2003A AMC 12 真题（P版）

1 答案 D

第 \(k\) 个正偶数是 \(2k\)，第 \(k\) 个正奇数是 \(2k-1\)，二者相差 \(1\)。

把 \(2003\) 对的差相加，得到 \(2003 \cdot 1 = 2003\)。

所以正确答案是 D。

English

The \(k\)th even counting number is \(2k\) and the \(k\)th odd counting number is \(2k-1,\) which differ by \(1.\)

Summing this difference over all \(2003\) pairs gives \(2003 \cdot 1 = 2003.\)

Thus, the correct answer is D.

2 答案 B

一件 T 恤价格为 \(4+5=9\) 美元。

每位成员需要两双袜子和两件 T 恤，费用为 \(2(4)+2(9)=26\) 美元。

因此成员人数为 \(2366 \div 26 = 91\)。

所以正确答案是 B。

English

A T-shirt costs \(4+5=9\) dollars.

Each member needs two pairs of socks and two shirts, costing \(2(4)+2(9)=26\) dollars.

The number of members is \(2366 \div 26 = 91.\)

Thus, the correct answer is B.

3 答案 D

原体积为 \(15 \cdot 10 \cdot 8 = 1200\)。

切去八个角立方体，每个体积为 \(3^3 = 27\)，总共切去 \(8 \cdot 27 = 216\)。

切去的比例为 \(\dfrac{216}{1200} = 0.18\)，即 \(18\%\)。

所以正确答案是 D。

English

The original volume is \(15 \cdot 10 \cdot 8 = 1200.\)

Eight corner cubes are removed, each of volume \(3^3 = 27,\) totaling \(8 \cdot 27 = 216.\)

The fraction removed is \(\dfrac{216}{1200} = 0.18,\) which is \(18\%.\)

Thus, the correct answer is D.

4 答案 A

Mary 总共走 \(2\) km，用时 \(30+10=40\) 分钟，即 \(\dfrac23\) 小时。

平均速度为 \(2 \div \dfrac23 = 3\) km/hr。

所以正确答案是 A。

English

Mary walks a total of \(2\) km in \(30+10=40\) minutes, which is \(\dfrac23\) hour.

Her average speed is \(2 \div \dfrac23 = 3\) km/hr.

Thus, the correct answer is A.

5 答案 E

\(\overline{AMC10}=100\cdot\overline{AMC}+10\)，且 \(\overline{AMC12}=100\cdot\overline{AMC}+12\)。

它们的和为 \(200\cdot\overline{AMC}+22=123422\)，所以 \(\overline{AMC}=617\)。

因此 \(A+M+C=6+1+7=14\)。

所以正确答案是 E。

English

Write \(\overline{AMC10}=100\cdot\overline{AMC}+10\) and \(\overline{AMC12}=100\cdot\overline{AMC}+12.\)

Their sum is \(200\cdot\overline{AMC}+22=123422,\) so \(\overline{AMC}=617.\)

Then \(A+M+C=6+1+7=14.\)

Thus, the correct answer is E.

6 答案 C

因为 \(x\heartsuit 0=|x-0|=|x|\)，选项 C 断言对所有 \(x\) 都有 \(|x|=x\)，但当 \(x\lt0\) 时并不成立。例如，\((-1)\heartsuit 0 = 1 \neq -1\)。

其他命题都直接来自绝对值的性质。

所以正确答案是 C。

English

Since \(x\heartsuit 0=|x-0|=|x|,\) statement (C) claims \(|x|=x\) for all \(x,\) which fails when \(x\lt0.\) For example, \((-1)\heartsuit 0 = 1 \neq -1.\)

Every other statement follows directly from properties of the absolute value.

Thus, the correct answer is C.

