problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $I$ be the incenter of $\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles , and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$ . Thus, $m + n = \boxed{923}$ | null | 923 |
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$ . By Heron's formula , the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$ . Since $\triangle ADE \sim \triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $A=\frac{1}{2}bh$ , we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$ . Thus, we have
Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$ | null | 923 |
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P$ be the incenter ; then it is be the intersection of all three angle bisectors . Draw the bisector $AP$ to where it intersects $BC$ , and name the intersection $F$
Using the angle bisector theorem , we know the ratio $BF:CF$ is $21:22$ , thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$ , giving $F$ a weight of $43$ . In the same manner, using another bisector, we find that $A$ has a weight of $20$ . So, now we know $P$ has a weight of $63$ , and the ratio of $FP:PA$ is $20:43$ . Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$ . So, $DE$ is $43/63$ the size of $BC$ . Multiplying this ratio by the length of $BC$ , we find $DE$ is $860/63 = m/n$ . Therefore, $m+n=\boxed{923}$ | null | 923 |
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\] | null | 923 |
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$
Applying the Angle Bisector Theorem on $\angle{A}$ we have \[\frac{AB}{BF}=\frac{AC}{CF}\] \[BF=BC\cdot(\frac{AB}{AB+AC})\] \[BF=\frac{420}{43}\] Since $BP$ is the angle bisector of $\angle{B}$ , we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives \[\frac{BA}{AP}=\frac{BF}{FP}\] \[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have \[\frac{DE}{BC}=\frac{AP}{AF}\] \[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\] Solving gets $DE=\frac{860}{63}$ . Thus $m+n=860+63=\boxed{923}$ | null | 923 |
d6b2092ac715a9fd182725d6c515b67d | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Label $P$ the point the angle bisector of $A$ intersects ${BC}$ . First we find ${BP}$ and ${PC}$ . By the Angle Bisector Theorem, $\frac{BP}{PC} = \frac{21}{22}$ and solving for each using the fact that ${BC} = 20$ , we see that ${BP} = \frac{420}{43}$ and $PC = \frac{440}{43}$
\[{AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}\] \[{AP} = 21*22 - \frac{440\cdot420}{43^2}\]
Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}$
Now because $\triangle{APE} ~ \triangle{ABC}$ , we can find ${DE}$ by $\frac{AO}{AP} \cdot 20$ . Dividing and simplifying, we see that $\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}$ . So the answer is $\boxed{923}$ | null | 923 |
3c1a3557fafe08b7ee600f98595fae58 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_8 | Call a positive integer $N$ 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double? | Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.
Given this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$
If $B$ in base $10$ is less than $2000$ , then $B$ as a number in base $7$ must be less than $2*7^3=686$
$686$ is non-existent in base $7$ , so we're gonna have to bump that down to $666_7$
This suggests that $A$ is less than $\frac{666}{2}=333$
Guess and check shows that $A<320$ , and checking values in that range produces $\boxed{315}$ | null | 315 |
3c1a3557fafe08b7ee600f98595fae58 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_8 | Call a positive integer $N$ 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double? | Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be \[abc\] in base 7. Then the number in expanded form is \[49a+7b+c\] in base 7 and \[100a+10b+c\] in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. \[98a+14b+2c=100a+10b+c\] which simplifies to \[2a=4b+c.\] The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, $b$ is $3$ and $c$ is $0$ . Therefore, the largest 7-10 double is 630 in base 7, or $\boxed{315}$ in base 10. | null | 315 |
58ab239cbcec74085b7cfd91b91ad59b | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_9 | In triangle $ABC$ $AB=13$ $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We let $[\ldots]$ denote area; then the desired value is
Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$ , we find that
and similarly that $\frac{[BDE]}{[ABC]} = q(1-p)$ and $\frac{[CEF]}{[ABC]} = r(1-q)$ . Thus, we wish to find \begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]} \\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*} We know that $p + q + r = \frac 23$ , and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}$ . Substituting, the answer is $\frac 1{45} - \frac 23 + 1 = \frac{16}{45}$ , and $m+n = \boxed{061}$ | null | 061 |
58ab239cbcec74085b7cfd91b91ad59b | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_9 | In triangle $ABC$ $AB=13$ $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By the barycentric area formula, our desired ratio is equal to \begin{align*} \begin{vmatrix} 1-p & p & 0 \\ 0 & 1-q & q \\ r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45} \end{align*}, so the answer is $\boxed{061}$ | null | 061 |
58ab239cbcec74085b7cfd91b91ad59b | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_9 | In triangle $ABC$ $AB=13$ $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Since the only conditions are that $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}$ , we can simply let one of the variables be equal to 0. In this case, let $p = 0$ . Then, $q + r = \frac{2}{3}$ and $q^2 + r^2$ $\frac{2}{5}$ . Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$ . Solving this system of equations, we get $q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}$ , and $r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}$ . Plugging back into $(1-q)(1-r)$ , we get $\frac{16}{45}$ , so the answer is $\boxed{061}$ | null | 061 |
e919338df3a25766f59838eee2d7fe5f | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10 | Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | The distance between the $x$ $y$ , and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore,
However, we have $3\cdot 4\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\frac {5\cdot 8\cdot 13 - 60}{60\cdot 59} = \frac {23}{177}\Longrightarrow m+n = \boxed{200}$ | null | 200 |
e919338df3a25766f59838eee2d7fe5f | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10 | Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. We group the points by parity of each individual coordinate -- that is, if $x$ is even or odd, $y$ is even or odd, and $z$ is even or odd. Note that to have something that works, the two points must have this same type of classification (otherwise, if one doesn't match, the resulting sum for the coordinates will be odd at that particular spot).
There are $12$ EEEs (the first position denotes the parity of $x,$ the second $y,$ and the third $z.$ ), $8$ EEOs, $12$ EOEs, $6$ OEEs, $8$ EOOs, $4$ OEOs, $6$ OOEs, and $4$ OOOs. Doing a sanity check, $12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,$ which is the total number of points.
Now, we can see that there are $12 \cdot 11$ ways to choose two EEEs (respective to order), $8 \cdot 7$ ways to choose two EEOs, and so on. Therefore, we get \[12\cdot11 + 8\cdot7 + 12\cdot11 + 6\cdot5 + 8\cdot7 + 4\cdot3 + 6\cdot5 + 4\cdot3 = 460\] ways to choose two points where order matters. There are $60 \cdot 59$ total ways to do this, so we get a final answer of \[\dfrac{460}{60 \cdot 59} = \dfrac{23}{3 \cdot 59} = \dfrac{23}{177},\] for our answer of $23 + 177 = \boxed{200}.$ | null | 200 |
e919338df3a25766f59838eee2d7fe5f | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10 | Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | Similarly to Solution 2, we note that there are $60$ points and that the parities of the two points' coordinates must be the same in order for the midpoint to be in $S$
Ignore the distinct points condition. The probability that the midpoint is in $S$ is then \[\left(\left(\frac 23\right)^2+\left(\frac 13\right)^2\right)\left(\left(\frac 24\right)^2+\left(\frac 24\right)^2\right)\left(\left(\frac 35\right)^2+\left(\frac 25\right)^2\right)=\frac{13}{90}.\]
Note that $\frac{13}{90}=\frac{520}{3600}$ . Since there are $3600$ total ways to choose $2$ points from $S$ , there must be $520$ pairs of points that have their midpoint in $S$ . Of these pairs, $60$ of them contain identical points (not distinct).
Subtracting these cases, our answer is $\frac{520-60}{3600-60}=\frac{23}{177}\implies\boxed{200}$ | null | 200 |
e919338df3a25766f59838eee2d7fe5f | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10 | Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. Note that in order for the midpoint of the line segment to be a lattice point, the lengths on the x, y, and z axis must be even numbers. We will define all segments by denoting the amount they extend in each dimension: $(x, y, z)$ . For example, the longest diagonal possible will be $(2,3,4)$ , the space diagonal of the box. Thus, any line segment must have dimensions that are even. For $x$ the segment may have a value of $0$ for $x$ , (in which case the segment would be two dimensional) or a value of $2$ . The same applies for $y$ , because although it is three units long the longest even integer is two. For $z$ the value may be $0$ $2$ , or $4$ . Notice that if a value is zero, then the segment will pertain to only two dimensions. If two values are zero then the line segment becomes one dimensional.
