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2009 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3:59

2009 AIME II真题

答案详细解析请参考文末

Problem 1

Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.

Problem 2

Suppose that $a$ , $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ , $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find $a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.$

Problem 3

In rectangle $ABCD$ , $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ .

Problem 4

A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ .

Problem 5

Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ , $C$ , and $D$ are all externally tangent to circle $E$ , which has radius $\dfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep}; draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5)); dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy]$

Problem 6

Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ .

Problem 7

Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ .

Problem 8

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m+n$ .

Problem 9

Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$ .

Problem 10

Four lighthouses are located at points $A$ , $B$ , $C$ , and $D$ . The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$ , the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$ , and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$ . To an observer at $A$ , the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$ , the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac{p\sqrt{r}}{q}$ , where $p$ , $q$ , and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p+q+r$ .

Problem 11

For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $mn$ .

Problem 12

From the set of integers $\{1,2,3,\dots,2009\}$ , choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$ . Find the maximum possible value of $k$ .

Problem 13

Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1,C_2,\dots,C_6$ . All chords of the form $\overline{AC_i}$ or $\overline{BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000$ .

Problem 14

The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}$ for $n \geq 0$ . Find the greatest integer less than or equal to $a_{10}$ .

Problem 15

Let $\overline{MN}$ be a diameter of a circle with diameter $1$ . Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\dfrac 35$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with the chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt t$ , where $r$ , $s$ , and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$ .

After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$ . The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$ . The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \boxed{114}$ .

First, we have: $x^{(\log_y z)^2} = x^{\left( (\log_y z)^2 \right) } = x^{(\log_y z) \cdot (\log_y z) } = \left( x^{\log_y z} \right)^{\log_y z}$

Now, let $x=y^w$ , then we have: $x^{\log_y z} = \left( y^w \right)^{\log_y z} = y^{w\log_y z} = y^{\log_y (z^w)} = z^w$

This is all we need to evaluate the given formula. Note that in our case we have $27=3^3$ , $49=7^2$ , and $\sqrt{11}=11^{1/2}$ . We can now compute:

$a^{(\log_3 7)^2} = \left( a^{\log_3 7} \right)^{\log_3 7} = 27^{\log_3 7} = (3^3)^{\log_3 7} = 7^3 = 343$

Similarly, we get $b^{(\log_7 11)^2} = (7^2)^{\log_7 11} = 11^2 = 121$

and $c^{(\log_{11} 25)^2} = (11^{1/2})^{\log_{11} 25} = 25^{1/2} = 5$

and therefore the answer is $343+121+5 = \boxed{469}$ .

$[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("$E$",E,N); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$F$",F,W); label("$100$",Q,W); [/asy]$ From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ , $\triangle FBA \sim \triangle BCA$ , and $\triangle FBA \sim \triangle ABE$ , so $\triangle ABE \sim \triangle BCA$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ . $AE=\frac{AD}{2}$ and $BC=AD$ , so $\frac {AD}{2AB}= \frac {AB}{AD}$ , or $(AD)^2=2(AB)^2$ , so $AD=AB \sqrt{2}$ , or $100 \sqrt{2}$ , so the answer is $\boxed{141}$

The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$ -st place), difference $d=-2$ , and number of terms $c$ . We can easily compute that this sum is equal to $c(n-c+1)$ .

Hence we have the equation $2009=c(n-c+1)$ , and we are looking for a solution $(n,c)$ , where both $n$ and $c$ are positive integers, $n\geq 2(c-1)$ , and $n$ is minimized. (The condition $n\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.)

The prime factorization of $2009$ is $2009=7^2 \cdot 41$ . Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$ :

$c=1$ , $n-c+1=2009$ , then $n=2009$ is a solution.
$c=7$ , $n-c+1=7\cdot 41=287$ , then $n=293$ is a solution.
$c=41$ , $n-c+1=7\cdot 7=49$ , then $n=89$ is a solution.
In each of the other three cases, $n$ will obviously be less than $2(c-1)$ , hence these are not valid solutions.

