Some unit cubes are stacked atop a flat 4 by 4 square. The figures show views of the stacks from two different sides. Find the maximum and minimum number of cubes that could be in the stacks. Also give top views of a maximum arrangement and a minimum arrangement with each stack marked with its height.
Diagramming the top view of the described figure, we have:
The apparent heights of each row and column are indicated at its relevant side. (We define a “row” as the group of stacks that are behind a single east-projected stack and a “column” as the group of stacks that are behind a single south-projected stack. Columns are labeled (c1, c2, …) from west to east, and rows are labeled (r1, r2, …) from north to south.)
To find the maximum number of blocks, we note that no stack in a given row or column can have a greater height than that indicated in its projection on the east and south sides.
Thus, we can determine the maximum possible height for a given stack by taking the maximum heights possible for the row and the column that contains it, and taking the greatest value that satisfies both the row’s and the column’s height restriction. In other words, when considering a given stack, we will take for its maximum possible height the lesser value of its two “projected” heights.
Following this procedure, we have:
This is the arrangement utilizing the maximum number of blocks. The numbers in each square represent the height of the stack occupying it.
The maximum number of blocks satisfying the given conditions is 38.
Redrawing the diagram for calculating the minimum arrangement,
each row and column must have at least one stack which has the given height of its projection.
To minimize the total number of cubes used, we want to use as few stacks as possible, preferably one per row or column, to generate the given projections’ heights.
Let us start with c1. We need at least one stack with height 4 in this column. Because the east projections of r1 and r3 also have height 4, we could satisfy one of these rows’ requirements simultaneously with this column’s requirements by placing our stack of four cubes on the intersection of c1 and either r1 or r3. In diagram below, we place this stack at (r1, c1).
(Note that the stack could also have been placed at (r3, c1). We are only finding one example of a minimum arrangement; however, there are other arrangements that will utilize the same, minimum number of cubes as this one will.)
Now, we will take c2, whose projection has height 1. Because all the rows have projected heights greater than 1, it does not matter which row we place a stack of height 1 in for c2. (Again, we are only finding one minimum arrangement.) We have:
We can satisfy both c3 and r3’s requirements by placing a stack with height four at their intersection:
Similarly, we can satisfy the projected height requirements for c4 and r2 by placing a stack with height 2 at their intersection:
Now, r4 has the only projection which has not been satisfied. There is no column with projected height 3; we can fulfill r4’s height requirement by placing a stack with height 3 in the intersection of r4 and either c1 or c3, which both have projected heights greater than 3:
This is one possible arrangement that fulfils the given requirements with the minimum possible number of cubes. There are 14 cubes used.
We can prove that this solution uses the fewest number of cubes possible; from the eastern projection, we know that there must be at least 4 + 2 + 4 + 3 = 13 blocks. Also, we know from the southern projection that there is one stack with height 1. So, there can be no less than 13 + 1 = 14 blocks.
Find four distinct positive integers, a, b, c, and d, such that each of the four sums a + b + c, a + b + d, a + c + d, and b + c + d is the square of an integer. Show that infinitely many quadruples (a, b, c, d) with this property can be created.
Let: a + b + c = k2 (1)
a + b + d = l2 (2)
a + c + d = m2 (3)
and b + c + d = n2 (4)
Adding (1), (2), (3), and (4),
or, a + b + c + d = .
Let = S
Then: S k2 = d
S l2 = c
S m2 = b
S n2 = a
Since a, b, c, and d are distinct, so are k, l, m, and n. (For example, if k = l it would imply that c = d.)
Now, if we can find k, l, m, n, such that:
Without loss of generality we can assume 0 < k < l < m < n
then we can find valid values for a, b, c, and d
Obviously, condition 2 can be satisfied if:
(a) k, l, m, and n are all multiples of three, or
one of is a multiple of three
(k2 ≡ 1 (mod 3) if k is not a multiple of 3)
For a simple case, let us take the case where k, l, m, and n are consecutive integers, and k is not divisible by 3.
That is, is given by , and k is not divisible by 3.
Then, in (k2, l2, m2, n2), exactly one of the four terms is divisible by 3, and is an integer.
We also need , or
k2 + (k +1)2 + (k + 2)2 > 2(k + 3)2
or 3k2 + 6k + 5 > 2k2 + 12k + 18
or k2 6k 13 > 0
k > 7.69… or k < -1.69…
k is a positive integer, so k > 7
Thus, the first possible consecutive is (8, 9, 10, 11)
So, k2 = 64
l2 = 81
m2 = 100
n2 = 121
S = = 122
d = S k2 = 58
c = S l2 = 41
b = S m2 = 22
a = S n2 = 1
(1, 22, 41, 58) is one example of
There are infinitely many ’s that are consecutive numbers where
k > 7 and is not divisible by 3.
So, there are infinitely many ’s, and thus there are infinitely many ’s.
