2005 AIME II 详解 | LIVE by Po-Shen Loh

1.

一个游戏使用一副由 $n$ 张不同卡片组成的牌，其中 $n$ 是整数且 $n \ge 6$ 。从中抽取 $6$ 张牌的组合数，是抽取 $3$ 张牌的组合数的 $6$ 倍。求 $n$ 。

A game uses a deck of $n$ different cards, where $n$ is an integer and $n \ge 6.$ The number of possible sets of $6$ cards that can be drawn from the deck is $6$ times the number of possible sets of $3$ cards that can be drawn. Find $n.$

答案

知识点：组合阶乘

难度评级：1890

小提示：

用阶乘写出比值 $\frac{\binom{n}{6}}{\binom{n}{3}}$ ；几乎所有因子都会约掉。

Write the ratio $\frac{\binom{n}{6}}{\binom{n}{3}}$ in factorials; almost everything cancels

大提示：

解 $(n-3)(n-4)(n-5) = 720$ ，方法是写成 $720 = 10 \cdot 9 \cdot 8$ 。

Solve $(n-3)(n-4)(n-5) = 720$ by writing $720 = 10 \cdot 9 \cdot 8$

解答：

题目条件为 $\binom{n}{6} = 6\binom{n}{3}$ 。将二项式系数相除， $\begin{aligned} \frac{\binom{n}{6}}{\binom{n}{3}} &= \frac{(n-3)(n-4)(n-5)}{6 \cdot 5 \cdot 4} \\ &= 6 \end{aligned}\text{，}$ 所以 $(n-3)(n-4)(n-5) = 720$ $= 10 \cdot 9 \cdot 8$ 。

由于乘积 $(n-3)(n-4)(n-5)$ 随 $n$ 增大而增大，唯一解是 $n - 3 = 10$ ，即 $n = 13$ 。

The condition says $\binom{n}{6} = 6\binom{n}{3}.$ Dividing the binomial coefficients, $\begin{aligned} \frac{\binom{n}{6}}{\binom{n}{3}} &= \frac{(n-3)(n-4)(n-5)}{6 \cdot 5 \cdot 4} \\ &= 6, \end{aligned}$ so $(n-3)(n-4)(n-5) = 720$ $= 10 \cdot 9 \cdot 8.$

Since the product $(n-3)(n-4)(n-5)$ is increasing in $n,$ the only solution is $n - 3 = 10,$ that is, $n = 13.$

2.

一家旅馆为三位客人各打包了一份早餐。每份早餐本应包含三种小面包：坚果、奶酪和水果口味各一个。准备早餐的人把这九个小面包分别包好；包好后，这些小面包彼此无法区分。然后她随机在每位客人的袋子里放入三个小面包。已知每位客人都拿到每种口味各一个小面包的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

A hotel packed a breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls, and, once they were wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability that each guest got one roll of each type is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

知识点：无放回抽样条件概率

难度评级：2170

小提示：

逐个抽取第一位客人袋中的三个小面包，求其中正好包含每种口味各一个的概率。

Find the chance the first guest’s bag has one roll of each type by drawing its three rolls one at a time

大提示：

在第一个袋子合格的条件下，第二个袋子从余下的六个小面包中抽取，其中每种口味各有两个；第三个袋子的内容随后自动确定。

Given the first bag is good, the second bag draws from six rolls, two of each type; the third bag is then forced

解答：

逐个装第一位客人的袋子。第一个小面包可以是任意口味；第二个必须避开与第一个同口味的剩余 $2$ 个小面包，成功概率为 $\frac{6}{8}$ ；第三个必须是剩余 $7$ 个中缺少口味的 $3$ 个之一。因此第一个袋子含有每种口味各一个的概率为 $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}$ 。

在此条件下，剩下六个小面包，每种口味各两个，同样的论证给出第二个袋子的概率为 $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ 。此时第三个袋子自动是每种口味各一个。所求概率为 $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ ，所以 $m + n = 9 + 70 = 79$ 。

Fill the first guest’s bag one roll at a time. The first roll can be anything; the second must avoid the $2$ remaining rolls of the first roll’s type, succeeding with probability $\frac{6}{8};$ and the third must be one of the $3$ rolls of the missing type among the remaining $7.$ So the first bag has one roll of each type with probability $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}.$

Given that, six rolls remain, two of each type, and the same argument gives $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ for the second bag. The third bag is then automatically one of each type. The probability is $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70},$ so $m + n = 9 + 70 = 79.$

3.

一个无穷等比级数的和为 $2005$ 。将原级数的每一项平方后得到一个新级数，其和是原级数和的 $10$ 倍。原级数的公比为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

An infinite geometric series has sum $2005.$ A new series, obtained by squaring each term of the original series, has sum $10$ times the sum of the original series. The common ratio of the original series is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：等比数列平方差

难度评级：2070

小提示：

平方后的级数仍是等比级数，首项为 $a^2$ ，公比为 $r^2$ 。

The squared series is also geometric, with first term $a^2$ and ratio $r^2$

大提示：

分解 $\frac{a^2}{1-r^2} = \frac{a}{1-r} \cdot \frac{a}{1+r}$ ，得到 $\frac{a}{1+r} = 10$ ，再与 $\frac{a}{1-r} = 2005$ 比较。

Factor $\frac{a^2}{1-r^2} = \frac{a}{1-r} \cdot \frac{a}{1+r}$ to get $\frac{a}{1+r} = 10,$ then compare with $\frac{a}{1-r} = 2005$

