AMC12 2009 A

There is $1$ hour and $60-34 = 26$ minutes between 10:34 AM and noon; and there is $1$ hour and $18$ minutes between noon and 1:18 PM. Hence the flight took $2$ hours and $26 + 18 = 44$ minutes,and 2+44=46s0 $h+m= 46\ \textbf{(A)}$.

We compute: \begin{align*} 1 + \frac {1}{1 + \frac {1}{1 + 1}} &= 1 + \frac {1}{1 + \frac {1}{1 + 1}} \\ &= 1 + \frac {1}{1 + \frac 12} \\ &= 1 + \frac {1}{\frac 32} \\ &= 1 + \frac 23 \\ &= \frac 53 \end{align*} This is choice $\boxed{\text{C}}$. Interesting sidenote: The continued fraction $1 + \frac {1}{1 + \frac {1}{1 + 1....}}$ is equal to the golden ratio, or $\frac{1+\sqrt{5}}{2}$.

We can rewrite the two given fractions as $\frac 3{12}$ and $\frac 9{12}$. (We multiplied all numerators and denominators by $3$.) Now it is obvious that the interval between them is divided into three parts by the fractions $\boxed{\frac 5{12}}$ and $\frac 7{12}$. The number we seek can be obtained as a weighted average of the two endpoints, where the closer one has weight $2$ and the further one $1$. We compute: \[\dfrac{ 2\cdot\frac 14 + 1\cdot\frac 34 }3 = \dfrac{ \frac 54 }3 = \boxed{\dfrac 5{12}}\]

Pre-Note: This solution is kinda just guessing, idk you decide. We can solve this problem by trying out numbers to get the answer choices and use the process of elimination to get the answer. Thinking for a few seconds, you can see that for 15 cents there are no possible ways to get it. Now you may be inclined to chose $\textbf{A}$, but we may have missed some way to get 15 cents, so we try the other answer choices. 25 cents can be made with 3 nickels and 1 dimes, 35 cents can be made with 3 dimes and 1 nickel, 45 cents can be made with 1 quarter, 1 dime, and 2 nickels, and finally, 55 cents can be made with 1 quarter and 3 dimes. Therefore the answer is $\boxed{\textbf{A}}$ MisW (talk) Minor Edits by yanchudeng Post-Note: This is my first time using LaTeX so it may look a little ugly.

Let the original cube have edge length $a$. Then its volume is $a^3$. The new box has dimensions $a-1$, $a$, and $a+1$, hence its volume is $(a-1)a(a+1) = a^3-a$. The difference between the two volumes is $a$. As we are given that the difference is $5$, we have $a=5$, and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{\text{(D)}}$.

We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{{E.}\ P^{2n} Q^m}$.

As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$. This solves to $x=4$. The first three terms then are $5$, $9$, and $13$. In general, the $n$th term is $1+4n$. Solving $1+4n=2009$, we get $n=\boxed{502}$.

The area of the outer square is $4$ times that of the inner square. Therefore the side of the outer square is $\sqrt 4 = 2$ times that of the inner square. Then the shorter side of the rectangle is $1/4$ of the side of the outer square, and the longer side of the rectangle is $3/4$ of the side of the outer square, hence their ratio is $\boxed{3}$.

As $f(x)=ax^2 + bx + c$, we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$. To compute $f(1)$, set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$.

By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$, and also $BD + CD > BC$, hence $BD > BC - CD = 17 - 5 = 12$. We get that $12 < BD < 14$, and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$.

Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$th triangular number is $\frac{n(n+1)}{2}$. Putting this together for $F_{20}$ this gives: $\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27 = 162$. Out of the numbers $1$ to $162$ the one with the largest sum of digits is $99$, and the sum is $9+9=18$. Hence the sum of digits will be at most $18$. Also, each number with this property is divisible by $6$, therefore it is divisible by $3$, and thus also its sum of digits is divisible by $3$. Thus, the number is divisible by $18$. We only have six possibilities left for the sum of the digits: $3$, $6$, $9$, $12$, $15$, and $18$, but since the number is divisible by $18$, the digits can only add to $9$ or $18$. This leads to the integers $18$, $36$, $54$, $72$, $90$, and $108$ being possibilities. We can check to see that $\boxed{1}$ solution: the number $54$ is the only solution that satisfies the conditions in the problem. We can write each integer between $1$ and $999$ inclusive as $\overline{abc}=100a+10b+c$ where $a,b,c\in\{0,1,\dots,9\}$ and $a+b+c>0$. The sum of digits of this number is $a+b+c$, hence we get the equation $100a+10b+c = 6(a+b+c)$. This simplifies to $94a + 4b - 5c = 0$. Clearly for $a>0$ there are no solutions, hence $a=0$ and we get the equation $4b=5c$. This obviously has only one valid solution $(b,c)=(5,4)$, hence the only solution is the number $54$! The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27 = 162$. Since the number is $6$ times the sum of its digits, it must be divisible by $6$, therefore also by $3$, therefore the sum of its digits must be divisible by $3$. With this in mind we can conclude that the number must be divisible by $18$, not just by $6$. Since the number is divisible by $18$, it is also divisible by $9$, therefore the sum of its digits is divisible by $9$, therefore the number is divisible by $54$, which leaves us with $54$, $108$ and $162$. Only $54$ is $6$ times its digits, hence the answer is $\boxed{1}$.

