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02841d381c965e9b346c41df7fd4cd82 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9 | In right triangle $ABC$ with right angle $C$ $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii . The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers . Find $p+q$ | It is known that $O_1O_2$ is parallel to AB. Thus, extending $O_1F$ and $GO_2$ to intersect at H yields similar triangles $O_1O_2H$ and BAC, so that $O_1O_2 = 2r$ $O_1H = \frac{16r}{17}$ , and $HO_2 = \frac{30r}{17}$ . It should be noted that $O_2G = r$ . Also, FHGC is a rectangle, and so AF = $\frac{47r}{17} - 30$ and similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: \[2r + \frac{47r}{17} - 30 + \frac{33r}{17} - 16 = 34\] Thus, r = $\frac{680}{57}$ , and the answer is $\boxed{737}.$ | null | 737 |
02841d381c965e9b346c41df7fd4cd82 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9 | In right triangle $ABC$ with right angle $C$ $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii . The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers . Find $p+q$ | Let the radius of the circle be $r$ . It can be seen that $\Delta FHO_{1}$ and $\Delta O_{2}GJ$ are similar to $\Delta ACB$ , and the length of the hypotenuses are $\frac{17}{8}r$ and $\frac {17}{15}r$ , respectively. Then, the entire length of $HJ$ is going to be $(\frac{17}{8}+\frac{17}{15}+2)r = \frac{631}{120}r$ . The length of the hypotenuse of $\Delta ACB$ is 34, so the length of the height to $AB$ is $\frac{16*30}{34} = \frac{240}{17}$ . Thus, the height to $\Delta HCJ$ is going to be $\frac{240}{17} + r$ $\Delta HCJ$ is similar to $\Delta ACB$ so we have the following: $\frac{\frac{631}{120}r}{34} = \frac{\frac{240}{17} + r}{\frac{240}{17}}$ . Cross multiplying and simplifying, we get that $r = \frac{680}{57}$ so the answer is $\boxed{737}$ .
~Leonard_my_dude~ | null | 737 |
988733704722df1e90257d57dac4f322 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | Consider the first column. There are ${6\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.
Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:
So there are $20(3+54+36) = 1860$ different shadings, and the solution is $\boxed{860}$ | null | 860 |
988733704722df1e90257d57dac4f322 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | We start by showing that every group of $6$ rows can be grouped into $3$ complementary pairs.
We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns $1$ and $2$ shaded. Note how if there is no complement to this, then all the other five rows must have at least one square in the first two columns shaded. That means that in total, the first two rows have $2+5=7$ squares shaded in- that is false since it should only be $6$ .
Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of $6$ rows can be grouped into $3$ complementary pairs.
Now we proceed with three cases.
Our answer is thus $720+1080+60=1860,$ leaving us with a final answer of $\boxed{860}.$ | null | 860 |
988733704722df1e90257d57dac4f322 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.
There are ${6 \choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a "1-row") within the first 3 columns and which rows have 2 (we'll call these "2-rows") within the first 3 columns. Next, we do some casework:
In total, we have ${6\choose3}(1+18+54+20)=20*93=1860$ . Thus our answer is $\boxed{860}$ | null | 860 |
988733704722df1e90257d57dac4f322 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | We can use generating functions. Suppose that the variables $a$ $b$ $c$ , and $d$ represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function $ab+ac+ad+bc+bd+cd$ , which we can write as $P(a,b,c,d)=(ab+cd)+(a+b)(c+d)$ . Therefore, $P(a,b,c,d)^6$ represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of $a^3b^3c^3d^3$ in $P(a,b,c,d)^6$
By the Binomial Theorem, \[P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}\] If we expand $(ab+cd)^k$ , then the powers of $a$ and $b$ are always equal. Therefore, to obtain terms of the form $a^3b^3c^3d^3$ , the powers of $a$ and $b$ in $(a+b)^{6-k}$ must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that $k$ must be even. We can use the same logic for $c$ and $d$ . Therefore, the coefficient of $a^3b^3c^3d^3$ in the following expression is the same as the coefficient of $a^3b^3c^3d^3$ in (1). \[\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}\] Now we notice that the only way to obtain terms of the form $a^3b^3c^3d^3$ is if we take the central term in the binomial expansion of $(ab+cd)^{2k}$ . Therefore, the terms that contribute to the coefficient of $a^3b^3c^3d^3$ in (2) are \[\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.\] This sum is $400+1080+360+20=1860$ so the answer is $\boxed{860}$ | null | 860 |
b8bfbb0b6f187635ded5a4f93604682a | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | $\left(k- \frac 12\right)^2=k^2-k+\frac 14$ and $\left(k+ \frac 12\right)^2=k^2+k+ \frac 14$ . Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\leq k^2+k$ . There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$ $44<\sqrt{2007}<45$ , so all numbers $1$ to $44$ have their full range. Summing this up with the formula for the sum of the first $n$ squares ( $\frac{n(n+1)(2n+1)}{6}$ ), we get $\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740$ . We need only consider the $740$ because we are working with modulo $1000$
Now consider the range of numbers such that $b(p)=45$ . These numbers are $\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981$ to $2007$ . There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$ , and $215+740= \boxed{955}$ , the answer. | null | 955 |
b8bfbb0b6f187635ded5a4f93604682a | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | Let $p$ be in the range of $a^2 \le p < (a+1)^2$ . Then, we need to find the point where the value of $b(p)$ flips from $k$ to $k+1$ . This will happen when $p$ exceeds $(a+\frac{1}{2})^2$ or $a(a+1)+\frac{1}{4}$ . Thus, if $a^2 \le p \le a(a+1)$ then $b(p)=a$ . For $a(a+1) < p < (a+1)^2$ , then $b(p)=a+1$ . There are $a+1$ terms in the first set of $p$ , and $a$ terms in the second set. Thus, the sum of $b(p)$ from $a^2 \le p <(a+1)^2$ is $2a(a+1)$ or $4\cdot\binom{a+1}{2}$ . For the time being, consider that $S = \sum_{p=1}^{44^2-1} b(p)$ . Then, the sum of the values of $b(p)$ is $4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)$ . We can collapse this to $4\binom{45}{3}=56760$ . Now, we have to consider $p$ from $44^2 \le p < 2007$ . Considering $p$ from just $44^2 \le p \le 1980$ , we see that all of these values have $b(p)=44$ . Because there are $45$ values of $p$ in that range, the sum of $b(p)$ in that range is $45\cdot44=1980$ . Adding this to $56760$ we get $58740$ or $740$ mod $1000$ . Now, take the range $1980 < p \le 2007$ . There are $27$ values of $p$ in this range, and each has $b(p)=45$ . Thus, that contributes $27*45=1215$ or $215$ to the sum. Finally, adding $740$ and $215$ we get $740+215=\boxed{955}$ | null | 955 |
a9b3a46a08fa47b2c1313ccdf272ab8c | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$ | Redefine the points in the same manner as the last time ( $\triangle AB'C'$ , intersect at $D$ $E$ , and $F$ ). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$
The area of $[\triangle AB'C'] = [\triangle ABC]$ . The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan (30 + 45)$ . The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $[\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$
The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles $AC' = AC = \frac{10}{\sin 15}$ . The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$ $AC'$ , in terms of the height of $\triangle ADC'$ , is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$ \begin{align*} [ADC'] &= \frac 12 AC' \cdot h \\ &= \frac 12 (10\sqrt{6} + 10\sqrt{2})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right) \\ &= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}} \\ &= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50 \end{align*}
The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$
Therefore, $[ADEF]$ $= (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right)$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and our answer is $\boxed{875}$ | null | 875 |
a9b3a46a08fa47b2c1313ccdf272ab8c | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$ | Call the points of the intersections of the triangles $D$ $E$ , and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\overline{AD}$ bisects $\angle EDE'$
Through HL congruency, we can find that $\triangle AED$ is congruent to $\triangle AE'D$ . This divides the region $AEDF$ (which we are trying to solve for) into two congruent triangles and an isosceles right triangle
Since $FE' = AE' = AE$ , we find that $[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}$
Now, we need to find $[AED] = [AE'D]$ . The acute angles of the triangles are $\frac{15}{2}$ and $90 - \frac{15}{2}$ . By repeated application of the half-angle formula , we can find that $\tan \frac{15}{2} = \sqrt{2} - \sqrt{3} + \sqrt{6} - 2$
The area of $[AED] = \frac 12 \left(20 \cos 15\right)^2 \left(\tan \frac{15}{2}\right)$ . Thus, $[AED] + [AE'D] = 2\left(\frac 12((5\sqrt{6} + 5\sqrt{2})^2 \cdot (\sqrt{2} - \sqrt{3} + \sqrt{6} - 2))\right)$ , which eventually simplifies to $500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$
Adding them together, we find that the solution is $[AEDF] = [AE'F] + [AED] + [AE'D]$ $= 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600=$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and the answer is $\boxed{875}$ | null | 875 |
a9b3a46a08fa47b2c1313ccdf272ab8c | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$ | From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection.
Now, we can equate the equations to find the intersections of all the points.
