2019 AMC 12B 详解
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所有题目均经美国数学协会（MAA）官方合法授权使用。
1.
Alicia 有两个容器。第一个装了 容积的水，第二个是空的。她把第一个容器中的水全部倒入第二个容器，此时第二个容器装了 容积的水。第一个容器的体积与第二个容器的体积之比是多少？
Alicia had two containers. The first was full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was full of water. What is the ratio of the volume of the first container to the volume of the second container?
2.
考虑命题：“如果 不是质数，那么 是质数。”下列哪个 的值是这个命题的反例？
Consider the statement, “If is not prime, then is prime.” Which of the following values of is a counterexample to this statement?
小提示：
反例要使命题的条件为真，但结论为假
A counterexample makes the hypothesis true but the conclusion false
大提示：
找一个不是质数的 ，且 也不是质数
Look for that is not prime while is also not prime
解答：
反例需要 不是质数（条件成立），且 不是质数（结论不成立）。
在选项中， 不是质数，且 也不是质数。质数 和 不满足条件，而 ， 都是质数。
所以正确答案是 E。
A counterexample needs not prime (so the hypothesis holds) and not prime (so the conclusion fails).
Among the choices, is not prime and is not prime. The primes and fail the hypothesis, and are prime.
Thus, E is the correct answer.
3.
下列哪个刚性变换（等距变换）把线段 映到线段 ，使得 的像是 ，且 的像是 ？
Which one of the following rigid transformations (isometries) maps the line segment onto the line segment so that the image of is and the image of is
关于 -轴反射
reflection in the -axis
绕原点逆时针旋转
counterclockwise rotation around the origin by
向右平移 个单位并向下平移 个单位
translation by units to the right and units down
关于 -轴反射
reflection in the -axis
绕原点顺时针旋转
clockwise rotation about the origin by
4.
正整数 满足方程 。 的各位数字之和是多少？
A positive integer satisfies the equation What is the sum of the digits of
5.
一家商店中每块糖果的价格都是整数美分。Casper 的钱刚好可以买 块红色糖果，或 块绿色糖果，或 块蓝色糖果，或 块紫色糖果。一块紫色糖果价格为 美分。 的最小可能值是多少？
Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either pieces of red candy, pieces of green candy, pieces of blue candy, or pieces of purple candy. A piece of purple candy costs cents. What is the smallest possible value of
小提示：
Casper 的总钱数必须能被每种糖果块数整除
Casper’s total money must be divisible by each candy count
大提示：
总钱数是 的倍数
The total is a multiple of
解答：
设 Casper 有 美分。因为他可以刚好买 、 或 块价格为整数美分的糖果，所以 是 的倍数。
紫色糖果每块 美分，所以 。最小的 是 ，得 。
所以正确答案是 B。
Let be Casper’s money in cents. Since he can exactly buy or whole-cent pieces, is a multiple of
Purple candy costs cents, so The smallest is giving
Thus, B is the correct answer.
6.
在一个给定平面中，点 和 相距 个单位。平面中有多少个点 ，使得 的周长为 个单位，且 的面积为 平方单位？
In a given plane, points and are units apart. How many points are there in the plane such that the perimeter of is units and the area of is square units?
无穷多个
infinitely many
小提示：
固定 后，周长条件使 在以 和 为焦点的椭圆上
With fixed, the perimeter condition puts on an ellipse with foci and
大提示：
底边 面积 需要高 ；与椭圆的最大高度比较
Area with base needs height compare that to the ellipse’s greatest height
解答：
周长条件给出 ，所以 在以 为焦点、长轴 的椭圆上。因此 且 ，所以半短轴为
若以 为底且面积为 ，则从 到底边的高必须为 。但椭圆上的最大可能高度是 ，所以不存在这样的 。
所以正确答案是 A。
The perimeter condition gives so lies on an ellipse with foci and major axis Thus and so the semi-minor axis is
For area with base the height from must be But the greatest possible height on the ellipse is so no such exists.
Thus, A is the correct answer.
7.