7 答案 B

设边长为 \(a\le b\le c\)，且 \(a+b+c=7\)。三角形不等式要求 \(c\lt a+b\)，所以 \(c\lt3.5\)，从而 \(c=3\)。

此时 \(a+b=4\)，且 \(a\le b\le3\) 得到 \(1\text{-}3\text{-}3\) 和 \(2\text{-}2\text{-}3\) 两种三角形。

所以正确答案是 B。

English

Let the sides be \(a\le b\le c\) with \(a+b+c=7.\) The triangle inequality requires \(c\lt a+b,\) so \(c\lt3.5,\) forcing \(c=3.\)

Then \(a+b=4\) with \(a\le b\le3,\) giving the triangles \(1\text{-}3\text{-}3\) and \(2\text{-}2\text{-}3.\)

Thus, the correct answer is B.

8 答案 E

\(60\) 的 \(12\) 个正因数为 \(1,2,3,4,5,6\)，以及 \(10,12,15,20,30,60\)。

其中小于 \(7\) 的有六个：\(1,2,3,4,5,6\)，所以概率为 \(\dfrac{6}{12}=\dfrac12\)。

所以正确答案是 E。

English

The number \(60\) has \(12\) positive factors: \(1,2,3,4,5,6,\) \(10,12,15,20,30,60.\)

Six of them are less than \(7,\) namely \(1,2,3,4,5,6,\) so the probability is \(\dfrac{6}{12}=\dfrac12.\)

Thus, the correct answer is E.

9 答案 D

关于 \(y=x\) 反射得到 \((3,2)\)，再关于坐标轴反射得到所有 \((\pm2,\pm3)\) 和 \((\pm3,\pm2)\)。

共有 \(8\) 个点，并且这个集合已经关于原点、两条坐标轴和 \(y=x\) 对称。

所以正确答案是 D。

English

Reflecting across \(y=x\) gives \((3,2),\) and reflecting across the axes gives all points \((\pm2,\pm3)\) and \((\pm3,\pm2).\)

There are \(8\) such points, and this set is already symmetric about the origin, both axes, and \(y=x.\)

Thus, the correct answer is D.

10 答案 D

三人的份额分别是 \(\dfrac12,\dfrac13,\dfrac16\)。

每个人都以为自己是第一个到的，所以在各自到达时，Al 留下当时糖果的 \(\dfrac12\)，Bert 留下 \(\dfrac23\)，Carl 留下 \(\dfrac56\)。

无论三人按什么顺序到达，无人领取的比例都是 \(\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18}\)。

所以正确答案是 D。

English

The shares are \(\dfrac12,\dfrac13,\dfrac16\) of the pile.

Each person assumes he is first, so Al leaves \(\dfrac12,\) Bert leaves \(\dfrac23,\) and Carl leaves \(\dfrac56\) of the candy present when he arrives.

The unclaimed fraction is \(\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18},\) regardless of the order.

Thus, the correct answer is D.

11 答案 C

令共同周长为 \(12\)，则正方形边长为 \(3\)，三角形边长为 \(4\)。

正方形的外接圆半径为 \(\dfrac{3\sqrt2}{2}\)，所以 \(A=\pi\left(\dfrac{3\sqrt2}{2}\right)^2=\dfrac{9\pi}{2}\)。

三角形的外接圆半径为 \(\dfrac{4}{\sqrt3}\)，所以 \(B=\pi\left(\dfrac{4}{\sqrt3}\right)^2=\dfrac{16\pi}{3}\)。

于是 \(\dfrac{A}{B}=\dfrac{\frac{9}{2}}{\frac{16}{3}}=\dfrac{27}{32}\)。

所以正确答案是 C。

English

Let the common perimeter be \(12,\) so the square has side \(3\) and the triangle has side \(4.\)

The square’s circumradius is \(\dfrac{3\sqrt2}{2},\) so \(A=\pi\left(\dfrac{3\sqrt2}{2}\right)^2=\dfrac{9\pi}{2}.\)

The triangle’s circumradius is \(\dfrac{4}{\sqrt3},\) so \(B=\pi\left(\dfrac{4}{\sqrt3}\right)^2=\dfrac{16\pi}{3}.\)

Then \(\dfrac{A}{B}=\dfrac{\frac{9}{2}}{\frac{16}{3}}=\dfrac{27}{32}.\)

Thus, the correct answer is C.

