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A triangle has sides of length $48$, $55$, and $73$. Let $a$ and $b$ be relatively prime positive integers such that $a/b$ is the length of the shortest altitude of the triangle. Find the value of $a+b$.
|
1. **Identify the type of triangle**: Given the sides of the triangle are \(48\), \(55\), and \(73\). We need to check if this is a right triangle. For a triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the largest side), it is a right triangle if \(a^2 + b^2 = c^2\).
\[
48^2 + 55^2 = 2304 + 3025 = 5329
\]
\[
73^2 = 5329
\]
Since \(48^2 + 55^2 = 73^2\), the triangle is a right triangle with the hypotenuse \(73\).
2. **Calculate the area of the triangle**: The area \(A\) of a right triangle can be calculated using the legs \(a\) and \(b\):
\[
A = \frac{1}{2} \times 48 \times 55 = \frac{1}{2} \times 2640 = 1320
\]
3. **Find the altitude to the hypotenuse**: The altitude \(h\) to the hypotenuse in a right triangle can be found using the area formula \(A = \frac{1}{2} \times \text{base} \times \text{height}\). Here, the base is the hypotenuse \(73\):
\[
1320 = \frac{1}{2} \times 73 \times h
\]
\[
2640 = 73h
\]
\[
h = \frac{2640}{73}
\]
4. **Simplify the fraction**: The fraction \(\frac{2640}{73}\) is already in its simplest form because \(2640\) and \(73\) are relatively prime (they have no common factors other than 1).
5. **Identify \(a\) and \(b\)**: Here, \(a = 2640\) and \(b = 73\).
6. **Calculate \(a + b\)**:
\[
a + b = 2640 + 73 = 2713
\]
The final answer is \(\boxed{2713}\)
|
2713
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The volume of a certain rectangular solid is $216\text{ cm}^3$, its total surface area is $288\text{ cm}^2$, and its three dimensions are in geometric progression. Find the sum of the lengths in cm of all the edges of this solid.
|
1. Let the dimensions of the rectangular solid be \(a/r\), \(a\), and \(ar\), where \(a\) is the middle term and \(r\) is the common ratio of the geometric progression.
2. The volume of the rectangular solid is given by:
\[
\frac{a}{r} \cdot a \cdot ar = a^3 = 216 \text{ cm}^3
\]
Solving for \(a\):
\[
a^3 = 216 \implies a = \sqrt[3]{216} = 6
\]
3. The total surface area of the rectangular solid is given by:
\[
2\left(\frac{a^2}{r} + a^2 + a^2 r\right) = 288 \text{ cm}^2
\]
Substituting \(a = 6\):
\[
2\left(\frac{6^2}{r} + 6^2 + 6^2 r\right) = 288
\]
Simplifying:
\[
2\left(\frac{36}{r} + 36 + 36r\right) = 288
\]
\[
\frac{36}{r} + 36 + 36r = 144
\]
Dividing by 36:
\[
\frac{1}{r} + 1 + r = 4
\]
Let \(x = r + \frac{1}{r}\):
\[
x + 1 = 4 \implies x = 3
\]
Solving the quadratic equation \(r + \frac{1}{r} = 3\):
\[
r^2 - 3r + 1 = 0
\]
Using the quadratic formula:
\[
r = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}
\]
4. The sum of the lengths of all the edges of the rectangular solid is:
\[
4\left(\frac{a}{r} + a + ar\right)
\]
Substituting \(a = 6\):
\[
4\left(\frac{6}{r} + 6 + 6r\right)
\]
Using \(r = \frac{3 + \sqrt{5}}{2}\) and \(r = \frac{3 - \sqrt{5}}{2}\), we find:
\[
\frac{6}{r} + 6 + 6r = \frac{6}{\frac{3 + \sqrt{5}}{2}} + 6 + 6 \cdot \frac{3 + \sqrt{5}}{2}
\]
Simplifying:
\[
\frac{6 \cdot 2}{3 + \sqrt{5}} + 6 + 3(3 + \sqrt{5})
\]
\[
\frac{12}{3 + \sqrt{5}} + 6 + 9 + 3\sqrt{5}
\]
Rationalizing the denominator:
\[
\frac{12(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} + 15 + 3\sqrt{5}
\]
\[
\frac{12(3 - \sqrt{5})}{9 - 5} + 15 + 3\sqrt{5}
\]
\[
\frac{12(3 - \sqrt{5})}{4} + 15 + 3\sqrt{5}
\]
\[
3(3 - \sqrt{5}) + 15 + 3\sqrt{5}
\]
\[
9 - 3\sqrt{5} + 15 + 3\sqrt{5} = 24
\]
Therefore:
\[
4 \times 24 = 96
\]
The final answer is \(\boxed{96}\).
|
96
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\phi(n)$ denote $\textit{Euler's phi function}$, the number of integers $1\leq i\leq n$ that are relatively prime to $n$. (For example, $\phi(6)=2$ and $\phi(10)=4$.) Let \[S=\sum_{d|2008}\phi(d),\] in which $d$ ranges through all positive divisors of $2008$, including $1$ and $2008$. Find the remainder when $S$ is divided by $1000$.
|
1. **Understanding Euler's Totient Function**: Euler's totient function, denoted as $\phi(n)$, counts the number of integers from $1$ to $n$ that are relatively prime to $n$. For example, $\phi(6) = 2$ because the numbers $1$ and $5$ are relatively prime to $6$.
2. **Given Problem**: We need to find the sum of $\phi(d)$ for all divisors $d$ of $2008$, and then find the remainder when this sum is divided by $1000$.
3. **Key Theorem**: The sum of Euler's totient function over all divisors of $n$ is equal to $n$. Mathematically, this is expressed as:
\[
\sum_{d|n} \phi(d) = n
\]
This theorem can be proven by considering the fractions $\frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}$ and noting that there are exactly $\phi(d)$ fractions with denominator $d$ for each divisor $d$ of $n$.
4. **Application to the Problem**: Applying the theorem to $n = 2008$, we get:
\[
\sum_{d|2008} \phi(d) = 2008
\]
5. **Finding the Remainder**: We need to find the remainder when $2008$ is divided by $1000$. This can be done using the modulo operation:
\[
2008 \mod 1000 = 8
\]
Conclusion:
\[
\boxed{8}
\]
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Alexis notices Joshua working with Dr. Lisi and decides to join in on the fun. Dr. Lisi challenges her to compute the sum of all $2008$ terms in the sequence. Alexis thinks about the problem and remembers a story one of her teahcers at school taught her about how a young Karl Gauss quickly computed the sum \[1+2+3+\cdots+98+99+100\] in elementary school. Using Gauss's method, Alexis correctly finds the sum of the $2008$ terms in Dr. Lisi's sequence. What is this sum?
|
To find the sum of the first 2008 terms in an arithmetic sequence, we need to use the formula for the sum of an arithmetic series:
\[ S_n = \frac{n}{2} \left( a + l \right) \]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( n \) is the number of terms,
- \( a \) is the first term,
- \( l \) is the last term.
Given:
- The sequence has 2008 terms (\( n = 2008 \)).
- The first term (\( a \)) is not explicitly given, but we can infer it from the context.
Let's assume the sequence starts at \( a = -1776 \) and has a common difference \( d = 11 \).
1. **Find the last term \( l \):**
The last term of an arithmetic sequence can be found using the formula:
\[ l = a + (n-1)d \]
Substituting the given values:
\[ l = -1776 + (2008-1) \cdot 11 \]
\[ l = -1776 + 2007 \cdot 11 \]
\[ l = -1776 + 22077 \]
\[ l = 20301 \]
2. **Calculate the sum \( S_{2008} \):**
Using the sum formula for an arithmetic series:
\[ S_{2008} = \frac{2008}{2} \left( -1776 + 20301 \right) \]
\[ S_{2008} = 1004 \left( 18525 \right) \]
\[ S_{2008} = 1004 \cdot 18525 \]
\[ S_{2008} = 18599100 \]
The final answer is \(\boxed{18599100}\).
|
18599100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Now Wendy wanders over and joins Dr. Lisi and her younger siblings. Thinking she knows everything there is about how to work with arithmetic series, she nearly turns right around to walk back home when Dr. Lisi poses a more challenging problem. "Suppose I select two distinct terms at random from the $2008$ term sequence. What's the probability that their product is positive?" If $a$ and $b$ are relatively prime positive integers such that $a/b$ is the probability that the product of the two terms is positive, find the value of $a+b$.
|
1. **Identify the sequence and its properties:**
- We have an arithmetic sequence with 2008 terms.
- Let's denote the sequence as \( a_1, a_2, a_3, \ldots, a_{2008} \).
2. **Determine the number of positive and negative terms:**
- Assume the sequence is such that it contains both positive and negative terms.
- Let \( n \) be the number of negative terms and \( p \) be the number of positive terms.
- Since the sequence has 2008 terms, we have \( n + p = 2008 \).
3. **Calculate the number of ways to choose two terms:**
- The total number of ways to choose 2 terms out of 2008 is given by the combination formula:
\[
\binom{2008}{2} = \frac{2008 \times 2007}{2}
\]
4. **Calculate the number of ways to choose two terms such that their product is positive:**
- The product of two terms is positive if both terms are either positive or both are negative.
- The number of ways to choose 2 positive terms is:
\[
\binom{p}{2} = \frac{p(p-1)}{2}
\]
- The number of ways to choose 2 negative terms is:
\[
\binom{n}{2} = \frac{n(n-1)}{2}
\]
5. **Calculate the total number of ways to choose two terms such that their product is positive:**
- The total number of favorable outcomes is:
\[
\binom{p}{2} + \binom{n}{2} = \frac{p(p-1)}{2} + \frac{n(n-1)}{2}
\]
6. **Calculate the probability:**
- The probability that the product of two randomly chosen terms is positive is:
\[
\frac{\binom{p}{2} + \binom{n}{2}}{\binom{2008}{2}} = \frac{\frac{p(p-1)}{2} + \frac{n(n-1)}{2}}{\frac{2008 \times 2007}{2}} = \frac{p(p-1) + n(n-1)}{2008 \times 2007}
\]
7. **Determine the values of \( n \) and \( p \):**
- Given that there are 161 negative terms, we have \( n = 161 \) and \( p = 2008 - 161 = 1847 \).
8. **Substitute the values of \( n \) and \( p \) into the probability formula:**
\[
\frac{1847 \times 1846 + 161 \times 160}{2008 \times 2007}
\]
9. **Simplify the expression:**
\[
1847 \times 1846 = 3404762
\]
\[
161 \times 160 = 25760
\]
\[
3404762 + 25760 = 3430522
\]
\[
\frac{3430522}{2008 \times 2007} = \frac{3430522}{4032048}
\]
10. **Simplify the fraction:**
- The greatest common divisor (GCD) of 3430522 and 4032048 is 2.
\[
\frac{3430522 \div 2}{4032048 \div 2} = \frac{1715261}{2016024}
\]
11. **Identify \( a \) and \( b \):**
- Here, \( a = 1715261 \) and \( b = 2016024 \).
12. **Calculate \( a + b \):**
\[
a + b = 1715261 + 2016024 = 3731285
\]
The final answer is \(\boxed{3731285}\)
|
3731285
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In order to save money on gas and use up less fuel, Hannah has a special battery installed in the family van. Before the installation, the van averaged $18$ miles per gallon of gas. After the conversion, the van got $24$ miles per gallong of gas.
Michael notes, "The amount of money we will save on gas over any time period is equal to the amount we would save if we were able to convert the van to go from $24$ miles per gallon to $m$ miles per gallon. It is also the same that we would save if we were able to convert the van to go from $m$ miles per gallon to $n$ miles per gallon."
Assuming Michael is correct, compute $m+n$. In this problem, assume that gas mileage is constant over all speeds and terrain and that the van gets used the same amount regardless of its present state of conversion.
|
1. **Initial Setup:**
- Before the conversion, the van averaged \(18\) miles per gallon (mpg).
- After the conversion, the van averaged \(24\) mpg.
2. **Calculate the improvement in fuel efficiency:**
- Before conversion: \(18\) mpg means \(6\) miles per \(\frac{6}{18} = \frac{1}{3}\) gallon.
- After conversion: \(24\) mpg means \(6\) miles per \(\frac{6}{24} = \frac{1}{4}\) gallon.
- The improvement in fuel efficiency is \(\frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}\) gallon per \(6\) miles.
3. **Determine the value of \(m\):**
- \(m\) mpg should improve the fuel efficiency by another \(\frac{1}{12}\) gallon per \(6\) miles.
- If \(m\) mpg means \(6\) miles per \(\frac{6}{m}\) gallons, then:
\[
\frac{1}{4} - \frac{6}{m} = \frac{1}{12}
\]
- Solving for \(m\):
\[
\frac{1}{4} - \frac{6}{m} = \frac{1}{12}
\]
\[
\frac{1}{4} - \frac{1}{12} = \frac{6}{m}
\]
\[
\frac{3}{12} - \frac{1}{12} = \frac{6}{m}
\]
\[
\frac{2}{12} = \frac{6}{m}
\]
\[
\frac{1}{6} = \frac{6}{m}
\]
\[
m = 36
\]
4. **Determine the value of \(n\):**
- \(n\) mpg should improve the fuel efficiency by another \(\frac{1}{12}\) gallon per \(6\) miles.
- If \(n\) mpg means \(6\) miles per \(\frac{6}{n}\) gallons, then:
\[
\frac{6}{m} - \frac{6}{n} = \frac{1}{12}
\]
- Since \(m = 36\):
\[
\frac{6}{36} - \frac{6}{n} = \frac{1}{12}
\]
\[
\frac{1}{6} - \frac{6}{n} = \frac{1}{12}
\]
\[
\frac{1}{6} - \frac{1}{12} = \frac{6}{n}
\]
\[
\frac{2}{12} - \frac{1}{12} = \frac{6}{n}
\]
\[
\frac{1}{12} = \frac{6}{n}
\]
\[
n = 72
\]
5. **Sum of \(m\) and \(n\):**
\[
m + n = 36 + 72 = 108
\]
The final answer is \(\boxed{108}\)
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find $a+b+c$, where $a,b,$ and $c$ are the hundreds, tens, and units digits of the six-digit number $123abc$, which is a multiple of $990$.
|
1. Given the six-digit number \(123abc\) is a multiple of \(990\), we need to find \(a + b + c\).
2. Since \(990 = 2 \times 3^2 \times 5 \times 11\), the number \(123abc\) must be divisible by \(2\), \(9\), and \(11\).
3. For \(123abc\) to be divisible by \(2\), the units digit \(c\) must be \(0\). Therefore, \(c = 0\).
4. Now, the number \(123ab0\) must be divisible by \(9\). A number is divisible by \(9\) if the sum of its digits is divisible by \(9\). Thus, we have:
\[
1 + 2 + 3 + a + b + 0 \equiv 6 + a + b \equiv 0 \pmod{9}
\]
This simplifies to:
\[
a + b \equiv 3 \pmod{9}
\]
5. Next, the number \(123ab0\) must be divisible by \(11\). A number is divisible by \(11\) if the alternating sum of its digits is divisible by \(11\). Thus, we have:
\[
(1 - 2 + 3 - a + b - 0) \equiv 1 + 3 + b - 2 - a \equiv 2 + b - a \equiv 0 \pmod{11}
\]
This simplifies to:
\[
b - a \equiv 9 \pmod{11}
\]
6. We now have two congruences:
\[
a + b \equiv 3 \pmod{9}
\]
\[
b - a \equiv 9 \pmod{11}
\]
7. Solving these congruences, we start with \(b - a \equiv 9 \pmod{11}\). This can be written as:
\[
b = a + 9
\]
8. Substitute \(b = a + 9\) into \(a + b \equiv 3 \pmod{9}\):
\[
a + (a + 9) \equiv 3 \pmod{9}
\]
\[
2a + 9 \equiv 3 \pmod{9}
\]
\[
2a \equiv 3 - 9 \pmod{9}
\]
\[
2a \equiv -6 \pmod{9}
\]
\[
2a \equiv 3 \pmod{9}
\]
9. To solve \(2a \equiv 3 \pmod{9}\), we find the multiplicative inverse of \(2\) modulo \(9\). The inverse of \(2\) modulo \(9\) is \(5\) because \(2 \times 5 \equiv 1 \pmod{9}\). Thus:
\[
a \equiv 3 \times 5 \pmod{9}
\]
\[
a \equiv 15 \pmod{9}
\]
\[
a \equiv 6 \pmod{9}
\]
10. Therefore, \(a = 6\). Substituting \(a = 6\) back into \(b = a + 9\):
\[
b = 6 + 9 = 15
\]
11. Since \(b\) must be a single digit, we need to re-evaluate our steps. We should check for other possible values of \(a\) and \(b\) that satisfy both congruences.
12. Re-evaluating, we find that \(a = 7\) and \(b = 5\) satisfy both conditions:
\[
a + b = 7 + 5 = 12 \equiv 3 \pmod{9}
\]
\[
b - a = 5 - 7 = -2 \equiv 9 \pmod{11}
\]
13. Therefore, \(a = 7\), \(b = 5\), and \(c = 0\).
14. The sum \(a + b + c = 7 + 5 + 0 = 12\).
The final answer is \(\boxed{12}\).
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Jerry's favorite number is $97$. He knows all kinds of interesting facts about $97$:
[list][*]$97$ is the largest two-digit prime.
[*]Reversing the order of its digits results in another prime.
[*]There is only one way in which $97$ can be written as a difference of two perfect squares.
[*]There is only one way in which $97$ can be written as a sum of two perfect squares.
[*]$\tfrac1{97}$ has exactly $96$ digits in the [smallest] repeating block of its decimal expansion.
[*]Jerry blames the sock gnomes for the theft of exactly $97$ of his socks.[/list]
A repunit is a natural number whose digits are all $1$. For instance, \begin{align*}&1,\\&11,\\&111,\\&1111,\\&\vdots\end{align*} are the four smallest repunits. How many digits are there in the smallest repunit that is divisible by $97?$
|
To determine the number of digits in the smallest repunit that is divisible by \(97\), we need to find the smallest integer \(n\) such that the repunit \(R_n\) (which consists of \(n\) digits all being \(1\)) is divisible by \(97\).
A repunit \(R_n\) can be expressed as:
\[ R_n = \frac{10^n - 1}{9} \]
We need \(R_n\) to be divisible by \(97\), which means:
\[ \frac{10^n - 1}{9} \equiv 0 \pmod{97} \]
This implies:
\[ 10^n - 1 \equiv 0 \pmod{97} \]
\[ 10^n \equiv 1 \pmod{97} \]
We are looking for the smallest positive integer \(n\) such that \(10^n \equiv 1 \pmod{97}\). This \(n\) is known as the order of \(10\) modulo \(97\).
By Fermat's Little Theorem, since \(97\) is a prime number, we know:
\[ 10^{96} \equiv 1 \pmod{97} \]
This tells us that \(10^n \equiv 1 \pmod{97}\) for some \(n\) that divides \(96\). We need to find the smallest such \(n\).
To do this, we check the divisors of \(96\) (which are \(1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96\)) to find the smallest \(n\) for which \(10^n \equiv 1 \pmod{97}\).
1. \(10^1 \equiv 10 \pmod{97}\) (not \(1\))
2. \(10^2 \equiv 100 \equiv 3 \pmod{97}\) (not \(1\))
3. \(10^3 \equiv 10 \cdot 3 \equiv 30 \pmod{97}\) (not \(1\))
4. \(10^4 \equiv 10 \cdot 30 \equiv 300 \equiv 9 \pmod{97}\) (not \(1\))
5. \(10^6 \equiv 10^2 \cdot 10^4 \equiv 3 \cdot 9 \equiv 27 \pmod{97}\) (not \(1\))
6. \(10^8 \equiv 10^4 \cdot 10^4 \equiv 9 \cdot 9 \equiv 81 \pmod{97}\) (not \(1\))
7. \(10^{12} \equiv 10^8 \cdot 10^4 \equiv 81 \cdot 9 \equiv 729 \equiv 51 \pmod{97}\) (not \(1\))
8. \(10^{16} \equiv 10^8 \cdot 10^8 \equiv 81 \cdot 81 \equiv 6561 \equiv 64 \pmod{97}\) (not \(1\))
9. \(10^{24} \equiv 10^{16} \cdot 10^8 \equiv 64 \cdot 81 \equiv 5184 \equiv 43 \pmod{97}\) (not \(1\))
10. \(10^{32} \equiv 10^{16} \cdot 10^{16} \equiv 64 \cdot 64 \equiv 4096 \equiv 21 \pmod{97}\) (not \(1\))
11. \(10^{48} \equiv 10^{24} \cdot 10^{24} \equiv 43 \cdot 43 \equiv 1849 \equiv 3 \pmod{97}\) (not \(1\))
12. \(10^{96} \equiv 1 \pmod{97}\) (this is \(1\))
Since \(10^{96} \equiv 1 \pmod{97}\) and no smaller \(n\) satisfies this condition, the smallest \(n\) is \(96\).
Conclusion:
\(\boxed{96}\)
|
96
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy's friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched out a blue piece of yarn between them. Then Wendy and Erin stretched out a red piece of yarn between them at about the same height so that the yarn would intersect if possible. If all possible positions of the students around the table are equally likely, let $m/n$ be the probability that the yarns intersect, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.
|
1. **Identify the problem**: We need to find the probability that two pieces of yarn stretched between four specific students (Lucy, Starling, Wendy, and Erin) intersect when the students are seated randomly around a circular table with 14 students.
2. **Simplify the problem**: The positions of the other 10 students do not affect whether the yarns intersect. Therefore, we only need to consider the relative positions of Lucy, Starling, Wendy, and Erin.
3. **Fix one student**: To simplify the counting, we can fix one student in a position. Let's fix Lucy at a specific position. This does not affect the probability because the table is circular and all positions are symmetric.
4. **Count the arrangements**: With Lucy fixed, we need to arrange the remaining three students (Starling, Wendy, and Erin) in the remaining 13 positions. However, since we are only interested in the relative positions of these four students, we can consider the positions of Starling, Wendy, and Erin relative to Lucy.
5. **Determine intersecting and non-intersecting cases**:
- The yarns will intersect if and only if the two pairs of students (Lucy-Starling and Wendy-Erin) are "crossed" in the circular arrangement.
- For the yarns to intersect, the students must be seated in such a way that one pair of students is separated by the other pair. For example, if Lucy is at position 1, Starling at position 3, Wendy at position 2, and Erin at position 4, the yarns will intersect.
6. **Count the intersecting arrangements**:
- Fix Lucy at position 1. We need to count the number of ways to place Starling, Wendy, and Erin such that the yarns intersect.
- There are 3! = 6 ways to arrange Starling, Wendy, and Erin around Lucy.
- Out of these 6 arrangements, exactly 3 will result in intersecting yarns. This is because for any fixed position of Lucy, half of the arrangements of the other three students will result in intersecting yarns.
7. **Calculate the probability**:
- The total number of ways to arrange the four students is 4! = 24.
- The number of intersecting arrangements is 3 (for each fixed position of Lucy) * 4 (since Lucy can be in any of the 4 positions) = 12.
- Therefore, the probability that the yarns intersect is:
\[
\frac{\text{Number of intersecting arrangements}}{\text{Total number of arrangements}} = \frac{12}{24} = \frac{1}{2}
\]
8. **Express the probability in simplest form**:
- The probability is already in simplest form, \(\frac{1}{2}\).
9. **Find \(m + n\)**:
- Here, \(m = 1\) and \(n = 2\).
- Therefore, \(m + n = 1 + 2 = 3\).
The final answer is \(\boxed{3}\).
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
As the Kubiks head out of town for vacation, Jerry takes the first driving shift while Hannah and most of the kids settle down to read books they brought along. Tony does not feel like reading, so Alexis gives him one of her math notebooks and Tony gets to work solving some of the problems, and struggling over others. After a while, Tony comes to a problem he likes from an old AMC 10 exam:
\begin{align*}&\text{Four distinct circles are drawn in a plane. What is the maximum}\\&\quad\,\,\text{number of points where at least two of the circles intersect?}\end{align*}
Tony realizes that he can draw the four circles such that each pair of circles intersects in two points. After careful doodling, Tony finds the correct answer, and is proud that he can solve a problem from late on an AMC 10 exam.
"Mom, why didn't we all get Tony's brain?" Wendy inquires before turning he head back into her favorite Harry Potter volume (the fifth year).
Joshua leans over to Tony's seat to see his brother's work. Joshua knows that Tony has not yet discovered all the underlying principles behind the problem, so Joshua challenges, "What if there are a dozen circles?"
Tony gets to work on Joshua's problem of finding the maximum number of points of intersections where at least two of the twelve circles in a plane intersect. What is the answer to this problem?
|
To solve the problem of finding the maximum number of points where at least two of the twelve circles intersect, we need to consider the number of intersection points formed by pairs of circles.
1. **Understanding Circle Intersections**:
- When two distinct circles intersect, they can intersect at most in 2 points.
- Therefore, for each pair of circles, there are at most 2 intersection points.
2. **Counting Pairs of Circles**:
- We need to determine the number of ways to choose 2 circles out of 12. This is a combination problem.
- The number of ways to choose 2 circles from 12 is given by the binomial coefficient:
\[
\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66
\]
3. **Calculating Total Intersection Points**:
- Since each pair of circles can intersect at most in 2 points, the total number of intersection points is:
\[
2 \times \binom{12}{2} = 2 \times 66 = 132
\]
Thus, the maximum number of points where at least two of the twelve circles intersect is \( \boxed{132} \).
|
132
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Alexis imagines a $2008\times 2008$ grid of integers arranged sequentially in the following way:
\[\begin{array}{r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r@{\hspace{20pt}}r}1,&2,&3,&\ldots,&2008\\2009,&2010,&2011,&\ldots,&4026\\4017,&4018,&4019,&\ldots,&6024\\\vdots&&&&\vdots\\2008^2-2008+1,&2008^2-2008+2,&2008^2-2008+3,&\ldots,&2008^2\end{array}\]
She picks one number from each row so that no two numbers she picks are in the same column. She them proceeds to add them together and finds that $S$ is the sum. Next, she picks $2008$ of the numbers that are distinct from the $2008$ she picked the first time. Again she picks exactly one number from each row and column, and again the sum of all $2008$ numbers is $S$. Find the remainder when $S$ is divided by $2008$.
|
1. **Understanding the Grid and the Problem:**
- The grid is a \(2008 \times 2008\) matrix with integers arranged sequentially from 1 to \(2008^2\).
- Alexis picks one number from each row such that no two numbers are in the same column, and sums them to get \(S\).
- She repeats this process with another set of 2008 numbers, ensuring no overlap with the first set, and again the sum is \(S\).
2. **Analyzing the Grid Modulo 2008:**
- Each row \(i\) (where \(1 \leq i \leq 2008\)) contains numbers that are congruent to \(i\) modulo 2008.
- For example, the first row modulo 2008 is \([1, 2, 3, \ldots, 2008]\), the second row is \([2009, 2010, \ldots, 4016]\) which is \([1, 2, \ldots, 2008]\) modulo 2008, and so on.
3. **Summing the Numbers:**
- When Alexis picks one number from each row such that no two numbers are in the same column, the numbers she picks are congruent to \([1, 2, 3, \ldots, 2008]\) modulo 2008 in some order.
- The sum of these numbers is:
\[
1 + 2 + 3 + \ldots + 2008 = \frac{2008 \cdot 2009}{2}
\]
- Calculating this sum modulo 2008:
\[
\frac{2008 \cdot 2009}{2} = 1004 \cdot 2009
\]
\[
1004 \cdot 2009 = 1004 \cdot (2008 + 1) = 1004 \cdot 2008 + 1004 \equiv 1004 \pmod{2008}
\]
4. **Considering the Second Set of Numbers:**
- When Alexis picks another set of 2008 numbers, ensuring no overlap with the first set, the sum of these numbers will also be congruent to \(1004\) modulo 2008.
- Therefore, the sum \(S\) for both sets of numbers is congruent to \(1004\) modulo 2008.
5. **Final Calculation:**
- Since both sums are congruent to \(1004\) modulo 2008, the remainder when \(S\) is divided by 2008 is:
\[
\boxed{1004}
\]
|
1004
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A triangle has sides of length $48$, $55$, and $73$. A square is inscribed in the triangle such that one side of the square lies on the longest side of the triangle, and the two vertices not on that side of the square touch the other two sides of the triangle. If $c$ and $d$ are relatively prime positive integers such that $c/d$ is the length of a side of the square, find the value of $c+d$.
|
1. **Calculate the area of the triangle:**
The sides of the triangle are \(a = 48\), \(b = 55\), and \(c = 73\). We can use Heron's formula to find the area of the triangle. First, we calculate the semi-perimeter \(s\):
\[
s = \frac{a + b + c}{2} = \frac{48 + 55 + 73}{2} = 88
\]
Now, using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{88(88-48)(88-55)(88-73)}
\]
\[
= \sqrt{88 \cdot 40 \cdot 33 \cdot 15}
\]
\[
= \sqrt{88 \cdot 40 \cdot 33 \cdot 15} = \sqrt{1742400} = 1320
\]
2. **Set up the equation for the area using the inscribed square:**
Let \(x\) be the side length of the inscribed square. The area of the triangle can also be expressed as the sum of the areas of the three smaller triangles formed by the square and the triangle:
\[
\text{Area} = x^2 + \frac{x(73-x)}{2} + \frac{x \left(\frac{48 \cdot 55}{73} - x\right)}{2}
\]
Simplifying the expression:
\[
1320 = x^2 + \frac{x(73-x)}{2} + \frac{x \left(\frac{2640}{73} - x\right)}{2}
\]
3. **Simplify and solve for \(x\):**
Combine the terms:
\[
1320 = x^2 + \frac{73x - x^2}{2} + \frac{2640x/73 - x^2}{2}
\]
\[
1320 = x^2 + \frac{73x}{2} - \frac{x^2}{2} + \frac{2640x}{146} - \frac{x^2}{2}
\]
\[
1320 = x^2 + \frac{73x}{2} - \frac{x^2}{2} + \frac{1320x}{73} - \frac{x^2}{2}
\]
\[
1320 = x^2 + \frac{73x}{2} + \frac{1320x}{73} - x^2
\]
\[
1320 = \frac{73x}{2} + \frac{1320x}{73}
\]
4. **Combine the fractions and solve for \(x\):**
\[
1320 = \frac{73 \cdot 73x + 2 \cdot 1320x}{2 \cdot 73}
\]
\[
1320 = \frac{5329x + 2640x}{146}
\]
\[
1320 = \frac{7969x}{146}
\]
\[
1320 \cdot 146 = 7969x
\]
\[
x = \frac{1320 \cdot 146}{7969}
\]
\[
x = \frac{192720}{7969}
\]
5. **Identify \(c\) and \(d\):**
Here, \(c = 192720\) and \(d = 7969\). Since \(c\) and \(d\) are relatively prime, we find:
\[
c + d = 192720 + 7969 = 200689
\]
The final answer is \(\boxed{200689}\).
|
200689
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One of Michael's responsibilities in organizing the family vacation is to call around and find room rates for hotels along the root the Kubik family plans to drive. While calling hotels near the Grand Canyon, a phone number catches Michael's eye. Michael notices that the first four digits of $987-1234$ descend $(9-8-7-1)$ and that the last four ascend in order $(1-2-3-4)$. This fact along with the fact that the digits are split into consecutive groups makes that number easier to remember.
Looking back at the list of numbers that Michael called already, he notices that several of the phone numbers have the same property: their first four digits are in descending order while the last four are in ascending order. Suddenly, Michael realizes that he can remember all those numbers without looking back at his list of hotel phone numbers. "Wow," he thinks, "that's good marketing strategy."
Michael then wonders to himself how many businesses in a single area code could have such phone numbers. How many $7$-digit telephone numbers are there such that all seven digits are distinct, the first four digits are in descending order, and the last four digits are in ascending order?
|
To solve the problem, we need to determine how many 7-digit telephone numbers exist such that all seven digits are distinct, the first four digits are in descending order, and the last four digits are in ascending order.
1. **Choose 7 distinct digits from 0 to 9:**
We need to select 7 distinct digits out of the 10 possible digits (0 through 9). The number of ways to choose 7 digits from 10 is given by the binomial coefficient:
\[
\binom{10}{7}
\]
2. **Arrange the chosen digits:**
Once we have chosen the 7 digits, we need to arrange them such that the first four digits are in descending order and the last four digits are in ascending order. Note that the middle digit will be shared between the two groups.
3. **Determine the position of the middle digit:**
The middle digit can be any one of the 7 chosen digits. Once we fix the middle digit, we have 3 digits left for the descending order and 3 digits left for the ascending order.
4. **Calculate the number of ways to choose the middle digit:**
There are 7 ways to choose the middle digit from the 7 chosen digits.
5. **Calculate the number of ways to arrange the remaining digits:**
For each choice of the middle digit, the remaining 6 digits must be split into two groups of 3. The first group of 3 digits will be arranged in descending order, and the second group of 3 digits will be arranged in ascending order. There is only one way to arrange 3 digits in descending order and one way to arrange 3 digits in ascending order.
6. **Combine the results:**
The total number of valid phone numbers is given by:
\[
\binom{10}{7} \times 7
\]
where \(\binom{10}{7}\) is the number of ways to choose 7 digits from 10, and 7 is the number of ways to choose the middle digit.
7. **Calculate the binomial coefficient:**
\[
\binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]
8. **Calculate the total number of valid phone numbers:**
\[
\binom{10}{7} \times 7 = 120 \times 7 = 840
\]
The final answer is \(\boxed{840}\).
|
840
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\triangle XOY$ be a right-angled triangle with $\angle XOY=90^\circ$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Find the length $XY$ given that $XN=22$ and $YM=31$.
|
1. Let $\triangle XOY$ be a right-angled triangle with $\angle XOY = 90^\circ$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. We are given that $XN = 22$ and $YM = 31$.
2. Let $OX = a$ and $OY = b$. Since $M$ and $N$ are midpoints, we have:
\[
MX = \frac{a}{2} \quad \text{and} \quad NY = \frac{b}{2}
\]
3. Using the distance formula, we can write the given distances as:
\[
XN = \sqrt{OX^2 + NY^2} = \sqrt{a^2 + \left(\frac{b}{2}\right)^2} = 22
\]
\[
YM = \sqrt{OY^2 + MX^2} = \sqrt{b^2 + \left(\frac{a}{2}\right)^2} = 31
\]
4. Squaring both equations, we get:
\[
a^2 + \left(\frac{b}{2}\right)^2 = 22^2 = 484
\]
\[
b^2 + \left(\frac{a}{2}\right)^2 = 31^2 = 961
\]
5. Simplifying the equations, we have:
\[
a^2 + \frac{b^2}{4} = 484
\]
\[
b^2 + \frac{a^2}{4} = 961
\]
6. To eliminate the fractions, multiply the first equation by 4:
\[
4a^2 + b^2 = 1936
\]
7. Now we have the system of equations:
\[
4a^2 + b^2 = 1936
\]
\[
a^2 + \frac{b^2}{4} = 484
\]
8. Multiply the second equation by 4:
\[
4a^2 + b^2 = 1936
\]
\[
4b^2 + a^2 = 3844
\]
9. Subtract the first equation from the second:
\[
4b^2 + a^2 - (4a^2 + b^2) = 3844 - 1936
\]
\[
3b^2 - 3a^2 = 1908
\]
\[
b^2 - a^2 = 636
\]
10. Add the equations $4a^2 + b^2 = 1936$ and $b^2 - a^2 = 636$:
\[
4a^2 + b^2 + b^2 - a^2 = 1936 + 636
\]
\[
3a^2 + 2b^2 = 2572
\]
11. Solve for $a^2$ and $b^2$:
\[
3a^2 + 2b^2 = 2572
\]
\[
b^2 - a^2 = 636
\]
12. Multiply the second equation by 2:
\[
2b^2 - 2a^2 = 1272
\]
13. Add the equations:
\[
3a^2 + 2b^2 + 2b^2 - 2a^2 = 2572 + 1272
\]
\[
a^2 + 4b^2 = 3844
\]
14. Solve for $b^2$:
\[
4b^2 = 3844 - a^2
\]
\[
b^2 = \frac{3844 - a^2}{4}
\]
15. Substitute $b^2$ back into $b^2 - a^2 = 636$:
\[
\frac{3844 - a^2}{4} - a^2 = 636
\]
\[
3844 - a^2 - 4a^2 = 2544
\]
\[
3844 - 5a^2 = 2544
\]
\[
5a^2 = 1300
\]
\[
a^2 = 260
\]
16. Substitute $a^2 = 260$ back into $b^2 - a^2 = 636$:
\[
b^2 - 260 = 636
\]
\[
b^2 = 896
\]
17. Finally, find the length $XY$:
\[
XY = \sqrt{a^2 + b^2} = \sqrt{260 + 896} = \sqrt{1156} = 34
\]
The final answer is $\boxed{34}$
|
34
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ be the two possible values of $\tan\theta$ given that \[\sin\theta + \cos\theta = \dfrac{193}{137}.\] If $a+b=m/n$, where $m$ and $n$ are relatively prime positive integers, compute $m+n$.
|
1. Given the equation \(\sin\theta + \cos\theta = \dfrac{193}{137}\), we start by squaring both sides to utilize the Pythagorean identity.
\[
(\sin\theta + \cos\theta)^2 = \left(\dfrac{193}{137}\right)^2
\]
\[
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \dfrac{193^2}{137^2}
\]
Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), we get:
\[
1 + 2\sin\theta\cos\theta = \dfrac{193^2}{137^2}
\]
\[
2\sin\theta\cos\theta = \dfrac{193^2}{137^2} - 1
\]
\[
2\sin\theta\cos\theta = \dfrac{193^2 - 137^2}{137^2}
\]
\[
2\sin\theta\cos\theta = \dfrac{(193 - 137)(193 + 137)}{137^2}
\]
\[
2\sin\theta\cos\theta = \dfrac{56 \cdot 330}{137^2}
\]
\[
2\sin\theta\cos\theta = \dfrac{18480}{18769}
\]
\[
\sin\theta\cos\theta = \dfrac{9240}{18769}
\]
2. Recall that \(\sin\theta\cos\theta = \dfrac{1}{2}\sin(2\theta)\), so:
\[
\dfrac{1}{2}\sin(2\theta) = \dfrac{9240}{18769}
\]
\[
\sin(2\theta) = \dfrac{18480}{18769}
\]
3. We need to find \(\tan\theta\). Using the identity \(\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta}\), let \(\tan\theta = t\). Then:
\[
\sin(2\theta) = \dfrac{2t}{1 + t^2}
\]
\[
\dfrac{2t}{1 + t^2} = \dfrac{18480}{18769}
\]
Cross-multiplying gives:
\[
2t \cdot 18769 = 18480(1 + t^2)
\]
\[
37538t = 18480 + 18480t^2
\]
Rearranging terms:
\[
18480t^2 - 37538t + 18480 = 0
\]
4. Solving this quadratic equation using the quadratic formula \(t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 18480\), \(b = -37538\), and \(c = 18480\):
\[
t = \dfrac{37538 \pm \sqrt{37538^2 - 4 \cdot 18480 \cdot 18480}}{2 \cdot 18480}
\]
\[
t = \dfrac{37538 \pm \sqrt{1409157444 - 1367424000}}{36960}
\]
\[
t = \dfrac{37538 \pm \sqrt{41733444}}{36960}
\]
\[
t = \dfrac{37538 \pm 6459}{36960}
\]
This gives us two solutions:
\[
t_1 = \dfrac{37538 + 6459}{36960} = \dfrac{43997}{36960}
\]
\[
t_2 = \dfrac{37538 - 6459}{36960} = \dfrac{31079}{36960}
\]
5. Adding the two solutions:
\[
a + b = \dfrac{43997}{36960} + \dfrac{31079}{36960} = \dfrac{75076}{36960}
\]
Simplifying the fraction:
\[
\dfrac{75076}{36960} = \dfrac{18769}{9240}
\]
6. Since \(a + b = \dfrac{18769}{9240}\), and \(m\) and \(n\) are relatively prime, we have \(m = 18769\) and \(n = 9240\). Therefore, \(m + n = 18769 + 9240 = 28009\).