使 ，，， 和 这五个数的中位数等于平均数的所有实数 的和是多少？
What is the sum of all real numbers for which the median of the numbers and is equal to the mean of those five numbers?
小提示：
平均数是 。
The mean is
大提示：
按中位数是 ，，还是 本身分情况
Split into cases by whether the median is or itself
解答：
平均数为 。
若 ，中位数是 ，所以 ，得 ，符合该范围。
若 ，中位数是 ，所以 ，得 ，不在范围内。若 ，中位数是 ，所以 ，得 ，不在范围内。
唯一解是 ，因此和为 。
所以正确答案是 A。
The mean is
If the median is so gives which is consistent.
If the median is so gives not in range. If the median is so gives not in range.
The only solution is so the sum is
Thus, A is the correct answer.
8.
令 。求下列和的值：
Let What is the value of the sum
小提示：
大提示：
将 与 配对，并比较它们的符号
Pair with and compare their signs
解答：
因为 ，所以 。
在这个和中，指标为 的项符号是 ，而指标为 的项数值相等，但符号是 ，恰好相反。
每一项都与它的配对项抵消，所以总和为 。
所以正确答案是 A。
Since we have
In the sum, the term with index has sign while the term with index equals it in value but has sign the opposite.
Every term cancels with its partner, so the total is
Thus, A is the correct answer.
9.
有多少个整数 ，使得边长为 、 和 的三角形面积为正？
For how many integral values of can a triangle of positive area be formed having side lengths and
小提示：
令 ，则 。
Let so
大提示：
对边长 应用三角形不等式
Apply the triangle inequality to sides
解答：
令 ，则 ，三边为 。
三角形不等式给出 （所以 ）以及 （所以 ）；第三个不等式自动成立。
因此 ，即 。整数 共有 个。
所以正确答案是 B。
Let Then and the sides are
The triangle inequalities give (so ) and (so ); the third inequality is automatic.
Thus i.e. The integers number
Thus, B is the correct answer.
10.
下图是一张地图，显示 座城市和连接某些城市对的 条道路。Paula 想从城市 出发，到城市 结束，恰好走过其中 条道路，且任何一段道路都不能走超过一次。（Paula 可以多次到访同一座城市。）Paula 有多少条不同路线可以选择？
The figure below is a map showing cities and roads connecting certain pairs of cities. Paula wishes to travel along exactly of those roads, starting at city and ending at city without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) How many different routes can Paula take?
小提示：
用掉 条道路中的 条，正好留下 条不用
Using of the roads leaves exactly roads unused
大提示：
只有当仅 和 的度数为奇数时才存在这样的迹；这会强制决定要删掉哪 条道路
A trail exists only if just and have odd degree; this forces which roads to drop
解答：
将上排四座城市依次命名为 ，中排为 ，下排为 。一条使用 条道路的路线是一条开放欧拉迹，所以在所用道路构成的图中，只有 和 的度数为奇数。
在完整地图中，需要改变度数奇偶性的顶点是 。因为只删除 条道路，这四条道路必须将这 个顶点两两配对。其中 只与 相邻，所以必须删除 ；随后 也被迫删除。同理， 只与 相邻，所以必须删除 ，随后删除 。
剩余图是一条链 两个 边形环各可沿两个方向遍历，其余部分都被迫确定。因此共有 条路线。
所以 E 是正确答案。
Name the four cities in the top row those in the middle row and those in the bottom row A route using roads is an open Euler trail, so in the used graph exactly and have odd degree.
In the full map, the vertices whose degree parity must change are Because only roads are removed, those roads must pair these vertices. Among them, is adjacent only to forcing to be removed; then is forced. Similarly is adjacent only to forcing and then
The remaining graph is a chain Each of the two -cycles can be traversed in either direction, and everything else is forced. Hence there are routes.
Thus, E is the correct answer.
11.
给定一个立方体，有多少对无序的棱可以确定一个平面？
How many unordered pairs of edges of a given cube determine a plane?
