2006A AMC 10 真题（P版）

五个三明治花费 \(5 \cdot 3 = 15\) 美元，八杯汽水花费 \(8 \cdot 2 = 16\) 美元。合计为 \(15 + 16 = 31\) 美元。

所以正确答案是 A。

English

Five sandwiches cost \(5 \cdot 3 = 15\) dollars and eight sodas cost \(8 \cdot 2 = 16\) dollars. Together they cost \(15 + 16 = 31\) dollars.

Thus, the correct answer is A.

内层运算给出 \(h \otimes h = h^3 - h\)。因此 \[ \begin{aligned} h \otimes (h^3 - h) &= h^3 - (h^3 - h) \\ &= h \end{aligned}\text{。}\]

所以正确答案是 C。

English

The inner operation gives \(h \otimes h = h^3 - h.\) Then \[ \begin{aligned} h \otimes (h^3 - h) &= h^3 - (h^3 - h) \\ &= h. \end{aligned} \]

Thus, the correct answer is C.

因为年龄比为 \(3 : 5\)，且 Alice 是 \(30\) 岁，所以 Mary 的年龄为 \(\dfrac{3}{5} \cdot 30 = 18\) 岁。

所以正确答案是 B。

English

Since the ratio is \(3 : 5\) and Alice is \(30,\) Mary is \(\dfrac{3}{5} \cdot 30 = 18\) years old.

Thus, the correct answer is B.

分钟从 \(00\) 到 \(59\)，所以分钟的最大数字和为 \(5 + 9 = 14\)，出现在 \(59\) 分。

对小时而言，单个数字 \(9\) 比 \(12\) 的 \(1 + 2 = 3\) 更大。最大总和为 \(9 + 14 = 23\)，出现在 \(9\!:\!59\)。

所以正确答案是 E。

English

The minutes run from \(00\) to \(59,\) so the largest digit sum for the minutes is \(5 + 9 = 14,\) at \(59\) minutes.

For the hour, the single digit \(9\) beats \(1 + 2 = 3\) from \(12.\) The largest total is \(9 + 14 = 23,\) occurring at \(9\!:\!59.\)

Thus, the correct answer is E.

披萨共有 \(8\) 块，所以每块披萨的普通价格是 \(\$1\)。Dave 吃了 \(5\) 块，并支付凤尾鱼的额外 \(\$2\)，总共 \(5 + 2 = 7\) 美元。

Doug 吃了 \(3\) 块，付 \(3\) 美元。因此 Dave 多付 \(7 - 3 = 4\) 美元。

所以正确答案是 D。

English

Each of the \(8\) slices costs \(\$1.\) Dave ate \(5\) slices and also pays the extra \(\$2\) for the anchovies, for a total of \(5 + 2 = 7\) dollars.

Doug ate \(3\) slices, paying \(3\) dollars. So Dave paid \(7 - 3 = 4\) dollars more.

Thus, the correct answer is D.

对等式两边取七次方根得 \((7x)^2 = 14x\)，所以 \(49x^2 = 14x\)。因为 \(x \neq 0\)，除以 \(x\) 得 \(49x = 14\)，于是 \(x = \dfrac{2}{7}\)。

所以正确答案是 B。

English

Taking the seventh root of both sides gives \((7x)^2 = 14x,\) so \(49x^2 = 14x.\) Since \(x \neq 0,\) divide by \(x\) to get \(49x = 14,\) hence \(x = \dfrac{2}{7}.\)

Thus, the correct answer is B.