We take these points and tie them together by shoelace, and the answer should come out to be $\boxed{875}$ | null | 875 |
b9f0a9ed121c8cd826d29f9aed107af6 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_13 | A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$ . A plane passes through the midpoints of $AE$ $BC$ , and $CD$ . The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$ . Find $p$
AIME I 2007-13.png | Note first that the intersection is a pentagon
Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,2\sqrt{2})$ . Using the coordinates of the three points of intersection $(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)$ , it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$ , and using the points we find that $2a = d \Longrightarrow d = \frac{a}{2}$ $-2b = d \Longrightarrow d = \frac{-b}{2}$ , and $-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}$ . It is then $x - y + 2\sqrt{2}z = 2$
Write the equation of the lines and substitute to find that the other two points of intersection on $\overline{BE}$ $\overline{DE}$ are $\left(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2}\right)$ . To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula $\sqrt{a^2 + b^2 + c^2}$ ), it is possible to find that the area of the triangle is $\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}$ . The trapezoid has area $\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}\left(2\sqrt{2} + 3\sqrt{2}\right) = \frac{5\sqrt{5}}{2}$ . In total, the area is $4\sqrt{5} = \sqrt{80}$ , and the solution is $\boxed{080}$ | null | 080 |
2405d03caf94de671ca19c7b09bb4093 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | Define a function $f(n)$ on the non-negative integers, as \[f(n) = \frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}\] We want $\left\lfloor f(2006) \right\rfloor$
Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \[.\phantom{------------} \frac{a_{n+1}}{a_{n}} = \frac{a_{n}}{a_{n-1}} + \frac{2007}{a_{n}a_{n-1}} \phantom{------------} (1)\] and dividing through by $a_{n}a_{n+1}$ , we get \[.\phantom{------------} \frac{a_{n-1}}{a_{n}} = \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} \phantom{------------} (2)\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \[\frac{a_{n+1}}{a_{n}} + \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} = \frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}} + \frac{2007}{a_{n}a_{n-1}}\] i.e. \[f(n)+ \frac{2007}{a_{n}a_{n+1}} = f(n-1) + \frac{2007}{a_{n}a_{n-1}}\] Summing over $n=1$ to $n=2006$ , we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with \[f(2006) + \frac{2007}{a_{2006}a_{2007}} = f(0) + \frac{2007}{a_{1}a_{0}}\] We have $f(0) = 2$ , and $2007/a_0a_1 = 2007/9 = 223$ . So \[f(2006) = 2 + 223 - \frac{2007}{a_{2006}a_{2007}} = 224 + \left( 1 - \frac{2007}{a_{2006}a_{2007}}\right)\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$ , this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$ . It follows that $\left\lfloor f(2006) \right\rfloor = \boxed{224}$ | null | 224 |
2405d03caf94de671ca19c7b09bb4093 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$
Add these two equations to get
This is an invariant . Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$ , the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$
We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$ . Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$ , notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that
$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $\boxed{224}$ | null | 224 |
2405d03caf94de671ca19c7b09bb4093 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \[\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\alpha b_n+\beta b_{n-1}$ for $n\ge 2$ . We wish to find $\alpha$ and $\beta$ such that $a_n=b_n$ for all $n\ge 1$ . To do this, we use the following matrix form of a linear recurrence relation
\[\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
When we take determinants, this equation becomes
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We want \[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=2007\] for all $n$ . Therefore, we replace the two matrices by $2007$ to find that
\[2007=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\cdot 2007\] \[1=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)=-\beta.\] Therefore, $\beta=-1$ . Computing that $a_3=672$ , and using the fact that $b_3=\alpha b_2-b_1$ , we conclude that $\alpha=225$ . Clearly, $a_1=b_1$ $a_2=b_2$ , and $a_3=b_3$ . We claim that $a_n=b_n$ for all $n\ge 1$ . We proceed by induction . If $a_k=b_k$ for all $k\le n$ , then clearly, \[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\] We also know by the definition of $b_{n+1}$ that
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$ . After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$ . Thinking of this as a linear equation in the variable $b_{n+1}$ , we already know that this has the solution $b_{n+1}=a_{n+1}$ . Therefore, by induction, $a_n=b_n$ for all $n\ge 1$ . We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$
It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\ge 3$ . Now
\[\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.\]
\[=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.\]
\[=225-\frac{2007}{a_{2005}a_{2006}}.\]
The sequence certainly grows fast enough such that $\frac{2007}{a_{2005}a_{2006}}<1$ . Therefore, the largest integer less than or equal to this value is $\boxed{224}$ | null | 224 |
2405d03caf94de671ca19c7b09bb4093 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$ $= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$ . Calculating the other values, $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$
The greatest number that does not exceed this is $\boxed{224}$ | null | 224 |
2405d03caf94de671ca19c7b09bb4093 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$ , and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$ , respectively. It will be clear why I decided to factor these expressions as I did momentarily.
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$ . For $n = 1$ , we get $\frac{1 + 224^2}{224}$ , the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$ . It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$ , while the second-largest term in the denominator is smaller than $224^2$ . Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$ , the degrees of both the numerator and denominator increase by $2$ , because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$ ). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two.
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$ , they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$ , whose $\lim_{k\to\infty}=$ $\boxed{224}$ | null | 224 |
31c506c0fa728b250785ae4ccc5508dc | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15 | Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime . Find $r$ | AIME I 2007-15.png
Denote the length of a side of the triangle $x$ , and of $\overline{AE}$ as $y$ . The area of the entire equilateral triangle is $\frac{x^2\sqrt{3}}{4}$ . Add up the areas of the triangles using the $\frac{1}{2}ab\sin C$ formula (notice that for the three outside triangles, $\sin 60 = \frac{\sqrt{3}}{2}$ ): $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}$ . This simplifies to $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)$ . Some terms will cancel out, leaving $y = \frac{5}{3}x - 22$
$\angle FEC$ is an exterior angle to $\triangle AEF$ , from which we find that $60 + \angle CED = 60 + \angle AFE$ , so $\angle CED = \angle AFE$ . Similarly, we find that $\angle EDC = \angle AEF$ . Thus, $\triangle AEF \sim \triangle CDE$ . Setting up a ratio of sides, we get that $\frac{5}{x-y} = \frac{y}{2}$ . Using the previous relationship between $x$ and $y$ , we can solve for $x$
Use the quadratic formula , though we only need the root of the discriminant . This is $\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}$ $= \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}$ . The answer is $\boxed{989}$ | null | 989 |
31c506c0fa728b250785ae4ccc5508dc | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15 | Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime . Find $r$ | First of all, assume $EC=x,BD=m, ED=a, EF=b$ , then we can find $BF=m-3, AE=2+m-x$ It is not hard to find $ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56$ , we apply LOC on $\triangle{DEF}, \triangle{BFD}$ , getting that $(m-3)^2+m^2-m(m-3)=a^2+b^2-ab$ , leads to $a^2+b^2=m^2-3m+65$ Apply LOC on $\triangle{CED}, \triangle{AEF}$ separately, getting $4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.$ Add those terms together and use the equality $a^2+b^2=m^2-3m+65$ , we can find: $2x^2-(2m+1)x+2m-42=0$
According to basic angle chasing, $\angle{A}=\angle{C}; \angle{AFE}=\angle{CED}$ , so $\triangle{AFE}\sim \triangle{CED}$ , the ratio makes $\frac{5}{x}=\frac{2+m-x}{2}$ , getting that $x^2-(2+m)x+10=0$ Now we have two equations with $m$ , and $x$ values for both equations must be the same, so we can solve for $x$ in two equations. $x=\frac{2m+1 \pm \sqrt{4m^2+4m+1-16m+336}}{4}; x=\frac{4+2m \pm \sqrt{4m^2+16m-144}}{4}$ , then we can just use positive sign to solve, simplifies to $3+\sqrt{4m^2+16m-144}=\sqrt{4m^2-12m+337}$ , getting $m=\frac{211-3\sqrt{989}}{10}$ , since the triangle is equilateral, $AB=BC=2+m=\frac{231-3\sqrt{989}}{10}$ , and the desired answer is $\boxed{989}$ | null | 989 |
432ee4c465ac9bee0934393368672ffc | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_1 | A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find $\frac{N}{10}$ | There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
Thus, $N = 2520 + 1200 = 3720$ , and $\frac{N}{10} = \boxed{372}$ | null | 372 |
5314dc761ad132a0f9e04962cbc87f7f | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_2 | Find the number of ordered triples $(a,b,c)$ where $a$ $b$ , and $c$ are positive integers $a$ is a factor of $b$ $a$ is a factor of $c$ , and $a+b+c=100$ | Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$ . The last condition reduces to $a(1 + x + y) = 100$ . Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$ . There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$ . Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}$ solutions of $(a,b,c)$ | null | 200 |
ea01585964791e26be04f8ad91708fd7 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3 | Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(A--E--B); draw(C--F--D); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy] | Drawing $EF$ , it clearly passes through the center of $ABCD$ . Letting this point be $P$ , we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}$ . Since $EF=EP+FP=2\cdot EP$ , the answer is $(17\sqrt{2})^2=\boxed{578}.$ | null | 578 |
ea01585964791e26be04f8ad91708fd7 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3 | Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(A--E--B); draw(C--F--D); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy] | We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$
First note that $\angle BAO = 45^{\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\angle EAB = \arccos{\left(\frac{12}{13}\right)}$ or $\arcsin{\left(\frac{5}{13}\right)}.$ Finally, we know that $AO =\frac{13\sqrt{2}}{2}.$ Now we apply laws of cosines on $\bigtriangleup AEO.$
We have $EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.$ We know that $\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.$ Thus we have $\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)$ which applying the cosine sum identity yields $\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.$
Note that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4 \cdot 12 \cdot 7 = \boxed{578}.$ | null | 578 |
345e8d86a35cee38c3d4e9a927c8e25a | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_4 | The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$ | Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.
Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):
Solve the system of equations with the first two equations to find that $(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)$ . Substitute this into the third equation to find that $1050 = 150 + 2m$ , so $m = \boxed{450}$ | null | 450 |
4a18182d0fe2729c1f8c5072039084a1 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_6 | An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd , and $a_{i}>a_{i+1}$ if $a_{i}$ is even . How many four-digit parity-monotonic integers are there? | This problem can be solved via recursion since we are "building a string" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications( $0$ can't be at the front and no digit is less than $9$ ). There are $4$ options to add no matter what(try some examples if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $\boxed{640}$ | null | 640 |
92756de1a3dbad416cfe7fc2077a63e1 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_7 | Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$
Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$ | Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$
The wording of this problem (which uses "exactly") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k$ . Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\leq 70k$ . Since $69k$ $3k^2+3k+1$ , and $70k$ are all integers, this inequality is equivalent to $69k\leq 3k^2+3k<70k$ . Since $k$ is positive, we can divide the inequality by $k$ to get $69 \leq 3k+3 < 70$ . Clearly the only $k$ that satisfies is $k=22$
Then, $\frac{n_{70}}{k}=k^2+69=484+69=\boxed{553}$ . (Remember we set $n_i$ equal to $k^3 + (i - 1)k$ !) | null | 553 |
0accae3267da63b9595de3ba952aef8b | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_8 | rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000. | Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$
FOIL this to get a quadratic, $-\frac 54x^2 + 502x - \frac{2003}4$ . Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$ . However, this gives a non-integral answer for $y$ . The closest two values that work are $(199,253)$ and $(203,248)$
We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$ . The solution is $\boxed{896}$ | null | 896 |
1a08d393efde70bc19999ebb21dcb64a | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_9 | Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$ | Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$ . Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$
Use the Two Tangent Theorem on $\triangle BEF$ . Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$ . By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$ , making $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$ . Also, $BX = BY$ $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$
Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$ . Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$ . Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$ , so $PQ = \boxed{259}$ | null | 259 |
ae4cc5d671889224ace9447b1b972146 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_10 | Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ $n$ , and $r$ are positive integers $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$ | Let $|S|$ denote the number of elements in a general set $S$ . We use complementary counting.
There is a total of $2^6$ elements in $P$ , so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$
Note that the number of $x$ -element subset of $S$ is $\binom{6}{x}$ . In general, for $0 \le |A| \le 6$ , in order for $B$ to be in neither $A$ nor $S-A$ $B$ must have at least one element from both $A$ and $S-A$ . In other words, $B$ must contain any subset of $A$ and $S-A$ except for the empty set $\{\}$ . This can be done in $\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)$ ways. As $|A|$ ranges from $0$ to $6$ , we can calculate the total number of unsuccessful outcomes to be \[\sum_{|A| = 0}^{6} \binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.\] So our desired answer is \[1 - \dfrac{2702}{2^{12}} = \dfrac{697}{2^{11}} \Rightarrow \boxed{710}.\] | null | 710 |
ae4cc5d671889224ace9447b1b972146 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_10 | Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ $n$ , and $r$ are positive integers $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$ | To begin with, we note that there are $2^6$ subsets of $S$ (which we can assume is $\{1,2,3,4,5,6\}$ ), including the null set. This gives a total of $(2^6)^2 = 2^{12}$ total possibilities for A and B.
Case 1: B is contained in A only.
If B has $0$ elements, which occurs in $\binom{6}{0}$ ways, A can be anything, giving us $\binom{6}{0} \cdot 2^6$ . If B has $1$ element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us $\binom{6}{1} \cdot 2^5$ . Summing, we end up with the binomial expansion of $(2 + 1)^6 = 3^6$
Case 2: B is contained in S-A only.
By symmetry, this case is the same as Case 1, once again giving us $3^6$ possibilities.
Case 3: B is contained in both.
We claim here that B can only be the null set. For contradiction, assume that there exists some element $x$ in B which satisfies this restriction. Then, A must contain $x$ as well, but we also know that $S-A$ contains $x$ , contradiction. Hence, B is the null set, whereas A can be anything. This gives us $2^6$ possibilities.
Since we have overcounted Case 3 in both of the other two cases, our final count is $2 \cdot 3^6 - 2^6$ . This gives us the probability $\frac{2 \cdot 3^6 - 2^6}{2^{12}}$ . Upon simplifying, we end up with $\frac{697}{2^{11}}$ , giving the desired answer of $\boxed{710}$ .
- Spacesam | null | 710 |
1f74767b69fa1aa39f927e9322791a82 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_11 | Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime . Find $a+b+c.$ | 2007 AIME II-11.png
If it weren’t for the small tube, the larger tube would travel $144\pi$ . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of $72 + 24 = 96$ . The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \triangle$
Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\frac{60}{360} = \frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\circ}$ central angle in the smaller circle indicates to take $\frac{120}{360} = \frac 13$ of the circumference. This adds up to $2 \cdot \frac 16 144\pi + \frac 13 48\pi = 64\pi$
The actual horizontal distance it takes can be found by using the $30-60-90 \triangle$ s. The missing leg is equal in length to $48\sqrt{3}$ . Thus, the total horizontal distance covered is $96\sqrt{3}$
Thus, we get $144\pi - 64\pi + 96\sqrt{3} = 80\pi + 96\sqrt{3}$ , and our answer is $\boxed{179}$ | null | 179 |
96a759ffca7d19e6f25cabe74f805957 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_12 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | Suppose that $x_0 = a$ , and that the common ratio between the terms is $r$
The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$ . Using the rules of logarithms , we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$ . Thus, $a^8r^{28} = 3^{308}$ . Since all of the terms of the geometric sequence are integral powers of $3$ , we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$ . We find that $8x + 28y = 308$ . The possible positive integral pairs of $(x,y)$ are $(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)$
The second condition tells us that $56 \le \log_3 (a + ar + \ldots ar^7) \le 57$ . Using the sum formula for a geometric series and substituting $x$ and $y$ , this simplifies to $3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}$ . The fractional part $\approx \frac{3^{8y}}{3^y} = 3^{7y}$ . Thus, we need $\approx 56 \le x + 7y \le 57$ . Checking the pairs above, only $(21,5)$ is close.
Our solution is therefore $\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}$ | null | 091 |
96a759ffca7d19e6f25cabe74f805957 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_12 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$ s and $0$ s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$ , and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\dots +(56-7r) = 308$ , and $r = 5$ . Thus $\log_3 (x_{14}) = \boxed{091}$ | null | 091 |
96a759ffca7d19e6f25cabe74f805957 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_12 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$ $x_1$ $x_2$ ..., as $3^n$ $3^{n+m}$ , and $3^{n+2m}$ ... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$ . Simplify to get $2n+7m=77$ . (1)
Now, rewrite the second given equation as $3^{56} \leq \left( \sum_{n=0}^{7}x_{n} \right) \leq 3^{57}$ . Obviously, $x_7$ , aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\leq56$ . Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$ . Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$ . (2)
Use equations 1 and 2 to get $m=5$ and $n=21$ $\log_{3}{x_{14}} = \log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}$ | null | 091 |
96a759ffca7d19e6f25cabe74f805957 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_12 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$ . As stated above, we then get that $a+a+n+...+a+7n=308$ , from which we simplify to $2a+7n=77$ . From here, we just go brute force using the second statement (that $3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}$ ). Rearranging the equation from earlier, we get \[n=11-\frac{2a}{7}\] from which it is clear that $a$ is a multiple of $7$ . Testing the first few values of $a$ , we get:
Case 1 ( $a=7,n=9$ )
The sequence is then $3^{7}+...+3^{70}$ , which breaks the upper bound.
Case 2 ( $a=14,n=7$ )
The sequence is then $3^{14}+...+3^{63}$ , which also breaks the upper bound.
Case 3 ( $a=21,n=5$ )
This is the first reasonable one, giving $3^{21}+...+3^{56}$ . It seems like this would break the upper bound, but from some testing we get: \[3^{21}+...+3^{56}<3^{57}\] \[1+...+3^{35}<3^{36}\] \[1+...+3^{30}<2*3^{35}\] \[3^{5}+...+3^{30}<2*3^{35}-1\] \[1+...+3^{25}<2*3^{30}-3^{-5}\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \[1<2*3^{5}\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$ , so $x_14=3^{a+14n}\rightarrow3^{91}$ . We then get the requested answer, $\log_3(3^{91})=\boxed{091}$ ~ amcrunner | null | 091 |
357cf01317f5e0cbca002dfa4a4f1146 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_13 | triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$
[asy] for (int i=0; i<12; ++i){ for (int j=0; j<i; ++j){ //dot((-j+i/2,-i)); draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle); } } [/asy] | Label each of the bottom squares as $x_0, x_1 \ldots x_9, x_{10}$
Through induction , we can find that the top square is equal to ${10\choose0}x_0 + {10\choose1}x_1 + {10\choose2}x_2 + \ldots {10\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the top position going only up is equal to the number of times it will be counted in the final sum.)
