AMC 12A Problem and Solution

2017 AMC 12A

Problem9

Let be the set of points

in the coordinate plane such that two of the three

quantities , , and are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of ?

Solution

If the two equal values are and , then . Also, because 3 is the common value. Solving for , we get . Therefore the portion of the line

where

is part of . This is a ray with an endpoint of

.

Similar to the process above, we assume that the two equal values are and . Solving the equation then . Also, because 3 is the common value. Solving for , we get . Therefore the portion of the line where is also part of . This is another ray with the same endpoint as the above ray:

.

If and are the two equal values, then . Solving the equation for , we get . Also because is one way to express the common value (using as the common value works as well). Solving for , we get . Therefore the portion of the line where is part of like the other two rays. The lowest possible value that can be achieved is also

.

, the answer is

.

Since is made up of three rays with common endpoint Solution by TheMathematicsTiger7

Problem10

Chloé chooses a real number uniformly at random from the interval Laurent chooses a real number uniformly at random from the interval probability that Laurent's number is greater than Chloe's number?

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. Independently, . What is the

2017 AMC 12A

Solution

Suppose Laurent's number is in the interval

. Then, by symmetry, the probability of

Laurent's number being greater is . Next, suppose Laurent's number is in the interval

. Then Laurent's number will be greater with probability . Since each

case is equally likely, the probability of Laurent's number being greater is answer is C.

, so the

Alternate Solution: Geometric Probability

Let be the number chosen randomly by Chloé. Because it is given that the number Chloé chooses is in the interval

,

. Next, let be the number chosen

randomly by Laurent. Because it is given that the number Laurent chooses is in the interval

,

. Since we are looking for when Laurent's number is greater

than Chloé's we write the equation . When these three inequalities are graphed the area captured by and represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area . The area captured by , , and represents the possibilities of

Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus making an area

. The simplified quotient of these two areas

is the probability Laurent's number is larger than Chloé's, which is

.

Problem11

Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?

Solution

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2017 AMC 12A

We know that the sum of the interior angles of the polygon is a multiple of . Note

that is

the answer is

and , so the angle Claire forgot

. Since the polygon is convex, the angle is

.

, so

Problem12

There are horses, named Horse 1, Horse 2, , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse runs one lap in exactly minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time , in minutes, at which all horses will again simultaneously be at the starting point is . Let be the least time, in minutes, such that at least of the horses are again at the starting point. What is the sum of the digits of ?

Solution

We know that Horse will be at the starting point after minutes if for the smallest such that at least of the numbers least positive integer divisors. We quickly see that that

. Thus, we are looking divide . Thus, has at

is the smallest number with at least positive integer divisors, and

.

are each numbers of horses. Thus, our answer is

Problem13

Driving at a constant speed, Sharon usually takes

minutes to drive from her house to her

mother's house. One day Sharon begins the drive at her usual speed, but after driving of the way, she hits a bad snowstorm and reduces her speed by miles per hour. This time the trip takes her a total of minutes. How many miles is the drive from Sharon's house to her mother's house?

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2017 AMC 12A

Solution

Let total distance be . Her speed in miles per minute is

. Then, the distance that she drove

miles per hour,

before hitting the snowstorm is . Her speed in snowstorm is reduced or miles per minute. Knowing it took her

minutes in total, we create equation:

Solving equation, we get

.

Problem14

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of chairs under these conditions?

Solution

Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields ways to seat the group. Next, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. ways to seat Alice times ways to fill the next chair times ways to fill the center chair times ways to fill the last two chairs yields ways to fill the chairs. Finally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields ways to fill the chairs. In total, there are

ways to fill the chairs, so the answer is .

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