2011-12 Fermat 10-22-11 AMC 8 Prep
2011-12 Fermat 10-22-11 AMC 8 Prep
2011-12 Fermat 10-22-11 AMC 8 Prep
Answer: (B)
Perimeter of a Triangle
• Note: the point of intersection of angle
bisectors of a triangle is the “incenter” of that
triangle.
• Answer: (E).
AMC 12 2001, Problem #11
AMC 12 2001, Problem #11
• The drawings can stop at drawing #2, drawing
#3, or, “worst case”, drawing #4.
• Suppose that we complete all 5 drawings,
regardless of the results of the first 4
drawings.
• We can define two mutually-exclusive results
of the first 4 drawings: A and B.
AMC 12 2001, Problem #11
• Result A: all the white chips have been drawn
during the first 4 drawings.
• Result B: all the red chips have been drawn
during the first 4 drawings.
• Then, the question in this problem can be
rephrased as:
“What is the probability of result A?”
AMC 12 2001, Problem #11
• Result A is possible if and only if the remaining
chip drawn in the 5th drawing is red.
• Thus, the initial question in this problem can
be further rephrased as:
“What is the probability that, as the result of
all 5 drawings, the chip drawn in the 5th
drawing is red?”
AMC 12 2001, Problem #11
• Experiment: doing all 5 drawings.
• All possible outcomes: any red or any white
chip in drawing #5.
• The desired event: any red chip in drawing #5.
• Total number of possible outcomes: 5.
• Total number of ways the desired event can
occur: 3.
AMC 12 2001, Problem #11
• The probability that a red chip is drawn in the
5th drawing is 3/5.
• Answer: (D).
AMC 8 2010, Problem #20
AMC 8 2010, Problem #20
• First of all, we need to calculate the minimum
number of people in the room.
• The fractions 2/5 and 3/4 correspond to the
numbers of people wearing gloves and hats
respectively.
• The minimum number of people in the room
equals the least common denominator of
these two fractions: 20.
AMC 8 2010, Problem #20
• Now, we can calculate the number of people
wearing gloves, 20 * 2/5 = 8,
• and the number of people wearing hats,
20 * 3/4 = 15.
• We can use the “worst case” method to
answer the question of the problem.
AMC 8 2010, Problem #20
• In the “best case”, 8 people wear both gloves
and a hat, 7 people wear hats and no gloves,
and 5 people wear neither gloves nor hats.
• However, in the “worst case”, 3 people must
wear both gloves and a hat; then 5 people
wear gloves and no hats, and 12 people wear
hats and no gloves.
• Answer: (A).
AMC 8 2005, Problem #24
AMC 8 2005, Problem #24
• Hint: start from the end and work backward.
• Since the only available operations are “+1”
and “*2”, the reverse operations are “-1” and
“/2”.
AMC 8 2005, Problem #24
• Clearly, we want to divide 200 by 2 and
continue to repeat this operation as long as
the result numbers are even.
• We get numbers 100, 50, 25.
• Since 25 is odd, the only choice is to subtract
1. We get 24.
• Now, we can start dividing numbers by 2
again. We get numbers 12, 6, 3.
AMC 8 2005, Problem #24
• 3 is odd. So, the last two reversed operations
result in numbers 2 and 1.
• Now, we can write the same numbers in the
ascending order and count the forward
operations: 1, 2, 3, 6, 12, 24, 25, 50, 100, 200.
• The number of operations is 9.
• Answer: (B)
AMC 8 2005, Problem #24
• Prove that 8 operations suggested in answer
(A) cannot result in 200.
• If all 8 operations are “*2”:
2*2*2*2*2*2*2*2 = 256 > 200
• If we replace any one operation “*2” with
“+1”:
3*64 = 192 < 200
AMC 8 2005, Problem #24
5*32 = 160 < 200
9*16 = 144 < 200
17*8 = 136 < 200
33*4 = 132 < 200
65*2 = 130 < 200
• It’s obvious that replacing one more “*2”
operation with “+1” in any of these cases will
only reduce the result.
AMC 12A 2006, Problem #23
AMC 12A 2006, Problem #23
• After dividing among 6 people, 4 coins are
left. Add 2 coins to make the total divisible by
6.
• After dividing among 5 people, 3 coins are
left. Again, add 2 coins to make the total
divisible by 5.
• The LCM of 6 and 5 is 30.
AMC 12A 2006, Problem #23
• The number of coins in the box is 30 - 2 = 28.
• 28 is divisible by 7.
• Answer: (A).
AMC 8 2005, Problem #8
AMC 8 2005, Problem #8
• This is an example of how the multiple-choice
answers can be used creatively.
• Since we need to choose the formula that has
odd values for all positive odd n and m, then
one counter example for a given formula
eliminates that formula.
AMC 8 2005, Problem #8
• If we simply calculate each formula in A, B, C,
D, and E for n = 1 and m = 1, then A, B, C, and
D produce even numbers and can be
eliminated.
• Only the formula in answer E produces odd
number that makes it the clear winner.
• Answer: (E).
AMC 8 2005, Problem #8
• Of course, it is not hard to solve this problem
using the rules of math for odd and even
numbers.
• Whichever method works better for you
during the test is fine, as long as it is correct.