2002B AMC 12 真题（P版）

1 答案 A

这九个数分别是 \(10^k-1\)，所以它们的和为 \(9+99+\cdots+999{,}999{,}999\)。把这个和除以 \(9\) 得到 M： \[ \begin{aligned} M &= 1+11+111+\cdots \\ &\quad {}+111{,}111{,}111 \\ &= 123{,}456{,}789\text{。} \end{aligned} \] 它的数字是 \(1\) 到 \(9\)，因此缺少的数字是 \(0\)。

所以正确答案是 A。

English

Each of the nine numbers is \(10^k-1,\) so their sum is \(9+99+\cdots+999{,}999{,}999.\) Dividing by \(9,\) \[ \begin{aligned} M &= 1+11+111+\cdots \\ &\quad {}+111{,}111{,}111 \\ &= 123{,}456{,}789. \end{aligned} \] Its digits are \(1\) through \(9,\) so the missing digit is \(0.\)

Thus, the correct answer is A.

2 答案 D

从前两项中提出 \(3x-2\)： \[ \begin{gathered} (3x-2)(4x+1) \\ {}-(3x-2)4x+1 \\ {}=(3x-2)(4x+1-4x)+1 \\ {}=3x-1\text{。} \end{gathered} \] 当 \(x=4\) 时，它等于 \(3\cdot4-1=11\)。

所以正确答案是 D。

English

Factor \(3x-2\) out of the first two terms: \[ \begin{gathered} (3x-2)(4x+1) \\ {}-(3x-2)4x+1 \\ {}=(3x-2)(4x+1-4x)+1 \\ {}=3x-1. \end{gathered} \] At \(x=4\) this equals \(3\cdot4-1=11.\)

Thus, the correct answer is D.

3 答案 B

分解得 \(n^2-3n+2=(n-1)(n-2)\)。当 \(n\ge4\) 时，两个因数都大于 \(1\)，所以结果是合数。检查 \(n=1,2,3\) 分别得到 \(0\)、\(0\)、\(2\)。只有 \(n=3\) 产生质数，因此这样的 \(n\) 恰有一个。

所以正确答案是 B。

English

Factor \(n^2-3n+2=(n-1)(n-2).\) For \(n\ge4\) both factors exceed \(1,\) so the value is composite. Checking \(n=1,2,3\) gives \(0,\) \(0,\) and \(2.\) Only \(n=3\) yields a prime, so there is exactly one such \(n.\)

Thus, the correct answer is B.

4 答案 E

因为 \(\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42}\)，所以和 \(\dfrac{41}{42}+\dfrac1n\) 严格在 \(0\) 和 \(2\) 之间，作为整数只能是 \(1\)。于是 \(\dfrac1n=\dfrac1{42}\) 得到 \(n=42\)。

现在 \(2\)、\(3\)、\(6\)、\(7\) 都整除 \(42\)，但 \(n\gt84\) 是假的。因此不正确的说法是 \(n\gt84\)。

所以正确答案是 E。

English

Since \(\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42},\) the sum \(\dfrac{41}{42}+\dfrac1n\) lies strictly between \(0\) and \(2,\) so it must equal \(1.\) Then \(\dfrac1n=\dfrac1{42},\) giving \(n=42.\)

Now \(2,\) \(3,\) \(6,\) and \(7\) all divide \(42,\) but \(n\gt84\) is false. The untrue statement is \(n\gt84.\)

Thus, the correct answer is E.

5 答案 D

五个内角的和是 \(540^\circ\)。把等差数列写成 \(x-2d\)、\(x-d\)、\(x\)、\(x+d\)、\(x+2d\)，它们的和为 \(5x=540\)，所以 \(x=108\)。

所以正确答案是 D。

English

The five interior angles sum to \(540^\circ.\) Writing the sequence as \(x-2d,\) \(x-d,\) \(x,\) \(x+d,\) \(x+2d,\) the sum is \(5x=540,\) so \(x=108.\)

Thus, the correct answer is D.

