1983 AMC 12 真题（P版）

第二个方程给出 \(x=16y\)。代入第一个方程，得到 \(8y=y^2\)，所以 \(y(y-8)=0\)。条件 \(x\ne0\) 排除 \(y=0\)，于是 \(y=8\)，且 \(x=128\)。

所以正确答案是 E。

English

The second equation gives \(x=16y.\) Substituting into the first gives \(8y=y^2,\) so \(y(y-8)=0.\) The condition \(x\ne0\) rules out \(y=0,\) leaving \(y=8\) and \(x=128.\)

Therefore, the correct answer is E.

与 \(P\) 相距 \(3\) 厘米的点构成一个以 \(P\) 为圆心的圆。两个不同的圆至多相交于两点，而且确实可能有两个交点。

所以正确答案是 B。

English

The points \(3\) cm from \(P\) form a circle centered at \(P.\) Two distinct circles intersect in at most two points, and two intersections are possible.

Therefore, the correct answer is B.

若 \(p\) 和 \(q\) 都是奇数，则 \(r=p+q\) 是大于 \(2\) 的偶数，因而不是素数。所以其中一个加数是唯一的偶素数 \(2\)。由于 \(p\lt q\)，这个加数就是 \(p\)。

所以正确答案是 A。

English

If both \(p\) and \(q\) were odd, then \(r=p+q\) would be even and greater than \(2,\) so it would not be prime. Thus one addend is the only even prime, \(2.\) Since \(p\lt q,\) that addend is \(p.\)

Therefore, the correct answer is A.

由平行方向和 \(60^\circ\) 角可知，在刚性运动意义下，各顶点可以取为 \[ \begin{aligned} A&=(0,0),\\ B&=(\tfrac{\sqrt3}{2},-\tfrac12),\\ C&=(\sqrt3,0),\\ D&=(\sqrt3,-1),\\ E&=(\tfrac{\sqrt3}{2},-\tfrac32),\\ F&=(0,-1) \end{aligned}\text{。}\]对这六个顶点应用鞋带公式，得到面积 \(\sqrt3\)。

所以正确答案是 D。

English

The parallel directions and \(60^\circ\) angles place the vertices, up to rigid motion, at \[ \begin{aligned} A&=(0,0),\\ B&=(\tfrac{\sqrt3}{2},-\tfrac12),\\ C&=(\sqrt3,0),\\ D&=(\sqrt3,-1),\\ E&=(\tfrac{\sqrt3}{2},-\tfrac32),\\ F&=(0,-1). \end{aligned} \] Applying the shoelace formula to these six vertices gives area \(\sqrt3.\)

Therefore, the correct answer is D.

由于 \(\sin A=\frac{2}{3}\)，所以 \(\cos A=\sqrt{1-\frac{4}{9}}=\frac{\sqrt5}{3}\)。又因 \(B=90^\circ-A\)，有 \(\tan B=\cot A\)，并且 \(\frac{\cos A}{\sin A}=\frac{\sqrt5}{2}\)。

所以正确答案是 D。

English

Since \(\sin A=\frac{2}{3},\) \(\cos A=\sqrt{1-\frac{4}{9}}=\frac{\sqrt5}{3}.\) Because \(B=90^\circ-A,\) we have \(\tan B=\cot A\) and \(\frac{\cos A}{\sin A}=\frac{\sqrt5}{2}.\)

Therefore, the correct answer is D.

最大指数来自 \(x^5\cdot x\cdot1=x^6\)。其余各项的指数都更小，而最小的可能指数为 \(5-1-3=1\)，所以乘积确实是一个多项式。它的次数为 \(6\)。

所以正确答案是 C。

English

The greatest exponent comes from \(x^5\cdot x\cdot1=x^6.\) Every other term has smaller exponent, and the smallest possible exponent is \(5-1-3=1,\) so the product is indeed a polynomial. Its degree is \(6.\)

Therefore, the correct answer is C.