Then the total number of possibilities will be $2 \cdot 2 \cdot 3$
Listing them out appears as follows:
$2,2,4$
$2,2,2$
$2,2,0$
$2,0,4$
$2,0,2$
$2,0,0$
$0,2,4$
$0,2,2$
$0,2,0$
$0,0,4$
$0,0,2$
$0,0,0$ * this value is a single point
Now, picture every line segment to be the space diagonal of a box. Allow this box to define the space the segment occupies. The question now transforms into "how many ways can we arrange this smaller box in the two by three by four?".
Notice that the amount an edge can shift inside the larger box is the length of an edge of the larger box (2, 3, or 4) minus the edge of the smaller box (also known as the edge), plus one. For example, (0, 2, 2) would be $3 \cdot 2 \cdot 3$ . Repeat this process.
$2,2,4$
$2,2,2$
$2,2,0$ 10
$2,0,4$
$2,0,2$ 12
$2,0,0$ 20
$0,2,4$
$0,2,2$ 18
$0,2,0$ 30
$0,0,4$ 12
$0,0,2$ 36
$0,0,0$ 60 * this won't be included, but notice that sixty the number of lattice points
Finally, we remember that there are four distinct space diagonals in a box, so we should multiply every value by four, right? Unfortunately we forgot to consider that some values have only one or two dimensions. They should be multiplied by one or two, respectively. This is because segments with two dimensions are the diagonals of a rectangle and thus have two orientations. Then any value on our list without any zeroes will be multiplied by four, and any value on our list with only one zero will be multiplied by two, and finally any value on our list with two zeroes will be multiplied by one:
$2,2,4$ 2 8
$2,2,2$ 6 24
$2,2,0$ 10 20
$2,0,4$ 4 8
$2,0,2$ 12 24
$2,0,0$ 20 20
$0,2,4$ 6 12
$0,2,2$ 18 36
$0,2,0$ 30 30
$0,0,4$ 12 12
$0,0,2$ 36 36
$0,0,0$ 60 * it's nice to point out that this value will be multiplied by zero
add every value on the rightmost side of each term and we will receive $230$ . Multiply by two because each segment can be flipped, to receive $460$ . There are $60 \cdot 59$ ways to choose two distinct points, so we get \[\dfrac{460}{60 \cdot 59} = \dfrac{23}{3 \cdot 59} = \dfrac{23}{177},\] for our answer of $23 + 177 = \boxed{200}$ | null | 200 |
21704a8c587cf4839fa3b71ea7eb9597 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_11 | In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$ | Let each point $P_i$ be in column $c_i$ . The numberings for $P_i$ can now be defined as follows. \begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*}
We can now convert the five given equalities. \begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{align} Equations $(1)$ and $(2)$ combine to form \[N = 24c_2 - 19\] Similarly equations $(3)$ $(4)$ , and $(5)$ combine to form \[117N +51 = 124c_3\] Take this equation modulo 31 \[24N+20\equiv 0 \pmod{31}\] And substitute for N \[24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}\] \[18 c_2 \equiv 2 \pmod{31}\]
Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$
The column values can also easily be found by substitution \begin{align*}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{align*} As these are all positive and less than $N$ $\boxed{149}$ is the solution. | null | 149 |
21704a8c587cf4839fa3b71ea7eb9597 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_11 | In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$ | If we express all the $c_i$ in terms of $N$ , we have \[24c_1=5N+23\] \[24c_2=N+19\] \[124c_3=117N+51\] \[124c_4=73N+35\] \[124c_5=89N+7\]
It turns out that there exists such an array satisfying the problem conditions if and only if \[N\equiv 149 \pmod{744}\]
In addition, the first two equation can be written $n = 5mod24$ , and chasing variables in the last three equation gives us $89n + 7 = 124e$ . With these two equations you may skip a lot of rewriting and testing. $\boxed{149}$ still appears as our answer. | null | 149 |
507bb68df5b5393577229202c5e901c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_12 | sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label("$I$",I,W); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W*-1); label("$D$",D,W*-1); [/asy]
The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$ , where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$
Thus $\frac{(I-C) \cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$
\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ \frac{44r -48}{28}&=-r\\ 44r-48&=-28r\\ 72r&=48\\ r&=\frac{2}{3} \end{align*}
Finally $2+3=\boxed{005}$ | null | 005 |
507bb68df5b5393577229202c5e901c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_12 | sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | Notice that we can split the tetrahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the tetrahedrons are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be $V$ and surface area be $F$ , using the volume formula for each pyramid(base times height divided by 3) we have $\dfrac{rF}{3}=V$ . The surface area of the pyramid is $\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$ . We know triangle ABC's side lengths, $\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},$ and $\sqrt{4^{2}+6^{2}}$ , so using the expanded form of heron's formula, \begin{align*}[ABC]&=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\ &=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}\\ &=\sqrt{196}\\ &=14\end{align*} Therefore, the surface area is $14+22=36$ , and the volume is $\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$ , and using the formula above that $\dfrac{rF}{3}=V$ , we have $12r=8$ and thus $r=\dfrac{2}{3}$ , so the desired answer is $2+3=\boxed{005}$ | null | 005 |
507bb68df5b5393577229202c5e901c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_12 | sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | First let us find the equation of the plane passing through $(6,0,0), (0,0,2), (0,4,0)$ . The "point-slope form" is $A(6-x1)+B(0-y1)+C(0-z1)=0.$ Plugging in $(0,0,2)$ gives $A(6)+B(0)+C(-2)=0.$ Plugging in $(0,4,0)$ gives $A(6)+B(-4)+C(0)=0.$ We can then use Cramer's rule/cross multiplication to get $A/(0-8)=-B/(0+12)=C/(-24)=k.$ Solve for A, B, C to get $2k, 3k, 6k$ respectively. We can then get $2k(x-x1)+3k(y-y1)+6k(z-z1)=0.$ Cancel out k on both sides. Next, let us substitute $(0,0,2)$ . We can then get $2x+3y+6z=12$ as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get $2x/7+2y/7+6z/7=12/7$ to be the normal form. Note that the point is going to be at $(r,r,r).$ We find the distance from $(r,r,r)$ to the plane as $2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})$ , which is $+/-(11r/7-12/7)$ . We take the negative value of this because if we plug in $(0,0,0)$ to the equation of the plane we get a negative value. We equate that value to r and we get the equation $-(11r/7-12/7)=r$ to solve $r={2/3}$ , so the answer is $\boxed{005}$ | null | 005 |
507bb68df5b5393577229202c5e901c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_12 | sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | Clearly, if the radius of the sphere is $r$ , the center of the sphere lies on $(r, r, r)$
We find the equation of plane $ABC$ to be $\frac16 x+\frac14 y+\frac12 z=1$ . From the definition of the insphere, it must be true that the distance from the center of the sphere to plane $ABC$ is equal to the length of the radius of the sphere. By point-to-plane, we have \[r=\frac{|\frac16 r+\frac14 r+\frac12 r-1|}{\sqrt{\left(\frac16\right)^2+\left(\frac14\right)^2+\left(\frac12\right)^2}} \implies r=\frac23,\] so the answer is $\boxed{005}$ | null | 005 |
ce7f62b86dd7a7d5926c7aaf7dd05c50 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13 | In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$ | Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ $AC$ , and $BD$ be the chords of the $d$ -degree arcs, and let $CD$ be the chord of the $3d$ -degree arc. Also let $x$ be equal to the chord length of the $3d$ -degree arc. Hence, the length of the chords, $AD$ and $BC$ , of the $2d$ -degree arcs can be represented as $x + 20$ , as given in the problem.
Using Ptolemy's theorem,
\[AB(CD) + AC(BD) = AD(BC)\] \[22x + 22(22) = (x + 20)^2\] \[22x + 484 = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \[x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}\] \[x = \frac{-18 + \sqrt{660}}{2}\]
$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$ | null | 174 |
ce7f62b86dd7a7d5926c7aaf7dd05c50 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13 | In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$ | Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, \[2R\sin z=22\] \[2R(\sin 2z-\sin 3z)=20\] Dividing the latter by the former, \[\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}\] \[4\cos^2z-2\cos z-\frac{1}{11}=0 (1)\] We want to find \[\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).\] From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174}$ | null | 174 |
ce7f62b86dd7a7d5926c7aaf7dd05c50 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13 | In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$ | Let $z=\frac{d}{2}$ $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z)\] \[20+a=2R\sin(2z)\] \[a=2R\sin(3z)\] Dividing the second from the first we get $\cos(z)=\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as follows: $a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer to be $\boxed{174}$ | null | 174 |
2e3eb9a57c7f57cbe2da416ce91c9d94 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$ -digit strings of $0$ 's and $1$ 's such that there are no two consecutive $1$ 's and no three consecutive $0$ 's.