6 答案 C

因为 \(a\) 和 \(b\) 是根， \[ \begin{gathered} x^2+ax+b \\ {}=(x-a)(x-b) \\ {}=x^2-(a+b)x+ab\text{。} \end{gathered} \] 比较系数得 \(a+b=-a\) 且 \(ab=b\)。

由于 \(b\neq0\)，第二个方程给出 \(a=1\)，再由 \(a+b=-a\) 得 \(b=-2\)。所以 \((a,b)=(1,-2)\)。

所以正确答案是 C。

English

Since \(a\) and \(b\) are the roots, \[ \begin{gathered} x^2+ax+b \\ {}=(x-a)(x-b) \\ {}=x^2-(a+b)x+ab. \end{gathered} \] Matching coefficients gives \(a+b=-a\) and \(ab=b.\)

As \(b\neq0,\) the second equation gives \(a=1,\) and then \(a+b=-a\) gives \(b=-2.\) So \((a,b)=(1,-2).\)

Thus, the correct answer is C.

7 答案 B

设三个整数为 \(n-1\)、\(n\)、\(n+1\)。由条件可得 \((n-1)n(n+1)=8\cdot3n\)，所以 \(n^2-1=24\)，进而 \(n=5\)。

三个整数是 \(4\)、\(5\)、\(6\)，平方和为 \(16+25+36=77\)。

所以正确答案是 B。

English

Let the integers be \(n-1,\) \(n,\) \(n+1.\) Then \((n-1)n(n+1)=8\cdot3n,\) so \(n^2-1=24\) and \(n=5.\)

The three integers \(4,\) \(5,\) \(6\) have squares summing to \(16+25+36=77.\)

Thus, the correct answer is B.

8 答案 D

七月有 \(31=4\cdot7+3\) 天，所以七月 \(1\) 日的星期会出现五次；既然星期一出现五次，星期一必在七月 \(1\)、\(2\) 或 \(3\) 日之一。八月中出现五次的是八月 \(1\)、\(2\)、\(3\) 日的星期，而八月 \(1\) 日比七月 \(1\) 日晚三天。

检查三种情况，八月出现五次的星期分别是星期四、星期五、星期六；星期三、星期四、星期五；以及星期二、星期三、星期四。

每种情况都包含星期四，所以正确答案是 D。

English

Since July has \(31=4\cdot7+3\) days, the weekday of July \(1\) occurs five times, so Monday falls on July \(1,\) \(2,\) or \(3.\) The days that occur five times in August are those of August \(1,\) \(2,\) \(3,\) and August \(1\) is three weekdays after July \(1.\)

Testing the three cases, the August weekdays occurring five times are Thursday through Saturday, Wednesday through Friday, and Tuesday through Thursday. Thursday appears in every case.

Thus, the correct answer is D.

9 答案 C

设 \(b=a+r\)、\(c=a+2r\)、\(d=a+3r\)。等比条件 \(b^2=ad\) 给出 \((a+r)^2=a(a+3r)\)，即 \(r^2=ar\)。因为数列递增，所以 \(r=a\)。

于是 \(d=a+3a=4a\)，所以 \(\dfrac{a}{d}=\dfrac14\)。

所以正确答案是 C。

English

Let \(b=a+r,\) \(c=a+2r,\) \(d=a+3r.\) The geometric condition \(b^2=ad\) gives \((a+r)^2=a(a+3r),\) i.e. \(r^2=ar,\) so \(r=a.\)

Then \(d=a+3a=4a\) and \(\dfrac{a}{d}=\dfrac14.\)

Thus, the correct answer is C.

10 答案 A

每个元素都比 \(3\) 的倍数多一，所以任意三个元素的和都是 \(3\) 的倍数。最小和为 \(1+4+7=12\)，最大和为 \(13+16+19=48\)，并且中间每个 \(3\) 的倍数都可以取得。

从 \(12\) 到 \(48\) 共有 \(13\) 个 \(3\) 的倍数。

所以正确答案是 A。

English

Every element is one more than a multiple of \(3,\) so any sum of three of them is a multiple of \(3.\) The smallest sum is \(1+4+7=12\) and the largest is \(13+16+19=48,\) and every multiple of \(3\) between them is attainable.

There are \(13\) multiples of \(3\) from \(12\) to \(48.\)

Thus, the correct answer is A.

