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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
1. The long and colorful traditions of Dressage go as far back as ancient Greece. Dressage, the highest expression
of horse training, is considered the art of equestrian sport (as well as at the Summer Olympics) and is used as the
groundwork for all other disciplines. With the disintegration of Ancient Greece, the art of riding slowly fell into
oblivion, until its revival during the Renaissance period. In the 18th century, classical Dressage reached its peak
with the creation of the world-famous Spanish Riding School in 1729 in Vienna, and laid the basis of the modern
discipline. If the Spanish riding school began with 9 horses, and grew by double each year after 1729, which of
the following would be an appropriate mathematical model to represent the number of horses as a function of time
t in years after the year 1729? (t=0 years corresponds to Midnight beginning January of 1730)
𝑡
𝑡
A. 𝐻(𝑡) = 9 ⁄2
D. 𝐻(𝑡) = 9 (2 ⁄1729 )
𝑡
B. 𝐻(𝑡) = 9 (2 ⁄2 )
C. 𝐻(𝑡) = 9(2
E. None of these Answers
𝑡)
2. Equestrian sports featured on the Olympic program of the Paris Games in 1900, with jumping events, and were
then withdrawn until the 1912 Games in Stockholm. Since then, this sport has been on the Olympic program with
remarkable regularity. How many distinguishable permutations can be made from the word “training”?
A. 40320
D. 5040
B. 20160
E. None of these Answers
C. 10080
3. The base of the stirrup and the length of the stirrup leather are
perpendicular lines (really line segments, but for the purpose of
this question, we will allow them to be lines). Which of the
following pairs of equations represent possible equations for a
stirrup and stirrup leather on a saddle?
A. 2𝑥 + 3𝑦 = 12
3𝑥 + 2𝑦 = 10
B. 3𝑥 + 3𝑦 = 12
2𝑥 + 2𝑦 = 10
C. 2𝑥 − 3𝑦 = 12
3𝑥 − 2𝑦 = 10
D. 2𝑥 + 3𝑦 = 12
3𝑥 − 2𝑦 = 10
E. None of these Answers
4. The stirrup of the saddle can be modeled by a parabolic equation opening down on a Cartesian plane with a vertex
at (−2,4) and through the point (5, −12). What is the focus of the stirrup’s parabola?
A. (−2, 207⁄64)
B. (−2, 15⁄16)
C. (−2, 113⁄16)
D. (17⁄16 , 4)
E. None of these Answers
5. At the 2008 Olympic Games in Beijing, the Netherlands’ Anky van Grunsven confirmed her talent by winning
gold for the third consecutive time in the individual dressage event. Her titles total eight medals - three gold and
five silver. She is joined by Germany’s Isabell Werth (five gold medals and three silver) and Reiner Klimke (six
gold and two bronze), who have also won eight medals each. If you were to randomly select an Olympian, and
then from that Olympian, randomly select a medal, what is the probability that you would select a gold medal?
A. 1⁄2
D. 3⁄4
E. None of these Answers
B. 7⁄12
C. 1⁄3
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
6. Use the horse feet placement diagram to answer the
following question. What letter shown below has the
same pattern as letter “W” if the same pattern
continues? (LF = Left Front, RF = Right Front, LH =
Left Hind (Back), RH = Right Hind)
A.
B.
C.
D.
E.
A
B
C
D
None of These Answers
7. When riders are walking the course before a jumping competition, they are doing math. Each stride they take is
3ft. An average horse’s canter stride is 9 – 12 ft. As they walk, the riders are making mathematical calculations,
determining how many strides their horse will take between jumps, what the best angle of approach will be for
each jump and assessing the height and width of every obstacle. A horse is 8 meters from an obstacle that is 1.8
meters high. If the obstacle is perpendicular to the ground, find the secant of the best angle of approach (angle of
elevation) for the obstacle.
A. 9⁄40
D. 41⁄40
E. None of these Answers
B. 4⁄3
C. 3⁄5
8. A zero in Equestrian competition is considered to be a perfect round where the rider and horse accumulate no
penalty points. At the London Olympics in 2012, Sweden’s Sara Algotsson needed a zero round for individual
gold but her horse Wega hit the last fence, handing victory to Germany’s Michael Jung. Which of the following is
not equal to zero? The usual restricted ranges of Arcfunctions are used.
A. 𝐶𝑜𝑠 −1 (0!)
B. 𝑆𝑖𝑛−1 (ln 1)
C. 𝑇𝑎𝑛−1 (∑∞
𝑛=1 0)
3 2
D. [
]
6 4
E. None of these Answers
9. Given the following data for penalty points versus hours practiced per day, determine the y value when x = 5 of
the inverse of this relation as well as whether the inverse is a function or not a function. All of the data points are
given for the set.
Hours
3
Penalties 1
A. 2, not a function
B. 2, function
C. 6, not a function
4
0
2
5
1
7
5
6
4
2
3.5
3
D. 6, function
E. None of these Answers
10. In the 2004 Summer Olympics in Athens, American Kimberly Severson and her horse Winsome A won a silver
medal in the Eventing division. If Winsome A was asked to trot in a perfect circle of diameter 12 feet, what is the
central angle in radians of the arc formed when the horse walks 36 inches?