小提示：
两条棱能确定一个平面，除非它们是异面直线
Two edges determine a plane unless they are skew
大提示：
分别数平行的棱对和共顶点的棱对
Count the pairs that are parallel and the pairs that share a vertex
解答：
两条棱恰好在共面时能确定一个平面，也就是它们平行或相交。
条棱分成 个方向，每个方向有 条平行棱，得到 对平行棱。共顶点的棱对有 对。
总数为 。
所以正确答案是 D。
Two edges determine a plane exactly when they are coplanar, that is, parallel or intersecting.
The edges split into directions of parallel edges, giving parallel pairs. Edges sharing a vertex give intersecting pairs.
The total is
Thus, D is the correct answer.
12.
如图，等腰直角三角形 的直角边长为 。在其斜边 上向外作直角三角形 ，其直角在 ，且两个三角形的周长相等。 是多少？
Right triangle with right angle at is constructed outwards on the hypotenuse of isosceles right triangle with leg length as shown, so that the two triangles have equal perimeters. What is
小提示：
的直角边为 ，斜边 ；设 并使用周长相等
has legs and hypotenuse set and use equal perimeters
大提示：
因为 ，使用
Since use
解答：
三角形 的周长为 ，且 。在 中设 ，则 ，周长相等给出
因此 ，所以 ，得 且 。
因为 ，令 则 ，所以 。又 ，因此
所以正确答案是 D。
Triangle has perimeter and In let so and equal perimeters give
Then so giving and
Since writing gives so With we get
Thus, D is the correct answer.
13.
一个红球和一个绿球随机且独立地被投入编号为正整数的箱子中。对每个球来说，投进第 号箱子的概率为 ，其中 ，，，。红球被投入编号比绿球更大的箱子的概率是多少？
A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin is for What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
小提示：
由对称性，红球编号更大和绿球编号更大的概率相等
By symmetry, red-higher and green-higher are equally likely
大提示：
先求平局概率：
Find the probability of a tie,
解答：
两球落入同一箱子的概率为
由对称性，红球编号更大和绿球编号更大的概率相等，所以各自概率为
所以正确答案是 C。
The probability the balls land in the same bin is
By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability
Thus, C is the correct answer.
14.
设 为 的所有正整数因数组成的集合。有多少个数可以表示为 中两个不同元素的乘积？
Let be the set of all positive integer divisors of How many numbers are the product of two distinct elements of
小提示：
因数形如 ，其中 ；乘积形如 ，其中 。
Divisors are with products are with
大提示：
所有 个乘积都能出现；去掉那些只能表示为某个因数乘以自身的值
All products arise; discard those obtainable only as a divisor times itself
解答：
因为 ，每个因数都形如 ，其中 。两个因数的乘积形如 ，其中 ，且每个这样的 都能达到，得到 个值。
我们需要两个不同的因数。一个值 只能表示为某个因数乘以自身，恰好发生在 和 都只有一种拆分方式时，也就是 。这 个角上的值（）不能使用两个不同因数表示。
数量为 。
所以正确答案是 C。
Since every divisor is with A product of two divisors is with and every such pair is attainable, giving values.
We need two distinct divisors. A value is forced to be a divisor times itself only when both and have a unique split, which happens exactly when Those corner values () cannot use two distinct divisors.
The count is
Thus, C is the correct answer.
15.