11 答案 E

\(A-B\) 和 \(A+B\) 奇偶性相同；因为两者都是质数，所以都是奇数，故 \(A\) 与 \(B\) 奇偶性相反。若 \(A\) 为偶数，则质数 \(A\) 必须等于 \(2\)，但正质数 \(B\) 满足 \(B\ge2\)，从而 \(A-B\le0\)。因此 \(A\) 是奇数，偶质数 \(B\) 为 \(2\)。

于是 \(A-2\)，\(A\)，和 \(A+2\) 是三个质数。其中一个能被 \(3\) 整除，所以它必须等于 \(3\)；这三个数是 \(3\)，\(5\)，和 \(7\)。它们再加上 \(2\) 的和为 \(2+3+5+7=17\)，也是质数。

所以正确答案是 E。

English

\(A-B\) and \(A+B\) have the same parity; being prime, both are odd, so \(A\) and \(B\) have opposite parity. If \(A\) were even, then the prime \(A\) would equal \(2,\) but the positive prime \(B\) would satisfy \(B\ge2\) and make \(A-B\le0.\) Hence \(A\) is odd and the even prime \(B\) is \(2.\)

Then \(A-2,\) \(A,\) \(A+2\) are three primes. One is divisible by \(3,\) so that one must equal \(3;\) the triple is \(3,\) \(5,\) \(7.\) Their sum together with \(2\) is \(2+3+5+7=17,\) a prime.

Thus, the correct answer is E.

12 答案 D

令 \(\dfrac{n}{20-n}=k^2\)。解得 \(n=\dfrac{20k^2}{k^2+1}\)。因为 \(k^2\) 和 \(k^2+1\) 互质，所以 \(k^2+1\) 必须整除 \(20\)，这只在 \(|k|=0,1,2,3\) 时发生。

\(k\) 的正负不改变 \(n\)；这些值给出 \(n=0\)，\(10\)，\(16\)，和 \(18\)，共四个值。

所以正确答案是 D。

English

Set \(\dfrac{n}{20-n}=k^2.\) Solving, \(n=\dfrac{20k^2}{k^2+1}.\) Since \(k^2\) and \(k^2+1\) are coprime, \(k^2+1\) must divide \(20,\) which happens only for \(|k|=0,1,2,3.\)

The signs of \(k\) do not change \(n;\) these give \(n=0,\) \(10,\) \(16,\) and \(18,\) which is four values.

Thus, the correct answer is D.

13 答案 B

\(n\)、\(n+1,\ldots\)、\(n+17\) 的和为 \(18n+\dfrac{17\cdot18}{2}=9(2n+17)\)。因为 \(9\) 已经是平方数，所以 \(2n+17\) 也必须是完全平方数。

使 \(2n+17\) 成为完全平方数的最小正整数 \(n\) 是 \(n=4\)，此时 \(2n+17=25\)，和为 \(9\cdot25=225\)。

所以正确答案是 B。

English

The sum of \(n,\) \(n+1,\ldots,\) \(n+17\) is \(18n+\dfrac{17\cdot18}{2}=9(2n+17).\) Since \(9\) is a perfect square, \(2n+17\) must be one too.

The smallest positive integer \(n\) making \(2n+17\) a perfect square is \(n=4,\) giving \(2n+17=25\) and a sum of \(9\cdot25=225.\)

Thus, the correct answer is B.

14 答案 D

每一对圆最多有 \(2\) 个交点，而四个圆共有 \(\binom{4}{2}=6\) 对，因此最多有 \(6\cdot2=12\) 个交点。

取四个位置一般的圆，使每一对圆都相交两次，且没有三个圆经过同一点，就能达到这个上界。

所以正确答案是 D。

English

Each pair of circles meets in at most \(2\) points, and there are \(\binom{4}{2}=6\) pairs, giving at most \(6\cdot2=12\) intersection points.

The bound is attainable by taking four circles in general position so that every pair crosses twice and no three pass through the same point.

Thus, the correct answer is D.