长方形的面积是 \(8 \cdot 18 = 144\)，所以拼成的正方形边长为 \(\sqrt{144} = 12\)。

设 \(E\) 和 \(F\) 分别为阶梯形切口的上端点和下端点。为了使两个全等部分拼成正方形，三个水平线段 \(DE\)、中间长为 \(y\) 的线段和 \(FB\) 必须相等。它们合起来横跨长方形的宽，所以 \(DE+y+FB=18\) 给出 \(3y=18\)，从而 \(y=6\)。的确，所得正方形的边长为 \(2y=12\)，与面积计算一致。

所以正确答案是 A。

English

The rectangle’s area is \(8 \cdot 18 = 144,\) so the square formed has side \(\sqrt{144} = 12.\)

Let \(E\) and \(F\) be the upper and lower endpoints of the stair-step cut. For the two congruent pieces to fit into a square, the three horizontal segments \(DE,\) the middle segment of length \(y,\) and \(FB\) must be equal. Together they span the rectangle’s width, so \(DE+y+FB=18\) gives \(3y=18\) and \(y=6.\) Indeed, the resulting square has side \(2y=12,\) agreeing with the area calculation.

Thus, the correct answer is A.

代入两个点得到 \(3 = 4 + 2b + c\) 和 \(3 = 16 + 4b + c\)。相减得 \(0 = 12 + 2b\)，所以 \(b = -6\)。

因此 \(c = 3 - 4 - 2(-6) = 11\)。

所以正确答案是 E。

English

Substituting the points gives \(3 = 4 + 2b + c\) and \(3 = 16 + 4b + c.\) Subtracting yields \(0 = 12 + 2b,\) so \(b = -6.\)

Then \(c = 3 - 4 - 2(-6) = 11.\)

Thus, the correct answer is E.

\(n\) 个连续整数的和等于 \(n\) 乘以它们的中位数。和为 \(15\) 时，\(n = 2\) 给出 \(7 + 8\)，\(n = 3\) 给出 \(4 + 5 + 6\)，\(n = 5\) 给出 \(1 + 2 + 3 + 4 + 5\)。

\(4\) 个连续整数不行，因为它们的和为偶数；\(6\) 个或更多连续正整数的和已经超过 \(15\)。共有 \(3\) 组。

所以正确答案是 C。

English

The sum of \(n\) consecutive integers equals \(n\) times their median. For a sum of \(15\): \(n = 2\) gives \(7 + 8,\) \(n = 3\) gives \(4 + 5 + 6,\) and \(n = 5\) gives \(1 + 2 + 3 + 4 + 5.\)

No set of \(4\) works (their sum is even), and \(6\) or more consecutive positive integers already exceed \(15.\) There are \(3\) such sets.

Thus, the correct answer is C.

令 \(k = \sqrt{120 - \sqrt{x}}\)。因为 \(\sqrt{x} \ge 0\)，需要 \(0 \le k \le \sqrt{120}\)，所以 \(k \in \{0, 1, \ldots, 10\}\)，共有 \(11\) 个可能。

每个 \(k\) 给出 \(\sqrt{x} = 120 - k^2\)，且 \(120 - k^2\) 为正并严格递减，所以所得 \(x = (120 - k^2)^2\) 互不相同。

所以正确答案是 E。

English

Let \(k = \sqrt{120 - \sqrt{x}}.\) Since \(\sqrt{x} \ge 0,\) we need \(0 \le k \le \sqrt{120},\) so \(k \in \{0, 1, \ldots, 10\},\) giving \(11\) values.

Each \(k\) yields \(\sqrt{x} = 120 - k^2,\) and since \(120 - k^2\) is positive and strictly decreasing, the resulting values \(x = (120 - k^2)^2\) are distinct.

Thus, the correct answer is E.

展开得 \(x^2 + 2xy + y^2 = x^2 + y^2\)，化简为 \(2xy = 0\)，也就是 \(xy = 0\)。

这恰好在 \(x = 0\) 或 \(y = 0\) 时成立，即两条坐标轴，所以图像是两条直线。

所以正确答案是 C。

English

Expanding, \(x^2 + 2xy + y^2 = x^2 + y^2,\) which reduces to \(2xy = 0,\) i.e. \(xy = 0.\)

This holds exactly when \(x = 0\) or \(y = 0,\) the two coordinate axes, so the graph is two lines.

Thus, the correct answer is C.