The last two digits of any $n$ -digit string can't be $11$ , so the only possibilities are $00$ $01$ , and $10$
Let $a_n$ be the number of $n$ -digit strings ending in $00$ $b_n$ be the number of $n$ -digit strings ending in $01$ , and $c_n$ be the number of $n$ -digit strings ending in $10$
If an $n$ -digit string ends in $00$ , then the previous digit must be a $1$ , and the last two digits of the $n-1$ digits substring will be $10$ . So \[a_{n} = c_{n-1}.\]
If an $n$ -digit string ends in $01$ , then the previous digit can be either a $0$ or a $1$ , and the last two digits of the $n-1$ digits substring can be either $00$ or $10$ . So \[b_{n} = a_{n-1} + c_{n-1}.\]
If an $n$ -digit string ends in $10$ , then the previous digit must be a $0$ , and the last two digits of the $n-1$ digits substring will be $01$ . So \[c_{n} = b_{n-1}.\]
Clearly, $a_2=b_2=c_2=1$ . Using the recursive equations and initial values: \[\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \multicolumn{19}{c}{}\\\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\\\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\\\hline \end{array}\]
As a result $a_{19}+b_{19}+c_{19}=\boxed{351}$ | null | 351 |
2e3eb9a57c7f57cbe2da416ce91c9d94 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | We split the problem into cases using the number of houses that get mail. Let "|" represent a house that gets mail, and "o" represent a house that doesn't. With a fixed number of |, an o can be inserted between 2 |'s or on the very left or right. There cannot be more than one o that is free to arrange to be placed between two |'s because no three o's can be adjacent, but there can be a maximum of two o's placed on the very left or right. Note that according to the Pigeonhole Principle , no more than 10 houses can get mail on the same day.
Case 1: 10 houses get mail. No 2 adjacent houses can get mail on the same day, so there must be an o between every two |. $10-1=9$ o's are fixed so we count the number of ways to insert $19 - 10 - 9 = 0$ o's to $10+1 = 11$ spots, or $\binom{11}{0} = 1$
Case 2: 9 houses get mail. In this case, $9-1 = 8$ o's are fixed so we count the number of ways to insert $19 - 9 - 8 = 2$ o's to $9+1=10$ spots. However, there is also the case where two o's are both on the very left / right. When both o's that are free to arrange are put on a side, there are $10-1=9$ spots left to insert $2-2=0$ o's. Hence the total number of ways in this case is $\binom{10}{2} + 2\binom{9}{0} = 47$
Case 3: 8 houses get mail. In this case, $8-1=7$ o's are fixed so we count the number of ways to insert $19-8-7=4$ o's to $8+1=9$ spots. When two o's are put to the very left / right, there are $9-1=8$ spots left to insert $4-2=2$ o's. We also need to take care of the case where two o's are on the very left and two o's are on the very right: we have $9-1-1=7$ spots to insert $4-2-2=0$ o's. Hence the total number of ways in this case is $\binom{9}{4} + 2\binom{8}{2} + \binom{7}{0} = 183$
Case 4: 7 houses get mail. In this case, $7-1=6$ o's are fixed so we count the number of ways to insert $19-7-6=6$ o's to $7+1=8$ spots. When two o's are put to the very left / right, there are $8-1=7$ spots left to insert $6-2=4$ o's. When two o's are on the very left and two o's are on the very right, we have $8-1-1=6$ spots to insert $6-2-2=2$ o's. Hence the total number of ways in this case is $\binom{8}{6} + 2\binom{7}{4} + \binom{6}{2} = 113$
Case 5: 6 houses get mail. We have to be careful in this case: $6-1=5$ o's are fixed so we are inserting $19-6-5=8$ o's to $6+1=7$ spots, which means that at least 1 of the 2 sides must have two o's. When 1 of the 2 sides have two o's, there are $7-1=6$ spots to insert $8-2=6$ o's. When both sides have two o's, there are $7-1-1=5$ spots to insert $8-2-2=4$ o's. Hence the total number of ways in this case is $2\binom{6}{6} + \binom{5}{4} = 7$
When less than 6 houses get(s) mail, it's again not possible since at least three o's must be together (again, according to the Pigeonhole Principle). Therefore, the desired answer is $1+47+183+113+7=\boxed{351}$ | null | 351 |
2e3eb9a57c7f57cbe2da416ce91c9d94 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | There doesn't seem to be anything especially noticeable about the number nineteen in this problem, meaning that we can replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candidate for recursion.
At first, it's not immediately clear how to relate the state of $n$ houses in general to that of $n - 1, n - 2,$ or $n - 3.$ We thus break it up into cases, based on whether the first house gets mail or not.
Let $p_n$ be the number of ways to distribute the mail to $n$ houses. Assume that the first house gets mail. Therefore, since no two adjacent houses get mail on the same day, the second house must not get mail. Starting from the third house, however, things start to look messy, and it looks like we have to break our recurrence down into even smaller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymore, we try to work backwards.
Once the mail carrier delivers the mail to the first and (lack of mail) to the second houses, have him deliver mail to the remaining $n - 3$ houses at the end of the row, skipping the third house. There are $p_{n - 3}$ ways to do this. Now, we see that the availability of mail at the third house is fixed -- if the fourth house doesn't receive mail, the third one must, and if the fourth house receives mail, the third one can't. Therefore, there are simply $p_{n-3}$ ways to deliver the mail if the first house gets mail.
If the first house doesn't get mail, then we use the same logic -- have the mail carrier skip the second house and deliver the remaining mail to the $n - 2$ houses in $p_{n-2}$ ways. Then, the availability of mail for the second house is fixed, so there are $p_{n - 2}$ ways to deliver the mail in this case.
We thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and cannot achieve both at the same time, we are confident about the validity of our relation: \[p_n = p_{n-2} + p_{n-3}.\] Now, we simply calculate $p_1, p_2,$ and $p_3.$ Then, it's off to the races for computation!
$p_1 = 2,$ because the first house can either gets mail or it doesn't -- there are no restrictions.
$p_2 = 3,$ because all of the possible deliveries are valid (of which there are $2 \cdot 2 = 4$ ) except the one where both houses receive mail.
$p_3 = 4,$ as there are $4$ possible ways (here, M represents that that house gets mail and N represents no mail): MNM, MNN, NNM, NMN.
Using our recurrence relation, we eventually get that $p_{19} = \boxed{351},$ and we're done. | null | 351 |
2e3eb9a57c7f57cbe2da416ce91c9d94 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | Let $w_n$ be the number of possible ways if the last house has mail, and $b_n$ be the number of possible ways if the last house does not have mail.
If the last house has mail, then, the next house can't have mail, meaning that $b_n = w_{n - 1}$
If the last house doesn't have mail, then the next house can either have mail or not have mail. If the next house has mail, then we simply count the number of ways that the row ends in a house with mail, so that means so far, our recursive rule is $w_n = b_{n - 1} + \text{something}$ . If the next house does not have mail, then the next house after that must have mail, meaning that $w_n = b_{n - 1} + b_{n - 2}$
Recursing all the way up to $b_{19}$ and $w_{19}$ , we get $100 + 251 = \boxed{351}$ | null | 351 |
2e3eb9a57c7f57cbe2da416ce91c9d94 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail.
$a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house.
$b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house.
Note that we only need list out values of $a_n$ as $b$ depends on $a$ $a_1=1, a_2=1, a_3=2, a_4=2, \ldots$
$a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\cdot a_{17}+a_{18}=65+2\cdot 86+114=\boxed{351}$ | null | 351 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | It is helpful to consider the cube $ABCDEFGH$ , where the vertices of the cube represent the faces of the octahedron, and the edges of the cube represent adjacent octahedral faces. Each assignment of the numbers $1,2,3,4,5,6,7$ , and $8$ to the faces of the octahedron corresponds to a permutation of $ABCDEFGH$ , and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem asks one to count octagonal circuits that can be formed by eight of these diagonals. Six of the diagonals are edges of tetrahedron $ACFH$ , six are edges of tetrahedron $DBEG$ , and four are $\textit{long}$ , joining a vertex of one tetrahedron to the diagonally opposite point from the other. Notice that each vertex belongs to exactly one long diagonal.