如图，线段 被点 和 三等分，使得 。三个半径为 的半圆 ，，和 的直径都在 上，并分别在 ，，和 处与直线 相切。一个半径为 的圆以 为圆心。图中阴影区域，即在该圆内但在三个半圆外的区域，其面积可表示为
其中 ，，，和 为正整数，且 与 互质。 是多少？
As shown in the figure, line segment is trisected by points and so that Three semicircles of radius and have their diameters on and are tangent to line at and respectively. A circle of radius has its center on The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form
where and are positive integers and and are relatively prime. What is
小提示：
取 ；半径为 的圆经过 和 ，且中间的半圆完全在它内部
Place the circle of radius then passes through and and the middle semicircle lies entirely inside it
大提示：
从圆的面积中减去三个半圆落在该圆内部的部分
Subtract from the circle’s area the parts of the three semicircles that fall inside it
解答：
设 ，，，，则三个半圆的圆心为 ，顶点为 ，，。该圆的圆心为 ，半径为 ，所以它经过 和 ，面积为 。
中间半圆 完全落在该圆内，去掉面积 。对于左边的半圆，重叠部分从 开始，包含它右侧的四分之一圆，但要除去落在大圆下方的那块区域。大圆与 轴相交于 。被除去的面积为 因此重叠部分的面积为 由对称性，右边的半圆贡献同样大的重叠面积。
阴影面积为 因此 ，所以 。
所以正确答案是 E。
Put so the semicircles are centered at and their tops are The circle has center and radius so it passes through and and has area
The middle semicircle lies entirely inside the circle, removing area For the left semicircle, the overlap starts at and includes its right-hand quarter-circle, except for the region below the large circle. The large circle meets the -axis at The excluded area is Therefore the overlap has area By symmetry, the right semicircle contributes the same overlap.
The shaded area is Hence so
Thus, E is the correct answer.
16.
一排睡莲叶依次编号为 到 。第 号和第 号睡莲叶上有捕食者，第 号睡莲叶上有一小块食物。青蛙 Fiona 从第 号睡莲叶出发；从任意一片睡莲叶出发，她有 的概率跳到下一片，也有同样的概率向前跳 片。Fiona 不落在第 号或第 号睡莲叶上而到达第 号的概率是多少？
There are lily pads in a row numbered to in that order. There are predators on lily pads and and a morsel of food on lily pad Fiona the frog starts on pad and from any given lily pad, has a chance to hop to the next pad, and an equal chance to jump pads. What is the probability that Fiona reaches pad without landing on either pad or pad
小提示：
每一步是 或 ，各自概率为 。
Each step is or each with probability
大提示：
追踪到达每片睡莲叶的概率，把第 号和第 号视为终止点
Track the probability of reaching each pad, treating pads and as dead ends
解答：
设 为在之前没有落到第 号或第 号的情况下落到第 号睡莲叶的概率。每片睡莲叶把概率 传到下一片，另一个 传到再下一片，而第 号和第 号不再向外传递概率。
因此 ，并且（跳过 ） ，然后（跳过 ） ，，。
最后
所以正确答案是 A。
Let be the probability of landing on pad without first landing on pad or Each pad sends probability to the next pad and two pads ahead, and pads and pass nothing on.
Then and (skipping ) then (skipping )
Finally
Thus, A is the correct answer.
17.
有多少个非零复数 满足：在复平面中，、、 所表示的点是一个等边三角形的三个不同顶点？
How many nonzero complex numbers have the property that and when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
无穷多个
infinitely many
小提示：
等边意味着 。
Equilateral means
大提示：
强制 ；然后要求 。
forces then require
解答：
这三个点构成等边三角形，当且仅当 。由 得 。
于是 ，所以需要 。写 ，则 ，所以 。
因此 ，给出四个 的值，并且所得顶点互不相同。
所以正确答案是 D。
The three points form an equilateral triangle iff From we get
Then so we need Writing so
This gives four values of all yielding distinct vertices.
Thus, D is the correct answer.
18.
四棱锥 的底面 是边长为 cm 的正方形，高 垂直于底面且长为 cm。点 在 上从 到 的三分之一处；点 在 上从 到 的三分之一处；点 在 上从 到 的三分之二处。 的面积是多少平方厘米？
Square pyramid has base which measures cm on a side, and altitude perpendicular to the base, which measures cm. Point lies on one third of the way from to point lies on one third of the way from to and point lies on two thirds of the way from to What is the area, in square centimeters, of
19.
