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2022AIME II 真题及答案

Category: AMC美国数学竞赛, 国际竞赛 Date: 2022年4月28日下午5:25

2022AIME II 真题及答案

2022AIME I 真题及答案

Problem1

Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution 1

Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$ , as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$ , as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$ , and since we are looking for the minimum amount of people, $x$ is $300$ . That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.

~eamo

Solution 2 (Kind of lame)

Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of adults must be a multiple of 12.

Call the amount of people in the beginning $x$ .Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12. Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.

Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.

Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.

So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\frac{11}{25}=154$ . Therefore the answer is $\boxed{154}$ .

I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.

~hastapasta

Problem2

Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution 1

Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.

This problem can be solved by using $2$ cases.

$\textbf{Case 1:}$ $C$ 's opponent for the semifinal is $A$

The probability $C$ 's opponent is $A$ is $\frac13$ . Therefore the probability $C$ wins the semifinal in this case is $\frac13 \cdot \frac13$ . The other semifinal game is played between $J$ and $S$ , it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\frac34$ , so the probability of $C$ winning the tournament in case 1 is $\frac13 \cdot \frac13 \cdot \frac34$

$\textbf{Case 2:}$ $C$ 's opponent for the semifinal is $J$ or $S$

It doesn't matter if $C$ 's opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$ 's opponent is $J$ or $S$ is $\frac23$ . Therefore the probability $C$ wins the semifinal in this case is $\frac23 \cdot \frac34$ . The other semifinal game is played between $A$ and $J$ or $S$ . In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.

$\textbf{Case 2.1:}$ $C$ 's opponent for the final is $A$

For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\frac34$ . Therefore, the probability that $C$ won $A$ in the final is $\frac34 \cdot \frac13$ .

$\textbf{Case 2.1:}$ $C$ 's opponent for the final is $J$ or $S$

For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\frac14$ . Therefore, the probability that $C$ won $J$ or $S$ in the final is $\frac14 \cdot \frac34$ .

In Case 2 the probability of $C$ winning the tournament is $\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)$

Adding case 1 and case 2 together we get $\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96}$ . So the answer is $29 + 96 = \boxed{\textbf{125}}$

~isabelchen

Problem3

A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$ ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$ . Because of the symmetrical property of the pyramid, we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.

Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$ , where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=(\frac{9}{2}-l)^2+(3\sqrt{2})^2$ according to pythagorean theorem.

Solve this equation will give us $l = \frac{17}{4}$ . Therefore, $m+n=\boxed{021}$

~DSAERF-CALMIT (https://binaryphi.site)

Solution 2

To start, we find the height of the pyramid. By the volume of a pyramid formula, we have $\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.$ Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$ , we have $x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.$ Taking the cross section of the pyramid and transforming the problem into $2$ -d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$ , $\frac{3\sqrt{17}}2$ , $6\sqrt2$ . This turns out to be easy by the formula $R=\frac{abc}{4A}$ , and through computing this value (the work has been left out) we find that $R=\frac{17}4$ , so our answer is $\boxed{\textbf{021}}$ .

~A1001

Problem4

There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that $\log_{20x} (22x)=\log_{2x} (202x).$ The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of logs, $\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}$ Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_{10} \frac{11}{101}$ . This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$ .

~ihatemath123

Solution 2

We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$ .

So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$

Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}$

Substitute $\textcircled{2}$ to $\textcircled{3}$ : $202x \cdot (10^v)=22x$

Thus, $v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})$ , where $m=11$ and $n=101$ .

Therefore, $m+n = \boxed{112}$ .

~DSAERF-CALMIT (https://binaryphi.site)

Solution 3

We have $\begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}$

We have $\begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}$

Because $\log_{20x} (22x)=\log_{2x} (202x)$ , we get $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .$

We denote this common value as $\lambda$ .

By solving the equality $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda$ , we get $\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}$ .

By solving the equality $\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda$ , we get $\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}$ .