It follows that an octagon cannot have two successive long diagonals. Also notice that an octagonal path can jump from one tetrahedron to the other only along one of the long diagonals. It follows that an octagon must contain either 2 long diagonals separated by 3 tetrahedron edges or 4 long diagonals alternating with tetrahedron edges.
To form an octagon that contains four long diagonals, choose two opposite edges from tetrahedron $ACFH$ and two opposite edges from tetrahedron $DBEG$ . For each of the three ways to choose a pair of opposite edges from tetrahedron $ACFH$ , there are two possible ways to choose a pair of opposite edges from tetrahedron $DBEG$ . There are 6 distinct octagons of this type and $8\cdot 2$ ways to describe each of them, making 96 permutations.
To form an octagon that contains exactly two of the long diagonals, choose a three-edge path along tetrahedron $ACFH$ , which can be done in $4! = 24$ ways. Then choose a three-edge path along tetrahedron $DBEG$ which, because it must start and finish at specified vertices, can be done in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8\cdot 24\cdot 2=384$ permutations of this type.
In all, there are $96 +384 = 480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\tfrac{480}{8!}=\tfrac{1}{84}$ , and $m+n=\boxed{085}$ | null | 085 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | Choose one face of the octahedron randomly and label it with $1$ . There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.
Clearly, the labels for the A-faces must come from the set $\{3,4,5,6,7\}$ , since these faces are all adjacent to $1$ . There are thus $5 \cdot 4 \cdot 3 = 60$ ways to assign the labels for the A-faces.
The labels for the B-faces and C-face are the two remaining numbers from the above set, plus $2$ and $8$ . The number on the C-face must not be consecutive to any of the numbers on the B-faces.
From here it is easiest to brute force the $10$ possibilities for the $4$ numbers on the B and C faces:
There is a total of $10$ possibilities. There are $3!=6$ permutations (more like "rotations") of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$ . There are $7!=5040$ ways to randomly fill in the rest of the octahedron. So the probability is $\frac {60}{5040} = \frac {1}{84}$ . The answer is $\boxed{085}$ | null | 085 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits ( $B$ for black and $W$ for white).
Type I: $BB-WWWW-BB$ . There are $4!$ ways to arrange the black vertices and consequently two of the white vertices and $2!$ ways to arrange the other two white vertices. Since the template has a period of $8$ , there are $4!\cdot 2!\cdot 8 = 384$ circuits of type I.
Type II: $B-WW-BB-WW-B$ . There are $4!$ ways to arrange the black vertices and consequently the white vertices. Since the template has a period of $4$ , there are $4! \cdot 4 = 96$ circuits of type II.
Thus, there are $384+96=480$ circuits satisfying the given condition, out of the $8!$ possible circuits. Therefore, the desired probability is $\frac{480}{8!} = \frac{1}{84}$ . The answer is $\boxed{085}$ | null | 085 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices.
The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in $5\cdot4\cdot3=60$ ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.
The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus $\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}$ , from which the answer is $\boxed{085}$ | null | 085 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [asy] import three; draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw((0,0,0)--(1,0,0)); draw((0,0,1)--(1,0,1)); for(int i = 0; i < 2; ++i) { for(int j = 0; j < 2; ++j) { for(int k = 0; k < 2; ++k) { dot((i,j,k)); } } } // dot((0,0,1),blue); // dot((0,1,0),green); // dot((1,0,0),red); draw((0,0,0)--(1,1,0)); draw((0,1,0)--(1,0,0)); draw((0,0,1)--(1,1,1)); draw((0,1,1)--(1,0,1)); [/asy]
The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$
Case 1: Four of the vertical edges are used. $6\cdot 2=12.$
Case 2: Two of the vertical edges are used. $4\cdot 3 \cdot 2\cdot 2=48.$
So, the probability is $\frac{60}{5040}=\frac{1}{84}.$ Therefore, our answer is $\boxed{085}$ | null | 085 |
04cb73c9d25de35394fe99fd17f4ffde | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | As with some of the previous solutions, consider the cube formed by connecting the centroids of the faces on the octahedron. We choose a random vertex(hence fixing the diagram), giving us $7!$ ways as our denominator. WLOG, we color this start vertex red, and we color all $3$ vertices adjacent to it blue. We repeat this for the other vertices.
Note that there is a 1-1 correspondence between the number of valid face numberings with rotational symmetry and the number of ways to move a particle to every vertex of the cube and returning to the start vertex with only diagonal moves. We can move along either short diagonals and long diagonals.
Next, note that we can only move from the red vertices to the blue vertices and back with long diagonals(since short diagonals keep us on the color we are currently on). Thus, it is easy to check that the only possible sequences of long and short diagonals are cyclic permutations of $SSSLSSSL$ and $SLSLSLSL$
Case 1: $SSSLSSSL$ .
There are $4$ cyclic permutations, and we can traverse the entirety of the red tetrahedron in $4$ steps in $3!$ ways. Then, after the move to the blue tetrahedron from one of the non-starting red vertices, we realize that our first step cannot be to the blue vertex opposite the starting red vertex. Hence, there are $2$ possibilities for our first step. However, once we make our first move, the path is fixed. This leaves us with $4! \cdot 2$ ways.
Case 2: $SLSLSLSL$ There are $2$ cyclic permutations in this case. After we choose one of the $3$ red vertices to go to from the starting vertex and move to the blue tetrahedron, we are once again left with $2$ choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our choices are fixed. This leaves us with $2 \cdot 3 \cdot 2$ ways.
Finally, summing and dividing by $7!$ gives us $\frac{60}{7!} = \frac{1}{84} \rightarrow \boxed{085}$ , as desired. - Spacesam | null | 085 |
0a23c0c24e8c353325e419bf25e70d46 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_1 | Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$ | The two-digit perfect squares are $16, 25, 36, 49, 64, 81$ . We try making a sequence starting with each one:
The largest is $81649$ , so our answer is $\boxed{816}$ | null | 816 |
4563accbbb63f0b236d81dd628a33222 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_2 | Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$ | Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cap F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$ , and $\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800$ . By the Principle of Inclusion-Exclusion
\[S+F- S \cap F = S \cup F = 2001\]
For $m = S \cap F$ to be smallest, $S$ and $F$ must be minimized.
\[1601 + 601 - m = 2001 \Longrightarrow m = 201\]
For $M = S \cap F$ to be largest, $S$ and $F$ must be maximized.