To answer the question we are asked, it is enough to compute $AC^2$ for two different angles, preferably for both extremes ($45$ and $60$ degrees). You can use the law of cosines to do so. Alternately, it is enough to compute $AC^2$ for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our $AC^2$ is the minimal or the maximal possible value of $AC^2$. Below we show a complete solution in which we also show that all possible values of $AC^2$ do indeed lie in the given interval. Let $C_1$ be the point the ship would reach if it turned $45^\circ$, and $C_2$ the point it would reach if it turned $60^\circ$. Obviously, $C_1$ is the furthest possible point from $A$, and $C_2$ is the closest possible point to $A$. Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$. We can find $AC_1^2$ and $AC_2^2$ as follows: Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$. The angles in the triangle $BD_1C_1$ are $45^\circ$, $45^\circ$, and $90^\circ$, hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$. Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$, $60^\circ$, and $90^\circ$, hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$. Hence we get: \[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\] \[AC_1^2 = AD_1^2 + D_1C_1^2 \]\[= (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 \]\[= 100 + 400/\sqrt 2 + 200 + 200 \]\[= 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\] Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{ \textbf{(D)}[700,800] }$. From the law of cosines, $500-400\cos120^\circ<AC^2<500-400\cos135^\circ\implies700<AC^2<500+200\sqrt{2}$. This is essentially the same solution as above. The answer is $\boxed{\textbf{(D)}}$.

Let's label the three points as $A=(0,0)$, $B=(1,1)$, and $C=(6m,0)$. Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$. Let $D$ be the point of intersection. The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the line $BC$. Hence they have equal areas if and only if $D$ is the midpoint of $BC$. The midpoint of the segment $BC$ has coordinates $\left( \frac{6m+1}2, \frac 12 \right)$. This point lies on the line $y=mx$ if and only if $\frac 12 = m \cdot \frac{6m+1}2$. This simplifies to $6m^2 + m - 1 = 0$. Using Vieta's formulas, we find that the sum of the roots is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$. For illustration, below are pictures of the situation for $m=1.5$, $m=0.5$, $m=1/3$, and $m=-1/2$. Note : In contests when you get problems like these, your AOPS knowledge will not be enough. AOPs books are just the basic, always prep for more.

We know that $i^x$ cycles every $4$ powers so we group the sum in $4$s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\] We can postulate that every group of $4$ is equal to $2-2i$. For 24 groups we thus, get $48-48i$ as our sum. We know the solution must lie near The next term is the $24*4+1=97$th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$) and our sum is now $48+49i$ so $n=97\Rightarrow\boxed{\mathbf{D}}$ is our answer

Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$. For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$, the distance between $C$ and $(3,0)$ must be exactly $r+1$. By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$, hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$. Simplifying, we obtain $r^2 - 8r + 8 = 0$. By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$. (We should actually solve for $r$ to verify that there are two distinct positive roots. In this case we get $r=4\pm 2\sqrt 2$. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.)

Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$. Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$. This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$. As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$. Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$.

The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$. For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$. For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$. For $k>4$ the value in the parentheses is odd, hence $N(k)=6$. This leaves the case $k=4$. We have $I_4 = 2^6 \left( 5^6 + 1 \right)$. The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$, we have $5^6 \equiv 1 \pmod 4$, and therefore $5^6 + 1 \equiv 2 \pmod 4$. Hence the largest power of $2$ that divides $5^6+1$ is $2^1$, and this gives us the desired maximum of the function $N$: $N(4) = \boxed{7}$.

