【同步首发】2018AMC12B官方真题及答案

2018 AMC 12B真题

答案解析请参考文末

Problem 1

Kate bakes 20-inch by 18-inch pan of cornbread. The cornbread is cut into pieces that measure 2 inches by 2 inches. How many pieces of cornbread does the pan contain?

Problem 2

Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes?

Problem 3

A line with slope 2 intersects a line with slope 6 at the point . What is the distance between the -intercepts of these two lines?

Problem 4

A circle has a chord of length , and the distance from the center of the circle to the chord is . What is the area of the circle?

Problem 5

How many subsets of contain at least one prime number?

Problem 6

Suppose cans of soda can be purchased from a vending machine for quarters. Which of the following expressions describes the number of cans of soda that can be purchased for dollars, where 1 dollar is worth 4 quarters?

Problem 7

What is the value of

Problem 8

Line segment is a diameter of a circle with . Point , not equal to or , lies on the circle. As point moves around the circle, the centroid (center of mass) of traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

Problem 9

What is

Problem 10

A list of positive integers has a unique mode, which occurs exactly times. What is the least number of distinct values that can occur in the list?

Problem 11

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point in the figure on the right. The box has base length and height . What is the area of the sheet of wrapping paper?

Problem 12

Side of has length . The bisector of angle meets at , and . The set of all possible values of is an open interval . What is ?

Problem 13

Square has side length . Point lies inside the square so that and . The centroids of , , , and are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

Problem 14

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 1 year older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

Problem 15

How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?

Problem 16

The solutions to the equation are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled and . What is the least possible area of

Problem 17

Let and be positive integers such thatand is as small as possible. What is ?

Problem 18

A function is defined recursively by andfor all integers . What is ?

Problem 19

Mary chose an even -digit number . She wrote down all the divisors of in increasing order from left to right: . At some moment Mary wrote as a divisor of . What is the smallest possible value of the next divisor written to the right of ?

Problem 20

Let be a regular hexagon with side length . Denote by , , and the midpoints of sides , , and , respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of and ?

Problem 21

In with side lengths , , and , let and denote the circumcenter and incenter, respectively. A circle with center is tangent to the legs and and to the circumcircle of . What is the area of ?

Problem 22

Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ?

Problem 23

Ajay is standing at point near Pontianak, Indonesia, latitude and longitude. Billy is standing at point near Big Baldy Mountain, Idaho, USA, latitude and longitude. Assume that Earth is a perfect sphere with center . What is the degree measure of ?

Problem 24

Let denote the greatest integer less than or equal to . How many real numbers satisfy the equation ?

Problem 25

Circles , , and each have radius and are placed in the plane so that each circle is externally tangent to the other two. Points , , and lie on , , and respectively such that and line is tangent to for each , where . See the figure below. The area of can be written in the form for positive integers and . What is ?