解答：

设原级数首项为 $a$ ，公比为 $r$ ，则 $\frac{a}{1-r} = 2005$ 。平方后的级数是首项 $a^2$ 、公比 $r^2$ 的等比级数，所以 $\begin{aligned} \frac{a^2}{1-r^2} &= \frac{a}{1-r} \cdot \frac{a}{1+r} \\ &= 2005 \cdot \frac{a}{1+r} \\ &= 10 \cdot 2005 \end{aligned}\text{，}$ 因此 $\frac{a}{1+r} = 10$ 。

两个方程相除，得 $\frac{1+r}{1-r} = \frac{2005}{10}$ ，所以 $2(1+r) = 401(1-r)$ ，进而 $403r = 399$ ， $r = \frac{399}{403}$ 。由于 $399 = 3 \cdot 7 \cdot 19$ 且 $403 = 13 \cdot 31$ ，该分数已最简，故 $m + n = 399 + 403 = 802$ 。

Let the original series have first term $a$ and ratio $r,$ so $\frac{a}{1-r} = 2005.$ The squared series is geometric with first term $a^2$ and ratio $r^2,$ so $\begin{aligned} \frac{a^2}{1-r^2} &= \frac{a}{1-r} \cdot \frac{a}{1+r} \\ &= 2005 \cdot \frac{a}{1+r} \\ &= 10 \cdot 2005, \end{aligned}$ which gives $\frac{a}{1+r} = 10.$

Dividing the two equations, $\frac{1+r}{1-r} = \frac{2005}{10},$ so $2(1+r) = 401(1-r),$ giving $403r = 399$ and $r = \frac{399}{403}.$ Since $399 = 3 \cdot 7 \cdot 19$ and $403 = 13 \cdot 31,$ the fraction is in lowest terms, and $m + n = 399 + 403 = 802.$

4.

求至少整除 $10^{10}$ 、 $15^7$ 、 $18^{11}$ 中一个数的正整数个数。

Find the number of positive integers that are divisors of at least one of $10^{10},$ $15^7,$ $18^{11}.$

答案

知识点：因数个数容斥原理最大公约数

难度评级：2230

小提示：

先由质因数分解数出每个数的因数个数，再修正重复计数。

Count the divisors of each number from its prime factorization, then fix the overcounting

大提示：

两个数的公共因数恰好是它们最大公因数的因数，例如 $\gcd(10^{10}, 15^7) = 5^7$ 。

The common divisors of two of the numbers are exactly the divisors of their gcd, e.g. $\gcd(10^{10}, 15^7) = 5^7$

解答：

由分解式 $10^{10} = 2^{10} 5^{10}$ 、 $15^7 = 3^7 5^7$ 以及 $18^{11} = 2^{11} 3^{22}$ ，它们的因数个数分别为 $11 \cdot 11 = 121$ 、 $8 \cdot 8 = 64$ 和 $12 \cdot 23 = 276$ 。

两个数的公共因数恰好是它们最大公因数的因数： $\gcd(10^{10}, 15^7) = 5^7$ 有 $8$ 个因数， $\gcd(10^{10}, 18^{11}) = 2^{10}$ 有 $11$ 个， $\gcd(15^7, 18^{11}) = 3^7$ 有 $8$ 个。只有 $1$ 同时整除三个数。

由容斥原理，所求个数为 $121 + 64 + 276$ $- 8 - 11 - 8 + 1$ $= 435$ 。

From the factorizations $10^{10} = 2^{10} 5^{10},$ $15^7 = 3^7 5^7,$ and $18^{11} = 2^{11} 3^{22},$ the divisor counts are $11 \cdot 11 = 121,$ $8 \cdot 8 = 64,$ and $12 \cdot 23 = 276.$

The divisors common to two of the numbers are exactly the divisors of their gcd: $\gcd(10^{10}, 15^7) = 5^7$ has $8$ divisors, $\gcd(10^{10}, 18^{11}) = 2^{10}$ has $11,$ and $\gcd(15^7, 18^{11}) = 3^7$ has $8.$ Only $1$ divides all three numbers.

By inclusion-exclusion, the count is $121 + 64 + 276$ $- 8 - 11 - 8 + 1$ $= 435.$

5.

求整数有序对 $(a, b)$ 的个数，使得 $\log_a b + 6\log_b a = 5$ ， $2 \le a \le 2005$ ，且 $2 \le b \le 2005$ 。

Determine the number of ordered pairs $(a, b)$ of integers such that $\log_a b + 6\log_b a = 5,$ $2 \le a \le 2005,$ and $2 \le b \le 2005.$

答案

知识点：对数二次方程区间内整数计数

难度评级：2310

小提示：

令 $x = \log_a b$ ；因为 $\log_b a = \frac{1}{x}$ ，方程变为 $x + \frac{6}{x} = 5$ 。

Set $x = \log_a b;$ since $\log_b a = \frac{1}{x},$ the equation becomes $x + \frac{6}{x} = 5$

大提示：

两种情形是 $b = a^2$ 和 $b = a^3$ ；分别数出有多少 $a \ge 2$ 能使 $b \le 2005$ 。

The two cases are $b = a^2$ and $b = a^3;$ count how many $a \ge 2$ keep $b \le 2005$ in each

解答：

令 $x = \log_a b$ 。因为 $\log_b a = \frac{1}{x}$ ，方程变为 $x + \frac{6}{x} = 5$ ，即 $x^2 - 5x + 6 = 0$ ，所以 $x = 2$ 或 $x = 3$ 。这意味着 $b = a^2$ 或 $b = a^3$ 。

对 $b = a^2 \le 2005$ ，需要 $2 \le a \le 44$ （因为 $44^2 = 1936$ 且 $45^2 = 2025$ ），得到 $43$ 个有序对。对 $b = a^3 \le 2005$ ，需要 $2 \le a \le 12$ （因为 $12^3 = 1728$ 且 $13^3 = 2197$ ），得到 $11$ 个有序对。

总共有 $43 + 11 = 54$ 个有序对。

Let $x = \log_a b.$ Since $\log_b a = \frac{1}{x},$ the equation becomes $x + \frac{6}{x} = 5,$ i.e. $x^2 - 5x + 6 = 0,$ so $x = 2$ or $x = 3.$ That means $b = a^2$ or $b = a^3.$

For $b = a^2 \le 2005$ we need $2 \le a \le 44$ (since $44^2 = 1936$ and $45^2 = 2025$ ), giving $43$ pairs. For $b = a^3 \le 2005$ we need $2 \le a \le 12$ (since $12^3 = 1728$ and $13^3 = 2197$ ), giving $11$ pairs.