The smallest valid solution is therefore $c=41$ , $n=\boxed{089}$ .

Let $X$ be the intersection of the circles with centers $B$ and $E$ , and $Y$ be the intersection of the circles with centers $C$ and $E$ . Since the radius of $B$ is $3$ , $AX =4$ . Assume $AE$ = $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ . $AC = 8$ , and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $2009 AIME II真题及答案解析$ , we obtain

$(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60$ .

The $2$ and the $\cos 60$ terms cancel out:

$p^2 + 12p +36 = p^2 + 64 - 8p$

$12p+ 36 = 64 - 8p$

$p =\frac {28}{20} = \frac {7}{5}$ . The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$ , so the answer is $27 + 5 = \boxed{032}$ .

We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$ s and 5 $B$ s, thereby showing that there are ${10\choose 5}$ such sets.

Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$ -th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$ s and 9 $B$ s. As no two numbers were consecutive, we know that in our string no two $A$ s are consecutive. We can now remove exactly one $B$ from between each pair of $A$ s to get a string with 5 $A$ s and 5 $B$ s. And clearly this is a bijection, as from each string with 5 $A$ s and 5 $B$ s we can reconstruct one original set by reversing the construction.

Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$ , and the answer is $1750 \bmod 1000 = \boxed{750}$ .

First, note that $(2n)!! = 2^n \cdot n!$ , and that $(2n)!! \cdot (2n-1)!! = (2n)!$ .

We can now take the fraction $\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$ .

Now we can recognize that $\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \choose i}$ , hence this fraction is $\dfrac{{2i\choose i}}{2^{2i}}$ , and our sum turns into $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$ .

Let $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ . Obviously $c$ is an integer, and $S$ can be written as $\dfrac{c}{2^{2\cdot 2009}}$ . Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\leq 2\cdot 2009$ .

In other words, we just showed that $b=1$ . To determine $a$ , we need to determine the largest power of $2$ that divides $c$ .

Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$ .

We can now return to the observation that $(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$ .

It immediately follows that $p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)$ , and hence $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)$ .

Obviously, for $i\in\{1,2,\dots,2009\}$ the function $f(i)=2\cdot 2009 - i - p(i!)$ is is a strictly decreasing function. Therefore $p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)$ .

We can now compute $p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001$ . Hence $p(c)=2009-2001=8$ .

And thus we have $a=2\cdot 2009 - p(c) = 4010$ , and the answer is $\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}$ .

Additionally, once you count the number of factors of $2$ in the summation, one can consider the fact that, since $b$ must be odd, it has to take on a value of $1,3,5,7,$ or $9$ (Because the number of $2$ s in the summation is clearly greater than $1000$ , dividing by $10$ will yield a number greater than $100$ , and multiplying this number by any odd number greater than $9$ will yield an answer $>999$ , which cannot happen on the AIME.) Once you calculate the value of $4010$ , and divide by $10$ , $b$ must be equal to $1$ , as any other value of $b$ will result in an answer $>999$ . This gives $\boxed{401}$ as the answer.

There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$ .

How to compute $p$ ?

Suppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\cdot (1/6)$ , as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens iff his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$ .

Thus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \cdot (1/6)$ .

Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$ . Hence:

$\begin{align*} p & = \sum_{t=1}^\infty \left(\frac 56\right)^{2t} \cdot \frac 16 \\ & = \frac 16 \cdot \left(\frac 56\right)^2 \cdot \sum_{x=0}^\infty \left(\frac{25}{36}\right)^x \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac 1{1 - \dfrac{25}{36}} \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac{36}{11} \\ & = \frac {25}{66} \end{align*}$

Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\cdot \frac{25}{66} = 1 - \frac{25}{33} = \frac 8{33}$ , and the answer is $8+33 = \boxed{041}$ .

It is actually reasonably easy to compute $m$ and $n$ exactly.