若标价为 \(L\)，由佣金相等可得 \(0.1(L-10)=0.2(L-20)\)。两边乘以 \(10\) 并求解，得到 \(L-10=2L-40\)，所以 \(L=30\)。

所以正确答案是 B。

English

If the list price is \(L,\) equality of commissions gives \(0.1(L-10)=0.2(L-20).\) Multiplying by \(10\) and solving gives \(L-10=2L-40,\) so \(L=30.\)

Therefore, the correct answer is B.

有 \[ \begin{aligned} f(-x)&=\frac{-x+1}{-x-1}\\ &=\frac{x-1}{x+1}=\frac1{f(x)} \end{aligned}\text{。}\]限制条件 \(x^2\ne1\) 保证所写的所有量都有定义。

所以正确答案是 A。

English

We have \[ \begin{aligned} f(-x)&=\frac{-x+1}{-x-1}\\ &=\frac{x-1}{x+1}=\frac1{f(x)}. \end{aligned} \] The restriction \(x^2\ne1\) makes all displayed quantities defined.

Therefore, the correct answer is A.

取 \(11\) 名女性和 \(10\) 名男性，所有人的年龄总和为 \(11\cdot34+10\cdot32=694\)。再除以总人数 \(21\)，得到 \(\frac{694}{21}=33\frac1{21}\)。

所以正确答案是 D。

English

Using \(11\) women and \(10\) men, the total of all ages is \(11\cdot34+10\cdot32=694.\) Dividing by \(21\) people gives \(\frac{694}{21}=33\frac1{21}.\)

Therefore, the correct answer is D.

由于 \(E\) 位于以 \(AB\) 为直径的圆上，所以 \(\angle AEB=90^\circ\)。又因为 \(ABC\) 是等边三角形，所以 \(\angle ABE=60^\circ\)。因此 \(ABE\) 是一个斜边 \(AB=2\) 的 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形，所以 \(AE=\sqrt3\)。

所以正确答案是 D。

English

Since \(E\) is on the circle with diameter \(AB,\) \(\angle AEB=90^\circ.\) Also \(\angle ABE=60^\circ\) because \(ABC\) is equilateral. Thus \(ABE\) is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle with hypotenuse \(AB=2,\) so \(AE=\sqrt3.\)

Therefore, the correct answer is D.

令 \(u=x-y\)、\(v=y\)，正弦和角公式给出 \[ \begin{aligned} &\sin(x-y)\cos y\\ &\quad+\cos(x-y)\sin y\\ &=\sin((x-y)+y)=\sin x \end{aligned}\text{。}\]

所以正确答案是 B。

English

Taking \(u=x-y\) and \(v=y,\) the sine addition identity gives \[ \begin{aligned} &\sin(x-y)\cos y\\ &\quad+\cos(x-y)\sin y\\ &=\sin((x-y)+y)=\sin x. \end{aligned} \]

Therefore, the correct answer is B.

原方程给出 \(\log_3(\log_2x)=1\)，所以 \(\log_2x=3\)，且 \(x=8\)。因此 \(x^{-\frac{1}{2}}=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2}\)，不在选项 A 至 D 中。

所以正确答案是 E。

English

The equation gives \(\log_3(\log_2x)=1,\) so \(\log_2x=3\) and \(x=8.\) Hence \(x^{-\frac{1}{2}}=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2},\) which is not among choices A–D.

Therefore, the correct answer is E.

由三个乘积可得 \[ \begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a} \end{aligned}\text{。}\]将它们的和通分到分母 \(abc\)，得到 \[ \begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc} \end{aligned}\text{。}\]

所以正确答案是 E。

English

From the three products, \[ \begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a}. \end{aligned} \] Putting their sum over denominator \(abc\) gives \[ \begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc}. \end{aligned} \]

Therefore, the correct answer is E.

