_2b_2 -h_1b_1}3</math> where <math>b_i</math> is the area of the base and <math>h_i</math> 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 <math>\frac{1}{16}</math> and the area of the larger back base is <math>\frac{1}4</math>
 +
 
 +
Now to find the height of the apex.
 +
 
 +
Extend the <math>DM</math> and (call the intersection of the plane with <math>FB</math> G) <math>NG</math> to meet at <math>x</math>. Now from similar triangles <math>XMG</math> and <math>XDN</math> we can easily find the total height of the triangle <math>XDN</math> to be <math>2</math>
 +
 
 +
Now straight from our formula, the volume is <math>\frac{7}{48}</math> Thus the answer is:
 +
 
 +
<math>1-\text{Volume} \Longrightarrow \boxed{089}</math>
 +
 
 +
== 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 <math>X</math> be the point where plane <math>DMN</math> intersects <math>FB</math>. Then <math>DMBCNX</math> can be split into triangular pyramid <math>DMBX</math> and quadrilateral pyramid <math>BCNXD</math>.
 +
 
 +
Pyramid <math>DMBX</math> has base <math>DMB</math> with area <math>\frac{1}2 \cdot 1 \div 2 = \frac{1}4</math>. The height is <math>BX = \frac{1}4</math>, so the volume of <math>DMBX</math> is <math>\frac{1}4 \cdot \frac{1}4 \div 3 = \frac{1}{48}</math>.
 +
 
 +
Similarly, pyramid <math>BCNXD</math> has base <math>BCNX</math> with area <math>(\frac{1}4 + \frac{1}2) \cdot 1 = \frac{3}8</math>. The height is <math>CD = 1</math>, so the volume of <math>BCNXD</math> is <math>\frac{3}8 \cdot 1 \div 3 = \frac{1}8</math>.
 +
 
 +
Adding up the volumes of <math>DMBX</math> and <math>BCNXD</math>, we find that the volume of <math>DMBCNX</math> is <math>\frac{1}{48} + \frac{6}{48} = \frac{7}{48}</math>. Therefore the volume of the larger solid is <math>1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}</math>
 +
 
 +
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
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2012|n=I|num-b=7|num-a=9}}
 +
 +
{{MAA Notice}}
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Revision as of 20:20, 24 January 2021

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 with formula

If you memorized the formula for a frustum, then this problem is very trivial.

The formula for 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

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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