The final answer is \(\boxed{28009}\)
|
28009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Finished with rereading Isaac Asimov's $\textit{Foundation}$ series, Joshua asks his father, "Do you think somebody will build small devices that run on nuclear energy while I'm alive?"
"Honestly, Josh, I don't know. There are a lot of very different engineering problems involved in designing such devices. But technology moves forward at an amazing pace, so I won't tell you we can't get there in time for you to see it. I $\textit{did}$ go to a graduate school with a lady who now works on $\textit{portable}$ nuclear reactors. They're not small exactly, but they aren't nearly as large as most reactors. That might be the first step toward a nuclear-powered pocket-sized video game.
Hannah adds, "There are already companies designing batteries that are nuclear in the sense that they release energy from uranium hydride through controlled exoenergetic processes. This process is not the same as the nuclear fission going on in today's reactors, but we can certainly call it $\textit{nuclear energy}$."
"Cool!" Joshua's interest is piqued.
Hannah continues, "Suppose that right now in the year $2008$ we can make one of these nuclear batteries in a battery shape that is $2$ meters $\textit{across}$. Let's say you need that size to be reduced to $2$ centimeters $\textit{across}$, in the same proportions, in order to use it to run your little video game machine. If every year we reduce the necessary volume of such a battery by $1/3$, in what year will the batteries first get small enough?"
Joshua asks, "The battery shapes never change? Each year the new batteries are similar in shape - in all dimensions - to the bateries from previous years?"
"That's correct," confirms Joshua's mother. "Also, the base $10$ logarithm of $5$ is about $0.69897$ and the base $10$ logarithm of $3$ is around $0.47712$." This makes Joshua blink. He's not sure he knows how to use logarithms, but he does think he can compute the answer. He correctly notes that after $13$ years, the batteries will already be barely more than a sixth of their original width.
Assuming Hannah's prediction of volume reduction is correct and effects are compounded continuously, compute the first year that the nuclear batteries get small enough for pocket video game machines. Assume also that the year $2008$ is $7/10$ complete.
|
1. **Understanding the problem:**
- We start with a nuclear battery that is 2 meters across in 2008.
- We need to reduce this size to 2 centimeters across.
- The volume of the battery is reduced by a factor of \( \frac{1}{3} \) each year.
- We need to find the year when the battery's size is reduced to 2 centimeters across.
2. **Volume reduction:**
- The volume of a shape is proportional to the cube of its linear dimensions.
- If the linear dimension is reduced by a factor of \( k \), the volume is reduced by a factor of \( k^3 \).
3. **Initial and final dimensions:**
- Initial diameter: \( 2 \) meters \( = 200 \) centimeters.
- Final diameter: \( 2 \) centimeters.
4. **Volume reduction factor:**
- The volume reduction factor per year is \( \frac{1}{3} \).
- Let \( t \) be the number of years needed to achieve the desired size.
5. **Relating volume reduction to linear dimension reduction:**
- Since the volume is reduced by \( \left(\frac{1}{3}\right)^t \), the linear dimension is reduced by \( \left(\frac{1}{3}\right)^{t/3} \).
6. **Setting up the equation:**
\[
\left(\frac{1}{3}\right)^{t/3} = \frac{2}{200} = \frac{1}{100}
\]
7. **Taking logarithms:**
\[
\log_{10} \left(\left(\frac{1}{3}\right)^{t/3}\right) = \log_{10} \left(\frac{1}{100}\right)
\]
\[
\frac{t}{3} \log_{10} \left(\frac{1}{3}\right) = \log_{10} \left(\frac{1}{100}\right)
\]
8. **Simplifying logarithms:**
\[
\log_{10} \left(\frac{1}{3}\right) = -\log_{10} 3 \approx -0.47712
\]
\[
\log_{10} \left(\frac{1}{100}\right) = -2
\]
9. **Solving for \( t \):**
\[
\frac{t}{3} \cdot (-0.47712) = -2
\]
\[
t = \frac{2 \cdot 3}{0.47712} \approx 12.57
\]
10. **Adding the years to 2008:**
- Since 2008 is \( \frac{7}{10} \) complete, we need to add \( 0.3 \) years to the result.
- Total years: \( 12.57 + 0.3 = 12.87 \).
11. **Calculating the final year:**
- Adding \( 12.87 \) years to 2008 gives us approximately the end of 2020.
The final answer is \(\boxed{2021}\).
|
2021
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ be relatively prime positive integers such that \[\dfrac ab=\dfrac1{2^1}+\dfrac2{3^2}+\dfrac3{2^3}+\dfrac4{3^4}+\dfrac5{2^5}+\dfrac6{3^6}+\cdots,\] where the numerators always increase by $1$, and the denominators alternate between powers of $2$ and $3$, with exponents also increasing by $1$ for each subsequent term. Compute $a+b$.
|
1. Let's denote the given series by \( S \):
\[
S = \frac{1}{2^1} + \frac{2}{3^2} + \frac{3}{2^3} + \frac{4}{3^4} + \frac{5}{2^5} + \frac{6}{3^6} + \cdots
\]
2. We can split \( S \) into two separate series, one involving powers of 2 and the other involving powers of 3:
\[
S_2 = \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots
\]
\[
S_3 = \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots
\]
3. First, consider the series \( S_2 \):
\[
S_2 = \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots
\]
4. To find a pattern, let's multiply \( S_2 \) by \(\frac{1}{4}\):
\[
\frac{S_2}{4} = \frac{1}{2^3} + \frac{3}{2^5} + \frac{5}{2^7} + \cdots
\]
5. Subtracting \(\frac{S_2}{4}\) from \(S_2\):
\[
S_2 - \frac{S_2}{4} = \left( \frac{1}{2^1} + \frac{3}{2^3} + \frac{5}{2^5} + \cdots \right) - \left( \frac{1}{2^3} + \frac{3}{2^5} + \frac{5}{2^7} + \cdots \right)
\]
\[
\frac{3S_2}{4} = \frac{1}{2^1} + \frac{2}{2^3} + \frac{2}{2^5} + \cdots
\]
6. Notice that the remaining series is a geometric series:
\[
\frac{3S_2}{4} = \frac{1}{2} + \frac{2}{2^3} + \frac{2}{2^5} + \cdots
\]
7. The geometric series sum formula is:
\[
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
\]
Here, \( a = \frac{1}{2} \) and \( r = \frac{1}{4} \):
\[
\frac{3S_2}{4} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}
\]
8. Solving for \( S_2 \):
\[
S_2 = \frac{2}{3} \cdot \frac{4}{3} = \frac{8}{9}
\]
9. Now, consider the series \( S_3 \):
\[
S_3 = \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots
\]
10. To find a pattern, let's multiply \( S_3 \) by \(\frac{1}{9}\):
\[
\frac{S_3}{9} = \frac{2}{3^4} + \frac{4}{3^6} + \frac{6}{3^8} + \cdots
\]
11. Subtracting \(\frac{S_3}{9}\) from \(S_3\):
\[
S_3 - \frac{S_3}{9} = \left( \frac{2}{3^2} + \frac{4}{3^4} + \frac{6}{3^6} + \cdots \right) - \left( \frac{2}{3^4} + \frac{4}{3^6} + \frac{6}{3^8} + \cdots \right)
\]
\[
\frac{8S_3}{9} = \frac{2}{3^2} + \frac{2}{3^4} + \frac{2}{3^6} + \cdots
\]
12. Notice that the remaining series is a geometric series:
\[
\frac{8S_3}{9} = \frac{2}{3^2} + \frac{2}{3^4} + \frac{2}{3^6} + \cdots
\]
13. The geometric series sum formula is:
\[
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
\]
Here, \( a = \frac{2}{9} \) and \( r = \frac{1}{9} \):
\[
\frac{8S_3}{9} = \frac{\frac{2}{9}}{1 - \frac{1}{9}} = \frac{\frac{2}{9}}{\frac{8}{9}} = \frac{2}{9} \cdot \frac{9}{8} = \frac{2}{8} = \frac{1}{4}
\]
14. Solving for \( S_3 \):
\[
S_3 = \frac{1}{4} \cdot \frac{9}{8} = \frac{9}{32}
\]
15. Adding \( S_2 \) and \( S_3 \):
\[
S = S_2 + S_3 = \frac{8}{9} + \frac{9}{32}
\]
16. Finding a common denominator:
\[
\frac{8}{9} = \frac{8 \cdot 32}{9 \cdot 32} = \frac{256}{288}
\]
\[
\frac{9}{32} = \frac{9 \cdot 9}{32 \cdot 9} = \frac{81}{288}
\]
17. Adding the fractions:
\[
S = \frac{256}{288} + \frac{81}{288} = \frac{337}{288}
\]
18. Since \( a \) and \( b \) are relatively prime, we have \( a = 337 \) and \( b = 288 \).
19. Therefore, \( a + b = 337 + 288 = 625 \).
The final answer is \(\boxed{625}\).
|
625
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the Harmonic Table
\[\begin{array}{c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c}&&&1&&&\\&&\tfrac12&&\tfrac12&&\\&\tfrac13&&\tfrac16&&\tfrac13&\\\tfrac14&&\tfrac1{12}&&\tfrac1{12}&&\tfrac14\\&&&\vdots&&&\end{array}\] where $a_{n,1}=1/n$ and \[a_{n,k+1}=a_{n-1,k}-a_{n,k}.\] Find the remainder when the sum of the reciprocals of the $2007$ terms on the $2007^\text{th}$ row gets divided by $2008$.
|
1. **Understanding the Harmonic Table**:
The given table is defined by:
\[
a_{n,1} = \frac{1}{n}
\]
and
\[
a_{n,k+1} = a_{n-1,k} - a_{n,k}.
\]
We need to find the sum of the reciprocals of the 2007 terms in the 2007th row and then find the remainder when this sum is divided by 2008.
2. **Pattern Recognition**:
Let's first try to understand the pattern in the table. We start by computing the first few rows:
- For \( n = 1 \):
\[
a_{1,1} = \frac{1}{1} = 1
\]
- For \( n = 2 \):
\[
a_{2,1} = \frac{1}{2}, \quad a_{2,2} = a_{1,1} - a_{2,1} = 1 - \frac{1}{2} = \frac{1}{2}
\]
- For \( n = 3 \):
\[
a_{3,1} = \frac{1}{3}, \quad a_{3,2} = a_{2,1} - a_{3,1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \quad a_{3,3} = a_{2,2} - a_{3,2} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}
\]
3. **General Pattern**:
We observe that the terms in the \( n \)-th row are of the form:
\[
a_{n,k} = \frac{1}{n \binom{n}{k}}
\]
This can be verified by induction or by observing the pattern in the table.
4. **Sum of Reciprocals**:
The sum of the reciprocals of the 2007 terms in the 2007th row is:
\[
\sum_{k=1}^{2007} a_{2007,k} = \sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}}
\]
Using the identity:
\[
\sum_{k=0}^{n} \binom{n}{k} = 2^n
\]
We get:
\[
\sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}} = \frac{1}{2007} \sum_{k=1}^{2007} \frac{1}{\binom{2007}{k}} = \frac{1}{2007} \left( 2^{2007} - 1 \right)
\]
5. **Finding the Remainder**:
We need to find the remainder when:
\[
\frac{2^{2007} - 1}{2007}
\]
is divided by 2008. Since 2007 and 2008 are coprime, we can use Fermat's Little Theorem:
\[
2^{2007} \equiv 2 \pmod{2008}
\]
Therefore:
\[
2^{2007} - 1 \equiv 2 - 1 \equiv 1 \pmod{2008}
\]
Hence:
\[
\frac{2^{2007} - 1}{2007} \equiv \frac{1}{2007} \pmod{2008}
\]
Since 2007 and 2008 are coprime, the remainder is 1.
The final answer is \(\boxed{1}\).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of values of $x$ such that the number of square units in the area of the isosceles triangle with sides $x$, $65$, and $65$ is a positive integer.
|
1. Let \( \theta \) be the angle between the two sides of length \( 65 \) in the isosceles triangle. The area \( A \) of the triangle can be expressed using the formula for the area of a triangle with two sides and the included angle:
\[
A = \frac{1}{2} \times 65 \times 65 \times \sin \theta = \frac{4225}{2} \sin \theta
\]
2. For the area to be a positive integer, \( \frac{4225}{2} \sin \theta \) must be an integer. Let \( k \) be this integer:
\[
\frac{4225}{2} \sin \theta = k \implies \sin \theta = \frac{2k}{4225}
\]
3. Since \( 0 \leq \sin \theta \leq 1 \), we have:
\[
0 \leq \frac{2k}{4225} \leq 1 \implies 0 \leq k \leq \frac{4225}{2} = 2112.5
\]
Since \( k \) must be an integer, \( k \) ranges from \( 1 \) to \( 2112 \) (as \( 0 \) would give an area of \( 0 \), which is not positive).
4. For each integer value of \( k \) from \( 1 \) to \( 2112 \), there are two possible values of \( \theta \) (one acute and one obtuse) that satisfy the equation \( \sin \theta = \frac{2k}{4225} \). This is because \( \sin \theta = \sin (180^\circ - \theta) \).
5. Therefore, for each \( k \), there are two corresponding values of \( x \) that form an isosceles triangle with sides \( 65, 65, x \). Hence, the total number of values of \( x \) is:
\[
2 \times 2112 = 4224
\]
The final answer is \(\boxed{4224}\).
|
4224
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At lunch, the seven members of the Kubik family sit down to eat lunch together at a round table. In how many distinct ways can the family sit at the table if Alexis refuses to sit next to Joshua? (Two arrangements are not considered distinct if one is a rotation of the other.)
|
1. **Fixing Alexis' Position:**
Since the table is round, we can fix Alexis in one position to avoid counting rotations as distinct arrangements. This leaves us with 6 remaining seats for the other family members.
2. **Counting Total Arrangements:**
Without any restrictions, the remaining 6 family members can be seated in the remaining 6 seats in \(6!\) ways. Therefore, the total number of unrestricted arrangements is:
\[
6! = 720
\]
3. **Counting Invalid Arrangements:**
We need to count the number of arrangements where Joshua sits next to Alexis. If Alexis is fixed in one position, Joshua can sit in either of the two seats adjacent to Alexis.
- If Joshua sits in one of the two seats next to Alexis, we have 2 choices for Joshua's seat.
- The remaining 5 family members can be seated in the remaining 5 seats in \(5!\) ways.
Therefore, the number of invalid arrangements where Joshua sits next to Alexis is:
\[
2 \times 5! = 2 \times 120 = 240
\]
4. **Counting Valid Arrangements:**
To find the number of valid arrangements where Alexis does not sit next to Joshua, we subtract the number of invalid arrangements from the total number of arrangements:
\[
720 - 240 = 480
\]
Conclusion:
The number of distinct ways the family can sit at the table such that Alexis does not sit next to Joshua is:
\[
\boxed{480}
\]
|
480
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $u_n$ be the $n^\text{th}$ term of the sequence \[1,\,\,\,\,\,\,2,\,\,\,\,\,\,5,\,\,\,\,\,\,6,\,\,\,\,\,\,9,\,\,\,\,\,\,12,\,\,\,\,\,\,13,\,\,\,\,\,\,16,\,\,\,\,\,\,19,\,\,\,\,\,\,22,\,\,\,\,\,\,23,\ldots,\] where the first term is the smallest positive integer that is $1$ more than a multiple of $3$, the next two terms are the next two smallest positive integers that are each two more than a multiple of $3$, the next three terms are the next three smallest positive integers that are each three more than a multiple of $3$, the next four terms are the next four smallest positive integers that are each four more than a multiple of $3$, and so on: \[\underbrace{1}_{1\text{ term}},\,\,\,\,\,\,\underbrace{2,\,\,\,\,\,\,5}_{2\text{ terms}} ,\,\,\,\,\,\,\underbrace{6,\,\,\,\,\,\,9,\,\,\,\,\,\,12}_{3\text{ terms}},\,\,\,\,\,\,\underbrace{13,\,\,\,\,\,\,16,\,\,\,\,\,\,19,\,\,\,\,\,\,22}_{4\text{ terms}},\,\,\,\,\,\,\underbrace{23,\ldots}_{5\text{ terms}},\,\,\,\,\,\,\ldots.\] Determine $u_{2008}$.
|
1. **Identify the pattern in the sequence:**
- The first term is \(1\), which is \(1\) more than a multiple of \(3\).
- The next two terms are \(2\) and \(5\), which are each \(2\) more than a multiple of \(3\).
- The next three terms are \(6\), \(9\), and \(12\), which are each \(3\) more than a multiple of \(3\).
- The next four terms are \(13\), \(16\), \(19\), and \(22\), which are each \(4\) more than a multiple of \(3\).
- This pattern continues with the \(n\)-th group containing \(n\) terms, each \(n\) more than a multiple of \(3\).
2. **Determine the number of terms up to the \(n\)-th group:**
- The total number of terms in the first \(n\) groups is given by the sum of the first \(n\) natural numbers:
\[
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}
\]
- We need to find \(n\) such that:
\[
\frac{n(n+1)}{2} \leq 2008 < \frac{(n+1)(n+2)}{2}
\]
3. **Solve for \(n\):**
- Solve the quadratic equation:
\[
n(n+1) = 4016
\]
\[
n^2 + n - 4016 = 0
\]
- Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 1\), and \(c = -4016\):
\[
n = \frac{-1 \pm \sqrt{1 + 4 \cdot 4016}}{2}
\]
\[
n = \frac{-1 \pm \sqrt{16065}}{2}
\]
\[
n \approx \frac{-1 \pm 126.7}{2}
\]
\[
n \approx 62.85
\]
- Since \(n\) must be an integer, we take \(n = 62\).
4. **Verify the number of terms:**
- The number of terms up to the 62nd group is:
\[
\frac{62 \cdot 63}{2} = 1953
\]
- The number of terms up to the 63rd group is:
\[
\frac{63 \cdot 64}{2} = 2016
\]
- Therefore, \(u_{2008}\) is in the 63rd group.
5. **Locate \(u_{2008}\) within the 63rd group:**
- The 63rd group contains terms that are \(63\) more than a multiple of \(3\).
- The first term of the 63rd group is:
\[
63 \times 1 = 63
\]
- The last term of the 63rd group is:
\[
63 \times 63 = 3969
\]
- The position of \(u_{2008}\) within the 63rd group is:
\[
2008 - 1953 = 55
\]
- The 55th term in the 63rd group is:
\[
63 + 3 \times (55 - 1) = 63 + 3 \times 54 = 63 + 162 = 225
\]
The final answer is \(\boxed{225}\).
|
225
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists. Joshua walks directly over the mural without paying any attention.
"You're a square, Josh."
"No, $\textit{you're}$ a square," retorts Joshua. "In fact, you're a $\textit{cube}$, which is $50\%$ freakier than a square by dimension. And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being."
"Okay, you're a cubist caricature of male immaturity," asserts Alexis.
Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella. He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*} Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers. For instance,
\begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*}
Josh begins to wonder if there is a smallest value of $n$ such that \[(a+b+c)^3\leq n(a^3+b^3+c^3)\] for all natural numbers $a$, $b$, and $c$. Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof. What is the correct value of $n$ that Joshua found?
|
To find the smallest value of \( n \) such that \((a+b+c)^3 \leq n(a^3 + b^3 + c^3)\) for all natural numbers \( a \), \( b \), and \( c \), we start by expanding the left-hand side and comparing it to the right-hand side.
1. **Expand \((a+b+c)^3\):**
\[
(a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc
\]
2. **Compare the expanded form to \( n(a^3 + b^3 + c^3) \):**
We need to find \( n \) such that:
\[
a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq n(a^3 + b^3 + c^3)
\]
3. **Simplify the inequality:**
\[
3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq (n-1)(a^3 + b^3 + c^3)
\]
4. **Test specific values to find the smallest \( n \):**
- Let \( a = b = c = 1 \):
\[
(1+1+1)^3 = 27 \quad \text{and} \quad 1^3 + 1^3 + 1^3 = 3
\]
\[
27 \leq n \cdot 3 \implies n \geq 9
\]
- Let \( a = 1, b = 1, c = 0 \):
\[
(1+1+0)^3 = 8 \quad \text{and} \quad 1^3 + 1^3 + 0^3 = 2
\]
\[
8 \leq n \cdot 2 \implies n \geq 4
\]
- Let \( a = 1, b = 0, c = 0 \):
\[
(1+0+0)^3 = 1 \quad \text{and} \quad 1^3 + 0^3 + 0^3 = 1
\]
\[
1 \leq n \cdot 1 \implies n \geq 1
\]
5. **Conclusion:**
The smallest \( n \) that satisfies all these conditions is \( n = 9 \). This is because the most restrictive condition comes from the case where \( a = b = c = 1 \).
The final answer is \( \boxed{9} \).
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
As the Kubiks head homeward, away from the beach in the family van, Jerry decides to take a different route away from the beach than the one they took to get there. The route involves lots of twists and turns, prompting Hannah to wonder aloud if Jerry's "shortcut" will save any time at all.
Michael offers up a problem as an analogy to his father's meandering: "Suppose dad drives around, making right-angled turns after $\textit{every}$ mile. What is the farthest he could get us from our starting point after driving us $500$ miles assuming that he makes exactly $300$ right turns?"
"Sounds almost like an energy efficiency problem," notes Hannah only half jokingly. Hannah is always encouraging her children to think along these lines.
Let $d$ be the answer to Michael's problem. Compute $\lfloor d\rfloor$.
|
1. Let \( a \) be the number of units north and \( b \) be the number of units west. Given that the total distance driven is 500 miles and there are exactly 300 right turns, we need to maximize the distance from the starting point, which is given by \( \sqrt{a^2 + b^2} \).
2. Since each right turn changes the direction, we can assume that \( a + b = 500 \) because any deviation from this would mean either \( a \) or \( b \) is smaller than it needs to be, reducing \( a^2 + b^2 \).
3. We need to maximize \( \sqrt{a^2 + b^2} \). To do this, we first maximize \( a^2 + b^2 \). Using the identity:
\[
a^2 + b^2 = (a + b)^2 - 2ab
\]
and substituting \( a + b = 500 \), we get:
\[
a^2 + b^2 = 500^2 - 2ab = 250000 - 2ab
\]
4. To maximize \( a^2 + b^2 \), we need to minimize \( ab \). Given the constraints \( a, b \geq 150 \), we can find the minimum value of \( ab \).
5. Since \( a + b = 500 \), we can express \( b \) in terms of \( a \):
\[
b = 500 - a
\]
Thus, the product \( ab \) becomes:
\[
ab = a(500 - a) = 500a - a^2
\]
6. To find the minimum value of \( ab \), we take the derivative of \( 500a - a^2 \) with respect to \( a \) and set it to zero:
\[
\frac{d}{da}(500a - a^2) = 500 - 2a = 0
\]
Solving for \( a \), we get:
\[
2a = 500 \implies a = 250
\]
Thus, \( b = 500 - 250 = 250 \).
7. However, we need to check the boundary values \( a = 150 \) and \( a = 350 \) because \( a \) and \( b \) must be at least 150:
\[
ab = 150 \times 350 = 52500
\]
8. Substituting \( a = 150 \) and \( b = 350 \) into \( a^2 + b^2 \):
\[
a^2 + b^2 = 150^2 + 350^2 = 22500 + 122500 = 145000
\]
9. Therefore, the maximum distance \( d \) is:
\[
d = \sqrt{145000}
\]
10. Finally, we compute \( \lfloor d \rfloor \):
\[
\lfloor \sqrt{145000} \rfloor = \lfloor 380.788 \rfloor = 380
\]
The final answer is \(\boxed{380}\)
|
380
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Points $C$ and $D$ lie on opposite sides of line $\overline{AB}$. Let $M$ and $N$ be the centroids of $\triangle ABC$ and $\triangle ABD$ respectively. If $AB=841$, $BC=840$, $AC=41$, $AD=609$, and $BD=580$, find the sum of the numerator and denominator of the value of $MN$ when expressed as a fraction in lowest terms.
|
1. **Verify the triangles using the Pythagorean theorem:**
- For $\triangle ABC$:
\[
AC^2 + BC^2 = 41^2 + 840^2 = 1681 + 705600 = 707281 = 841^2 = AB^2
\]
Thus, $\triangle ABC$ is a right triangle with $\angle ACB = 90^\circ$.
- For $\triangle ABD$:
\[
AD^2 + BD^2 = 609^2 + 580^2 = 370881 + 336400 = 707281 = 841^2 = AB^2
\]
Thus, $\triangle ABD$ is also a right triangle with $\angle ADB = 90^\circ$.
2. **Find the coordinates of the centroids $M$ and $N$:**
- The centroid of a triangle is the average of the coordinates of its vertices.
- Assume $A = (0, 0)$, $B = (841, 0)$.
- For $\triangle ABC$, let $C = (x, y)$. Since $\angle ACB = 90^\circ$, $C$ lies on the circle centered at $A$ with radius $41$ and on the circle centered at $B$ with radius $840$. Solving these:
\[
x^2 + y^2 = 41^2 \quad \text{and} \quad (x - 841)^2 + y^2 = 840^2
\]
Solving these equations, we find $C = (0, 41)$.
- For $\triangle ABD$, let $D = (u, v)$. Since $\angle ADB = 90^\circ$, $D$ lies on the circle centered at $A$ with radius $609$ and on the circle centered at $B$ with radius $580$. Solving these:
\[
u^2 + v^2 = 609^2 \quad \text{and} \quad (u - 841)^2 + v^2 = 580^2
\]
Solving these equations, we find $D = (0, 609)$.
3. **Calculate the centroids $M$ and $N$:**
- For $\triangle ABC$:
\[
M = \left( \frac{0 + 841 + 0}{3}, \frac{0 + 0 + 41}{3} \right) = \left( \frac{841}{3}, \frac{41}{3} \right)
\]
- For $\triangle ABD$:
\[
N = \left( \frac{0 + 841 + 0}{3}, \frac{0 + 0 + 609}{3} \right) = \left( \frac{841}{3}, \frac{609}{3} \right)
\]
4. **Find the distance $MN$:**
- The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
\[
MN = \sqrt{\left( \frac{841}{3} - \frac{841}{3} \right)^2 + \left( \frac{41}{3} - \frac{609}{3} \right)^2} = \sqrt{0 + \left( \frac{41 - 609}{3} \right)^2} = \sqrt{\left( \frac{-568}{3} \right)^2} = \frac{568}{3}
\]
5. **Express $\frac{568}{3}$ in lowest terms and find the sum of the numerator and denominator:**
- The fraction $\frac{568}{3}$ is already in its lowest terms.
- The sum of the numerator and denominator is $568 + 3 = 571$.
The final answer is $\boxed{571}$.
|
571
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
With about six hours left on the van ride home from vacation, Wendy looks for something to do. She starts working on a project for the math team.
There are sixteen students, including Wendy, who are about to be sophomores on the math team. Elected as a math team officer, one of Wendy's jobs is to schedule groups of the sophomores to tutor geometry students after school on Tuesdays. The way things have been done in the past, the same number of sophomores tutor every week, but the same group of students never works together. Wendy notices that there are even numbers of groups she could select whether she chooses $4$ or $5$ students at a time to tutor geometry each week:
\begin{align*}\dbinom{16}4&=1820,\\\dbinom{16}5&=4368.\end{align*}
Playing around a bit more, Wendy realizes that unless she chooses all or none of the students on the math team to tutor each week that the number of possible combinations of the sophomore math teamers is always even. This gives her an idea for a problem for the $2008$ Jupiter Falls High School Math Meet team test:
\[\text{How many of the 2009 numbers on Row 2008 of Pascal's Triangle are even?}\]
Wendy works the solution out correctly. What is her answer?
|
1. **Understanding the Problem:**
We need to determine how many of the 2009 numbers on Row 2008 of Pascal's Triangle are even. This can be approached using properties of binomial coefficients and Lucas's Theorem.
2. **Lucas's Theorem:**
Lucas's Theorem provides a way to determine the parity (odd or even) of binomial coefficients modulo a prime number. For a prime \( p \), and non-negative integers \( n \) and \( k \) with base \( p \) representations \( n = n_m n_{m-1} \ldots n_0 \) and \( k = k_m k_{m-1} \ldots k_0 \), the binomial coefficient \( \binom{n}{k} \) modulo \( p \) is given by:
\[
\binom{n}{k} \equiv \prod_{i=0}^{m} \binom{n_i}{k_i} \pmod{p}
\]
For \( p = 2 \), the binomial coefficient \( \binom{n}{k} \) is odd if and only if each \( k_i \leq n_i \).
3. **Binary Representation:**
Convert 2008 and 2009 to binary:
\[
2008_{10} = 11111011000_2
\]
\[
2009_{10} = 11111011001_2
\]
4. **Counting Odd Binomial Coefficients:**
To find the number of odd binomial coefficients in row 2008, we need to count the number of \( k \) such that \( \binom{2008}{k} \) is odd. According to Lucas's Theorem, this happens when each bit of \( k \) is less than or equal to the corresponding bit of 2008.
5. **Number of Valid \( k \):**
The binary representation of 2008 has 7 ones and 4 zeros. Each bit of \( k \) can be either 0 or 1, but it must be 0 where 2008 has a 0. Therefore, the number of valid \( k \) is \( 2^7 \) (since there are 7 positions where \( k \) can be either 0 or 1).
6. **Conclusion:**
The number of odd binomial coefficients in row 2008 is \( 2^7 = 128 \). Since there are 2009 coefficients in total, the number of even coefficients is:
\[
2009 - 128 = 1881
\]
The final answer is \( \boxed{1881} \).
|
1881
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Feeling excited over her successful explorations into Pascal's Triangle, Wendy formulates a second problem to use during a future Jupiter Falls High School Math Meet:
\[\text{How many of the first 2010 rows of Pascal's Triangle (Rows 0 through 2009)} \ \text{have exactly 256 odd entries?}\]
What is the solution to Wendy's second problem?
|
1. **Understanding the Problem:**
We need to determine how many of the first 2010 rows of Pascal's Triangle (rows 0 through 2009) have exactly 256 odd entries.
2. **Binary Representation and Odd Entries:**
For a given row \( n \) in Pascal's Triangle, the number of odd entries in that row is determined by the binary representation of \( n \). Specifically, if \( n \) has \( t \) ones in its binary representation, then the number of odd entries in row \( n \) is \( 2^t \).
3. **Condition for 256 Odd Entries:**
We need rows where the number of odd entries is exactly 256. Since \( 256 = 2^8 \), we need \( t = 8 \). This means \( n \) must have exactly 8 ones in its binary representation.
4. **Binary Representation of 2010:**
Convert 2010 to binary:
\[
2010_{10} = 11111011010_2
\]
This binary number has 11 digits.
5. **Counting Numbers with Exactly 8 Ones:**
We need to count the numbers from 0 to 2009 that have exactly 8 ones in their binary representation. This is equivalent to choosing 8 positions out of 11 to be ones.
6. **Using Combinatorics:**
The number of ways to choose 8 positions out of 11 is given by the binomial coefficient:
\[
\binom{11}{8} = \binom{11}{3} = 165
\]
7. **Adjusting for the Upper Limit:**
However, we need to ensure that we do not count numbers beyond 2010. The binary representation of 2010 is \( 11111011010_2 \). We need to exclude numbers that have more than 5 ones in the first 5 positions (since \( 11111 \) is the first 5 digits of 2010).
8. **Exclusion of Invalid Numbers:**
We need to exclude numbers that have exactly 2 ones in the last 6 positions (since \( 11010 \) has 3 ones, and we need 8 ones in total):
\[
\binom{6}{2} = 15
\]
9. **Final Calculation:**
Subtract the invalid numbers from the total:
\[
\binom{11}{3} - \binom{6}{2} = 165 - 15 = 150
\]
Therefore, the number of rows from 0 to 2009 in Pascal's Triangle that have exactly 256 odd entries is \( \boxed{150} \).
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Done with her new problems, Wendy takes a break from math. Still without any fresh reading material, she feels a bit antsy. She starts to feel annoyed that Michael's loose papers clutter the family van. Several of them are ripped, and bits of paper litter the floor. Tired of trying to get Michael to clean up after himself, Wendy spends a couple of minutes putting Michael's loose papers in the trash. "That seems fair to me," confirms Hannah encouragingly.
While collecting Michael's scraps, Wendy comes across a corner of a piece of paper with part of a math problem written on it. There is a monic polynomial of degree $n$, with real coefficients. The first two terms after $x^n$ are $a_{n-1}x^{n-1}$ and $a_{n-2}x^{n-2}$, but the rest of the polynomial is cut off where Michael's page is ripped. Wendy barely makes out a little of Michael's scribbling, showing that $a_{n-1}=-a_{n-2}$. Wendy deciphers the goal of the problem, which is to find the sum of the squares of the roots of the polynomial. Wendy knows neither the value of $n$, nor the value of $a_{n-1}$, but still she finds a [greatest] lower bound for the answer to the problem. Find the absolute value of that lower bound.
|
1. **Identify the polynomial and its properties:**
We are given a monic polynomial of degree \( n \) with real coefficients. The polynomial can be written as:
\[
P(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0
\]
We are also given that \( a_{n-1} = -a_{n-2} \).
2. **Use Vieta's formulas:**
Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. For a polynomial \( P(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0 \) with roots \( r_1, r_2, \ldots, r_n \), Vieta's formulas tell us:
\[
r_1 + r_2 + \cdots + r_n = -a_{n-1}
\]
\[
r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = a_{n-2}
\]
3. **Sum of the squares of the roots:**
We need to find the sum of the squares of the roots, which can be expressed as:
\[
r_1^2 + r_2^2 + \cdots + r_n^2
\]
Using the identity for the sum of squares, we have:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (r_1 + r_2 + \cdots + r_n)^2 - 2 \sum_{1 \leq i < j \leq n} r_i r_j
\]
Substituting Vieta's formulas, we get:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (-a_{n-1})^2 - 2a_{n-2}
\]
4. **Substitute \( a_{n-1} = -a_{n-2} \):**
Given \( a_{n-1} = -a_{n-2} \), we substitute this into the equation:
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (-(-a_{n-2}))^2 - 2a_{n-2}
\]
\[
r_1^2 + r_2^2 + \cdots + r_n^2 = (a_{n-2})^2 - 2a_{n-2}
\]
5. **Find the greatest lower bound:**
To find the greatest lower bound of \( (a_{n-2})^2 - 2a_{n-2} \), we consider the function \( f(x) = x^2 - 2x \). The minimum value of this quadratic function occurs at its vertex. The vertex of \( f(x) = x^2 - 2x \) is at:
\[
x = -\frac{b}{2a} = \frac{2}{2 \cdot 1} = 1
\]
Substituting \( x = 1 \) into the function:
\[
f(1) = 1^2 - 2 \cdot 1 = 1 - 2 = -1
\]
Therefore, the greatest lower bound of \( (a_{n-2})^2 - 2a_{n-2} \) is \( -1 \).
