Download OHMIO 2014 Ciphering Problems Practice Find the exact simplified

OHMIO 2014 Ciphering Problems
Practice
Find the exact simplified value of
15 + 2 15 + 2 15 + 2 15 + ...
1. Farmer Dan is making a fence in the shape of a regular hexagon of side length 60 feet. As
shown, when he is halfway done, he ties his horse to a corner with a rope of length 90 feet. How
much square footage can the horse cover, assuming there are no obstructions other than the 3 pieces
of completed fencing?
2.For 0 < x < p in radians, find the exact simplified value of the sum of the solutions to the
equation (cos2x + 3sin2x)2 = 2 .
3. The answer is a four-letter mathematical term determined by the following code
(A =1, B = 2, C = 3, …, Y = 25, Z = 26), where each letter can be found by solving the four
problems below in order:
Letter 1: The solution to (2 + 2i)8 = 4 x , where i = -1 .
Letter 2: The number of positive integer factors of 2500.
Letter 3: The number of integer solutions to the inequality x + 3 > 4 - 3x .
æp
ö
Letter 4: The maximum value of f x = 6 - 3sin3 ç x -14÷ .
è 20
ø
()
4.The probability it is raining given that I have an umbrella is
umbrella given that it is not raining is
2
while the probability that I have an
3
3
. The probability that it is not raining and I do not have an
4
1
. The probability that it is raining and I have an umbrella can be expressed as the
10
m
simplified fraction . Find m + n.
n
umbrella is
5. From the point (-3, 1), a segment is drawn tangent to the graph of x 2 + y2 = 3+ 2x + 8y . What is
the exact length of this segment?
2014
6. The sum
å k ×i
k
can be written in the form a + bi , where i = -1 . Find the value of a - b .
k=0
7. Let f (x) = log2 (sin x) + log 2 (cos x) + log 2 (cos(2x)) + log2 (cos(4x)) + log 2 (cos(8x)) + log2 (sin(80x)) .
Determine the exact value of f (
p
32
) as a simplified fraction.
é 1 2 ù
é 2 -1
8. Let A = ê
ú and B = ê
ë 3 4 û
ë 1 -1
ù
-1
T
ú . What is the determinant of the matrix AB + BA ?
û
Tie Breaker 1
The sum of two real numbers is 7 and the sum of their cubes exceeds the cube of their sum by 1260.
Find the sum of the squares of the two original numbers.
Tie Breaker 2
3 f (x + k) - 3 f (x)
and f (x) = sin(2x) - e4 x . Determine the value of g'(0) .
k®0
2k
Let g(x) = lim
OHMIO 2014 Ciphering Solutions
Quick Answers:
5p
3
P. 5
1. 6000p
2.
6. -2015
7. -4
8. -18
3. FOCI
4. 8
TB1. 169
TB2. -24
5.
5
Practice
Find the exact simplified value of
15 + 2 15 + 2 15 + 2 15 + ...
2
æ xö
Solution: Let x = 2 15 + 2 15 + 2 15 + ... , so ç ÷ - 15 = x or x 2 - 4x - 60 = 0. By factoring,
è 2ø
( x -10 ) (x + 6) = 0 for which x = 10. Half the value of x is desired which equals
5.
1. Farmer Dan is making a fence in the shape of a regular hexagon of side length 60 feet. As
shown, when he is halfway done, he ties his horse to a corner with a rope of length 90 feet. How
much square footage can the horse cover, assuming there are no obstructions other than the 3 pieces
of completed fencing?
Solution: By freely rotating before reaching a side of the hexagon, the horse will have covered an
2
area of p 90 2 = 5400p . If the horse rotates counterclockwise passed the far right side, an area
3
1
of p 30 2 = 450p . Lastly if the horse rotates clockwise passed the far left side, an area of
2
1
p 30 2 = 150p . The total area covered by the horse is 6000p square feet .
6
(
(
)
)
(
)
2.For 0 < x < p in radians, find the exact simplified value of the sum of the solutions to the
equation (cos2x + 3sin2x)2 = 2 .
1
3
p
p
p
Solution: cos 2x + 3 sin 2x = 2( cos2x +
sin 2x) = 2(sin cos2x + cos sin 2x) = 2sin(2x + ).
2
2
6
6
6
2
p
p
p
2
Thus, cos2x + 3sin 2x = 4 sin 2 (2x + ) = 2 , so sin(2x + ) = ±
. Letting u = 2x + ,
6
2
6
6
p p
p p
p 7p 13p 19p
2
when u = + k and thus x =
+ k . For 0 < x < p , x = , ,
,
sinu = ±
2
4 2
24 4
24 24 24 24
40p
5p
whose sum is
.
=
24
3
(
)
3. The answer is a four-letter mathematical term determined by the following code (A =1, B = 2, C
= 3, …, Y = 25, Z = 26), where each letter can be found by solving the four problems below in
order:
Letter 1: The solution to (2 + 2i)8 = 4 x , where i = -1 .
Letter 2:The number of positive integer factors of 2500.
Letter 3:The number of integer solutions to the inequality x + 3 > 4 - 3x .
æp
ö
Letter 4: The maximum value of f x = 6 - 3sin3 ç x -14÷ .
è 20
ø
()
Solution: The word is “FOCI”.