Examine the equation $\mod 3$ . All of the coefficients from $x_2 \ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$ ). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \equiv 0 \mod 3$ . Reduce to find that $x_0 + x_1 + x_9 + x_{10} \equiv 0 \mod 3$ . Out of $x_0,\ x_1,\ x_9,\ x_{10}$ , either all are equal to $0$ , or three of them are equal to $1$ . This gives ${4\choose0} + {4\choose3} = 1 + 4 = 5$ possible combinations of numbers that work.
The seven terms from $x_2 \ldots x_8$ can assume either $0$ or $1$ , giving us $2^7$ possibilities. The answer is therefore $5 \cdot 2^7 = \boxed{640}$ | null | 640 |
c630af3affff6e75ed3e37a1cebf61b8 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | If the leading term of $f(x)$ is $ax^m$ , then the leading term of $f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}$ , and the leading term of $f(2x^3 + x) = 2^max^{3m}$ . Hence $2^ma^2 = 2^ma$ , and $a = 1$ . Because $f(0) = 1$ , the product of all the roots
of $f(x)$ is $\pm 1$ . If $f(\lambda) = 0$ , then $f(2\lambda^3 + \lambda) = 0$ . Assume that there exists a root $\lambda$ with $|\lambda| \neq 1$ . Then there must be such a root $\lambda_1$ with $|\lambda_1| > 1$ . Then \[|2\lambda^3 + \lambda| \ge 2|\lambda|^3 - |\lambda| > 2|\lambda| - |\lambda| = |\lambda|.\] But then $f(x)$ would have infinitely many roots, given by $\lambda_{k+1} = 2\lambda_k^3 + \lambda_k$ , for $k \ge 1$ .
Therefore $|\lambda| = 1$ for all of the roots of the polynomial. Thus $\lambda\overline{\lambda}=1$ , and $(2\lambda^3 + \lambda)\overline{(2\lambda^3 + \lambda)} = 1$ .
Solving these equations simultaneously for $\lambda = a + bi$ yields $a = 0$ $b^2 = 1$ , and so $\lambda^2 = -1$ . Because the polynomial has real coefficients, the polynomial must have the form $f(x) = (1 + x^2)^n$ for some integer $n \ge 1$ . The condition $f(2) + f(3) = 125$ implies $n = 2$ , giving $f(5) = \boxed{676}$ | null | 676 |
c630af3affff6e75ed3e37a1cebf61b8 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.
Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\neq1$ . Clearly, $f(-2)\neq-1$ , because that would imply the existence of a real root.)
The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have
\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\] \[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\]
Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$
Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = \boxed{676}$ | null | 676 |
c630af3affff6e75ed3e37a1cebf61b8 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | Let $r$ be a root of $f(x).$ This means that $f(r)f(2r^2)=f(2r^3+r).$ In other words, $2r^3+r$ is a root of $f(x)$ too. Since $f(x)$ can't have infinitely many roots, \[Q(x)=P(P(\dotsb P(P(r)) \dotsb))\] is cyclic, where $P(x)=2x^3+x.$ Now, we will do casework.
Case 1: $\deg f\geq1$
Subcase 1: $|r|>1$
This means that \[|2r^3+r|\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\cdot1^2-1)=|r|.\] It follows that $|2r^3+r|>|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 2: $|r|<1$
This means that $|2r^3+r|\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.$ It follows that $|2r^3+r|<|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 3: $|r|=1.$
Since $|r|$ is not greater than or less than 1, $|r|=1.$ This means that all the roots of the polynomial have a magnitude of $1.$ More specifically, $|2r^3+r|$ has a magnitude of one. Since this would mean an equality condition from the triangle inequality, $2r^3$ and $r$ are collinear with the origin in the complex plane. In other words, $\frac{2r^3}{r}=\pm c\Leftrightarrow cr=2r^3\Leftrightarrow 2r^2=c\Leftrightarrow r=\pm\sqrt{\pm\frac{c}{2}},$ for some real constant $c.$ Now, from $|r|=1,$ we find that $\left|\pm\sqrt{\pm\frac{c}{2}}\right|=1\Leftrightarrow \sqrt{\pm\frac{c}{2}}=1\Leftrightarrow \pm\frac{c}{2}=1\Leftrightarrow c=\pm2.$ Putting this back into the equation, we find that $r=1,-1,i,-i.$ Now, this means that $2r^3+r=3,-3,i,-i.$ $3$ and $-3$ obviously doesn't have a magnitude of $1.$ Thus, $i,-i$ are the only possible roots of the polynomial. Since roots come in conjugate pairs, $f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,$ works for all constants $n\neq0.$
Case 2: $\deg f=0.$
This means that $f(x)=c,$ for some constant $c.$ In other words, $c^2=c.$ We can easily find that this means that $c=0,1.$ Combining all the cases, we conclude that $f(x)=(x^2+1)^n,0,1$ are the only polynomials that satisfy this equation.
Now, we can test! $f(x)=0,1$ obviously don't satisfy $f(2)+f(3)=125.$ Thus, $f(x)=(x^2+1)^n.$ Substituting, we find that $5^n+10^n=125\Leftrightarrow n=2.$ We conclude that $f(5)=(5^2+1)^2=26^2=\boxed{676}.$ | null | 676 |
446f1628a652daf20e060f2b5acbf415 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [asy] defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p); [/asy]
First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . The semiperimeter is $21$ , so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$
Now consider the incenter $I$ of $\triangle ABC$ . Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $O_A$ $O_B$ , and $O_C$ . Let the center of the circle tangent to those three circles be $O$ . The homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ maps $\triangle ABC$ to $\triangle XYZ$ ; since $OO_A = OO_B = OO_C = 2r$ $O$ is the circumcenter of $\triangle XYZ$ and $\mathcal{H}$ therefore maps the circumcenter of $\triangle ABC$ to $O$ . Thus, $2r = R \cdot \frac{4 - r}{4}$ , where $R$ is the circumradius of $\triangle ABC$ . Substituting $R = \frac{abc}{4[ABC]} = \frac{65}{8}$ $r = \frac{260}{129}$ and the answer is $\boxed{389}$ | null | 389 |
446f1628a652daf20e060f2b5acbf415 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2007 AIME II-15b.gif
Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$
The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$
Cut and combine the triangles, as shown. Then solve for $4u$
The solution is $260+129=\boxed{389}$ | null | 389 |
446f1628a652daf20e060f2b5acbf415 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Let $A'$ $B'$ $C'$ , and $O$ be the centers of circles $\omega_{A}$ $\omega_{B}$ $\omega_{C}$ $\omega$ , respectively, and let $x$ be their radius.
Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$
Since $OA'=OB'=OC'=2x$ $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$ . Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$ . Then the inradius of $A'B'C'=r-x$ , so now we compute r. Computing the inradius by $A=rs$ , we find that the inradius of $ABC$ is $4$ . Additionally, using the circumradius formula $R=\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\frac{65}{8}$ . Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$
\[\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}\]
Solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$ | null | 389 |
446f1628a652daf20e060f2b5acbf415 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is $2r$ . Now denoting $AB=13;BC=14;AC=15$ , and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular to $BC$ . It is easy to know that $tanB=\frac{12}{5}$ , so $tan\angle OBJ=\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$ . So we can see that $JK=ON=14-\frac{7x}{2}$ . Obviously, $\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}$ . After solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$ . ~bluesoul | null | 389 |
446f1628a652daf20e060f2b5acbf415 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [asy] defaultpen(fontsize(12)+0.8); size(300); pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p); dot("$H$",(7,0),down,p); [/asy]
Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$ , respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$ . Since $AB||O_AO_B$ $BC||O_BO_C$ , and $CA||O_CO_A$ , we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]
That means $\angle ABC = \angle O_AO_BO_C$ $\angle BAC = \angle O_BO_AO_C$ , and $\angle ACB = \angle O_AO_CO_B$ . Thus, $\triangle ABC \sim \triangle O_AO_BO_C$ . We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear ; same apply with $B$ and $C$ ).
Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$ . To start off, we know the circumradius of $O_AO_BO_C$ is $O$ , since $OO_A = OO_B = OO_C = 2r$ . Since $O_AO_B = 13x$ $O_BO_C = 14x$ $O_CO_A = 15x$ , we can get an relationship involving $x$ and $r$ via another way to find the circumradius:
\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]
Take notice of the inradius of $ABC$ . We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$ . Let the tangency point of the incircle and side $BC$ be $H$ . We know $IH = 4$ . We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$ . To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$ . We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$ , which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).
Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$ . That gives $\boxed{389}$ as our final answer. | null | 389 |
c5202ac8596a5f6345d2ec26185a8975 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_1 | In quadrilateral $ABCD$ $\angle B$ is a right angle diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ $AB=18$ $BC=21$ , and $CD=14$ . Find the perimeter of $ABCD$ | From the problem statement, we construct the following diagram:
Using the Pythagorean Theorem
Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$
Plugging in the given information:
So the perimeter is $18+21+14+31=84$ , and the answer is $\boxed{084}$ | null | 084 |
a11c6f9b27be7ced1a4fcae205004605 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_2 | Let set $\mathcal{A}$ be a 90- element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$ | The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$ . The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$ . All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$
Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element subset of $\{1,2,3,\ldots,100\}$ , and let $T$ be the sum of the elements of $\mathcal{B}$ . Note that the number of possible $S$ is the number of possible $T=5050-S$ . The smallest possible $T$ is $1+2+ \ldots +10 = 55$ and the largest is $91+92+ \ldots + 100 = 955$ , so the number of possible values of T, and therefore S, is $955-55+1=\boxed{901}$ | null | 901 |
5296f041b4767da56c85a8565d3db598 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_3 | Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer. | Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, \[N_0 + a_n\cdot 10^n = 29N_0\] \[a_n \cdot 10^n = 28N_0.\] The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives \[7 \cdot 10^n = 28N_0\] or \[10^n = 4N_0.\] Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then \[N = 7 \cdot 10^2 + 25 = \boxed{725},\] and indeed, $725 = 29 \cdot 25.$ $\square$ | null | 725 |
5296f041b4767da56c85a8565d3db598 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_3 | Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer. | Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits.
Let $N=\overline{abc}$ , so $N=100a+10b+c$ . Thus, $N'=\overline{bc}=10b+c$ . By the constraints of the problem, we see that $N=29N'$ , so \[100a+10b+c=29(10b+c).\] Now, we subtract and divide to get \[100a=28(10b+c)\] \[25a=70b+7c.\] Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$ . Thus, $c=5$ . Now, we plug that into the equation: \[25a=70b+7(5)\] \[25a=70b+35\] \[5a=14b+7.\] By the same line of reasoning as earlier, $a=7$ . We again plug that into the equation to get \[35=14b+7\] \[b=2.\] Now, since $a=7$ $b=2$ , and $c=5$ , our number $N=100a+10b+c=\boxed{725}$ | null | 725 |
58d1d3d5fd4a307672ef8ab6c2f0af52 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_4 | Let $N$ be the number of consecutive $0$ 's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$ | A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: $96$ of the numbers $1!,\ 2!,\ 3!,\ 100!$ have a factor of $5$ $91$ have a factor of $10$ $86$ have a factor of $15$ . And so on. This gives us an initial count of $96 + 91 + 86 + \ldots + 1$ . Summing this arithmetic series of $20$ terms, we get $970$ . However, we have neglected some powers of $5$ - every $n!$ term for $n\geq25$ has an additional power of $5$ dividing it, for $76$ extra; every n! for $n\geq 50$ has one more in addition to that, for a total of $51$ extra; and similarly there are $26$ extra from those larger than $75$ and $1$ extra from $100$ . Thus, our final total is $970 + 76 + 51 + 26 + 1 = 1124$ , and the answer is $\boxed{124}$ | null | 124 |
3af40de8c4347508ecff1551e4986fb6 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5 | The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$ | We begin by equating the two expressions:
\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]
Squaring both sides yields:
\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]
Since $a$ $b$ , and $c$ are integers, we can match coefficients:
\begin{align*} 2ab\sqrt{6} &= 104\sqrt{6} \\ 2ac\sqrt{10} &=468\sqrt{10} \\ 2bc\sqrt{15} &=144\sqrt{15}\\ 2a^2 + 3b^2 + 5c^2 &=2006 \end{align*}
Solving the first three equations gives: \begin{eqnarray*}ab &=& 52\\ ac &=& 234\\ bc &=& 72 \end{eqnarray*}
Multiplying these equations gives $(abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}$ | null | 936 |
3af40de8c4347508ecff1551e4986fb6 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5 | The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$ | We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\sqrt{2}$ $y=\sqrt{3}$ , and $z=\sqrt{5}$ . Since
\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]
we attempt to rewrite the radicand in this form:
\[2006+2(52xy+234xz+72yz)\]
Factoring, we see that $52=13\cdot4$ $234=13\cdot18$ , and $72=4\cdot18$ . Setting $p=13$ $q=4$ , and $r=18$ , we see that
\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]
so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$ . Square rooting gives us $13\sqrt{2}+4\sqrt{3}+18\sqrt{5}$ and our answer is $13\cdot4\cdot18=\boxed{936}$ | null | 936 |
546213b88bb71ae76cea08de1277330c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$ | Numbers of the form $0.\overline{abc}$ can be written as $\frac{abc}{999}$ . There are $10\times9\times8=720$ such numbers. Each digit will appear in each place value $\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\frac{45\times72\times111}{999}= \boxed{360}$ | null | 360 |
546213b88bb71ae76cea08de1277330c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$ | Alternatively, for every number, $0.\overline{abc}$ , there will be exactly one other number, such that when they are added together, the sum is $0.\overline{999}$ , or, more precisely, 1. As an example, $.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1$
Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$ | null | 360 |
dfbcf3847d0382c1466513442905d7a6 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy] | Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof
Let the set of parallel lines be perpendicular to the x-axis , such that they cross it at $0, 1, 2 \ldots$ . The base of region $\mathcal{A}$ is on the line $x = 1$ . The bigger base of region $\mathcal{D}$ is on the line $x = 7$ .
Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to $\frac{1}{2}bh$
\[\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}\]
Solve this to find that $s = \frac{5}{6}$
Using the same reasoning as above, we get $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$ , which is $\boxed{408}$ | null | 408 |
dfbcf3847d0382c1466513442905d7a6 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy] | Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$ . Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\left(\frac{x+1}{x}\right)}^2$ . Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$ . Repeating this process, we get that the area of B is $2x+3$ , the area of C is $2x+7$ , and the area of D is $2x+11$
We can now use the given condition that the ratio of C and B is $\frac{11}{5}$
$\frac{11}{5} = \frac{2x+7}{2x+3}$ gives us $x = \frac{1}{6}$
So now we compute the ratio of D and A, which is $\frac{(2)(\frac{1}{6}) + 11}{(\frac{1}{6})^2} = \boxed{408.}$ | null | 408. |
dfbcf3847d0382c1466513442905d7a6 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy] | Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.$ Therefore the ratio we seek is $\frac{66(ay)}{11xy} =\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \implies ab = 6bx \implies a = 6x$ so the final ratio is $6 \cdot 68 = \boxed{408}.$ | null | 408 |
4f49f8307e21769ed7e49b6cf6f463a4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_8 | Hexagon $ABCDEF$ is divided into five rhombuses $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent , and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$ . Given that $K$ is a positive integer , find the number of possible values for $K$
[asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy] | Let $x$ denote the common side length of the rhombi.
Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$ . Then $x^2\sin(y)=\sqrt{2006}$ . We also see that $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$ . Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$ $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$ , so $K$ can be any integer between 1 and 89, inclusive. Thus the number of positive values for $K$ is $\boxed{089}$ | null | 089 |
4f49f8307e21769ed7e49b6cf6f463a4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_8 | Hexagon $ABCDEF$ is divided into five rhombuses $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent , and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$ . Given that $K$ is a positive integer , find the number of possible values for $K$
[asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy] | Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where $w*h=\sqrt{2006}$ . The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$ . Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$ . Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$ . Combine like terms to get $2 * \sqrt{2006-h^4}$ . This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/4$ . There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of $n^2$ for $2006-h^4$ . Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$ , but we must also test $44.5^2$ , because the product of two will make it an integer. $44.5^2$ is also less than $2006$ , so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us $44*2+1=$ $\boxed{089}$ | null | 089 |
4f49f8307e21769ed7e49b6cf6f463a4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_8 | Hexagon $ABCDEF$ is divided into five rhombuses $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent , and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$ . Given that $K$ is a positive integer , find the number of possible values for $K$
[asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy] | [asy] size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); label("$A$",A,2*N); label("$B$",B,2*N); label("$C$",C,2*E); label("$D$",D,2*S); label("$E$",EE,2*S); label("$F$",F,2*W); label("$G$",(0.47,-1.55),NW); label("$H$",(3.73,-1.55),NE); label("$I$",I,2*N); label("$J$",J,2*S); label("$K$",K,2*SW); draw(F--C); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); draw((2.1,0)--(2.1,-3.2)); [/asy] To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas.
In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$
Denote the length $HC=a$ and $KH=b.$ We have \[KI=\sqrt{a^2-b^2}\] by the Pythagorean Theorem, which implies \[[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}\] and \[[GIHJ]=2b\sqrt{a^2-b^2}.\] The first equation yields \[\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.\] Plugging into the second, we have \[[GIHJ]=2\sqrt{2006}\frac{b}{a}.\] The maximal value of $\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means \[\frac{b}{a}\leq 1.\] Plugging back up, we have \[[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.\] We have \[\lfloor \sqrt{8024} \rfloor = 89,\] thus our answer is \[89-1+1=\boxed{089}.\] | null | 089 |
ae3c4eac32924ca9eda44233781ef33d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_9 | The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | \[\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) \\ = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66})\]
So our question is equivalent to solving $\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers $a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so $a^{2}r^{11}=2^{1003}$
The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$
\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\ 2^{2x}\cdot 2^{11y}&=&2^{1003}\\ 2x+11y&=&1003\\ y&=&\frac{1003-2x}{11} \end{eqnarray*}
For $y$ to be an integer, the numerator must be divisible by $11$ . This occurs when $x=1$ because $1001=91*11$ . Because only even integers are being subtracted from $1003$ , the numerator never equals an even multiple of $11$ . Therefore, the numerator takes on the value of every odd multiple of $11$ from $11$ to $1001$ . Since the odd multiples are separated by a distance of $22$ , the number of ordered pairs that work is $1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46$ . (We must add 1 because both endpoints are being included.) So the answer is $\boxed{046}$ | null | 046 |
ae3c4eac32924ca9eda44233781ef33d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_9 | The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | Using the above method, we can derive that $a^{2}r^{11} = 2^{1003}$ .