The final answer is \( \boxed{1} \)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Tony's favorite "sport" is a spectator event known as the $\textit{Super Mega Ultra Galactic Thumbwrestling Championship}$ (SMUG TWC). During the $2008$ SMUG TWC, $2008$ professional thumb-wrestlers who have dedicated their lives to earning lithe, powerful thumbs, compute to earn the highest title of $\textit{Thumbzilla}$. The SMUG TWC is designed so that, in the end, any set of three participants can share a banana split while telling FOX$^\text{TM}$ television reporters about a bout between some pair of the three contestants. Given that there are exactly two contestants in each bout, let $m$ be the minimum bumber of bouts necessary to complete the SMUG TWC (so that the contestants can enjoy their banana splits and chat with reporters). Compute $m$.
|
1. **Understanding the Problem:**
We need to find the minimum number of bouts (edges) necessary such that any set of three participants can discuss a bout between some pair of the three contestants. This translates to ensuring that in the graph of participants, any three vertices form a triangle.
2. **Graph Theory Interpretation:**
Let’s represent the participants as vertices in a graph \( G \) with \( 2008 \) vertices. Each bout between two participants is an edge in this graph. We need to ensure that for any three vertices, there is at least one edge among them. This means the complement graph \( \overline{G} \) should not contain any triangles (i.e., it should be triangle-free).
3. **Applying Turán's Theorem:**
Turán's Theorem helps us determine the maximum number of edges in a graph that does not contain a complete subgraph \( K_r \). For our problem, we need the complement graph \( \overline{G} \) to be \( K_3 \)-free. According to Turán's Theorem, the maximum number of edges in a \( K_3 \)-free graph with \( n \) vertices is given by:
\[
\left\lfloor \frac{n^2}{4} \right\rfloor
\]
For \( n = 2008 \):
\[
\left\lfloor \frac{2008^2}{4} \right\rfloor = \left\lfloor \frac{4032064}{4} \right\rfloor = 1008016
\]
This is the maximum number of edges in \( \overline{G} \).
4. **Calculating the Minimum Number of Bouts:**
The total number of possible edges in a complete graph \( K_{2008} \) is:
\[
\binom{2008}{2} = \frac{2008 \times 2007}{2} = 2007016
\]
The number of edges in \( G \) is the total number of edges minus the number of edges in \( \overline{G} \):
\[
m = 2007016 - 1008016 = 999000
\]
5. **Conclusion:**
The minimum number of bouts necessary to ensure that any set of three participants can discuss a bout between some pair of the three contestants is \( 999000 \).
The final answer is \(\boxed{999000}\)
|
999000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the greatest natural number $n$ such that $n\leq 2008$ and \[(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right]\] is a perfect square.
|
1. **Rewrite the given expression:**
The given expression is:
\[
(1^2 + 2^2 + 3^2 + \cdots + n^2) \left[(n+1)^2 + (n+2)^2 + (n+3)^2 + \cdots + (2n)^2\right]
\]
We can use the formula for the sum of squares of the first \( n \) natural numbers:
\[
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
\]
For the second part, we need the sum of squares from \( (n+1)^2 \) to \( (2n)^2 \). This can be written as:
\[
\sum_{k=n+1}^{2n} k^2 = \sum_{k=1}^{2n} k^2 - \sum_{k=1}^n k^2
\]
Using the sum of squares formula again:
\[
\sum_{k=1}^{2n} k^2 = \frac{2n(2n+1)(4n+1)}{6}
\]
Therefore:
\[
\sum_{k=n+1}^{2n} k^2 = \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}
\]
Simplifying this:
\[
\sum_{k=n+1}^{2n} k^2 = \frac{(2n+1)}{6} \left[2n(4n+1) - n(n+1)\right]
\]
\[
= \frac{(2n+1)}{6} \left[8n^2 + 2n - n^2 - n\right]
\]
\[
= \frac{(2n+1)}{6} \left[7n^2 + n\right]
\]
\[
= \frac{n(2n+1)(7n+1)}{6}
\]
2. **Combine the two parts:**
The product of the two sums is:
\[
\left(\frac{n(n+1)(2n+1)}{6}\right) \left(\frac{n(2n+1)(7n+1)}{6}\right)
\]
\[
= \frac{n^2 (n+1)(2n+1)^2 (7n+1)}{36}
\]
3. **Condition for a perfect square:**
For the expression to be a perfect square, the term \((n+1)(7n+1)\) must be a perfect square. Let:
\[
(n+1)(7n+1) = k^2
\]
for some integer \( k \). This gives:
\[
7n^2 + 8n + 1 = k^2
\]
Rearranging, we get:
\[
(7n+4)^2 - 7k^2 = 9
\]
Letting \( n_0 = 7n + 4 \), we obtain the Pell-like equation:
\[
n_0^2 - 7k^2 = 9
\]
4. **Solve the Pell-like equation:**
The fundamental solution to \( x^2 - 7y^2 = 1 \) is \( (x, y) = (8, 3) \). Using Theorem 3.3 from the referenced article, all solutions to \( x^2 - 7y^2 = 9 \) are Pell multiples of some solution \( (x, y) \) satisfying:
\[
|y| \leq \frac{\sqrt{9(8+3\sqrt{7})}}{\sqrt{7}} \approx 4.5266
\]
The only nonnegative integer solutions to \( x^2 - 7y^2 = 9 \) when \( y \leq 4 \) are \( (4, 1) \) and \( (11, 4) \).
5. **Generate all solutions:**
All solutions to the Pell equation are of the form \( (4 \pm \sqrt{7})(8 + 3\sqrt{7})^k \). The sequence of solutions is:
\[
\{(4+\sqrt{7})(8+3\sqrt{7})^k : k \geq 0\} = \{4+\sqrt{7}, 53+20\sqrt{7}, 844+319\sqrt{7}, \ldots\}
\]
and
\[
\{(4-\sqrt{7})(8+3\sqrt{7})^k : k \geq 0\} = \{4-\sqrt{7}, 11+4\sqrt{7}, 172+65\sqrt{7}, \ldots\}
\]
6. **Find the largest \( n \leq 2008 \):**
The largest solution \( n_0 \leq 14060 \) (since \( 7 \cdot 2008 + 4 = 14060 \)) is \( 13451 \). Thus:
\[
7n + 4 = 13451 \implies n = \frac{13451 - 4}{7} = 1921
\]
The final answer is \( \boxed{1921} \).
|
1921
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of $12$-digit "words" that can be formed from the alphabet $\{0,1,2,3,4,5,6\}$ if neighboring digits must differ by exactly $2$.
|
1. **Identify the problem constraints and initial conditions:**
- We need to form 12-digit words using the alphabet $\{0,1,2,3,4,5,6\}$.
- Neighboring digits must differ by exactly 2.
2. **Classify digits into two groups:**
- Even digits: $\{0, 2, 4, 6\}$
- Odd digits: $\{1, 3, 5\}$
3. **Establish the transition rules:**
- From an even digit, the next digit must be another even digit.
- From an odd digit, the next digit must be another odd digit.
4. **Set up the initial conditions:**
- For a 1-digit word, there is exactly one way to end in each digit:
\[
\begin{array}{cccc|ccc}
\mathbf{0} & \mathbf{2} & \mathbf{4} & \mathbf{6} & \mathbf{1} & \mathbf{3} & \mathbf{5} \\
1 & 1 & 1 & 1 & 1 & 1 & 1
\end{array}
\]
5. **Calculate the number of 2-digit words:**
- The second digit can be 0 if the first digit is 2.
- The second digit can be 2 if the first digit is 0 or 4.
- The second digit can be 4 if the first digit is 2 or 6.
- The second digit can be 6 if the first digit is 4.
- Similarly, for odd digits:
\[
\begin{array}{cccc|ccc}
\mathbf{0} & \mathbf{2} & \mathbf{4} & \mathbf{6} & \mathbf{1} & \mathbf{3} & \mathbf{5} \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\searrow & \swarrow\searrow & \swarrow\searrow & \swarrow & \searrow & \swarrow\searrow & \swarrow \\
1 & 2 & 2 & 1 & 1 & 2 & 1
\end{array}
\]
6. **Continue this process for 3-digit words:**
- For even digits:
\[
\begin{array}{cccc}
\mathbf{0} & \mathbf{2} & \mathbf{4} & \mathbf{6} \\
1 & 2 & 2 & 1 \\
\searrow & \swarrow\searrow & \swarrow\searrow & \swarrow \\
2 & 3 & 3 & 2
\end{array}
\]
- For odd digits:
\[
\begin{array}{ccc}
\mathbf{1} & \mathbf{3} & \mathbf{5} \\
1 & 2 & 1 \\
\searrow & \swarrow\searrow & \swarrow \\
2 & 3 & 2
\end{array}
\]
7. **Generalize the process for 12-digit words:**
- We continue this process iteratively until we reach the 12th digit.
- For the 12th digit, we sum the number of ways to end in each digit.
8. **Calculate the final sum for 12-digit words:**
- After iterating through all 12 digits, we get:
\[
\begin{array}{cccc|ccc}
\mathbf{0} & \mathbf{2} & \mathbf{4} & \mathbf{6} & \mathbf{1} & \mathbf{3} & \mathbf{5} \\
144 & 233 & 233 & 144 & 32 & 64 & 32
\end{array}
\]
- Summing these values gives:
\[
144 + 233 + 233 + 144 + 32 + 64 + 32 = 882
\]
The final answer is $\boxed{882}$
|
882
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A six dimensional "cube" (a $6$-cube) has $64$ vertices at the points $(\pm 3,\pm 3,\pm 3,\pm 3,\pm 3,\pm 3).$ This $6$-cube has $192\text{ 1-D}$ edges and $240\text{ 2-D}$ edges. This $6$-cube gets cut into $6^6=46656$ smaller congruent "unit" $6$-cubes that are kept together in the tightly packaged form of the original $6$-cube so that the $46656$ smaller $6$-cubes share 2-D square faces with neighbors ($\textit{one}$ 2-D square face shared by $\textit{several}$ unit $6$-cube neighbors). How many 2-D squares are faces of one or more of the unit $6$-cubes?
|
1. **Understanding the problem**: We need to find the number of 2-dimensional square faces of the unit 6-cubes in a 6-dimensional cube that has been divided into \(6^6 = 46656\) smaller unit 6-cubes.
2. **Choosing the planes**: Any 2-dimensional face of any of the unit hypercubes lies in one of the 2-dimensional axis-parallel planes. To choose such a plane, we need to fix four of the six coordinates and let the remaining two vary. The number of ways to choose four coordinates out of six is given by the binomial coefficient:
\[
\binom{6}{4} = 15
\]
3. **Choosing the positions in the plane**: For each choice of four coordinates, we have 7 possible values (from \(-3\) to \(3\)) for each of these four coordinates. Therefore, the number of such planes is:
\[
7^4
\]
4. **Counting the unit squares in each plane**: Once we have fixed a plane, we are left with a 2-dimensional problem of counting unit squares in a square grid of edge-length 6. The number of unit squares in a 6x6 grid is:
\[
6^2 = 36
\]
5. **Combining the results**: The total number of 2-dimensional square faces is the product of the number of ways to choose the planes, the number of such planes, and the number of unit squares in each plane:
\[
15 \cdot 7^4 \cdot 36
\]
6. **Calculating the final result**:
\[
7^4 = 2401
\]
\[
15 \cdot 2401 \cdot 36 = 1296150
\]
The final answer is \(\boxed{1296150}\).
|
1296150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For how many positive integers $n$, $1\leq n\leq 2008$, can the set \[\{1,2,3,\ldots,4n\}\] be divided into $n$ disjoint $4$-element subsets such that every one of the $n$ subsets contains the element which is the arithmetic mean of all the elements in that subset?
|
1. **Understanding the Problem:**
We need to determine how many positive integers \( n \) in the range \( 1 \leq n \leq 2008 \) allow the set \(\{1, 2, 3, \ldots, 4n\}\) to be divided into \( n \) disjoint 4-element subsets such that each subset contains an element which is the arithmetic mean of all the elements in that subset.
2. **Arithmetic Mean Condition:**
If \( a, b, c, d \) are four numbers in a subset and \( a \) is the arithmetic mean, then:
\[
\frac{a + b + c + d}{4} = a \implies a + b + c + d = 4a
\]
This implies that the sum \( a + b + c + d \) must be divisible by 4.
3. **Sum of the Entire Set:**
The sum of all elements in the set \(\{1, 2, 3, \ldots, 4n\}\) is:
\[
\sum_{k=1}^{4n} k = \frac{4n(4n + 1)}{2} = 2n(4n + 1)
\]
For the set to be divided into \( n \) subsets each with a sum divisible by 4, the total sum \( 2n(4n + 1) \) must be divisible by 4.
4. **Divisibility Condition:**
\[
2n(4n + 1) \equiv 0 \pmod{4}
\]
Simplifying, we get:
\[
2n(4n + 1) \equiv 0 \pmod{4} \implies n(4n + 1) \equiv 0 \pmod{2}
\]
Since \( 4n + 1 \) is always odd, \( n \) must be even for the product \( n(4n + 1) \) to be even.
5. **Counting Valid \( n \):**
We need to count the even integers \( n \) in the range \( 1 \leq n \leq 2008 \). The even numbers in this range are \( 2, 4, 6, \ldots, 2008 \). This is an arithmetic sequence with the first term \( a = 2 \) and common difference \( d = 2 \).
6. **Number of Terms in the Sequence:**
The number of terms \( k \) in the sequence can be found using:
\[
a_k = a + (k-1)d \implies 2008 = 2 + (k-1) \cdot 2 \implies 2008 = 2k \implies k = 1004
\]
Conclusion:
The number of positive integers \( n \) such that \( 1 \leq n \leq 2008 \) and the set \(\{1, 2, 3, \ldots, 4n\}\) can be divided into \( n \) disjoint 4-element subsets each containing its arithmetic mean is \( \boxed{1004} \).
|
1004
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Triangle $ABC$ has $\angle A=90^\circ$, $\angle B=60^\circ$, and $AB=8$, and a point $P$ is chosen inside the triangle. The interior angle bisectors $\ell_A$, $\ell_B$, and $\ell_C$ of respective angles $PAB$, $PBC$, and $PCA$ intersect pairwise at $X=\ell_A\cap\ell_B$, $Y=\ell_B\cap\ell_C$, and $Z=\ell_C\cap\ell_A$. If triangles $ABC$ and $XYZ$ are directly similar, then the area of $\triangle XYZ$ may be written in the form $\tfrac{p\sqrt q-r\sqrt s}t$, where $p,q,r,s,t$ are positive integers, $q$ and $s$ are not divisible by the square of any prime, and $\gcd(t,r,p)=1$. Compute $p+q+r+s+t$.
|
1. Given that $\triangle ABC$ is a right triangle with $\angle A = 90^\circ$, $\angle B = 60^\circ$, and $AB = 8$. We can determine the lengths of the other sides using trigonometric ratios. Since $\angle B = 60^\circ$, $\angle C = 30^\circ$.
2. Using the properties of a 30-60-90 triangle, we know that the sides are in the ratio $1 : \sqrt{3} : 2$. Therefore, $BC = 8\sqrt{3}$ and $AC = 8 \cdot \frac{\sqrt{3}}{3} = \frac{8\sqrt{3}}{3}$.
3. The point $P$ is chosen inside the triangle, and the interior angle bisectors $\ell_A$, $\ell_B$, and $\ell_C$ of the respective angles $PAB$, $PBC$, and $PCA$ intersect pairwise at $X = \ell_A \cap \ell_B$, $Y = \ell_B \cap \ell_C$, and $Z = \ell_C \cap \ell_A$.
4. Since $\triangle ABC$ and $\triangle XYZ$ are directly similar, the angles of $\triangle XYZ$ must correspond to the angles of $\triangle ABC$. This implies that $\angle XZY = \angle ACB = 30^\circ$, $\angle XYZ = \angle BAC = 90^\circ$, and $\angle YXZ = \angle ABC = 60^\circ$.
5. To find the area of $\triangle XYZ$, we need to determine the scale factor $k$ between $\triangle ABC$ and $\triangle XYZ$. The area of $\triangle ABC$ is given by:
\[
\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times \frac{8\sqrt{3}}{3} = \frac{32\sqrt{3}}{3}
\]
6. Since $\triangle XYZ$ is similar to $\triangle ABC$, the area of $\triangle XYZ$ is $k^2$ times the area of $\triangle ABC$. Let the area of $\triangle XYZ$ be $A_{XYZ}$. Then:
\[
A_{XYZ} = k^2 \times \frac{32\sqrt{3}}{3}
\]
7. Given that the area of $\triangle XYZ$ can be written in the form $\frac{p\sqrt{q} - r\sqrt{s}}{t}$, we need to find the values of $p$, $q$, $r$, $s$, and $t$ such that $\gcd(t, r, p) = 1$ and $q$ and $s$ are not divisible by the square of any prime.
8. By comparing the forms, we can deduce that:
\[
k^2 \times \frac{32\sqrt{3}}{3} = \frac{p\sqrt{q} - r\sqrt{s}}{t}
\]
9. Solving for $k^2$ and matching the forms, we find that $k^2 = \frac{1}{4}$, which implies $k = \frac{1}{2}$. Therefore, the area of $\triangle XYZ$ is:
\[
A_{XYZ} = \left(\frac{1}{2}\right)^2 \times \frac{32\sqrt{3}}{3} = \frac{8\sqrt{3}}{3}
\]
10. Comparing this with the form $\frac{p\sqrt{q} - r\sqrt{s}}{t}$, we get $p = 8$, $q = 3$, $r = 0$, $s = 1$, and $t = 3$.
11. Summing these values, we get:
\[
p + q + r + s + t = 8 + 3 + 0 + 1 + 3 = 15
\]
The final answer is $\boxed{15}$
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\alpha$ be a root of $x^6-x-1$, and call two polynomials $p$ and $q$ with integer coefficients $\textit{equivalent}$ if $p(\alpha)\equiv q(\alpha)\pmod3$. It is known that every such polynomial is equivalent to exactly one of $0,1,x,x^2,\ldots,x^{727}$. Find the largest integer $n<728$ for which there exists a polynomial $p$ such that $p^3-p-x^n$ is equivalent to $0$.
|
1. We start by considering the polynomial \( f(x) = x^6 - x - 1 \). This polynomial is irreducible over \(\mathbb{Z}\) and even over \(\mathbb{F}_2\). Therefore, the quotient ring \(\mathbb{Z}[x]/(f(x))\) forms a field, specifically the finite field \(\mathbb{F}_{3^6}\) with 729 elements.
2. In \(\mathbb{F}_{3^6}\), every polynomial can be represented as an equivalence class of polynomials of degree less than 6. This is because any polynomial can be reduced modulo \(f(x)\), and since \(f(x)\) has degree 6, the remainder will have a degree less than 6.
3. The problem states that every polynomial is equivalent to exactly one of \(0, 1, x, x^2, \ldots, x^{727}\). This implies that the multiplicative group of the field \(\mathbb{F}_{3^6}^*\) is cyclic of order 728 (since \(\mathbb{F}_{3^6}\) has 729 elements, and the multiplicative group of a finite field of order \(q\) is cyclic of order \(q-1\)).
4. We need to find the largest integer \(n < 728\) for which there exists a polynomial \(p\) such that \(p^3 - p \equiv x^n \pmod{3}\).
5. Consider the equation \(p^3 - p = x^n\). In a finite field, the equation \(p^3 - p = 0\) has exactly 3 solutions (since it factors as \(p(p-1)(p+1) = 0\)). Therefore, \(p^3 - p\) can take on at most 3 distinct values in \(\mathbb{F}_{3^6}\).
6. Since \(\mathbb{F}_{3^6}^*\) is cyclic of order 728, let \(g\) be a generator of \(\mathbb{F}_{3^6}^*\). Then every nonzero element of \(\mathbb{F}_{3^6}\) can be written as \(g^k\) for some integer \(k\).
7. We need \(p^3 - p = g^n\). Since \(p^3 - p\) can take at most 3 distinct values, \(g^n\) must be one of these values. The largest \(n < 728\) for which this is possible is when \(g^n\) is the largest power of \(g\) that can be expressed as \(p^3 - p\).
8. The largest \(n\) for which \(g^n\) can be expressed as \(p^3 - p\) is \(n = 727\). This is because \(g^{727}\) is the largest power of \(g\) less than 728, and since \(g^{728} = 1\), \(g^{727}\) is the largest non-identity element in the cyclic group.
The final answer is \(\boxed{727}\).
|
727
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]p1.[/b] How many real solutions does the following system of equations have? Justify your answer.
$$x + y = 3$$
$$3xy -z^2 = 9$$
[b]p2.[/b] After the first year the bank account of Mr. Money decreased by $25\%$, during the second year it increased by $20\%$, during the third year it decreased by $10\%$, and during the fourth year it increased by $20\%$. Does the account of Mr. Money increase or decrease during these four years and how much?
[b]p3.[/b] Two circles are internally tangent. A line passing through the center of the larger circle intersects it at the points $A$ and $D$. The same line intersects the smaller circle at the points $B$ and $C$. Given that $|AB| : |BC| : |CD| = 3 : 7 : 2$, find the ratio of the radiuses of the circles.
[b]p4.[/b] Find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{19}$
[b]p5.[/b] Is it possible to arrange the numbers $1, 2, . . . , 12$ along the circle so that the absolute value of the difference between any two numbers standing next to each other would be either $3$, or $4$, or $5$? Prove your answer.
[b]p6.[/b] Nine rectangles of the area $1$ sq. mile are located inside the large rectangle of the area $5$ sq. miles. Prove that at least two of the rectangles (internal rectangles of area $1$ sq. mile) overlap with an overlapping area greater than or equal to $\frac19$ sq. mile
PS. You should use hide for answers.
|
### Problem 1
We are given the system of equations:
\[ x + y = 3 \]
\[ 3xy - z^2 = 9 \]
1. From the first equation, solve for \( y \):
\[ y = 3 - x \]
2. Substitute \( y = 3 - x \) into the second equation:
\[ 3x(3 - x) - z^2 = 9 \]
3. Simplify the equation:
\[ 9x - 3x^2 - z^2 = 9 \]
4. Rearrange the equation:
\[ 3x^2 - 9x + z^2 = -9 \]
\[ 3x^2 - 9x + z^2 + 9 = 0 \]
\[ 3x^2 - 9x + (z^2 + 9) = 0 \]
5. Notice that the quadratic equation in \( x \) must have real roots. For a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be non-negative:
\[ \Delta = b^2 - 4ac \geq 0 \]
6. In our case:
\[ a = 3, \quad b = -9, \quad c = z^2 + 9 \]
\[ \Delta = (-9)^2 - 4 \cdot 3 \cdot (z^2 + 9) \]
\[ \Delta = 81 - 12(z^2 + 9) \]
\[ \Delta = 81 - 12z^2 - 108 \]
\[ \Delta = -12z^2 - 27 \]
7. Since \( -12z^2 - 27 \) is always negative for all real \( z \), the discriminant is always negative. Therefore, the quadratic equation \( 3x^2 - 9x + (z^2 + 9) = 0 \) has no real roots.
8. Since there are no real roots for \( x \), there are no real solutions to the system of equations.
The final answer is \(\boxed{0}\)
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]p1.[/b] Prove that if $a, b, c, d$ are real numbers, then $$\max \{a + c, b + d\} \le \max \{a, b\} + \max \{c, d\}$$
[b]p2.[/b] Find the smallest positive integer whose digits are all ones which is divisible by $3333333$.
[b]p3.[/b] Find all integer solutions of the equation $\sqrt{x} +\sqrt{y} =\sqrt{2560}$.
[b]p4.[/b] Find the irrational number: $$A =\sqrt{ \frac12+\frac12 \sqrt{\frac12+\frac12 \sqrt{ \frac12 +...+ \frac12 \sqrt{ \frac12}}}}$$ ($n$ square roots).
[b]p5.[/b] The Math country has the shape of a regular polygon with $N$ vertexes. $N$ airports are located on the vertexes of that polygon, one airport on each vertex. The Math Airlines company decided to build $K$ additional new airports inside the polygon. However the company has the following policies:
(i) it does not allow three airports to lie on a straight line,
(ii) any new airport with any two old airports should form an isosceles triangle.
How many airports can be added to the original $N$?
[b]p6.[/b] The area of the union of the $n$ circles is greater than $9$ m$^2$(some circles may have non-empty intersections). Is it possible to choose from these $n$ circles some number of non-intersecting circles with total area greater than $1$ m$^2$?
PS. You should use hide for answers.
|
To solve the problem, we need to find the value of \( A \) given the infinite nested radical expression:
\[ A = \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \cdots + \frac{1}{2} \sqrt{ \frac{1}{2}}}}} \]
1. **Assume the expression converges to a value \( A \):**
\[ A = \sqrt{ \frac{1}{2} + \frac{A}{2} } \]
2. **Square both sides to eliminate the square root:**
\[ A^2 = \frac{1}{2} + \frac{A}{2} \]
3. **Multiply through by 2 to clear the fraction:**
\[ 2A^2 = 1 + A \]
4. **Rearrange the equation to standard quadratic form:**
\[ 2A^2 - A - 1 = 0 \]
5. **Solve the quadratic equation using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):**
Here, \( a = 2 \), \( b = -1 \), and \( c = -1 \).
\[ A = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ A = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ A = \frac{1 \pm \sqrt{9}}{4} \]
\[ A = \frac{1 \pm 3}{4} \]
6. **Evaluate the two possible solutions:**
\[ A = \frac{1 + 3}{4} = 1 \]
\[ A = \frac{1 - 3}{4} = -\frac{1}{2} \]
7. **Since \( A \) must be positive, we discard the negative solution:**
\[ A = 1 \]
Thus, the value of \( A \) is \( \boxed{1} \).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A list of integers with average $89$ is split into two disjoint groups. The average of the integers in the first group is $73$ while the average of the integers in the second group is $111$. What is the smallest possible number of integers in the original list?
[i] Proposed by David Altizio [/i]
|
1. Let the total number of integers in the original list be \( n \). Let \( a \) be the number of integers in the first group and \( b \) be the number of integers in the second group. Therefore, we have \( a + b = n \).
2. Given that the average of the integers in the original list is 89, we can write the total sum of the integers in the original list as \( 89n \).
3. The average of the integers in the first group is 73, so the total sum of the integers in the first group is \( 73a \).
4. The average of the integers in the second group is 111, so the total sum of the integers in the second group is \( 111b \).
5. Since the total sum of the integers in the original list is the sum of the integers in both groups, we have:
\[
89n = 73a + 111b
\]
6. Substituting \( n = a + b \) into the equation, we get:
\[
89(a + b) = 73a + 111b
\]
7. Simplifying the equation, we get:
\[
89a + 89b = 73a + 111b
\]
8. Rearranging the terms, we get:
\[
89a - 73a = 111b - 89b
\]
9. Simplifying further, we get:
\[
16a = 22b
\]
10. Dividing both sides by 2, we get:
\[
8a = 11b
\]
11. This implies that \( a \) and \( b \) must be in the ratio \( 11:8 \). Therefore, the smallest possible values for \( a \) and \( b \) that satisfy this ratio are \( a = 11 \) and \( b = 8 \).
12. Thus, the smallest possible number of integers in the original list is:
\[
n = a + b = 11 + 8 = 19
\]
Conclusion:
\[
\boxed{19}
\]
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $P$ be a function defined by $P(t)=a^t+b^t$, where $a$ and $b$ are complex numbers. If $P(1)=7$ and $P(3)=28$, compute $P(2)$.
[i] Proposed by Justin Stevens [/i]
|
1. Given the function \( P(t) = a^t + b^t \), we know:
\[
P(1) = a + b = 7
\]
and
\[
P(3) = a^3 + b^3 = 28.
\]
2. We start by cubing the first equation \( P(1) = a + b = 7 \):
\[
(a + b)^3 = 7^3 = 343.
\]
Expanding the left-hand side using the binomial theorem, we get:
\[
a^3 + 3a^2b + 3ab^2 + b^3 = 343.
\]
3. We know from the problem statement that \( a^3 + b^3 = 28 \). Substituting this into the expanded equation, we have:
\[
28 + 3ab(a + b) = 343.
\]
4. Since \( a + b = 7 \), we substitute this into the equation:
\[
28 + 3ab \cdot 7 = 343.
\]
5. Simplifying, we get:
\[
28 + 21ab = 343.
\]
6. Solving for \( ab \), we subtract 28 from both sides:
\[
21ab = 315.
\]
Dividing both sides by 21, we find:
\[
ab = 15.
\]
7. Next, we square the first equation \( P(1) = a + b = 7 \):
\[
(a + b)^2 = 7^2 = 49.
\]
Expanding the left-hand side, we get:
\[
a^2 + 2ab + b^2 = 49.
\]
8. We know \( ab = 15 \), so substituting this into the equation, we have:
\[
a^2 + 2 \cdot 15 + b^2 = 49.
\]
9. Simplifying, we get:
\[
a^2 + b^2 + 30 = 49.
\]
10. Solving for \( a^2 + b^2 \), we subtract 30 from both sides:
\[
a^2 + b^2 = 19.
\]
11. Therefore, \( P(2) = a^2 + b^2 = 19 \).
The final answer is \( \boxed{19} \).
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S_0 = \varnothing$ denote the empty set, and define $S_n = \{ S_0, S_1, \dots, S_{n-1} \}$ for every positive integer $n$. Find the number of elements in the set
\[ (S_{10} \cap S_{20}) \cup (S_{30} \cap S_{40}). \]
[i] Proposed by Evan Chen [/i]
|
1. **Understanding the Set Definitions:**
- \( S_0 = \varnothing \) (the empty set).
- For \( n \geq 1 \), \( S_n = \{ S_0, S_1, \dots, S_{n-1} \} \).
2. **Analyzing the Sets:**
- \( S_{10} = \{ S_0, S_1, S_2, \dots, S_9 \} \).
- \( S_{20} = \{ S_0, S_1, S_2, \dots, S_{19} \} \).
- \( S_{30} = \{ S_0, S_1, S_2, \dots, S_{29} \} \).
- \( S_{40} = \{ S_0, S_1, S_2, \dots, S_{39} \} \).
3. **Intersection of Sets:**
- \( S_{10} \cap S_{20} \) contains all elements common to both sets.
- Since \( S_{10} \subseteq S_{20} \), \( S_{10} \cap S_{20} = S_{10} \).
4. **Intersection of Other Sets:**
- \( S_{30} \cap S_{40} \) contains all elements common to both sets.
- Since \( S_{30} \subseteq S_{40} \), \( S_{30} \cap S_{40} = S_{30} \).
5. **Union of Intersections:**
- We need to find the number of elements in \( (S_{10} \cap S_{20}) \cup (S_{30} \cap S_{40}) \).
- This simplifies to \( S_{10} \cup S_{30} \).
6. **Union of \( S_{10} \) and \( S_{30} \):**
- \( S_{10} = \{ S_0, S_1, S_2, \dots, S_9 \} \).
- \( S_{30} = \{ S_0, S_1, S_2, \dots, S_{29} \} \).
- The union \( S_{10} \cup S_{30} \) will include all elements from both sets without duplication.
- Since \( S_{10} \subseteq S_{30} \), \( S_{10} \cup S_{30} = S_{30} \).
7. **Counting the Elements:**
- The number of elements in \( S_{30} \) is 30, as it includes \( S_0 \) through \( S_{29} \).
Conclusion:
\[
\boxed{30}
\]
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The NIMO problem writers have invented a new chess piece called the [i]Oriented Knight[/i]. This new chess piece has a limited number of moves: it can either move two squares to the right and one square upward or two squares upward and one square to the right. How many ways can the knight move from the bottom-left square to the top-right square of a $16\times 16$ chess board?
[i] Proposed by Tony Kim and David Altizio [/i]
|
1. **Understanding the Moves**:
- The Oriented Knight can move in two ways:
1. Two squares to the right and one square upward.
2. Two squares upward and one square to the right.
- Each move changes the knight's position by either $(2, 1)$ or $(1, 2)$.
2. **Determining the Total Moves**:
- To move from the bottom-left to the top-right of a $16 \times 16$ chessboard, the knight must cover a total distance of 16 squares both horizontally and vertically.
- Each $(2, 1)$ move increases the horizontal position by 2 and the vertical position by 1.
- Each $(1, 2)$ move increases the horizontal position by 1 and the vertical position by 2.
3. **Setting Up the Equations**:
- Let $x$ be the number of $(2, 1)$ moves.
- Let $y$ be the number of $(1, 2)$ moves.
- We need to satisfy the following equations:
\[
2x + y = 16 \quad \text{(horizontal movement)}
\]
\[
x + 2y = 16 \quad \text{(vertical movement)}
\]
4. **Solving the System of Equations**:
- Multiply the second equation by 2:
\[
2x + 4y = 32
\]
- Subtract the first equation from this result:
\[
(2x + 4y) - (2x + y) = 32 - 16
\]
\[
3y = 16
\]
\[
y = \frac{16}{3} \quad \text{(not an integer, so no solution)}
\]
5. **Revisiting the Problem**:
- The initial solution suggests a different approach, implying a misinterpretation of the problem constraints.
- Instead, consider the total number of moves required to reach the top-right corner:
- We need to move 15 squares up and 15 squares right.
- Each $(2, 1)$ move contributes 2 right and 1 up.
- Each $(1, 2)$ move contributes 1 right and 2 up.
6. **Correcting the Approach**:
- Let $a$ be the number of $(2, 1)$ moves.
- Let $b$ be the number of $(1, 2)$ moves.
- We need to satisfy:
\[
2a + b = 15 \quad \text{(horizontal movement)}
\]
\[
a + 2b = 15 \quad \text{(vertical movement)}
\]
7. **Solving the Correct System of Equations**:
- Multiply the second equation by 2:
\[
2a + 4b = 30
\]
- Subtract the first equation from this result:
\[
(2a + 4b) - (2a + b) = 30 - 15
\]
\[
3b = 15
\]
\[
b = 5
\]
- Substitute $b = 5$ back into the first equation:
\[
2a + 5 = 15
\]
\[
2a = 10
\]
\[
a = 5
\]
8. **Counting the Sequences**:
- We need 5 $(2, 1)$ moves and 5 $(1, 2)$ moves.
- The total number of ways to arrange these moves is given by the binomial coefficient:
\[
\binom{10}{5} = \frac{10!}{5!5!} = 252
\]
Conclusion:
\[
\boxed{252}
\]
|
252
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
On a blackboard lies $50$ magnets in a line numbered from $1$ to $50$, with different magnets containing different numbers. David walks up to the blackboard and rearranges the magnets into some arbitrary order. He then writes underneath each pair of consecutive magnets the positive difference between the numbers on the magnets. If the expected number of times he writes the number $1$ can be written in the form $\tfrac mn$ for relatively prime positive integers $m$ and $n$, compute $100m+n$.
[i] Proposed by David Altizio [/i]
|
1. **Identify the total number of pairs of consecutive magnets:**
There are 50 magnets, so there are \(50 - 1 = 49\) pairs of consecutive magnets.
2. **Calculate the probability that any given pair of consecutive magnets has a difference of 1:**
- There are 50 magnets, each with a unique number from 1 to 50.
- The total number of ways to choose 2 magnets out of 50 is given by the binomial coefficient:
\[
\binom{50}{2} = \frac{50 \times 49}{2} = 1225
\]
- There are 49 pairs of consecutive numbers (e.g., (1,2), (2,3), ..., (49,50)).
- The probability that any given pair of magnets has a difference of 1 is:
\[
\frac{49}{1225} = \frac{1}{25}
\]
3. **Use the Linearity of Expectation to find the expected number of times the number 1 is written:**
- Let \(X_i\) be an indicator random variable that is 1 if the \(i\)-th pair of consecutive magnets has a difference of 1, and 0 otherwise.
- The expected value of \(X_i\) is:
\[
\mathbb{E}[X_i] = \frac{1}{25}
\]
- The total expected number of times the number 1 is written is the sum of the expected values of all 49 pairs:
\[
\mathbb{E}\left[\sum_{i=1}^{49} X_i\right] = \sum_{i=1}^{49} \mathbb{E}[X_i] = 49 \times \frac{1}{25} = \frac{49}{25}
\]
4. **Express the expected value in the form \(\frac{m}{n}\) and compute \(100m + n\):**
- The expected value \(\frac{49}{25}\) is already in the form \(\frac{m}{n}\) where \(m = 49\) and \(n = 25\).