模 \(10\) 时，\[ 3^{1001}13^{1003}\equiv3^{2004}\equiv1\text{，}\]因为 \(3\) 的幂每 \(4\) 次循环一次。另外，\(7^{1002}\equiv7^2\equiv9\pmod {10}\)。因此乘积的个位数字为 \(9\)。

所以正确答案是 E。

English

Modulo \(10,\) \[ 3^{1001}13^{1003}\equiv3^{2004}\equiv1 \] because powers of \(3\) repeat every \(4.\) Also \(7^{1002}\equiv7^2\equiv9\pmod {10}.\) Thus the product has units digit \(9.\)

Therefore, the correct answer is E.

和为 \(6\) 的有序三元组为 \((1,2,3)\) 的六个排列和三元组 \((2,2,2)\)。它们的可能性相同，七个三元组中恰有一个含有三个 \(2\)。因此条件概率为 \(\frac{1}{7}\)。

所以正确答案是 C。

English

The ordered triples with sum \(6\) are the six permutations of \((1,2,3)\) and the triple \((2,2,2).\) They are equally likely, and exactly one of the seven has three \(2\)’s. The conditional probability is therefore \(\frac{1}{7}.\)

Therefore, the correct answer is C.

一位数提供 \(9\) 位，两位数提供 \(90\cdot2=180\) 位，共 \(189\) 位。因此所求数字是三位数部分中的第 \(1983-189=1794\) 位。由于 \(1794=598\cdot3\)，它是第 \(598\) 个三位数的末位数字；该三位数为 \(100+597=697\)。所以这位数字是 \(7\)。

所以正确答案是 D。

English

The one-digit integers contribute \(9\) digits and the two-digit integers contribute \(90\cdot2=180,\) for \(189\) total. Thus the desired digit is the \(1983-189=1794\)th digit among the three-digit integers. Since \(1794=598\cdot3,\) it is the last digit of the \(598\)th three-digit integer, namely \(100+597=697.\) That digit is \(7.\)

Therefore, the correct answer is D.

若 \(F=re^{i\theta}\)，则 \(\frac{1}{F}=r^{-1}e^{-i\theta}\)。由于 \(F\) 位于第一象限且在单位圆外，其倒数位于第四象限的单位圆内，辐角与原来关于实轴对称。只有点 \(C\) 具有这些性质。

所以正确答案是 C。

English

If \(F=re^{i\theta},\) then \(\frac{1}{F}=r^{-1}e^{-i\theta}.\) Because \(F\) is outside the unit circle in quadrant I, its reciprocal is inside the unit circle in quadrant IV, with the reflected argument. Only point \(C\) has those properties.

Therefore, the correct answer is C.

令 \(t=x^2+1\)，则 \[ \begin{aligned} x^4+5x^2+3 &=(t-1)^2\\ &\quad+5(t-1)+3\\ &=t^2+3t-1 \end{aligned}\text{。}\]因此 \(f(t)=t^2+3t-1\)。代入 \(t=x^2-1\)，得到 \[ \begin{aligned} f(x^2-1) &=(x^2-1)^2\\ &\quad+3(x^2-1)-1\\ &=x^4+x^2-3 \end{aligned}\text{。}\]

所以正确答案是 B。

English

With \(t=x^2+1,\) \[ \begin{aligned} x^4+5x^2+3 &=(t-1)^2\\ &\quad+5(t-1)+3\\ &=t^2+3t-1. \end{aligned} \] Hence \(f(t)=t^2+3t-1.\) Substituting \(t=x^2-1\) gives \[ \begin{aligned} f(x^2-1) &=(x^2-1)^2\\ &\quad+3(x^2-1)-1\\ &=x^4+x^2-3. \end{aligned} \]

Therefore, the correct answer is B.