In total there are $43 + 11 = 54$ ordered pairs.

6.

一叠 $2n$ 张卡片从上到下连续编号为 $1$ 到 $2n$ 。取走上面的 $n$ 张卡片，保持顺序，形成牌堆 $A$ 。剩下的卡片形成牌堆 $B$ 。现在把卡片重新叠成一叠，方法是分别从牌堆 $B$ 和牌堆 $A$ 的顶部交替取牌。在这个过程中，编号为 $(n + 1)$ 的卡片是新牌堆的底牌，编号为 $1$ 的卡片放在它上面，如此继续，直到牌堆 $A$ 和 $B$ 都用完。如果重新叠牌后，每个牌堆中至少有一张卡片仍占据它在原来整叠牌中的位置，则称这叠牌为神奇牌堆。例如，八张卡片形成一叠神奇牌，因为编号为 $3$ 和编号为 $6$ 的卡片保留了原来的位置。求在编号为 $131$ 的卡片保留原位置的神奇牌堆中，卡片的张数。

The cards in a stack of $2n$ cards are numbered consecutively from $1$ through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are now restacked into a single stack by taking cards alternately from the tops of pile $B$ and pile $A,$ respectively. In this process, card number $(n + 1)$ is the bottom card of the new stack, card number $1$ is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is called magical. For example, eight cards form a magical stack because cards number $3$ and number $6$ retain their original positions. Find the number of cards in the magical stack in which card number $131$ retains its original position.

答案

知识点：排列奇偶性一次方程

难度评级：2560

小提示：

重新叠牌后，牌堆 $A$ 的卡片以相反顺序占据从顶部数的奇数位置，牌堆 $B$ 的卡片占据偶数位置。

After restacking, pile $A$ ’s cards fill the odd positions from the top in reverse order, and pile $B$ ’s fill the even positions

大提示：

卡片 $131$ 必须来自牌堆 $A$ ，而牌堆 $A$ 中原位置为 $i$ 的卡片会落到位置 $2(n-i)+1$ 。

Card $131$ must come from pile $A,$ and a pile- $A$ card at position $i$ lands at position $2(n-i)+1$

解答：

从底部往上读，新牌堆为 $n+1,\ 1,\ n+2,\ 2,\ \ldots,\ 2n,\ n$ 。因此牌堆 $B$ 的卡片以相反顺序占据从顶部数的偶数位置，牌堆 $A$ 的卡片以相反顺序占据奇数位置：原位置 $i \le n$ （牌堆 $A$ ）的卡片移到位置 $2(n - i) + 1$ ，而原位置 $i \gt n$ （牌堆 $B$ ）的卡片移到位置 $2(2n - i) + 2$ 。

由于 $131$ 是奇数，编号为 $131$ 的卡片若要保持原位置，只能来自牌堆 $A$ ，因而 $131 = 2(n - 131) + 1$ ，解得 $n = 196$ 。确实， $131 \le 196$ ，且这叠牌是神奇的，因为牌堆 $B$ 中编号为 $262$ 的卡片也保持不动： $2(2n - 262) + 2 = 262$ 。这叠牌共有 $2n = 392$ 张。

The new stack, read from the bottom up, is $n+1,\ 1,\ n+2,\ 2,\ \ldots,\ 2n,\ n.$ So pile $B$ ’s cards occupy the even positions from the top in reverse order, and pile $A$ ’s cards occupy the odd positions in reverse order: a card at original position $i \le n$ (pile $A$ ) moves to position $2(n - i) + 1,$ while a card at position $i \gt n$ (pile $B$ ) moves to position $2(2n - i) + 2.$

Since $131$ is odd, card $131$ can keep its position only if it comes from pile $A,$ so $131 = 2(n - 131) + 1,$ which gives $n = 196.$ Indeed $131 \le 196,$ and the stack is magical because card $262$ from pile $B$ also stays fixed: $2(2n - 262) + 2 = 262.$ The stack has $2n = 392$ cards.

7.

设 $\begin{aligned} x &= \scriptsize \frac{4}{(\sqrt{5} + 1)(\sqrt[4]{5} + 1)(\sqrt[8]{5} + 1)(\sqrt[16]{5} + 1)} \end{aligned}\text{。}$ 求 $(x + 1)^{48}$ 。

Let $\begin{aligned} x &= \scriptsize \frac{4}{(\sqrt{5} + 1)(\sqrt[4]{5} + 1)(\sqrt[8]{5} + 1)(\sqrt[16]{5} + 1)}. \end{aligned}$ Find $(x + 1)^{48}.$

