The faces of 27 unit cubes are painted red, white, and blue in such a manner that we can assemble them into three different configurations: a red 3 x 3 x 3 cube, a white 3 x 3 x 3 cube, and a blue 3 x 3 x 3 cube. Determine, with proof, the number of unit cubes on whose faces all three colors appear.
Let a “unitface” be defined as the face of a unit cube.
Each unit cube has six unitfaces, so with 27 unit cubes we have a total of 162 unitfaces.
Each 3 x 3 x 3 cube has 6(9) = 54 unitfaces showing.
Since the unit cubes can be arranged in the 3 x 3 x 3 cube with red, white, and blue faces showing, respectively, there must be at least 54 unitfaces of each color. However, since the total number of unitfaces is 162 (= 54 + 54 + 54), we must have exactly 54 unitfaces of each color.
Taking the red 3 x 3 x 3 cube, all of the red unitfaces must be showing, and none of the “inner faces” (the faces not showing) may be red.
Also, there are exactly 8 “corner” unit cubes with 3 red faces, 12 “edge” unit cubes with 2 red faces, 6 “central” unit cubes with one red face, and 1 “inner” unit cube with no red faces.
Thus, exactly one unit cube contains no red faces; it must have 3 white faces and 3 blue faces.
Similarly, taking the white and blue 3 x 3 x 3 cubes, exactly one unit cube contains no white faces, and exactly one unit cube contains no blue faces.
Thus, we have exactly three unit cubes that contain only two different colors among their faces; the rest have all 3 different colors among their faces.
24 cubes must contain all three colors among their faces.
We can easily show that one can have this arrangement having:
3 cubes with (
18 cubes with (3
each of the type
6 cubes with (6
of the type
(Note: “
Thus, for a 3 x 3 x 3 red cube, we will have 8 corner pieces
(
3 (
3 x 3 x 3 configurations.
For any positive integer n, let s(n) denote the sum of the digits of n in base 10. Find, with proof, the largest n for which n = 7s(n).
Let a, b, c, d, … be the digits of n when n is written in base 10. (a is the unit digit, b the ten, c, the hundred, etc.)
Then,
n = a + 10b + 100c + 1000d + …
and
s(n) = a + b + c + d + …
If n = 7s(n), then
a + 10b + 100c + 1000d + … = 7a + 7b + 7c + 7d + …
or
6a = 3b + 93c + 993d + …
Since a < 9, the left side of this is 6a < 54.
This implies that c = d = e = … = 0.
So, 6a = 3b, or
2a = b
Since b < 9,
a < (9 / 2) < 4.
The maximum value of n occurs when a = 4 and b = 8.
So, the largest value for n = 7s(n) is n = 84.
How many circles in the plane contain at least three of the nine points (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)? Rigorously verify that no circle was skipped or counted more than once in the result.
For easier reference, let us label the nine points as follows:
We have 9 points, and so we have = 84 sets of 3 points.
A circle can be circumscribed about any triangular set of 3 points.
A circle cannot be circumscribed about 3 collinear points, so we ignore the sets
{ABC}, {DEF}, {GHI}, {ADG}, {BEH}, {CFI}, {AEI}, and {CEG}
We have 84 8 = 76 triangular sets of points.
Some of these triangles will be circumscribed by a common circle.
Any set of four points that forms a rectangle (rectangles include squares) will be circumscribed by a single circle. There are a total of 10 such rectangular sets:
{ABED}, {BCFE}, {EFIH}. {DEHG}, {ACFD}, {DFIG}, {ABHG}, {BCIH}, {DBFH}, and {ACIG}
Also, each of the 4point sets {ABFI}, {CFHG}, {IHDA}, and {GDBC} is circumscribed by a single circle. (The opposite angles of this figure 45° and 135° sum to 180°.)
We can prove that there are no circles which pass through 5 (or more) points.
Let us divide the nine points into 3 groups:
Group A, the 4 corner points {A, C, G, I},
Group B, the 4 “edge” points {B, D, F, H},
and Group C, the central point {E}.
Since Group A is a square, a circle passing through any 3 of its points is a unique circle (which circumscribes the square). Similarly, a single circle circumscribes the points in group B. So, we cannot select 3 (or more) points from any group, and thus it is not possible for any circle to go through 6 or more points, and the only possibility for a 5 point circle is a 221 arrangement or a circle passing through the center point (E) and two corners. These arrangements are contradictions because they will either form straight lines (diagonal AEI or CEG) or a three point circle (e.g. ACE).
For each circumscribable set of four points, we are counting the same circle four times; to obtain the number of unique triangles formed, we ignore 3 of the triangles formed for each of these four point sets.
This leaves us with a total of 76 3(10 + 4) = 34 circles.
(I have also systematically counted all the three point combinations and confirmed that the answer is indeed 34.)
1 
ABC 

29 
BCD 
BCDG circle 
57 
CEH 

2 
ABD 
ABDE circle 
30 
BCE 
BCEF circle 
58 
CEI 

3 
ABE 
Same as #2 
31 
BCF 
Same as #30 
59 
CFG 
CFGH circle 
4 
ABF 
ABFI circle 
32 
BCG 
Same as #29 
60 
CFH 
Same as #59 
5 
ABG 
ABGH circle 
33 
BCH 
BCHI circle 
61 
CFI 