由角平分线定理，可令 \(CD=t\)、\(DB=2t\)。再令 \(AD=d\)。三角形 \(ACD\) 与 \(ABD\) 在 \(A\) 处的角均为 \(60^\circ\)，对它们应用余弦定理，得到 \[ \begin{aligned} t^2&=9+d^2-3d,\\ 4t^2&=36+d^2-6d \end{aligned}\text{。}\]用第二个方程减去第一个方程的四倍，得到 \(0=-3d^2+6d\)。由于 \(d\gt0\)，所以 \(d=2\)。

所以正确答案是 A。

English

By the angle bisector theorem, write \(CD=t\) and \(DB=2t.\) Let \(AD=d.\) The law of cosines in triangles \(ACD\) and \(ABD,\) whose angles at \(A\) are both \(60^\circ,\) gives \[ \begin{aligned} t^2&=9+d^2-3d,\\ 4t^2&=36+d^2-6d. \end{aligned} \] Subtracting four times the first equation from the second yields \(0=-3d^2+6d.\) Since \(d\gt0,\) \(d=2.\)

Therefore, the correct answer is A.

令第一组根为 \(u=\tan\alpha\) 和 \(v=\tan\beta\)。则 \(u+v=p\)，且 \(uv=q\)。第二组根为 \(\frac{1}{u}\) 和 \(\frac{1}{v}\)，所以 \[ \begin{aligned} r&=\frac1u+\frac1v=\frac pq,\\ s&=\frac1{uv}=\frac1q \end{aligned}\text{。}\]因此 \(rs=\frac{p}{q^2}\)。

所以正确答案是 C。

English

Let the first roots be \(u=\tan\alpha\) and \(v=\tan\beta.\) Then \(u+v=p\) and \(uv=q.\) The second roots are \(\frac{1}{u}\) and \(\frac{1}{v},\) so \[ \begin{aligned} r&=\frac1u+\frac1v=\frac pq,\\ s&=\frac1{uv}=\frac1q. \end{aligned} \] Therefore \(rs=\frac{p}{q^2}.\)

Therefore, the correct answer is C.

由于 \(100\gt99\)，选项 A 为正；由于 \(324\lt325\)，选项 C 为负；又因 \(2601\gt2600\)，选项 D 为正而 E 为负。两个正数满足 \[ \begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}} \end{aligned}\text{。}\]第二个分母较大，所以选项 D 是较小的正数。

所以正确答案是 D。

English

Since \(100\gt99,\) choice A is positive; since \(324\lt325,\) choice C is negative; and since \(2601\gt2600,\) choice D is positive while E is negative. The two positive values satisfy \[ \begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}}. \end{aligned} \] The second denominator is larger, so choice D is the smaller positive number.

Therefore, the correct answer is D.

交点对应于方程的实根：\[ x^2+2(b-a)x+(1-2ab)=0\text{。}\]没有实根当且仅当其判别式为负：\[ \begin{aligned} &4(b-a)^2-4(1-2ab)\\ &=4(a^2+b^2-1)\\ &\lt0 \end{aligned}\text{。}\]因此 \(S\) 是 \(ab\) 平面内单位圆的内部，面积为 \(\pi\)。

所以正确答案是 B。

English

Intersections correspond to roots of \[ x^2+2(b-a)x+(1-2ab)=0. \] There are no real roots exactly when its discriminant is negative: \[ \begin{aligned} &4(b-a)^2-4(1-2ab)\\ &=4(a^2+b^2-1)\\ &\lt0. \end{aligned} \] Thus \(S\) is the interior of the unit circle in the \(ab\)-plane, with area \(\pi.\)

Therefore, the correct answer is B.

各圆心都位于 \(L_1\) 与 \(L_2\) 的角平分线上。任意两个相邻圆的构型都相似，因此相邻半径之比为常数。所以五个半径组成等比数列。若中间半径为 \(m\)，五项等比数列的对称性给出 \(m^2=8\cdot18\)，所以 \(m=12\)。

所以正确答案是 A。

English

The centers lie on the angle bisector of \(L_1\) and \(L_2.\) Similarity of the configuration for any two consecutive circles shows that consecutive radii have a constant ratio. Thus the five radii form a geometric sequence. If the middle radius is \(m,\) symmetry of a five-term geometric sequence gives \(m^2=8\cdot18,\) so \(m=12.\)

Therefore, the correct answer is A.