12 答案 E

在 \(3,4,5,6\) 中，\(4\) 只能整除 \(4\)，\(5\) 只能整除 \(5\)，所以两端必须是 \(R4,B4,\dots,B5,R5\)。

又因为 \(2\) 只能整除 \(4\) 和 \(6\)，所以下一张是 \(R2,B6\)；而 \(3\) 只能整除 \(3\) 和 \(6\)，完整顺序为 \(R4\)，\(B4\)，\(R2\)，\(B6\)，\(R3\)，\(B3\)，\(R1\)，\(B5\)，\(R5\)。

中间三张是 \(6,3,3\)，和为 \(12\)。

所以正确答案是 E。

English

Since \(4\) divides only \(4\) and \(5\) divides only \(5\) among \(3,4,5,6,\) the two ends must be \(R4,B4,\dots,B5,R5.\)

Because \(2\) divides only \(4\) and \(6,\) the next card is \(R2,B6,\) and since \(3\) divides only \(3\) and \(6,\) the full stack is \(R4,\) \(B4,\) \(R2,\) \(B6,\) \(R3,\) \(B3,\) \(R1,\) \(B5,\) \(R5.\)

The middle three cards are \(6,3,3,\) which sum to \(12.\)

Thus, the correct answer is E.

13 答案 E

缺少一个面的立方体有 \(5\) 个面，所以第五个正方形必须折到四正方形拼块尚未占据的面上。追踪每个连接位置折到的面可知，位置 \(1\)，\(2\)，和 \(3\) 会折到已占据的面并与之重叠。位置 \(4\)，\(5\)，\(6\)，\(7\)，\(8\)，和 \(9\) 都会折到未占据的面。因此九个位置中有 \(6\) 个可行。

所以正确答案是 E。

English

A cube missing one face has \(5\) faces, so the fifth square must fold onto a face not already occupied by the four-square piece. Tracing the face reached by each attachment, positions \(1,\) \(2,\) and \(3\) fold onto a face already occupied and overlap it. Each of positions \(4,\) \(5,\) \(6,\) \(7,\) \(8,\) and \(9\) folds onto an unoccupied face. Therefore \(6\) of the nine positions work.

Thus, the correct answer is E.

14 答案 D

正方形 \(ABCD\) 边长为 \(4\)。由 \(90^\circ\) 旋转对称性可知，\(KLMN\) 也是正方形。

每个边长为 \(4\) 的等边三角形高为 \(2\sqrt3\)，所以对角线 \(KM=4+2(2\sqrt3)=4+4\sqrt3\)。

对角线为 \(d\) 的正方形面积是 \(\dfrac12 d^2\)，因此 \([KLMN]=\dfrac12(4+4\sqrt3)^2\) \(=32+16\sqrt3\)。

所以正确答案是 D。

English

The square \(ABCD\) has side \(4.\) By the \(90^\circ\) rotational symmetry, \(KLMN\) is also a square.

Each equilateral triangle on a side of length \(4\) has height \(2\sqrt3,\) so the diagonal \(KM=4+2(2\sqrt3)=4+4\sqrt3.\)

A square with diagonal \(d\) has area \(\dfrac12 d^2,\) so \([KLMN]=\dfrac12(4+4\sqrt3)^2\) \(=32+16\sqrt3.\)

Thus, the correct answer is D.

15 答案 C

小半圆的直径是大圆半径 \(1\) 中长为 \(1\) 的弦，所以它在圆心处张成 \(60^\circ\)。

由该弦和小圆弧围成的区域等于一个面积 \(\dfrac{\sqrt3}{4}\) 的等边三角形加上小半圆面积 \(\dfrac12\pi\left(\dfrac12\right)^2=\dfrac{\pi}{8}\)。

减去大圆的 \(60^\circ\) 扇形面积 \(\dfrac16\pi(1)^2=\dfrac{\pi}{6}\)，得月牙面积 \[\dfrac{\sqrt3}{4}+\dfrac{\pi}{8}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{4}-\dfrac{\pi}{24}\text{。}\]

所以正确答案是 C。

English

The small semicircle’s diameter is a chord of length \(1\) in the large circle of radius \(1,\) so it subtends a \(60^\circ\) angle at the center.