Raashan、Sylvia 和 Ted 玩下面的游戏。每人初始有 。每隔 秒铃响一次，此时当前有钱的每个玩家 同时独立随机地选择另外两名玩家之一，并给该玩家 。铃响 次后，每个玩家都有 的概率是多少？（例如，Raashan 和 Ted 可以都决定给 Sylvia ，而 Sylvia 可以决定把她的一美元给 Ted，这时 Raashan 有 ，Sylvia 有 ，Ted 有 ，第一轮游戏结束。第二轮中 Raashan 没有钱可给，但 Sylvia 和 Ted 可能互相选择给对方 ，则第二轮结束时持有金额不变。）
Raashan, Sylvia, and Ted play the following game. Each starts with A bell rings every seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives to that player. What is the probability that after the bell has rung times, each player will have (For example, Raashan and Ted may each decide to give to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have Sylvia will have and Ted will have and that is the end of the first round of play. In the second round Raashan has no money to give, but Sylvia and Ted might choose each other to give their to, and the holdings will be the same at the end of the second round.)
小提示：
可能的状态只有 和 的排列
The only states are and permutations of
大提示：
证明从任一种状态出发，下一状态为 的概率都是 。
Show that from either kind of state the next state is with probability
解答：
从 出发，三名玩家各自把钱给另外两人之一，所以有 个等可能结果；只有 种循环赠送模式会回到 ，概率为 。
从 状态出发，没钱的玩家不给钱；检查另外两人的 个等可能选择，恰好一个会得到 ，概率仍为 。
因此任意一次铃响后，状态为 的概率都是 ，包括铃响 次之后。
所以正确答案是 B。
From each of the three players gives to one of two others, so there are equally likely outcomes; only the cyclic gift patterns return to a probability of
From a state the broke player gives nothing, and checking the equally likely choices of the other two shows exactly one yields again probability
So after any ring the probability of is including after rings.
Thus, B is the correct answer.
20.
点 和 在平面中的圆 上。假设 在 和 处的切线相交于 -轴上的一点。 的面积是多少？
Points and lie on circle in the plane. Suppose that the tangent lines to at and intersect at a point on the -axis. What is the area of
小提示：
相等的切线长说明交点到 和 的距离相等
Equal tangent lengths put the intersection point equidistant from and
大提示：
先找出 -轴上的交点，再用 和 定位圆心 。
Find that point on the -axis, then use and to locate the center
解答：
设切线交点为 。相等切线长给出 ，所以 ，得 ，即 。
圆心 满足 和 。因为 且 ，得到 和 ，所以 。
于是 ，所以面积为 。
所以正确答案是 C。
Let be the intersection. Equal tangent lengths give so yielding and
The center satisfies and With and these give and so
Then so the area is
Thus, C is the correct answer.
21.
有多少个实系数二次多项式满足：根的集合等于系数的集合？（说明：若多项式为 ，，根为 和 ，则要求 。）
How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is and the roots are and then the requirement is that )
无穷多个
infinitely many
小提示：
因为 至多只有两个值，所以至少两个系数相等
Since has at most two values, at least two coefficients are equal
大提示：
由韦达定理， 且 ；逐一处理哪些系数相等的情况
By Vieta, and work through each equal-coefficient case
解答：
如果三个系数都等于同一个值 ，多项式就是 ，其根不等于 。因此系数集合和根集合都有两个不同的值，所以恰有两个系数相同。由韦达定理， 且 。
先设 且 。根为 ，所以韦达定理给出 且 。因而 ，得到 和 。
如果 且 ，同一个乘积方程给出 ，而和的方程只留下 。这给出 。
最后，如果 且 ，乘积方程给出 ，所以 。和的方程变为 。这个严格递增的三次函数有唯一实根 ，因此恰好再得到一个多项式 。所以共有 个多项式。
所以 B 是正确答案。
If all three coefficients had one value the polynomial would be whose roots do not equal Thus the coefficient and root sets both have two distinct values, so exactly two coefficients coincide. By Vieta’s formulas, and
First suppose and The roots are so Vieta gives and Hence producing and
If and the same product equation gives while the sum equation leaves only This gives
Finally, if and the product equation gives so The sum equation becomes This strictly increasing cubic has one real root producing exactly one more polynomial, Therefore there are polynomials.