安排 I 中，狗拴在一条边的中点，可以扫过半径为 \(8\) 的半圆，面积为 \(\frac12 \pi \cdot 8^2 = 32\pi\)。绳子正好够到角，所以不会绕过角。

安排 II 中，狗拴在离角 \(4\) 英尺处。它同样扫过 \(32\pi\) 的半圆；绳子到达角后还剩 \(4\) 英尺，可以扫过半径为 \(4\) 的四分之一圆，面积为 \(\frac14 \pi \cdot 4^2 = 4\pi\)。

因此 II 的面积为 \(36\pi\)，比 I 多 \(4\pi\)。

所以正确答案是 C。

English

In arrangement I the dog is tied at the middle of a side and sweeps a half-disk of radius \(8\): area \(\frac12 \pi \cdot 8^2 = 32\pi.\) The rope reaches exactly to the corners, so nothing wraps.

In arrangement II the dog is tied \(4\) feet from a corner. It sweeps the same \(32\pi\) half-disk, and after the rope reaches the corner, \(4\) feet remain to sweep a quarter-disk of radius \(4\): \(\frac14 \pi \cdot 4^2 = 4\pi.\)

So II gives \(36\pi,\) exceeding I by \(4\pi.\)

Thus, the correct answer is C.

玩家只有在第一次掷到偶数（概率 \(\frac12\)）且第二次与第一次相同（概率 \(\frac16\)）时获胜，所以获胜概率为 \(\frac12 \cdot \frac16 = \frac{1}{12}\)。

公平游戏满足 \(\frac{1}{12} x = 5\)，所以 \(x = 60\)。

所以正确答案是 D。

English

The player wins only if the first roll is even (probability \(\frac12\)) and the second roll matches it (probability \(\frac16\)), so the win probability is \(\frac12 \cdot \frac16 = \frac{1}{12}.\)

For a fair game, \(\frac{1}{12} x = 5,\) so \(x = 60.\)

Thus, the correct answer is D.

最上面的圆环贡献完整外径 \(20\) 厘米。因为圆环厚 \(1\) 厘米，每个圆环会比上方圆环的顶部低 \(2\) 厘米，所以每个较低圆环增加的是它的外径减去 \(2\)。

外径依次为 \(20, 19, \ldots, 3\)，所以下方各环增加的距离为 \(17, 16, \ldots, 1\)。总距离为 \[ \begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173 \end{gathered}\text{。}\]

所以正确答案是 B。

English

The top ring contributes its full outside diameter, \(20\) cm. Because the rings are \(1\) cm thick, each ring hangs \(2\) cm below the top of the ring above it, so each lower ring adds its outside diameter minus \(2.\)

The outside diameters run \(20, 19, \ldots, 3,\) so the added distances are \(17, 16, \ldots, 1.\) The total is \[ \begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173. \end{gathered} \]

Thus, the correct answer is B.

Odell 一圈为 \(2\pi(50) = 100\pi\) 米，速度为每分钟 \(250\) 米，用时 \(\frac{100\pi}{250} = 0.4\pi\) 分钟。Kershaw 一圈为 \(2\pi(60) = 120\pi\) 米，速度为每分钟 \(300\) 米，也用时 \(\frac{120\pi}{300} = 0.4\pi\) 分钟。

两人的单圈用时相同。由于他们反向跑，在 \(t = \frac{k}{2}(0.4\pi)\) 时相遇，其中 \(k = 1, 2, \ldots\)。条件 \(t \le 30\) 给出 \(k \le \frac{60}{0.4\pi} = \frac{150}{\pi} \approx 47.7\)，所以他们共相遇 \(47\) 次。

所以正确答案是 D。

English

Odell’s lap is \(2\pi(50) = 100\pi\) m at \(250\) m/min, taking \(\frac{100\pi}{250} = 0.4\pi\) min. Kershaw’s lap is \(2\pi(60) = 120\pi\) m at \(300\) m/min, also \(\frac{120\pi}{300} = 0.4\pi\) min.