The region bounded by that chord and the small arc is an equilateral triangle of area \(\dfrac{\sqrt3}{4}\) topped by the small semicircle of area \(\dfrac12\pi\left(\dfrac12\right)^2=\dfrac{\pi}{8}.\)

Subtracting the \(60^\circ\) sector of the large circle, \(\dfrac16\pi(1)^2=\dfrac{\pi}{6},\) gives the lune area \[\dfrac{\sqrt3}{4}+\dfrac{\pi}{8}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{4}-\dfrac{\pi}{24}.\]

Thus, the correct answer is C.

16 答案 C

\(\triangle ABP\)、\(\triangle ACP\)、\(\triangle BCP\) 的底边都是等边三角形的边，长度相同，所以它们的面积分别与 \(P\) 到对应边的距离成正比。

由等边三角形的三重对称性，\(\triangle ABP\) 成为最大面积三角形的概率与另外两个相同，因此概率为 \(\dfrac13\)。

所以正确答案是 C。

English

The triangles \(\triangle ABP,\) \(\triangle ACP,\) and \(\triangle BCP\) have equal bases (the sides of the equilateral triangle), so their areas are proportional to the distances from \(P\) to those sides.

By the threefold symmetry of the equilateral triangle, \(\triangle ABP\) is the largest with the same probability as each of the other two, so that probability is \(\dfrac13.\)

Thus, the correct answer is C.

17 答案 B

取 \(D=(0,0)\)，\(C=(4,0)\)，\(A=(0,4)\)。以 \(M=(2,0)\) 为圆心的圆为 \((x-2)^2+y^2=4\)，以 \(A\) 为圆心的圆为 \(x^2+(y-4)^2=16\)。

解这两个方程，得到交点 \(P=\left(\dfrac{16}{5},\dfrac85\right)\)。

因为 \(\overline{AD}\) 在 \(y\)-轴上，点 \(P\) 到 \(\overline{AD}\) 的距离就是它的 \(x\)-坐标，即 \(\dfrac{16}{5}\)。

所以正确答案是 B。

English

Place \(D=(0,0),\) \(C=(4,0),\) and \(A=(0,4).\) The circle centered at \(M=(2,0)\) is \((x-2)^2+y^2=4,\) and the circle centered at \(A\) is \(x^2+(y-4)^2=16.\)

Solving these equations gives the intersection \(P=\left(\dfrac{16}{5},\dfrac85\right).\)

Since \(\overline{AD}\) lies on the \(y\)-axis, the distance from \(P\) to \(\overline{AD}\) is its \(x\)-coordinate, \(\dfrac{16}{5}.\)

Thus, the correct answer is B.

18 答案 B

\(n=100q+r=(q+r)+99q\)，且 \(99\) 能被 \(11\) 整除，所以 \(q+r\equiv n\pmod{11}\)。

因此 \((q+r)\) 是 \(11\) 的倍数，当且仅当 \(n\) 是 \(11\) 的倍数。

在 \(5\) 位数中，\(11\) 的倍数个数为 \(\left\lfloor\dfrac{99999}{11}\right\rfloor\) \(-\left\lfloor\dfrac{9999}{11}\right\rfloor\) \(=9090-909=8181\)。

所以正确答案是 B。

English

Since \(n=100q+r=(q+r)+99q\) and \(99\) is divisible by \(11,\) we have \(q+r\equiv n\pmod{11}.\)

So \((q+r)\) is a multiple of \(11\) exactly when \(n\) is a multiple of \(11.\)

Among the \(5\)-digit numbers, the count of multiples of \(11\) is \(\left\lfloor\dfrac{99999}{11}\right\rfloor\) \(-\left\lfloor\dfrac{9999}{11}\right\rfloor\) \(=9090-909=8181.\)

Thus, the correct answer is B.