In any regular polygon with side length $2$, consider the isosceles triangle formed by the center of the polygon $S$ and two consecutive vertices $X$ and $Y$. We are given that $XY=2$. Obviously $SX=SY=r$, where $r$ is the radius of the circumcircle. Let $T$ be the midpoint of $XY$. Then $XT=TY=1$, and $TS=\rho$, where $\rho$ is the radius of the incircle. Applying the Pythagorean theorem on the triangle $STX$, we get that $\rho^2 + 1 = r^2$. Then the area between the circumcircle and the incircle can be computed as $\pi r^2 - \pi \rho^2 = \pi r^2 - \pi (r^2 - 1) = \pi$. Hence $A=\pi$, $B=\pi$, and therefore $\boxed{\textbf{(C) }A=B}$.

Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$.

From the three zeroes, we have $p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)$. Then $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c$. Let's do each factor case by case: - $x^4 - (2009 + 9002\pi i) = 0$: Clearly, all the fourth roots are going to be complex. - $x^4 - 2009 = 0$: The real roots are $\pm \sqrt [4]{2009}$, and there are two complex roots. - $x^4 - 9002 = 0$: The real roots are $\pm \sqrt [4]{9002}$, and there are two complex roots. So the answer is $4 + 2 + 2 = 8\ \mathbf{(C)}$.

Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to precisely $\frac{1}{3}$ of the sides of the net. Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is $\frac{1}{2}$. Thus, the answer is$\frac {3\sqrt {3}}{8}$, and $a + b + c = 14\ \mathbf{(E)}$. (Can somebody clarify this and provide a diagram?)

The two quadratics are $180^{\circ}$ rotations of each other about $(50,0)$. Since we are only dealing with differences of roots, we can translate them to be symmetric about $(0,0)$. Now $x_3 = - x_2 = 75$ and $x_4 = - x_1$. Say our translated versions of $f$ and $g$ are $p$ and $q$, respectively, so that $p(x) = - q( - x)$. Let $x_3 = 75$ be a root of $p$ and $x_2 = - 75$ a root of $q$ by symmetry. Note that since they each contain each other's vertex, $x_1$, $x_2$, $x_3$, and $x_4$ must be roots of alternating polynomials, so $x_1$ is a root of $p$ and $x_4$ a root of $q$ \[p(x) = a(x - 75)(x - x_1)\] \[q(x) = - a(x + 75)(x + x_1)\] The vertex of $p(x)$ is half the sum of its roots, or $\frac {75 + x_1}{2}$. We are told that the vertex of one quadratic lies on the other, so \begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{2}\right) \\ & = & - \frac {a}{4}(x_1 - 75)^2 \\ - \frac {a}{4}(x_1 - 75)^2 & = & q\left(\frac {75 + x_1}{2}\right) \\ & = & - a\left(\frac {x_1 + 225}{2}\right)\left(\frac {3x_1 + 75}{2}\right) \\ & = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*} Let $x_1 = 75u$ and divide through by $75^2$, since it will drastically simplify computations. We know $u < - 1$ and that $(u - 1)^2 = (3u + 1)(u + 3)$, or \begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\ & = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\ & = & 2u^2 + 12u + 2 \\ & = & u^2 + 6u + 1 \end{eqnarray*} So $u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2$. Since $u < - 1$, $u = - 3 - 2\sqrt2$. The answer is $x_4 - x_1 = (-x_1) - x_1 = - 150u = 450 + 300\sqrt {2}$, and $450 + 300 + 2 = 752\ \mathbf{(D)}$.

Testing the first two (or three) positive integers instead of 2009, $k$ seems to always be 4 more. Put E and go on to tackle #25 :)

Consider another sequence $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$, and $0 \leq \theta_n < 180$. The given recurrence becomes \begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\ \tan{\theta_{n + 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\ \tan{\theta_{n + 2}} & = \tan(\theta_{n + 1} + \theta_n) \end{align*} It follows that $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$. Since $\theta_1 = 45, \theta_2 = 30$, all terms in the sequence $\{\theta_1, \theta_2, \theta_3...\}$ will be a multiple of $15$. Now consider another sequence $\{b_1, b_2, b_3...\}$ such that $b_n = \theta_n/15$, and $0 \leq b_n < 12$. The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$. As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the first few terms of the sequence until it starts to repeat. $\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$ Note that $b_{25} = b_1 = 3$ and $b_{26} = b_2 = 2$. Thus $\{b_n\}$ has a period of $24$: $b_{n + 24} = b_n$. It follows that $b_{2009} = b_{17} = 0$ and $\theta_{2009} = 15 b_{2009} = 0$. Thus $a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$ Our answer is $|a_{2009}| = \boxed{\textbf{(A)}\ 0}$.