By equating these two equations, we get $\frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1} = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} .$

Therefore, $\begin{align*} \log_{20x} (22x) & = \lambda \\ & = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\ & = \frac{\log_k \frac{11}{101}}{\log_k 10} \\ & = \log_{10} \frac{11}{101} . \end{align*}$

Therefore, the answer is $11 + 101 = \boxed{\textbf{(112) }}$ .

~Steven Chen (www.professorchenedu.com)

Solution 4

By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-b}{c-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{20}=\log\frac{11}{101}$ . The requested sum is $11+101=\boxed{112}.$

~MathIsFun286

Problem5

Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.

Solution 1

Let $a$ , $b$ , and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$ . $a - b = p_1$ $b - c = p_2$ $a - c = p_3$

$p_3 = a - c = a - b + b - c = p_1 + p_2$ . Because $p_3$ is the sum of two primes, $p_1$ and $p_2$ , $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_3 = p_2 + 2$ . There are only $8$ primes less than $20$ : $2, 3, 5, 7, 11, 13, 17, 19$ . Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \in \{ 3, 5, 11, 17 \}$ .

Once $a$ is determined, $b = a+2$ and $c = b + p_2$ . There are $18$ values of $a$ where $a+2 \le 20$ , and $4$ values of $p_2$ . Therefore the answer is $18 \cdot 4 = \boxed{\textbf{072}}$

Problem6

Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

To find the greatest value of $x_{76} - x_{16}$ , $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$ . The other numbers between $x_{16}$ and $x_{76}$ equal to $0$ . Let $a = x_{76}$ , $b = x_{16}$ . Substituting $a$ and $b$ into $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ we get: $25a - 16b = 1$ $25a + 16b = 0$ $a = \frac{1}{50}$ , $b = -\frac{1}{32}$

$x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}$ . $m+n = \boxed{\textbf{841}}$

~isabelchen

Solution 2

Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.

Since the sum of the absolute values of all the numbers is $1$ , $|s_N|+|s_P|=1$ .

Since the sum of all the numbers is $0$ , $s_N=-s_P\implies |s_N|=|s_P|$ .

Therefore, $|s_N|=|s_P|=\frac 12$ , so $s_N=-\frac 12$ and $s_P=\frac 12$ since $s_N$ is negative and $s_P$ is positive.

To maximize $x_{76}-x_{16}$ , we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible.

Note that $x_{76}+x_{77}+\cdots+x_{100}=\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order.

Similarly, $x_{1}+x_{2}+\cdots+x_{16}=-\frac 12$ is less than or equal to $16x_{16}$ .

So we now know that $\frac 1{50}$ is the best we can do for $x_{76}$ , and $-\frac 1{32}$ is the least we can do for $x_{16}$ .

Finally, the maximum value of $x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}$ , so the answer is $\boxed{841}$ .

(Indeed, we can easily show that $x_1=x_2=\cdots=x_{16}=-\frac 1{32}$ , $x_{17}=x_{18}=\cdots=x_{75}=0$ , and $x_{76}=x_{77}=\cdots=x_{100}=\frac 1{50}$ works.)

~inventivedant

Problem7

A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution 1

$[asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); [/asy]$

$r_1 = O_1A = 24$ , $r_2 = O_2B = 6$ , $AG = BO_2 = r_2 = 6$ , $O_1G = r_1 - r_2 = 24 - 6 = 18$ , $O_1O_2 = r_1 + r_2 = 30$

$\triangle O_2BD \sim \triangle O_1GO_2$ , $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$ , $\frac{O_2D}{30} = \frac{6}{18}$ , $O_2D = 10$

$CD = O_2D + r_1 = 10 + 6 = 16$ ,

$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$

$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{\textbf{192}}$

~isabelchen

Solution 2

Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$ . Since $\triangle{APC} \sim \triangle{BPD}$ , we have $\frac{AP}{AP+30}=\frac14 \implies AP=10.$ Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$ . Then, $P = (-10, 0)$ , and if $C = (x, y)$ , we have $(x+10)^2+y^2=64,$ $x^2+y^2=36.$ Combining these and solving, we get $(x, y)=\left(-\frac{18}5, \frac{24}5\right)$ . Notice now that $P$ , $C$ , and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\frac{-\frac{18}5+10}{\frac{24}5}=\frac34$ . Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$ . By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$ , $(6, 12)$ , and $(6, -12)$ is $\frac12|-120-0-72-72+0-120|=\boxed{\textbf{192}}.$

Problem8

Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ , $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ .

Solution 1

We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$ , $5$ , and/or $6$ which are also multiples of $4$ , $5$ , and/or $6$ when $1$ is added to them.

We begin by noting that the LCM of $4$ , $5$ , and $6$ is $60$ . We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ ( $600$ / $60$ = $10$ ).

After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$ , $5$ , $15$ , $24$ , $35$ , $44$ , $54$ , and $55$ . This gives us $8$ numbers. $8$ * $10$ = $\boxed{080}$ .

Soon after the test was administered, a formal request was made to also accept $\boxed{081}$ as an answer and MAA decided to honor this request. The gist of this request stated that the phrasing of the first part of the question could reasonably be interpreted to mean that one is given the condition to begin with that the integer is less than or equal to $600$ . In this case, if one was told that the values of $\left\lfloor \frac n4\right\rfloor$ , $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ were $150$ , $120$ , and $100$ respectively, then the only possible choice for $n$ would be $600$ as $601$ , $602$ , and $603$ do not meet the condition as stated in the first part of the problem. ~burkinafaso

Solution 1.5

This is Solution 1 with a slick element included. Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$ , $4$ , and/or $5$ and $60k+l+1$ is a multiple of $3$ , $4$ , and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$ . But keeping the same conditions in mind for $k$ and $l$ , we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution! Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$ . One only has to determine the number of values between $1$ and $30$ and then multiply by $20$ . The values of $n$ that work between $1$ and $30$ are $4$ , $5$ , $15$ , and $24$ . This gives us $4$ numbers. $4$ * $20$ = $\boxed{080}$ .

Solution 2

1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:

1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$

2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$

3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$

4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$

Solving for each case, we see that there are $30$ possibilities for cases 1 and 3 each, and $20$ possibilities for cases 2 and 4 each. However, we over-counted the cases where

1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$

2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$

Each case has $10$ possibilities.

Adding all the cases and correcting for over-counting, we get $30 + 20 + 30 + 20 - 10 - 10 = \boxed {080}.$

~Lucasfunnyface

Solution 2 Supplement

Here is a detailed solution for Solution 2.

 , 
, , , , , , , , , 30 integers.

 , 
, , , , , , , , , 20 integers.

 , 
, , , , , , , , , 30 integers.

 , 
, , , , , , , , , 20 integers.

Over-counted cases:

 , 
, , , , , , , , , , , 10 integers.

 , 
, , , , , , , , , , , 10 integers.

$30 + 20 + 30 + 20 - 10 - 10 = \boxed{\textbf{080}}$

~isabelchen

Solution 3

The problem is the same as asking how many unique sets of values of $\lfloor\frac{n}{4}\rfloor$ , $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ can be produced by one and only one value of $n$ for positive integers $n$ less than or equal to 600.

Seeing that we are dealing with the unique values of the floor function, we ought to examine when it is about to change values, for instance, when $n$ is close to a multiple of 4 in $\lfloor\frac{n}{4}\rfloor$ .

For a particular value of $n$ , let $a$ , $b$ , and $c$ be the original values of $\lfloor\frac{n}{4}\rfloor$ , $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ , respectively.

Notice when $n$ $\equiv0\mod4$ and $n$ $\equiv4\mod5$ , the value of $\lfloor\frac{n-1}{4}\rfloor$ will be 1 less than the original $a$ . The value of $\lfloor\frac{n+1}{5}\rfloor$ will be 1 greater than the original value of $b$ .

More importantly, this means that no other value less than or greater than $n$ will be able to produce the set of original values of $a$ , $b$ , and $c$ , since they make either $a$ or $b$ differ by at least 1.

Generalizing, we find that $n$ must satisfy:

Where $j$ and $k$ are pairs of distinct values of 4, 5, and 6.

Plugging in the values of $j$ and $k$ , finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are $\boxed{080}$ solutions of $n$ .

