【同步首發(fā)】2018AMC12B官方真題及答案
2018 AMC 12B真題
答案解析請參考文末
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 that
and is as small as possible. What is 
?
Problem 18
A function 
is defined recursively by 
and
for 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 
?
- The area of the pan is
= 
. Since the area of each piece is 
, there are 
pieces. Thus, the answer is 
. - 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. - 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 
- 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 that
The area of a circle is 
, so the answer is 
- 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 
- The unit price for a can of soda (in quarters) is
. Thus, the number of cans which can be bought for 
dollars (
quarters) is
- Change of base makes this
- 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 
. - 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:
- 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
= 
- 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 
. - 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 
. - 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:
- 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 
- 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:
- 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 
- We claim that, between any two fractions
and 
, if 
, the fraction with smallest denominator between them is 
. To prove this, we see that
which reduces to 
. We can easily find that 
, giving an answer of 
. 
Thus, 
. 
- 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 
. 
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
- Let the triangle have coordinates
Then the coordinates of the incenter and circumcenter are 
and 
respectively. If we let 
then satisfies
Now the area of our triangle can be calculated with the Shoelace Theorem. The answer turns out to be 
- Suppose our polynomial is equal to
Then we are given that
If we let 
then we have
The number of solutions to this equation is simply 
by stars and bars, so our answer is 
- Suppose that Earth is a unit sphere with center
We can let
The angle 
between these two vectors satisfies 
yielding 
or 
- 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 
. - 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 
gives
which simplifies to 
. The positive solution is 
. Then 
, and the required area is
The requested sum is 
.
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