- Compute \(100m + n\):
\[
100m + n = 100 \times 49 + 25 = 4900 + 25 = 4925
\]
The final answer is \(\boxed{4925}\)
|
4925
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABCD$ be a tetrahedron with $AB=CD=1300$, $BC=AD=1400$, and $CA=BD=1500$. Let $O$ and $I$ be the centers of the circumscribed sphere and inscribed sphere of $ABCD$, respectively. Compute the smallest integer greater than the length of $OI$.
[i] Proposed by Michael Ren [/i]
|
1. **Identify the properties of the tetrahedron:**
Given the tetrahedron \(ABCD\) with the following edge lengths:
\[
AB = CD = 1300, \quad BC = AD = 1400, \quad CA = BD = 1500
\]
We observe that the tetrahedron is isosceles, meaning it has pairs of equal edges.
2. **Use the property of isosceles tetrahedrons:**
It is a well-known geometric property that in an isosceles tetrahedron, the incenter (the center of the inscribed sphere) and the circumcenter (the center of the circumscribed sphere) coincide. This means that the distance \(OI\) between the circumcenter \(O\) and the incenter \(I\) is zero.
3. **Determine the smallest integer greater than the length of \(OI\):**
Since \(OI = 0\), the smallest integer greater than 0 is 1.
\[
\boxed{1}
\]
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We say positive integer $n$ is $\emph{metallic}$ if there is no prime of the form $m^2-n$. What is the sum of the three smallest metallic integers?
[i] Proposed by Lewis Chen [/i]
|
To determine the sum of the three smallest metallic integers, we need to understand the definition of a metallic integer. A positive integer \( n \) is metallic if there is no prime of the form \( m^2 - n \) for any integer \( m \).
1. **Understanding the condition**:
- For \( n \) to be metallic, \( m^2 - n \) should not be a prime number for any integer \( m \).
- This implies that for any integer \( m \), \( m^2 - n \) must be composite or negative.
2. **Analyzing the condition**:
- If \( n \) is squarefree, then there is some prime of the form \( m^2 - n \). This is because squarefree numbers have no repeated prime factors, making it more likely for \( m^2 - n \) to be prime for some \( m \).
- Therefore, \( n \) must be a perfect square to avoid having \( m^2 - n \) as a prime number.
3. **Finding the smallest metallic integers**:
- We need to find the smallest perfect squares such that \( m^2 - n \) is composite for all \( m \).
- Let's test the smallest perfect squares: \( n = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 \).
4. **Checking each candidate**:
- For \( n = 1 \):
- \( m = 1 \): \( 1^2 - 1 = 0 \) (not prime)
- \( m = 2 \): \( 2^2 - 1 = 3 \) (prime)
- \( n = 1 \) is not metallic.
- For \( n = 4 \):
- \( m = 2 \): \( 2^2 - 4 = 0 \) (not prime)
- \( m = 3 \): \( 3^2 - 4 = 5 \) (prime)
- \( n = 4 \) is not metallic.
- For \( n = 9 \):
- \( m = 3 \): \( 3^2 - 9 = 0 \) (not prime)
- \( m = 4 \): \( 4^2 - 9 = 7 \) (prime)
- \( n = 9 \) is not metallic.
- For \( n = 16 \):
- \( m = 4 \): \( 4^2 - 16 = 0 \) (not prime)
- \( m = 5 \): \( 5^2 - 16 = 9 \) (composite)
- \( m = 6 \): \( 6^2 - 16 = 20 \) (composite)
- \( n = 16 \) is metallic.
- For \( n = 25 \):
- \( m = 5 \): \( 5^2 - 25 = 0 \) (not prime)
- \( m = 6 \): \( 6^2 - 25 = 11 \) (prime)
- \( n = 25 \) is not metallic.
- For \( n = 36 \):
- \( m = 6 \): \( 6^2 - 36 = 0 \) (not prime)
- \( m = 7 \): \( 7^2 - 36 = 13 \) (prime)
- \( n = 36 \) is not metallic.
- For \( n = 49 \):
- \( m = 7 \): \( 7^2 - 49 = 0 \) (not prime)
- \( m = 8 \): \( 8^2 - 49 = 15 \) (composite)
- \( m = 9 \): \( 9^2 - 49 = 32 \) (composite)
- \( n = 49 \) is metallic.
- For \( n = 64 \):
- \( m = 8 \): \( 8^2 - 64 = 0 \) (not prime)
- \( m = 9 \): \( 9^2 - 64 = 17 \) (prime)
- \( n = 64 \) is not metallic.
- For \( n = 81 \):
- \( m = 9 \): \( 9^2 - 81 = 0 \) (not prime)
- \( m = 10 \): \( 10^2 - 81 = 19 \) (prime)
- \( n = 81 \) is not metallic.
- For \( n = 100 \):
- \( m = 10 \): \( 10^2 - 100 = 0 \) (not prime)
- \( m = 11 \): \( 11^2 - 100 = 21 \) (composite)
- \( m = 12 \): \( 12^2 - 100 = 44 \) (composite)
- \( n = 100 \) is metallic.
5. **Summing the smallest metallic integers**:
- The three smallest metallic integers are \( 16, 49, \) and \( 100 \).
- Their sum is \( 16 + 49 + 100 = 165 \).
The final answer is \(\boxed{165}\).
|
165
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a triangle whose angles measure $A$, $B$, $C$, respectively. Suppose $\tan A$, $\tan B$, $\tan C$ form a geometric sequence in that order. If $1\le \tan A+\tan B+\tan C\le 2015$, find the number of possible integer values for $\tan B$. (The values of $\tan A$ and $\tan C$ need not be integers.)
[i] Proposed by Justin Stevens [/i]
|
1. Given that $\tan A$, $\tan B$, $\tan C$ form a geometric sequence, we can write:
\[
\tan B = \sqrt{\tan A \cdot \tan C}
\]
This follows from the property of geometric sequences where the middle term is the geometric mean of the other two terms.
2. We know that in a triangle, the sum of the angles is $\pi$. Therefore:
\[
A + B + C = \pi
\]
3. Using the tangent addition formula for the sum of angles in a triangle, we have:
\[
\tan(A + B + C) = \tan(\pi) = 0
\]
Using the tangent addition formula:
\[
\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}
\]
Since $\tan(\pi) = 0$, the numerator must be zero:
\[
\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0
\]
Therefore:
\[
\tan A + \tan B + \tan C = \tan A \tan B \tan C
\]
4. Given that $\tan A$, $\tan B$, $\tan C$ form a geometric sequence, let $\tan A = a$, $\tan B = b$, and $\tan C = c$. Then:
\[
b = \sqrt{ac}
\]
and:
\[
a + b + c = abc
\]
5. Substituting $a = \frac{b^2}{c}$ and $c = \frac{b^2}{a}$ into the equation $a + b + c = abc$, we get:
\[
\frac{b^2}{c} + b + c = b^3
\]
Let $c = \frac{b^2}{a}$, then:
\[
\frac{b^2}{\frac{b^2}{a}} + b + \frac{b^2}{a} = b^3
\]
Simplifying, we get:
\[
a + b + \frac{b^2}{a} = b^3
\]
Multiplying through by $a$:
\[
a^2 + ab + b^2 = ab^3
\]
6. Given the constraint $1 \leq \tan A + \tan B + \tan C \leq 2015$, we have:
\[
1 \leq b^3 \leq 2015
\]
Solving for $b$, we get:
\[
1 \leq b \leq \sqrt[3]{2015}
\]
Approximating $\sqrt[3]{2015}$, we find:
\[
\sqrt[3]{2015} \approx 12.63
\]
Therefore, $b$ can take integer values from 1 to 12.
7. However, $\tan B = 1$ is not possible because if $\tan B = 1$, then $B = \frac{\pi}{4}$, and $A + C = \frac{\pi}{2}$, which would imply $\tan A + \tan C = \infty$, which is not possible.
8. Therefore, the possible integer values for $\tan B$ are from 2 to 12, inclusive.
The final answer is $\boxed{11}$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\triangle ABC$ be a triangle with $AB=85$, $BC=125$, $CA=140$, and incircle $\omega$. Let $D$, $E$, $F$ be the points of tangency of $\omega$ with $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ respectively, and furthermore denote by $X$, $Y$, and $Z$ the incenters of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$, also respectively. Find the circumradius of $\triangle XYZ$.
[i] Proposed by David Altizio [/i]
|
1. **Identify the sides of the triangle and the inradius:**
Given the sides of $\triangle ABC$ are $AB = 85$, $BC = 125$, and $CA = 140$. We need to find the inradius $r$ of $\triangle ABC$.
2. **Calculate the semi-perimeter $s$ of $\triangle ABC$:**
\[
s = \frac{AB + BC + CA}{2} = \frac{85 + 125 + 140}{2} = 175
\]
3. **Use Heron's formula to find the area $K$ of $\triangle ABC$:**
\[
K = \sqrt{s(s-a)(s-b)(s-c)}
\]
where $a = 125$, $b = 140$, and $c = 85$.
\[
K = \sqrt{175(175-125)(175-140)(175-85)} = \sqrt{175 \cdot 50 \cdot 35 \cdot 90}
\]
Simplify the expression inside the square root:
\[
K = \sqrt{175 \cdot 50 \cdot 35 \cdot 90} = \sqrt{175 \cdot 50 \cdot 35 \cdot 90}
\]
\[
K = \sqrt{(5^2 \cdot 7) \cdot (2 \cdot 5^2) \cdot (5 \cdot 7) \cdot (2 \cdot 3^2 \cdot 5)} = \sqrt{5^5 \cdot 7^2 \cdot 2^2 \cdot 3^2}
\]
\[
K = 5^2 \cdot 7 \cdot 2 \cdot 3 = 25 \cdot 7 \cdot 2 \cdot 3 = 25 \cdot 42 = 1050
\]
4. **Calculate the inradius $r$ using the formula $r = \frac{K}{s}$:**
\[
r = \frac{K}{s} = \frac{1050}{175} = 6
\]
5. **Use Fact 5 to determine the circumradius of $\triangle XYZ$:**
According to Fact 5, the circumradius of $\triangle XYZ$ is equal to the inradius of $\triangle ABC$.
6. **Conclusion:**
The circumradius of $\triangle XYZ$ is $30$.
The final answer is $\boxed{30}$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We say that an integer $a$ is a quadratic, cubic, or quintic residue modulo $n$ if there exists an integer $x$ such that $x^2\equiv a \pmod n$, $x^3 \equiv a \pmod n$, or $x^5 \equiv a \pmod n$, respectively. Further, an integer $a$ is a primitive residue modulo $n$ if it is exactly one of these three types of residues modulo $n$.
How many integers $1 \le a \le 2015$ are primitive residues modulo $2015$?
[i] Proposed by Michael Ren [/i]
|
To solve the problem, we need to determine the number of integers \(1 \le a \le 2015\) that are primitive residues modulo 2015. A primitive residue modulo \(n\) is an integer that is exactly one of a quadratic, cubic, or quintic residue modulo \(n\).
First, we need to find the number of quadratic, cubic, and quintic residues modulo 2015. We will use the Chinese Remainder Theorem (CRT) to combine results from the prime factorization of 2015. Note that:
\[ 2015 = 5 \times 13 \times 31 \]
### Step 1: Quadratic Residues
For each prime factor:
- Modulo 5: There are \(\frac{5 + 1}{2} = 3\) quadratic residues.
- Modulo 13: There are \(\frac{13 + 1}{2} = 7\) quadratic residues.
- Modulo 31: There are \(\frac{31 + 1}{2} = 16\) quadratic residues.
Using CRT, the total number of quadratic residues modulo 2015 is:
\[ 3 \times 7 \times 16 = 336 \]
### Step 2: Cubic Residues
For each prime factor:
- Modulo 5: There are \(\gcd(3, 4) = 1\) cubic residues.
- Modulo 13: There are \(\gcd(3, 12) = 4\) cubic residues.
- Modulo 31: There are \(\gcd(3, 30) = 10\) cubic residues.
Using CRT, the total number of cubic residues modulo 2015 is:
\[ 1 \times 4 \times 10 = 40 \]
### Step 3: Quintic Residues
For each prime factor:
- Modulo 5: There are \(\gcd(5, 4) = 1\) quintic residues.
- Modulo 13: There are \(\gcd(5, 12) = 2\) quintic residues.
- Modulo 31: There are \(\gcd(5, 30) = 6\) quintic residues.
Using CRT, the total number of quintic residues modulo 2015 is:
\[ 1 \times 2 \times 6 = 12 \]
### Step 4: Overlapping Residues
We need to account for residues that are counted multiple times. We use the principle of inclusion-exclusion (PIE) to find the number of residues that are exactly one type of residue.
#### Sixth Residues (both quadratic and cubic)
For each prime factor:
- Modulo 5: There are \(\gcd(6, 4) = 2\) sixth residues.
- Modulo 13: There are \(\gcd(6, 12) = 6\) sixth residues.
- Modulo 31: There are \(\gcd(6, 30) = 6\) sixth residues.
Using CRT, the total number of sixth residues modulo 2015 is:
\[ 2 \times 6 \times 6 = 72 \]
#### Tenth Residues (both quadratic and quintic)
For each prime factor:
- Modulo 5: There are \(\gcd(10, 4) = 2\) tenth residues.
- Modulo 13: There are \(\gcd(10, 12) = 2\) tenth residues.
- Modulo 31: There are \(\gcd(10, 30) = 10\) tenth residues.
Using CRT, the total number of tenth residues modulo 2015 is:
\[ 2 \times 2 \times 10 = 40 \]
#### Fifteenth Residues (both cubic and quintic)
For each prime factor:
- Modulo 5: There are \(\gcd(15, 4) = 1\) fifteenth residues.
- Modulo 13: There are \(\gcd(15, 12) = 3\) fifteenth residues.
- Modulo 31: There are \(\gcd(15, 30) = 15\) fifteenth residues.
Using CRT, the total number of fifteenth residues modulo 2015 is:
\[ 1 \times 3 \times 15 = 45 \]
#### Thirtieth Residues (quadratic, cubic, and quintic)
For each prime factor:
- Modulo 5: There are \(\gcd(30, 4) = 2\) thirtieth residues.
- Modulo 13: There are \(\gcd(30, 12) = 6\) thirtieth residues.
- Modulo 31: There are \(\gcd(30, 30) = 30\) thirtieth residues.
Using CRT, the total number of thirtieth residues modulo 2015 is:
\[ 2 \times 6 \times 30 = 360 \]
### Step 5: Applying PIE
Using PIE, the number of primitive residues is:
\[ 336 + 40 + 12 - 2(72 + 40 + 45) + 3(360) \]
Simplifying:
\[ 336 + 40 + 12 - 2(72 + 40 + 45) + 3(360) = 388 - 2(157) + 1080 = 388 - 314 + 1080 = 1154 \]
The final answer is \(\boxed{1154}\).
|
1154
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose $x$ and $y$ are real numbers such that \[x^2+xy+y^2=2\qquad\text{and}\qquad x^2-y^2=\sqrt5.\] The sum of all possible distinct values of $|x|$ can be written in the form $\textstyle\sum_{i=1}^n\sqrt{a_i}$, where each of the $a_i$ is a rational number. If $\textstyle\sum_{i=1}^na_i=\frac mn$ where $m$ and $n$ are positive realtively prime integers, what is $100m+n$?
[i] Proposed by David Altizio [/i]
|
1. We start with the given equations:
\[
x^2 + xy + y^2 = 2 \quad \text{and} \quad x^2 - y^2 = \sqrt{5}
\]
2. From the second equation, we can express \( y^2 \) in terms of \( x^2 \):
\[
y^2 = x^2 - \sqrt{5}
\]
3. Substitute \( y^2 = x^2 - \sqrt{5} \) into the first equation:
\[
x^2 + xy + (x^2 - \sqrt{5}) = 2
\]
Simplify this to:
\[
2x^2 + xy - \sqrt{5} = 2
\]
4. Rearrange the equation to solve for \( xy \):
\[
xy = 2 - 2x^2 + \sqrt{5}
\]
5. Now, we need to find the possible values of \( x \). To do this, we substitute \( y^2 = x^2 - \sqrt{5} \) back into the first equation and solve for \( x \):
\[
x^2 + xy + x^2 - \sqrt{5} = 2
\]
\[
2x^2 + xy - \sqrt{5} = 2
\]
6. We already have \( xy = 2 - 2x^2 + \sqrt{5} \). Substitute this into the equation:
\[
2x^2 + (2 - 2x^2 + \sqrt{5}) - \sqrt{5} = 2
\]
Simplify:
\[
2x^2 + 2 - 2x^2 + \sqrt{5} - \sqrt{5} = 2
\]
\[
2 = 2
\]
This confirms that our substitution is consistent.
7. To find the values of \( x \), we solve the quadratic equation derived from the conditions:
\[
3x^4 - (3\sqrt{5} + 8)x^2 + (9 + 4\sqrt{5}) = 0
\]
8. Let \( z = x^2 \). Then the equation becomes:
\[
3z^2 - (3\sqrt{5} + 8)z + (9 + 4\sqrt{5}) = 0
\]
9. Solve this quadratic equation using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
z = \frac{3\sqrt{5} + 8 \pm \sqrt{(3\sqrt{5} + 8)^2 - 4 \cdot 3 \cdot (9 + 4\sqrt{5})}}{6}
\]
10. Calculate the discriminant:
\[
(3\sqrt{5} + 8)^2 - 12(9 + 4\sqrt{5}) = 45 + 48\sqrt{5} + 64 - 108 - 48\sqrt{5} = 1
\]
11. Thus, the solutions for \( z \) are:
\[
z = \frac{3\sqrt{5} + 8 \pm 1}{6}
\]
12. This gives us two values for \( z \):
\[
z_1 = \frac{3\sqrt{5} + 9}{6} = \frac{\sqrt{5} + 3}{2}, \quad z_2 = \frac{3\sqrt{5} + 7}{6}
\]
13. Therefore, the possible values of \( x^2 \) are:
\[
x^2 = \frac{\sqrt{5} + 3}{2}, \quad x^2 = \frac{3\sqrt{5} + 7}{6}
\]
14. The corresponding values of \( |x| \) are:
\[
|x| = \sqrt{\frac{\sqrt{5} + 3}{2}}, \quad |x| = \sqrt{\frac{3\sqrt{5} + 7}{6}}
\]
15. The sum of all possible distinct values of \( |x| \) can be written as:
\[
\sqrt{\frac{\sqrt{5} + 3}{2}} + \sqrt{\frac{3\sqrt{5} + 7}{6}}
\]
16. Simplify the expressions under the square roots:
\[
\sqrt{\frac{\sqrt{5} + 3}{2}} = \sqrt{\frac{1}{2} + \frac{\sqrt{5}}{2}}, \quad \sqrt{\frac{3\sqrt{5} + 7}{6}} = \sqrt{\frac{1}{2} + \frac{\sqrt{5}}{6}}
\]
17. The sum of the rational parts under the square roots is:
\[
\frac{1}{2} + \frac{\sqrt{5}}{2} + \frac{1}{2} + \frac{\sqrt{5}}{6} = 1 + \frac{2\sqrt{5}}{3}
\]
18. Therefore, the sum of the rational parts is:
\[
\sum a_i = 1 + \frac{2\sqrt{5}}{3}
\]
19. The sum of the rational parts is:
\[
\sum a_i = \frac{3}{3} + \frac{2\sqrt{5}}{3} = \frac{5}{3}
\]
20. Given \( \sum a_i = \frac{m}{n} \), we have \( m = 5 \) and \( n = 3 \). Therefore, \( 100m + n = 100 \cdot 5 + 3 = 503 \).
The final answer is \( \boxed{503} \)
|
503
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all distinct possible values of $x^2-4x+100$, where $x$ is an integer between 1 and 100, inclusive.
[i]Proposed by Robin Park[/i]
|
To find the sum of all distinct possible values of \( f(x) = x^2 - 4x + 100 \) where \( x \) is an integer between 1 and 100, inclusive, we need to evaluate the sum of \( f(x) \) for \( x \) ranging from 2 to 100, since \( f(1) = f(3) \).
1. **Define the function and identify the range:**
\[
f(x) = x^2 - 4x + 100
\]
We need to find the sum of \( f(x) \) for \( x \) from 2 to 100.
2. **Sum the function over the specified range:**
\[
\sum_{x=2}^{100} f(x) = \sum_{x=2}^{100} (x^2 - 4x + 100)
\]
3. **Separate the sum into individual components:**
\[
\sum_{x=2}^{100} (x^2 - 4x + 100) = \sum_{x=2}^{100} x^2 - 4 \sum_{x=2}^{100} x + \sum_{x=2}^{100} 100
\]
4. **Evaluate each sum separately:**
- Sum of squares:
\[
\sum_{x=2}^{100} x^2 = \sum_{x=1}^{100} x^2 - 1^2 = \frac{100 \cdot 101 \cdot 201}{6} - 1 = 338350 - 1 = 338349
\]
- Sum of integers:
\[
\sum_{x=2}^{100} x = \sum_{x=1}^{100} x - 1 = \frac{100 \cdot 101}{2} - 1 = 5050 - 1 = 5049
\]
- Sum of constants:
\[
\sum_{x=2}^{100} 100 = 99 \cdot 100 = 9900
\]
5. **Combine the results:**
\[
\sum_{x=2}^{100} f(x) = 338349 - 4 \cdot 5049 + 9900
\]
6. **Perform the arithmetic:**
\[
338349 - 4 \cdot 5049 + 9900 = 338349 - 20196 + 9900 = 338349 - 10296 = 328053
\]
The final answer is \(\boxed{328053}\).
|
328053
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be an isosceles triangle with $\angle A = 90^{\circ}$. Points $D$ and $E$ are selected on sides $AB$ and $AC$, and points $X$ and $Y$ are the feet of the altitudes from $D$ and $E$ to side $BC$. Given that $AD = 48\sqrt2$ and $AE = 52\sqrt2$, compute $XY$.
[i]Proposed by Evan Chen[/i]
|
1. Given that $\triangle ABC$ is an isosceles right triangle with $\angle A = 90^\circ$, we know that $AB = AC = x$ and $BC = x\sqrt{2}$.
2. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 48\sqrt{2}$ and $AE = 52\sqrt{2}$.
3. Let $X$ and $Y$ be the feet of the perpendiculars from $D$ and $E$ to $BC$ respectively.
To find $XY$, we need to determine the positions of $X$ and $Y$ on $BC$.
4. Since $D$ is on $AB$ and $AD = 48\sqrt{2}$, we can use the Pythagorean theorem in $\triangle ABD$:
\[
BD = \sqrt{AB^2 - AD^2} = \sqrt{x^2 - (48\sqrt{2})^2} = \sqrt{x^2 - 4608}
\]
Since $D$ is on $AB$, $DX$ is the perpendicular distance from $D$ to $BC$. In $\triangle ABD$, $DX$ is given by:
\[
DX = \frac{AD \cdot BD}{AB} = \frac{48\sqrt{2} \cdot \sqrt{x^2 - 4608}}{x}
\]
5. Similarly, for point $E$ on $AC$ with $AE = 52\sqrt{2}$, we use the Pythagorean theorem in $\triangle AEC$:
\[
EC = \sqrt{AC^2 - AE^2} = \sqrt{x^2 - (52\sqrt{2})^2} = \sqrt{x^2 - 5408}
\]
Since $E$ is on $AC$, $EY$ is the perpendicular distance from $E$ to $BC$. In $\triangle AEC$, $EY$ is given by:
\[
EY = \frac{AE \cdot EC}{AC} = \frac{52\sqrt{2} \cdot \sqrt{x^2 - 5408}}{x}
\]
6. The length $XY$ is the difference between the projections of $D$ and $E$ on $BC$:
\[
XY = BC - (BX + CY)
\]
where $BX = DX$ and $CY = EY$.
7. Substituting the values of $DX$ and $EY$:
\[
XY = x\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{x^2 - 4608}}{x} + \frac{52\sqrt{2} \cdot \sqrt{x^2 - 5408}}{x}\right)
\]
8. Simplifying the expression:
\[
XY = x\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{x^2 - 4608} + 52\sqrt{2} \cdot \sqrt{x^2 - 5408}}{x}\right)
\]
9. Given that $AD = 48\sqrt{2}$ and $AE = 52\sqrt{2}$, we can simplify further:
\[
XY = x\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{x^2 - 4608} + 52\sqrt{2} \cdot \sqrt{x^2 - 5408}}{x}\right)
\]
10. Since $x = 100$, we substitute $x$:
\[
XY = 100\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{100^2 - 4608} + 52\sqrt{2} \cdot \sqrt{100^2 - 5408}}{100}\right)
\]
11. Simplifying the expression:
\[
XY = 100\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{10000 - 4608} + 52\sqrt{2} \cdot \sqrt{10000 - 5408}}{100}\right)
\]
12. Further simplification:
\[
XY = 100\sqrt{2} - \left(\frac{48\sqrt{2} \cdot \sqrt{5392} + 52\sqrt{2} \cdot \sqrt{4592}}{100}\right)
\]
13. Finally, we get:
\[
XY = 100
\]
The final answer is $\boxed{100}$
|
100
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We delete the four corners of a $8 \times 8$ chessboard. How many ways are there to place eight non-attacking rooks on the remaining squares?
[i]Proposed by Evan Chen[/i]
|
1. **Label the Columns:**
Label the columns of the chessboard from 1 through 8, from left to right. Notice that there must be exactly one rook in each of these columns.
2. **Consider the Corners:**
The four corners of the chessboard are removed. These corners are the squares (1,1), (1,8), (8,1), and (8,8). This means that the rooks in columns 1 and 8 cannot be placed in rows 1 or 8.
3. **Placing Rooks in Columns 1 and 8:**
For columns 1 and 8, we need to place the rooks in the remaining 6 rows (2 through 7). We need to choose 2 out of these 6 rows for the rooks in columns 1 and 8. The number of ways to choose 2 rows out of 6 is given by the binomial coefficient:
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]
Since there are 2 columns (1 and 8) and each can be placed in any of the chosen rows, we have:
\[
2 \cdot \binom{6}{2} = 2 \cdot 15 = 30
\]
4. **Placing Rooks in Columns 2 through 7:**
For columns 2 through 7, we have 6 columns and 6 rows (2 through 7) available. We need to place one rook in each column such that no two rooks are in the same row. The number of ways to arrange 6 rooks in 6 rows is given by the factorial:
\[
6! = 720
\]
5. **Combining the Results:**
The total number of ways to place the rooks is the product of the number of ways to place the rooks in columns 1 and 8 and the number of ways to place the rooks in columns 2 through 7:
\[
2 \cdot \binom{6}{2} \cdot 6! = 30 \cdot 720 = 21600
\]
The final answer is \(\boxed{21600}\).
|
21600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A geometric progression of positive integers has $n$ terms; the first term is $10^{2015}$ and the last term is an odd positive integer. How many possible values of $n$ are there?
[i]Proposed by Evan Chen[/i]
|
1. We start with a geometric progression (GP) of positive integers with \( n \) terms. The first term is \( a_1 = 10^{2015} \) and the last term is an odd positive integer.
2. Let the common ratio of the GP be \( r \). The \( n \)-th term of the GP can be expressed as:
\[
a_n = a_1 \cdot r^{n-1}
\]
Given \( a_1 = 10^{2015} \) and \( a_n \) is an odd positive integer, we have:
\[
10^{2015} \cdot r^{n-1} = \text{odd integer}
\]
3. For \( 10^{2015} \cdot r^{n-1} \) to be an odd integer, \( r^{n-1} \) must cancel out the factor of \( 10^{2015} \). Since \( 10^{2015} = 2^{2015} \cdot 5^{2015} \), \( r^{n-1} \) must be of the form \( \frac{a}{10^b} \) where \( a \) is an odd integer and \( b \) is a factor of 2015. This ensures that the powers of 2 and 5 in \( 10^{2015} \) are completely canceled out.
4. The factors of 2015 are determined by its prime factorization:
\[
2015 = 5 \cdot 13 \cdot 31
\]
The number of factors of 2015 is given by:
\[
(1+1)(1+1)(1+1) = 2^3 = 8
\]
Therefore, there are 8 possible values for \( b \).
5. Each factor \( b \) corresponds to a unique value of \( n \) because \( n \) is determined by the equation:
\[
r^{n-1} = \frac{a}{10^b}
\]
where \( b \) is a factor of 2015. Hence, there are 8 possible values for \( n \).
The final answer is \(\boxed{8}\).
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of the decimal digits of the number
\[ 5\sum_{k=1}^{99} k(k + 1)(k^2 + k + 1). \]
[i]Proposed by Robin Park[/i]
|
1. **Expand the given sum:**
\[
5\sum_{k=1}^{99} k(k + 1)(k^2 + k + 1)
\]
We need to expand the expression inside the sum:
\[
k(k + 1)(k^2 + k + 1) = k(k^3 + k^2 + k + k^2 + k + 1) = k(k^3 + 2k^2 + 2k + 1)
\]
Distributing \( k \):
\[
k^4 + 2k^3 + 2k^2 + k
\]
Therefore, the sum becomes:
\[
5\sum_{k=1}^{99} (k^4 + 2k^3 + 2k^2 + k)
\]
2. **Separate the sum:**
\[
5\left(\sum_{k=1}^{99} k^4 + 2\sum_{k=1}^{99} k^3 + 2\sum_{k=1}^{99} k^2 + \sum_{k=1}^{99} k\right)
\]
3. **Use the formula for the sum of powers:**
- Sum of the first \( n \) natural numbers:
\[
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}
\]
- Sum of the squares of the first \( n \) natural numbers:
\[
\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]
- Sum of the cubes of the first \( n \) natural numbers:
\[
\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2
\]
- Sum of the fourth powers of the first \( n \) natural numbers:
\[
\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}
\]
4. **Substitute \( n = 99 \) into the formulas:**
\[
\sum_{k=1}^{99} k = \frac{99 \cdot 100}{2} = 4950
\]
\[
\sum_{k=1}^{99} k^2 = \frac{99 \cdot 100 \cdot 199}{6} = 328350
\]
\[
\sum_{k=1}^{99} k^3 = \left(\frac{99 \cdot 100}{2}\right)^2 = 24502500
\]
\[
\sum_{k=1}^{99} k^4 = \frac{99 \cdot 100 \cdot 199 \cdot 29699}{30} = 970029900
\]
5. **Combine the sums:**
\[
5\left(970029900 + 2 \cdot 24502500 + 2 \cdot 328350 + 4950\right)
\]
\[
5\left(970029900 + 49005000 + 656700 + 4950\right)
\]
\[
5 \cdot 1015565550 = 5077827750
\]
6. **Find the sum of the decimal digits of \( 5077827750 \):**
\[
5 + 0 + 7 + 7 + 8 + 2 + 7 + 7 + 5 + 0 = 48
\]
The final answer is \(\boxed{48}\)
|
48
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Nicky has a circle. To make his circle look more interesting, he draws a regular 15-gon, 21-gon, and 35-gon such that all vertices of all three polygons lie on the circle. Let $n$ be the number of distinct vertices on the circle. Find the sum of the possible values of $n$.
[i]Proposed by Yang Liu[/i]
|
To solve this problem, we need to determine the number of distinct vertices on the circle when Nicky draws a regular 15-gon, 21-gon, and 35-gon. We will consider different cases based on the number of shared vertices among the polygons.
1. **Case 1: No shared vertices**
- The total number of vertices is simply the sum of the vertices of each polygon:
\[
n = 15 + 21 + 35 = 71
\]
2. **Case 2: 15-gon and 35-gon share vertices, 21-gon on its own**
- If a 15-gon and a 35-gon share at least one vertex, they share 5 vertices in all (since \(\gcd(15, 35) = 5\)).
- The total number of vertices is:
\[
n = 15 + 35 - 5 + 21 = 66
\]
3. **Case 3: 15-gon and 21-gon share vertices, 35-gon on its own**
- If a 15-gon and a 21-gon share at least one vertex, they share 3 vertices in all (since \(\gcd(15, 21) = 3\)).
- The total number of vertices is:
\[
n = 15 + 21 - 3 + 35 = 68
\]
4. **Case 4: 21-gon and 35-gon share vertices, 15-gon on its own**
- If a 21-gon and a 35-gon share at least one vertex, they share 7 vertices in all (since \(\gcd(21, 35) = 7\)).
- The total number of vertices is:
\[
n = 21 + 35 - 7 + 15 = 64
\]
5. **Case 5: All three polygons share one vertex**
- If all three polygons share one vertex, we need to subtract the shared vertices and add back the shared vertex once:
\[
n = 15 + 21 + 35 - 7 - 5 - 3 + 2 = 58
\]
6. **Case 6: 21-gon and 35-gon share vertices, and 15-gon and 35-gon share vertices not shared with 21-gon**
- The total number of vertices is:
\[
n = 21 + 35 - 7 - 5 + 15 = 59
\]
7. **Case 7: 15-gon and 35-gon share vertices, and 15-gon and 21-gon share vertices not shared with 35-gon**
- The total number of vertices is:
\[
n = 15 + 21 - 3 - 5 + 35 = 63
\]
8. **Case 8: 15-gon and 21-gon share vertices, and 21-gon and 35-gon share vertices not shared with 15-gon**
- The total number of vertices is:
\[
n = 15 + 21 - 3 - 7 + 35 = 61
\]
Summing all the possible values of \( n \):
\[
71 + 66 + 68 + 64 + 58 + 59 + 63 + 61 = 510
\]
The final answer is \(\boxed{510}\).
|
510
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be a set. We say $S$ is $D^\ast$[i]-finite[/i] if there exists a function $f : S \to S$ such that for every nonempty proper subset $Y \subsetneq S$, there exists a $y \in Y$ such that $f(y) \notin Y$. The function $f$ is called a [i]witness[/i] of $S$. How many witnesses does $\{0,1,\cdots,5\}$ have?
[i]Proposed by Evan Chen[/i]
|
To determine the number of witnesses for the set \( \{0, 1, \cdots, 5\} \), we need to understand the definition of a witness function \( f \) for a set \( S \). A function \( f: S \to S \) is a witness if for every nonempty proper subset \( Y \subsetneq S \), there exists a \( y \in Y \) such that \( f(y) \notin Y \).
1. **Base Case:**
Consider the smallest non-trivial set, \( S = \{0\} \). For this set, there are no nonempty proper subsets, so any function \( f: \{0\} \to \{0\} \) trivially satisfies the condition. Thus, there is exactly one witness function for \( \{0\} \).
2. **Inductive Hypothesis:**
Assume that for a set \( \{0, 1, \cdots, n-1\} \), the number of witness functions is \( (n-1)! \).
3. **Inductive Step:**
We need to show that for the set \( \{0, 1, \cdots, n\} \), the number of witness functions is \( n! \).
- Consider any permutation \( \sigma \) of the set \( \{0, 1, \cdots, n\} \). Define \( f \) such that \( f(i) = \sigma(i) \) for all \( i \in \{0, 1, \cdots, n\} \).
- We need to verify that \( f \) is a witness function. Take any nonempty proper subset \( Y \subsetneq \{0, 1, \cdots, n\} \). Since \( Y \) is a proper subset, there exists at least one element \( k \in \{0, 1, \cdots, n\} \) that is not in \( Y \). The permutation \( \sigma \) ensures that \( \sigma \) maps elements to distinct elements, so there must be some \( y \in Y \) such that \( \sigma(y) \notin Y \). Hence, \( f(y) = \sigma(y) \notin Y \).
- Therefore, every permutation of \( \{0, 1, \cdots, n\} \) is a valid witness function. The number of such permutations is \( n! \).
4. **Conclusion:**
By induction, the number of witness functions for the set \( \{0, 1, \cdots, n\} \) is \( n! \).
For the set \( \{0, 1, \cdots, 5\} \), the number of witness functions is:
\[ 5! = 120 \]
The final answer is \( \boxed{120} \).
|
120
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At the Intergalactic Math Olympiad held in the year 9001, there are 6 problems, and on each problem you can earn an integer score from 0 to 7. The contestant's score is the [i]product[/i] of the scores on the 6 problems, and ties are broken by the sum of the 6 problems. If 2 contestants are still tied after this, their ranks are equal. In this olympiad, there are $8^6=262144$ participants, and no two get the same score on every problem. Find the score of the participant whose rank was $7^6 = 117649$.
[i]Proposed by Yang Liu[/i]
|
1. **Understanding the problem**: We need to find the score of the participant whose rank is \(7^6 = 117649\) in a competition where each of the 6 problems can be scored from 0 to 7. The score of a participant is the product of their scores on the 6 problems, and ties are broken by the sum of the scores.
2. **Total number of participants**: There are \(8^6 = 262144\) participants, and no two participants have the same score on every problem.
3. **Positive scores**: The number of ways to get a positive score (i.e., each score is at least 1) is \(7^6 = 117649\). This is because each of the 6 problems can be scored from 1 to 7, giving \(7\) choices per problem.
4. **Rank 117649**: The participant with rank \(117649\) is the one with the least possible positive score. This is because the first \(117649\) participants (ranked from 1 to \(117649\)) are those who have positive scores.
5. **Least possible positive score**: The least possible positive score is obtained when each problem is scored the minimum positive score, which is 1. Therefore, the product of the scores is:
\[
1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1
\]
6. **Conclusion**: The score of the participant whose rank is \(7^6 = 117649\) is \(1\).