Solution 3 Supplement

By Chinese Remainder Theorem, the general solution of systems of $2$ linear congruences is:

, , 
Find  and  such that , 
Then

$lcm[4, 5, 6] = 60$ , we solve the number of values for $n \le 60$ , then multiply by $10$ to get the number of values for $n \le 600$ . We are going to solve the following $6$ systems of linear congruences:


No solution


No solution

$n \in \{ 4, 5, 15, 24, 35, 44, 54, 55 \}$ , there are $8$ values for $n \le 60$ . For $n \le 600$ , the answer is $8 \cdot 10 = \boxed{\textbf{080}}$ .

~isabelchen

Sidenote

For solving a system of linear congruences, see https://youtu.be/-a88u99nmkw

Solution 4

Observe that if $1 \le n \le 60$ such that n is a solution to the desired equation, so is $n + 60\cdot m$ , where m is an integer, $0 \le m \le 9$ . So we only need to consider n from 1 to 60. As shown in Solution 2, there are 4 cases which we will split into 2 main cases:

Case 1: $4 \mid n$ or $6 \mid n$ , $5 \mid (n+1)$
Case 2: $4 \mid (n+1)$ or $6 \mid (n+1)$ , $5 \mid n$

There are 4 values of n where $1 \le n \le 12$ satisfying $4 \mid n$ or $6 \mid n$ .

I claim that there are 4 values of $n \le 60$ satisfying Case 1. Suppose x is one value of n satisfying $4 \mid n$ or $6 \mid n$ , and $n \le 12$ . Hence the solutions satisfying $4 \mid n$ or $6 \mid n$ , $n \le 60$ are of the form $x + 12m$ , so the values of $n + 1$ are $x + 12m + 1 \equiv x + 2m + 1 \equiv 0$ (mod 5), so $2m \equiv 4 + 4x$ (mod 5) and hence the value of m is unique since $0 \le m \le 4$ to satisfy $1 \le n \le 60$ and 2 and 5 are relatively prime.

A similar approach can be used to show the same for Case 2, that there are 4 values of $n \le 60$ .

Hence our answer is $(4+4)*10 = \fbox{080}$ .

~Bxiao31415

Problem9

Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ . $[asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]$

Solution 1

We can use recursion to solve this problem:

1. Fix 7 points on $\ell_A$ , then put one point $B_1$ on $\ell_B$ . Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$ . For example, $f(1) = 6$ because there are 6 regions.

2. Now, put the second point $B_2$ on $\ell_B$ . Join $A_1~A_7$ and $B_2$ will create $7$ new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between $B_2$ and $A_1$ intersect lines $\overline{B_1A_2}$ , $\overline{B_1A_3}$ , ..., $\overline{B_1A_7}$ at $6$ points $\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: $f(2) = f(1) + 7 + (6+5+...+1) = 34.$

3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting $\overline{A_2B_1}, \overline{A_2B_2};$ $\overline{A_3B_1}, \overline{A_3B_2};$ $...$ $\overline{A_7B_1}, \overline{A_7B_2}$ at $12$ points, creating $12$ regions, etc. Thus, we have: $f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.$

Yes, you might already notice that: $f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.$

5. (Finally) we have $f(4) = 153$ , and $f(5)=244$ . Therefore, the answer is $\boxed{244}$ .

Note: we could deduce a general formula of this recursion: $f(n+1)=f(n)+N_a+\frac{n\cdot (N_a) \cdot (N_a-1)}{2}$ , where $N_a$ is the number of points on $\ell_A$ .

~DSAERF-CALMIT (https://binaryphi.site)

Solution 2

We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$ . Now let's focus on $f(m+1, n)-f(m, n)$ , which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$ . This new point, call it #m, when connected to point #1 on $\ell_B$ , crosses $m*(n-1)$ lines, thus making additional $m*(n-1)+1$ bounded regions; when connected to point #2 on $\ell_B$ , it crosses $m*(n-2)$ lines, thus making additional $m*(n-2)+1$ bounded regions; etc. By simple algebra/recursion methods, we see

$f(m+1, n)-f(m, n)=m*\frac{n(n-1)}{2} +n$

Notice $f(1,n)=n-1$ . Not very difficult to figure out:

$f(m, n)=\frac{m(m-1)n(n-1)}{4} +mn-1$

The fact that $f(3,2)=8$ makes us more confident about the formula. Now plug in $m=5, n=7$ , we get the final answer of $\boxed{244}$ .