Now, think about what happens when r is an even power of 2. Then $a^{2}$ must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so $2^{1}$ $2^{3}$ $2^{5}$ .... all work for r, until r hits $2^{93}$ , when it gets greater than $2^{1003}$ , so the greatest value for r is $2^{91}$ . All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields $\boxed{046}$ | null | 046 |
ae3c4eac32924ca9eda44233781ef33d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_9 | The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | Using the method from Solution 1, we get $\log_8a^{12}r^{66}=2006 \implies a^{12}r^{66}=8^{2006}=2^{6018}$
Since $a$ and $r$ both have to be powers of $2$ , we can rewrite this as $12x+66y=6018$
$6018 \equiv 66 \equiv 6\pmod{12}$ . So, when we subtract $12$ from $6018$ , the result is divisible by $66$ . Evaluating that, we get $(1,91)$ as a valid solution. Since $66 \cdot 2 = 12 \cdot 11$ , when we add $11$ to the value of $a$ , we can subtract $2$ from the value of $r$ to keep the equation valid. Using this, we get $(1,91),(12,89),(23,87), \cdots (541,1)$ . In order to count the number of ordered pairs, we can simply count the number of $y$ values. Every odd number from $1$ to $91$ is included, so we have $\boxed{046}$ solutions. | null | 046 |
e80516d0fc1bc04e14c333d1d4d8a98d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_10 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ [asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label("x", (7,0)); label("y", (0,7));[/asy] | The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\left(1,\frac 12\right)$ and $\left(\frac 32,2\right)$ , which can be easily solved to be $6x = 2y + 5$ . Thus, $a^2 + b^2 + c^2 = \boxed{065}$ | null | 065 |
e80516d0fc1bc04e14c333d1d4d8a98d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_10 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ [asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label("x", (7,0)); label("y", (0,7));[/asy] | This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: $6x = 2y + 5$ . The answer is then $6^2 + 2^2 + 5^2 = \boxed{065}$ . This solution is an alternate explanation to solution 1. | null | 065 |
0e35af9b5f503001e27c94c8d4642d81 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_11 | A collection of 8 cubes consists of one cube with edge length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:
Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000? | We proceed recursively . Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$ . How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$ ? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldots, m$ . Given a tower using blocks $1, 2, \ldots, m$ (with $m \geq 2$ ), we can insert the block of size $m + 1$ in exactly 3 places: at the beginning, immediately following the block of size $m - 1$ or immediately following the block of size $m$ . Thus, there are 3 times as many towers using blocks of size $1, 2, \ldots, m, m + 1$ as there are towers using only $1, 2, \ldots, m$ . There are 2 towers which use blocks $1, 2$ , so there are $2\cdot 3^6 = 1458$ towers using blocks $1, 2, \ldots, 8$ , so the answer is $\boxed{458}$ | null | 458 |
a49730c5d974905b733f6f21a4e77d44 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_12 | Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$ , where $x$ is measured in degrees and $100< x< 200.$ | Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ $\cos 3x = 0$ $\cos 4x = 0$ , or $\cos 5x = 0$
Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$ | null | 906 |
1c69744c1fbbf9ccc259f6b26556b1e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | Given $g : x \mapsto \max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \cdots + g(2^n)$ . Define $S = \{2, 4, \ldots, 2^n\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$
Thus \begin{align*} S_n &= 2^0\cdot2^n + 2^0\cdot2^{n-1} + 2^1\cdot2^{n-2} + \cdots + 2^{n-2}\cdot2^1\\ &= 2^n + (n-1)2^{n-1}\\ &= 2^{n-1}(n+1).\end{align*} Let $2^k$ be the highest power of $2$ that divides $n+1$ . Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$ . For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$ , we have five choices for $k$ , namely $k=0,2,4,6,8$
If $k=0$ , then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square.
In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^{k}$ is an odd perfect square.
If $k=2$ , then $n<1000$ implies that $\frac{n+1}{4} \le 250$ , so we have $n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2$
If $k=4$ , then $n<1000$ implies that $\frac{n+1}{16} \le 62$ , so $n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2$
If $k=6$ , then $n<1000$ implies that $\frac{n+1}{64}\le 15$ , so $n+1=64,64\cdot 3^2$
If $k=8$ , then $n<1000$ implies that $\frac{n+1}{256}\le 3$ , so $n+1=256$
Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\cdot 3^2 \cdot 5^2 - 1 = \boxed{899}$ | null | 899 |
1c69744c1fbbf9ccc259f6b26556b1e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even.
so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$ . However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$ | null | 899 |
1c69744c1fbbf9ccc259f6b26556b1e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations.
We first simplify all the messy notation in the $S_n$ term. Note that the problem asks us to find the smallest value of $n<1000$ such that there exists an integer $k$ that satisfies
\[k^2 = g(2) + g(4) + \cdots + g(2^n)\]
Since there is no obvious way to approach this problem, we start by experimenting with small values of $n$ to evaluate some $S_n$
We play with these values:
\[S_1 = g(2) = 2\]
\[S_2 = g(2) + g(4) = 2+4 = 6\]
\[S_3 = g(2) + g(4) + g(6) + g(8) = 16\]
\[S_4 = g(2) + g(4) + g(6) + g(8) + g(10) +g(12)+g(14)+g(16) = 40\]
We are certainly not going to expand all of this out... so let's look for patterns from these $4$ values!
Using a little bit of ingenuity, we note that
\[S_2 = 2+4 = S_1 + 4\]
\[S_3 = 2+4+2+8 = 8+8 = S_2 + S_1 + 8\]
\[S_4 = 2+4+2+8+2+4+2+16 = S_3 + S_2 + S_1 + 16\]
Aha! We see powers of two in each of our terms! Therefore, we can say that
\[S_2 = S_1 + 2^2\]
\[S_3 = S_2+S_1 + 2^3\]
\[S_4 = S_3 + S_2 + S_1 + 2^4\]
Hooray! We have a recursion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it.
On the actual AIME, go with whatever patterns you see, because most likely those are the patterns that the test-makers want the students to see.
So we may generalize a formula for $S_n$
\[S_n = 2^n + S_{n-1} + S_{n-2} + \cdots + S_2 + S_1\]
Uh oh... this formula is not in closed form. Looks like we'll have to use our recursion to develop one manually. We do so by using our recursion for $S_{n-1}$
\[S_n = 2^n + (2^{n-1} + S_{n-2} + S_{n-3} + \cdots + S_2 + S_1) + S_{n-2} + S_{n-3} + \cdots + S_2 + S_1\]
\[S_n = 2^n + 2^{n-1} + 2 (S_{n-2} + S_{n-3} + \cdots + S_2 + S_1\]
Let's simplify a bit further, where we use our recursion for $S_{n-2}$
\[S_n = 2^n + 2^{n-1} +(2S_{n-2}) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
\[S_n = 2^n + 2^{n-1} + 2(2^{n-2}) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
\[S_n = 2^n + 2^{n-1} + 2^{n-1} + 4(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form:
\[S_n = 2^n + \underbrace{2^{n-1}}_{n-2} + 2^{n-2}(S_1)\]
\[S_n = 2^n + 2^{n-1}(n-2) + 2^{n-1}\]
\[S_n = 2^n + 2^{n-1}(n-1)\]
\[S_n = 2^{n-1}(2 + (n-1)\]
\[S_n = 2^{n-1}(n+1)\]
We have our closed form, so now we can find the largest value of $n$ such that $S_n$ is a perfect square.
In order for $S_n$ to be a perfect square, we must have $n-1$ even and $n+1$ be a perfect square.
Since $n<1000$ , we have $n+1 < 1001$ . We first try $n+1 = 31^2 = 961$ (since it is the smallest square below $1000$ , which gives us $n=960$ . But $n-1$ isn't even, so we discard this value.
Next, we try the second smallest value, which is $n = 30^2 = 900$ , which tells us that $n=899$ $n-1$ is indeed even, and $n+1$ is a perfect square, so the largest value of $n$ such that $S_n$ is a perfect square is $899$
Our answer is $\boxed{899}$ | null | 899 |
1c69744c1fbbf9ccc259f6b26556b1e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | First, we set intervals. Say that a number $N$ falls strictly within $[2^k, 2^{k+1}]$
$N<2^{k+1}=2^k+2^k$
We can generalize this:
If a number is in the form $N=2^k+2^{R}O$ where $O$ is a positive odd number, $R<k$
$N<2^{k+1}=2^k+2^k\Longrightarrow O<2^{k-R}$ so there are $2^{k-R-1}$ numbers that satisfy this form.