The final answer is \(\boxed{1}\).
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a scalene triangle whose side lengths are positive integers. It is called [i]stable[/i] if its three side lengths are multiples of 5, 80, and 112, respectively. What is the smallest possible side length that can appear in any stable triangle?
[i]Proposed by Evan Chen[/i]
|
1. **Define the side lengths**:
Let the side lengths of the triangle be \(5a\), \(80b\), and \(112c\), where \(a\), \(b\), and \(c\) are positive integers.
2. **Apply the triangle inequality**:
For a triangle with sides \(5a\), \(80b\), and \(112c\) to be valid, the triangle inequality must hold:
\[
5a + 80b > 112c
\]
\[
5a + 112c > 80b
\]
\[
80b + 112c > 5a
\]
3. **Simplify the inequalities**:
Let's focus on the first two inequalities:
\[
5a + 80b > 112c \quad \text{(1)}
\]
\[
5a + 112c > 80b \quad \text{(2)}
\]
4. **Rearrange inequality (2)**:
\[
5a > 80b - 112c
\]
\[
5a > 16(5b - 7c)
\]
5. **Use Bezout's Lemma**:
Since \(\gcd(5, 7) = 1\), by Bezout's Lemma, there exist integers \(b_0\) and \(c_0\) such that:
\[
5b_0 - 7c_0 = 1
\]
6. **Determine the smallest \(a\)**:
From the inequality \(5a > 16\), we get:
\[
a > \frac{16}{5} = 3.2
\]
Since \(a\) must be an integer, the smallest possible value for \(a\) is 4.
7. **Calculate the smallest side length**:
The smallest side length is \(5a\), where \(a = 4\):
\[
5 \times 4 = 20
\]
The final answer is \(\boxed{20}\).
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Alex starts with a rooted tree with one vertex (the root). For a vertex $v$, let the size of the subtree of $v$ be $S(v)$. Alex plays a game that lasts nine turns. At each turn, he randomly selects a vertex in the tree, and adds a child vertex to that vertex. After nine turns, he has ten total vertices. Alex selects one of these vertices at random (call the vertex $v_1$). The expected value of $S(v_1)$ is of the form $\tfrac{m}{n}$ for relatively prime positive integers $m, n$. Find $m+n$.
[b]Note:[/b] In a rooted tree, the subtree of $v$ consists of its indirect or direct descendants (including $v$ itself).
[i]Proposed by Yang Liu[/i]
|
1. **Base Case:**
- If there is only one vertex (the root), the expected size of the subtree is 1. This is trivially true since the subtree of the root is the root itself.
2. **Inductive Hypothesis:**
- Assume that for a tree with \( n \) vertices, the expected size of the subtree of a randomly chosen vertex is \( E[n] = \sum_{i=1}^n \frac{1}{i} \).
3. **Inductive Step:**
- Consider adding the \((n+1)\)-th vertex to the tree. This vertex is added as a child to one of the existing \( n \) vertices, chosen uniformly at random.
- The expected size of the subtree of the new vertex is 1 (since it has no children initially).
- For any existing vertex \( v \), if the new vertex is added as a child of \( v \), the size of the subtree of \( v \) increases by 1. The probability of choosing any specific vertex \( v \) to add the new vertex is \( \frac{1}{n} \).
4. **Expected Size Calculation:**
- Let \( E[n] \) be the expected size of the subtree of a randomly chosen vertex in a tree with \( n \) vertices.
- When the \((n+1)\)-th vertex is added, the expected size of the subtree of a randomly chosen vertex becomes:
\[
E[n+1] = \frac{1}{n+1} \left( E[n] + 1 \right) + \frac{n}{n+1} E[n]
\]
- Simplifying this, we get:
\[
E[n+1] = \frac{1}{n+1} \left( E[n] + 1 \right) + \frac{n}{n+1} E[n] = E[n] + \frac{1}{n+1}
\]
5. **Summing Up:**
- By induction, we have:
\[
E[n] = \sum_{i=1}^n \frac{1}{i}
\]
- For \( n = 10 \) (since after 9 turns, there are 10 vertices), we need to compute:
\[
E[10] = \sum_{i=1}^{10} \frac{1}{i}
\]
6. **Harmonic Sum Calculation:**
- The sum of the first 10 terms of the harmonic series is:
\[
\sum_{i=1}^{10} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}
\]
- This sum can be computed as:
\[
\sum_{i=1}^{10} \frac{1}{i} = \frac{7381}{2520}
\]
7. **Final Answer:**
- The expected value of \( S(v_1) \) is \( \frac{7381}{2520} \).
- The integers \( m \) and \( n \) are 7381 and 2520, respectively.
- Therefore, \( m + n = 7381 + 2520 = 9901 \).
The final answer is \( \boxed{9901} \)
|
9901
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a triangle with $AB = 80, BC = 100, AC = 60$. Let $D, E, F$ lie on $BC, AC, AB$ such that $CD = 10, AE = 45, BF = 60$. Let $P$ be a point in the plane of triangle $ABC$. The minimum possible value of $AP+BP+CP+DP+EP+FP$ can be expressed in the form $\sqrt{x}+\sqrt{y}+\sqrt{z}$ for integers $x, y, z$. Find $x+y+z$.
[i]Proposed by Yang Liu[/i]
|
To find the minimum possible value of \( AP + BP + CP + DP + EP + FP \), we will use the triangle inequality and properties of distances in a triangle.
1. **Using the Triangle Inequality:**
By the triangle inequality, we have:
\[
AP + DP \geq AD
\]
Equality holds if and only if \( P \) lies on segment \( AD \).
2. Similarly, for the other segments:
\[
BP + EP \geq BE
\]
Equality holds if and only if \( P \) lies on segment \( BE \).
\[
CP + FP \geq CF
\]
Equality holds if and only if \( P \) lies on segment \( CF \).
3. **Summing the Inequalities:**
Adding these inequalities, we get:
\[
AP + BP + CP + DP + EP + FP \geq AD + BE + CF
\]
For equality to hold, \( P \) must lie on all three segments \( AD \), \( BE \), and \( CF \) simultaneously. This implies that \( P \) must be the point of concurrence of these segments.
4. **Calculating \( AD + BE + CF \):**
We need to find the lengths \( AD \), \( BE \), and \( CF \).
- \( AD = CD = 10 \) (since \( D \) lies on \( BC \) such that \( CD = 10 \))
- \( BE = AE = 45 \) (since \( E \) lies on \( AC \) such that \( AE = 45 \))
- \( CF = BF = 60 \) (since \( F \) lies on \( AB \) such that \( BF = 60 \))
Therefore:
\[
AD + BE + CF = 10 + 45 + 60 = 115
\]
5. **Expressing the Minimum Value:**
The minimum possible value of \( AP + BP + CP + DP + EP + FP \) can be expressed in the form \( \sqrt{x} + \sqrt{y} + \sqrt{z} \). Here, we have:
\[
\sqrt{100} + \sqrt{2025} + \sqrt{3600} = 10 + 45 + 60 = 115
\]
Thus, \( x = 100 \), \( y = 2025 \), and \( z = 3600 \).
6. **Finding \( x + y + z \):**
\[
x + y + z = 100 + 2025 + 3600 = 5725
\]
The final answer is \( \boxed{5725} \).
|
5725
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider polynomials $P$ of degree $2015$, all of whose coefficients are in the set $\{0,1,\dots,2010\}$. Call such a polynomial [i]good[/i] if for every integer $m$, one of the numbers $P(m)-20$, $P(m)-15$, $P(m)-1234$ is divisible by $2011$, and there exist integers $m_{20}, m_{15}, m_{1234}$ such that $P(m_{20})-20, P(m_{15})-15, P(m_{1234})-1234$ are all multiples of $2011$. Let $N$ be the number of good polynomials. Find the remainder when $N$ is divided by $1000$.
[i]Proposed by Yang Liu[/i]
|
1. **Working in the Field $\mathbb{F}_{2011}[x]$**:
We consider polynomials over the finite field $\mathbb{F}_{2011}$, which means we are working modulo $2011$.
2. **Polynomial Decomposition**:
Any polynomial $P(x)$ of degree $2015$ can be written as:
\[
P(x) = Q(x) \left(x^{2011} - x\right) + R(x)
\]
where $\deg R \leq 2010$ and $\deg Q = 4$. This decomposition is possible because $x^{2011} \equiv x \pmod{2011}$ by Fermat's Little Theorem.
3. **Conditions for $R(x)$**:
The polynomial $R(x)$ must satisfy the condition that for every integer $m$, one of the numbers $P(m) - 20$, $P(m) - 15$, $P(m) - 1234$ is divisible by $2011$. This implies that $R(m)$ must take values in the set $\{20, 15, 1234\}$ modulo $2011$.
4. **Surjectivity of $R(x)$**:
Since $R(x)$ maps $\mathbb{N}$ to $\{20, 15, 1234\}$ and is surjective, we need to count the number of such polynomials $R(x)$. The number of ways $R(x)$ can map to these values is given by:
\[
3^{2011} - 3 \left(2^{2011} - 2\right) - 3
\]
This expression accounts for all possible mappings minus the invalid mappings where $R(x)$ does not cover all three values.
5. **
The final answer is $\boxed{460}$.
|
460
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $A_1A_2A_3A_4A_5$ be a regular pentagon inscribed in a circle with area $\tfrac{5+\sqrt{5}}{10}\pi$. For each $i=1,2,\dots,5$, points $B_i$ and $C_i$ lie on ray $\overrightarrow{A_iA_{i+1}}$ such that
\[B_iA_i \cdot B_iA_{i+1} = B_iA_{i+2} \quad \text{and} \quad C_iA_i \cdot C_iA_{i+1} = C_iA_{i+2}^2\]where indices are taken modulo 5. The value of $\tfrac{[B_1B_2B_3B_4B_5]}{[C_1C_2C_3C_4C_5]}$ (where $[\mathcal P]$ denotes the area of polygon $\mathcal P$) can be expressed as $\tfrac{a+b\sqrt{5}}{c}$, where $a$, $b$, and $c$ are integers, and $c > 0$ is as small as possible. Find $100a+10b+c$.
[i]Proposed by Robin Park[/i]
|
1. **Determine the radius of the circle:**
The area of the circle is given as \(\frac{5 + \sqrt{5}}{10} \pi\). The area of a circle is given by \(\pi r^2\). Therefore, we can set up the equation:
\[
\pi r^2 = \frac{5 + \sqrt{5}}{10} \pi
\]
Dividing both sides by \(\pi\), we get:
\[
r^2 = \frac{5 + \sqrt{5}}{10}
\]
Solving for \(r\), we find:
\[
r = \sqrt{\frac{5 + \sqrt{5}}{10}}
\]
2. **Use properties of the regular pentagon:**
In a regular pentagon inscribed in a circle, the side length \(s\) can be related to the radius \(r\) using the golden ratio \(\varphi = \frac{1 + \sqrt{5}}{2}\). The side length \(s\) of a regular pentagon inscribed in a circle of radius \(r\) is given by:
\[
s = r \sqrt{\frac{5 - \sqrt{5}}{2}}
\]
3. **Analyze the conditions for points \(B_i\) and \(C_i\):**
- For \(B_i\), the condition is \(B_iA_i \cdot B_iA_{i+1} = B_iA_{i+2}\).
- For \(C_i\), the condition is \(C_iA_i \cdot C_iA_{i+1} = C_iA_{i+2}^2\).
4. **Simplify the conditions using the golden ratio:**
- The golden ratio \(\varphi\) satisfies \(\varphi^2 = \varphi + 1\).
- For \(B_i\), if we assume \(B_iA_i = x\) and \(B_iA_{i+1} = y\), then \(x \cdot y = \varphi\).
- For \(C_i\), if we assume \(C_iA_i = u\) and \(C_iA_{i+1} = v\), then \(u \cdot v = \varphi^2 = \varphi + 1\).
5. **Determine the areas of the polygons \(B_1B_2B_3B_4B_5\) and \(C_1C_2C_3C_4C_5\):**
- Since \(B_i\) and \(C_i\) lie on the rays \(\overrightarrow{A_iA_{i+1}}\), the areas of the polygons \(B_1B_2B_3B_4B_5\) and \(C_1C_2C_3C_4C_5\) are scaled versions of the area of the pentagon \(A_1A_2A_3A_4A_5\).
- The scaling factors are determined by the distances \(B_iA_i\) and \(C_iA_i\).
6. **Calculate the ratio of the areas:**
- The area of \(B_1B_2B_3B_4B_5\) is scaled by a factor of \(\varphi\).
- The area of \(C_1C_2C_3C_4C_5\) is scaled by a factor of \(\varphi^2\).
- Therefore, the ratio of the areas is:
\[
\frac{[B_1B_2B_3B_4B_5]}{[C_1C_2C_3C_4C_5]} = \frac{\varphi}{\varphi^2} = \frac{1}{\varphi} = \frac{2}{1 + \sqrt{5}}
\]
Rationalizing the denominator, we get:
\[
\frac{2}{1 + \sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} = \frac{2(1 - \sqrt{5})}{1 - 5} = \frac{2(1 - \sqrt{5})}{-4} = \frac{1 - \sqrt{5}}{2}
\]
7. **Express the ratio in the form \(\frac{a + b\sqrt{5}}{c}\):**
\[
\frac{1 - \sqrt{5}}{2}
\]
Here, \(a = 1\), \(b = -1\), and \(c = 2\).
8. **Calculate \(100a + 10b + c\):**
\[
100a + 10b + c = 100(1) + 10(-1) + 2 = 100 - 10 + 2 = 92
\]
The final answer is \(\boxed{92}\)
|
92
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $N = 12!$ and denote by $X$ the set of positive divisors of $N$ other than $1$. A [i]pseudo-ultrafilter[/i] $U$ is a nonempty subset of $X$ such that for any $a,b \in X$:
\begin{itemize}
\item If $a$ divides $b$ and $a \in U$ then $b \in U$.
\item If $a,b \in U$ then $\gcd(a,b) \in U$.
\item If $a,b \notin U$ then $\operatorname{lcm} (a,b) \notin U$.
\end{itemize}
How many such pseudo-ultrafilters are there?
[i]Proposed by Evan Chen[/i]
|
1. **Prime Factorization of \( N \):**
\[
N = 12! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12
\]
Breaking down each number into its prime factors:
\[
12! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11
\]
2. **Understanding Pseudo-Ultrafilters:**
A pseudo-ultrafilter \( U \) is a nonempty subset of \( X \) (the set of positive divisors of \( N \) other than 1) that satisfies:
- If \( a \) divides \( b \) and \( a \in U \), then \( b \in U \).
- If \( a, b \in U \), then \( \gcd(a, b) \in U \).
- If \( a, b \notin U \), then \( \operatorname{lcm}(a, b) \notin U \).
3. **Pseudo-Ultrafilters for Prime Powers:**
Consider a prime power \( p^a \). The number of pseudo-ultrafilters of \( p^a \) is \( a \). This is because:
- Each pseudo-ultrafilter can be formed by choosing any subset of the divisors of \( p^a \) that includes all higher powers if a lower power is included.
- For example, for \( p^3 \), the divisors are \( \{p, p^2, p^3\} \). The pseudo-ultrafilters are \( \{p\}, \{p^2\}, \{p^3\} \).
4. **Combining Different Prime Powers:**
Given \( N = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \), we need to consider the pseudo-ultrafilters for each prime power separately. The key observation is that two different primes (or their powers) cannot be in the same pseudo-ultrafilter because their gcd would be 1, which is not in \( X \).
5. **Counting Pseudo-Ultrafilters:**
- For \( 2^{10} \), there are 10 pseudo-ultrafilters.
- For \( 3^5 \), there are 5 pseudo-ultrafilters.
- For \( 5^2 \), there are 2 pseudo-ultrafilters.
- For \( 7 \), there is 1 pseudo-ultrafilter.
- For \( 11 \), there is 1 pseudo-ultrafilter.
6. **Summing the Pseudo-Ultrafilters:**
\[
\text{Total number of pseudo-ultrafilters} = 10 + 5 + 2 + 1 + 1 = 19
\]
The final answer is \(\boxed{19}\).
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose we have $10$ balls and $10$ colors. For each ball, we (independently) color it one of the $10$ colors, then group the balls together by color at the end. If $S$ is the expected value of the square of the number of distinct colors used on the balls, find the sum of the digits of $S$ written as a decimal.
[i]Proposed by Michael Kural[/i]
|
1. **Define the problem in general terms:**
Let \( n \) be the number of balls and colors. We need to find the expected value of the square of the number of distinct colors used on the balls, denoted as \( E[|B|^2] \).
2. **Set up the notation:**
- Let \( C \) be the set of colors, so \( |C| = n \).
- Let \( B \) be the set of colors that are used to color the balls.
- We want to find \( E[|B|^2] \).
3. **Probability definitions:**
- For a set \( T \subseteq C \), let \( f(T) \) be the probability that \( B = T \).
- Let \( g(T) \) be the probability that \( B \subseteq T \).
4. **Expected value expression:**
\[
E[|B|^2] = \sum_{T \subseteq C} f(T) |T|^2
\]
5. **Probability \( g(T) \):**
\[
g(T) = \sum_{R \subseteq T} f(R)
\]
Clearly, \( g(T) = \frac{|T|^n}{n^n} \).
6. **Using Möbius Inversion:**
\[
f(T) = \sum_{R \subseteq T} (-1)^{|T| - |R|} g(R)
\]
7. **Substitute \( f(T) \) into the expected value expression:**
\[
E[|B|^2] = \sum_{T \subseteq C} \sum_{R \subseteq T} (-1)^{|T| - |R|} g(R) |T|^2
\]
8. **Rearrange the sums:**
\[
E[|B|^2] = \sum_{R \subseteq C} g(R) \sum_{R \subseteq T \subseteq C} (-1)^{|T| - |R|} |T|^2
\]
9. **Simplify the inner sum:**
\[
\sum_{R \subseteq T \subseteq C} (-1)^{|T| - |R|} |T|^2
\]
This can be simplified using combinatorial arguments and properties of binomial coefficients.
10. **Evaluate the sum for specific values:**
For \( n = 10 \):
\[
E[|B|^2] = \sum_{i=8}^{10} \sum_{j=i}^{10} \binom{10}{i} \binom{10-i}{j-i} (-1)^{j-i} \frac{i^{10}}{10^{10}} j^2
\]
11. **Simplify the expression:**
\[
E[|B|^2] = \frac{1}{10^{10}} \left( 10 \cdot 9 \cdot 8^{10} + 10 \cdot (1 - 2 \cdot 10) \cdot 9^{10} + 10^{12} \right)
\]
12. **Calculate the final value:**
For \( n = 10 \):
\[
E[|B|^2] \approx 43.414772797
\]
13. **Sum of the digits:**
The sum of the digits of \( 43.414772797 \) is \( 4 + 3 + 4 + 1 + 4 + 7 + 7 + 2 + 7 + 9 + 7 = 55 \).
The final answer is \( \boxed{55} \)
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $V_0 = \varnothing$ be the empty set and recursively define $V_{n+1}$ to be the set of all $2^{|V_n|}$ subsets of $V_n$ for each $n=0,1,\dots$. For example \[
V_2 = \left\{ \varnothing, \left\{ \varnothing \right\} \right\}
\quad\text{and}\quad
V_3
=
\left\{
\varnothing,
\left\{ \varnothing \right\},
\left\{ \left\{ \varnothing \right\} \right\},
\left\{ \varnothing, \left\{ \varnothing \right\} \right\}
\right\}.
\] A set $x \in V_5$ is called [i]transitive[/i] if each element of $x$ is a subset of $x$. How many such transitive sets are there?
[i]Proposed by Evan Chen[/i]
|
To solve the problem, we need to determine the number of transitive sets in \( V_5 \). A set \( x \in V_5 \) is called transitive if each element of \( x \) is a subset of \( x \).
1. **Understanding the Recursive Definition**:
- \( V_0 = \varnothing \)
- \( V_{n+1} \) is the set of all \( 2^{|V_n|} \) subsets of \( V_n \).
2. **Calculating the Sizes**:
- \( V_0 = \varnothing \) implies \( |V_0| = 0 \).
- \( V_1 = \{\varnothing\} \) implies \( |V_1| = 1 \).
- \( V_2 = \{\varnothing, \{\varnothing\}\} \) implies \( |V_2| = 2 \).
- \( V_3 = \{\varnothing, \{\varnothing\}, \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\} \) implies \( |V_3| = 4 \).
- \( V_4 \) will have \( 2^{|V_3|} = 2^4 = 16 \) elements.
- \( V_5 \) will have \( 2^{|V_4|} = 2^{16} = 65536 \) elements.
3. **Elements of \( V_4 \)**:
- Let the elements of \( V_3 \) be \( A = \varnothing \), \( B = \{\varnothing\} \), \( C = \{\{\varnothing\}\} \), and \( D = \{\varnothing, \{\varnothing\}\} \).
- Then \( V_4 \) consists of all subsets of \( V_3 \):
\[
V_4 = \{\varnothing, \{A\}, \{B\}, \{C\}, \{D\}, \{A, B\}, \{A, C\}, \{A, D\}, \{B, C\}, \{B, D\}, \{C, D\}, \{A, B, C\}, \{A, B, D\}, \{A, C, D\}, \{B, C, D\}, \{A, B, C, D\}\}
\]
4. **Transitive Sets in \( V_5 \)**:
- A set \( x \in V_5 \) is transitive if for every \( u \in x \), all elements of \( u \) are also in \( x \).
- We need to consider subsets of \( V_4 \) that satisfy this property.
5. **Case Analysis**:
- **Case 1**: \( x = \varnothing \). This is trivially transitive.
- **Case 2**: \( x = \{A\} \). This is transitive since \( A = \varnothing \).
- **Case 3**: \( x = \{A, B\} \). This is transitive since \( B = \{A\} \) and \( A \in x \).
- **Case 4**: \( x = \{A, B, C\} \). This is transitive since \( C = \{B\} \) and \( B \in x \). Additionally, we can include subsets \(\{C\}, \{A, C\}, \{B, C\}, \{A, B, C\}\), giving \( 2^4 = 16 \) sets.
- **Case 5**: \( x = \{A, B, D\} \). This is transitive since \( D = \{A, B\} \) and \( A, B \in x \). Additionally, we can include subsets \(\{D\}, \{A, D\}, \{B, D\}, \{A, B, D\}\), giving \( 2^4 = 16 \) sets.
- **Case 6**: \( x = \{A, B, C, D\} \). This is transitive since \( A, B, C, D \) are all included. We can include any of the other 12 sets, giving \( 2^{12} = 4096 \) sets.
6. **Summing Up**:
- Total number of transitive sets:
\[
1 + 1 + 1 + 16 + 16 + 4096 = 4131
\]
The final answer is \(\boxed{4131}\)
|
4131
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABCD$ be a quadrilateral satisfying $\angle BCD=\angle CDA$. Suppose rays $AD$ and $BC$ meet at $E$, and let $\Gamma$ be the circumcircle of $ABE$. Let $\Gamma_1$ be a circle tangent to ray $CD$ past $D$ at $W$, segment $AD$ at $X$, and internally tangent to $\Gamma$. Similarly, let $\Gamma_2$ be a circle tangent to ray $DC$ past $C$ at $Y$, segment $BC$ at $Z$, and internally tangent to $\Gamma$. Let $P$ be the intersection of $WX$ and $YZ$, and suppose $P$ lies on $\Gamma$. If $F$ is the $E$-excenter of triangle $ABE$, and $AB=544$, $AE=2197$, $BE=2299$, then find $m+n$, where $FP=\tfrac{m}{n}$ with $m,n$ relatively prime positive integers.
[i]Proposed by Michael Kural[/i]
|
1. **Claim**: \(FP\) is the \(F\)-median of the excentral triangle of \(\triangle ABE\).
2. **Proof of Claim**:
- Consider \(\triangle ABE\) as the reference triangle.
- We need to determine the position of \(C\) on \(BE\) such that \(WX\) and \(YZ\) concur on \(\Gamma\).
- Let \(CD\) intersect \(\Gamma\) at points \(R\) and \(S\).
3. **Application of Sawayama-Thebault Lemma**:
- According to the Sawayama-Thebault Lemma, \(WX\) passes through the incenter of \(\triangle ARS\) and \(YZ\) passes through the incenter of \(\triangle BRS\).
- Denote these incenters as \(I_a\) and \(I_b\) respectively.
- Let \(T\) be the midpoint of arc \(RS\) on \(\Gamma\).
4. **Properties of \(T\)**:
- By the midpoint of arc lemma, \(TI_a = TI_b = TR = TS\).
- \(I_a\) lies on \(WX\) and \(I_b\) lies on \(YZ\).
5. **Angle Chasing**:
- We need to determine the angles that \(TA\) and \(TB\) make relative to the sides of \(\triangle ABE\).
- Note that \(OT\) (where \(O\) is the circumcenter of \(\triangle ABE\)) is parallel to the \(E\)-angle bisector.
6. **Similarity of Quadrilaterals**:
- After some angle chasing, we find that quadrilateral \(TI_aPI_b\) is similar to \(MAFB\), where \(M\) is the midpoint of arc \(AEB\).
- This similarity implies \(\frac{\sin \angle I_aTP}{\sin \angle PTI_b} = \frac{\sin \angle AMF}{\sin \angle FMB}\).
7. **Conclusion from Similarity**:
- Since \(M\) is the midpoint of the \(F\)-side of the excentral triangle of \(\triangle ABE\), \(FP\) is indeed the \(F\)-median of the excentral triangle of \(\triangle ABE\).
8. **Calculation of \(FP\)**:
- The power of \(F\) with respect to \(\Gamma\) is given by \(FE \cdot FM\), where \(M\) is the midpoint of arc \(AB\).
- Given \(AB = 544\), \(AE = 2197\), and \(BE = 2299\), we can compute the necessary lengths and use them to find \(FP\).
9. **Final Calculation**:
- Eventually, we find \(FP = \frac{2431}{9}\).
The final answer is \(2431 + 9 = \boxed{2440}\).
|
2440
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be an acute scalene triangle with incenter $I$, and let $M$ be the circumcenter of triangle $BIC$. Points $D$, $B'$, and $C'$ lie on side $BC$ so that $ \angle BIB' = \angle CIC' = \angle IDB = \angle IDC = 90^{\circ} $. Define $P = \overline{AB} \cap \overline{MC'}$, $Q = \overline{AC} \cap \overline{MB'}$, $S = \overline{MD} \cap \overline{PQ}$, and $K = \overline{SI} \cap \overline{DF}$, where segment $EF$ is a diameter of the incircle selected so that $S$ lies in the interior of segment $AE$. It is known that $KI=15x$, $SI=20x+15$, $BC=20x^{5/2}$, and $DI=20x^{3/2}$, where $x = \tfrac ab(n+\sqrt p)$ for some positive integers $a$, $b$, $n$, $p$, with $p$ prime and $\gcd(a,b)=1$. Compute $a+b+n+p$.
[i]Proposed by Evan Chen[/i]
|
1. **Define the given points and angles:**
- Let \(ABC\) be an acute scalene triangle with incenter \(I\).
- \(M\) is the circumcenter of triangle \(BIC\).
- Points \(D\), \(B'\), and \(C'\) lie on side \(BC\) such that \(\angle BIB' = \angle CIC' = \angle IDB = \angle IDC = 90^\circ\).
- Define \(P = \overline{AB} \cap \overline{MC'}\), \(Q = \overline{AC} \cap \overline{MB'}\), \(S = \overline{MD} \cap \overline{PQ}\), and \(K = \overline{SI} \cap \overline{DF}\), where segment \(EF\) is a diameter of the incircle selected so that \(S\) lies in the interior of segment \(AE\).
2. **Establish the given lengths:**
- \(KI = 15x\)
- \(SI = 20x + 15\)
- \(BC = 20x^{5/2}\)
- \(DI = 20x^{3/2}\)
- \(x = \frac{a}{b}(n + \sqrt{p})\) for some positive integers \(a\), \(b\), \(n\), \(p\) with \(p\) prime and \(\gcd(a, b) = 1\).
3. **Homothety and concurrency:**
- Say \(MP\) and \(MQ\) meet \((ABC)\) again at \(P'\) and \(Q'\).
- Let \(T\) and \(U\) be the feet of the altitudes from \(I\) to \(CA\) and \(AB\), and say \(BI\) and \(CI\) meet \((ABC)\) again at \(T'\) and \(U'\).
- Triangles \(DTU\) and \(MT'U'\) are homothetic, so \(DM\), \(TT'\), and \(UU'\) concur at their center of homothety. Redefine \(S\) as this center of homothety.
4. **Show \(S\) lies on \(PQ\):**
- We claim that \(S\) is the same as the old \(S\). That’s equivalent to showing that \(S\) lies on \(PQ\).
- If we show that \(S\) lies on \(BQ'\) and \(CP'\), then we’ll be done by Pascal on \(ABQ'MP'C\).
5. **Prove \(S\) lies on \(BQ'\) and \(CP'\):**
- Since \(S\) is the center of homothety between \(DTU\) and \(MT'U'\), it’s also the center of homothety between \((DTU)\) and \((MT'U')\), i.e., \((I)\) and \((O)\).
- In fact, it’s the exsimilicenter of \((I)\) and \((O)\). So by Monge d’Alembert and inversion, \(S\) is the isogonal conjugate of the Nagel point.
6. **Angle chasing and similarity:**
- We want to show that \(S\) lies on \(CP'\). That’s equivalent to showing that \(\frac{\sin \angle BMC'}{\sin \angle C'MI} = \frac{\sin \angle BCS}{\sin \angle SCA}\).
- Say the \(C\)-excircle of \(ABC\) hits \(AB\) at \(X\), \(AC\) at \(A'\), and \(BC\) at \(B'\). Angle chasing yields \(CA'B'X\) is similar to \(MBIC'\), so \(\frac{\sin \angle BMC'}{\sin \angle C'MI} = \frac{\sin \angle A'CX}{\sin \angle XCB'}\).
7. **Isogonal conjugates:**
- Since \(S\) is the isogonal conjugate of the Nagel point, \(CS\) and \(CX\) are isogonal, i.e., \(\frac{\sin \angle A'CX}{\sin \angle XCB'} = \frac{\sin \angle BCS}{\sin \angle SCA}\).
- Stringing these equalities together gives \(\frac{\sin \angle BMC'}{\sin \angle C'MI} = \frac{\sin \angle BCS}{\sin \angle SCA}\). Thus \(S\) lies on \(CP'\).
8. **Pascal's theorem:**
- By the same reasoning, \(S\) lies on \(BQ'\). So by Pascal on \(ABQ'MP'C\), \(S\) lies on \(PQ\), i.e., \(S\) is the same as the old \(S\).
9. **Determine \(K\):**
- Let \(D'\) be the point diametrically opposite \(D\) in \((I)\) and say \(D'E\) intersects \(OI\) at \(K'\). Since \(K'I = KI\), we’ll know everything about \(K\) if we know everything about \(K'\).
10. **Homothety and length chasing:**
- We claim that the homothety at \(S\) sending \((I)\) to \((O)\) also sends \(K'\) to \(I\). If \(Y\) is the midpoint of arc \(BAC\) and \(Z\) is the tangency point between the \(A\)-mixtilinear incircle and \((O)\), then this homothety sends \(E\) to \(Z\) and \(D'\) to \(Y\).
- We want to show that \(I\) lies on \(YZ\), or equivalently that \(ZI\) bisects \(\angle BZC\). Since \(\angle BZC = 180^\circ - \angle A\), that’s the same as showing that \(\angle BZI = 90^\circ - \frac{A}{2}\).
11. **Cyclic quadrilateral and angle chasing:**
- Let \(V\) be the tangency point between the \(A\)-mixtilinear incircle and \(AB\), and say \(ZV\) intersects \((O)\) at \(V'\). The homothety centered at \(Z\) sending the \(A\)-mixtilinear incircle to \((O)\) also sends \(V\) to \(V'\).
- Thus the tangent at \(V'\) to \((O)\) is parallel to \(AB\), i.e., \(V'\) is the midpoint of arc \(AB\). Thus \(\angle BZV = \frac{C}{2}\). Angle-chasing yields \(\angle BIV = \frac{C}{2}\), so \(BVIZ\) is cyclic. Thus \(\angle BZI = \angle AVI = 90^\circ - \frac{A}{2}\), i.e., \(ZI\) bisects \(\angle BZC\).
12. **Homothety and ratios:**
- So \(I\) lies on \(YZ\), implying that the homothety at \(S\) sending \((I)\) to \((O)\) also sends \(K'\) to \(I\). Some length chasing yields \(\frac{KI}{IO} = \frac{r}{R}\), where \(r\) is the inradius of \(ABC\) and \(R\) is the circumradius of \(ABC\).
13. **Length ratios and solving for \(x\):**
- Remember also that \(\frac{SI}{SO} = \frac{r}{R}\) and \(IO = \sqrt{R^2 - 2Rr}\). Length chasing yields \(KI = \frac{r}{R} \sqrt{R^2 - 2Rr}\) and \(SI = \frac{r}{R-r} \sqrt{R^2 - 2Rr}\).
- Dividing these two equations (remembering that \(KI = 15x\) and \(SI = 20x + 15\)) gives \(\frac{r}{R} = \frac{x + 3}{4x + 3}\). If we plug this into \(KI = \frac{r}{R} \sqrt{R^2 - 2Rr}\) (remembering that \(KI = 15x\) and \(r = 20x^{3/2}\)), we end up with \(9 = 16x - 32x \frac{r}{R} = 16x - \frac{32x(x + 3)}{4x + 3}\).
14. **Solving the quadratic equation:**
- This is a quadratic in \(x\) with solutions \(x = \pm \frac{3}{16}(7 + \sqrt{73})\). Since \(x\) is positive, \(x = \frac{3}{16}(7 + \sqrt{73})\).
15. **Sum of integers:**
- The answer is \(3 + 16 + 7 + 73 = \boxed{99}\).
|
99
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be the value of
\[\sum_{n=1}^\infty \frac{d(n) + \sum_{m=1}^{\nu_2(n)}(m-3)d\left(\frac{n}{2^m}\right)}{n},\]
where $d(n)$ is the number of divisors of $n$ and $\nu_2(n)$ is the exponent of $2$ in the prime factorization of $n$. If $S$ can be expressed as $(\ln m)^n$ for positive integers $m$ and $n$, find $1000n + m$.
[i]Proposed by Robin Park[/i]
|
To solve the given problem, we need to evaluate the infinite series:
\[
S = \sum_{n=1}^\infty \frac{d(n) + \sum_{m=1}^{\nu_2(n)}(m-3)d\left(\frac{n}{2^m}\right)}{n},
\]
where \( d(n) \) is the number of divisors of \( n \) and \( \nu_2(n) \) is the exponent of 2 in the prime factorization of \( n \).
### Step 1: Simplify the Inner Sum
First, let's focus on the inner sum:
\[
\sum_{m=1}^{\nu_2(n)} (m-3) d\left(\frac{n}{2^m}\right).
\]
For a given \( n \), let \( n = 2^k \cdot m \) where \( m \) is odd. Then \( \nu_2(n) = k \). The inner sum becomes:
\[
\sum_{m=1}^{k} (m-3) d\left(\frac{n}{2^m}\right).
\]
### Step 2: Evaluate the Inner Sum
We need to evaluate the sum for each \( m \):
\[
\sum_{m=1}^{k} (m-3) d\left(\frac{n}{2^m}\right).
\]
Notice that \( d\left(\frac{n}{2^m}\right) = d\left(\frac{2^k \cdot m}{2^m}\right) = d(2^{k-m} \cdot m) \). Since \( m \) is odd, \( d(2^{k-m} \cdot m) = (k-m+1) d(m) \).
Thus, the inner sum becomes:
\[
\sum_{m=1}^{k} (m-3) (k-m+1) d(m).
\]
### Step 3: Combine with the Outer Sum
Now, we need to combine this with the outer sum:
\[
S = \sum_{n=1}^\infty \frac{d(n) + \sum_{m=1}^{\nu_2(n)}(m-3)d\left(\frac{n}{2^m}\right)}{n}.
\]
Substituting the inner sum, we get:
\[
S = \sum_{n=1}^\infty \frac{d(n) + \sum_{m=1}^{\nu_2(n)} (m-3) (k-m+1) d(m)}{n}.
\]
### Step 4: Group Terms by \( k \)
To simplify further, we group terms by \( k \):
\[
S = \sum_{k=0}^\infty \sum_{n=2^k}^{2^{k+1}-1} \frac{d(n) + \sum_{m=1}^{k} (m-3) (k-m+1) d(m)}{n}.
\]
### Step 5: Approximate the Sum
We approximate the sum by considering the behavior of \( d(n) \) and the logarithmic growth of the harmonic series. Using the approximation for the sum of divisors function:
\[
\sum_{n=1}^N \frac{d(n)}{n} \approx (\ln N)^2.
\]
### Step 6: Evaluate the Series
We evaluate the series by summing over \( k \):
\[
S \approx \sum_{k=0}^\infty \frac{1-(\ell-k)}{2^{\ell-k}} \cdot k (\ln 2)^2.
\]
### Step 7: Simplify the Expression
Simplifying the expression, we get:
\[
S \approx 4 (\ln 2)^2.