答案

知识点：裂项相消平方差根式

难度评级：2340

小提示：

令 $y = \sqrt[16]{5}$ ，并将分子和分母同乘以 $y - 1$ 。

Let $y = \sqrt[16]{5}$ and multiply the numerator and denominator by $y - 1$

大提示：

反复使用平方差公式会把分母化简为 $y^{16} - 1 = 4$ 。

Repeated difference of squares collapses the denominator to $y^{16} - 1 = 4$

解答：

令 $y = \sqrt[16]{5}$ 。分子和分母同乘以 $y - 1$ ，反复使用平方差公式后，分母中的各项会逐次相消： $\begin{aligned} x &= \scriptsize \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} \\ &= \frac{4(y-1)}{y^{16} - 1} \\ &= \frac{4(y-1)}{4} \\ &= y - 1 \end{aligned}\text{。}$

因此 $x + 1 = y = 5^{\frac{1}{16}}$ ，且 $(x+1)^{48} = 5^{\frac{48}{16}} = 5^3 = 125$ 。

Let $y = \sqrt[16]{5}.$ Multiplying the numerator and denominator by $y - 1$ telescopes the denominator by repeated difference of squares: $\begin{aligned} x &= \scriptsize \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} \\ &= \frac{4(y-1)}{y^{16} - 1} \\ &= \frac{4(y-1)}{4} \\ &= y - 1. \end{aligned}$

Hence $x + 1 = y = 5^{\frac{1}{16}},$ and $(x+1)^{48} = 5^{\frac{48}{16}} = 5^3 = 125.$

8.

圆 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 外切，并且它们都内切于圆 $\mathcal{C}_3$ 。 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 的半径分别为 $4$ 和 $10$ ，且三个圆的圆心共线。 $\mathcal{C}_3$ 的一条弦同时也是 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 的一条公外切线。已知这条弦的长度为 $\frac{m\sqrt{n}}{p}$ ，其中 $m$ 、 $n$ 和 $p$ 是正整数， $m$ 与 $p$ 互质，且 $n$ 不被任何质数的平方整除，求 $m + n + p$ 。

Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are externally tangent, and they are both internally tangent to circle $\mathcal{C}_3.$ The radii of $\mathcal{C}_1$ and $\mathcal{C}_2$ are $4$ and $10,$ respectively, and the centers of the three circles are all collinear. A chord of $\mathcal{C}_3$ is also a common external tangent of $\mathcal{C}_1$ and $\mathcal{C}_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

答案

知识点：相切圆弦勾股定理

难度评级：2710

小提示：

相切条件迫使 $\mathcal{C}_3$ 的半径为 $14$ ，且其圆心位于连接另外两个圆心的线段上，距 $\mathcal{C}_1$ 的圆心为 $10$ 。

Tangency forces $\mathcal{C}_3$ to have radius $14,$ with its center on the segment joining the other two centers, $10$ from $\mathcal{C}_1$ ’s center

大提示：

共线圆心到切线的距离线性变化：中间圆心的距离为 $4 + \frac{10}{14}(10 - 4)$ 。然后使用半弦构成的直角三角形。

Distances from the collinear centers to the tangent line vary linearly: the middle one is $4 + \frac{10}{14}(10 - 4).$ Then use the half-chord right triangle.

解答：

设 $P_1$ 、 $P_2$ 、 $P_3$ 为三个圆的圆心， $R$ 为 $\mathcal{C}_3$ 的半径。外切给出 $P_1P_2 = 4 + 10 = 14$ ，内切给出 $P_3P_1 = R - 4$ 和 $P_3P_2 = R - 10$ 。由于圆心共线， $(R - 4) + (R - 10) = 14$ ，所以 $R = 14$ ，且 $P_3$ 位于 $\overline{P_1P_2}$ 上，满足 $P_3P_1 = 10$ 和 $P_3P_2 = 4$ 。

向弦作垂线 $P_1X$ ， $P_2Y$ ， $P_3Z$ ，于是 $P_1X = 4$ ， $P_2Y = 10$ ，且 $Z$ 是弦的中点。沿着直线 $P_1P_2$ 移动时，到切线的距离线性变化，而 $P_3$ 位于从 $P_1$ 到 $P_2$ 的 $\frac{10}{14}$ 处，所以 $P_3Z = 4 + \frac{10}{14}(10 - 4) = \frac{58}{7}\text{。}$

半弦长为 $\sqrt{14^2 - \left(\frac{58}{7}\right)^2}$ $= \frac{\sqrt{9604 - 3364}}{7}$ $= \frac{4\sqrt{390}}{7}$ ，所以弦长为 $\frac{8\sqrt{390}}{7}$ 。由于 $390 = 2 \cdot 3 \cdot 5 \cdot 13$ 无平方因子且 $\gcd(8, 7) = 1$ ，答案为 $m + n + p = 8 + 390 + 7 = 405$ 。

Let $P_1,$ $P_2,$ $P_3$ be the centers of the circles and $R$ the radius of $\mathcal{C}_3.$ External tangency gives $P_1P_2 = 4 + 10 = 14,$ and internal tangency gives $P_3P_1 = R - 4$ and $P_3P_2 = R - 10.$ Since the centers are collinear, $(R - 4) + (R - 10) = 14,$ so $R = 14$ and $P_3$ lies on $\overline{P_1P_2}$ with $P_3P_1 = 10$ and $P_3P_2 = 4.$

Drop perpendiculars $P_1X,$ $P_2Y,$ $P_3Z$ to the chord, so $P_1X = 4,$ $P_2Y = 10,$ and $Z$ is the midpoint of the chord. The distance from a point moving along line $P_1P_2$ to the tangent line changes linearly, and $P_3$ is $\frac{10}{14}$ of the way from $P_1$ to $P_2,$ so $P_3Z = 4 + \frac{10}{14}(10 - 4) = \frac{58}{7}.$

The half-chord is $\sqrt{14^2 - \left(\frac{58}{7}\right)^2}$ $= \frac{\sqrt{9604 - 3364}}{7}$ $= \frac{4\sqrt{390}}{7},$ so the chord has length $\frac{8\sqrt{390}}{7}.$ Since $390 = 2 \cdot 3 \cdot 5 \cdot 13$ is squarefree and $\gcd(8, 7) = 1,$ the answer is $m + n + p = 8 + 390 + 7 = 405.$

9.