Their periods are equal. Running in opposite directions, they meet at times \(t = \frac{k}{2}(0.4\pi)\) for \(k = 1, 2, \ldots\) Requiring \(t \le 30\) gives \(k \le \frac{60}{0.4\pi} = \frac{150}{\pi} \approx 47.7,\) so they pass \(47\) times.

Thus, the correct answer is D.

设 \(O, O'\) 为小圆和大圆的圆心，设 \(D\) 为小圆与 \(AC\) 的切点。沿 \(AC\) 截出的直角三角形相似，所以 \(\frac{AO}{1} = \frac{AO + 3}{2}\)，得到 \(AO = 3\) 且 \(AO' = 6\)。

切线长 \(AD = \sqrt{AO^2 - 1^2}\) \(= \sqrt{3^2 - 1^2}\) \(= 2\sqrt2\)。设 \(F\) 为 \(BC\) 的中点，则 \(AF = AO' + 2 = 8\)。

因为 \(\triangle ADO \sim \triangle AFC\)，有 \(\frac{FC}{1} = \frac{AF}{2\sqrt2} = \frac{8}{2\sqrt2} = 2\sqrt2\)。因此 \(BC = 4\sqrt2\)，面积为 \(\frac12 \cdot BC \cdot AF\) \(= \frac12 \cdot 4\sqrt2 \cdot 8\) \(= 16\sqrt2\)。

所以正确答案是 D。

English

Let \(O, O'\) be the centers of the small and large circles, and let \(D\) be the point where the small circle touches \(AC.\) The right triangles cut off along \(AC\) are similar, so \(\frac{AO}{1} = \frac{AO + 3}{2},\) giving \(AO = 3\) and \(AO' = 6.\)

The tangent length is \(AD = \sqrt{AO^2 - 1^2}\) \(= \sqrt{3^2 - 1^2}\) \(= 2\sqrt2.\) Let \(F\) be the midpoint of \(BC\); then \(AF = AO' + 2 = 8.\)

Since \(\triangle ADO \sim \triangle AFC,\) we get \(\frac{FC}{1} = \frac{AF}{2\sqrt2} = \frac{8}{2\sqrt2} = 2\sqrt2.\) Thus \(BC = 4\sqrt2,\) and the area is \(\frac12 \cdot BC \cdot AF\) \(= \frac12 \cdot 4\sqrt2 \cdot 8\) \(= 16\sqrt2.\)

Thus, the correct answer is D.

设 \(A = (0, 0)\)、\(D = (3, 0)\)、\(H = (0, 2)\)，则 \(B = (1, 0)\)、\(C = (2, 0)\)、\(G = (1, 2)\)、\(F = (2, 2)\)、\(E = (3, 2)\)。

图中线段交于 \(W = (1.5, 1.5)\)，\(X = (1, 1)\)，\(Y = (1.5, 0.5)\)，\(Z = (2, 1)\)。这些点形成一个正方形，其互相垂直的对角线 \(WY\) 和 \(XZ\) 的长度都为 \(1\)。

面积为 \(\frac12 \cdot 1 \cdot 1 = \frac12\)。

所以正确答案是 A。

English

Set \(A = (0, 0),\) \(D = (3, 0),\) \(H = (0, 2),\) so \(B = (1, 0),\) \(C = (2, 0),\) \(G = (1, 2),\) \(F = (2, 2),\) and \(E = (3, 2).\)

The drawn segments meet at \(W = (1.5, 1.5),\) \(X = (1, 1),\) \(Y = (1.5, 0.5),\) and \(Z = (2, 1).\) These form a square whose perpendicular diagonals \(WY\) and \(XZ\) each have length \(1.\)

Its area is \(\frac12 \cdot 1 \cdot 1 = \frac12.\)

Thus, the correct answer is A.