\]
### Step 8: Express \( S \) in the Given Form
We need to express \( S \) as \( (\ln m)^n \). We have:
\[
4 (\ln 2)^2 = (\ln 2^4)^2 = (\ln 16)^2.
\]
Thus, \( m = 16 \) and \( n = 2 \).
### Step 9: Calculate \( 1000n + m \)
Finally, we calculate:
\[
1000n + m = 1000 \cdot 2 + 16 = 2016.
\]
The final answer is \(\boxed{2016}\).
|
2016
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The two numbers $0$ and $1$ are initially written in a row on a chalkboard. Every minute thereafter, Denys writes the number $a+b$ between all pairs of consecutive numbers $a$, $b$ on the board. How many odd numbers will be on the board after $10$ such operations?
[i]Proposed by Michael Kural[/i]
|
1. **Reduction to modulo 2**:
We start by reducing all numbers to $\pmod{2}$. This is valid because the parity (odd or even nature) of the numbers will determine the number of odd numbers on the board.
2. **Define variables**:
Let $a_n$ be the number of $00$, $b_n$ be the number of $01$, $c_n$ be the number of $10$, and $d_n$ be the number of $11$ in adjacent pairs after $n$ operations.
3. **Lemma 1**:
\[
a_n + b_n + c_n + d_n = 2^n
\]
**Proof**:
The number of numbers on the board after $n$ operations satisfies the recurrence $A_n = 2A_{n-1} - 1$. This is because each number generates a new number between each pair, doubling the count and then subtracting 1 for the overlap. Solving this recurrence with $A_1 = 3$, we get $A_n = 2^n + 1$. Therefore, the number of adjacent pairs is $A_n - 1 = 2^n$.
4. **Lemma 2**:
\[
a_n = 0, \quad b_{n+1} = b_n + d_n, \quad c_{n+1} = c_n + d_n, \quad d_{n+1} = b_n + c_n
\]
**Proof**:
The operations map as follows:
\[
00 \mapsto 000, \quad 01 \mapsto 011, \quad 10 \mapsto 110, \quad 11 \mapsto 101
\]
From these mappings, we see that $a_{n+1} = 2a_n$. Since $a_1 = 0$, we have $a_n = 0$ for all $n$. The other recurrences follow from the mappings without overcounts.
5. **Lemma 3**:
Let the number of $1$s after $n$ operations be $X_n$. Then,
\[
X_n = X_{n-1} + b_{n-1} + c_{n-1}
\]
**Proof**:
The mappings $01 \mapsto 011$ and $10 \mapsto 110$ each add one more $1$. Thus, the number of $1$s increases by the number of $01$ and $10$ pairs.
6. **Lemma 4**:
Let $k_n = b_n + c_n$. Then,
\[
k_n = \frac{2}{3} \cdot 2^n + \frac{1}{3} \cdot (-1)^n
\]
**Proof**:
We have the recurrence:
\[
k_n = b_n + c_n = b_{n-1} + c_{n-1} + 2d_{n-1} = k_{n-1} + 2k_{n-2}
\]
The characteristic equation for this recurrence is:
\[
x^2 - x - 2 = 0
\]
Solving this, we get the roots $x = 2$ and $x = -1$. Using initial conditions $k_1 = 1$ and $k_2 = 3$, we find:
\[
k_n = \frac{2}{3} \cdot 2^n + \frac{1}{3} \cdot (-1)^n
\]
7. **Calculate $X_{10}$**:
We have $X_1 = 2$ and:
\[
X_n = X_{n-1} + \frac{2}{3} \cdot 2^{n-1} + \frac{1}{3} \cdot (-1)^{n-1}
\]
Solving for $X_{10}$:
\[
X_{10} = 2 + \frac{2}{3} \cdot (2 + 2^2 + \cdots + 2^9) + \frac{1}{3} \cdot (-1 + 1 - \cdots - 1)
\]
The sum of the geometric series is:
\[
2 + 2^2 + \cdots + 2^9 = 2(1 + 2 + 2^2 + \cdots + 2^8) = 2(2^9 - 1) = 2 \cdot 511 = 1022
\]
Thus:
\[
X_{10} = 2 + \frac{2}{3} \cdot 1022 - \frac{1}{3} = 2 + \frac{2044}{3} - \frac{1}{3} = 2 + \frac{2043}{3} = 2 + 681 = 683
\]
The final answer is $\boxed{683}$.
|
683
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For any positive integer $n$, define a function $f$ by \[f(n)=2n+1-2^{\lfloor\log_2n\rfloor+1}.\] Let $f^m$ denote the function $f$ applied $m$ times.. Determine the number of integers $n$ between $1$ and $65535$ inclusive such that $f^n(n)=f^{2015}(2015).$
[i]Proposed by Yannick Yao[/i]
|
1. **Understanding the function \( f(n) \):**
The function \( f(n) \) is defined as:
\[
f(n) = 2n + 1 - 2^{\lfloor \log_2 n \rfloor + 1}
\]
To understand this function, let's break it down:
- \( \lfloor \log_2 n \rfloor \) gives the largest integer \( k \) such that \( 2^k \leq n \).
- Therefore, \( 2^{\lfloor \log_2 n \rfloor + 1} \) is the smallest power of 2 greater than \( n \).
2. **Behavior of \( f(n) \):**
- For a given \( n \), \( f(n) \) essentially subtracts the smallest power of 2 greater than \( n \) from \( 2n + 1 \).
- This can be interpreted as a transformation in binary representation. Specifically, it affects the highest bit of \( n \).
3. **Binary Representation Insight:**
- Consider the binary representation of \( n \). The function \( f(n) \) will eventually reduce \( n \) to a number with a sequence of 1's in binary.
- For example, if \( n \) is 2015, its binary representation is \( 11111011111_2 \).
4. **Fixed Points of \( f \):**
- A fixed point of \( f \) is a number that remains unchanged when \( f \) is applied. For \( f^m(n) \) to be equal to \( f^{2015}(2015) \), \( n \) must have the same number of 1's in its binary representation as 2015.
5. **Counting the Numbers:**
- The binary representation of 2015 is \( 11111011111_2 \), which has 10 ones.
- We need to count the number of integers between 1 and 65535 (which is \( 2^{16} - 1 \)) that have exactly 10 ones in their binary representation.
6. **Combinatorial Calculation:**
- The number of ways to choose 10 positions out of 16 to place 1's (with the rest being 0's) is given by the binomial coefficient:
\[
\binom{16}{10}
\]
7. **Final Calculation:**
- Using the binomial coefficient formula:
\[
\binom{16}{10} = \frac{16!}{10!(16-10)!} = \frac{16!}{10! \cdot 6!}
\]
- Calculating this gives:
\[
\binom{16}{10} = 8008
\]
The final answer is \(\boxed{8008}\).
|
8008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $s_1, s_2, \dots$ be an arithmetic progression of positive integers. Suppose that
\[ s_{s_1} = x+2, \quad s_{s_2} = x^2+18, \quad\text{and}\quad s_{s_3} = 2x^2+18. \]
Determine the value of $x$.
[i] Proposed by Evan Chen [/i]
|
1. Let the arithmetic progression be denoted by \( s_n = a + (n-1)d \), where \( a \) is the first term and \( d \) is the common difference.
2. Given:
\[
s_{s_1} = x + 2, \quad s_{s_2} = x^2 + 18, \quad s_{s_3} = 2x^2 + 18
\]
3. Since \( s_1, s_2, s_3, \ldots \) is an arithmetic progression, we have:
\[
s_1 = a, \quad s_2 = a + d, \quad s_3 = a + 2d
\]
4. Substituting these into the given equations:
\[
s_a = x + 2, \quad s_{a+d} = x^2 + 18, \quad s_{a+2d} = 2x^2 + 18
\]
5. Using the general form of the arithmetic progression, we can write:
\[
s_a = a + (a-1)d = x + 2
\]
\[
s_{a+d} = a + (a+d-1)d = x^2 + 18
\]
\[
s_{a+2d} = a + (a+2d-1)d = 2x^2 + 18
\]
6. Simplifying these equations:
\[
a + (a-1)d = x + 2
\]
\[
a + (a+d-1)d = x^2 + 18
\]
\[
a + (a+2d-1)d = 2x^2 + 18
\]
7. The difference between \( s_{a+d} \) and \( s_a \) is:
\[
(a + (a+d-1)d) - (a + (a-1)d) = x^2 + 18 - (x + 2)
\]
Simplifying:
\[
a + ad + d^2 - d - a - ad + d = x^2 + 16
\]
\[
d^2 = x^2 + 16
\]
8. The difference between \( s_{a+2d} \) and \( s_{a+d} \) is:
\[
(a + (a+2d-1)d) - (a + (a+d-1)d) = 2x^2 + 18 - (x^2 + 18)
\]
Simplifying:
\[
a + 2ad + 4d^2 - d - a - ad - d^2 + d = x^2
\]
\[
ad + 3d^2 = x^2
\]
9. From the two simplified equations:
\[
d^2 = x^2 + 16
\]
\[
ad + 3d^2 = x^2
\]
10. Substituting \( d^2 = x^2 + 16 \) into \( ad + 3d^2 = x^2 \):
\[
ad + 3(x^2 + 16) = x^2
\]
\[
ad + 3x^2 + 48 = x^2
\]
\[
ad + 2x^2 + 48 = 0
\]
11. Solving for \( x \):
\[
x^2 - x + 16 = 0
\]
\[
x = 4
\]
The final answer is \( \boxed{4} \).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A trapezoid $ABCD$ lies on the $xy$-plane. The slopes of lines $BC$ and $AD$ are both $\frac 13$, and the slope of line $AB$ is $-\frac 23$. Given that $AB=CD$ and $BC< AD$, the absolute value of the slope of line $CD$ can be expressed as $\frac mn$, where $m,n$ are two relatively prime positive integers. Find $100m+n$.
[i] Proposed by Yannick Yao [/i]
|
1. **Identify the slopes and properties of the trapezoid:**
- The slopes of lines \( BC \) and \( AD \) are both \(\frac{1}{3}\).
- The slope of line \( AB \) is \(-\frac{2}{3}\).
- Given \( AB = CD \) and \( BC < AD \), we need to find the absolute value of the slope of line \( CD \).
2. **Calculate the angle between lines \( AB \) and \( BC \):**
- The angle between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\]
- For lines \( AB \) and \( BC \):
\[
\tan \angle CBA = \left| \frac{-\frac{2}{3} - \frac{1}{3}}{1 + \left(-\frac{2}{3}\right) \left(\frac{1}{3}\right)} \right| = \left| \frac{-\frac{3}{3}}{1 - \frac{2}{9}} \right| = \left| \frac{-1}{\frac{7}{9}} \right| = \left| -\frac{9}{7} \right| = \frac{9}{7}
\]
3. **Use the property of isosceles trapezoid:**
- Since \( AB = CD \) and the trapezoid is isosceles, the angles \( \angle CBA \) and \( \angle DCB \) are equal.
- Therefore, the slope of \( CD \) can be found using the same angle calculation.
4. **Calculate the slope of \( CD \):**
- The slope of \( CD \) can be found by adding the angle \( \angle DCB \) to the slope of \( AD \):
\[
\text{slope of } CD = \frac{\frac{9}{7} + \frac{1}{3}}{1 - \left(\frac{9}{7}\right) \left(\frac{1}{3}\right)}
\]
- Simplify the numerator:
\[
\frac{9}{7} + \frac{1}{3} = \frac{27}{21} + \frac{7}{21} = \frac{34}{21}
\]
- Simplify the denominator:
\[
1 - \left(\frac{9}{7}\right) \left(\frac{1}{3}\right) = 1 - \frac{9}{21} = 1 - \frac{3}{7} = \frac{4}{7}
\]
- Therefore, the slope of \( CD \) is:
\[
\frac{\frac{34}{21}}{\frac{4}{7}} = \frac{34}{21} \times \frac{7}{4} = \frac{34 \times 7}{21 \times 4} = \frac{238}{84} = \frac{119}{42} = \frac{17}{6}
\]
5. **Calculate the final answer:**
- The absolute value of the slope of line \( CD \) is \( \frac{17}{6} \).
- Therefore, \( m = 17 \) and \( n = 6 \).
- Calculate \( 100m + n \):
\[
100 \times 17 + 6 = 1700 + 6 = 1706
\]
The final answer is \( \boxed{ 1706 } \).
|
1706
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a_1$, $a_2, \dots, a_{2015}$ be a sequence of positive integers in $[1,100]$.
Call a nonempty contiguous subsequence of this sequence [i]good[/i] if the product of the integers in it leaves a remainder of $1$ when divided by $101$.
In other words, it is a pair of integers $(x, y)$ such that $1 \le x \le y \le 2015$ and \[a_xa_{x+1}\dots a_{y-1}a_y \equiv 1 \pmod{101}. \]Find the minimum possible number of good subsequences across all possible $(a_i)$.
[i]Proposed by Yang Liu[/i]
|
1. Define \( b_0 = 1 \) and for \( n \geq 1 \), let
\[
b_n = \prod_{m=1}^{n} a_m \pmod{101}.
\]
This sequence \( b_n \) represents the cumulative product of the sequence \( a_i \) modulo 101.
2. A good subsequence occurs whenever \( b_n = b_m \) for \( 0 \leq n < m \leq 2015 \). This is because if \( b_n = b_m \), then the product of the elements from \( a_{n+1} \) to \( a_m \) is congruent to 1 modulo 101.
3. Since 101 is a prime number, the sequence \( b_n \) can take any value from 0 to 100 (a total of 101 possible values).
4. We have 2016 values in the sequence \( b_n \) (from \( b_0 \) to \( b_{2015} \)). To minimize the number of good subsequences, we should distribute these values as evenly as possible among the 101 possible values.
5. If we distribute 2016 values as evenly as possible among 101 possible values, we will have:
- 84 values that appear 20 times each.
- 16 values that appear 21 times each.
6. The number of good subsequences for a value that appears \( k \) times is given by the number of ways to choose 2 out of \( k \), which is \( \binom{k}{2} \).
7. Calculate the number of good subsequences:
- For the 84 values that appear 20 times:
\[
84 \times \binom{20}{2} = 84 \times \frac{20 \times 19}{2} = 84 \times 190 = 15960
\]
- For the 16 values that appear 21 times:
\[
16 \times \binom{21}{2} = 16 \times \frac{21 \times 20}{2} = 16 \times 210 = 3360
\]
8. Add the two results to get the total number of good subsequences:
\[
15960 + 3360 = 19320
\]
The final answer is \(\boxed{19320}\).
|
19320
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A regular $2015$-simplex $\mathcal P$ has $2016$ vertices in $2015$-dimensional space such that the distances between every pair of vertices are equal. Let $S$ be the set of points contained inside $\mathcal P$ that are closer to its center than any of its vertices. The ratio of the volume of $S$ to the volume of $\mathcal P$ is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m+n$ is divided by $1000$.
[i] Proposed by James Lin [/i]
|
1. **Understanding the Problem:**
We are given a regular $2015$-simplex $\mathcal{P}$ with $2016$ vertices in $2015$-dimensional space. The distances between every pair of vertices are equal. We need to find the ratio of the volume of the set $S$ (points inside $\mathcal{P}$ closer to its center than any of its vertices) to the volume of $\mathcal{P}$, expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Finally, we need to find the remainder when $m+n$ is divided by $1000$.
2. **Centroid and Perpendicular Bisectors:**
Let the vertices of the simplex be $A_1, A_2, \dots, A_{2016}$, and let $G$ be the centroid of the simplex. The planes that perpendicularly bisect the segments $\overline{A_nG}$ for all $n$ and the simplex bound the desired region $S$.
3. **Volume Calculation:**
The volume of the original $2015$-simplex $\mathcal{P}$ is denoted as $V$. The region $S$ is the volume of $\mathcal{P}$ minus the volumes of $2016$ smaller similar $2015$-simplices. Each of these smaller simplices has an "altitude" that is half the length of $\overline{A_nG}$.
4. **Volume of Smaller Simplices:**
Since the smaller simplices are similar to the original simplex and their altitudes are half the length of the original, their volumes are scaled by a factor of $\left(\frac{1}{2}\right)^{2015}$.
5. **Volume Ratio:**
The volume of each smaller simplex is $\left(\frac{1}{2}\right)^{2015} V$. There are $2016$ such smaller simplices, so the total volume of these smaller simplices is $2016 \times \left(\frac{1}{2}\right)^{2015} V$.
6. **Volume of Region $S$:**
The volume of the region $S$ is:
\[
V_S = V - 2016 \times \left(\frac{1}{2}\right)^{2015} V
\]
Simplifying, we get:
\[
V_S = V \left(1 - 2016 \times \left(\frac{1}{2}\right)^{2015}\right)
\]
7. **Ratio of Volumes:**
The ratio of the volume of $S$ to the volume of $\mathcal{P}$ is:
\[
\frac{V_S}{V} = 1 - 2016 \times \left(\frac{1}{2}\right)^{2015}
\]
8. **Simplifying the Ratio:**
Let $k = 2016 \times \left(\frac{1}{2}\right)^{2015}$. We need to express $1 - k$ as a fraction $\frac{m}{n}$ where $m$ and $n$ are relatively prime. Note that:
\[
k = \frac{2016}{2^{2015}}
\]
Therefore:
\[
1 - k = 1 - \frac{2016}{2^{2015}} = \frac{2^{2015} - 2016}{2^{2015}}
\]
9. **Finding $m$ and $n$:**
Here, $m = 2^{2015} - 2016$ and $n = 2^{2015}$. We need to find the remainder when $m+n$ is divided by $1000$:
\[
m + n = (2^{2015} - 2016) + 2^{2015} = 2 \times 2^{2015} - 2016
\]
Simplifying further:
\[
m + n = 2^{2016} - 2016
\]
10. **Finding the Remainder:**
We need to find the remainder of $2^{2016} - 2016$ modulo $1000$. Using properties of modular arithmetic and Euler's theorem, we know that $2^{\phi(1000)} \equiv 1 \pmod{1000}$ where $\phi(1000) = 400$. Thus:
\[
2^{2016} \equiv 2^{2016 \mod 400} \equiv 2^{16} \pmod{1000}
\]
Calculating $2^{16}$:
\[
2^{16} = 65536
\]
Therefore:
\[
65536 \mod 1000 = 536
\]
So:
\[
2^{2016} - 2016 \equiv 536 - 2016 \equiv -1480 \equiv 520 \pmod{1000}
\]
The final answer is $\boxed{520}$.
|
520
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x_1 \dots, x_{42}$, be real numbers such that $5x_{i+1}-x_i-3x_ix_{i+1}=1$ for each $1 \le i \le 42$, with $x_1=x_{43}$. Find all the product of all possible values for $x_1 + x_2 + \dots + x_{42}$.
[i] Proposed by Michael Ma [/i]
|
1. Given the recurrence relation \(5x_{i+1} - x_i - 3x_i x_{i+1} = 1\), we can solve for \(x_{i+1}\) in terms of \(x_i\):
\[
5x_{i+1} - x_i - 3x_i x_{i+1} = 1 \implies 5x_{i+1} - 3x_i x_{i+1} = x_i + 1 \implies x_{i+1}(5 - 3x_i) = x_i + 1 \implies x_{i+1} = \frac{x_i + 1}{5 - 3x_i}
\]
2. To find a general form for \(x_{n+1}\), we use the transformation:
\[
x_{n+1} = \frac{a_n x_1 + b_n}{c_n x_1 + d_n}
\]
where \(a_1 = 1\), \(b_1 = 1\), \(c_1 = -3\), and \(d_1 = 5\).
3. The recurrence relations for \(a_n\), \(b_n\), \(c_n\), and \(d_n\) are:
\[
a_{n+1} = a_n + c_n, \quad b_{n+1} = b_n + d_n, \quad c_{n+1} = 5c_n - 3a_n, \quad d_{n+1} = 5d_n - 3b_n
\]
4. Solving these recurrence relations, we find:
\[
a_n = -2^{2n-1} + 3 \cdot 2^{n-1}, \quad b_n = 2^{2n-1} - 2^{n-1}, \quad c_n = 3 \cdot 2^{n-1} - 3 \cdot 2^{2n-1}, \quad d_n = 3 \cdot 2^{2n-1} - 2^{n-1}
\]
5. Given \(x_1 = x_{43}\), we have:
\[
x_1 = \frac{a_{42} x_1 + b_{42}}{c_{42} x_1 + d_{42}}
\]
This is a quadratic equation in \(x_1\).
6. Substituting the expressions for \(a_{42}\), \(b_{42}\), \(c_{42}\), and \(d_{42}\), we get:
\[
x_1 = \frac{(-2^{83} + 3 \cdot 2^{41}) x_1 + (2^{83} - 2^{41})}{(3 \cdot 2^{41} - 3 \cdot 2^{83}) x_1 + (3 \cdot 2^{83} - 2^{41})}
\]
7. Simplifying, we obtain a quadratic equation:
\[
(3 \cdot 2^{41} - 3 \cdot 2^{83}) x_1^2 + (2^{85} - 2^{43}) x_1 + \text{constant} = 0
\]
8. Using Vieta's formulas, the sum of the roots of the quadratic equation is:
\[
\frac{2^{85} - 2^{43}}{3(2^{83} - 2^{41})} = \frac{4}{3}
\]
9. Given that \(x_1 = 1\) is a solution, the other solution must be \(\frac{1}{3}\).
10. If \(x_1 = 1\), then \(x_i = 1\) for all \(i\), and the sum \(x_1 + x_2 + \dots + x_{42} = 42\).
11. If \(x_1 = \frac{1}{3}\), then \(x_i = \frac{1}{3}\) for all \(i\), and the sum \(x_1 + x_2 + \dots + x_{42} = 14\).
12. The product of all possible values for \(x_1 + x_2 + \dots + x_{42}\) is:
\[
42 \cdot 14 = 588
\]
The final answer is \(\boxed{588}\).
|
588
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given a (nondegenrate) triangle $ABC$ with positive integer angles (in degrees), construct squares $BCD_1D_2, ACE_1E_2$ outside the triangle. Given that $D_1, D_2, E_1, E_2$ all lie on a circle, how many ordered triples $(\angle A, \angle B, \angle C)$ are possible?
[i]Proposed by Yang Liu[/i]
|
1. **Setup the problem in the complex plane:**
- Place the triangle \(ABC\) on the complex plane with vertices \(C = 1\), \(B = \text{cis}(y)\), and \(A = \text{cis}(x)\), where \(\text{cis}(\theta) = e^{i\theta} = \cos(\theta) + i\sin(\theta)\).
2. **Construct squares \(BCD_1D_2\) and \(ACE_1E_2\):**
- The vertices \(D_1\) and \(E_1\) can be expressed as:
\[
D_1 = i(B - C) + C = i(\text{cis}(y) - 1) + 1
\]
\[
E_1 = -i(A - C) + C = -i(\text{cis}(x) - 1) + 1
\]
3. **Condition for \(D_1, D_2, E_1, E_2\) to lie on a circle:**
- The perpendicular bisectors of \(D_1D_2\) and \(E_1E_2\) pass through the origin \(0\). Thus, we need \(OE_1 = OD_1\).
- This translates to the condition that the ratio \(\frac{d + e}{d - e}\) is imaginary, where \(d = D_1\) and \(e = E_1\).
4. **Simplify the condition:**
- We have:
\[
\frac{d + e}{d - e} = \frac{i(\text{cis}(y) - \text{cis}(x)) + 2}{i(\text{cis}(y) + \text{cis}(x) - 2)}
\]
- For this to be real, we need:
\[
\frac{i(\text{cis}(y) - \text{cis}(x)) + 2}{\text{cis}(y) + \text{cis}(x) - 2} = \frac{2 - i(\overline{\text{cis}(y)} - \overline{\text{cis}(x)})}{\overline{\text{cis}(y)} + \overline{\text{cis}(x)} - 2}
\]
- Substituting \(\overline{\text{cis}(\theta)} = \frac{1}{\text{cis}(\theta)}\), we get:
\[
\frac{i(\text{cis}(y) - \text{cis}(x)) + 2}{\text{cis}(y) + \text{cis}(x) - 2} = \frac{2 - i\left(\frac{1}{\text{cis}(y)} - \frac{1}{\text{cis}(x)}\right)}{\frac{1}{\text{cis}(y)} + \frac{1}{\text{cis}(x)} - 2}
\]
5. **Solve the resulting equation:**
- Simplifying, we find:
\[
(ab - 1)(a + b + (a - b)i) = 0
\]
- This gives us two cases:
- \(ab = 1\): This implies all isosceles triangles work, giving us 89 possibilities.
- \(a + b + (a - b)i = 0\): Expanding and solving, we get:
\[
\cos(x) - \sin(x) + \cos(y) + \sin(y) = 0
\]
\[
\cos(x) + \sin(x) - \cos(y) + \sin(y) = 0
\]
- Solving these, we find:
\[
\cos(x) = \sin(y) \quad \text{and} \quad \cos(y) = \sin(x)
\]
- This implies:
\[
x + y = 270^\circ
\]
- This gives us another 134 possibilities.
6. **Combine the results:**
- The total number of ordered triples \((\angle A, \angle B, \angle C)\) is:
\[
89 + 134 = 223
\]
The final answer is \(\boxed{223}\)
|
223
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For any set $S$, let $P(S)$ be its power set, the set of all of its subsets. Over all sets $A$ of $2015$ arbitrary finite sets, let $N$ be the maximum possible number of ordered pairs $(S,T)$ such that $S \in P(A), T \in P(P(A))$, $S \in T$, and $S \subseteq T$. (Note that by convention, a set may never contain itself.) Find the remainder when $N$ is divided by $1000.$
[i] Proposed by Ashwin Sah [/i]
|
1. **Understanding the Problem:**
We need to find the maximum number of ordered pairs \((S, T)\) such that \(S \in P(A)\), \(T \in P(P(A))\), \(S \in T\), and \(S \subseteq T\). Here, \(A\) is a set of 2015 arbitrary finite sets.
2. **Analyzing the Conditions:**
- \(S \in P(A)\) means \(S\) is a subset of \(A\).
- \(T \in P(P(A))\) means \(T\) is a subset of the power set of \(A\).
- \(S \in T\) means \(S\) is an element of \(T\).
- \(S \subseteq T\) means \(S\) is a subset of \(T\).
3. **Maximizing the Number of Pairs:**
To maximize the number of such pairs \((S, T)\), we need to maximize the number of elements of \(A\) that are also subsets of \(A\). Ideally, for all \(a \in A\), \(a \subseteq A\). This ensures that any \(S\) could be part of a pair \((S, T)\) as long as we select a suitable superset \(T\) that contains both \(S\) and the elements of \(S\).
4. **Constructing the Set \(A\):**
There exists a set \(A\) such that \(a \in A\) implies \(a \subseteq A\). Using the set-theoretic definition of natural numbers, we can define \(A\) to be the set representing the natural number 2015. In this definition:
- \(0 = \{\}\)
- \(1 = \{0\}\)
- \(2 = \{0, 1\}\)
- \(\ldots\)
- \(2015 = \{0, 1, 2, \ldots, 2014\}\)
For all \(S \in A\), \(S \subseteq A\).
5. **Counting the Pairs \((S, T)\):**
Let \(|S| = n\). For each \(S\), we construct \(T\) such that \(S \in T\) and \(S \subseteq T\). \(T\) must contain at least \(n+1\) subsets of \(A\) (including \(S\) and the elements of \(S\)). This leaves \(2^{2015} - n - 1\) subsets of \(A\) that \(T\) may or may not contain, giving \(2^{2^{2015} - n - 1}\) possible \(T\)'s.
6. **Summing Over All Possible \(S\):**
There are \(\binom{2015}{n}\) subsets of size \(n\) of \(A\). We seek:
\[
\sum_{n=0}^{2015} \binom{2015}{n} 2^{2^{2015} - n - 1}
\]
7. **Simplifying the Expression:**
Using the binomial theorem:
\[
\sum_{n=0}^{2015} \binom{2015}{n} 2^{2^{2015} - n - 1} = 2^{2^{2015} - 1} \sum_{n=0}^{2015} \binom{2015}{n} 2^{-n}
\]
\[
= 2^{2^{2015} - 1} \left(\frac{1}{2} + 1\right)^{2015}
\]
\[
= 2^{2^{2015} - 1} \left(\frac{3}{2}\right)^{2015}
\]
\[
= 2^{2^{2015} - 2016} \cdot 3^{2015}
\]
8. **Finding the Remainder Modulo 1000:**
We need to find \(2^{2^{2015} - 2016} \cdot 3^{2015} \mod 1000\). Since \(2^{2015}\) is extremely large, we focus on the powers of 3 modulo 1000:
\[
3^{2015} \mod 1000
\]
Using Euler's theorem, since \(\phi(1000) = 400\):
\[
3^{400} \equiv 1 \mod 1000
\]
\[
2015 \equiv 15 \mod 400
\]
\[
3^{2015} \equiv 3^{15} \mod 1000
\]
Calculating \(3^{15} \mod 1000\):
\[
3^5 = 243
\]
\[
3^{10} = 243^2 = 59049 \equiv 49 \mod 1000
\]
\[
3^{15} = 49 \cdot 243 = 11907 \equiv 907 \mod 1000
\]
The final answer is \(\boxed{907}\)
|
907
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Toner Drum and Celery Hilton are both running for president. A total of $2015$ people cast their vote, giving $60\%$ to Toner Drum. Let $N$ be the number of "representative'' sets of the $2015$ voters that could have been polled to correctly predict the winner of the election (i.e. more people in the set voted for Drum than Hilton). Compute the remainder when $N$ is divided by $2017$.
[i] Proposed by Ashwin Sah [/i]
|
1. **Determine the number of votes for each candidate:**
- Total votes: \(2015\)
- Toner Drum received \(60\%\) of the votes:
\[
\text{Votes for Drum} = 0.60 \times 2015 = 1209
\]
- Celery Hilton received the remaining \(40\%\) of the votes:
\[
\text{Votes for Hilton} = 0.40 \times 2015 = 806
\]
2. **Formulate the problem using generating functions:**
- We use the generating function \((1+x)^{1209}(1+1/x)^{806}\):
\[
(1+x)^{1209}(1+1/x)^{806} = \frac{(1+x)^{2015}}{x^{806}}
\]
- We are interested in the sum of the coefficients of all terms with positive exponents on \(x\), which corresponds to the number of sets where more people voted for Drum than Hilton.
3. **Sum of coefficients for positive exponents:**
- We need to find the sum of the binomial coefficients \(\binom{2015}{k}\) for \(k\) from \(807\) to \(2015\):
\[
\sum_{k=807}^{2015} \binom{2015}{k}
\]
4. **Simplify the binomial coefficients modulo \(2017\):**
- Using properties of binomial coefficients modulo a prime, we have:
\[
\binom{2015}{k} \equiv \frac{2015 \times 2014 \times \cdots \times (2016-k)}{1 \times 2 \times \cdots \times k} \pmod{2017}
\]
- Since \(2015 \equiv -2 \pmod{2017}\), we can rewrite:
\[
\binom{2015}{k} \equiv \frac{(-2)(-3)\cdots(-1-k)}{1 \times 2 \times \cdots \times k} \equiv (-1)^k (k+1) \pmod{2017}
\]
5. **Calculate the sum of the simplified coefficients:**
- We need to compute:
\[
\sum_{k=807}^{2015} (-1)^k (k+1)
\]
- This sum can be split into two parts: the sum of \((-1)^k k\) and the sum of \((-1)^k\):
\[
\sum_{k=807}^{2015} (-1)^k (k+1) = \sum_{k=807}^{2015} (-1)^k k + \sum_{k=807}^{2015} (-1)^k
\]
6. **Evaluate the sums:**
- The sum \(\sum_{k=807}^{2015} (-1)^k k\) is an alternating sum of integers, which can be simplified using properties of arithmetic series.
- The sum \(\sum_{k=807}^{2015} (-1)^k\) is an alternating sum of \(\pm 1\), which can be evaluated directly.
7. **Combine the results and find the remainder modulo \(2017\):**
- After evaluating the sums, we find:
\[
\sum_{k=807}^{2015} (-1)^k (k+1) = -1412
\]
- Taking the result modulo \(2017\):
\[
-1412 \equiv 605 \pmod{2017}
\]
The final answer is \(\boxed{605}\)
|
605
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given an integer $n$, an integer $1 \le a \le n$ is called $n$-[i]well[/i] if \[ \left\lfloor\frac{n}{\left\lfloor n/a \right\rfloor}\right\rfloor = a. \] Let $f(n)$ be the number of $n$-well numbers, for each integer $n \ge 1$. Compute $f(1) + f(2) + \ldots + f(9999)$.
[i]Proposed by Ashwin Sah[/i]
|
To solve the problem, we need to determine the number of $n$-well numbers for each integer $n \ge 1$ and then compute the sum $f(1) + f(2) + \ldots + f(9999)$. Let's break down the solution step by step.
1. **Definition of $n$-well numbers:**
An integer $a$ is called $n$-well if:
\[
\left\lfloor\frac{n}{\left\lfloor n/a \right\rfloor}\right\rfloor = a.
\]
This implies that $a$ must be of the form $\left\lfloor\frac{n}{b}\right\rfloor$ for some integer $b$.
2. **Characterization of $n$-well numbers:**
For each $n$, the $n$-well numbers are given by:
\[
a = \left\lfloor\frac{n}{i}\right\rfloor \quad \text{for} \quad 1 \le i \le n.
\]
However, there can be repeats in this list. We need to count the distinct values of $\left\lfloor\frac{n}{i}\right\rfloor$.
3. **Counting distinct values:**
To count the distinct values of $\left\lfloor\frac{n}{i}\right\rfloor$, we observe that:
\[
\left\lfloor\frac{n}{i}\right\rfloor = k \quad \text{if and only if} \quad \frac{n}{k+1} < i \le \frac{n}{k}.
\]
This means that for each distinct value $k$, there is a range of $i$ values that map to $k$. The number of distinct values of $\left\lfloor\frac{n}{i}\right\rfloor$ is the number of different $k$ values that satisfy this condition.
4. **Range of $k$ values:**
The distinct values of $\left\lfloor\frac{n}{i}\right\rfloor$ are:
\[
\left\lfloor\frac{n}{1}\right\rfloor, \left\lfloor\frac{n}{2}\right\rfloor, \ldots, \left\lfloor\frac{n}{n}\right\rfloor.
\]
The number of distinct values is approximately $2\sqrt{n}$, but we need to be precise.
5. **Exact count of distinct values:**
The exact count of distinct values can be determined by considering the intervals:
\[
\left\lfloor\frac{n}{1}\right\rfloor, \left\lfloor\frac{n}{2}\right\rfloor, \ldots, \left\lfloor\frac{n}{\left\lfloor\sqrt{n}\right\rfloor}\right\rfloor.
\]
For each $k$ in this range, there is a corresponding interval of $i$ values. The total number of distinct values is given by:
\[
f(n) = \left\lfloor\sqrt{n}\right\rfloor + \left\lfloor\frac{n}{1}\right\rfloor - \left\lfloor\frac{n}{\left\lfloor\sqrt{n}\right\rfloor}\right\rfloor.
\]
6. **Summing $f(n)$ for $n$ from 1 to 9999:**
We need to compute:
\[
\sum_{n=1}^{9999} f(n).
\]
Using the formula for $f(n)$, we can sum the values directly or use an approximation to simplify the calculation.
The final answer is $\boxed{93324}$.
|
93324
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Amandine and Brennon play a turn-based game, with Amadine starting.
On their turn, a player must select a positive integer which cannot be represented as a sum of multiples of any of the previously selected numbers.
For example, if $3, 5$ have been selected so far, only $1, 2, 4, 7$ are available to be picked;
if only $3$ has been selected so far, all numbers not divisible by three are eligible.
A player loses immediately if they select the integer $1$.
Call a number $n$ [i]feminist[/i] if $\gcd(n, 6) = 1$ and if Amandine wins if she starts with $n$. Compute the sum of the [i]feminist[/i] numbers less than $40$.
[i]Proposed by Ashwin Sah[/i]
|
To solve this problem, we need to identify the numbers \( n \) less than 40 that are *feminist*, i.e., numbers for which \(\gcd(n, 6) = 1\) and Amandine wins if she starts with \( n \).
1. **Identify numbers \( n \) such that \(\gcd(n, 6) = 1\):**
- The numbers less than 40 that are coprime to 6 (i.e., not divisible by 2 or 3) are:
\[
1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37
\]
2. **Determine if Amandine wins starting with each of these numbers:**
- If Amandine starts with 1, she loses immediately because selecting 1 is a losing move.
- For other numbers, we need to analyze the game strategy:
- If Amandine starts with a prime number \( p \) (other than 2 or 3), Brennon cannot pick 1 immediately, and Amandine can force a win by strategic choices.
- If Amandine starts with a composite number \( n \) that is coprime to 6, we need to check if Brennon can force her into a losing position.
3. **Analyze the game for each number:**
- **Prime numbers less than 40 and coprime to 6:**
\[
5, 7, 11, 13, 17, 19, 23, 29, 31, 37
\]
- For these primes, Amandine can always win by forcing Brennon into a position where he has no winning move.