有多少个正整数 $n$ 不超过 $1000$ ，使得 $\begin{aligned} &(\sin t + i \cos t)^n \\ &= \sin nt + i \cos nt \end{aligned}$ 对所有实数 $t$ 都成立？

For how many positive integers $n$ less than or equal to $1000$ is $\begin{aligned} &(\sin t + i \cos t)^n \\ &= \sin nt + i \cos nt \end{aligned}$ true for all real $t?$

答案

知识点：棣莫弗定理复数三角恒等式

难度评级：2460

小提示：

写成 $\sin t + i\cos t = i(\cos t - i\sin t)$ 且 $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt)$ 。

Write $\sin t + i\cos t = i(\cos t - i\sin t)$ and $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt)$

大提示：

由棣莫弗定理，方程化为 $i^n = i$ ，因此 $n$ 必须比 $4$ 的某个倍数大 $1$ 。

By de Moivre the equation reduces to $i^n = i,$ so $n$ must be $1$ more than a multiple of $4$

解答：

因为 $\sin t + i\cos t = i(\cos t - i \sin t)$ 且 $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt)$ ，由棣莫弗定理（应用于角 $-t$ ）可得 $\begin{aligned} &(\sin t + i \cos t)^n \\ &= i^n(\cos t - i\sin t)^n \\ &= i^n(\cos nt - i \sin nt) \end{aligned}\text{。}$

所以该等式对所有实数 $t$ 成立，当且仅当 $i^n = i$ ，也就是 $n \equiv 1 \pmod 4$ 。数值 $n = 1, 5, 9, \ldots, 997$ 给出不超过 $1000$ 的 $250$ 个正整数。

Since $\sin t + i\cos t = i(\cos t - i \sin t)$ and $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt),$ de Moivre’s theorem (applied to angle $-t$ ) gives $\begin{aligned} &(\sin t + i \cos t)^n \\ &= i^n(\cos t - i\sin t)^n \\ &= i^n(\cos nt - i \sin nt). \end{aligned}$

So the equation holds for all real $t$ exactly when $i^n = i,$ that is, when $n \equiv 1 \pmod 4.$ The values $n = 1, 5, 9, \ldots, 997$ give exactly $250$ positive integers up to $1000.$

10.

已知 $\mathcal{O}$ 是一个正八面体， $\mathcal{C}$ 是以 $\mathcal{O}$ 各面的中心为顶点的立方体，且 $\mathcal{O}$ 的体积与 $\mathcal{C}$ 的体积之比为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

Given that $\mathcal{O}$ is a regular octahedron, that $\mathcal{C}$ is the cube whose vertices are the centers of the faces of $\mathcal{O},$ and that the ratio of the volume of $\mathcal{O}$ to that of $\mathcal{C}$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

知识点：立体几何体积坐标几何

难度评级：2450

小提示：

将八面体的顶点放在 $(\pm 1, 0, 0)$ 、 $(0, \pm 1, 0)$ 、 $(0, 0, \pm 1)$ ，并把它看成两个四棱锥。

Place the octahedron’s vertices at $(\pm 1, 0, 0),$ $(0, \pm 1, 0),$ $(0, 0, \pm 1)$ and view it as two square pyramids

大提示：

每个面的重心是其三个顶点的平均值，所以立方体顶点为 $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right)$ 。

Each face centroid is the average of its three vertices, so the cube’s vertices are $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right)$

解答：

将八面体的顶点放在 $(\pm 1, 0, 0)$ 、 $(0, \pm 1, 0)$ 、 $(0, 0, \pm 1)$ 。它由两个四棱锥沿着顶点为 $(\pm 1, 0, 0)$ 和 $(0, \pm 1, 0)$ 的正方形拼合而成；该正方形面积为 $2$ ，每个四棱锥的高为 $1$ ，所以 $V_{\mathcal{O}} = 2 \cdot \frac{1}{3} \cdot 2 \cdot 1 = \frac{4}{3}\text{。}$

每个面的重心是该面三个顶点的平均值，例如 $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ ，因此立方体顶点为 $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right)$ 。它的边长为 $\frac{2}{3}$ ，体积为 $\frac{8}{27}$ 。

比值为 $\frac{\frac{4}{3}}{\frac{8}{27}} = \frac{9}{2}$ ，所以 $m + n = 9 + 2 = 11$ 。

Place the octahedron’s vertices at $(\pm 1, 0, 0),$ $(0, \pm 1, 0),$ $(0, 0, \pm 1).$ It is two square pyramids glued along the square with vertices $(\pm 1, 0, 0)$ and $(0, \pm 1, 0),$ which has area $2,$ and each pyramid has height $1,$ so $V_{\mathcal{O}} = 2 \cdot \frac{1}{3} \cdot 2 \cdot 1 = \frac{4}{3}.$

Each face centroid is the average of that face’s three vertices, e.g. $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right),$ so the cube has vertices $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right).$ Its edge is $\frac{2}{3}$ and its volume is $\frac{8}{27}.$

The ratio is $\frac{\frac{4}{3}}{\frac{8}{27}} = \frac{9}{2},$ so $m + n = 9 + 2 = 11.$

11.