因为两个字母必须相邻，把它们看作一个整体。车牌于是由 \(4\) 个数字加上这个字母整体组成，共 \(5\) 个对象，而字母整体可以占 \(5\) 个位置。

数字有 \(10^4\) 种选择，两个字母有 \(26^2\) 种选择，所以总数为 \(5 \cdot 10^4 \cdot 26^2\)。

所以正确答案是 C。

English

Since the two letters must be adjacent, treat them as one block. A plate is then \(4\) digits plus this block—\(5\) objects—and the block can occupy \(5\) positions.

There are \(10^4\) choices for the digits and \(26^2\) for the two letters, so the total is \(5 \cdot 10^4 \cdot 26^2.\)

Thus, the correct answer is C.

设三个角为 \(n - d\)，\(n\)，\(n + d\)。它们的和为 \(3n = 180\)，所以 \(n = 60\)。

角度是互不相同的正整数，所以 \(d \ge 1\)，且 \(n - d \gt 0\) 迫使 \(d \lt 60\)。因此 \(d \in \{1, 2, \ldots, 59\}\)，给出 \(59\) 个互不相似的三角形。

所以正确答案是 C。

English

Let the angles be \(n - d,\) \(n,\) \(n + d.\) Their sum is \(3n = 180,\) so \(n = 60.\)

The measures are distinct positive integers, so \(d \ge 1,\) and \(n - d \gt 0\) forces \(d \lt 60.\) Thus \(d \in \{1, 2, \ldots, 59\},\) giving \(59\) non-similar triangles.

Thus, the correct answer is C.

按模 \(5\) 的余数给整数分组。可能余数只有 \(5\) 种，但有 \(6\) 个整数，所以根据鸽巢原理，必有两个整数余数相同。

它们的差就是 \(5\) 的倍数。这一定发生，所以概率为 \(1\)。

所以正确答案是 E。

English

Group the integers by their remainder modulo \(5.\) There are only \(5\) possible remainders but \(6\) integers, so by the Pigeonhole Principle two share a remainder.

Their difference is then a multiple of \(5.\) This always happens, so the probability is \(1.\)

Thus, the correct answer is E.

四位正整数共有 \(9000\) 个。若避开 \(2\) 和 \(3\)，首位可从 \(\{1, 4, 5, 6, 7, 8, 9\}\) 中选择，有 \(7\) 种；其余每位可从 \(\{0, 1, 4, 5, 6, 7, 8, 9\}\) 中选择，有 \(8\) 种，共有 \(7 \cdot 8^3 = 3584\) 个。

因此至少有一位是 \(2\) 或 \(3\) 的有 \(9000 - 3584 = 5416\) 个。

所以正确答案是 E。

English

There are \(9000\) four-digit integers. For those avoiding \(2\) and \(3,\) the leading digit is one of \(\{1, 4, 5, 6, 7, 8, 9\}\) (\(7\) choices) and each remaining digit is one of \(\{0, 1, 4, 5, 6, 7, 8, 9\}\) (\(8\) choices): \(7 \cdot 8^3 = 3584.\)

So \(9000 - 3584 = 5416\) have at least one \(2\) or \(3.\)

Thus, the correct answer is E.

可结清的债务可写为 \(D = 300p + 210g\)，其中 \(p\)、\(g\) 为整数，负值表示收到找零。因为 \(D = 30(10p + 7g)\)，且 \(\gcd(10, 7) = 1\)，所以 \(10p + 7g\) 可以是任意整数，故 \(D\) 可以是任意 \(30\) 的倍数。

最小的正数是 \(30\)，因为 \(30 = 300(-2) + 210(3)\)，也就是交出 \(3\) 只山羊并找回 \(2\) 只猪。

所以正确答案是 C。

English

A resolvable debt is \(D = 300p + 210g\) for integers \(p,\) \(g,\) where a negative value means change received. Since \(D = 30(10p + 7g)\) and \(\gcd(10, 7) = 1,\) the value \(10p + 7g\) can be any integer, so \(D\) is any multiple of \(30.\)

The smallest positive one is \(30,\) achieved by \(30 = 300(-2) + 210(3)\) (give \(3\) goats, receive \(2\) pigs).

Thus, the correct answer is C.