15 答案 D

设 \(a\) 是首位数字，\(x\) 是删去首位后得到的三位数，则 \(N=1000a+x\)。条件 \(N=9x\) 给出 \(1000a=8x\)，即 \(x=125a\)。

当 \(a=1,\ldots,7\) 时，\(x\) 是三位数；而 \(a=8\) 时 \(x=1000\)。所以共有 \(7\) 个这样的数。

所以正确答案是 D。

English

Let \(a\) be the leading digit and \(x\) the three-digit number after removing it, so \(N=1000a+x.\) The condition \(N=9x\) gives \(1000a=8x,\) i.e. \(x=125a.\)

For \(a=1,\ldots,7\) this makes \(x\) a three-digit number, while \(a=8\) gives \(x=1000.\) So there are \(7\) such numbers.

Thus, the correct answer is D.

16 答案 C

乘积是 \(3\) 的倍数，当且仅当至少一枚骰子掷出 \(3\) 或 \(6\)。八面骰避开 \(3,6\) 的概率为 \(\dfrac68=\dfrac34\)，六面骰避开它们的概率为 \(\dfrac46=\dfrac23\)。

因此两枚骰子都没有出现 \(3\) 的倍数的概率是 \(\dfrac34\cdot\dfrac23=\dfrac12\)，所求概率为 \(1-\dfrac12=\dfrac12\)。

所以正确答案是 C。

English

The product is a multiple of \(3\) if and only if at least one die shows \(3\) or \(6.\) The octahedral die avoids \(3,6\) with probability \(\dfrac68=\dfrac34,\) and the six-sided die avoids them with probability \(\dfrac46=\dfrac23.\)

So neither shows a multiple of \(3\) with probability \(\dfrac34\cdot\dfrac23=\dfrac12,\) and the answer is \(1-\dfrac12=\dfrac12.\)

Thus, the correct answer is C.

17 答案 B

设 Andy 的草坪面积为 \(A\)，则 Beth 的面积是 \(\dfrac A2\)，Carlos 的面积是 \(\dfrac A3\)。若 Carlos 的割草速度为 \(R\)，则 Beth 的速度为 \(2R\)，Andy 的速度为 \(3R\)。

三人的用时分别为 \(\dfrac{A}{3R}\)、\(\dfrac{\frac{A}{2}}{2R}=\dfrac{A}{4R}\)、\(\dfrac{\frac{A}{3}}{R}=\dfrac{A}{3R}\)。其中 \(\dfrac{A}{4R}\) 最小。

所以 Beth 最先完成，正确答案是 B。

English

Let Andy’s lawn have area \(A;\) then Beth’s is \(\dfrac A2\) and Carlos’ is \(\dfrac A3.\) With Carlos’ rate \(R,\) Beth mows at \(2R\) and Andy at \(3R.\)

Their times are \(\dfrac{A}{3R},\) \(\dfrac{\frac{A}{2}}{2R}=\dfrac{A}{4R},\) and \(\dfrac{\frac{A}{3}}{R}=\dfrac{A}{3R}\) respectively. Beth’s time \(\dfrac{A}{4R}\) is the smallest, so Beth finishes first.

Thus, the correct answer is B.

18 答案 C

到 \((0,0)\) 比到 \((3,1)\) 更近的点位于这两点连线垂直平分线的原点一侧，该垂直平分线为 \(3x+y=5\)。

在矩形中，这一区域是一个梯形，其平行边长分别为 \(\dfrac53\)（在 \(y=0\) 处）和 \(\dfrac43\)（在 \(y=1\) 处），面积为 \(\dfrac12\left(\dfrac53+\dfrac43\right)=\dfrac32\)。矩形面积为 \(2\)，所以概率是 \(\dfrac{\frac{3}{2}}{2}=\dfrac34\)。

所以正确答案是 C。

English

The points closer to \((0,0)\) than to \((3,1)\) lie on the origin side of the perpendicular bisector of that segment, the line \(3x+y=5.\)

Within the rectangle, this region is a trapezoid whose parallel sides have lengths \(\dfrac53\) (at \(y=0\)) and \(\dfrac43\) (at \(y=1\)), so its area is \(\dfrac12\left(\dfrac53+\dfrac43\right)=\dfrac32.\) The rectangle has area \(2,\) so the probability is \(\dfrac{\frac{3}{2}}{2}=\dfrac34.\)

Thus, the correct answer is C.