设 $m$ 是正整数，且 $a_0$ 、 $a_1$ 、 $\ldots$ 、 $a_m$ 是一列实数，满足 $a_0 = 37$ 、 $a_1 = 72$ 、 $a_m = 0$ ，并且对 $k = 1$ 、 $2$ 、 $\ldots$ 、 $m - 1$ 都有 $a_{k+1} = a_{k-1} - \frac{3}{a_k}\text{。}$ 求 $m$ 。

Let $m$ be a positive integer, and let $a_0,$ $a_1,$ $\ldots,$ $a_m$ be a sequence of real numbers such that $a_0 = 37,$ $a_1 = 72,$ $a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac{3}{a_k}$ for $k = 1,$ $2,$ $\ldots,$ $m - 1.$ Find $m.$

答案

知识点：递推等差数列代数变形

难度评级：2520

小提示：

将递推式乘以 $a_k$ ，可看出乘积 $a_k a_{k-1}$ 每一步减少 $3$ 。

Multiply the recurrence by $a_k$ to see that the products $a_k a_{k-1}$ drop by $3$ at each step

大提示：

$a_1 a_0 = 2664 = 3 \cdot 888$ ，且 $a_m = 0$ 恰好在乘积 $a_m a_{m-1}$ 达到 $0$ 时发生。

$a_1 a_0 = 2664 = 3 \cdot 888,$ and $a_m = 0$ exactly when the product $a_m a_{m-1}$ reaches $0$

解答：

将递推式乘以 $a_k$ ，得 $a_{k+1} a_k = a_k a_{k-1} - 3$ ，所以乘积 $b_k = a_k a_{k-1}$ 构成公差为 $-3$ 的等差数列。由于 $b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888$ ，得 $\begin{aligned} b_k &= 2664 - 3(k - 1) \\ &= 3(889 - k) \end{aligned}\text{。}$

因此当 $k \le 888$ 时 $b_k \gt 0$ ，所以 $a_{889}$ 之前没有任何一项为零（递推式也不会除以零），而 $b_{889} = a_{889} a_{888} = 0$ 且 $a_{888} \ne 0$ 。因此 $a_{889} = 0$ ，所以 $m = 889$ 。

Multiplying the recurrence by $a_k$ gives $a_{k+1} a_k = a_k a_{k-1} - 3,$ so the products $b_k = a_k a_{k-1}$ form an arithmetic sequence with common difference $-3.$ Since $b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888,$ we get $\begin{aligned} b_k &= 2664 - 3(k - 1) \\ &= 3(889 - k). \end{aligned}$

Thus $b_k \gt 0$ for $k \le 888,$ so no term before $a_{889}$ can vanish (and the recurrence never divides by zero), while $b_{889} = a_{889} a_{888} = 0$ with $a_{888} \ne 0.$ Hence $a_{889} = 0,$ and $m = 889.$

12.

正方形 $ABCD$ 的中心为 $O$ ， $AB = 900$ 。点 $E$ 和 $F$ 在 $\overline{AB}$ 上，且 $AE \lt BF$ ， $E$ 在 $A$ 与 $F$ 之间， $m\angle EOF = 45^\circ$ ，并且 $EF = 400$ 。已知 $BF = p + q\sqrt{r}$ ，其中 $p$ 、 $q$ 、 $r$ 是正整数，且 $r$ 不被任何质数的平方整除，求 $p + q + r$ 。

Square $ABCD$ has center $O,$ $AB = 900,$ $E$ and $F$ are on $\overline{AB}$ with $AE \lt BF$ and $E$ between $A$ and $F,$ $m\angle EOF = 45^\circ,$ and $EF = 400.$ Given that $BF = p + q\sqrt{r},$ where $p,$ $q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$

答案

知识点：三角恒等式韦达定理正方形（几何）

难度评级：3060

小提示：

令 $G$ 为 $\overline{AB}$ 的中点；则 $OG = 450$ ，且射线 $OG$ 将 $45^\circ$ 角分成 $\alpha + \beta$ 。

Let $G$ be the midpoint of $\overline{AB};$ then $OG = 450,$ and ray $OG$ splits the $45^\circ$ angle into $\alpha + \beta$

大提示：

由 $\tan\alpha + \tan\beta = \frac{8}{9}$ 和 $\tan(\alpha + \beta) = 1$ ，得 $\tan\alpha\tan\beta = \frac{1}{9}$ ；然后解一个二次方程。

From $\tan\alpha + \tan\beta = \frac{8}{9}$ and $\tan(\alpha + \beta) = 1,$ get $\tan\alpha\tan\beta = \frac{1}{9};$ then solve a quadratic

解答：

令 $G$ 为 $\overline{AB}$ 的中点，则 $OG \perp AB$ 且 $OG = 450$ 。设 $\alpha = \angle EOG$ 、 $\beta = \angle FOG$ ，它们位于射线 $OG$ 的两侧，则 $EG = 450\tan\alpha$ ， $FG = 450\tan\beta$ ，且 $\alpha + \beta = 45^\circ$ 。由 $EG + FG = EF = 400$ ，得 $\tan\alpha + \tan\beta = \frac{8}{9}$ 。

正切加法公式给出 $1 = \tan 45^\circ = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\text{，}$ 所以 $\tan\alpha\tan\beta = 1 - \frac{8}{9} = \frac{1}{9}$ 。因此 $\tan\alpha$ 和 $\tan\beta$ 是 $9t^2 - 8t + 1 = 0$ 的根，即 $\frac{4 \pm \sqrt{7}}{9}$ 。

由于 $AE = 450 - EG$ 且 $BF = 450 - FG$ ，条件 $AE \lt BF$ 意味着 $EG \gt FG$ ，因此 $\tan\beta = \frac{4 - \sqrt{7}}{9}$ 。于是 $BF = 450 - 450 \cdot \frac{4 - \sqrt{7}}{9}$ $= 250 + 50\sqrt{7}$ ，且 $p + q + r = 250 + 50 + 7 = 307$ 。