\[1700 + 800 - M = 2001 \Longrightarrow M = 499\]
Therefore, the answer is $M - m = 499 - 201 = \boxed{298}$ | null | 298 |
60d884f2e8d4d6f1c38d71bd68eaf942 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3 | Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$ | We find that $x_5 = 267$ by the recursive formula. Summing the recursions
\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \end{align*}
yields $x_{n} = -x_{n-5}$ . Thus $x_n = (-1)^k x_{n-5k}$ . Since $531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5$ , it follows that
\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.\] | null | 898 |
60d884f2e8d4d6f1c38d71bd68eaf942 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3 | Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$ | The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$ , we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$
Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}$
Thus,
\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.\] | null | 898 |
60d884f2e8d4d6f1c38d71bd68eaf942 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3 | Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$ | Calculate the first few terms:
\[211,375,420,523,267,-211,-375,-420,-523,\dots\]
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously $211+420+267=\boxed{898}$ | null | 898 |
8a9fcf3ff510eb8897d149680ade57f3 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_4 | Let $R = (8,6)$ . The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$ , respectively, such that $R$ is the midpoint of $\overline{PQ}$ . The length of $PQ$ equals $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | The coordinates of $P$ can be written as $\left(a, \frac{15a}8\right)$ and the coordinates of point $Q$ can be written as $\left(b,\frac{3b}{10}\right)$ . By the midpoint formula, we have $\frac{a+b}2=8$ and $\frac{15a}{16}+\frac{3b}{20}=6$ . Solving for $b$ gives $b= \frac{80}{7}$ , so the point $Q$ is $\left(\frac{80}7, \frac{24}7\right)$ . The answer is twice the distance from $Q$ to $(8,6)$ , which by the distance formula is $\frac{60}{7}$ . Thus, the answer is $\boxed{067}$ | null | 067 |
f72b3c83fbf6e76b003e3de3caa95276 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_6 | Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$ | Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem , the radius of $\odot O = OC = a\sqrt{2}$
Now consider right triangle $OGI$ , where $I$ is the midpoint of $\overline{GH}$ . Then, by the Pythagorean Theorem,
\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}
Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$ , and the answer is $10n + m = \boxed{251}$ | null | 251 |
f72b3c83fbf6e76b003e3de3caa95276 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_6 | Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$ | Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$ . Let $DF$ $b$ and diameter $HI$ go through $J$ the midpoint of $EF$ . Since a diameter always bisects a chord perpendicular to it, $DJ$ $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$ $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$ $\sqrt{2} / 2$ , half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$ . Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$ , and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$ , meaning the ratio of areas $((a-2b)/a)^2$ $(1/5)^2$ $1/25$ $m/n.$ Then $10n + m = 25 * 10 + 1 = \boxed{251}$ | null | 251 |
fde51bbcd0f43e27a937c8b13ee51e42 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ | Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter , and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$ . Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles .) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$ , and so $RS = 60, UQ = 30$
Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$ . Since the ratio of corresponding lengths of similar figures are the same, we have
\[\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.\]
Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$ , respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$ . Therefore, the answer is $n = \boxed{725}$ | null | 725 |
fde51bbcd0f43e27a937c8b13ee51e42 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ | We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$
By the Two Tangent Theorem , we find that $A_{1}Q = 60$ $A_{1}R = 90$ . Using the similar triangles, $RA_{2} = 45$ $QA_{3} = 20$ , so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$ . Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$ | null | 725 |
fde51bbcd0f43e27a937c8b13ee51e42 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ | The radius of an incircle is $r=A_t/\text{semiperimeter}$ . The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$ . The radius, therefore, is equal to $\frac{5400}{180} = 30$ . Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$ $\frac{1}{2}(90) = 45$ , and $\frac{1}{2}\times150 = 75$ . The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$
Defining $P$ as $(0,0)$ $C_2$ is equal to $(60+15,15)$ or $(75,15)$ . Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$ $\frac{1}{3}\times120$ $\frac{1}{3}\times150$ or $30,40,50$ . The radius of $C_3$ by using the formula mentioned at the beginning is $10$ . Using $P$ as $(0,0)$ $C_3$ is equal to $(10, 60+10)$ or $(10,70)$ . Using the distance formula, the distance between $C_2$ and $C_3$ $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$ , thus $n$ is $\boxed{725}$ | null | 725 |
fde51bbcd0f43e27a937c8b13ee51e42 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ | We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$ . We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$ , as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$ , which is $15$ less than the square plus the inradius of $C_3$ . Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$ | null | 725 |
e94247119f9ec6586f8f41714f2589c3 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_8 | A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$ | Iterating the condition $f(3x) = 3f(x)$ , we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$ . We know the definition of $f(x)$ from $1 \le x \le 3$ , so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$ . Indeed,
\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]
We now need the smallest $x$ such that $f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186$ . The range of $f(x),\ 1 \le x \le 3$ , is $0 \le f(x) \le 1$ . So when $1 \le \frac{x}{3^k} \le 3$ , we have $0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1$ . Multiplying by $3^k$ $0 \le 186 \le 3^k$ , so the smallest value of $k$ is $k = 5$ . Then,
\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]
Because we forced $1 \le \frac{x}{3^5} \le 3$ , so
\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243.\]
We want the smaller value of $x = \boxed{429}$ | null | 429 |
e94247119f9ec6586f8f41714f2589c3 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_8 | A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$ | First, we start by graphing the function when $1\leq{x}\leq3$ , which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$ . Similarly, using $f(3x)=3f(x)$ , we get a dilation of our initial figure by a factor of 3 for the next interval and so on.
Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$ . First, we compute $f(2001)$ . The nearest intersection point is $(1458,729)$ when $a=7$ . Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$ , we compute $f(2001)=729-543=186$ . However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$ , namely $(486,243)$ . Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$ , we get $x=486-57=\boxed{429}$ | null | 429 |
61b92791df83290f2d931d04f0569fc1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9 | Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We can use complementary counting , counting all of the colorings that have at least one red $2\times 2$ square.
By the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.
There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$ , and $417+512=\boxed{929}$ | null | 929 |
61b92791df83290f2d931d04f0569fc1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9 | Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We consider how many ways we can have 2*2 grid
$(1)$ : All the grids are red-- $1$ case
$(2)$ : One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases
$(3)$ : Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases.
Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\binom 9 2-8-2=26$ cases
$(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\cdot (\binom 5 3)-4=36$
$(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\cdot \binom 5 4=20$
$(6)$ Five unit squares are blue, $4$ cases in all
Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.
In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\frac{95}{512}=\frac{417}{512}$ leads to $\boxed{929}$ | null | 929 |
61b92791df83290f2d931d04f0569fc1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9 | Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | \[\begin{array}{|c|c|c|} \hline C_{11} & C_{12} & C_{13}\\ \hline C_{21} & C_{22} & C_{23}\\ \hline C_{31} & C_{32} & C_{33}\\ \hline \end{array}\]
Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square
Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$ . To make sure there are no more 2-by-2 red squares, $C_{31} \text{and} C_{32}$ can't both be red and $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \text{and} C_{32}$ and $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$ , there are $3 \cdot 3 \cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \cdot 4 = 68$ coloring methods.
Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares
Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ . To make sure there are no more 2-by-2 red squares, $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ , there are $3 \cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \cdot 4 = 20$ coloring methods.
Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$ . For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 2 = 2$ coloring methods.
Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods.
Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares
Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$ . For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 4 = 4$ coloring methods.
Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares
If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method.
In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \[\frac{m}{n}=1-\frac{95}{2^9}=\frac{417}{512}\]
\[m+n=417+512=\boxed{929}\] | null | 929 |
3bf05b8fd2bb27c72bb3cd57e3969ed0 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_10 | How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$ | Observation: We see that there is a pattern with $10^k \pmod{1001}$ \[10^0 \equiv 1 \pmod{1001}\] \[10^1 \equiv 10 \pmod{1001}\] \[10^2 \equiv 100 \pmod{1001}\] \[10^3 \equiv -1 \pmod{1001}\] \[10^4 \equiv -10 \pmod{1001}\] \[10^5 \equiv -100 \pmod{1001}\] \[10^6 \equiv 1 \pmod{1001}\] \[10^7 \equiv 10 \pmod{1001}\] \[10^8 \equiv 100 \pmod{1001}\]
So, this pattern repeats every 6.
Also, $10^j-10^i \equiv 0 \pmod{1001}$ , so $10^j \equiv 10^i \pmod{1001}$ , and thus, \[j \equiv i \pmod{6}\] . Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$ | null | 784 |
3bf05b8fd2bb27c72bb3cd57e3969ed0 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_10 | How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$ | Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$ ; therefore, the order of 10 modulo $7,11$ , and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$ | null | 784 |
3bf05b8fd2bb27c72bb3cd57e3969ed0 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_10 | How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$ | $10^j - 10^i \equiv 0 \pmod{1001} \iff 10^{j - i} - 1 \equiv 0 \pmod{1001} \iff 10^{j - i} \equiv 1 \pmod{1001} \iff j \equiv i \pmod 6$ . If $j \equiv i \equiv n \pmod 6$ for $n = 0, 1, 2, 3$ , there are $17$ choices for each value of $n$ , yielding $4 \cdot \dbinom{17}{2} = 544$ . However, if $n = 4, 5$ , there are only $16$ choices, giving us $2 \cdot \dbinom{16}{2} = 240$ . So, our final answer is $544 + 240 = \boxed{784}$ .