First, note that if $4x+3y+2z=2009$ , then $y$ must be odd. Let $y=2y'-1$ . We get $4x + 6y' - 3 + 2z = 2009$ , which simplifies to $2x + 3y' + z = 1006$ . For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$ such that the equality holds. Hence we need to count the pairs $(x,y')$ .

For a fixed $y'$ , $x$ can be at most $\left\lfloor \dfrac{1005-3y'}2 \right\rfloor$ . Hence the number of solutions is

$\begin{align*} m & = \sum_{y'=1}^{334} \left\lfloor \dfrac{1005-3y'}2 \right\rfloor \\ & = 501 + 499 + 498 + 496 + \cdots + 6 + 4 + 3 + 1 \\ & = 1000 + 994 + \cdots + 10 + 4 \\ & = 83834 \end{align*}$

Similarly, we can compute that $n=82834$ , hence $(m-n)\bmod 1000 = 1000\bmod 1000 = \boxed{000}$ .

10.

Let $O$ be the intersection of $BC$ and $AD$ . By the Angle Bisector Theorem, $\frac {5}{BO}$ = $\frac {13}{CO}$ , so $BO$ = $5x$ and $CO$ = $13x$ , and $BO$ + $OC$ = $BC$ = $12$ , so $x$ = $\frac {2}{3}$ , and $OC$ = $\frac {26}{3}$ . Let $P$ be the foot of the altitude from $D$ to $OC$ . It can be seen that triangle $DOP$ is similar to triangle $AOB$ , and triangle $DPC$ is similar to triangle $ABC$ . If $DP$ = $15y$ , then $CP$ = $36y$ , $OP$ = $10y$ , and $OD$ = $5y\sqrt {13}$ . Since $OP$ + $CP$ = $46y$ = $\frac {26}{3}$ , $y$ = $\frac {13}{69}$ , and $AD$ = $\frac {60\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$ ). The answer is $60$ + $13$ + $23$ = $\boxed{096}$ .

11.

We have $\log m - \log k = \log \left( \frac mk \right)$ , hence we can rewrite the inequality as follows: $- \log n < \log \left( \frac mk \right) < \log n$ We can now get rid of the logarithms, obtaining: $\frac 1n < \frac mk < n$ And this can be rewritten in terms of $k$ as $\frac mn < k < mn$

From $k<mn$ it follows that the $50$ solutions for $k$ must be the integers $mn-1, mn-2, \dots, mn-50$ . This will happen if and only if the lower bound on $k$ is in a suitable range -- we must have $mn-51 \leq \frac mn < mn-50$ .

Obviously there is no solution for $n=1$ . For $n>1$ the left inequality can be rewritten as $m\leq\dfrac{51n}{n^2-1}$ , and the right one as $m > \dfrac{50n}{n^2-1}$ .

Remember that we must have $m\geq n$ . However, for $n\geq 8$ we have $\dfrac{51n}{n^2-1} < n$ , and hence $m<n$ , which is a contradiction. This only leaves us with the cases $n\in\{2,3,4,5,6\}$ .

For $n=2$ we have $\dfrac{100}3 < m \leq \dfrac{102}3$ with a single integer solution $m=\dfrac{102}3=34$ .
For $n=3$ we have $\dfrac{150}8 < m \leq \dfrac{153}8$ with a single integer solution $m=\dfrac{152}8=19$ .
For $n=4,5,6$ our inequality has no integer solutions for $m$ .

Therefore the answer is $34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}$ .

12.

Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\cdots+2k=k(2k+1)$ . On the other hand, as the sum of each pair is distinct and at most equal to $2009$ , the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \cdots + (2009-(k-1)) = k(4019-k)/2$ .

Hence we get a necessary condition on $k$ : For a solution to exist, we must have $k(4019-k)/2 \geq k(2k+1)$ . As $k$ is positive, this simplifies to $(4019-k)/2 \geq 2k+1$ , whence $5k\leq 4017$ , and as $k$ is an integer, we have $k\leq \lfloor 4017/5\rfloor = 803$ .