For all numbers that satisfy this form, $g(N)=2^R$ . The sum of $g(N)$ for all $N$ in this form and interval is $2^{k-1}$ $R$ can vary between $1$ and $k-1$ , so the total sum is $\underbrace{2^{k-1}+2^{k-1}\cdots 2^{k-1}}_{k-1}=(k-1)2^{k-1}$ . The domain of the function we are trying to find is all even integers in the interval $[2^1, 2^n]$ so there are $n-1$ values of $k$ . Now we have the sum $\sum_{k=1}^{n-1}{(k-1)2^{k-1}}=\sum_{i=1}^{n-2}{k\cdot2^{k}}$ . However, we did not consider powers of two yet(since our interval was strictly between powers of 2), so we have to add $\sum_{k=1}^{n}{2^k}$ .
Note that $\sum_{i=1}^{n-2}{k\cdot2^{k}}=(2^1+2^2\cdots 2^{n-2})+(2^2+2^3\cdots 2^{n-2})\cdots +(2^{n-3}+2^{n-2})+2^{n-2}=2^1(2^{n-2}-1)+2^2(2^{n-3}-1)\cdots +2^{n-3}(2^2-1)+2^{n-2}(2^1-1)=(n-2)(2^{n-1})-\sum_{k=1}^{n-2}{2^k}=(n-2)(2^{n-1})-2(2^{n-2}-1)=(n-3)(2^{n-1})+2$ . Adding $\sum_{k=1}^{n}{2^k}=2(2^n-1)$ , we get $(n+1)(2^{n-1})$ .
If this sum is a perfect square, $n\not\equiv0\pmod2$ , since that would make $2^{n-1}$ not a perfect square, and $n+1$ odd so it cannot contribute a factor of 2 to make the power of 2 a perfect square.
We want the least odd number less than $1000$ such that $(n+1)$ is an even perfect square. The greatest even square less than $1000$ is $30^2=900$ so $n+1=900 \Longrightarrow n=\boxed{899}$ | null | 899 |
91a423b774b9ac758d9cebbf8917e71a | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_14 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$ | Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields \[BO=\sqrt{TB^2-TO^2}=3\] Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and \[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\] Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields \[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}\] Apply Law of Cosines on $\bigtriangleup TBC$ we have \[BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC\] \[(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC\] \[\cos BTC=\frac{23}{50}\] Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have \[SB^2=ST^2+BT^2-2(ST)(BT)\cos STB\] \[SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}\] Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields \[SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}\] Let the perpendicular from $T$ hits $SBC$ at $P$ . Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$ . Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \[TP^2=TS^2-SP^2=TM^2-PM^2\] \[4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2\] Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$
Finally, by Pythagorean Theorem, \[TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}\] so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$ | null | 183 |
ca4c6cfa47bb9ce18c4261f3e3bd8a87 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_15 | Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | Suppose $b_{i} = \frac {x_{i}}3$ .
We have \[\sum_{i = 1}^{2006}b_{i}^{2} = \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\] So \[\sum_{i = 0}^{2005}b_{i} = \frac {b_{2006}^{2} - 2006}2\] Now \[\sum_{i = 1}^{2006}b_{i} = \frac {b_{2006}^{2} + 2b_{2006} - 2006}2\] Therefore \[\left|\sum_{i = 1}^{2006}b_{i}\right| = \left|\frac {(b_{2006} + 1)^{2} - 2007}2\right|\geq \left|\frac{2025 - 2007}{2}\right| = 9\] This lower bound can be achieved if we use $b_1 = -1$ $b_2 = 0$ $b_3 = -1$ $b_4 = 0$ , and so on until $b_{1962} = 0$ , after which we let $b_k = b_{k - 1} + 1$ so that $b_{2006} = 44$ . So \[\left|\sum_{i = 1}^{2006}x_{i}\right|\geq \boxed{027}\] | null | 027 |
ca4c6cfa47bb9ce18c4261f3e3bd8a87 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_15 | Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | We know $|x_k| = |x_{k - 1} + 3|$ . We get rid of the absolute value by squaring both sides: ${x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9$ . So we set this up:
\begin{align*} {x_1}^2 - {x_0}^2 & = 6{x_0} + 9 \\ {x_2}^2 - {x_1}^2 & = 6{x_1} + 9 \\ & \vdots \\ {x_{2007}}^2 - {x_{2006}}^2 & = 6{x_{2006}} + 9 \end{align*} There are $2007$ equations. Sum them. We get: ${x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}$
So $|x_1 + x_2 + \cdots + x_{2006}| = \tfrac 16 \left|{x_{2007}}^2 - 9\cdot{2007}\right|$
We know $3\ |\ x_{2007}$ and we want to minimize $\left|{x_{2007}}^2 - 9\cdot{2007}\right|$ , so $x_{2007}$ must be $3\cdot{45}$ for it to be minimal ( $45^2 = 2025$ which is closest to $2007$ ). We can achieve this with $x_k=3k$ till $x_{45}=135$ and then alternating $x_{46}=132$ $x_{47}=135$ and so on ... Then $x_{2k}=132$ and $x_{2k+1}=135$ for all $k>22$ . Since $2007$ is odd, we have $x_{2007}=135$
This means that $|x_1 + x_2 + \cdots + x_{2006}| = \tfrac 16 \left|9(2025 - 2007)\right| = \boxed{027}$ | null | 027 |
ca4c6cfa47bb9ce18c4261f3e3bd8a87 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_15 | Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | Playing around with a couple numbers, we see that we can generate the sequence $0, 3, -6, 3, -6, \cdots$ , and we can also generate the sequence $3, 6, 9, 12, \cdots$ after each $-6$ value. Thus, we will apply this to try and find some bounds. We can test if the first $1000$ pairs of numbers each sum up to $-3$ , and the rest form an arithmetic sequence, if the first $990$ pairs sum up to $-3$ , and so on. When we get to $980$ , we find that $980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303$ . If we shift the number of pairs up by $1$ , we get $981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}$ | null | 027 |
7d205f60a44af588c766e1c810027d50 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_1 | In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$ | Let the side length be called $x$ , so $x=AB=BC=CD=DE=EF=AF$
2006 II AIME-1.png
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$ . Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$ ,
and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation
Therefore, $AB$ is $\boxed{046}$ | null | 046 |
7d205f60a44af588c766e1c810027d50 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_1 | In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$ | Because $\angle B$ $\angle C$ $\angle E$ , and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$ . Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$ . Then $BF=x\sqrt2$ . Thus \begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*} so $x^2=2116$ , and $x=\boxed{046}$ | null | 046 |
5d8afe8885328d03835ca343dda2b23f | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_2 | The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ | By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$ , we have that
Also,
Combining these two inequalities:
\[6.25 < n < 900\]
Thus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$ | null | 893 |
ee9123cfb515df2793ad87243b94d4e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_3 | Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$ | Note that the product of the first $100$ positive odd integers can be written as $1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}$
Hence, we seek the number of threes in $200!$ decreased by the number of threes in $100!.$
There are
$\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97$
threes in $200!$ and
$\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48$
threes in $100!$
Therefore, we have a total of $97-48=\boxed{049}$ threes. | null | 049 |
ee9123cfb515df2793ad87243b94d4e4 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_3 | Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$ | We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\equiv 2 \pmod{3}$ in set S will be divisible by 3. In other words, every number $\equiv 3 \pmod{6}$ . This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. $33 + 11 + 4 + 1 = \boxed{49}$ | null | 49 |
d14ce79cf734b9a93b4685f376cde17c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_4 | Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which
An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations. | Clearly, $a_6=1$ . Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$ , and let the remaining $6$ be $a_7$ through ${a_{12}}$ . It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\ldots,a_{12})$ . Thus, there will be ${11 \choose 5}=\boxed{462}$ such ordered 12-tuples. | null | 462 |
d14ce79cf734b9a93b4685f376cde17c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_4 | Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which
An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations. | Clearly, $a_6=1$ , and either $a_1$ or $a_{12}$ is 12.
Case 1: $a_1 = 12$
In this case, there are 4 empty spaces between $a_1$ and $a_6$ , and 6 empty spaces between $a_6$ and $a_{12}$ $\binom{10}{4}$ is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.
Case 2: $a_{12} = 12$
In this case, there are 5 empty spaces between $a_1$ and $a_6$ , and 5 empty spaces between $a_6$ and $a_{12}$ $\binom{10}{5}$ is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.