Let $G$ be the midpoint of $\overline{AB},$ so $OG \perp AB$ and $OG = 450.$ With $\alpha = \angle EOG$ and $\beta = \angle FOG$ on either side of ray $OG,$ we have $EG = 450\tan\alpha,$ $FG = 450\tan\beta,$ and $\alpha + \beta = 45^\circ.$ From $EG + FG = EF = 400,$ we get $\tan\alpha + \tan\beta = \frac{8}{9}.$

The tangent addition formula gives $1 = \tan 45^\circ = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta},$ so $\tan\alpha\tan\beta = 1 - \frac{8}{9} = \frac{1}{9}.$ Hence $\tan\alpha$ and $\tan\beta$ are the roots of $9t^2 - 8t + 1 = 0,$ namely $\frac{4 \pm \sqrt{7}}{9}.$

Since $AE = 450 - EG$ and $BF = 450 - FG,$ the condition $AE \lt BF$ means $EG \gt FG,$ so $\tan\beta = \frac{4 - \sqrt{7}}{9}.$ Then $BF = 450 - 450 \cdot \frac{4 - \sqrt{7}}{9}$ $= 250 + 50\sqrt{7},$ and $p + q + r = 250 + 50 + 7 = 307.$

13.

设 $P(x)$ 是一个整系数多项式，满足 $P(17) = 10$ 且 $P(24) = 17$ 。已知方程 $P(n) = n + 3$ 有两个不同的整数解 $n_1$ 和 $n_2$ ，求乘积 $n_1 \cdot n_2$ 。

Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17) = 10$ and $P(24) = 17.$ Given that the equation $P(n) = n + 3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1 \cdot n_2.$

答案

知识点：多项式整除性因数

难度评级：2760

小提示：

考虑 $S(x) = P(x) - x - 3$ ，它在 $x = 17$ 和 $x = 24$ 处都等于 $-10$ 。

Consider $S(x) = P(x) - x - 3,$ which equals $-10$ at both $x = 17$ and $x = 24$

大提示：

若 $S(n) = 0$ ，则 $(n-17)(n-24)$ 整除 $10$ ；这两个因子都是整数且相差 $7$ 。

If $S(n) = 0,$ then $(n-17)(n-24)$ divides $10;$ those two factors are integers differing by $7$

解答：

令 $S(x) = P(x) - x - 3$ ，则 $S(17) = S(24) = -10$ 。由于 $S(x) + 10$ 有整数系数并在 $17$ 和 $24$ 处为零， $\begin{aligned} S(x) &= -10 \\ &\quad {}+ (x - 17)(x - 24)Q(x) \end{aligned}$ 其中 $Q$ 是某个整系数多项式。

若某个整数 $n$ 满足 $P(n) = n + 3$ ，则 $(n-17)(n-24)Q(n) = 10$ ，所以 $(n-17)(n-24)$ 整除 $10$ 。因子 $n - 17$ 和 $n - 24$ 是相差 $7$ 的整数，其乘积整除 $10$ ，所以它们为 $\{2, -5\}$ 或 $\{5, -2\}$ ，得到 $n = 19$ 和 $n = 22$ 。这两个值确实能同时成为解，例如取 $P(x) = x - 7$ $- (x-17)(x-24)$ 。

因此 $n_1 \cdot n_2 = 19 \cdot 22 = 418$ 。

Let $S(x) = P(x) - x - 3,$ so $S(17) = S(24) = -10.$ Since $S(x) + 10$ has integer coefficients and vanishes at $17$ and $24,$ $\begin{aligned} S(x) &= -10 \\ &\quad {}+ (x - 17)(x - 24)Q(x) \end{aligned}$ for some polynomial $Q$ with integer coefficients.

If $P(n) = n + 3$ for an integer $n,$ then $(n-17)(n-24)Q(n) = 10,$ so $(n-17)(n-24)$ divides $10.$ The factors $n - 17$ and $n - 24$ are integers differing by $7$ whose product divides $10,$ so they are $\{2, -5\}$ or $\{5, -2\},$ giving $n = 19$ and $n = 22.$ Both occur, for example, for $P(x) = x - 7$ $- (x-17)(x-24).$

Hence $n_1 \cdot n_2 = 19 \cdot 22 = 418.$

14.

在三角形 $ABC$ 中， $AB = 13$ 、 $BC = 15$ 、 $CA = 14$ 。点 $D$ 在 $\overline{BC}$ 上，且 $CD = 6$ 。点 $E$ 在 $\overline{BC}$ 上，使得 $\angle BAE \cong \angle CAD$ 。已知 $BE = \frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数，求 $q$ 。

In triangle $ABC,$ $AB = 13,$ $BC = 15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD = 6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE \cong \angle CAD.$ Given that $BE = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers, find $q.$

答案

知识点：面积比三角学比与比例

难度评级：3060

小提示：

比较 $\frac{BD}{DC}$ 与 $\frac{BE}{EC}$ ：用 $\frac{1}{2}xy\sin\theta$ 计算以 $A$ 为顶点的三角形面积，并把每个比值写成面积之比。

Compare $\frac{BD}{DC}$ and $\frac{BE}{EC}$ by writing each as a ratio of triangle areas $\frac{1}{2}xy\sin\theta$ at vertex $A$

大提示：

相等的角会配对：两个比值相乘得到 $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}$ 。

The equal angles pair up: multiplying the two ratios gives $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}$

解答：

线段 $AD$ 将对边分成的比满足 $\begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD} \end{aligned}\text{，}$ 同理， $\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}$ 。