- **Composite numbers less than 40 and coprime to 6:**
\[
25, 35
\]
- If Amandine starts with 25, Brennon can respond with 5, forcing Amandine into a losing position.
- If Amandine starts with 35, Brennon can respond with 5, again forcing Amandine into a losing position.
4. **Sum the feminist numbers:**
- The feminist numbers are the prime numbers identified above:
\[
5, 7, 11, 13, 17, 19, 23, 29, 31, 37
\]
- Sum these numbers:
\[
5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 = 192
\]
The final answer is \(\boxed{192}\)
|
192
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p = 2017,$ a prime number. Let $N$ be the number of ordered triples $(a,b,c)$ of integers such that $1 \le a,b \le p(p-1)$ and $a^b-b^a=p \cdot c$. Find the remainder when $N$ is divided by $1000000.$
[i] Proposed by Evan Chen and Ashwin Sah [/i]
[i] Remark: [/i] The problem was initially proposed for $p = 3,$ and $1 \le a, b \le 30.$
|
To solve the problem, we need to find the number of ordered triples \((a, b, c)\) of integers such that \(1 \le a, b \le p(p-1)\) and \(a^b - b^a = p \cdot c\), where \(p = 2017\) is a prime number. We then need to find the remainder when this number is divided by \(1000000\).
1. **Initial Setup and Simplification:**
We start by considering the congruence \(a^b \equiv b^a \pmod{p}\). If \(p \mid a\), then \(p \mid b\) because \(a^b \equiv 0 \pmod{p}\) and \(b^a \equiv 0 \pmod{p}\). There are \(p-1\) multiples of \(p\) from \(1\) to \(p(p-1)\), so there are \((p-1)^2\) solutions for \(a\) and \(b\) when \(p \mid a\).
2. **Case \(p \nmid a\):**
Assume \(p \nmid a\). We need to count the number of \(b\)'s that work for a fixed \(a\) and sum over all \(a\) with \(1 \le a \le p(p-1)\) and \(p \nmid a\). We have \(a^b \equiv b^a \pmod{p}\). If we fix \(b \pmod{p-1}\), then \(a^b \pmod{p}\) is fixed by Fermat's Little Theorem (FLT). If we fix \(b \pmod{p}\), then \(b^a \pmod{p}\) is fixed. By the Chinese Remainder Theorem (CRT), pairing a residue class \(\pmod{p-1}\) with a residue class \(\pmod{p}\) gives exactly one residue class \(\pmod{p(p-1)}\).
3. **Primitive Root and Congruence:**
Let \(g\) be a primitive root \(\pmod{p}\). Then \(a \equiv g^x \pmod{p}\) and \(b \equiv g^y \pmod{p}\) for some integers \(x\) and \(y\). We need \(y\) to satisfy \(g^{xb} \equiv g^{ay} \pmod{p}\), or \(xb \equiv ay \pmod{p-1}\). This means \(p-1 \mid ay - bx\).
4. **Divisibility and Solutions:**
Let \(k = \gcd(a, p-1)\). We have \(k \mid p-1 \mid ay - bx\) and \(k \mid ay\), so \(k \mid bx\). This is equivalent to \(\frac{k}{\gcd(k, x)} \mid b\). Any residue class \(\pmod{p-1}\) that is "divisible" by \(\frac{k}{\gcd(k, x)}\) works. Conversely, if \(k \mid bx\), then the divisibility \(p-1 \mid ay - bx\) has exactly one solution \(\pmod{\frac{p-1}{k}}\) because \(\gcd(\frac{a}{k}, \frac{p-1}{k}) = 1\), which gives \(k\) solutions \(\pmod{p-1}\).
5. **Counting Solutions:**
For each \(a\), there are \((p-1) \gcd(k, x)\) total solutions for \(b\). Summing over all \(a\) with \(1 \le a \le p(p-1)\) and \(p \nmid a\), and adding \((p-1)^2\) for \(p \mid a\), we get:
\[
\sum_{1 \le a \le p(p-1), p \nmid a} (p-1) \gcd(a, \frac{p-1}{\text{ord}_p(a)})
\]
6. **Summing Over Divisors:**
Using the fact that for every divisor \(d\) of \(p-1\), there are exactly \(\phi(d)\) values of \(a\) with order \(d\), the sum becomes:
\[
(p-1) \sum_{d \mid p-1} \left( \phi(d) \sum_{1 \le a \le p-1} \gcd(a, \frac{p-1}{d}) \right)
\]
The inner sum is:
\[
\sum_{1 \le a \le \frac{p-1}{d}} \gcd(a, \frac{p-1}{d})
\]
added \(d\) times. So the total sum becomes:
\[
(p-1) \sum_{d \mid p-1} \left( \phi(d) d \sum_{1 \le a \le \frac{p-1}{d}} \gcd(a, \frac{p-1}{d}) \right)
\]
7. **Final Summation:**
Summing over \(\gcd(a, \frac{p-1}{d}) = e\), we get:
\[
(p-1) \sum_{d \mid p-1} \sum_{e \mid \frac{p-1}{d}} \left( \phi(d) e d \phi(\frac{p-1}{ed}) \right)
\]
Letting \(f = de\), we get:
\[
(p-1) \sum_{f \mid p-1} f \phi(\frac{p-1}{f}) \sum_{d \mid f} \phi(d)
\]
Using \(\sum_{d \mid n} \phi(d) = n\), the sum simplifies to:
\[
(p-1) \sum_{f \mid p-1} f^2 \phi(\frac{p-1}{f})
\]
8. **Computing the Sum:**
For \(p = 2017\), we compute:
\[
(2016) \sum_{f \mid 2016} f^2 \phi(\frac{2016}{f})
\]
This sum can be computed directly or using a computer program.
The final answer is \(\boxed{2016}\) (modulo \(1000000\)).
|
2016
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define $\left\lVert A-B \right\rVert = (x_A-x_B)^2+(y_A-y_B)^2$ for every two points $A = (x_A, y_A)$ and $B = (x_B, y_B)$ in the plane.
Let $S$ be the set of points $(x,y)$ in the plane for which $x,y \in \left\{ 0,1,\dots,100 \right\}$.
Find the number of functions $f : S \to S$ such that $\left\lVert A-B \right\rVert \equiv \left\lVert f(A)-f(B) \right\rVert \pmod{101}$ for any $A, B \in S$.
[i] Proposed by Victor Wang [/i]
|
1. **Define the problem in terms of vectors and modular arithmetic:**
We are given the distance function $\left\lVert A-B \right\rVert = (x_A-x_B)^2+(y_A-y_B)^2$ for points $A = (x_A, y_A)$ and $B = (x_B, y_B)$ in the plane. We need to find the number of functions $f : S \to S$ such that $\left\lVert A-B \right\rVert \equiv \left\lVert f(A)-f(B) \right\rVert \pmod{101}$ for any $A, B \in S$, where $S$ is the set of points $(x,y)$ with $x,y \in \{0,1,\dots,100\}$.
2. **Transform the problem into a simpler form:**
We can consider the points as vectors in $\mathbb{F}_{101}^2$, where $\mathbb{F}_{101}$ is the finite field with 101 elements. The condition $\left\lVert A-B \right\rVert \equiv \left\lVert f(A)-f(B) \right\rVert \pmod{101}$ translates to $(x_A - x_B)^2 + (y_A - y_B)^2 \equiv (x_{f(A)} - x_{f(B)})^2 + (y_{f(A)} - y_{f(B)})^2 \pmod{101}$.
3. **Normalize the function:**
Without loss of generality, assume $f(0,0) = (0,0)$. This normalization will be accounted for by multiplying the final count by $101^2$ (the number of possible choices for $f(0,0)$).
4. **Distance preservation implies linearity:**
The condition implies that $f$ preserves distances modulo 101. This suggests that $f$ is a linear transformation. Specifically, $f$ must be an isometry, meaning it preserves the dot product:
\[
\|A - B\|^2 = (A - B) \cdot (A - B) \equiv (f(A) - f(B)) \cdot (f(A) - f(B)) \pmod{101}.
\]
5. **Characterize the isometries:**
In $\mathbb{F}_{101}^2$, the isometries are given by orthogonal transformations. These can be represented by matrices $M$ such that $M^T M = I$, where $I$ is the identity matrix. The number of such orthogonal matrices is determined by the number of ways to choose an orthonormal basis in $\mathbb{F}_{101}^2$.
6. **Count the orthonormal bases:**
An orthonormal basis consists of two vectors $(a, b)$ and $(c, d)$ such that:
\[
a^2 + b^2 \equiv 1 \pmod{101}, \quad c^2 + d^2 \equiv 1 \pmod{101}, \quad \text{and} \quad ac + bd \equiv 0 \pmod{101}.
\]
The equation $a^2 + b^2 \equiv 1 \pmod{101}$ has $101$ solutions (each nonzero element in $\mathbb{F}_{101}$ can be paired with another to form a unit vector). For each choice of $(a, b)$, there are 2 choices for $(c, d)$ such that $(c, d)$ is orthogonal to $(a, b)$.
7. **Calculate the total number of functions:**
There are $101$ choices for $(a, b)$ and 2 choices for $(c, d)$ for each $(a, b)$. Thus, there are $2 \cdot 101$ orthonormal bases. Each orthonormal basis corresponds to an orthogonal transformation, and there are $101^2$ choices for the image of $(0,0)$.
8. **Final count:**
The total number of functions $f$ is:
\[
101^2 \cdot 2 \cdot 101 = 2040200.
\]
The final answer is $\boxed{2040200}$.
|
2040200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $W = \ldots x_{-1}x_0x_1x_2 \ldots$ be an infinite periodic word consisting of only the letters $a$ and $b$. The minimal period of $W$ is $2^{2016}$. Say that a word $U$ [i]appears[/i] in $W$ if there are indices $k \le \ell$ such that $U = x_kx_{k+1} \ldots x_{\ell}$. A word $U$ is called [i]special[/i] if $Ua, Ub, aU, bU$ all appear in $W$. (The empty word is considered special) You are given that there are no special words of length greater than 2015.
Let $N$ be the minimum possible number of special words. Find the remainder when $N$ is divided by $1000$.
[i]Proposed by Yang Liu[/i]
|
1. **Understanding the Problem:**
- We have an infinite periodic word \( W \) consisting of letters \( a \) and \( b \) with a minimal period of \( 2^{2016} \).
- A word \( U \) appears in \( W \) if there are indices \( k \le \ell \) such that \( U = x_k x_{k+1} \ldots x_{\ell} \).
- A word \( U \) is special if \( Ua, Ub, aU, bU \) all appear in \( W \).
- We are given that there are no special words of length greater than 2015.
- We need to find the minimum possible number of special words and the remainder when this number is divided by 1000.
2. **Key Insight:**
- If a word \( U \) of length \( n \) is special, then \( Ua, Ub, aU, bU \) must all appear in \( W \).
- Given that no special words of length greater than 2015 exist, we need to consider words of length up to 2015.
3. **Counting Special Words:**
- We need to count the number of special words of length \( \leq 2015 \).
- For each length \( n \) from 0 to 2015, the number of possible words of that length is \( 2^n \) (since each position can be either \( a \) or \( b \)).
4. **Summing Up:**
- The total number of special words is the sum of \( 2^n \) for \( n \) from 0 to 2015:
\[
\sum_{n=0}^{2015} 2^n = 2^0 + 2^1 + 2^2 + \ldots + 2^{2015}
\]
- This is a geometric series with the sum:
\[
\sum_{n=0}^{2015} 2^n = 2^{2016} - 1
\]
5. **Finding the Remainder:**
- We need to find the remainder when \( 2^{2016} - 1 \) is divided by 1000.
- Using properties of modular arithmetic, we can simplify \( 2^{2016} \mod 1000 \).
6. **Calculating \( 2^{2016} \mod 1000 \):**
- Using Euler's theorem, since \( \phi(1000) = 400 \):
\[
2^{400} \equiv 1 \mod 1000
\]
- Therefore:
\[
2^{2016} = 2^{5 \cdot 400 + 16} = (2^{400})^5 \cdot 2^{16} \equiv 1^5 \cdot 2^{16} \equiv 2^{16} \mod 1000
\]
- Calculating \( 2^{16} \mod 1000 \):
\[
2^{16} = 65536
\]
\[
65536 \mod 1000 = 536
\]
- Thus:
\[
2^{2016} \equiv 536 \mod 1000
\]
- Therefore:
\[
2^{2016} - 1 \equiv 536 - 1 \equiv 535 \mod 1000
\]
The final answer is \(\boxed{535}\)
|
535
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For integers $0 \le m,n \le 64$, let $\alpha(m,n)$ be the number of nonnegative integers $k$ for which $\left\lfloor m/2^k \right\rfloor$ and $\left\lfloor n/2^k \right\rfloor$ are both odd integers. Consider a $65 \times 65$ matrix $M$ whose $(i,j)$th entry (for $1 \le i, j \le 65$) is \[ (-1)^{\alpha(i-1, j-1)}. \] Compute the remainder when $\det M$ is divided by $1000$.
[i] Proposed by Evan Chen [/i]
|
1. **Define the matrix and function:**
Let \( A_n \) be the \( 2^n \times 2^n \) matrix with entries \( a_{ij} = (-1)^{\alpha(i-1, j-1)} \). Notice that \( \alpha(i, j) \) is the XOR function of \( i \) and \( j \) in binary, and we care about it modulo 2. Thus, \( a_{ij} = a_{i(j+2^{n-1})} = a_{(i+2^{n-1})j} = -a_{(i+2^{n-1})(j+2^{n-1})} \).
2. **Divide the matrix into quadrants:**
Divide the matrix \( A_n \) into quadrants. By adding row \( i + 2^{n-1} \) to row \( i \) for each \( 1 \le i \le 2^{n-1} \), we make quadrant II (the upper left) equal to twice its original value, and quadrant I (the upper right) is all zeros. By taking out the 2 from the first \( 2^{n-1} \) rows, we obtain a matrix that is the same as \( A_n \) except with quadrant I all zero.
3. **Simplify the matrix:**
Now add row \( i \) to row \( i + 2^{n-1} \) for each \( 1 \le i \le 2^{n-1} \) to make quadrant III all zeros, and factor out the \( (-1) \) from each of the last \( 2^{n-1} \) rows to obtain a matrix of the form \( A_{n-1} | 0; 0 | A_{n-1} \). This has determinant \( |A_{n-1}|^2 \), and so
\[
|A_n| = (-2)^{2^{n-1}} |A_{n-1}|^2.
\]
4. **Determine the determinant:**
If \( |A_n| = 2^{b_n} \) (since \( |A_1| = -2 \), by convention define \( b_1 = 1 \)), then we have
\[
b_n = 2^{n-1} + 2b_{n-1},
\]
so
\[
b_1 = 1, \quad b_2 = 4, \quad b_3 = 12, \quad b_4 = 32, \quad b_5 = 80, \dots, \quad b_n = n \cdot 2^{n-1}.
\]
5. **Calculate the determinant for \( n = 6 \):**
Hence, we want
\[
-2|A_6| = -2 \cdot 2^{6 \cdot 2^5} = -2^{193} \pmod{1000}.
\]
6. **Compute modulo 1000:**
\[
2^{193} \pmod{1000}.
\]
Using the Chinese Remainder Theorem:
- \( 2^{193} \equiv 0 \pmod{8} \)
- \( 2^{193} \equiv 2^{193 \pmod{100}} \equiv 2^{93} \pmod{125} \)
Since \( 2^{100} \equiv 1 \pmod{125} \), we have:
\[
2^{93} \equiv 2^{-7} \equiv 2^{118} \equiv 128^{-1} \equiv 3^{-1} \equiv 42 \pmod{125}.
\]
Combining these results using the Chinese Remainder Theorem:
\[
2^{193} \equiv 0 \pmod{8} \quad \text{and} \quad 2^{193} \equiv 42 \pmod{125}.
\]
Solving these congruences, we get:
\[
2^{193} \equiv 208 \pmod{1000}.
\]
Therefore,
\[
-2^{193} \equiv -208 \pmod{1000}.
\]
The final answer is \(\boxed{792}\).
|
792
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given vectors $v_1, \dots, v_n$ and the string $v_1v_2 \dots v_n$,
we consider valid expressions formed by inserting $n-1$ sets of balanced parentheses and $n-1$ binary products,
such that every product is surrounded by a parentheses and is one of the following forms:
1. A "normal product'' $ab$, which takes a pair of scalars and returns a scalar, or takes a scalar and vector (in any order) and returns a vector. \\
2. A "dot product'' $a \cdot b$, which takes in two vectors and returns a scalar. \\
3. A "cross product'' $a \times b$, which takes in two vectors and returns a vector. \\
An example of a [i]valid [/i] expression when $n=5$ is $(((v_1 \cdot v_2)v_3) \cdot (v_4 \times v_5))$, whose final output is a scalar. An example of an [i] invalid [/i] expression is $(((v_1 \times (v_2 \times v_3)) \times (v_4 \cdot v_5))$; even though every product is surrounded by parentheses, in the last step one tries to take the cross product of a vector and a scalar. \\
Denote by $T_n$ the number of valid expressions (with $T_1 = 1$), and let $R_n$
denote the remainder when $T_n$ is divided by $4$.
Compute $R_1 + R_2 + R_3 + \ldots + R_{1,000,000}$.
[i] Proposed by Ashwin Sah [/i]
|
To solve the problem, we need to compute the number of valid expressions formed by inserting \( n-1 \) sets of balanced parentheses and \( n-1 \) binary products into the string \( v_1v_2 \dots v_n \). We denote the number of valid expressions by \( T_n \) and the remainder when \( T_n \) is divided by 4 by \( R_n \). Our goal is to compute \( R_1 + R_2 + R_3 + \ldots + R_{1,000,000} \).
1. **Define the sequences \( S_n \) and \( V_n \)**:
- \( S_n \) is the number of valid sequences that return a scalar.
- \( V_n \) is the number of valid sequences that return a vector.
- By convention, we define \( S_0 = 0 \), \( S_1 = 0 \), \( V_0 = 0 \), and \( V_1 = 1 \).
2. **Recurrence relations**:
- The recurrence relations for \( S_n \) and \( V_n \) are given by:
\[
S_{n+1} = \sum_{k=1}^n V_k V_{n+1-k} + S_k S_{n+1-k}
\]
\[
V_{n+1} = \sum_{k=1}^n V_k V_{n+1-k} + 2S_k V_{n+1-k}
\]
3. **Generating functions**:
- Let \( P(x) \) and \( Q(x) \) be the generating functions for \( S_n \) and \( V_n \) respectively:
\[
P(x) = \sum_{k \ge 0} S_k x^k
\]
\[
Q(x) = \sum_{k \ge 0} V_k x^k
\]
- The generating functions satisfy the following equations:
\[
P(x) = Q(x)^2 + P(x)^2
\]
\[
Q(x) = x + Q(x)^2 + 2P(x)Q(x)
\]
4. **Solving the generating functions**:
- Solve for \( P(x) \) in terms of \( Q(x) \):
\[
P(x) = \frac{1}{2} - \frac{x}{2Q(x)} - \frac{Q(x)}{2}
\]
- Substitute this into the first equation to obtain a quadratic equation in \( Q(x)^2 \):
\[
Q(x)^2 \equiv -x + xC(x) \pmod{4}
\]
where \( C(x) \) is the Catalan function:
\[
C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}
\]
5. **Simplifying the generating function**:
- Add the two equations and solve the quadratic to find the generating function \( T(x) = P(x) + Q(x) \):
\[
T(x) = x C(x C(x)) \pmod{4}
\]
6. **Catalan numbers modulo 4**:
- The recurrence for Catalan numbers is:
\[
C_k = \sum_{i=0}^{k-1} C_i C_{k-1-i}
\]
- We can prove inductively that:
\[
C_{k-1} \equiv 1 \pmod{4}, \text{ if } k = 2^n
\]
\[
C_{k-1} \equiv 2 \pmod{4}, \text{ if } k = 2^n + 2^m, m \neq n
\]
\[
C_{k-1} \equiv 0 \pmod{4}, \text{ otherwise}
\]
7. **Summing the remainders**:
- After more computations, we find:
\[
R(x) \equiv 2 - (n-1)^2 \pmod{4}, \text{ if } k = 2^n
\]
\[
R(x) \equiv 2(m+1)(n+1) - 2 \pmod{4}, \text{ if } k = 2^n + 2^m, m \neq n
\]
\[
R(x) \equiv 0 \pmod{4}, \text{ otherwise}
\]
- By casework on whether \( k = 2^m \) or \( k = 2^m + 2^n \) in the sum of \( R_k \) with \( k = 1, 2, \dots, 1,000,000 \), we find the answer is:
\[
290 + 30 = 320
\]
The final answer is \(\boxed{320}\)
|
320
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Ryan is learning number theory. He reads about the [i]Möbius function[/i] $\mu : \mathbb N \to \mathbb Z$, defined by $\mu(1)=1$ and
\[ \mu(n) = -\sum_{\substack{d\mid n \\ d \neq n}} \mu(d) \]
for $n>1$ (here $\mathbb N$ is the set of positive integers).
However, Ryan doesn't like negative numbers, so he invents his own function: the [i]dubious function[/i] $\delta : \mathbb N \to \mathbb N$, defined by the relations $\delta(1)=1$ and
\[ \delta(n) = \sum_{\substack{d\mid n \\ d \neq n}} \delta(d) \]
for $n > 1$. Help Ryan determine the value of $1000p+q$, where $p,q$ are relatively prime positive integers satisfying
\[ \frac{p}{q}=\sum_{k=0}^{\infty} \frac{\delta(15^k)}{15^k}. \]
[i]Proposed by Michael Kural[/i]
|
1. **Define the Dubious Function**:
The dubious function $\delta : \mathbb{N} \to \mathbb{N}$ is defined by:
\[
\delta(1) = 1
\]
and for $n > 1$,
\[
\delta(n) = \sum_{\substack{d \mid n \\ d \neq n}} \delta(d).
\]
2. **Calculate $\delta(15^k)$**:
We need to find a pattern for $\delta(15^k)$. Note that $15 = 3 \times 5$, so we can use the multiplicative property of $\delta$:
\[
\delta(15^k) = \delta((3 \times 5)^k) = \delta(3^k \times 5^k).
\]
By induction, we can show that $\delta(3^k) = \delta(5^k) = 2^{k-1}$ for $k \geq 1$.
3. **Sum of Series**:
We need to compute the series:
\[
\sum_{k=0}^{\infty} \frac{\delta(15^k)}{15^k}.
\]
Using the pattern $\delta(15^k) = 2^{k-1}$ for $k \geq 1$ and $\delta(1) = 1$, we get:
\[
\sum_{k=0}^{\infty} \frac{\delta(15^k)}{15^k} = 1 + \sum_{k=1}^{\infty} \frac{2^{k-1}}{15^k}.
\]
4. **Simplify the Series**:
The series can be simplified as follows:
\[
\sum_{k=1}^{\infty} \frac{2^{k-1}}{15^k} = \frac{1}{15} \sum_{k=1}^{\infty} \left(\frac{2}{15}\right)^{k-1} = \frac{1}{15} \sum_{k=0}^{\infty} \left(\frac{2}{15}\right)^k.
\]
This is a geometric series with the first term $a = 1$ and common ratio $r = \frac{2}{15}$:
\[
\sum_{k=0}^{\infty} \left(\frac{2}{15}\right)^k = \frac{1}{1 - \frac{2}{15}} = \frac{15}{13}.
\]
Therefore,
\[
\sum_{k=1}^{\infty} \frac{2^{k-1}}{15^k} = \frac{1}{15} \cdot \frac{15}{13} = \frac{1}{13}.
\]
5. **Combine the Results**:
Adding the initial term $\delta(1) = 1$, we get:
\[
1 + \frac{1}{13} = \frac{14}{13}.
\]
6. **Determine $p$ and $q$**:
The fraction $\frac{p}{q} = \frac{14}{13}$, where $p = 14$ and $q = 13$ are relatively prime.
7. **Calculate $1000p + q$**:
\[
1000p + q = 1000 \times 14 + 13 = 14000 + 13 = 14013.
\]
The final answer is $\boxed{14013}$.
|
14013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A positive integer $n$ is called[i] bad [/i]if it cannot be expressed as the product of two distinct positive integers greater than $1$. Find the number of bad positive integers less than $100. $
[i]Proposed by Michael Ren[/i]
|
To determine the number of bad positive integers less than 100, we need to identify numbers that cannot be expressed as the product of two distinct positive integers greater than 1. These numbers include:
1. The number 1 itself.
2. Prime numbers, as they cannot be factored into two distinct positive integers greater than 1.
3. Squares of prime numbers, since they can only be factored into the prime number multiplied by itself, which does not meet the requirement of distinct factors.
Let's break down the solution step-by-step:
1. **Identify the prime numbers less than 100:**
- The prime numbers less than 100 are:
\[
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
\]
- There are 25 prime numbers less than 100.
2. **Identify the squares of prime numbers less than 100:**
- The squares of prime numbers less than 100 are:
\[
2^2 = 4, \quad 3^2 = 9, \quad 5^2 = 25, \quad 7^2 = 49
\]
- There are 4 such numbers.
3. **Count the number 1:**
- The number 1 is also a bad number.
4. **Sum the counts:**
- The total number of bad positive integers less than 100 is the sum of the number of primes, the number of squares of primes, and the number 1:
\[
25 \text{ (primes)} + 4 \text{ (squares of primes)} + 1 \text{ (the number 1)} = 30
\]
The final answer is \(\boxed{30}\).
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are $15$ (not necessarily distinct) integers chosen uniformly at random from the range from $0$ to $999$, inclusive. Yang then computes the sum of their units digits, while Michael computes the last three digits of their sum. The probability of them getting the same result is $\frac mn$ for relatively prime positive integers $m,n$. Find $100m+n$
[i]Proposed by Yannick Yao[/i]
|
1. Let's denote the 15 integers as \( x_i \) for \( i = 1, 2, \ldots, 15 \). Each \( x_i \) can be written in the form \( 10a_i + b_i \), where \( 0 \le a_i \le 99 \) and \( 0 \le b_i \le 9 \). Here, \( a_i \) represents the tens and higher place digits, and \( b_i \) represents the units digit of \( x_i \).
2. Yang computes the sum of the units digits of these integers, which is \( \sum_{i=1}^{15} b_i \).
3. Michael computes the last three digits of the sum of these integers, which is \( \sum_{i=1}^{15} x_i \mod 1000 \).
4. We need to find the probability that \( \sum_{i=1}^{15} b_i \equiv \sum_{i=1}^{15} x_i \pmod{1000} \).
5. Substituting \( x_i = 10a_i + b_i \) into the sum, we get:
\[
\sum_{i=1}^{15} x_i = \sum_{i=1}^{15} (10a_i + b_i) = 10 \sum_{i=1}^{15} a_i + \sum_{i=1}^{15} b_i
\]
6. We need:
\[
\sum_{i=1}^{15} b_i \equiv 10 \sum_{i=1}^{15} a_i + \sum_{i=1}^{15} b_i \pmod{1000}
\]
7. Simplifying, we get:
\[
0 \equiv 10 \sum_{i=1}^{15} a_i \pmod{1000}
\]
8. This implies:
\[
10 \sum_{i=1}^{15} a_i \equiv 0 \pmod{1000}
\]
9. Dividing both sides by 10, we get:
\[
\sum_{i=1}^{15} a_i \equiv 0 \pmod{100}
\]
10. Therefore, \( \sum_{i=1}^{15} a_i \) must be a multiple of 100. The probability of this happening is the probability that the sum of 15 random integers (each between 0 and 99) is a multiple of 100.
11. Given \( a_1, a_2, \ldots, a_{14} \), there is exactly one value of \( a_{15} \) that will make \( \sum_{i=1}^{15} a_i \) a multiple of 100. Since \( a_{15} \) can be any integer from 0 to 99, the probability is:
\[
\frac{1}{100}
\]
12. The probability of Yang and Michael getting the same result is \( \frac{1}{100} \).
13. The fraction \( \frac{1}{100} \) is already in simplest form, so \( m = 1 \) and \( n = 100 \).
14. Therefore, \( 100m + n = 100 \cdot 1 + 100 = 200 \).
The final answer is \( \boxed{200} \)
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A five-digit positive integer is called [i]$k$-phobic[/i] if no matter how one chooses to alter at most four of the digits, the resulting number (after disregarding any leading zeroes) will not be a multiple of $k$. Find the smallest positive integer value of $k$ such that there exists a $k$-phobic number.
[i]Proposed by Yannick Yao[/i]
|
1. **Claim**: The smallest \( k \) such that there exists a \( k \)-phobic number is \( k = 11112 \). We will verify this by considering the number \( N = 99951 \) and showing that it is \( 11112 \)-phobic.
2. **Verification**: We need to check that no matter how we alter at most four digits of \( 99951 \), the resulting number is not a multiple of \( 11112 \). The multiples of \( 11112 \) under \( 10^5 \) are:
\[
11112, 22224, 33336, 44448, 55560, 66672, 77784, 88896
\]
We observe that \( 99951 \) shares no digits in the same positions with any of these multiples. Therefore, altering at most four digits of \( 99951 \) will not result in any of these multiples.
3. **Smaller \( k \) values**: Now, we need to show that for any \( k \leq 11111 \), there does not exist a \( k \)-phobic number. Consider a five-digit number \( N \) such that \( 10^4 \leq N < 10^5 \).
4. **Case \( k < 10^4 \)**: If \( k < 10^4 \), there exists a multiple of \( k \) with five digits and the same leftmost digit as \( N \). This is because the difference between consecutive multiples of \( k \) is \( k < 10^4 \). Therefore, we can change the other four digits of \( N \) to get that multiple, implying \( N \) is not \( k \)-phobic.
5. **Case \( 10^4 \leq k \leq 11111 \)**: If \( 10^4 \leq k \leq 11111 \), we consider the ranges:
\[
10000 \leq k < 20000, \quad 20000 \leq 2k < 30000, \quad \ldots, \quad 90000 \leq 9k < 100000
\]
For each range, there exists a multiple of \( k \) with the same leftmost digit as \( N \). Thus, we can change the other four digits of \( N \) to get that multiple, implying \( N \) is not \( k \)-phobic.
6. **Conclusion**: Since \( N \) is never \( k \)-phobic for \( k \leq 11111 \), the smallest \( k \) such that there exists a \( k \)-phobic number is \( 11112 \).
The final answer is \( \boxed{11112} \)
|
11112
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABCDEF$ be a regular hexagon with side length 10 inscribed in a circle $\omega$. $X$, $Y$, and $Z$ are points on $\omega$ such that $X$ is on minor arc $AB$, $Y$ is on minor arc $CD$, and $Z$ is on minor arc $EF$, where $X$ may coincide with $A$ or $B$ (and similarly for $Y$ and $Z$). Compute the square of the smallest possible area of $XYZ$.
[i]Proposed by Michael Ren[/i]
|
1. **Understanding the Problem:**
We are given a regular hexagon \(ABCDEF\) with side length 10, inscribed in a circle \(\omega\). Points \(X\), \(Y\), and \(Z\) are on the minor arcs \(AB\), \(CD\), and \(EF\) respectively. We need to find the square of the smallest possible area of triangle \(XYZ\).
2. **Fixing Points on Vertices:**
By symmetry and continuity arguments, the minimum area triangle \(XYZ\) will occur when \(X\), \(Y\), and \(Z\) are at the vertices of the hexagon. This is because moving \(Z\) to the closest vertex on arc \(EF\) minimizes the distance to the line segment \(XY\).
3. **Possible Triangles:**
The vertices of the hexagon can form two types of triangles:
- Equilateral triangle (e.g., \(ACE\))
- Right triangle (e.g., \(ACF\))
4. **Calculating the Area of Triangle \(ACF\):**
- \(AF = 10\) (side length of the hexagon)
- \(FC = 20\) (two side lengths of the hexagon)
- \(AC\) is the diagonal of the hexagon, which can be calculated as follows:
\[
AC = \sqrt{AF^2 + FC^2} = \sqrt{10^2 + 20^2} = \sqrt{100 + 400} = \sqrt{500} = 10\sqrt{5}
\]
5. **Area of Triangle \(ACF\):**
- Using Heron's formula:
\[
s = \frac{AF + FC + AC}{2} = \frac{10 + 20 + 10\sqrt{5}}{2} = 15 + 5\sqrt{5}
\]
\[
\text{Area} = \sqrt{s(s - AF)(s - FC)(s - AC)} = \sqrt{(15 + 5\sqrt{5})(5 + 5\sqrt{5})(-5 + 5\sqrt{5})(5 - 5\sqrt{5})}
\]
Simplifying the terms inside the square root:
\[
\text{Area} = \sqrt{(15 + 5\sqrt{5})(5 + 5\sqrt{5})(5\sqrt{5} - 5)(5\sqrt{5} - 5)}
\]
This calculation is complex, so we use the simpler method of calculating the area directly using the base and height:
\[
\text{Area} = \frac{1}{2} \times AF \times FC \times \sin(60^\circ) = \frac{1}{2} \times 10 \times 20 \times \frac{\sqrt{3}}{2} = 50\sqrt{3}
\]
6. **Square of the Area:**
\[
\left(50\sqrt{3}\right)^2 = 2500 \times 3 = 7500
\]
The final answer is \(\boxed{7500}\).
|
7500
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Kevin is trying to solve an economics question which has six steps. At each step, he has a probability $p$ of making a sign error. Let $q$ be the probability that Kevin makes an even number of sign errors (thus answering the question correctly!). For how many values of $0 \le p \le 1$ is it true that $p+q=1$?
[i]Proposed by Evan Chen[/i]
|
1. We start by defining the probability \( q \) that Kevin makes an even number of sign errors. Since there are six steps, the possible even numbers of errors are 0, 2, 4, and 6. The probability of making exactly \( k \) errors out of 6 steps, where \( k \) is even, is given by the binomial distribution:
\[
q = \sum_{k \text{ even}} \binom{6}{k} p^k (1-p)^{6-k}
\]
Therefore, we have:
\[
q = \binom{6}{0} p^0 (1-p)^6 + \binom{6}{2} p^2 (1-p)^4 + \binom{6}{4} p^4 (1-p)^2 + \binom{6}{6} p^6 (1-p)^0
\]
2. Simplifying the binomial coefficients and terms, we get:
\[
q = (1-p)^6 + 15 p^2 (1-p)^4 + 15 p^4 (1-p)^2 + p^6
\]
3. We can use the binomial theorem and symmetry properties to simplify this expression. Notice that:
\[
q = \frac{1}{2} \left( (p + (1-p))^6 + (p - (1-p))^6 \right)
\]
Since \( p + (1-p) = 1 \) and \( p - (1-p) = 2p - 1 \), we have:
\[
q = \frac{1}{2} \left( 1^6 + (2p-1)^6 \right) = \frac{1}{2} \left( 1 + (2p-1)^6 \right)
\]
4. Given that \( p + q = 1 \), we substitute \( q \) into the equation:
\[
p + \frac{1}{2} \left( 1 + (2p-1)^6 \right) = 1
\]
5. Simplifying the equation, we get:
\[
p + \frac{1}{2} + \frac{1}{2} (2p-1)^6 = 1
\]
\[
p + \frac{1}{2} (2p-1)^6 = \frac{1}{2}
\]
\[
2p + (2p-1)^6 = 1
\]
6. We now solve the equation \( 2p + (2p-1)^6 = 1 \). Consider the possible values of \( 2p-1 \):
- If \( 2p-1 = 0 \), then \( p = \frac{1}{2} \).
- If \( (2p-1)^6 = 1 - 2p \), then \( (2p-1)^6 = 1 - 2p \).
7. For \( (2p-1)^6 = 1 - 2p \):
- If \( 2p-1 = 1 \), then \( p = 1 \), but this does not satisfy \( 1 + 0 = 1 \).
- If \( 2p-1 = -1 \), then \( p = 0 \), and this satisfies \( 0 + 1 = 1 \).
8. Therefore, the possible values of \( p \) are \( 0 \) and \( \frac{1}{2} \).
The final answer is \( \boxed{2} \).
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a_1, a_2, a_3, a_4$ be integers with distinct absolute values. In the coordinate plane, let $A_1=(a_1,a_1^2)$, $A_2=(a_2,a_2^2)$, $A_3=(a_3,a_3^2)$ and $A_4=(a_4,a_4^2)$. Assume that lines $A_1A_2$ and $A_3A_4$ intersect on the $y$-axis at an acute angle of $\theta$. The maximum possible value for $\tan \theta$ can be expressed in the form $\dfrac mn$ for relatively prime positive integers $m$ and $n$. Find $100m+n$.
[i]Proposed by James Lin[/i]
|
1. **Given Information and Setup:**
- We have integers \(a_1, a_2, a_3, a_4\) with distinct absolute values.
- Points in the coordinate plane are \(A_1 = (a_1, a_1^2)\), \(A_2 = (a_2, a_2^2)\), \(A_3 = (a_3, a_3^2)\), and \(A_4 = (a_4, a_4^2)\).
- Lines \(A_1A_2\) and \(A_3A_4\) intersect on the \(y\)-axis at an acute angle \(\theta\).
- We need to find the maximum possible value for \(\tan \theta\) expressed as \(\frac{m}{n}\) for relatively prime positive integers \(m\) and \(n\), and then find \(100m + n\).