因为 \(AC \perp CD\)，所以 \(CE = \sqrt{AE^2 - AC^2}\) \(= \sqrt{25 - 9}\) \(= 4\)。

由于 \(\triangle ACE \sim \triangle BDE\)，有 \(\frac{DE}{CE} = \frac{BD}{AC}\)，所以 \(DE = 4 \cdot \frac{8}{3} = \frac{32}{3}\)。

因此 \(CD = CE + DE\) \(= 4 + \frac{32}{3}\) \(= \frac{44}{3}\)。

所以正确答案是 B。

English

Since \(AC \perp CD,\) we have \(CE = \sqrt{AE^2 - AC^2}\) \(= \sqrt{25 - 9}\) \(= 4.\)

Because \(\triangle ACE \sim \triangle BDE,\) \(\frac{DE}{CE} = \frac{BD}{AC},\) so \(DE = 4 \cdot \frac{8}{3} = \frac{32}{3}.\)

Then \(CD = CE + DE\) \(= 4 + \frac{32}{3}\) \(= \frac{44}{3}.\)

Thus, the correct answer is B.

六个面中心形成一个正八面体，可看成两个全等的正方形棱锥共用底面。相邻面中心距离为 \(\frac{\sqrt2}{2}\)，所以正方形底面面积为 \(\left(\frac{\sqrt2}{2}\right)^2 = \frac12\)。

每个棱锥高为 \(\frac12\)，体积为 \(\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}\)。八面体体积为 \(2 \cdot \frac{1}{12} = \frac16\)。

所以正确答案是 B。

English

The six face centers form a regular octahedron, viewed as two congruent square pyramids sharing a base. Adjacent face centers are \(\frac{\sqrt2}{2}\) apart, so the square base has area \(\left(\frac{\sqrt2}{2}\right)^2 = \frac12.\)

Each pyramid has height \(\frac12,\) so its volume is \(\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}.\) The octahedron has volume \(2 \cdot \frac{1}{12} = \frac16.\)

Thus, the correct answer is B.

经过 \(7\) 次移动共有 \(3^7 = 2187\) 条等可能路径。成功路径会恰好访问每个顶点一次。

用二进制三元组标记立方体的顶点，使相邻顶点恰好有一个坐标不同。从起点出发，第一步有 \(3\) 种选择；若虫子不返回起点，第二步有 \(2\) 种选择。由对称性，可将前两步固定为 \(000\to100\to110\)。

成功的后续路线恰好为 \[ \begin{aligned} 110&\to111\to101\\ &\to001\to011\to010 \end{aligned}\text{、}\]\[ \begin{aligned} 110&\to010\to011\\ &\to001\to101\to111 \end{aligned} \] 和 \[ \begin{aligned} 110&\to010\to011\\ &\to111\to101\to001 \end{aligned}\text{。}\]因此，每一组允许的前两步都有 \(3\) 条成功的后续路线，共有 \(3\cdot2\cdot3=18\) 条成功路径。

概率为 \(\frac{18}{2187} = \frac{2}{243}\)。

所以正确答案是 C。

English

After \(7\) moves there are \(3^7 = 2187\) equally likely walks. A successful walk visits every vertex exactly once.

Label the cube’s vertices by binary triples, with adjacent vertices differing in one coordinate. There are \(3\) choices for the first move and \(2\) for the second move if the bug is not to return to its starting point. By symmetry, fix these first moves as \(000\to100\to110.\)

The successful continuations are exactly \[ \begin{aligned} 110&\to111\to101\\ &\to001\to011\to010, \end{aligned} \] \[ \begin{aligned} 110&\to010\to011\\ &\to001\to101\to111, \end{aligned} \] and \[ \begin{aligned} 110&\to010\to011\\ &\to111\to101\to001. \end{aligned} \] Thus each allowed pair of first moves has \(3\) successful continuations, giving \(3\cdot2\cdot3=18\) successful walks.

The probability is \(\frac{18}{2187} = \frac{2}{243}.\)

Thus, the correct answer is C.