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2012 AIME II真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2018年5月5日下午3:48

2012 AIME II真题

答案解析请参考文末

Problem 1

Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ .

Problem 2

Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ , $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ .

Problem 3

At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements.

Problem 4

Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$ , where $p$ , $q$ , and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$ .

Problem 5

In the accompanying figure, the outer square $S$ has side length $40$ . A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S$ . From each midpoint of a side of $S$ , segments are drawn to the two closest vertices of $S'$ . The result is a four-pointed starlike figure inscribed in $S$ . The star figure is cut out and then folded to form a pyramid with base $S'$ . Find the volume of this pyramid.

$[asy] pair S1 = (20, 20), S2 = (-20, 20), S3 = (-20, -20), S4 = (20, -20); pair M1 = (S1+S2)/2, M2 = (S2+S3)/2, M3=(S3+S4)/2, M4=(S4+S1)/2; pair Sp1 = (7.5, 7.5), Sp2=(-7.5, 7.5), Sp3 = (-7.5, -7.5), Sp4 = (7.5, -7.5); draw(S1--S2--S3--S4--cycle); draw(Sp1--Sp2--Sp3--Sp4--cycle); draw(Sp1--M1--Sp2--M2--Sp3--M3--Sp4--M4--cycle); [/asy]$

Problem 6

Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ .

Problem 7

Let $S$ be the increasing sequence of positive integers whose binary representation has exactly $8$ ones. Let $N$ be the 1000th number in $S$ . Find the remainder when $N$ is divided by $1000$ .

Problem 8

The complex numbers $z$ and $w$ satisfy the system $z + \frac{20i}w = 5+i$ $w+\frac{12i}z = -4+10i$ Find the smallest possible value of $\vert zw\vert^2$ .

Problem 9

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Problem 10

Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$ .

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .

Problem 11

Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 12

For a positive integer $p$ , define the positive integer $n$ to be $p$ -safe if $n$ differs in absolute value by more than $2$ from all multiples of $p$ . For example, the set of $10$ -safe numbers is $\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}$ . Find the number of positive integers less than or equal to $10,000$ which are simultaneously $7$ -safe, $11$ -safe, and $13$ -safe.

Problem 13

Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ , $AD_1E_2$ , $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ , with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$ .

Problem 14

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$ .

Problem 15

Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ , $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{034}$

Call the common $r.$ Now since the $n$ th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \cdot r^{14} = b_1 \cdot r^{10} \longrightarrow r^4 = \frac{99}{27} = \frac{11}{3}.$ But $a_9$ equals $a_1 \cdot r^8 = a_1 \cdot (r^4)^2,$ so $a_9 = 27 \cdot \left(\frac{11}{3}\right)^2 = \boxed{363.}$

There are two cases:

Case 1: One man and one woman is chosen from each department.

Case 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department.

For the first case, in each department there are ${{2}\choose{1}} \times {{2}\choose{1}} = 4$ ways to choose one man and one woman. Thus there are $4^3 = 64$ total possibilities conforming to case 1.

For the second case, there is only ${{2}\choose{2}} = 1$ way to choose two professors of the same gender from a department, and again there are $4$ ways to choose one man and one woman. Thus there are $1 \cdot 1 \cdot 4 = 4$ ways to choose two men from one department, two women from another department, and one man and one woman from the third department. However, there are $3! = 6$ different department orders, so the total number of possibilities conforming to case 2 is $4 \cdot 6 = 24$ .

Summing these two values yields the final answer: $64 + 24 = \boxed{088}$ .

$[asy]draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label("$D$", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label("$a$", (0.6,0),dir(-90)); label("$b$", (3.6,0),dir(-90)); label("$c$", (6,0.7),dir(0));[/asy]$

Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.

Observe that $\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$ , and $\dfrac{b+c}{6.2}=1$ or $b+c=6.2$ . Subtracting the second equation from the first gives $2a=2.4$ , or $a=1.2$ .

Now, let us solve $b$ and $c$ . Note that $\dfrac{\sqrt{b^2+c^2}}{5}=1$ , or $b^2+c^2=25$ . We also have $b+c=6.2$ .