~Puck_0 | null | 784 |
113c9208dee84b63794cd66f32f96294 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_11 | Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle , the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\frac{6!}{3!3!}$ ways for the first, and $\frac{6!}{2!2!2!}$ $\frac{6!}{4!}$ , and $1$ ways for the rest, respectively, out of a total of $3^6$ . This gives a probability of $141/729$ . Then the desired answer is $\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}$ , so the answer is $m+n = \boxed{341}$ | null | 341 |
113c9208dee84b63794cd66f32f96294 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_11 | Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways
Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases
Last case, it only wins $1$ game so the rest games must be all draw, $1$ game
The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$ | null | 341 |
d4dd32e54ba5a59ce17c50cb0ea0ab6a | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_12 | Given a triangle , its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$ , replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $P_{3}$ is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | On the first construction, $P_1$ , four new tetrahedra will be constructed with side lengths $\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\left(\frac 12\right)^3 = \frac 18$ . The total volume added here is then $\Delta P_1 = 4 \cdot \frac 18 = \frac 12$
We now note that for each midpoint triangle we construct in step $P_{i}$ , there are now $6$ places to construct new midpoint triangles for step $P_{i+1}$ . The outward tetrahedron for the midpoint triangle provides $3$ of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other $3$ . This is because if you read this question carefully, it asks to add new tetrahedra to each face of $P_{i}$ which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of $\frac 18$ . Thus we have the recursion $\Delta P_{i+1} = \frac{6}{8} \Delta P_i$ , and so $\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1$
The volume of $P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}$ , and $m+n=\boxed{101}$ . Note that the summation was in fact a geometric series | null | 101 |
dd87f7f3ccc0a66fa860cfe009552cf7 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_13 | In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$ . Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$ , we know that $\triangle ABD \sim \triangle DCE$ . Hence $\angle ADB = \angle DEC$ , and $\triangle BDE$ is isosceles . Then $BD = BE = 10$
Using the similarity, we have:
\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]
The answer is $m+n = \boxed{069}$ | null | 069 |
dd87f7f3ccc0a66fa860cfe009552cf7 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_13 | In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Draw a line from $B$ , parallel to $\overline{AD}$ , and let it meet $\overline{CD}$ at $M$ . Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$ . Now since $ADMB$ is an isosceles trapezoid, $MD=8$ . By the similarity, we have $MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}$ , hence $CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}$ | null | 069 |
dd87f7f3ccc0a66fa860cfe009552cf7 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_13 | In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Since $\angle{BAD}=\angle{ADM}$ , if we extend AB and DC, they must meet at one point to form a isosceles triangle $\triangle{ADM}$ .Now, since the problem told that $\angle{ABD}=\angle{BCD}$ , we can imply that $\angle{DBM}=\angle{BCM}$ Since $\angle{M}=\angle{M}$ , so $\triangle{CBM}\sim\triangle{BDM}$ . Assume the length of $BM=x$ ;Since $\frac{BC}{MB}=\frac{DB}{MD}$ we can get $\frac{6}{x}=\frac{10}{8+x}$ , we get that $x=12$ .So $AM=DM=20$ similarly, we use the same pair of similar triangle we get $\frac{CM}{BM}=\frac{BM}{DM}$ , we get that $CM=\frac{36}{5}$ . Finally, $CD=MD-MC=\frac{64}{5}\implies 64+5=69=\boxed{069}$ ~bluesoul | null | 069 |
dd87f7f3ccc0a66fa860cfe009552cf7 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_13 | In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Denote $\angle{BAD}=\angle{CDA}=x$ , and $\angle{ABD}=\angle{BCD}=y$ . Note that $\angle{ADB}=180^\circ-x-y$ , and $\angle{DBC}=360^\circ-2x-2y$ . This motivates us to draw the angle bisector of $\angle{DBC}$ because $\angle{DBC} = 2 \angle{ADB}$ , so we do so and consider the intersection with $CD$ as $E$ . By the angle bisector theorem, we have $\frac{CE}{DE} = \frac{BC}{BD} = \frac{3}{5}$ , so we write $CE=3z$ and $DE=5z$ . We also know that $\angle{EBC}=\angle{ADB}$ and $\angle{BCE}=\angle{DBA}$ , so $\triangle{ADB} \sim \triangle{EBC}$ . Hence, $\frac{CE}{BC}=\frac{AB}{BD}$ , so we have $3z=\frac{24}{5}$ . As $CD=8z$ , it must be that $CD=\frac{64}{5}$ , so the final answer is $\boxed{069}$ | null | 069 |
5c1a1dada574542ce9f72c18a8365da5 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_14 | There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ | $z$ can be written in the form $\text{cis\,}\theta$ . Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$
Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i$ , or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i$
Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$ , we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
Again setting up equations ( $Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$ ) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$
Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed{840}$ degrees. We only want the sum of a certain number of theta, not all of it. | null | 840 |
5c8b431cb89e9d9b9c00060c5c300bfc | https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_15 | Let $EFGH$ $EFDC$ , and $EHBC$ be three adjacent square faces of a cube , for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ $J$ , and $K$ , be the points on $\overline{EF}$ $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$ , and containing the edges, $\overline{IJ}$ $\overline{JK}$ , and $\overline{KI}$ . The surface area of $S$ , including the walls of the tunnel, is $m + n\sqrt {p}$ , where $m$ $n$ , and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$ | Set the coordinate system so that vertex $E$ , where the drilling starts, is at $(8,8,8)$ . Using a little visualization (involving some similar triangles , because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$ , and $(0,1,0)$ to $(2,2,0)$ , and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$ , and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$ . Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$ . The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$ . Around the corner $E$ we're missing an area of $6$ , the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$ . Thus the the total surface area is $372 + 39\sqrt {6}$ , and the answer is $372 + 39 + 6 = \boxed{417}$ | null | 417 |
cdcf34586b457fc70674a9a1ebe1f8b9 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_1 | Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$ | If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization , then that factor will end in a $0$ . Therefore, we have left to consider the case when the two factors have the $2$ s and the $5$ s separated, so we need to find the first power of 2 or 5 that contains a 0.
For $n = 1:$ \[2^1 = 2 , 5^1 = 5\] $n = 2:$ \[2^2 = 4 , 5 ^ 2 =25\] $n = 3:$ \[2^3 = 8 , 5 ^3 = 125\]
and so on, until,
$n = 8:$ $2^8 = 256$ $5^8 = 390625$
We see that $5^8$ contains the first zero, so $n = \boxed{8}$ | null | 8 |
6b7c814d3d8c7479502452ec0a5e6587 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_2 | Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$ | Since $A = (u,v)$ , we can find the coordinates of the other points: $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$ . Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$ . Since $u,v$ are positive, $u+3v>u$ , and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$ . Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$ | null | 021 |
6b7c814d3d8c7479502452ec0a5e6587 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_2 | Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$ | We find the coordinates like in the solution above: $A = (u,v)$ $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . Then we apply the Shoelace Theorem \[A = \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\] \[\frac{1}{2}(2u^2 + 6uv) = 451\] \[u(u + 3v) = 451\]
This means that $(u,v) = (11, 10)$ or $(1,150)$ , but since $v < u$ , then the answer is $\boxed{021}$ | null | 021 |
52e95d5423678cacea4ea69c3a368328 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_3 | In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$ | Using the binomial theorem $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$
Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$ , and $a+b=\boxed{667}$ | null | 667 |
a5c70b124a5ce5efe19047a8540d1607 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy] | Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.
The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}
Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$
We can guess that $a_1 = 2$ . (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$ $a_6=25$ $a_8 = 33$ , which gives us $l=61,w=69$ . These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$ | null | 260 |
a5c70b124a5ce5efe19047a8540d1607 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy] | We can just list the equations: \begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \begin{align*} s_4 &= 2s_1 + s_2 \\ s_5 &= 3s_1 +2s_2 \\ s_6 &= 5s_1 + 3s_2 \\ s_7 &= 4s_1 + 4s_2 \\ s_8 &= 4s_1 + 5s_2 \\ s_9 &= 3s_1 + 6s_2 \\ \end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \[2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.\] Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.$ ~peelybonehead | null | 260 |
a5c70b124a5ce5efe19047a8540d1607 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy] | We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$ | null | 260 |
f916cf799f2a5b305f0a85f372011d73 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_5 | Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$ | If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$ . The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.
Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$ . Then, $a + b = 25$ , and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$ $50|ab$ . It follows that $5|a,b$ , so there are 2 pairs of $a$ and $b: (20,5),(15,10)$
Thus, $m + n = \boxed{026}$ | null | 026 |
f916cf799f2a5b305f0a85f372011d73 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_5 | Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$ | Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2.