6 
ABH 
Same as #5 
34 
BCI 
Same as #33 
62 
CGH 
Same as #59 
7 
ABI 
Same as #4 
35 
BDE 
Same as #2 
63 
CGI 
Same as #11 
8 
ACD 
ACDF circle 
36 
BDF 
BDFH circle 
64 
CHI 
Same as #33 
9 
ACE 

37 
BDG 
Same as #29 
65 
DEF 

10 
ACF 
Same as #8 
38 
BDH 
Same as #36 
66 
DEG 
DEGH circle 
11 
ACG 
ACGI circle 
39 
BDI 

67 
DEH 
Same as #66 
12 
ACH 

40 
BEF 
Same as #30 
68 
DEI 

13 
ACI 
Same as #11 
41 
BEG 

69 
DFG 
DFGI circle 
14 
ADE 
Same as #2 
42 
BEH 

70 
DFH 
Same as #36 
15 
ADF 
Same as #8 
43 
BEI 

71 
DFI 
Same as #69 
16 
ADG 

44 
BFG 

72 
DGH 
Same as #66 
17 
ADH 
ADHI circle 
45 
BFH 
Same as #36 
73 
DGI 
Same as #69 
18 
ADI 
Same as #17 
46 
BFI 
Same as #4 
74 
DHI 
Same as #17 
19 
AEF 

47 
BGH 
Same as #5 
75 
EFG 

20 
AEG 

48 
BGI 

76 
EFH 
EFHI circle 
21 
AEH 

49 
BHI 
Same as #33 
77 
EFI 
Same as #76 
22 
AEI 

50 
CDE 

78 
EGH 
Same as #66 
23 
AFG 

51 
CDF 
Same as #8 
79 
EGI 

24 
AFH 

52 
CDG 
Same as #29 
80 
EHI 
Same as #76 
25 
AFI 
Same as #4 
53 
CDH 

81 
FGH 
Same as #59 
26 
AGH 
Same as #5 
54 
CDI 

82 
FGI 
Same as #69 
27 
AGI 
Same as #11 
55 
CEF 
Same as #30 
83 
FHI 
Same as #76 
28 
AHI 
Same as #17 
56 
CEG 

84 
GHI 

The systematic count of unique circles (in bold above) is 34.
In how many ways can one choose three angle sizes, α, β, and γ, with α < β < γ from the set of integral degrees, 1°, 2°, 3°, …, 178°, such that those angle sizes correspond to the angles of a nondegenerate triangle? How many of the resulting triangles are acute, right, and obtuse, respectively?
γ is the largest (or equal to another of the largest) angle of a triangle; when 91° < γ < 178°, we have an obtuse triangle; when γ = 90°, we have a right triangle, and when 60° < γ < 89°, we have an acute triangle. Obviously, γ cannot be less than 60°.
For obtuse triangles:
(Even, γ = 2n; 92° < 2n < 178°, or 46° < n < 89°)
For each γ we have unique sets {α, β}.
Since γ = 2n, for each even γ we have 90 n obtuse triangles.
(Odd, γ = 2n 1; 91° < (2n 1) < 177°, or 46° < n < 89°)
For each γ we have unique pairs {α, β}.
Since γ = 2n 1, for each odd γ we also have 90 n obtuse triangles.
Total number of obtuse triangles:
= 1980 unique obtuse triangles.
For right triangles:
When γ = 90°, we have = 45 pairs {α, β}. So, we have 45 unique right triangles.
For acute triangles:
(Even, γ = 2n; 60° < 2n < 88°, or 30° < n < 44°)
For each even γ, we have or 3n 89 acute triangles.
(Odd, γ = 2n + 1; 61° < (2n + 1) < 89°, or 30° < n < 44°)
For each odd γ, we have or 3n 88 acute triangles.
Total number of acute triangles:
=
675 unique acute triangles.
Clearly draw or describe a convex polyhedron that has exactly three pentagons among its faces and the fewest edges possible. Prove that the number of edges is a minimum.
A pentagon has five corners. If we ignore duplicate corners, three pentagons will have 15 corners.
Each pair of pentagons on a polyhedron can have at most two common corners. (Three common corners would make them lie on the same plane.) With three pentagons, at most 6 corners may be shared.
So, a polyhedron with three pentagons among its faces must have at least 15 6 = 9 corners.
A minimum of three edges meet at each corner of a polyhedron. Thus, since there are a minimum of 9 corners, we must have at least 9 * 3 = 27 “meeting edges.” This is doublecounting all of the edges since each edge has two corners, and is counted twice.
So, the number of edges must be greater than (27 / 2), and since it must be an integer, it must be greater than or equal to 14.
Below is an example of a 14edged polyhedron with three pentagonal faces:
(I have constructed this figure with cardboard to confirm its validity.)
The figure above contains 9 corners, 14 edges, and 7 faces (3 pentagons, 3 triangles, and 1 quadrilateral).
There may be other solutions, but no other polyhedron will have a lesser number of edges.