从任意一个周长为 \(P\)、面积为 \(K\) 的直角三角形开始。将每条边缩放 \(k=\frac{P}{K}\) 倍后，周长为 \(kP\)，面积为 \(k^2K\)，两者相等。互不相似的直角三角形形状有无穷多个，所以所得三角形也互不全等。

所以正确答案是 E。

English

Start with any right triangle having perimeter \(P\) and area \(K.\) Scaling every side by \(k=\frac{P}{K}\) produces perimeter \(kP\) and area \(k^2K,\) which are equal. Infinitely many nonsimilar right-triangle shapes exist, and the resulting triangles are therefore noncongruent.

Therefore, the correct answer is E.

由于 \(a=\log_{60}3\) 且 \(b=\log_{60}5\)，\[ \begin{aligned} 1-a-b &=\log_{60}\frac{60}{15}=\log_{60}4,\\ 1-b &=\log_{60}\frac{60}{5}=\log_{60}12 \end{aligned}\text{。}\]因此指数为 \(\frac12\log_{12}4\)，且 \[ 12^{(\frac{1}{2})\log_{12}4}=\sqrt4=2\text{。}\]

所以正确答案是 B。

English

Since \(a=\log_{60}3\) and \(b=\log_{60}5,\) \[ \begin{aligned} 1-a-b &=\log_{60}\frac{60}{15}=\log_{60}4,\\ 1-b &=\log_{60}\frac{60}{5}=\log_{60}12. \end{aligned} \] Therefore the exponent is \(\frac12\log_{12}4,\) and \[ 12^{(\frac{1}{2})\log_{12}4}=\sqrt4=2. \]

Therefore, the correct answer is B.

容斥原理给出 \[ \begin{aligned} p&=P(A\cap B)\\ &=\frac34+\frac23-P(A\cup B) \end{aligned}\text{。}\]由于 \(\frac{3}{4}\le P(A\cup B)\le1\)，可得 \[ \frac5{12}\le p\le\frac23\text{。}\]两个端点都能取到，所以这是必定成立的最小区间。

所以正确答案是 D。

English

Inclusion-exclusion gives \[ \begin{aligned} p&=P(A\cap B)\\ &=\frac34+\frac23-P(A\cup B). \end{aligned} \] Since \(\frac{3}{4}\le P(A\cup B)\le1,\) it follows that \[ \frac5{12}\le p\le\frac23. \] Both endpoints can occur, so this is the smallest necessary interval.

Therefore, the correct answer is D.

以球体与地面的接触点为 \((0,0)\)，圆心为 \((0,r)\)。影子边界处的太阳光线经过影子端点 \((10,0)\)，斜率为 \(-\frac{1}{2}\)，所以方程为 \(x+2y-10=0\)。相切意味着点 \((0,r)\) 到这条直线的距离等于 \(r\)：\[ \frac{10-2r}{\sqrt5}=r\text{。}\]因此 \(r=\frac{10}{2+\sqrt5}=10\sqrt5-20\)。

所以正确答案是 E。

English

Take the sphere’s ground-contact point as \((0,0)\) and its center as \((0,r).\) The limiting sun ray passes through the shadow endpoint \((10,0)\) and has slope \(-\frac{1}{2},\) so its equation is \(x+2y-10=0.\) Tangency means the distance from \((0,r)\) to this line equals \(r:\) \[ \frac{10-2r}{\sqrt5}=r. \] Hence \(r=\frac{10}{2+\sqrt5}=10\sqrt5-20.\)

Therefore, the correct answer is E.