19 答案 D

三个方程相加得 \(2(ab+bc+ca)=484\)，所以 \(ab+bc+ca=242\)。用它分别减去原来的三个方程，得到 \(bc=90\)、\(ca=80\)、\(ab=72\)。

相乘得 \((abc)^2=90\cdot80\cdot72=720^2\)。又因 \(abc\gt0\)，所以 \(abc=720\)。

所以正确答案是 D。

English

Adding the three equations gives \(2(ab+bc+ca)=484,\) so \(ab+bc+ca=242.\) Subtracting each original equation from this yields \(bc=90,\) \(ca=80,\) and \(ab=72.\)

Multiplying, \((abc)^2=90\cdot80\cdot72=720^2,\) and since \(abc\gt0,\) we get \(abc=720.\)

Thus, the correct answer is D.

20 答案 B

设 \(OM=a\)，\(ON=b\)。因为 \(M,N\) 是中点，得到 \(XN^2=(2a)^2+b^2=361\)，且 \(YM^2=a^2+(2b)^2=484\)。

两式相加得 \(5(a^2+b^2)=845\)，所以 \(a^2+b^2=169\)。于是 \(MN=\sqrt{a^2+b^2}=13\)。又因为 \(MN\) 是连接 \(M\)、\(N\) 的中位线，\(XY=2\,MN=26\)。

所以正确答案是 B。

English

Let \(OM=a\) and \(ON=b.\) Since \(M,N\) are midpoints, \(XN^2=(2a)^2+b^2=361\) and \(YM^2=a^2+(2b)^2=484.\)

Adding, \(5(a^2+b^2)=845,\) so \(a^2+b^2=169.\) Then \(MN=\sqrt{a^2+b^2}=13,\) and since \(MN\) is the midsegment joining \(M\) and \(N,\) we have \(XY=2\,MN=26.\)

Thus, the correct answer is B.

21 答案 A

因为 \(2002=11\cdot13\cdot14\)，且 \(11\)、\(13\)、\(14\) 两两互质，所以 \(a_n=11\) 当 \(n\) 是 \(182\) 的倍数，\(a_n=13\) 当 \(n\) 是 \(154\) 的倍数，\(a_n=14\) 当 \(n\) 是 \(143\) 的倍数。（没有 \(n\lt2002\) 能被三者同时整除。）

当 \(n\le2001\) 时，\(182\) 的倍数有 \(10\) 个，\(154\) 的倍数有 \(12\) 个，\(143\) 的倍数有 \(13\) 个。所以和为 \[ \begin{gathered} 11\cdot10+13\cdot12+14\cdot13 \\ {}=110+156+182 \\ {}=448 \end{gathered}\text{。}\]

所以正确答案是 A。

English

Since \(2002=11\cdot13\cdot14\) with \(11,\) \(13,\) \(14\) pairwise coprime, \(a_n=11\) when \(n\) is divisible by \(182,\) \(a_n=13\) when \(n\) is divisible by \(154,\) and \(a_n=14\) when \(n\) is divisible by \(143\) (and no \(n\lt2002\) is divisible by all three).

For \(n\le2001\) there are \(10\) multiples of \(182,\) \(12\) of \(154,\) and \(13\) of \(143.\) So the sum is \[ \begin{gathered} 11\cdot10+13\cdot12+14\cdot13 \\ {}=110+156+182 \\ {}=448. \end{gathered} \]

Thus, the correct answer is A.

22 答案 B

由换底公式，\(a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n\)。\[ \begin{gathered} b-c \\ {}=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14} \end{gathered}\text{。}\]

该分数等于 \(\dfrac{120}{240240}=\dfrac{1}{2002}\)，因此 \(b-c=\log_{2002}\dfrac{1}{2002}=-1\)。

所以正确答案是 B。

English

By change of base, \(a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n.\) So \[ \begin{gathered} b-c \\ {}=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}. \end{gathered} \]

The fraction equals \(\dfrac{120}{240240}=\dfrac{1}{2002},\) so \(b-c=\log_{2002}\dfrac{1}{2002}=-1.\)

Thus, the correct answer is B.

