2018 AMC 12B ANSWER KEY答案解析

Problem 1

Solution 1

The area of the pan is = . Since the area of each piece is , there are pieces. Thus, the answer is .

Solution 2

By dividing the each of the dimensions by , we get a grid which makes pieces. Thus, the answer is .

 

Problem 2

Solution

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.

 

Problem 3

Solution 

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

 

Problem 4

Solution

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 triangle is , so the answer is

 

Problem 5

Solution 1

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

Solution 2

We can construct our subset by choosing which primes are included and which composites are included. There are ways to select the primes (total subsets minus the empty set) and ways to select the composites. Thus, there are ways to choose a subset of the eight numbers, or .

 

Problem 6

Solution 

The unit price for a can of soda (in quarters) is . Thus, the number of cans which can be bought for dollars ( quarters) is

 

Problem 7

Solution 1

Change of base makes this

Solution 2

Using the chain rule for logarithms (), we get .

 

Problem 8

Solution

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 .

 

Problem 9

Solution 1

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:

Solution 2

Solution 3

 

Problem 10

 

 

Problem 11

Problem 12

Solution

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 .

 

Problem 13

Solution 1 (Drawing an Accurate Diagram)

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:

Solution 2

The centroid of a triangle is of the way from a vertex to the midpoint of the opposing side. Thus, the length of any diagonal of this quadrilateral is . The diagonals are also parallel to sides of the square, so they are perpendicular to each other, and so the area of the quadrilateral is , .

Solution 3

The midpoints of the sides of the square form another square, with side length and area . Dilating the corners of this square through point by a factor of results in the desired quadrilateral (also a square). The area of this new square is of the area of the original dilated square. Thus, the answer is

Solution 4

We put the diagram on a coordinate plane. The coordinates of the square are and the coordinates of point P are By using the centroid formula, we find that the coordinates of the centroids are and Shifting the coordinates down by does not change its area, and we ultimately get that the area is equal to the area covered by which has an area of

 

Problem 14

Solution 1

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

Solution 2

Here's a different way of saying your solution.

If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has 9 factors. Therefore, the difference between Chloe and Zoe's age is 36, so Chloe is 37, and Joey is 38. The common factor that will divide both of their ages is 37, so Joey will be 74. 7 + 4 =

Solution 3

Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)

Let denote Chloe's age, denote Joey's age, and denote Zoe's age, where is the number of years from now. We are told that is a multiple of exactly nine times. Because is at and will increase until greater than , it will hit every natural number less than , including every factor of . For to be an integral multiple of , the difference must also be a multiple of , which happens if is a factor of . Therefore, has nine factors. The smallest number that has nine positive factors is (we want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's). We also know and . Thus,By our above logic, the next time is a multiple of will occur when is a factor of . Because is prime, the next time this happens is at , when .

 

Problem 15

Solution 1 (For Dummies)

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:

Solution 2

There are choices for the last digit (), and choices for the first digit (exclude ). We know what the second digit mod is, so there are choices for it (pick from one of the sets ). The answer is (Plasma_Vortex)

 

Problem 16

Solution

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

 

Problem 17

Solution 1

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 .

Solution 2 (requires justification)

Assume that the difference results in a fraction of the form . Then,

Also assume that the difference results in a fraction of the form . Then,

Solving the system of equations yields and . Therefore, the answer is

Solution 3

Cross-multiply the inequality to get

Then,

Since , are integers, is an integer. To minimize , start from , which gives . This limits to be greater than , so test values of starting from . However, to do not give integer values of .

Once , it is possible for to be equal to , so could also be equal to The next value, , is not a solution, but gives . Thus, the smallest possible value of is , and the answer is .

Solution 4

Graph the regions and . Note that the lattice point is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is and the answer is .

Remark: This also gives an intuitive geometric proof of the mediant using vectors.

Solution 5 (Using answer choices to prove mediant)

As the other solutions do, the mediant is between the two fractions, with a difference of . Suppose that the answer was not , then the answer must be or as otherwise would be negative. Then, the possible fractions with lower denominator would be for and for which are clearly not anywhere close to

 

Problem 18

Solution 1

Thus, .

Solution 2 (A Bit Bashy)

Start out by listing some terms of the sequence.

Notice that whenever is an odd multiple of , and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. The closest odd multiple of to is , so we have

Solution 3(Bashy Pattern Finding)

Writing out the first few values, we get: . Examining, we see that every number where has , , and . The greatest number that's 1 (mod 6) and less is , so we have

 

Problem 19

Solution 1

Prime factorizing gives you . Looking at the answer choices, is the smallest number divisible by or .

Solution 2

Let the next largest divisor be . Suppose . Then, as , therefore, However, because ,. Therefore, . Note that . Therefore, the smallest the gcd can be is and our answer is .

Problem 20

Solution 1

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

Solution 2

Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). We know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of for a total area of (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be giving us an area of per triangle, and a total area of . Adding the two areas together, we get . Finding the total area, we get . Taking the complement, we get

Solution 3 (Trig)

Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is and the trapezoid is isosceles, we know that the angle opposite is , and thus the side length of this triangle is . So the area of this triangle is Now let's find the area of the smaller triangles. Notice, triangle cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then and the sum of the areas is Therefore, the area of the convex hexagon is

 

Problem 21

Solution 1

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

Solution 2

Notice that we can let . If , then and . Using shoelace formula, we get .

 

Problem 22

Solution

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

 

Problem 23

Solution

Suppose that Earth is a unit sphere with center We can letThe angle between these two vectors satisfies yielding or

 

Problem 24

Solution

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 .

Alternative, Bashy Solution

Same as the first solution, .

We can write as . Expanding everything, we get a quadratic in in terms of :

We use the quadratic formula to solve for {x}:

Since , we get an inequality which we can then solve. After simplifying a lot, we get that .

Solving over the integers, , and since is an integer, there are solutions. Each value of should correspond to one value of , so we are done.

Another Solution

Let where is the integer portion of and is the decimal portion. We can then rewrite the problem below:

From here, we get

Solving for ...

Because , we know that cannot be less than or equal to nor greater than or equal to . Therefore:

There are 199 elements in this range, so the answer is

Yet Another Solution

As in the first solution, we can write .

Now obviously, .

So, x can get be .

Hence, the answer is .

-Pi_3.14_Squared

 

Problem 25

Solution 1

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 .

Solution 2

Let and be the centers of and respectively and draw , , and . Note than and are both right. Furthermore, since is equilateral, and . Mark as the base of the altitude from to . By special right triangles, and . since and , we can find find . Thus, . This makes . This makes the answer .