作 \(DE\)。由 \[ \begin{aligned} [ABE]&=[ADE]+[DBE],\\ [DBEF]&=[FDE]+[DBE] \end{aligned}\text{，}\]可知面积相等意味着 \([ADE]=[FDE]\)。它们共有底 \(DE\)，所以 \(AF\parallel DE\)，从而三角形 \(DBE\) 与 \(ABC\) 相似。因此 \(\frac{BE}{BC}=\frac{DB}{AB}=\frac{3}{5}\)。三角形 \(ABE\) 与 \(ABC\) 共有从 \(A\) 到 \(BC\) 的高，所以 \([ABE]=(\frac{3}{5})[ABC]=6\)。

所以正确答案是 C。

English

Draw \(DE.\) From \[ \begin{aligned} [ABE]&=[ADE]+[DBE],\\ [DBEF]&=[FDE]+[DBE], \end{aligned} \] equality implies \([ADE]=[FDE].\) Their common base \(DE\) then gives \(AF\parallel DE,\) so triangles \(DBE\) and \(ABC\) are similar. Hence \(\frac{BE}{BC}=\frac{DB}{AB}=\frac{3}{5}.\) Triangles \(ABE\) and \(ABC\) share the altitude from \(A\) to \(BC,\) so \([ABE]=(\frac{3}{5})[ABC]=6.\)

Therefore, the correct answer is C.

令 \(D=(0,0)\)、\(A=(1,0)\)、\(B=(1,1)\)、\(C=(0,1)\)，并令 \(P=(x,y)\)。则 \[ \begin{aligned} &(x-1)^2+y^2\\ &\quad+(x-1)^2+(y-1)^2\\ &=x^2+(y-1)^2 \end{aligned}\text{，}\]化简得 \((x-2)^2+y^2=2\)。因此 \(P\) 位于一个圆上，该圆的圆心与 \(D\) 相距 \(2\)，半径为 \(\sqrt2\)。所以它到 \(D\) 的最大距离为 \(2+\sqrt2\)。

所以正确答案是 C。

English

Place \(D=(0,0),\) \(A=(1,0),\) \(B=(1,1),\) and \(C=(0,1),\) with \(P=(x,y).\) Then \[ \begin{aligned} &(x-1)^2+y^2\\ &\quad+(x-1)^2+(y-1)^2\\ &=x^2+(y-1)^2, \end{aligned} \] which simplifies to \((x-2)^2+y^2=2.\) Thus \(P\) lies on a circle centered \(2\) units from \(D\) with radius \(\sqrt2.\) Its greatest distance from \(D\) is \(2+\sqrt2.\)

Therefore, the correct answer is C.

由于 \(\overset{\frown}{MA}=40^\circ\)，所以圆心角 \(\angle MCA=40^\circ\)，从而 \(\angle ACP=140^\circ\)。于是 \(\angle APC=180^\circ-140^\circ\) \({}-10^\circ=30^\circ\)。在三角形 \(ACP\) 和 \(BCP\) 中应用正弦定理，并利用 \(AC=BC\)，得到 \(\sin\angle BPC=\sin30^\circ\)。由于 \(A\) 与 \(B\) 不同，所以 \(\angle BPC=150^\circ\)。因此 \[ \begin{aligned} \angle BCN &=180^\circ-150^\circ-10^\circ\\ &=20^\circ \end{aligned}\text{，}\]所以 \(\overset{\frown}{BN}=20^\circ\)。

所以正确答案是 C。

English

Since \(\overset{\frown}{MA}=40^\circ,\) the central angle \(\angle MCA=40^\circ,\) so \(\angle ACP=140^\circ.\) Hence \(\angle APC=180^\circ-140^\circ\)\({}-10^\circ=30^\circ.\) Applying the law of sines in triangles \(ACP\) and \(BCP,\) and using \(AC=BC,\) gives \(\sin\angle BPC=\sin30^\circ.\) Because \(A\) and \(B\) are distinct, \(\angle BPC=150^\circ.\) Therefore \[ \begin{aligned} \angle BCN &=180^\circ-150^\circ-10^\circ\\ &=20^\circ, \end{aligned} \] so \(\overset{\frown}{BN}=20^\circ.\)

Therefore, the correct answer is C.