$210 + 252 = \boxed{462}$ ordered 12-tuples. | null | 462 |
9cd7763611ae055a6723acf986df73bc | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | Without loss of generality , assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$ , totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$ . Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$
\[A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}\]
Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so
\begin{align*}A(1)\cdot A(6)+A(1)\cdot A(6)&=\frac{5}{96}\\ A(1)\cdot A(6)&=\frac{5}{192}\end{align*}
Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$ , so:
\[A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}\]
Combining the equations:
\begin{align*}A(6)\left(\frac{1}{3}-A(6)\right)&=\frac{5}{192}\\ 0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\ A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*} We know that $A(6)>\frac{1}{6}$ , so it can't be $\frac{1}{8}$ . Therefore, the probability is $\frac{5}{24}$ and the answer is $5+24=\boxed{29}$ | null | 29 |
9cd7763611ae055a6723acf986df73bc | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | We have that the cube probabilities to land on its faces are $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}+x$ $\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \[4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}\] multiplying by 288 we get: \[32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] dividing by 16 and rearranging we get: \[\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}\] so the probability F which is greater than $\frac{1}{6}$ is equal $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}$ | null | 29 |
9cd7763611ae055a6723acf986df73bc | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is \[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\] Let the probability of obtaining face $F$ be $(1/6)+x$ . Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$ . Therefore
\begin{align*} {{47}\over{288}}&= 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) \left({1\over6}-x\right)\cr&= {4\over36}+2\left({1\over36}-x^2\right)\cr&= {1\over6}-2x^2. \end{align*}
Then $2x^2=1/288$ , and so $x=1/24$ . The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$ , and $m+n=\boxed{29}$ | null | 29 |
fc21682659f6861d498067bf5d06a0a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | [asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); [/asy] Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel , by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$ . Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$
$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$ . Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$ , so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$ . Since $\triangle AEF$ is equilateral $EF = AE = \sqrt{6} - \sqrt{2}$ $\triangle CEF$ is a $45-45-90 \triangle$ , so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$ . Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$ , so $(3 - \sqrt{3})s = 2 - \sqrt{3}$ . Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$ , and $a + b + c = 3 + 3 + 6 = \boxed{12}$ | null | 12 |
fc21682659f6861d498067bf5d06a0a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | Suppose $\overline{AB} = \overline{AD} = x.$ Note that $\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\angle BAE = \angle DAF = 15.$ Note that if $\overline{AD} = x$ and $\angle BAE = 15$ , then $\overline{AA'}=\frac{x}{\tan(15)}.$ Also note that \[AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x\] Using the fact $\tan(15) = 2-\sqrt{3}$ , this yields \[x = \frac{1}{3+\sqrt{3}} = \frac{3-\sqrt{3}}{6} \rightarrow 3 + 3 + 6 = \boxed{12}\] | null | 12 |
fc21682659f6861d498067bf5d06a0a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\frac{x}{1-x}$ , therefore $CE=\frac{1-2x}{1-x}$ . Then $AE=\sqrt{2}*\frac{1-2x}{1-x}$ . Using Pythagorean Theorem on $\triangle{ABE}$ yields $\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}$ . This means $6x^2-6x+1=0$ , and it's clear we take the smaller root: $x=\frac{3-\sqrt{3}}{6}$ . Answer: $\boxed{12}$ | null | 12 |
fc21682659f6861d498067bf5d06a0a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | Since $AEF$ is equilateral, $AE=EF$ . Let $BE=x$ . By the Pythagorean theorem $1+x^2=2(1-x)^2$ . Simplifying, we get $x^2-4x+1=0$ . By the quadratic formula, the roots are $2 \pm \sqrt{3}$ . Since $x<1$ , we discard the root with the "+", giving $x=2-\sqrt{3}$ [asy] real n; n=0.26794919243; real m; m=0.2113248654; draw((0,0)--(0,n)--(1,0)--(0,0)); draw((0,m)--(m,m)--(m,0)); label((0,0), "$B$",SW); label((0,n), "$E$",SW); label((0,m), "$M$",SW); label((1,0), "$A$",SW); label((m,0), "$N$",SW); label((m,m), "$K$",NE); [/asy] Let the side length of the square be s. Since $MEK$ is similar to $ABE$ $s=\frac{2-\sqrt{3}-s}{2-\sqrt{3}}$ . Solving, we get $s=\frac{3-\sqrt{3}}{6}$ and the final answer is $\boxed{012}$ | null | 012 |
dd00df1243139ab16fac6fc02dd7fdcd | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting ).
Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$ . Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs. | null | 738 |
dd00df1243139ab16fac6fc02dd7fdcd | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | We proceed by casework on the number of digits of $a.$
Case 1: Both $a$ and $b$ have three digits
We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \cdot 8 \cdot 9 = 576$ possible values in this case.
Case 2: $a$ (or $b$ ) has two digits
If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$ ). There thus are $8 \cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \cdot 2 = 144$ possible values in this case.
Case 3: $a$ (or $b$ ) has one digit
This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \cdot 2 = 18$ possible values for this case.
Adding them all up, we get $576 + 144 + 18 = \boxed{738},$ and we're done. | null | 738 |
dd00df1243139ab16fac6fc02dd7fdcd | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | For every $a \in [1, 999]$ $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits.
First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have:
However, we have still overlooked the cases where $1000 - a$ contains zero digits:
Adding up the cases with $a \in [1, 999]$ with no zero digits and removing the cases with $1000 - a$ with zero digits gives us \[(9^1 + 9^2 + 9^3) - 9^2 = \boxed{738}\] | null | 738 |
35ae5aaf31e93f6f02f2c11a9bcd5f3a | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy] | If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
There are 6 possible colors for the center triangle.
Thus, in total we have $6\cdot(20 + 30 + 6) = \boxed{336}$ total possibilities. | null | 336 |
35ae5aaf31e93f6f02f2c11a9bcd5f3a | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy] | We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.
Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.
Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.
Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.
Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.
Now, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\boxed{336}$ distinguishable ways to color the triangle. | null | 336 |
35ae5aaf31e93f6f02f2c11a9bcd5f3a | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy] | There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \[6 \binom{5 + 3}{3} = 6\binom{8}{3} = \boxed{336}.\] | null | 336 |
34fd4d35ee281f291c891f4b29ea411d | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_9 | Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope , and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$ | 2006 II AIME-9.png
Call the centers $O_1, O_2, O_3$ , the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$ , and $s_2$ on $\mathcal{C}_2$ ), and the intersection of each common internal tangent to the X-axis $r, s$ $\triangle O_1r_1r \sim \triangle O_2r_2r$ since both triangles have a right angle and have vertical angles, and the same goes for $\triangle O_2s_2s \sim \triangle O_3s_1s$ . By proportionality , we find that $O_1r = 4$ ; solving $\triangle O_1r_1r$ by the Pythagorean theorem yields $r_1r = \sqrt{15}$ . On $\mathcal{C}_3$ , we can do the same thing to get $O_3s_1 = 4$ and $s_1s = 4\sqrt{3}$
The vertical altitude of each of $\triangle O_1r_1r$ and $\triangle O_3s_1s$ can each by found by the formula $c \cdot h = a \cdot b$ (as both products equal twice of the area of the triangle). Thus, the respective heights are $\frac{\sqrt{15}}{4}$ and $2\sqrt{3}$ . The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: $\sqrt{15 - \frac{15}{16}} = \frac{15}{4}$ , and by 30-60-90: $6$
From this information, the slope of each tangent can be uncovered. The slope of $t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{4}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}$ . The slope of $t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}$
The equation of $t_1$ can be found by substituting the point $r (4,0)$ into $y = \frac{1}{\sqrt{15}}x + b$ , so $y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}$ . The equation of $t_2$ , found by substituting point $s (16,0)$ , is $y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$ . Putting these two equations together results in the desired $\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}$ $\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}$ $= \frac{76 - 12\sqrt{5}}{4}$ $= 19 - 3\sqrt{5}$ . Thus, $p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}$ | null | 027 |
850b6fe534612f3227f18641b975ddca | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$
Of these three cases ( $|A| > |B|, |A| < |B|, |A|=|B|$ ), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ${5\choose k}$ ways to $A$ to have $k$ victories, and ${5\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$
Thus $p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}$ . The desired probability is the sum of the cases when $|A| \ge |B|$ , so the answer is $\frac{126}{512} + \frac{193}{512} = \frac{319}{512}$ , and $m+n = \boxed{831}$ | null | 831 |
850b6fe534612f3227f18641b975ddca | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.
Summing these 6 cases, we get $\frac{638}{1024}$ , which simplifies to $\frac{319}{512}$ , so our answer is $319 + 512 = \boxed{831}$ | null | 831 |
850b6fe534612f3227f18641b975ddca | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We can apply the concept of generating functions here.
The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$ . Thus, the total generating function for number of games he wins is
The generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$ .
Thus, the generating function for $A$ is
$1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$
The probability that $B$ wins 0 games is $\frac{1}{32}$ . Since the coefficients for all $x^{n}$ where
$n \geq 1$ sums to 32, the probability that $A$ wins more games is $\frac{32}{32}$
Thus, the probability that $A$ has more wins than $B$ is $\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}$
Thus, $319 + 512 = \boxed{831}$ | null | 831 |
850b6fe534612f3227f18641b975ddca | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | After the first game, there are $10$ games we care about-- those involving $A$ or $B$ . There are $3$ cases of these $10$ games: $A$ wins more than $B$ $B$ wins more than $A$ , or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity . There are therefore $\frac{1024-252}{2} = 386$ possibilities for each of the other two cases.
If $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\frac{638}{1024} = \frac{319}{512}$ , and $m+n=\boxed{831}$ | null | 831 |
34022b0112a8cfaa6b22f6f8751b58e8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11 | sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | Define the sum as $s$ . Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:
Thus $s = \frac{a_{28} + a_{30}}{2}$ , and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\boxed{834}$ .− | null | 834 |