Thus, B is the correct answer.
22.
递归定义数列：，且
对所有非负整数 成立。设 为满足
的最小正整数。 位于下列哪个区间？
Define a sequence recursively by and
for all nonnegative integers Let be the least positive integer such that
In which of the following intervals does lie?
小提示：
令 ；则 。
Let then
大提示：
因为 递减趋向 ，这个比值始终在 与 之间
Since decreases toward the ratio stays between and
解答：
令 。简单计算得到
从 开始，各项保持为正并递减。因为 从 递减趋向 ，每个比值 都严格介于 和 之间。
因此 被夹在 与 之间。求解 可知 大约在 到 之间，落在 中。
所以正确答案是 C。
Let A short computation gives
Starting from the terms stay positive and decrease. Because decreases from toward each ratio lies strictly between and
Hence is squeezed between and Solving puts between about and which lies in
Thus, C is the correct answer.
23.
有多少个由 和 组成的长度为 的序列，满足以 开头、以 结尾、不含两个连续的 ，且不含三个连续的 ？
How many sequences of s and s of length are there that begin with a end with a contain no two consecutive s, and contain no three consecutive s?
小提示：
相邻两个 之间是一段一个或两个 组成的块
Between consecutive s there is a block of one or two s
大提示：
若有 个零，则有 个大小为 或 的块，其大小和为 。
With zeros there are blocks of size or whose sizes sum to
解答：
没有两个 相邻，所以 之间由若干 的块隔开，每块大小为 或 （不能为 ）。若有 个零，则有 个这样的块，合计 个一。
大小为 的块数为 ，它必须满足 ，即 。
对 求和 得到 。
所以正确答案是 C。
No two s are adjacent, so the s are separated by blocks of s, each of size or (never ). If there are zeros, there are such blocks summing to ones.
The number of size- blocks is which must satisfy i.e.
Summing over gives
Thus, C is the correct answer.
24.
令 。设 表示复平面中所有形如 的点，其中 ，，且 。 的面积是多少？
Let Let denote all points in the complex plane of the form where and What is the area of
小提示：
是沿 三个方向的线段的闵可夫斯基和
is the Minkowski sum of three segments along
大提示：
带状多边形的面积等于所有生成向量对的 。
A zonogon’s area equals over the generating vectors
解答：
当 在 中变化时，集合 是沿向量 ，， 的三个单位线段的闵可夫斯基和。
这是一个带状多边形，其面积等于所有向量对的叉积大小之和。每一对都给出 。
因此面积为 。
所以正确答案是 C。
As range over the set is the Minkowski sum of the three unit segments along
This is a zonogon whose area is the sum of the cross-product magnitudes over pairs. Each pair gives
Therefore the area is
Thus, C is the correct answer.
25.
设 是一个凸四边形，且 、。假设 ，，和 的重心构成一个等边三角形的顶点。 的面积最大可能值是多少？
Let be a convex quadrilateral with and Suppose that the centroids of and form the vertices of an equilateral triangle. What is the maximum possible value of the area of
小提示：
三个重心之间的差向量为 。
The differences of the three centroids are
大提示：
因此 是边长为 的等边三角形；用 表示面积并最大化
So is equilateral with side write the area using and maximize
解答：
三个重心分别为 ，，。它们两两之差为 ，所以重心三角形为等边三角形迫使 ；也就是说， 是边长为 的等边三角形。
沿 分割，其中 。由余弦定理，，所以
表达式 的最大值为 ，所以最大面积是 。当 时取等号；构造具有这个角的 ，并在 另一侧作等边 ，可以得到凸四边形，所以最大值能够达到。
所以 C 是正确答案。
The centroids are Their pairwise differences are so an equilateral centroid triangle forces that is, is equilateral with side
Splitting along where By the Law of Cosines so
The expression has maximum so the greatest area is Equality occurs at constructing with that angle and placing equilateral on the opposite side of produces a convex quadrilateral, so the maximum is attainable.
Thus, C is the correct answer.