1. The area of the pan is = . Since the area of each piece is , there are pieces. Thus, the answer is .
2. Let Sam drive at exactly mph in the first half hour, mph in the second half hour, and mph in the third half hour.Due to , and that min is half an hour, he covered miles in the first mins.SImilarly, he covered miles in the nd half hour period.The problem states that Sam drove miles in min, so that means that he must have covered miles in the third half hour period., so .Therefore, Sam was driving miles per hour in the third half hour.
3. Using the slope-intercept form, we get the equations and . Simplifying, we get and . Letting in both equations and solving for gives the -intercepts: and , respectively. Thus the distance between them is
4. The shortest segment that connects the center of the circle to a chord is the perpendicular bisector of the chord. Applying the Pythagorean theorem, we find thatThe area of a circle is , so the answer is
5. Since an element of a subset is either in or out, the total number of subsets of the 8 element set is . However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are subsets with at least 1 prime so the answer is
6. The unit price for a can of soda (in quarters) is . Thus, the number of cans which can be bought for dollars ( quarters) is
7. Change of base makes this
8. Draw the Median connecting C to the center O of the circle. Note that the centroid is of the distance from O to C. Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius .The area of this circle is .
9. We can start by writing out the first couple of terms:Looking at the second terms in the parentheses, we can see that occurs times. It goes horizontally and exists times vertically. Looking at the first terms in the parentheses, we can see that occurs times. It goes vertically and exists times horizontally.Thus, we have:This gives us:
10. To minimize the number of values, we want to maximize the number of times they appear. So, we could have 223 numbers appear 9 times, 1 number appear once, and the mode appear 10 times, giving us a total of =
11. Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is . The area of the rectangle that is by is . The combined figure of the two triangles with base is a square with as its diagonal. Using the Pythagorean Theorem, each side of this square is . Thus, the area is the side length squared which is . Similarly, the combined figure of the two triangles with base is a square with area . Adding all of these together, we get . Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting .
12. Let . Then by Angle Bisector Theorem, we have . Now, by the triangle inequality, we have three inequalities.
    • , so . Solve this to find that , so .
    • , so . Solve this to find that , so .
    • The third inequality can be disregarded, because has no real roots.

    Then our interval is simply to get .

13. We can draw an accurate diagram by using centimeters and scaling everything down by a factor of . The centroid is the intersection of the three medians in a triangle.After connecting the centroids, we see that the quadrilateral looks like a square with side length of . However, we scaled everything down by a factor of , so the length is . The area of a square is , so the area is:
14. Let Joey's age be , Chloe's age be , and we know that Zoe's age is .We know that there must be values such that where is an integer.Therefore, and . Therefore, we know that, as there are solutions for , there must be solutions for . We know that this must be a perfect square. Testing perfect squares, we see that , so . Therefore, . Now, since , by similar logic, , so and Joey will be and the sum of the digits is
15. Analyze that the three-digit integers divisible by start from . In the 's, it starts from . In the 's, it starts from . We see that the units digits is and Write out the 1- and 2-digit multiples of starting from and Count up the ones that meet the conditions. Then, add up and multiply by , since there are three sets of three from to Then, subtract the amount that started from , since the 's all contain the digit .We get:This gives us:
16. The answer is the same if we consider Now we just need to find the area of the triangle bounded by and This is just
17. We claim that, between any two fractions and , if , the fraction with smallest denominator between them is . To prove this, we see thatwhich reduces to . We can easily find that , giving an answer of .
18. Thus, .
19. Prime factorizing gives you . The desired answer needs to be a multiple of or , because if it is not a multiple of or , the LCM, or the least possible value for , will not be more than 4 digits. Looking at the answer choices, is the smallest number divisible by or . Checking, we can see that would be .
20. The desired area (hexagon ) consists of an equilateral triangle () and three right triangles (, , and ).Notice that (not shown) and are parallel. divides transversals and into a ratio. Thus, it must also divide transversal and transversal into a ratio. By symmetry, the same applies for and as well as and In , we see that and . Our desired area becomes
21. Let the triangle have coordinates Then the coordinates of the incenter and circumcenter are and respectively. If we let then satisfiesNow the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be
22. Suppose our polynomial is equal toThen we are given thatIf we let then we haveThe number of solutions to this equation is simply by stars and bars, so our answer is
23. Suppose that Earth is a unit sphere with center We can letThe angle between these two vectors satisfies yielding or
24. This rewrites itself to .Graphing and we see that the former is a set of line segments with slope from to with a hole at , then to with a hole at etc.Here is a graph of and for visualization.Now notice that when then graph has a hole at which the equation passes through and then continues upwards. Thus our set of possible solutions is bounded by . We can see that intersects each of the lines once and there are lines for an answer of .
25. Let be the center of circle for , and let be the intersection of lines and . Because , it follows that is a triangle. Let; then and . The Law of Cosines in giveswhich simplifies to . The positive solution is . Then , and the required area isThe requested sum is .