19 答案 D

把抛物线写成顶点式 \(y=a(x-h)^2+k\)。关于 \(x\) 轴反射后，方程为 \(y=-a(x-h)^2-k\)。

向相反方向平移后，可写成 \(f(x)=a(x-h+5)^2+k\) 和 \(g(x)=-a(x-h-5)^2-k\)。

两式相加后平方项抵消，得到 \((f+g)(x)=20a(x-h)\)。因为 \(a\neq0\)，它是一条非水平直线。

所以正确答案是 D。

English

Write the parabola in vertex form \(y=a(x-h)^2+k.\) Its reflection about the \(x\)-axis is \(y=-a(x-h)^2-k.\)

Shifting in opposite directions gives \(f(x)=a(x-h+5)^2+k\) and \(g(x)=-a(x-h-5)^2-k.\)

Adding, the squared terms cancel and \((f+g)(x)=20a(x-h),\) which is a non-horizontal line since \(a\neq0.\)

Thus, the correct answer is D.

20 答案 A

设第一段有 \(k\) 个 B 和 \(5-k\) 个 C。剩下的 \(k\) 个 C 必须放在第二段（因为第三段没有 C），于是第二段有 \(5-k\) 个 A。

那么第三段含有剩下的 \(k\) 个 A 和 \(5-k\) 个 B。

对每个 \(k\)，第一段中的 \(k\) 个 B、第二段中的 \(k\) 个 C、第三段中的 \(k\) 个 A 的位置共有 \(\binom{5}{k}^3\) 种选法，所以总数为 \(\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3\)。

所以正确答案是 A。

English

Suppose the first block holds \(k\) B’s and \(5-k\) C’s. The remaining \(k\) C’s must go in the second block (since the third has no C’s), forcing \(5-k\) A’s there.

Then the third block contains the remaining \(k\) A’s and \(5-k\) B’s.

For each \(k,\) the \(k\) B’s in the first block, \(k\) C’s in the second, and \(k\) A’s in the third can be placed in \(\binom{5}{k}^3\) ways, so the total is \(\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3.\)

Thus, the correct answer is A.

21 答案 D

因为 \((0,0)\) 是截距，\(P(0)=e=0\)，所以 \(P(x)\) \(=x\left(x^4+ax^3+bx^2+cx+d\right)\)。

其余四个截距都是非零且互不相同的根，而 \(d\) 等于这四个非零根的乘积，因此不可能为零。

通过适当选择根，\(a,b,c\) 都可能为零，但 \(d\neq0\)。

所以正确答案是 D。

English

Since \((0,0)\) is an intercept, \(P(0)=e=0,\) so \(P(x)\) \(=x\left(x^4+ax^3+bx^2+cx+d\right).\)

The four remaining intercepts are nonzero and distinct, and \(d\) equals their product, which is therefore nonzero.

Any of \(a,b,c\) can be zero for suitable choices of those roots, but \(d\neq0.\)

Thus, the correct answer is D.

22 答案 C

两物体相距 \(12\) 步，所以只能在每个物体各走 \(6\) 步后，在反对角线 \(x+y=6\) 上相遇。

把 \(A\) 的六步路径与 \(B\) 反向看的六步路径配对，相遇路径对与从 \((0,0)\) 到 \((5,7)\) 的单调路径一一对应，共 \(\binom{12}{5}\) 条。

概率为 \(\dfrac{\binom{12}{5}}{2^{12}}=\dfrac{792}{4096}\approx0.19\)，最接近 \(0.20\)。

所以正确答案是 C。

English

The objects are \(12\) steps apart, so they can only meet after each takes \(6\) steps, on the anti-diagonal \(x+y=6.\)

Pairing \(A\)’s six-step path with \(B\)’s reversed six-step path matches meeting pairs one-to-one with the \(\binom{12}{5}\) monotone walks from \((0,0)\) to \((5,7).\)

The probability is \(\dfrac{\binom{12}{5}}{2^{12}}=\dfrac{792}{4096}\approx0.19,\) which is closest to \(0.20.\)

Thus, the correct answer is C.

