Sprint, Target, and Team Questions Sampler
1. What is the sum of the reciprocals of all positive integers who have no prime divisors greater than 3?
2. For positive integers x, y, and z, find the value of (logyx)(logzy)(logxz).
3. A bag contains four balls. Three are red, and the fourth is either red or blue, selected with equal
probability. A ball is randomly drawn from the bag, and it is red. What is the probability that one of
the balls still in the bag is blue?
4. Three teachers and six students sit in a line of chairs, with each teacher sitting between 2 students.
How many arrangements are possible?
5. Points A and B lie on circle O, and C lies on line AB. The shortest distance from C to the circle is 5.
CA = 10 and CB = 6. Exactly two distinct circles meet these criteria. What is the sum of their
diameters?
6. Lines AB and CD meet at point E.
Lines AD and BC meet at F, as shown. ∠AEC = y° ,
∠AFC = x , x + y = 180, and DE = EC = AE * cos( x) . Compute the ratio AD in terms of x.
o
BC
A
C
E
D
B
F
7. From 186 feet high, on the top of the Leaning Tower of Pisa, Lisa drops a ball to the ground.
Each time the ball hits the ground, it rebounds to one-half of its previous height. How many
feet will the ball travel before coming to a rest?
8. Assume that an Earth day is exactly 24 hours and an Earth year is exactly 365 days. In a
distant galaxy, a planet named Ono takes 48 Earth-hours to revolve once around its own axis
and takes 584 Ono-days to orbit once around its sun. When Yoko is 16 Earth years old, what
is his age in Ono years?
9. Find the equation of the set of points equidistant from the origin and (12, 9).
10. Simplify
Sin2A + Sin4A + Sin6A + Sin 8A
.
Cos2A + Cos4A + Cos6A + Cos8A
Solutions
1. The expansion of the following product yield all of the required reciprocals
(1 +
1 1 1
1 1
1
1
1
3
+ + + ...)(1 + + +
+ ...) =
•
= 2• = 3.
2 4 8
3 9 27
1−1 2 1−1 3
2
3
2. (logyx)(logzy)(logxz) = (logyx)(logxz)(logzy) = (logyz)(logzy) = 1
1
3. There are three possible outcomes to this situation. First, we can have that the fourth ball in
the bag is red. In this case, a red ball is drawn no matter what. The probability of this
outcome is 12 . Second, we can have the fourth ball in the bag be blue, but draw a red ball.
The probability of this is 12 * 43 = 83 . Third, we can have the fourth ball blue, and the blue ball
drawn. However, we know that a red ball was drawn, so we must disregard this third case.
3
The probability that there is a blue ball in the bag is
3
8
3
8
+ 1
=
2
7
3/7
4. Of the 5 spaces between the 6 students, there are C(5,3)=10 ways to select places for the 3
teachers. Then there are 6! ways to seat the students and 3! ways to seat the teachers.
10*6!*3! = 43200.
43200
5. Point C can be either in the interior or the exterior of circle O. Let’s first assume that C is in
the exterior of the circle. Let D be the closest point on the circle to C. Then CD lies along a
diameter of the circle. Let E be the other point where CD intersects the circle. By the Power
of a Point Theorem, CD * CE = CB * CA . Letting d be the diameter of the circle, we have
5(d + 5) = 6*10 . Solving this equation gives d=7. If point C is in the interior of the circle,
the Power of a Point Theorem still applies and yields the equation: 5(d − 5) = 6*10 and d =
17. 7+17 = 24.
24
6. First, note that ∠AED = 180o − y o = x o . Since DE / AE = cos x° , ∠ADE must be a right
angle. This makes ∠EAD = 90° − x . Since ∠AFB = x°, then ∠AFB + ∠EAD = 90o , and
∠ABF is also a right angle. This makes triangles ADE and CBE similar! So
AD AE
1
=
=
= sec( x) .
BC EC cos( x)
sec(x)
7. 186 + 93 + 93 + 46.5 + 46.5 + …..= 186 + 2*93/(1 - 1/2) = 558 ft
558 ft
8. 16 EY * 365 ED/1 EY * 24 EH/1 ED * 1 OD/48 EH * 1 OY/584 OD = 5 OY
5
9. This set of points is the perpendicular line that bisect the segment from the origin to (12, 9).
This line’s slope is then equal to −12/9 = −4/3 and its equation is y – 9/2 = −4/3(x – 6), or
equivalently, y = −4/3x + 25/2.
−4/3x + 25/2
10. Rewriting the numerator: (Sin 8A + Sin2A ) + (Sin 6A + Sin 4A)
= 2 Sin 5A Cos3A + 2 Sin 5A CosA = 2 Sin5A (Cos3A + CosA)
Similarly for the denominator, (Cos8A + Cos2A) + (Cos 6A + Cos 4A)
2 Cos5A (Cos3A + CosA). Cancelling gives sin5A/Cos5A = Tan5A.
Tan5A