Since there are $25$ marbles in the box:
$w_1 + w_2 + b_1 + b_2 = 25$
From the fact that there is a $\frac{27}{50}$ chance of drawing one black marble from each box:
$\frac{b_1 \cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \frac{27}{50} = \frac{54}{100} = \frac{81}{150}$
Thinking of the numerator and denominator separately, if $\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \cdot b_2 = 27$ . Since $b_1 < 25$ , this forces the variables to be $3$ and $9$ in some permutation. Without loss of generality, let $b_1 = 3$ and $b_2 = 9$
The denominator becomes: $(3 + w_1)(9 + w_2) = 50$
Since there have been $12$ black marbles used, there must be $13$ white marbles. Substituting that in:
$(3 + w_1)(9 + (13 - w_1)) = 50$
$(3 + w_1)(22 - w_1) = 50$
Since the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$ , the quantity $3 + w_1$ must equal one of those. However, since $w_1 < 13$ , testing $2$ and $7$ for $w_1$ does not give a correct product. Thus, $\frac{27}{50}$ must be a reduced form of the actual fraction.
First assume that the fraction was reduced from $\frac{54}{100}$ , yielding the equations $b_1\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$ .
Factoring $b_1 \cdot b_2 = 54$ and saying WLOG that $b_1 < b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$ . Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$
Since $w_1 + w_2 = 4$ , the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble. Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$ , so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$
Thus, the chance of drawing 2 white marbles is $\frac{w_1 \cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \frac{4}{100} = \frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \boxed{026}.$ | null | 026 |
f916cf799f2a5b305f0a85f372011d73 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_5 | Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$ | We know that $\frac{27}{50} = \frac{b_1}{t_1} \cdot \frac{b_2}{t_2}$ , where $b_1$ and $b_2$ are the number of black marbles in the first and the second box respectively, and $t_1$ and $t_2$ is the total number of marbles in the first and the second boxes respectively. So, $t_1 + t_2 = 25$ . Then, we can realize that $\frac{27}{50} = \frac{9}{10} \cdot \frac{3}{5} = \frac{9}{10} \cdot \frac{9}{15}$ , which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes $\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}$ . The answer is $1 + 25 = \boxed{026}$ | null | 026 |
1d040528fc345db092ea66594ef8ade8 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | \begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}
Because $y > x$ , we only consider $+2$
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can be is $\sqrt{10^6} - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$ , an integer). Then $\sqrt{x} = 997$ , and we continue this downward until $\sqrt{y} = 3$ , in which case $\sqrt{x} = 1$ . The number of pairs of $(\sqrt{x},\sqrt{y})$ , and so $(x,y)$ is then $\boxed{997}$ | null | 997 |
1d040528fc345db092ea66594ef8ade8 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Let $a^2$ $x$ and $b^2$ $y$ , where $a$ and $b$ are positive.
Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2ab + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$ , then we can use all positive integers less than 1000 for $a$ and $b$
We know that because $x < y$ , we get $a < b$
We can count even and odd pairs separately to make things easier*:
Odd: \[(1,3) , (3,5) , (5,7) . . . (997,999)\]
Even: \[(2,4) , (4,6) , (6,8) . . . (996,998)\]
This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs. | null | 997 |
1d040528fc345db092ea66594ef8ade8 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]
Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$ , so the problem asks for solutions of \[(x-y-4)^2 = 16y\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$ $y$ must be from $1^2$ to $999^2$ , giving at most 999 options for $y$
However if $y = 1^2$ , you get $(x-5)^2 = 16$ , which has solutions $x = 9$ and $x = 1$ . Both of those solutions are not less than $y$ , so $y$ cannot be equal to 1. If $y = 2^2 = 4$ , you get $(x - 8)^2 = 64$ , which has 2 solutions, $x = 16$ , and $x = 0$ . 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$ , you get exactly 1 solution for $x$ , and that gives a total of $999 - 2 = \boxed{997}$ pairs. | null | 997 |
1d040528fc345db092ea66594ef8ade8 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Rearranging our conditions to
\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
\[(k-1)^2=x-1.\]
There, a unique value of $x, y$ is formed for every value of $k$ . However, we must have
\[y<10^6 \implies (k+1)^2< 10^6-1\]
and
\[x=(k-1)^2+1>0.\]
Therefore, there are only $\boxed{997}$ pairs of $(x,y).$ | null | 997 |
42236750eeb984df57f1626703983b11 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | We can rewrite $xyz=1$ as $\frac{1}{z}=xy$
Substituting into one of the given equations, we have \[x+xy=5\] \[x(1+y)=5\] \[\frac{1}{x}=\frac{1+y}{5}.\]
We can substitute back into $y+\frac{1}{x}=29$ to obtain \[y+\frac{1+y}{5}=29\] \[5y+1+y=145\] \[y=24.\]
We can then substitute once again to get \[x=\frac15\] \[z=\frac{5}{24}.\] Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$ , so $m+n=\boxed{005}$ | null | 005 |
42236750eeb984df57f1626703983b11 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | Let $r = \frac{m}{n} = z + \frac {1}{y}$
\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}
Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$ . So $m + n = 1 + 4 = \boxed{5}$ | null | 5 |
42236750eeb984df57f1626703983b11 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | (Hybrid between 1/2)
Because $xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz,$ and $\hspace{0.05cm}\frac{1}{z} = xy$ . Substituting and factoring, we get $x(y+1) = 5$ $\hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$ , but $xyz$ is $1$ , and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$ . And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$ , and if we make a substitution for $xyz$ , and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.
Now, lets add the 3 equations $x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$ , so $k = \frac{1}{4}$ which gives an answer of $1+4 = \boxed{005}$ | null | 005 |
42236750eeb984df57f1626703983b11 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$ . Then, since $xyz=1$ , we have $\tfrac 1y-5z=-1$ and $\tfrac 1z-29x=-1$ . Then, since we know that $\tfrac 1z+x=5$ , we can subtract these two equations to get that $30x=6\implies x=5$ . The result follows that $z=\tfrac 5{24}$ and $y=24$ , so $z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14$ , and the requested answer is $1+4=\boxed{005}.$ | null | 005 |
42236750eeb984df57f1626703983b11 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | Rewrite the equations in terms of x.
$x+\frac{1}{z}=5$ becomes $z=\frac{1}{x+5}$
$y+\frac{1}{x}=29$ becomes $y=29-\frac{1}{x}$
Now express $xyz=1$ in terms of x.
$\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1$
This evaluates to $29x-1=5-x$ , giving us $x=\frac{1}{5}$ . We can now plug x into the other equations to get $y=24$ and $z=\frac{5}{24}$
Therefore, $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$
$1+4=\boxed{5}$ , and we are done.
~MC413551 | null | 5 |
91dcb4ed463e4128c2c9e122d86097fc | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_8 | A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$ | The scale factor is uniform in all dimensions, so the volume of the liquid is $\left(\frac{3}{4}\right)^{3}$ of the container. The remaining section of the volume is $\frac{1-\left(\frac{3}{4}\right)^{3}}{1}$ of the volume, and therefore $\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height when the vertex is at the top.
So, the liquid occupies $\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height, or $12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)$ . Thus $m+n+p=\boxed{052}$ | null | 052 |
91dcb4ed463e4128c2c9e122d86097fc | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_8 | A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$ | (Computational) The volume of a cone can be found by $V = \frac{\pi}{3}r^2h$ . In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'$
By similar triangles , we find that the dimensions of the liquid in the first cone to the entire cone is $\frac{3}{4}$ , and that $r' = \frac{rh'}{h}$ ; equating,
\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\ \frac{37}{64}r^2h &= \frac{r^2}{h^2}(h')^3 \\ h' &= \sqrt[3]{\frac{37}{64} \cdot 12^3} = 3\sqrt[3]{37}\end{align*}
Thus the answer is $12 - h' = 12-3\sqrt[3]{37}$ , and $m+n+p=\boxed{052}$ | null | 052 |
91dcb4ed463e4128c2c9e122d86097fc | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_8 | A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$ | From the formula $V=\frac{\pi r^2h}{3}$ , we can find that the volume of the container is $100\pi$ . The cone formed by the liquid is similar to the original, but scaled down by $\frac{3}{4}$ in all directions, so its volume is $100\pi*\frac{27}{64}=\frac{675\pi}{16}$ . The volume of the air in the container is the volume of the container minus the volume of the liquid, which is $\frac{925\pi}{16}$ , which is $\frac{37}{64}$ of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have $\sqrt[3]{\frac{37}{64}}=\frac{\sqrt[3]{37}}{4}$ of the height of the container. This means that the height of the liquid is $12\left(1-\frac{\sqrt[3]{37}}{4}\right)=12-3\sqrt[3]{37}$ inches, so our answer is $\boxed{052}$ . Solution by Zeroman | null | 052 |
91dcb4ed463e4128c2c9e122d86097fc | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_8 | A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$ | We find that the volume of the cone is $100\pi$
The volume of the cone with height 9 is $\frac{675}{16}\pi$
The difference between the two volumes is $\frac{925}{16}\pi$ . Note that this is the volume of the cone essentially 'on top of' the frustum described in the problem when the liquid is held with the base horizontal.
