Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 AHSME Problems/Problem 30"

(→‎Solution)
m (→‎Solution 2)
 
(16 intermediate revisions by 4 users not shown)
Line 4: Line 4:
 
<math> \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }  </math>
 
<math> \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }  </math>
  
−
==Solution==
+
==Solution 1==
  
 
Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
 
Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
Line 10: Line 10:
 
Now consider the diagonal from <math>(0,0,0)</math> to <math>(3,3,3)</math>. The midpoint of this diagonal is at <math>\left(\frac 32,\frac 32,\frac 32\right)</math>. The plane that passes through this point and is orthogonal to the diagonal has the equation <math>x+y+z=\frac 92</math>.
 
Now consider the diagonal from <math>(0,0,0)</math> to <math>(3,3,3)</math>. The midpoint of this diagonal is at <math>\left(\frac 32,\frac 32,\frac 32\right)</math>. The plane that passes through this point and is orthogonal to the diagonal has the equation <math>x+y+z=\frac 92</math>.
  
−
The unit cube with opposite corners at <math>(x,y,z)</math> and <math>(x+1,y+1,z+1)</math> is intersected by this plane if and only if <math>x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3</math>. Therefore the cube is intersected by this plane if and only if <math>x+y+z\in\{1,2,3,4\}</math>.
+
The unit cube with opposite corners at <math>(x,y,z)</math> and <math>(x+1,y+1,z+1)</math> is intersected by this plane if and only if <math>x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3</math>. Therefore the cube is intersected by this plane if and only if <math>x+y+z\in\{2,3,4\}</math>.
  
−
There are three cubes with <math>x+y+z=1</math> and six cubes with <math>x+y+z=2</math>. The cases <math>x+y+z\in\{3,4\}</math> are symmetric. Therefore there are <math>2(3+6)=\boxed{18}</math> intersected cubes.
+
There are six cubes such that <math>x+y+z=2</math>: permutations of <math>(1,1,0)</math> and <math>(2,0,0)</math>. <br/>
 +
Symmetrically, there are six cubes such that <math>x+y+z=4</math>. <br/>
 +
Finally, there are seven cubes such that <math>x+y+z=3</math>: permutations of <math>(2,1,0)</math> and the central cube <math>(1,1,1)</math>.
 +
 
 +
That gives a total of <math>\boxed{19}</math> intersected cubes.
 +
 
 +
Note that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.
 +
 
 +
==Solution 2==
 +
 
 +
Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of <math>3\sqrt{3}</math>, the altitude of the top vertex of the newly placed cube is <math>3\sqrt{3}</math>. The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into <math>\frac{3\sqrt{3}}{2}</math>, so that the height of the plane is <math>\frac{3\sqrt{3}}{2}</math>.
 +
 
 +
By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.
 +
 
 +
We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be <math>h</math>. The base of the pyramid is an equilateral triangle with side length of <math>\sqrt{3^2+3^2}=3\sqrt{2}</math>. The height of the triangle is <math>\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}</math>. The distance of the center of the triangle to the vertex is <math> \frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}</math>. Therefore, <math>h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}</math>. The altitude of the pyramid at the bottom of the cube is also <math>h</math>. The altitude in the middle is <math>3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}</math>.
 +
 
 +
The altitude of the vertex at the top is <math>3\sqrt{3}</math>. The altitude of the second highest <math>3</math> vertices are all <math>2\sqrt{3}</math>. The altitude of the third highest <math>3</math> vertices are all <math>\sqrt{3}</math>. The altitude of the bottom-most vertex is <math>0</math>.
 +
 
 +
By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is <math>\sqrt{3}</math>. The altitude of the second highest <math>3</math> vertices are all <math>\frac{2\sqrt{3}}{3}</math>. The altitude of the third highest <math>3</math> vertices are all <math>\frac{\sqrt{3}}{3}</math>. The altitude of the bottom-most vertex is <math>0</math>.
 +
 
 +
The length of the space diagonal of a unit cube is <math>\sqrt{3}</math>. The highest vertex of the bottom-most unit cube has an altitude of <math>\sqrt{3}</math>. As <math>\sqrt{3} < \frac{3\sqrt{3}}{2}</math>, therefore, the plane will not pass through the unit cube at the bottom.
 +
 
 +
For the next <math>3</math> cubes from the bottom, the altitude of their highest vertex is <math>\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}</math>. As <math>\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}</math>, therefore, the plane will not pass through the next <math>3</math> unit cubes.
 +
 
 +
For the next <math>3</math> cubes from the bottom, the altitude of their highest vertex is <math>\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}</math>. As <math>\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}</math>, therefore, the plane will pass through the next <math>3</math> unit cubes.
 +
 
 +
So at the bottom half of the cube, there are only <math>1+3 = 4</math> unit cubes that the plane does not passes through. By symmetry, the plane will not pass through <math>4</math> unit cubes at the top half of the cube.
 +
 
 +
Thus, the plane does not pass through <math>4+4 = 8</math> unit cubes, it passes through <math>27-8=\boxed{\textbf{(D) } 19}</math> unit cubes.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1995|num-b=29|after=Final Question}}
 
{{AHSME box|year=1995|num-b=29|after=Final Question}}
 +
{{MAA Notice}}
 +
 +
<noinclude>[[Category:3D Geometry Problems]]</noinclude>

Latest revision as of 04:52, 30 September 2023

Problem

A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is

$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$

Solution 1

Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.

Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$. The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$. The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$.

The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$. Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$.

There are six cubes such that $x+y+z=2$: permutations of $(1,1,0)$ and $(2,0,0)$.
Symmetrically, there are six cubes such that $x+y+z=4$.
Finally, there are seven cubes such that $x+y+z=3$: permutations of $(2,1,0)$ and the central cube $(1,1,1)$.

That gives a total of $\boxed{19}$ intersected cubes.

Note that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.

Solution 2

Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\sqrt{3}$, the altitude of the top vertex of the newly placed cube is $3\sqrt{3}$. The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$, so that the height of the plane is $\frac{3\sqrt{3}}{2}$.

By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.

We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$. The base of the pyramid is an equilateral triangle with side length of $\sqrt{3^2+3^2}=3\sqrt{2}$. The height of the triangle is $\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}$. The distance of the center of the triangle to the vertex is $\frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}$. Therefore, $h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}$. The altitude of the pyramid at the bottom of the cube is also $h$. The altitude in the middle is $3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}$.

The altitude of the vertex at the top is $3\sqrt{3}$. The altitude of the second highest $3$ vertices are all $2\sqrt{3}$. The altitude of the third highest $3$ vertices are all $\sqrt{3}$. The altitude of the bottom-most vertex is $0$.

By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\sqrt{3}$. The altitude of the second highest $3$ vertices are all $\frac{2\sqrt{3}}{3}$. The altitude of the third highest $3$ vertices are all $\frac{\sqrt{3}}{3}$. The altitude of the bottom-most vertex is $0$.

The length of the space diagonal of a unit cube is $\sqrt{3}$. The highest vertex of the bottom-most unit cube has an altitude of $\sqrt{3}$. As $\sqrt{3} < \frac{3\sqrt{3}}{2}$, therefore, the plane will not pass through the unit cube at the bottom.

For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}$. As $\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}$, therefore, the plane will not pass through the next $3$ unit cubes.

For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}$. As $\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}$, therefore, the plane will pass through the next $3$ unit cubes.

So at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.

Thus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\boxed{\textbf{(D) } 19}$ unit cubes.

~isabelchen

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Final Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems/Problem_30&oldid=198832"