23 答案 C

设 \(M\) 为 \(BC\) 的中点，令 \(AM=2a\)，并设 \(\theta=\angle AMB\)，于是 \(\angle AMC=180^\circ-\theta\)。因 \(BM=CM=a\)，故 \(BC=2a\)。在 \(\triangle ABM\) 和 \(\triangle AMC\) 中用余弦定理，得到 \[a^2+4a^2-4a^2\cos\theta=1\text{，}\]\[a^2+4a^2+4a^2\cos\theta=4\text{。}\]

两式相加得 \(10a^2=5\)，所以 \(a=\dfrac{\sqrt2}{2}\)，从而 \(BC=2a=\sqrt2\)。

所以正确答案是 C。

English

Let \(M\) be the midpoint of \(BC,\) set \(AM=2a,\) and let \(\theta=\angle AMB,\) so \(\angle AMC=180^\circ-\theta.\) With \(BM=CM=a,\) so that \(BC=2a,\) the Law of Cosines in \(\triangle ABM\) and \(\triangle AMC\) gives \[a^2+4a^2-4a^2\cos\theta=1,\] \[a^2+4a^2+4a^2\cos\theta=4.\]

Adding, \(10a^2=5,\) so \(a=\dfrac{\sqrt2}{2}\) and \(BC=2a=\sqrt2.\)

Thus, the correct answer is C.

24 答案 E

任意四边形的面积不超过 \(\tfrac12\,d_1 d_2\)，其中 \(d_1,d_2\) 为对角线长；等号恰在对角线垂直时成立。这里 \[ \begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002 \end{gathered}\text{。}\]

因此等号成立，两条对角线垂直并在 \(P\) 相交。于是 \[ \begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113} \end{aligned}\text{，}\]\[ \begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51 \end{aligned}\text{。}\]

周长为 \(144+4\sqrt{113}=4\left(36+\sqrt{113}\right)\)。

所以正确答案是 E。

English

For any quadrilateral, the area is at most \(\tfrac12\,d_1 d_2\) where \(d_1,d_2\) are the diagonals, with equality exactly when they are perpendicular. Here \[ \begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002. \end{gathered} \]

Equality forces the diagonals to be perpendicular and to intersect at \(P.\) Then \[ \begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned} \] \[ \begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned} \]

The perimeter is \(144+4\sqrt{113}=4\left(36+\sqrt{113}\right).\)

Thus, the correct answer is E.

25 答案 E

配方得 \[ \begin{gathered} f(x)+f(y) \\ {}=(x+3)^2+(y+3)^2-16\text{，} \end{gathered} \] 所以第一个条件表示以 \((-3,-3)\) 为圆心、半径为 \(4\) 的圆盘。

又 \[ \begin{gathered} f(x)-f(y) \\ {}=(x-y)(x+y+6)\text{，} \end{gathered} \] 因此第二个条件 \((x-y)(x+y+6)\le0\) 表示由过 \((-3,-3)\)、斜率分别为 \(1\) 和 \(-1\) 的两条垂直直线所确定的两个半平面。这些直线把圆盘分成两半。

因此 \(R\) 的面积为 \(\tfrac12\pi\cdot4^2=8\pi\approx25.13\)，最接近 \(25\)。

所以正确答案是 E。

English

Completing the square, \[ \begin{gathered} f(x)+f(y) \\ {}=(x+3)^2+(y+3)^2-16, \end{gathered} \] so the first condition is the disk of radius \(4\) centered at \((-3,-3).\)

Also \[ \begin{gathered} f(x)-f(y) \\ {}=(x-y)(x+y+6), \end{gathered} \] so the second condition \((x-y)(x+y+6)\le0\) describes two half-planes bounded by the perpendicular lines through \((-3,-3)\) of slopes \(1\) and \(-1.\) These cut the disk into two equal halves.

Thus \(R\) has area \(\tfrac12\pi\cdot4^2=8\pi\approx25.13,\) which is closest to \(25.\)

Thus, the correct answer is E.