因为 $\angle BAE = \angle CAD$ ，也有 $\angle BAD = \angle CAE$ （二者都是这个公共角加上 $\angle EAD$ ），所以两个比值相乘时所有正弦都约去： $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}$ 。代入 $BD = 9$ 、 $DC = 6$ 、 $AB = 13$ 和 $AC = 14$ ，得 $\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}\text{。}$

因而 $BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}$ 。由于 $463$ 是质数，且不整除 $2535 = 3 \cdot 5 \cdot 13^2$ ，该分数已最简，所以 $q = 463$ 。

A cevian $AD$ splits the opposite side in the ratio $\begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD}, \end{aligned}$ and similarly $\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}.$

Since $\angle BAE = \angle CAD,$ we also have $\angle BAD = \angle CAE$ (each is that common angle plus $\angle EAD$ ), so multiplying the two ratios cancels all the sines: $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}.$ With $BD = 9,$ $DC = 6,$ $AB = 13,$ and $AC = 14,$ this gives $\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.$

Hence $BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}.$ Since $463$ is prime and does not divide $2535 = 3 \cdot 5 \cdot 13^2,$ the fraction is in lowest terms, and $q = 463.$

15.

设 $\omega_1$ 和 $\omega_2$ 分别表示圆 $x^2 + y^2 + 10x - 24y - 87 = 0$ 与 $x^2 + y^2 - 10x - 24y + 153 = 0$ 。在使直线 $y = ax$ 经过某个内切于 $\omega_1$ 且外切于 $\omega_2$ 的圆的圆心的所有正数 $a$ 中，令 $m$ 为最小值。已知 $m^2 = \frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数，求 $p + q$ 。

Let $\omega_1$ and $\omega_2$ denote the circles $x^2 + y^2 + 10x - 24y - 87 = 0$ and $x^2 + y^2 - 10x - 24y + 153 = 0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y = ax$ contains the center of a circle that is internally tangent to $\omega_1$ and externally tangent to $\omega_2.$ Given that $m^2 = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers, find $p + q.$

答案

知识点：椭圆相切圆配方法二次方程

难度评级：3160

小提示：

配方：两个圆的圆心为 $(-5, 12)$ 和 $(5, 12)$ ，半径为 $16$ 和 $4$ 。将两个相切距离条件相加。

Complete the square: the circles have centers $(-5, 12)$ and $(5, 12)$ with radii $16$ and $4.$ Add the two tangency distance conditions.

大提示：

圆心的轨迹是焦点为 $(\pm 5, 12)$ 的椭圆；代入 $y = ax$ ，并要求判别式非负。

The centers trace an ellipse with foci $(\pm 5, 12);$ substitute $y = ax$ and require the discriminant to be nonnegative

解答：

配方得 $\omega_1\colon (x+5)^2 + (y-12)^2 = 256$ 和 $\omega_2\colon (x-5)^2 + (y-12)^2 = 16$ ，圆心分别为 $F_1 = (-5, 12)$ 与 $F_2 = (5, 12)$ ，半径分别为 $16$ 和 $4$ 。若一个圆的圆心为 $P$ 、半径为 $r$ ，且它内切于 $\omega_1$ 、外切于 $\omega_2$ ，则 $PF_1 = 16 - r$ 且 $PF_2 = 4 + r$ ，所以 $PF_1 + PF_2 = 20$ 。

因此 $P$ 位于焦点为 $F_1$ 和 $F_2$ 、长轴为 $20$ 的椭圆上：半长轴为 $10$ ，中心到焦点的距离为 $5$ ，所以半短轴平方为 $100 - 25 = 75$ ，得到 $\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1\text{，}$ 即 $3x^2 + 4y^2 - 96y + 576 = 300$ 。代入 $y = ax$ ，得到 $(3 + 4a^2)x^2 - 96ax + 276 = 0$ 。

直线 $y = ax$ 含有这样的圆心，当且仅当这个二次方程有实根，即 $(96a)^2 - 4 \cdot 276\,(3 + 4a^2) \ge 0$ ，化简为 $4800a^2 \ge 3312$ ，所以 $a^2 \ge \frac{69}{100}$ 。最小的正 $a$ 满足 $m^2 = \frac{69}{100}$ ，因而 $p + q = 69 + 100 = 169$ 。

Completing the square gives $\omega_1\colon (x+5)^2 + (y-12)^2 = 256$ and $\omega_2\colon (x-5)^2 + (y-12)^2 = 16,$ with centers $F_1 = (-5, 12)$ and $F_2 = (5, 12)$ and radii $16$ and $4.$ If a circle with center $P$ and radius $r$ is internally tangent to $\omega_1$ and externally tangent to $\omega_2,$ then $PF_1 = 16 - r$ and $PF_2 = 4 + r,$ so $PF_1 + PF_2 = 20.$

Thus $P$ lies on the ellipse with foci $F_1$ and $F_2$ and major axis $20:$ the semimajor axis is $10,$ the center-to-focus distance is $5,$ so the semiminor axis squared is $100 - 25 = 75,$ giving $\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1,$ i.e. $3x^2 + 4y^2 - 96y + 576 = 300.$ Substituting $y = ax$ yields $(3 + 4a^2)x^2 - 96ax + 276 = 0.$

The line $y = ax$ contains such a center exactly when this quadratic has a real root, i.e. $(96a)^2 - 4 \cdot 276\,(3 + 4a^2) \ge 0,$ which simplifies to $4800a^2 \ge 3312,$ so $a^2 \ge \frac{69}{100}.$ The smallest positive such $a$ has $m^2 = \frac{69}{100},$ and $p + q = 69 + 100 = 169.$