If we now find a solution with $k=803$ , we can be sure that it is optimal.

From the proof it is clear that we don't have much "maneuvering space", if we want to construct a solution with $k=803$ . We can try to use the $2k$ smallest numbers: $1$ to $2\cdot 803 = 1606$ . When using these numbers, the average sum will be $1607$ . Hence we can try looking for a nice systematic solution that achieves all sums between $1607-401=1206$ and $1607+401=2008$ , inclusive.

Such a solution indeed does exist, here is one:

Partition the numbers $1$ to $1606$ into four sequences:

$A=1,3,5,7,\dots,801,803$
$B=1205,1204,1203,1202,\dots,805,804$
$C=2,4,6,8,\dots,800,802$
$D=1606,1605,1604,1603,\dots,1207,1206$

Sequences $A$ and $B$ have $402$ elements each, and the sums of their corresponding elements are $1206,1207,1208,1209,\dots,1606,1607$ . Sequences $C$ and $D$ have $401$ elements each, and the sums of their corresponding elements are $1608,1609,1610,1611,\dots,2007,2008$ .

Thus we have shown that there is a solution for $k=\boxed{803}$ and that for larger $k$ no solution exists.

13.

Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\ldots, C_6$ are 6 of the 14th roots of unity. Let $\omega=\text{cis}\frac{360^{\circ}}{14}$ ; then $C_1,\ldots, C_6$ correspond to $\omega,\ldots, \omega^6$ . Let $C_1',\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\omega^8\ldots, \omega^{13}$ . Then the lengths of the segments are $|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|$ . Noting that $B$ represents 1 in the complex plane, the desired product is $\begin{align*} BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6\\ &= |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| \end{align*}$

for $x=1$ . However, the polynomial $(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$ . Hence $(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})=\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\cdots +x^2+1.$ Thus the product is $|x^{12}+\cdots +x^2+1|=7$ ( $x=1$ ) when the radius is 1, and the product is $2^{12}\cdot 7=28672$ . Thus the answer is $\boxed {672}$ .

14.

We can now simply start to compute the values $b_i$ by hand:

$\begin{align*} b_1 & = \frac 35 \\ b_2 & = \frac 45\cdot \frac 35 + \frac 35 \sqrt{1 - \left(\frac 35\right)^2} = \frac{24}{25} \\ b_3 & = \frac 45\cdot \frac {24}{25} + \frac 35 \sqrt{1 - \left(\frac {24}{25}\right)^2} = \frac{96}{125} + \frac 35\cdot\frac 7{25} = \frac{117}{125} \\ b_4 & = \frac 45\cdot \frac {117}{125} + \frac 35 \sqrt{1 - \left(\frac {117}{125}\right)^2} = \frac{468}{625} + \frac 35\cdot\frac {44}{125} = \frac{600}{625} = \frac{24}{25} \end{align*}$

We now discovered that $b_4=b_2$ . And as each $b_{i+1}$ is uniquely determined by $b_i$ , the sequence becomes periodic. In other words, we have $b_3=b_5=b_7=\cdots=\frac{117}{125}$ , and $b_2=b_4=\cdots=b_{10}=\cdots=\frac{24}{25}$ .

Therefore the answer is

$\begin{align*} \lfloor a_{10} \rfloor & = \left\lfloor 2^{10} b_{10} \right\rfloor = \left\lfloor \dfrac{1024\cdot 24}{25} \right\rfloor = \left\lfloor \dfrac{1025\cdot 24}{25} - \dfrac{24}{25} \right\rfloor \\ & = \left\lfloor 41\cdot 24 - \dfrac{24}{25} \right\rfloor = 41\cdot 24 - 1 = \boxed{983} \end{align*}$

15.

Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$ . Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$ . Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$ . Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem.

By the Angle Bisector Formula, $\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)$ $\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)$

As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$ , and finally $VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$ . Taking the derivative of $VW$ with respect to $\alpha$ , we arrive at $VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}$ Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$ . Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$ , we get