Download Math Contest 6

Math Contest 6
Calulators are not allowed.
Name:
Area 1:
First Period Instructor:
1. Leah and Nicole are riding a Ferris wheel. This particular Ferris wheel has radius 20 feet
and revolves at a constant rate of one revolution per minute. How many seconds does it
take Leah and Nicole to travel from the bottom of the wheel to a point 10 vertical feet
from the bottom?
1.
Solution: In the figure below, the center of the wheel is at O, and the riders (Leah
and Nicole) travel from A to B. Since AC = 10 and OB = OA = 20, the point C is
the midpoint of OA. In the right �OCB, we have OC half of the length of the hypotenuse OB, so m∠COB = 60◦ . Since the wheel turns through an angle of 360◦ in
60
60 seconds, the time required to turn through an angle of 60◦ is 60( 360
) = 10 seconds.
2. A window has 9 panes in the form of a 3x3 grid. In how many ways can one color 6 of
these panes yellow so that the window looks the same from the inside and outside the
house?
2.
Solution: If we first color all the panes yellow and then color the remaining 3 panes
white (to represent unpainted) we can solve an easier problem. The solution has
to be mirror symmetric, if we reflect along the vertical middle line. In the central
column there could be either 3 or 1 white panes. For 3, there is only one possibility.
For 1 white pane in the middle there are 3 ∗ 3 = 9 possibilities: 3 for the middle one,
and 3 for the white panel in the left column; then we have to color a pane in the
similar position on the right. Thus the total number is 1 + 9 = 10.
3. Consider all three digit numbers. How many three-digit numbers satisfy the property
that the middle digit is the average of the first and the last digits?
3.
Solution: The first and last digits must be both odd or both even from their average
to be an integer. There are 5 ∗ 5 = 25 odd-odd combinations for the first and last
digits. There are 4*5=20 even-even combinations that do not use zero as the first
digit. Hence there are 45.
4. For this problem we will consider happy numbers. A happy number is defined by the
following process. Starting with any positive integer replace the number by the sum of
the squares of its digits and repeat. If this process ends in 1 we call the number happy.
For example: 7 → 72 = 49 → 42 + 92 = 97 · · · . Find all two digit happy numbers that
reach 1 in exactly 2 steps.
4.
Solution: We can represent our two digit number as xy where x is the first digit
and y is the second digit. We also know that 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9. This means
that the largest possible number attained from our first is 162 = 92 + 92 . From this
we know in order to reach 1 in exactly two steps that the sum of the digits must
equal either 10 or 100. Since 10 becomes 12 + 02 = 1 and 12 + 02 + 02 = 1. We need
only pairs of perfect squares that sum to 10 and 100. We get that 10 = 1 + 9 and
100 = 64 + 36 thus the answers are 86, 68, 31, 13.
5. The sides of a triangle are consecutive integers and the area of the triangle is also an
integer. What is the smallest perimeter possible such that the triangle is not a right
triangle?
5.
Page 2
Solution: The answer can be found using and alternative
� formula for the area of a
triangle know as Heron’s formula. Heron’s formula is s(s − a)(s − b)(s − c) where
s = a+b+c
is the semiperimeter of the triangle.
2
Let the
sides
be b − 1, b, b + 1, so the area is
�
�
K =
3b 3b
( − (b − 1))( 3b
− (b))( 3b
2 2
2
2
�
b
to 4 3(b − 2)(b + 2), so it
− (b + 1)) =
3b b+2 b b−2
( 2 )( 2 )( 2 ).
2
This simplifies
further
looks like b has to be a multiple of 4 or one of
b − 2 or b + 2 must be a perfect square of 3 times a perfect square. If b = 4, then
the area is an integer, (6), but the triangle is then a 3 − 4 − 5 triangle but is a right
triangle.
The next value that
�
� works, is b = 1414, with the area
14
14
3(14 − 2)(14 + 2) = 4 3(12)(16) = 4 (6)(4) = 84. Thus the answer is 13 +
4
14 + 15 = 42.
6. We consider the equation x2 + kx + 1 = 0. A single fair die is rolled to determine the
value of the middle coefficient, k. The value for k is the number of dots on the upper
face of the die. What is the probability that the equation will have real, unequal root?
6.
√
2
Solution: Using the quadratic formula, −b± 2ab −4ac , we can write all solutions to the
given equation. We now need to determine when this is real and produces unequal
roots. This happens when b2 − 4ac > 0. Using the given equation we then get that
k 2 − 4 ∗ 1 ∗ 1 > 0, thus k must be greater than 2. So k can equal 3, 4, 5, or 6. Thus
the probability is 46 = 23
Page 3