Letter 1: Note that (1+ i)2 = 1+ 2i -1 = 2i so (1+ i)8 = (2i)4 = 16 and
(2 + 2i)8 = 28 (16) = 4 4 × 4 2 = 4 6 so x = 6 → F
Letter 2: Note 2500 = 54 · 22, so there are a total of (4+1)(2+1) = 15 positive integer factors. 15 →
O
Letter 3:Equality holds when x + 3 = 4 - 3x ® x =
1
7
or x + 3 = 3x - 4 ® x = . The solution set is
4
2
1 7
( , ) for which there are three integer solutions, x = 1, 2, or 3. 3 → C
4 2
æp
ö
p
Letter 4: Function f is maximized when 3sin 3 ç x -14÷ is minimized. Letting u =
x -14 , note
è 20
ø
20
-1 £ sinu £1, so -1 £ sin 3 u £ 1, so -3 £ 3sin 3 u £ 3 . The maximum of f is then 6 - (-3) = 9 → I
4.The probability it is raining given that I have an umbrella is
umbrella given that it is not raining is
2
while the probability that I have an
3
3
. The probability that it is not raining and I do not have an
4
1
. The probability that it is raining and I have an umbrella can be expressed as the
10
m
simplified fraction . Find m + n.
n
umbrella is
Solution: Let A be the event of rain and Bbe the event of having an umbrella. Recall by definition
P(A Ç B)
P ( A | B) =
. Now consider the following Venn Diagram:
P(B)
x
2
y
3
æ 1ö 3
= ® x = 2y,
= ® y = 3z, x = 6z = 6 ç ÷ =
è 10 ø 5
x+y 3
y+z 4
m + n = 3+ 5 = 8
5. From the point (-3, 1), a segment is drawn tangent to the graph of x 2 + y2 = 3+ 2x + 8y . What is
the exact length of this segment?
Solution: Complete the square to write the equation of the circle as (x -1)2 + (y - 4)2 = 20 for
which the square of the radius is 20. The square of the distance between the given point and the
center of the circle is (1+ 3)2 + (4 -1)2 = 25 . Since the tangent segment and radius make a right
angle, the desired tangent segment’s length is
2014
6. The sum
å k ×i
k
25 - 20 =
5
can be written in the form a + bi , where i = -1 . Find the value of a - b .
k=0
Solution: Powers of i cycle in a period of 4: 1, i, -1, -i. The first four terms of the sum are 0, i, -2, 3i, whose sum is –2 – 2i. The next four terms of the sum are 4, 5i, -6, -7i, whose sum is also –2 –
ê 2014 ú
2i. There will be ê
= 503 sums of –2 – 2i with the terms 2012 – 2013i – 2014 remaining in
ë 4 úû
the original sum. The final value is 503(–2 – 2i) + 2012 + 2013i – 2014 = -1008 + 1007i. Thus, a =
-1008, b = 1007 and a – b = -2015 .
7. Let
f (x) = log2 (sin x) + log 2 (cos x) + log 2 (cos(2x)) + log2 (cos(4x)) + log 2 (cos(8x)) + log2 (sin(80x)) .
p
Determine the exact value of f ( ) as a simplified fraction.
32
Solution: By a log property, f (x) = log2 (sin x × cos x ×cos(2x)× cos(4x)×cos(8x)×sin(80x)) .
Now a double angle trig identity states sin(2u) = 2sinu cosu . We will use this repeatedly…
1
1
First, sin x cos x cos(2x) = sin(2x)cos(2x) = sin(4x) .
2
4
1
1
Then sin x cos x cos(2x)cos(4x) = sin(4x)cos(4x) = sin(8x) .
4
8
1
1
Then sin x cos x cos(2x)cos(4x)cos(8x) = sin(8x)cos(8x) = sin(16x) .
8
16
1
p
1
p
5p
1
Hence f (x) = log 2 ( sin(16x)sin(80x)) and f ( ) = log 2 ( sin( )sin( )) = log 2 ( ) = -4
16
32
16
2
2
16
é 1 2 ù
é 2 -1 ù
-1
T
8. Let A = ê
ú and B = ê
ú . What is the determinant of the matrix AB + BA ?
3
4
1
-1
ë
û
ë
û
é 1 2 ù é 1 -1 ù é 2 -1 ù é 1 3 ù
Solution: AB-1 + BAT = ê
úê
ú+ê
úê
ú
3
4
1
-2
ë
ûë
û ë 1 -1 û ë 2 4 û
é 3 -5 ù é 0 2 ù é 3 -3 ù
=ê
ú+ê
ú=ê
ú.
ë 7 -11 û ë -1 -1 û ë 6 -12 û
The determinant is 3(-12) – 6(-3) = -36 + 18 = -18 .
Tie Breaker 1
The sum of two real numbers is 7 and the sum of their cubes exceeds the cube of their sum by 1260.
Find the sum of the squares of the two original numbers.
ìx + y = 7
Solution: Let the two numbers be x and y. We have: í 3
.
3
3
x
+
y
=
(x
+
y)
+1260
î
3
3
2
2
Factor a sum of cubes, x + y = (x + y)(x - xy + y )and substitute x + y = 7 into the second
equation to see 7(x 2 - xy + y2 ) = 7 3 +1260 or upon dividing by 7, x 2 - xy + y2 = 49 +180 . Since
(x + y)2 = x 2 + 2xy + y2 = 72 = 49 , then 3xy = -180 or xy = -60 . Thus
49 = x 2 + 2(-60) + y2 ® x 2 + y2 = 169 .
Tie Breaker 2
3 f (x + k) - 3 f (x)
and f (x) = sin(2x) - e4 x . Determine the value of g'(0) .
k®0
2k
Let g(x) = lim
f (x + k) - f (x)
3
3
so g(x) = f '(x) = (2 cos(2x) - 4e4 x ) .
k®0
k
2
2
Upon simplifying and taking another derivative, g'(x) = -6sin(2x) - 24e4 x and g'(0) = -24 .
Solution: By definition, f '(x) = lim