A. 0.5
D. 6
B. 1
E. None of these Answers
C. 𝜋⁄2
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
11. Theorem: All horses are the same color.
Proof: Base case: 1 horse. With just 1 horse, all horses have the same color.
We'll show that if it is true for any group of N horses, that all have the same color, then it is true for any
group of N+1 horses.
Given any set of N+1 horses, if you exclude the last horse, you get a set of N horses. By a previous step
these N horses all have the same color. But by excluding the first horse in the pack of N+1 horses, you
can conclude that the last N horses also have the same color. Therefore all N+1 horses have the same
color.
Clearly not all horses have the same color. So what's wrong with this proof?
A. The argument in this proof by mathematical induction breaks down in going from n=1 to n=2, because
the first 1 horse and the last 1 horse have no horses in common, and therefore may not all have the same
color.
B. The argument in this proof by mathematical deduction breaks down in going from n=1 to n=2, because
the first 1 horse and the last 1 horse have no horses in common, and therefore may not all have the same
color.
C. This argument fails because every two column proof needs a reason to justify the steps.
D. The argument in this proof by mathematical induction breaks down when you assume 1 horse has the
same color as itself.
E. None of these Answers
12. In the 2004 Summer Olympics in Athens, American Chris Kappler and his horse Royal Kaliber won a silver
medal in the Jumping division. If Chris’ motions while jumping can be described by the sinusoidal function
below, what is the sum of the period and phase shift of his function?
𝜋
3
𝑠(𝑡) = −3 cos (2𝜋𝑡 − ) +
2
𝜋
A.
B.
C.
3
𝜋
3
𝜋
3
4
5
4
+1
D.
+2
E. None of these Answers
13. The English Equestrian team asked the Swiss team to solve this trigonometric puzzle. Can you help the Swiss
team simplify this expression into a single trigonometric function? Restrict the x-values to where defined.
2csc(2𝑥)
sec(𝑥)
A.
B.
C.
D.
E.
sin(𝑥)
sec(𝑥)
cos(𝑥)
csc(𝑥)
None of these Answers
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
14. In the Jumping events at the Summer Olympics, horsemen and women compete against each other on a level
playing field in a thrilling show of skill and courage, where every obstacle presents its own challenge. One
jumper’s height as a function of time is given below. Someone noticed that a jumper was at ground level at the
same time as shown by the zeroes on the graph of h(t). Find all times when the jumper will be at ground level.
ℎ(𝑡) = −3𝑡 3 + 9 −
51
33
𝑡 + 𝑡2
2
2
A. 𝑡 = ±1, ± 1⁄2
B. 𝑡 = 1⁄2 , 1
C. 𝑡 = 1⁄2 , 2,3
D. 𝑡 = 3, ± 1⁄2
E. None of these Answers
15. Many of the obstacles in equestrian events are made out of wood (logs to be exact!). Solve this logarithmic
𝑥
equation with base 4 in the spirit of the equestrian events who depend on them.
3log 𝑥⁄ (729⁄64) = 18
4
A.
B.
C.
16.
A.
B.
C.
D.
E.
𝑥>0
3
D. 12
6
E. None of these Answers
9
Refusal is when the horse chooses not to jump an obstacle. If a horse has a known probability of refusal of 10%,
what is the probability that a horse refuses for the first time in an event on the fourth obstacle?
0.00729
0.0729
0.09
0.1
None of these Answers
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
17. In Stockholm 1912, the obstacles in sequence were first introduced, inspired by Italian Captain Federico Caprilli,
the father of modern equestrian sports. In Mathematics, we have many discoverers who are also known as
“Fathers” of their respective disciplines. Who is widely considered to be the Father of Trigonometry? Hint: He
was a Greek astronomer who lived between 190-120 B.C. and also developed the first accurate star map.
Hint: The answer to the question is the largest of the following values:
𝐴 = 𝑐𝑜𝑠 15𝜋⁄4
A.
B.
C.
18.
A.
B.
C.
19.
𝐵 = 𝑠𝑖𝑛 −3𝜋⁄4
𝐶 = 𝑡𝑎𝑛 3𝜋⁄4
𝐷 = 𝑡𝑎𝑛(2016𝜋)
𝐸 = 𝑠𝑖𝑛 −𝜋⁄3
Hipparchus
D. Galileo
Pythagoras
E. None of these Answers
Archimedes
At the London 2012 Games, Canada’s Ian Millar, aged 65, became the rider to have participated in the most
Olympic Games: 10 in all, from 1972 to 2012. In honor of Ian, perform the following operation with complex
numbers:
(2 − 2𝑖)10
32768
D. -32768i
1024i
E. None of these Answers
-1024
An equestrian is trying to judge whether a tree is an appropriate height for her horse to jump. Using a
measurement tool in hand, she knows the angle of elevation is 30° . What is the greatest horizontal distance in feet
away from the tree the equestrian can be for the horse to be able to jump the tree if its maximum jump is 3√6
feet?