23 答案 B

该乘积为 \[1! \cdot 2! \cdots 9! = 2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}\text{。}\]

一个平方因数形如 \(2^{2a}3^{2b}5^{2c}7^{2d}\)，其中 \(0\le a\le15\)，\(0\le b\le6\)，\(0\le c\le2\)，且 \(0\le d\le1\)。

选择数为 \(16\cdot7\cdot3\cdot2=672\)。

所以正确答案是 B。

English

The product is \[1! \cdot 2! \cdots 9! = 2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}.\]

A perfect-square divisor has the form \(2^{2a}3^{2b}5^{2c}7^{2d}\) with \(0\le a\le15,\) \(0\le b\le6,\) \(0\le c\le2,\) and \(0\le d\le1.\)

The number of choices is \(16\cdot7\cdot3\cdot2=672.\)

Thus, the correct answer is B.

24 答案 B

展开得 \[ \begin{aligned} &\log_a\dfrac ab+\log_b\dfrac ba \\ &\quad {}=(1-\log_a b)+(1-\log_b a) \\ &\quad {}=2-\left(\log_a b+\log_b a\right) \end{aligned}\text{。}\]

令 \(c=\log_a b\gt0\)。由 AM-GM，\(c+\dfrac1c\ge2\)，所以原式至多为 \(0\)。

当 \(c=1\)，即 \(a=b\) 时取等号，因此最大值为 \(0\)。

所以正确答案是 B。

English

Expand: \[ \begin{aligned} &\log_a\dfrac ab+\log_b\dfrac ba \\ &\quad {}=(1-\log_a b)+(1-\log_b a) \\ &\quad {}=2-\left(\log_a b+\log_b a\right). \end{aligned} \]

Let \(c=\log_a b\gt0.\) Since \(c+\dfrac1c\ge2\) by AM-GM, the expression is at most \(0.\)

Equality holds when \(c=1,\) that is, when \(a=b,\) so the largest value is \(0.\)

Thus, the correct answer is B.

25 答案 C

若 \(a=0\) 则 \(f(x)=\sqrt{bx}\) 的定义域和值域都是 \([0,\infty)\)，所以 \(a=0\) 可行。

若 \(a\gt0\)，定义域为 \(({-}\infty,-b/a]\cup[0,\infty)\)，而值域为 \([0,\infty)\)，二者不相同，因此没有这样的 \(b\)。

若 \(a\lt0\)，定义域是 \([0,-\frac{b}{a}]\)，值域是 \(\left[0,\dfrac{b}{2\sqrt{-a}}\right]\)。令右端点相等，得 \(-\dfrac ba=\dfrac{b}{2\sqrt{-a}}\)，所以 \(2\sqrt{-a}=-a\)，从而 \(a=-4\)。

因此共有 \(2\) 个 \(a\) 值，所以正确答案是 C。

English

If \(a=0,\) then \(f(x)=\sqrt{bx}\) has domain and range both \([0,\infty),\) so \(a=0\) works.

If \(a\gt0,\) the domain is \(({-}\infty,-b/a]\cup[0,\infty),\) while the range is \([0,\infty),\) so no such \(b\) exists.

If \(a\lt0,\) the domain is \([0,-\frac{b}{a}]\) and the range is \(\left[0,\dfrac{b}{2\sqrt{-a}}\right].\) Equating the right endpoints gives \(-\dfrac ba=\dfrac{b}{2\sqrt{-a}},\) so \(2\sqrt{-a}=-a,\) giving \(a=-4.\)

Thus there are \(2\) values of \(a,\) and the correct answer is C.