We can express the volume as $x^2\pi\cdot\frac{12}{5}x\cdot\frac{1}{3}$ , where x is the radius of the cone.
Solving this equation, we get $x=\frac{5\sqrt[3]{37}}{4}$ . The height of this cone is $\frac{12}{5}$ of the radius.
Then we subtract the value from the height, 12, to get our answer: $12-3\sqrt[3]{37}$
Therefore, our answer is $12+3+37=\boxed{52}$ . ~MC413551 | null | 52 |
8ea1ca667d173fd240f34b10d3a68770 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_9 | The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ | Since $\log ab = \log a + \log b$ , we can reduce the equations to a more recognizable form:
\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{eqnarray*}
Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT , the above equations become (*)
\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1 \end{eqnarray*}
Small note from different author: $-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.$
From here, multiplying the three equations gives
\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}
Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$ . (Note from different author if you are confused on this step: if $\pm$ is positive then $\log y = \log 2 + 1 = \log 2 + \log 10 = \log 20,$ so $y=20.$ if $\pm$ is negative then $\log y = 1 - \log 2 = \log 10 - \log 2 = \log 5,$ so $y=5.$ ) This gives $y_1 = 20, y_2 = 5$ , and the answer is $y_1 + y_2 = \boxed{025}$ | null | 025 |
8ea1ca667d173fd240f34b10d3a68770 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_9 | The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ | Subtracting the second equation from the first equation yields \begin{align*} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{align*} If $1-\log y=0$ then $y=10$ . Substituting into the first equation yields $\log20000=4$ which is not possible.
If $\log\frac{x}{z}=0$ then $\frac{x}{z}=1\Longrightarrow x=z$ . Substituting into the third equation gets \begin{align*} \log x^2-(\log x)(\log x) &= 0 \\ \log x^2-\log x^x &= 0 \\ \log x^{2-x} &= 0 \\ x^{2-x} &= 1 \\ \end{align*} Thus either $x=1$ or $2-x=0\Longrightarrow x=2$ . (Note that here $x\neq-1$ since logarithm isn't defined for negative number.)
Substituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$ , respectively. Thus $y_1+y_2=\boxed{025}$ | null | 025 |
8ea1ca667d173fd240f34b10d3a68770 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_9 | The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ | Let $a = \log x$ $b = \log y$ and $c = \log z$ . Then the given equations become:
\begin{align*} \log 2 + a + b - ab = 1 \\ \log 2 + b + c - bc = 1 \\ a+c = ac \\ \end{align*}
Equating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \implies{a = c}$ . Plugging this result into the third equation, we get $c = 0$ or $2$ . Substituting each of these values of $c$ into the second equation, we get $b = 1 - \log 2$ and $b = 1 + \log 2$ . Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$ , respectively, so our answer is $\boxed{025}$ | null | 025 |
1cc7339b237e94773cd0530fc7fcae2a | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10 | sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$ | Let the sum of all of the terms in the sequence be $\mathbb{S}$ . Then for each integer $k$ $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}
Now, substituting for $x_{50}$ , we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$ , and the answer is $75+98=\boxed{173}$ | null | 173 |
1cc7339b237e94773cd0530fc7fcae2a | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10 | sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$ | Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$
In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\dfrac{50}{2}$
Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$ | null | 173 |
c99e8fe112a30629f59867a437bb6062 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$ , it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$ , where $-3 \le x,y \le 3$ . Thus every number in the form of $a/b$ will be expressed one time in the product
\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]
Using the formula for a geometric series , this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$ , and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$ | null | 248 |
c99e8fe112a30629f59867a437bb6062 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | Essentially, the problem asks us to compute \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}\] which is pretty easy: \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2} \bigg) = 2480 + \frac{437}{1000}\] so our answer is $\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}$ | null | 248 |
c99e8fe112a30629f59867a437bb6062 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | The sum is equivalent to $\sum_{i | 10^6}^{} \frac{i}{1000}$ Therefore, it's the sum of the factors of $10^6$ divided by $1000$ . The sum is $\frac{127 \times 19531}{1000}$ by the sum of factors formula. The answer is therefore $\boxed{248}$ after some computation.
- whatRthose | null | 248 |
c99e8fe112a30629f59867a437bb6062 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor.
To begin with the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the "partner" of any divisor.
Now we now the amount to multiply the numerator if given number is in the denominator. Now we simply combine and reduce these groups. If the powers of 2 are on the denominator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property application. There is nothing complicated about this except to be careful.
Add all powers of 2: 15
Add their partners: 1875
Add all powers of 5: 156
Add their partners: 1248
Then, follow this formula: (sum of powers * sum opposite power's partners)+(sum of powers * sum opposite power's partners)
Or, $156*1875+15*1248$ $=311220$
Now, divide by 1000 to compensate for the denominator. $311.22$
Finally, we have to calculate the other list of fractions with 1 and another divisor. e.g. 1 - 250, 1 - 20 etc. (these all count)
This time we need to list all divisors of 1000, including 1. Remove all powers of 2 or 5, because we already included those in the other list. Now, notice there are two cases. 1: 1 is in the denominator, making the fraction an integer. 2: 1 is in the numerator.
Adding all the integers in the first case gives us 2169. The second case can actually be discarded, but still can be found. Realize that if we include the powers of 2 and 5, then the second case is (sum of all divisors)/1000. Remove all partners of powers of 2 and 5, and we'll get exactly 217/1000, or 0.22.
Finally, add all the numbers together: $311.22 + 2169 + 0.22 = 2480.44$
And divide by 10: $248.044$
After an odyssey of bashing: $\boxed{248}$ | null | 248 |
d544ef36f0513a8cb0e7469b5e9895bd | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_12 | Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$ | \begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}
Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm
\[f(x) = f(352 + x)\]
So we need only to consider one period $f(0), f(1), ... f(351)$ , which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.
But we also know that $f(x) = f(46 - x) = f(398 - x)$ , so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated. This gives a total of
\[352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{177}\] | null | 177 |
f4a619ec75f326e55aba0dc87a3f9aed | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_13 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$ . All these circles are homothetic with respect to a center at $(5,0)$
Now consider the circle at $(0,0)$ . Draw a line tangent to it at $A$ and passing through $B (5,0)$ . By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$ . Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$ , so the slope of line $AB$ is $\frac{-7}{24}$ . Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$
This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$ . The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$ . By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$ | null | 731 |
d0cdd5adf9dc5527f01a6184eea8d22e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Let $\angle QPB=x^\circ$ . Because $\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\angle PAQ$ has the same measure. Because $\angle BPC$ is exterior to $\triangle BPA$ its measure is $3x$ . Let $\angle PBC = y^\circ$ . It follows that $\angle ACB = x+y$ and that $4x+2y=180^\circ$ . Two of the angles of triangle $APQ$ have measure $2x$ , and thus the measure of $\angle APQ$ is $2y$ . It follows that $AQ=2\cdot AP\cdot \sin y$ . Because $AB=AC$ and $AP=QB$ , it also follows that $AQ=PC$ . Now apply the Law of Sines to triangle $PBC$ to find \[\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}\] because $AP=BC$ . Hence $\sin 3x = \tfrac 12$ . Since $4x<180$ , this implies that $3x=30$ , i.e. $x=10$ . Thus $y=70$ and \[r=\frac{10+70}{2\cdot 70}=\frac 47,\] which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$ | null | 571 |
d0cdd5adf9dc5527f01a6184eea8d22e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$ . Then $PQBR$ is a rhombus , so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid . Since $\overline{PB}$ bisects $\angle QBR$ , it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$ . Thus $R$ lies on the perpendicular bisector of $\overline{BC}$ , and $BC = BR = RC$ . Hence $\triangle BCR$ is an equilateral triangle
Now $\angle ABR = \angle BAC = \angle ACR$ , and the sum of the angles in $\triangle ABC$ is $\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}$ . Then $\angle APQ = 140^{\circ}$ and $\angle ACB = 80^{\circ}$ , so the answer is $\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}$ | null | 571 |