Round 1: Quick Questions
Answer: 8
 S must only contain the points: (1, 2), (2, 1)
together with their reflections in the other
three quadrants,
 i.e. (1, 2), (2, 1) in all possible ways.
Note: reflection about the x- then the y-axes
amounts to a reflection about the origin.
Answer: 1/5
 There are 5 different orders of the 4 red
cards and the black card, namely:
BRRRR, RBRRR, RRBRR, RRRBR, RRRRB
 And they are all equally likely.
Answer: 1
 These terms are the x-coordinates of the
four non-horizontal (counterclockwise) unit
vectors of a regular pentagon.
Answer: 12
 The equation in the first quadrant is the line
segment x/3+y/2=1 joining (3, 0) & (0, 2).
 The above line segment is reflected about
the x- and the y- axis forming a rhombus
with Area = 4(Triangle Area).
Answer: 3, 4, and 6
 (See figure)
 > 6 sides is impossible,
since a cube has 6 faces.
 5 (pentagon) is impossible, since two sides
must occur on opposite faces and must
then be parallel.
Solution:
1)
4E = A + 10u
2) 4D + u = B + 10x
3) 4C + x = C + 10y
4) 4B + y = D + 10z
5) 4A + z = E
From (5) A = 1 or 2.
From (1) A  1, so A = 2.
Solution:
1)
4E = 2 + 10u
2) 4D + u = B + 10x
3) 4C + x = C + 10y
4) 4B + y = D + 10z
5) 42 + z = E
From (5) E = 8 or 9.
From (1) E  9, so E = 8, u = 3, and z = 0.
Solution:
1)
2)
3)
4)
5)
4D + 3 = B + 10x
4C + x = C + 10y
4B + y = D
From (4) B = 0 or 1.
From (2) B  0, so B = 1.
Solution:
1)
2)
3)
4)
4D + 2 = 10x
4C + x = C + 10y
4+y =D
5)
From (2) D = 7, so x = 3, y = 3.
Solution:
1)
2)
3)
4C + 3 = C + 30
4)
5)
From (3) C = 9. Thus, ABCDE = 21978
Answer: 50 kg
 The invariant here is the non-water part,
initially 1% and now 2%.
 Thus,
100 kg (1%) = New_Weight (2%)
Answer: The are infinitely many ways, say:
2/3 = 1/4 + 1/6 + 1/7 + 1/12 + 1/42
 It’s clear that 2/3 = 1/2 + 1/6.
 Now use the observation
1
1
1


n n  1 n(n  1)
 to write any fraction as the sum of two
smaller ones
Answer: Let f = (1 + x)(1 + y)(1 + z), and
g = xyz. By restricting our search to a
suitable bounded closed subset of “g = 3”,
we know that a minimum of f exists, and
must then be at a critical point satisfying
fx/gx = fy/gy = fz/gz (Lagrange Multiplier).
The first equation leads to x = y, and we can
use symmetry to get: x = y = z = 31/3.
Thus, the minimum of f is (1+ 31/3)3.