We have a system of equations: $\left\{\begin{array}{l}b+c=6.2\\ b^2+c^2=25\end{array}\right.$

Squaring the first equation gives $b^2+2bc+c^2=38.44$ , and subtracting the second from this gives $2bc=13.44$ . Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$ , or $b-c=3.4$ . Now we have the following two equations:

$\left\{\begin{array}{l}b+c=6.2\\ b-c=3.4\end{array}\right.$

Adding the equations and dividing by two gives $b=4.8$ , and it follows that $c=1.4$ .

The ratios we desire are therefore $1.4:6:4.8=7:30:24$ , and our answer is $7+30+24=\boxed{061}$ .

Note that in our diagram, we labeled the part of the bottom $b$ and the side $c$ . However, these labels are interchangeable. We can cancel out the case where the side is $4.8$ and the part of the bottom is $1.4$ by noting a restriction of the problem: "...a rectangular field whose longer side runs due west." If we had the side be $4.8$ , then the entire bottom would be $1.2+1.4=2.6$ , clearly less than $4.8$ and

The volume of this pyramid can be found by the equation $V=\frac{1}{3}bh$ , where $b$ is the base and $h$ is the height. The base is easy, since it is a square and has area $15^2=225$ .

To find the height of the pyramid, the height of the four triangles is needed, which will be called $h^\prime$ . By drawing a line through the middle of the larger square, we see that its length is equal to the length of the smaller rectangle and two of the triangle's heights. Then $40=2h^\prime +15$ , which means that $h^\prime=12.5$ .

When the pyramid is made, you see that the height is the one of the legs of a right triangle, with the hypotenuse equal to $h^\prime$ and the other leg having length equal to half of the side length of the smaller square, or $7.5$ . So, the Pythagorean Theorem can be used to find the height.

$h=\sqrt{12.5^2-7.5^2}=\sqrt{100}=10$

Finally, $V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750.}$

Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have

$|z^3| * |z^2 - (1+2i)|$

$=|z|^3 * |z^2 - (1+2i)|$

$=125|z^2 - (1+2i)|$

Thus we only need to maximize the value of $|z^2 - (1+2i)|$ .

To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$ . The unit vector in the complex plane in the desired direction is $\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i$ . Furthermore, we know that the magnitude of $z^2$ is $25$ , because the magnitude of $z$ is $5$ . From this information, we can find that $z^2 = \sqrt{5} (-5 - 10i)$

Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$ . Finally, $c+d = -375 + 500 = \boxed{125}$

Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is $11111111$ , which is the only way to choose 8 1's out of 8 spaces, or $\binom{8}{8}$ . What about 9 spaces? Well, all told, there are $\binom{9}{8}=9$ , which includes the first 1. Similarly, for 10 spaces, there are $\binom{10}{8}=45,$ which includes the first 9. For 11 spaces, there are $\binom{11}{8}=165$ , which includes the first 45. You're getting the handle. For 12 spaces, there are $\binom{12}{8}=495$ , which includes the first 165; for 13 spaces, there are $\binom{13}{8}=13 \cdot 99 > 1000$ , so we now know that $N$ has exactly 13 spaces, so the $2^{12}$ digit is 1.

Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the $1000-495=505th$ number. Well, $\binom{11}{7}=330$ , so we know that the $2^{11}$ digit also is 1, and we're left with finding the $505-330=175th$ number with 11 spaces and 6 1's. Now $\binom{10}{6}=210,$ which is too big, but $\binom{9}{6}=84.$ Thus, the $2^9$ digit is 1, and we're now looking for the $175-84=91st$ number with 9 spaces and 5 1's. Continuing the same process, $\binom{8}{5}=56$ , so the $2^8$ digit is 1, and we're left to look for the $91-56=35th$ number with 8 spaces and 4 1's. But here $\binom{7}{4}=35$ , so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of $N$ must be $01111000$ , and to summarize, $N=1101101111000$ in base $2$ . Therefore, $N = 8+16+32+64+256+512+2048+4096 \equiv 32 \pmod{1000}$ , and the answer is $032$ .