9√2
D. 3√2
E. None of these Answers
9√6
A.
B.
C. 3√6
20. The path as a function of distance above or below an arbitrary point P as a function of time for a show horse is
shown below. The x-axis represents time in seconds and the y-axis represents distance in feet. Time 𝑡 < 0 is
before Everett started watching the horse and time 𝑡 ≥ 0 is after Everett started watching the horse. On what
interval(s) is the horse’s movements decreasing? Assume the x-axis scale to be in single units. For this question,
endpoints of an interval need not be included.
A.
B.
C.
D.
E.
[−∞, −5] ∪ (1,4)
(−∞, −5) ∪ (1,4)
(−5, −3) ∪ (−1,1) ∪ (4,5)
(−∞, −5) ∪ (1, ∞)
None of these Answers
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
Info to cite: Su, Francis E., et al. "All Horses are the Same Color." Math Fun Facts.
http://www.math.hmc.edu/funfacts
Solutions:
1C
2C
3D
4A
5B
The initial value is 9, and the common ratio is 2. The time says years after 1729, so you do not need to
account for the 1729 in the problem.
8!
The total number of letters factorial over the number of repeats factorial in the denominator: 2!2!
Perpendicular lines and line segments have slopes that are negative reciprocals of each other.
−2⁄ 𝑎𝑛𝑑 3⁄
3
2
First, you must use the picture to see that a stirrup is a parabola facing down, which makes the equation a
𝑦 = −𝑎(𝑥 − ℎ)2 + 𝑘 model. By plugging in the vertex and point given, we can solve for the coefficient
a. −12 = −𝑎(5 − (−2))2 + 4; a=16/49 which makes 4p=49/16. Going p in the direction of the opening
of the parabola, I find that the focus is at (−2, 207⁄64).
1
6E
7D
8D
9B
10 A
11 A
12 D
13 D
1
3
3
8
1
3
5
8
First we must randomly select an Olympian, then randomly select a gold medal: ( ) ( ) + ( ) ( ) +
6
(3) (8) = 7/12
The pattern repeats every 8th letter and W is the 23rd letter, which makes it the same as letter G.
The triangle follows the Pythagorean triple 9-40-41, with a multiple of 1/5. Secant = hypotenuse /
adjacent, so 41/40.
The brackets on choice D represent a matrix, not a determinant.
For the inverse, you use the y-value of the original relation as the x value. Then the corresponding yvalue (x of the original) is found on the table. Because none of the x-values of the inverse (y-values of the
original) are repeated, then it is a function.
I first found the circumference, and then set up a proportion:
3 𝑓𝑡
12𝜋𝑓𝑡
=
𝑥 𝑟𝑎𝑑
2𝜋 𝑟𝑎𝑑
to solve for the angle.
Proof by mathematical induction failed when you assumed 2 horses would have to be the same color.
𝜋
−(− )⁄
1
2𝜋
2𝜋
−𝑐
2
Period: ⁄𝑏 = ⁄2𝜋 = 1 Phase Shift: ⁄𝑏 =
2𝜋 = 4 Sum = 5/41
2csc(2𝑥)
2cos(𝑥)
2cos(𝑥)
1
= 2 cos(𝑥) csc(2𝑥) =
=
=
= 𝑐𝑠𝑐𝑥
sec(𝑥)
sin(2𝑥) 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥
14 C
51
33
ℎ(𝑡) = −3𝑡 3 + 9 − 𝑡 + 𝑡 2
2
2
1
1
= −3(𝑡 − 2) (𝑡 − ) (𝑡 − 3); 𝑡 = 2 , , 3
2
2
15 B
1/6
𝑥 6
3 3 𝑥
3log 𝑥⁄4 729⁄64 = 18; log 𝑥⁄4 729⁄64 = 6; ( ) = 729⁄64 ; (729⁄64) = ; = ; 𝑥 = 6
4
2 2 4
16 B
17 A
18 D
(0.9)(0.9)(0.9)(0.1) = Jump and Jump and Jump and Refuse = 0.0729
Hipparchus
(2 − 2𝑖)10 = 210 (1 − 𝑖)10 The binomial expansion can be done a few ways, but I am going to
choose to do it with complex numbers in trigonometric form: 1 − 𝑖 = √2𝑐𝑖𝑠 − 45°;
10
(1 − 𝑖)10 = (√2𝑐𝑖𝑠 − 45°) ; 32𝑐𝑖𝑠 − 450° = −32𝑖
1024(−32) = −32768𝑖
19 A
By using the 30-60-90 triangle, we can find that the side opposite the 60 degree angle will be multiplied
by √3. (3√6)(√3) = 9√2
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PreCalculus Individual – January 16, 2016 – FAMAT State-Wide Competition hosted by Vero Beach HS
20 B or E
This can be seen on the graph where the graph goes down when reading it from left to right.
7