[For example, k = 8 gave the above solution for .
k = 10 gives:
S = 178,
a = 178 132 = 9
b = 178 122 = 34
c = 178 112 = 57
d = 178 102 = 78
(9, 34, 57, 78)
k = 11 gives:
S = 210
a = 210 142 = 14
b = 210 132 = 41
c = 210 122 = 66
d = 210 112 = 89
(14, 41, 66, 89)
and so on.]
For a set of points in a plane, we construct the perpendicular bisectors of the line segments connecting every pair of those points and we count the number of points in which these perpendicular bisectors intersect each other. If we start with twelve points, the maximum possible number of intersection points is 1705. What is the maximum possible number of intersection points if we start with thirteen points?
We assume that no two bisectors are common to each other, as this would result in “infinite points of intersection”.
If we start with x points, we will have = connecting lines, and thus also perpendicular bisectors.
If we have n oblique lines with no restrictions on their placement, the maximum number of points of intersection that they can create is = .
In our case, any 3 of the starting points form a triangle, and thus their perpendicular bisectors will all intersect at a single point. This means that, for every set of three starting points, we will have a single point of intersection instead of 3 (given by ).
Thus, we must subtract from the total possible number of points of intersection given by the formula when n = to get the correct answer.
Therefore, the maximum number of points of intersection for the perpendicular bisectors of lines is:
So, when x = 12,
= = 1705
and when x = 13,
= = 2431
A transposition of a vector is created by switching
exactly two entries of the vector. For
example, (1, 5, 3, 4, 2, 6, 7) is a transposition of (1, 2, 3, 4, 5, 6,
7). Find the vector X if S = (0, 0, 1,
1, 0, 1, 1), T = (0, 0, 1, 1, 1, 1, 0), U = (1, 0, 1, 0, 1, 1, 0), and
V = (1, 1, 0, 1, 0, 1, 0) are all transpositions of X. Describe your method for finding X.
Entries cannot be added or removed; only switched; therefore, X must have four 1’s and three 0’s.
For the first two entries, S has (0, 0, …) and V has (1, 1, …). Since there is only one transposition, the only possibility is that:
X must have one 0 and one 1 among its first two entries.
This leaves two 0’s and three 1’s in the remaining five entries of X.
Since the transposition from X to S (and also T) involves a 0 from the last five entries to the first two, any 0 in S (and also T) in the last five entries tells us that there is a 0 in its corresponding position in X.
The fifth entry of S is a 0, and the seventh entry of T is a 0. Thus, the fifth and seventh entries of X are 0’s. Since there are exactly two 0’s among the last five entries, all other entries among the last five entries of X are 1’s.
X can be either
(0, 1, 1, 1, 0, 1, 0)
or (1, 0, 1, 1, 0, 1, 0)
If X is (0, 1, 1, 1, 0, 1, 0), exactly one pair of
its entries can be transposed to
U = (1, 0, 1, 0, 1, 1, 0).
The first two entries of X are transposed with respect to U. However, so are the third and fourth entries. This contradicts the condition that only one pair of entries can be transposed between U and X.
Thus, X cannot equal (0, 1, 1, 1, 0, 1, 0)
Thus, X = (1, 0, 1, 1, 0, 1, 0)
(X and S have their first and seventh terms transposed.
X and T have their first and fifth terms transposed.
X and U have their fourth and fifth terms transposed.
X and V have their second and third terms transposed.)
As illustrated below, we can dissect every triangle ABC into four pieces so that piece 1 is a triangle similar to the original triangle, while the other three pieces can be assembled into a triangle also similar to the original triangle. Determine the ratios of the sizes of the three triangles and verify that the construction works.
(Tri. X) (Tri. Y) (Tri. Z)
If , then:
, so .
Similarly, , so .
Also, and , so .
(If you extend to meet at K then FKC = KFG, hence )
Opposite sides of quadrilateral DBFH are parallel, that is, DBFH is a parallelogram.
Also, opposite sides of quadrilateral GFCE are parallel, so GFCE is a parallelogram.
Let GH = a.
In , and are common. So, GH = GD, and thus
GD = a.
DH = DG + GH, so
DH = 2a.
DH and BF are opposite sides of , so , and thus
BF = 2a.
In , and are common. So, HE = BF, and thus
HE = 2a.
GE = GH + HE, so
GE = 3a.
GE and FC are opposite sides of , so , and thus
FC = 3a.
DE = DH + HE, so
DE = 4a.
BC = BF + FC, so
BC = 5a.
Thus, the ratio of corresponding side lengths for and is , or 3 : 4.
the ratio of corresponding side lengths for and is , or 4 : 5
and so the ratio of side lengths of , , is 3 : 4 : 5,
and the ratio of their areas is (32 : 42 : 52) :: ().
(Note that 9 + 16 = 25, as it should be.)
It is easy to verify that this construction will always work, as we can draw parallel to (D is determined by the required ratio, that is AD : AB = 4 : 5), take a point F by dividing at a ratio of 2 : 3, and draw lines (parallel to ) and (parallel to ). This will ensure that all the angles will be such that the pieces will fit and the triangles will indeed be symmetric. (This is easy to check, using the properties of parallel lines).