Multiplying the two equations together gives us $zw + 32i - \frac{240}{zw} = -30 + 46i$ and multiplying by $zw$ then gives us a quadratic in $zw$ : $(zw)^2 + (30-14i)zw - 240 =0.$ Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040.}$

Examine the first term in the expression we want to evaluate, $\frac{\sin 2x}{\sin 2y}$ , separately from the second term, $\frac{\cos 2x}{\cos 2y}$ .

The First Term

Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$ , we have:

$\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}$

The Second Term

Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2. Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$ , we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$ .

Equation 1 then becomes:

$\sin x = 3\sin y$ .

Equation 2 then becomes:

$\cos x = \frac{1}{2} \cos y$

Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:

$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$

Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:

$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$

Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$ .

So, $\sin^2 y = \frac{3}{35}$ .

Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$ ), we can solve for $\sin^2 x$ :

$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$

Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$ , we can solve for $\frac{\cos 2x}{\cos 2y}$ .

$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$

$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$

Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$ .

Now Back to the Solution!

Finally, $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$ .

So, the answer is $49+58=\boxed{107}$ .

10.

We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational.

Let $x = a + \frac{b}{c}$ where a,b,c are nonnegative integers and $0 \le b < c$ (essentially, x is a mixed number). Then,

$n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}$ Here it is sufficient for $\frac{ab}{c}$ to be an integer. We can use casework to find values of n based on the value of a:

$a = 0 \implies$ nothing because n is positive

$a = 1 \implies \frac{b}{c} = \frac{0}{1}$

$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$

The pattern continues up to $a = 31$ . Note that if $a = 32$ , then $n > 1000$ . However if $a = 31$ , the largest possible x is $31 + 30/31$ , in which $n$ is still less than $1000$ . Therefore the number of positive integers for n is equal to $1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}$

11.

After evaluating the first few values of $f_k (x)$ , we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$ . Since $1001 \equiv 2 \mod 3$ , $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$ . We set this equal to $x-3$ , i.e.

$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$ . The answer is thus $5+3 = \boxed{008.}$

12.

We see that a number $n$ is $p$ -safe if and only if the residue of $n \mod p$ is greater than $2$ and less than $p-2$ ; thus, there are $p-5$ residues $\mod p$ that a $p$ -safe number can have. Therefore, a number $n$ satisfying the conditions of the problem can have $2$ different residues $\mod 7$ , $6$ different residues $\mod 11$ , and $8$ different residues $\mod 13$ . The Chinese Remainder Theorem states that for a number $x$ that is $a$ (mod b) $c$ (mod d) $e$ (mod f) has one solution if $gcd(b,d,f)=1$ . For example, in our case, the number $n$ can be: 3 (mod 7) 3 (mod 11) 7 (mod 13) so since $gcd(7,11,13)$ =1, there is 1 solution for n for this case of residues of $n$ .

This means that by the Chinese Remainder Theorem, $n$ can have $2\cdot 6 \cdot 8 = 96$ different residues mod $7 \cdot 11 \cdot 13 = 1001$ . Thus, there are $960$ values of $n$ satisfying the conditions in the range $0 \le n < 10010$ . However, we must now remove any values greater than $10000$ that satisfy the conditions. By checking residues, we easily see that the only such values are $10007$ and $10006$ , so there remain $\fbox{958}$ values satisfying the conditions of the problem.

13.

Note that there are only two possible locations for points $D_1$ and $D_2$ , as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$ , so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect. Let $D_1$ be the point on the opposite side of $\overline{AB}$ from $C$ , and $D_2$ the point on the same side of $\overline{AB}$ as $C$ .

Let $\theta$ be the measure of angle $BAD_1$ (also the measure of angle $BAD_2$ ); by the Law of Cosines, $\begin{align*}\sqrt{11}^2 &= \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot \cos\theta\\ 11 &= 222(1 - \cos\theta)\end{align*}$

There are two equilateral triangles with $\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$ , and $E_2$ be the third vertex that is nearest to $C$ .

Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta$ ; by the Law of Cosines, $\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) (E_1C)\cos(120 + \theta)\\ &= 111 + 111 - 222\cos(120 + \theta)\end{align*}$ Angle $E_2AC = \theta$ ; by the Law of Cosines, $\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) (E_2C)\cos\theta\\ &= 111 + 111 - 222\,\cos\theta\end{align*}$

There are two equilateral triangles with $\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$ , and $E_4$ be the third vertex that is nearest to $C$ .

Angle $E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta$ ; by the Law of Cosines, $\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) (E_3C)\cos(120 - \theta)\\ &= 111 + 111 - 222\cos(120 - \theta)\end{align*}$ Angle $E_4AC = \theta$ ; by the Law of Cosines, $\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) (E_4C)\cos\theta \\ &= 111 + 111 - 222\cos\theta\end{align*}$

The solution is: $\begin{align*} \sum_{k=1}^4(CE_k)^2 &= (E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2\\ &= 222(1 - \cos(120 + \theta)) + 222(1 - \cos(120 - \theta)) + 222(1 - \cos\theta) + 222(1 - \cos\theta)\\ &= 222((1 - (\cos120\cos\theta - \sin120\sin\theta)) + (1 - (\cos120\cos\theta + \sin120\sin\theta)) + 2(1 -\cos\theta))\\ &= 222(1 - \cos120\cos\theta + \sin120\sin\theta + 1 - \cos120\cos\theta - \sin120\sin\theta + 2 - 2\cos\theta)\\ &= 222(1 + \frac{1}{2}\cos\theta + 1 + \frac{1}{2}\cos\theta + 2 - 2\cos\theta)\\ &= 222(4 - \cos\theta)\\ &= 666 + 222(1 - \cos\theta) \end{align*}$ Substituting $11$ for $222(1 - \cos\theta)$ gives the solution $666 + 11 = \framebox{677}.$

14.

Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)

Case 1: To create our groups of three, there are $\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}$ . In general, the number of ways we can arrange people within the rings to count properly is $\dfrac{(n-1)!}{2}$ , since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has $\dfrac{(3-1)!}{2}=1$ arrangements. Therefore, for this case, there are $(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!})(1)^3=280$

Case 2: For three and six, there are $\dbinom{9}{6}=84$ sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is $\dfrac{(6-1)!}{2}=60$ . This means there are $(84)(1)(60)=5040$ arrangements.

Case 3: For four and five, there are $\dbinom{9}{5}=126$ sets for the rings. Within the five, there are $\dfrac{4!}{2}=12$ , and within the four there are $\dfrac{3!}{2}=3$ arrangements. This means the total is $(126)(12)(3)=4536$ .

Case 4: For the nine case, there is $\dbinom{9}{9}=1$ arrangement for the ring. Within it, there are $\dfrac{8!}{2}=20160$ arrangements.

Summing the cases, we have $280+5040+4536+20160=30016 \to \boxed{016}$ .

15.

Use the angle bisector theorem to find $CD=\frac{21}{8}$ , $BD=\frac{35}{8}$ , and use the Stewart's Theorem to find $AD=\frac{15}{8}$ . Use Power of the Point to find $DE=\frac{49}{8}$ , and so $AE=8$ . Use law of cosines to find $\angle CAD = \frac{\pi} {3}$ , hence $\angle BAD = \frac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$ .

I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:

$AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.$ (1)

$AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.$ Adding these two and simplifying we get:

$EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF$ (2). Ah, but $\angle AFE = \angle ACE$ (since $F$ lies on $\omega$ ), and we can find $cos \angle ACE$ using the law of cosines:

$AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE$ , and plugging in $AE = 8, AC = 3, BE = BC = 7,$ we get $\cos \angle ACE = -1/7 = \cos \angle AFE$ .

Also, $\angle AEF = \angle DEF$ , and $\angle DFE = \pi/2$ (since $F$ is on the circle $\gamma$ with diameter $DE$ ), so $\cos \angle AEF = EF/DE = 8 \cdot EF/49$ .