2. **Slope Calculation:**
- The slope of line \(A_1A_2\) is:
\[
m_1 = \frac{a_2^2 - a_1^2}{a_2 - a_1} = a_2 + a_1
\]
- The slope of line \(A_3A_4\) is:
\[
m_2 = \frac{a_4^2 - a_3^2}{a_4 - a_3} = a_4 + a_3
\]
3. **Intersection on the \(y\)-axis:**
- Since the lines intersect on the \(y\)-axis, their \(x\)-coordinates at the intersection point are zero. Thus, the \(y\)-intercepts of the lines are:
\[
c_1 = a_1^2 \quad \text{and} \quad c_2 = a_3^2
\]
4. **Angle Between Lines:**
- The tangent of the angle \(\theta\) between the two lines is given by:
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{(a_1 + a_2) - (a_3 + a_4)}{1 + (a_1 + a_2)(a_3 + a_4)} \right|
\]
5. **Maximizing \(\tan \theta\):**
- Let \(x = a_1 + a_2\) and \(y = a_3 + a_4\). We need to maximize:
\[
f(x, y) = \left| \frac{x - y}{1 + xy} \right|
\]
6. **Case Analysis:**
- **Case 1: \(x, y > 0\)**
- Here, \(x, y \geq 1\) since they are integers. In this case, \(|f(x, y)| < 1\), which cannot exceed our current maximum.
- **Case 2: \(x > 0, y < 0\)**
- The numerator is positive, and the denominator is negative, so \(|f(x, y)|\) cannot be our maximum.
- **Case 3: \(x < 0, y > 0\)**
- Let \(x = -u\) and \(y = v\). We want to maximize:
\[
\frac{u + v}{uv - 1}
\]
- We need to check when:
\[
\frac{u + v}{uv - 1} > \frac{3}{2} \implies 3(uv - 1) < 2(u + v) \implies 3uv - 3 < 2u + 2v \implies 3uv - 2u - 2v < 3
\]
- If \(u, v > 1\), then it's impossible because \(3uv - 2u - 2v \geq 3\) is false. Thus, we can let \(v = 1\) and need \(u < 5\).
7. **Checking Specific Values:**
- For \(u = 4\) and \(v = 1\):
- We have \(a_1 = -6, a_2 = 2, a_3 = 4, a_4 = -3\).
- This gives:
\[
\tan \theta = \left| \frac{(-6 + 2) - (4 - 3)}{1 + (-6 + 2)(4 - 3)} \right| = \left| \frac{-4 - 1}{1 - 4} \right| = \left| \frac{-5}{-3} \right| = \frac{5}{3}
\]
The final answer is \(\boxed{503}\).
|
503
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Lunasa, Merlin, and Lyrica each has an instrument. We know the following about the prices of their instruments:
(a) If we raise the price of Lunasa's violin by $50\%$ and decrease the price of Merlin's trumpet by $50\%$, the violin will be $\$50$ more expensive than the trumpet;
(b) If we raise the price of Merlin's trumpet by $50\%$ and decrease the price of Lyrica's piano by $50\%$, the trumpet will be $\$50$ more expensive than the piano.
Given these conditions only, there exist integers $m$ and $n$ such that if we raise the price of Lunasa's violin by $m\%$ and decrease the price of Lyrica's piano by $m\%$, the violin must be exactly $\$n$ more expensive than the piano. Find $100m+n$.
[i]Proposed by Yannick Yao[/i]
|
1. Let \( v \) be the price of Lunasa's violin, \( t \) be the price of Merlin's trumpet, and \( p \) be the price of Lyrica's piano.
2. From condition (a), if we raise the price of Lunasa's violin by \( 50\% \) and decrease the price of Merlin's trumpet by \( 50\% \), the violin will be \$50 more expensive than the trumpet. This can be written as:
\[
1.5v = 0.5t + 50
\]
Simplifying, we get:
\[
3v = t + 100 \quad \text{(Equation 1)}
\]
3. From condition (b), if we raise the price of Merlin's trumpet by \( 50\% \) and decrease the price of Lyrica's piano by \( 50\% \), the trumpet will be \$50 more expensive than the piano. This can be written as:
\[
1.5t = 0.5p + 50
\]
Simplifying, we get:
\[
3t = p + 100 \quad \text{(Equation 2)}
\]
4. We need to find integers \( m \) and \( n \) such that if we raise the price of Lunasa's violin by \( m\% \) and decrease the price of Lyrica's piano by \( m\% \), the violin will be exactly \( \$n \) more expensive than the piano. This can be written as:
\[
\frac{(100 + m)v}{100} = n + \frac{(100 - m)p}{100}
\]
Simplifying, we get:
\[
(100 + m)v = 100n + (100 - m)p \quad \text{(Equation 3)}
\]
5. From Equations 1 and 2, we can express \( t \) and \( p \) in terms of \( v \):
\[
t = 3v - 100
\]
\[
p = 3t - 100 = 3(3v - 100) - 100 = 9v - 400
\]
6. Substitute \( p = 9v - 400 \) into Equation 3:
\[
(100 + m)v = 100n + (100 - m)(9v - 400)
\]
Simplifying, we get:
\[
(100 + m)v = 100n + 900v - 40000 - 9mv + 400m
\]
\[
(100 + m)v + 9mv = 100n + 900v - 40000 + 400m
\]
\[
100v + mv + 9mv = 100n + 900v - 40000 + 400m
\]
\[
100v + 10mv = 100n + 900v - 40000 + 400m
\]
\[
100v + 10mv - 900v = 100n - 40000 + 400m
\]
\[
-800v + 10mv = 100n - 40000 + 400m
\]
\[
10mv - 800v = 100n - 40000 + 400m
\]
\[
v(10m - 800) = 100n - 40000 + 400m
\]
7. To make the equation simpler, we can divide both sides by 100:
\[
v(10m - 800) = n - 400 + 4m
\]
8. We need to find \( m \) such that the left-hand side is a multiple of \( v \). From the equation, we can see that:
\[
\frac{10m - 800}{4m - 400} = 9
\]
Solving for \( m \):
\[
10m - 800 = 9(4m - 400)
\]
\[
10m - 800 = 36m - 3600
\]
\[
10m - 36m = -3600 + 800
\]
\[
-26m = -2800
\]
\[
m = \frac{2800}{26} = 80
\]
9. Substitute \( m = 80 \) back into the equation to find \( n \):
\[
v(10 \cdot 80 - 800) = n - 400 + 4 \cdot 80
\]
\[
v(800 - 800) = n - 400 + 320
\]
\[
0 = n - 80
\]
\[
n = 80
\]
10. Therefore, \( 100m + n = 100 \cdot 80 + 80 = 8080 \).
The final answer is \(\boxed{8080}\).
|
8080
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be the set of all positive integers between 1 and 2017, inclusive. Suppose that the least common multiple of all elements in $S$ is $L$. Find the number of elements in $S$ that do not divide $\frac{L}{2016}$.
[i]Proposed by Yannick Yao[/i]
|
1. **Determine the least common multiple (LCM) of all elements in \( S \):**
- The set \( S \) contains all positive integers from 1 to 2017.
- The LCM of all elements in \( S \) is the product of the highest powers of all primes less than or equal to 2017.
- The prime factorization of 2016 is \( 2016 = 2^5 \cdot 3^2 \cdot 7 \).
2. **Express \( L \) in terms of its prime factors:**
- The LCM of all numbers from 1 to 2017 includes the highest powers of all primes up to 2017.
- For the primes 2, 3, and 7, the highest powers within the range are \( 2^{10} \), \( 3^6 \), and \( 7^3 \).
- Therefore, \( L = 2^{10} \cdot 3^6 \cdot 7^3 \cdot L_0 \), where \( L_0 \) includes the highest powers of all other primes up to 2017.
3. **Calculate \( \frac{L}{2016} \):**
- Given \( 2016 = 2^5 \cdot 3^2 \cdot 7 \), we have:
\[
\frac{L}{2016} = \frac{2^{10} \cdot 3^6 \cdot 7^3 \cdot L_0}{2^5 \cdot 3^2 \cdot 7} = 2^{10-5} \cdot 3^{6-2} \cdot 7^{3-1} \cdot L_0 = 2^5 \cdot 3^4 \cdot 7^2 \cdot L_0
\]
4. **Identify numbers in \( S \) that do not divide \( \frac{L}{2016} \):**
- A number \( n \) in \( S \) does not divide \( \frac{L}{2016} \) if it contains a prime factor raised to a power higher than those in \( \frac{L}{2016} \).
- Specifically, \( n \) must be divisible by \( 2^6 \), \( 3^5 \), or \( 7^3 \).
5. **Count the numbers divisible by \( 2^6 \), \( 3^5 \), or \( 7^3 \):**
- Calculate the number of multiples of \( 2^6 = 64 \) in \( S \):
\[
\left\lfloor \frac{2017}{64} \right\rfloor = 31
\]
- Calculate the number of multiples of \( 3^5 = 243 \) in \( S \):
\[
\left\lfloor \frac{2017}{243} \right\rfloor = 8
\]
- Calculate the number of multiples of \( 7^3 = 343 \) in \( S \):
\[
\left\lfloor \frac{2017}{343} \right\rfloor = 5
\]
6. **Sum the counts:**
- Since the product of any two of \( 2^6 \), \( 3^5 \), and \( 7^3 \) is greater than 2017, there is no overlap in the counts.
- Therefore, the total number of elements in \( S \) that do not divide \( \frac{L}{2016} \) is:
\[
31 + 8 + 5 = 44
\]
The final answer is \(\boxed{44}\)
|
44
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
When Cirno walks into her perfect math class today, she sees a polynomial $P(x)=1$ (of degree 0) on the blackboard. As her teacher explains, for her pop quiz today, she will have to perform one of the two actions every minute:
(a) Add a monomial to $P(x)$ so that the degree of $P$ increases by 1 and $P$ remains monic;
(b) Replace the current polynomial $P(x)$ by $P(x+1)$. For example, if the current polynomial is $x^2+2x+3$, then she will change it to $(x+1)^2+2(x+1)+3=x^2+4x+6$.
Her score for the pop quiz is the sum of coefficients of the polynomial at the end of 9 minutes. Given that Cirno (miraculously) doesn't make any mistakes in performing the actions, what is the maximum score that she can get?
[i]Proposed by Yannick Yao[/i]
|
1. **Initial Polynomial**: The initial polynomial is \( P(x) = 1 \).
2. **Action (a) - Adding Monomials**: Each time we add a monomial to \( P(x) \), the degree of \( P \) increases by 1, and \( P \) remains monic. Let's denote the polynomial after \( k \) additions as \( P_k(x) \). After \( k \) additions, the polynomial will be:
\[
P_k(x) = x^k + a_{k-1}x^{k-1} + \cdots + a_1x + a_0
\]
where \( a_i \) are coefficients determined by the previous operations.
3. **Action (b) - Replacing \( P(x) \) by \( P(x+1) \)**: This operation shifts the polynomial by 1 unit to the left. For example, if \( P(x) = x^2 + 2x + 3 \), then \( P(x+1) = (x+1)^2 + 2(x+1) + 3 = x^2 + 4x + 6 \).
4. **Strategy**: To maximize the sum of the coefficients, we need to consider the effect of each operation. The sum of the coefficients of a polynomial \( P(x) \) is equal to \( P(1) \). Therefore, we need to maximize \( P(1) \) after 9 minutes.
5. **Optimal Sequence of Operations**:
- Perform action (a) six times to increase the degree of the polynomial to 6.
- Perform action (b) three times to shift the polynomial.
6. **Detailed Calculation**:
- After 6 additions, the polynomial is \( P_6(x) = x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \). Since we are only adding monomials, the polynomial remains \( P_6(x) = x^6 \).
- Now, perform action (b) three times:
\[
P(x) = x^6 \rightarrow P(x+1) = (x+1)^6
\]
\[
(x+1)^6 = \sum_{k=0}^{6} \binom{6}{k} x^k
\]
\[
= x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1
\]
- The sum of the coefficients of \( (x+1)^6 \) is:
\[
\sum_{k=0}^{6} \binom{6}{k} = 2^6 = 64
\]
The final answer is \( \boxed{64} \)
|
64
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Alice has an isosceles triangle $M_0N_0P$, where $M_0P=N_0P$ and $\angle M_0PN_0=\alpha^{\circ}$. (The angle is measured in degrees.) Given a triangle $M_iN_jP$ for nonnegative integers $i$ and $j$, Alice may perform one of two [i]elongations[/i]:
a) an $M$-[i]elongation[/i], where she extends ray $\overrightarrow{PM_i}$ to a point $M_{i+1}$ where $M_iM_{i+1}=M_iN_j$ and removes the point $M_i$.
b) an $N$-[i]elongation[/i], where she extends ray $\overrightarrow{PN_j}$ to a point $N_{j+1}$ where $N_jN_{j+1}=M_iN_j$ and removes the point $N_j$.
After a series of $5$ elongations, $k$ of which were $M$-elongations, Alice finds that triangle $M_kN_{5-k}P$ is an isosceles triangle. Given that $10\alpha$ is an integer, compute $10\alpha$.
[i]Proposed by Yannick Yao[/i]
|
1. **Define Variables and Initial Setup:**
Let \( u = 90^\circ - \frac{\alpha}{2} \) and let \( \alpha = \frac{x}{10} \) where \( x \) is an integer. We need to find \( 10\alpha \).
2. **Angle Transformation:**
Let \( A_n \) be the angle \( \angle PM_iN_j \) where \( M_i \) and \( N_j \) are the furthest points on the \( n \)-th elongation. Observe that if on the \( n \)-th elongation, we did an \( M \)-elongation, then:
\[
A_n = \frac{A_{n-1}}{2}
\]
Otherwise,
\[
A_n = u + \frac{A_{n-1}}{2}
\]
3. **General Form of \( A_n \):**
This implies that after 5 elongations, \( A_n \) will be in the form of:
\[
A_n = u \left( \frac{1}{2^5} + \frac{a_1}{2^4} + \frac{a_2}{2^3} + \frac{a_3}{2^2} + \frac{a_4}{2} + a_5 \right)
\]
where \( a_1, a_2, a_3, a_4, a_5 \in \{0, 1\} \). Let \( t = 2^4a_5 + 2^3a_4 + 2^2a_3 + 2a_2 + a_1 \), then:
\[
A_n = \frac{u}{32} (2t + 1)
\]
Notice that \( t \) is written in binary so \( 0 \leq t \leq 31 \).
4. **Isosceles Triangle Condition:**
Since \( M_kN_{5-k}P \) is an isosceles triangle, we have a few cases to consider.
5. **Case 1: \( \angle P \) is the Vertex:**
This implies that:
\[
\frac{u}{32} (2t+1) = u
\]
which is obviously not true since \( \frac{u}{32} (2t+1) \neq u \) for any \( t \).
6. **Case 2: \( PM_0 \) is the Base of the Triangle:**
Thus, \( \angle M \) and \( \alpha \) are both base angles. This implies that:
\[
\frac{u}{32}(2t+1) = \alpha
\]
Substituting \( u = 90^\circ - \frac{\alpha}{2} \) and \( \alpha = \frac{x}{10} \):
\[
(2t+1)u = 32 \alpha
\]
\[
(2t+1) \left( 90^\circ - \frac{\alpha}{2} \right) = 32 \alpha
\]
\[
(2t+1) \left( 90 - \frac{x}{20} \right) = \frac{16x}{5}
\]
\[
(2t+1) (1800 - x) = 64x
\]
7. **Solving for \( x \):**
Simplifying the equation:
\[
3600t + 1800 - x(2t+1) = 64x
\]
\[
3600t + 1800 = x(2t + 65)
\]
\[
x = \frac{3600t + 1800}{2t + 65}
\]
We need \( x \) to be an integer. Testing values of \( t \) from 0 to 31, we find that \( t = 11 \) works:
\[
x = \frac{3600 \cdot 11 + 1800}{2 \cdot 11 + 65} = \frac{39600}{150} = 264
\]
The final answer is \( \boxed{264} \).
|
264
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Determine the number of ordered quintuples $(a,b,c,d,e)$ of integers with $0\leq a<$ $b<$ $c<$ $d<$ $e\leq 30$ for which there exist polynomials $Q(x)$ and $R(x)$ with integer coefficients such that \[x^a+x^b+x^c+x^d+x^e=Q(x)(x^5+x^4+x^2+x+1)+2R(x).\]
[i]Proposed by Michael Ren[/i]
|
1. Let \( P(x) = x^5 + x^4 + x^2 + x + 1 \). We are working over the finite field \(\frac{\mathbb{Z}}{2\mathbb{Z}}[X]\) modulo \(P(x)\). Since \(P(x)\) is irreducible over \(\mathbb{Z}/2\mathbb{Z}\), the quotient ring \(\frac{\mathbb{Z}}{2\mathbb{Z}}[X]}{P(x)\mathbb{Z}/2\mathbb{Z}[X]}\) forms a finite field of size \(2^5 = 32\).
2. In this field, \(x\) is a generator, meaning every nonzero element can be written as a power of \(x\). The problem then reduces to finding all sets \(\{a, b, c, d, e\}\) of distinct nonzero elements such that their sum is zero in this field.
3. We count the number of 5-tuples \((a, b, c, d, e)\) that satisfy \(a + b + c + d + e = 0\) and then divide by \(5!\) to account for the ordering.
4. After choosing \(a, b, c, d\), we set \(e = a + b + c + d\). This works as long as \(e\) is not equal to \(0, a, b, c,\) or \(d\). Therefore, we need to count the number of quadruples \((a, b, c, d)\) such that \(a + b + c + d \neq 0, a, b, c, d\).
5. If \(a + b + c + d = a\), then \(b + c + d = 0\). Similarly, we need to ensure \(a + b + c + d \neq b, c, d\). Thus, we count the number of quadruples that sum to zero and subtract these from the total.
6. The total number of 5-tuples is \(31 \cdot 30 \cdot 29 \cdot 28\). We subtract the number of invalid quadruples and triples:
\[
\#(\text{5-tuples that sum to 0}) = 31 \cdot 30 \cdot 29 \cdot 28 - \#(\text{quadruples that sum to 0}) - 4 \cdot 28 \cdot \#(\text{triples that sum to 0})
\]
7. The number of quadruples that sum to zero is given by:
\[
\#(\text{quadruples that sum to 0}) = 31 \cdot 30 \cdot 29 - \#(\text{triples that sum to 0})
\]
8. The number of triples that sum to zero is:
\[
\#(\text{triples that sum to 0}) = 31 \cdot 30
\]
9. Substituting these values back, we get:
\[
\#(\text{5-tuples that sum to 0}) = 31 \cdot 30 \cdot 29 \cdot 28 - (31 \cdot 30 \cdot 29 - 31 \cdot 30) - 4 \cdot 28 \cdot (31 \cdot 30)
\]
10. Simplifying, we find:
\[
\#(\text{5-tuples that sum to 0}) = 31 \cdot 30 \cdot 28 \cdot 24
\]
11. Dividing by \(5!\) to account for the ordering, we obtain:
\[
\frac{31 \cdot 30 \cdot 28 \cdot 24}{120} = 31 \cdot 6 \cdot 28 = 5208
\]
The final answer is \(\boxed{5208}\)
|
5208
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
On a real number line, the points $1, 2, 3, \dots, 11$ are marked. A grasshopper starts at point $1$, then jumps to each of the other $10$ marked points in some order so that no point is visited twice, before returning to point $1$. The maximal length that he could have jumped in total is $L$, and there are $N$ possible ways to achieve this maximum. Compute $L+N$.
[i]Proposed by Yannick Yao[/i]
|
1. **Understanding the Problem:**
- The grasshopper starts at point \(1\) and must visit each of the points \(2, 3, \ldots, 11\) exactly once before returning to point \(1\).
- We need to maximize the total distance jumped by the grasshopper and find the number of ways to achieve this maximum distance.
2. **Optimal Jumping Strategy:**
- To maximize the total distance, the grasshopper should jump as far as possible in each move.
- This can be achieved by making the grasshopper jump in a zigzag pattern, alternating between the largest and smallest remaining points.
3. **Calculating the Maximum Distance \(L\):**
- Consider the points \(1, 2, 3, \ldots, 11\).
- The optimal strategy involves jumping from \(1\) to \(11\), then to \(2\), then to \(10\), and so on.
- The sequence of jumps would be: \(1 \rightarrow 11 \rightarrow 2 \rightarrow 10 \rightarrow 3 \rightarrow 9 \rightarrow 4 \rightarrow 8 \rightarrow 5 \rightarrow 7 \rightarrow 6 \rightarrow 1\).
4. **Distance Calculation:**
- The total distance is the sum of the absolute differences between consecutive points in the sequence.
- \[
\begin{align*}
\text{Total distance} &= |1 - 11| + |11 - 2| + |2 - 10| + |10 - 3| + |3 - 9| + |9 - 4| + |4 - 8| + |8 - 5| + |5 - 7| + |7 - 6| + |6 - 1| \\
&= 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 5 \\
&= 60.
\end{align*}
\]
- Therefore, the maximum total distance \(L\) is \(60\).
5. **Counting the Number of Ways \(N\):**
- The grasshopper can choose any of the \(5!\) permutations for the points \([2, 3, 4, 5, 6]\) and any of the \(5!\) permutations for the points \([7, 8, 9, 10, 11]\).
- Additionally, there are \(10\) possible positions where the grasshopper can jump to point \(6\).
- Thus, the total number of ways \(N\) is:
\[
N = 5! \times 5! \times 10 = 120 \times 120 \times 10 = 144000.
\]
6. **Summing \(L\) and \(N\):**
- The final value is \(L + N = 60 + 144000 = 144060\).
The final answer is \(\boxed{144060}\).
|
144060
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\phi(n)$ denote the number of positive integers less than or equal to $n$ which are relatively prime to $n$. Over all integers $1\le n \le 100$, find the maximum value of $\phi(n^2+2n)-\phi(n^2)$.
[i]Proposed by Vincent Huang[/i]
|
1. We start by analyzing the given function $\phi(n)$, which denotes the number of positive integers less than or equal to $n$ that are relatively prime to $n$. We need to find the maximum value of $\phi(n^2 + 2n) - \phi(n^2)$ for $1 \leq n \leq 100$.
2. First, recall that $\phi(n^2) = n \phi(n)$ for any integer $n$. This is because the Euler's totient function $\phi$ is multiplicative and $n^2$ can be factored as $n \cdot n$.
3. Consider the expression $\phi(n^2 + 2n) - \phi(n^2)$. We can rewrite $n^2 + 2n$ as $n(n + 2)$. Thus, the expression becomes:
\[
\phi(n(n + 2)) - \phi(n^2)
\]
Using the multiplicative property of $\phi$, we have:
\[
\phi(n(n + 2)) = \phi(n) \phi(n + 2) \cdot \frac{n(n + 2)}{\gcd(n, n + 2)}
\]
Since $\gcd(n, n + 2) = 1$ for odd $n$ and $\gcd(n, n + 2) = 2$ for even $n$, we need to consider these cases separately.
4. **Case 1: $n$ is odd.**
\[
\phi(n(n + 2)) = \phi(n) \phi(n + 2)
\]
Therefore, the expression becomes:
\[
\phi(n(n + 2)) - \phi(n^2) = \phi(n) \phi(n + 2) - n \phi(n) = \phi(n) (\phi(n + 2) - n)
\]
For this to be maximized, $\phi(n + 2)$ should be as large as possible. If $n + 2$ is a prime number $p$, then $\phi(p) = p - 1$. Thus, we need to find the maximum value of $\phi(n)$ for $n = p - 2$ where $p$ is a prime number less than or equal to 102.
5. **Case 2: $n$ is even.**
\[
\phi(n(n + 2)) = \phi(n) \phi(n + 2) \cdot 2
\]
Therefore, the expression becomes:
\[
\phi(n(n + 2)) - \phi(n^2) = 2 \phi(n) \phi(n + 2) - n \phi(n) = \phi(n) (2 \phi(n + 2) - n)
\]
For this to be maximized, $2 \phi(n + 2)$ should be as large as possible. If $n + 2$ is a power of 2, then $\phi(n + 2) = n + 1$. Thus, we need to find the maximum value of $2 \phi(n)$ for $n = 2^k - 2$ where $2^k - 2 \leq 100$.
6. We now compute $\phi(n)$ for various values of $n$:
- For $n = 95$, $n + 2 = 97$ (a prime), $\phi(95) = 72$.
- For $n = 99$, $n + 2 = 101$ (a prime), $\phi(99) = 60$.
- For $n = 87$, $n + 2 = 89$ (a prime), $\phi(87) = 56$.
- For $n = 81$, $n + 2 = 83$ (a prime), $\phi(81) = 54$.
- For $n = 77$, $n + 2 = 79$ (a prime), $\phi(77) = 60$.
7. Comparing these values, the maximum value of $\phi(n)$ for odd $n$ is $\phi(95) = 72$.
8. For even $n$, the maximum value of $2 \phi(n)$ is $2 \phi(62) = 60$, which is not larger than 72.
Therefore, the maximum value of $\phi(n^2 + 2n) - \phi(n^2)$ for $1 \leq n \leq 100$ is $\boxed{72}$.
|
72
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p$ be an odd prime number less than $10^5$. Granite and Pomegranate play a game. First, Granite picks a integer $c \in \{2,3,\dots,p-1\}$.
Pomegranate then picks two integers $d$ and $x$, defines $f(t) = ct + d$, and writes $x$ on a sheet of paper.
Next, Granite writes $f(x)$ on the paper, Pomegranate writes $f(f(x))$, Granite writes $f(f(f(x)))$, and so on, with the players taking turns writing.
The game ends when two numbers appear on the paper whose difference is a multiple of $p$, and the player who wrote the most recent number wins. Find the sum of all $p$ for which Pomegranate has a winning strategy.
[i]Proposed by Yang Liu[/i]
|
1. **Define the game and the function:**
- Granite picks an integer \( c \in \{2, 3, \dots, p-1\} \).
- Pomegranate picks two integers \( d \) and \( x \), and defines the function \( f(t) = ct + d \).
- The sequence starts with \( x \), and each player alternately writes \( f \) applied to the last number written.
2. **Determine the condition for the game to end:**
- The game ends when two numbers appear on the paper whose difference is a multiple of \( p \). This means there exist integers \( m \) and \( n \) such that:
\[
f^m(x) \equiv f^n(x) \pmod{p}
\]
where \( f^k \) denotes the function \( f \) applied \( k \) times.
3. **Express \( f^k(x) \):**
- We can express \( f^k(x) \) as:
\[
f^k(x) = c^k x + d \sum_{i=0}^{k-1} c^i
\]
Using the formula for the sum of a geometric series, we get:
\[
\sum_{i=0}^{k-1} c^i = \frac{c^k - 1}{c - 1}
\]
Therefore:
\[
f^k(x) = c^k x + d \frac{c^k - 1}{c - 1}
\]
4. **Set up the congruence condition:**
- For the game to end, we need:
\[
c^m x + d \frac{c^m - 1}{c - 1} \equiv c^n x + d \frac{c^n - 1}{c - 1} \pmod{p}
\]
Simplifying, we get:
\[
c^m x - c^n x \equiv d \left( \frac{c^n - 1}{c - 1} - \frac{c^m - 1}{c - 1} \right) \pmod{p}
\]
\[
(c^m - c^n) x \equiv d \frac{c^n - c^m}{c - 1} \pmod{p}
\]
\[
(c^m - c^n) \left( x + \frac{d}{c - 1} \right) \equiv 0 \pmod{p}
\]
5. **Analyze the condition:**
- For the above congruence to hold, either \( c^m \equiv c^n \pmod{p} \) or \( x + \frac{d}{c - 1} \equiv 0 \pmod{p} \).
- If \( x + \frac{d}{c - 1} \equiv 0 \pmod{p} \), Granite can choose \( x = \frac{d}{1 - c} \) to win.
- Otherwise, we need \( c^m \equiv c^n \pmod{p} \), which implies \( c^{m-n} \equiv 1 \pmod{p} \). This means the order of \( c \) modulo \( p \) must divide \( m - n \).
6. **Determine the winning strategy:**
- Pomegranate wins if the order of \( c \) modulo \( p \) is even. This is because Pomegranate can always choose \( d \) and \( x \) such that the sequence will eventually repeat with an even period.
7. **Identify primes \( p \) where Pomegranate has a winning strategy:**
- If \( p-1 \) has an odd factor, there exists a number whose order modulo \( p \) is odd, and Granite can force a win.
- Therefore, we need \( p-1 \) to be a power of 2, i.e., \( p \) must be of the form \( 2^n + 1 \).
8. **List primes of the form \( 2^n + 1 \) less than \( 10^5 \):**
- The primes of this form are \( 3, 5, 17, 257, 65537 \).
9. **Sum these primes:**
- The sum is:
\[
3 + 5 + 17 + 257 + 65537 = 65819
\]
The final answer is \(\boxed{65819}\)
|
65819
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\mathbb{Z}_{\geq 0}$ be the set of nonnegative integers. Let $f: \mathbb{Z}_{\geq0} \to \mathbb{Z}_{\geq0}$ be a function such that, for all $a,b \in \mathbb{Z}_{\geq0}$: \[f(a)^2+f(b)^2+f(a+b)^2=1+2f(a)f(b)f(a+b).\]
Furthermore, suppose there exists $n \in \mathbb{Z}_{\geq0}$ such that $f(n)=577$. Let $S$ be the sum of all possible values of $f(2017)$. Find the remainder when $S$ is divided by $2017$.
[i]Proposed by Zack Chroman[/i]
|
1. **Given Function and Transformation:**
We start with the given function \( f: \mathbb{Z}_{\geq 0} \to \mathbb{Z}_{\geq 0} \) satisfying:
\[
f(a)^2 + f(b)^2 + f(a+b)^2 = 1 + 2f(a)f(b)f(a+b)
\]
for all \( a, b \in \mathbb{Z}_{\geq 0} \). We define a new function \( g(x) = 2f(x) \). Substituting \( g(x) \) into the equation, we get:
\[
\left(\frac{g(a)}{2}\right)^2 + \left(\frac{g(b)}{2}\right)^2 + \left(\frac{g(a+b)}{2}\right)^2 = 1 + 2 \left(\frac{g(a)}{2}\right) \left(\frac{g(b)}{2}\right) \left(\frac{g(a+b)}{2}\right)
\]
Simplifying, we obtain:
\[
g(a)^2 + g(b)^2 + g(a+b)^2 = 4 + g(a)g(b)g(a+b)
\]
2. **Applying Lemma:**
From the lemma, if \( x, y, z \) are positive reals such that:
\[
x^2 + y^2 + z^2 - xyz = 4
\]
then \( x = a + \frac{1}{a}, y = b + \frac{1}{b}, z = c + \frac{1}{c} \) for \( a, b, c > 0 \) and \( abc = 1 \).
3. **Function Transformation:**
We assume \( g(x) = h(x) + \frac{1}{h(x)} \) where \( h(x) > 1 \). Substituting into the equation, we get:
\[
\left(h(a) + \frac{1}{h(a)}\right)^2 + \left(h(b) + \frac{1}{h(b)}\right)^2 + \left(h(a+b) + \frac{1}{h(a+b)}\right)^2 = 4 + \left(h(a) + \frac{1}{h(a)}\right) \left(h(b) + \frac{1}{h(b)}\right) \left(h(a+b) + \frac{1}{h(a+b)}\right)
\]
Simplifying, we find:
\[
h(a+b) = h(a)h(b)
\]
This implies \( h(x) = h(1)^x \).
4. **Determining \( h(1) \):**
Let \( h(1) = c \). Then \( h(x) = c^x \). Given \( f(n) = 577 \), we have:
\[
g(n) = 2f(n) = 2 \times 577 = 1154
\]
Thus:
\[
g(n) = c^n + \frac{1}{c^n} = 1154
\]
Solving for \( c^n \), we get:
\[
c^n = 577 + 408\sqrt{2}
\]
5. **Finding \( f(2017) \):**
We need to find \( f(2017) \). We know:
\[
g(2017) = c^{2017} + \frac{1}{c^{2017}}
\]
Using the properties of \( c \), we have:
\[
c = 3 + 2\sqrt{2}
\]
Therefore:
\[
g(2017) = (3 + 2\sqrt{2})^{2017} + (3 - 2\sqrt{2})^{2017}
\]
and:
\[
f(2017) = \frac{(3 + 2\sqrt{2})^{2017} + (3 - 2\sqrt{2})^{2017}}{2}
\]
6. **Modulo Calculation:**
We need to find the remainder when the sum of all possible values of \( f(2017) \) is divided by 2017. The possible values of \( f(2017) \) are:
\[
\frac{(3 + 2\sqrt{2})^{2017} + (3 - 2\sqrt{2})^{2017}}{2}, \frac{(17 + 12\sqrt{2})^{2017} + (17 - 12\sqrt{2})^{2017}}{2}, \frac{(577 + 408\sqrt{2})^{2017} + (577 - 408\sqrt{2})^{2017}}{2}
\]
When expanded and taken modulo 2017, the terms with \( \sqrt{2} \) cancel out, and we are left with:
\[
3^{2017}, 17^{2017}, 577^{2017}
\]
By Fermat's Little Theorem, since 2017 is a prime:
\[
a^{2017} \equiv a \pmod{2017}
\]
Therefore:
\[
3^{2017} \equiv 3 \pmod{2017}, \quad 17^{2017} \equiv 17 \pmod{2017}, \quad 577^{2017} \equiv 577 \pmod{2017}
\]
Summing these, we get:
\[
3 + 17 + 577 = 597
\]
The final answer is \(\boxed{597}\).
|
597
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a triangle, not right-angled, with positive integer angle measures (in degrees) and circumcenter $O$. Say that a triangle $ABC$ is [i]good[/i] if the following three conditions hold:
(a) There exists a point $P\neq A$ on side $AB$ such that the circumcircle of $\triangle POA$ is tangent to $BO$.
(b) There exists a point $Q\neq A$ on side $AC$ such that the circumcircle of $\triangle QOA$ is tangent to $CO$.
(c) The perimeter of $\triangle APQ$ is at least $AB+AC$.
Determine the number of ordered triples $(\angle A, \angle B,\angle C)$ for which $\triangle ABC$ is good.
[i]Proposed by Vincent Huang[/i]
|
1. **Define the problem and setup:**
Let $ABC$ be a triangle with positive integer angle measures (in degrees) and circumcenter $O$. We need to determine the number of ordered triples $(\angle A, \angle B, \angle C)$ for which $\triangle ABC$ is *good* based on the given conditions.
2. **Identify the conditions:**
- (a) There exists a point $P \neq A$ on side $AB$ such that the circumcircle of $\triangle POA$ is tangent to $BO$.
- (b) There exists a point $Q \neq A$ on side $AC$ such that the circumcircle of $\triangle QOA$ is tangent to $CO$.
- (c) The perimeter of $\triangle APQ$ is at least $AB + AC$.
3. **Analyze the geometric properties:**
- Let $O_B$ and $O_C$ be the centers of the circumcircles of $\triangle POA$ and $\triangle QOA$, respectively.
- $O_B$ is the intersection of the perpendicular from $O$ to $BO$ and the perpendicular bisector of $AO$.
- $O_C$ is the intersection of the perpendicular from $O$ to $CO$ and the perpendicular bisector of $AO$.
4. **Calculate angles:**
- $\angle OAO_C = \angle AOO_C = 2B - 90^\circ$
- $\angle OAC = 90^\circ - B$
- Therefore, $\angle QAO_C = \angle OAO_C - \angle OAC = 3B - 180^\circ$
- $\angle AO_CQ = 540^\circ - 6B$
- $\angle OO_CQ = \angle AO_CQ - \angle AO_CO = 540^\circ - 6B - (360^\circ - 4B) = 180^\circ - 2B$
- Hence, $\angle O_COQ = B$
- $\angle QOC = 90^\circ - B$
- $\triangle OQC$ is isosceles.
5. **Apply the triangle inequality:**
- By the triangle inequality, $AP + AQ + PQ \leq AP + AQ + PO + OQ = AB + AC$
- By hypothesis, we must have equality, so $P$, $O$, and $Q$ are collinear.
6. **Determine angle $A$:**
- This collinearity condition implies:
\[
\angle QOC + \angle BOC + \angle POB = 180^\circ
\]
\[
90^\circ - B + 2A + 90^\circ - C = 180^\circ
\]
\[
A = 60^\circ
\]
7. **Check constraints on angles $B$ and $C$:**
- We need $\angle B > 30^\circ$ and $\angle C > 30^\circ$ for points $P$ and $Q$ to lie on the interior of segments $\overline{AB}$ and $\overline{AC}$.
- This is because $\angle BOP < \angle BOA$ implies $90^\circ - C < 2C \Rightarrow C > 30^\circ$.
- Similarly, $\angle B > 30^\circ$.
8. **Count the number of valid triples:**
- Since $\angle A = 60^\circ$ and $\angle B, \angle C > 30^\circ$, we can choose $\angle B = 31^\circ, 32^\circ, \ldots, 89^\circ$.
- This gives $59$ possibilities for $\angle B$ and correspondingly $\angle C = 180^\circ - 60^\circ - \angle B$.
The final answer is $\boxed{59}$
|
59
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
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