Problem10

Find the remainder when $\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}$ is divided by $1000$ .

Solution

To solve this problem, we need to use the following result:

$\sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1} .$

Now, we use this result to solve this problem.

We have $\begin{align*} \sum_{i=3}^{40} \binom{\binom{i}{2}}{2} & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \\ & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( i \left( i - 1 \right) - 2 \right) \\ & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( \left( i - 2 \right) \left( i - 3 \right) + 4 \left( i - 2 \right) \right) \\ & = 3 \left( \sum_{i=3}^{40} \binom{i}{4} + \sum_{i=3}^{40} \binom{i}{3} \right) \\ & = 3 \left( \binom{41}{5} - \binom{3}{5} + \binom{41}{4} - \binom{3}{4} \right) \\ & = 3 \left( \binom{41}{5} + \binom{41}{4} \right) \\ & = 3 \cdot \frac{41 \cdot 40 \cdot 39 \cdot 38}{5!} \left( 37 + 5 \right) \\ & = 3 \cdot 41 \cdot 13 \cdot 38 \cdot 42 \\ & = 38 \cdot 39 \cdot 41 \cdot 42 \\ & = \left( 40 - 2 \right) \left( 40 - 1 \right) \left( 40 + 1 \right) \left( 40 + 2 \right) \\ & = \left( 40^2 - 2^2 \right) \left( 40^2 - 1^2 \right) \\ & = \left( 40^2 - 4 \right) \left( 40^2 - 1 \right) \\ & = 40^4 - 40^2 \cdot 5 + 4 . \end{align*}$

Therefore, modulo 1000, $\sum_{i=3}^{40} \binom{\binom{i}{2}}{2} \equiv \boxed{\textbf{(004) }}$ .

~Steven Chen (www.professorchenedu.com)

Solution 2 (similar to solution 1)

Doing simple algebra calculation will give the following equation: $\begin{align*} \binom{\binom{n}{2}}{2} = \frac{\frac{n(n-1)}{2} \cdot (\frac{n(n-1)}{2}-1)}{2} \\ = \frac{n(n-1)(n^2-n-2)}{8} \\ = \frac{(n+1)n(n-1)(n-2)}{8} \\ = \frac{(n+1)!}{8\cdot (n-3)!} = 3 \cdot \frac{(n+1)!}{4!\cdot (n-3)!} \\ = 3 \binom{n+1}{4} \end{align*}$

Next, by using Hockey-Stick Identity, we have: $3 \cdot \sum_{i=3}^{40} \binom{n+1}{4} = 3 \binom{42}{5} = 42 \cdot 41 \cdot 39 \cdot 38$ $=(40^2-2^2)(40^2-1^2) \equiv \boxed{004} (mod 1000)$

~DSAERF-CALMIT (https://binaryphi.site)

Solution 3

Since $40$ seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from $1$ term: $3$ , $18$ , $63$ , $168$ , $378$ , and $756$ . Notice that these are just $3 \cdot \dbinom50$ , $3 \cdot \dbinom61$ , $3 \cdot \dbinom72$ , $3 \cdot \dbinom83$ , $3 \cdot \dbinom94$ , $3 \cdot \dbinom{10}5$ . It's clear that this pattern continues up to $38$ terms, noticing that the "indexing" starts with $\dbinom32$ instead of $\dbinom12$ . Thus, the value of the sum is $3 \cdot \dbinom{42}{37}=2552004 \equiv \boxed{\textbf{004}} \pmod{1000}$ .

~A1001

Problem11

Let $ABCD$ be a convex quadrilateral with $AB=2$ , $AD=7$ , and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}$ . Find the square of the area of $ABCD$ .

Solution 1

2022 AIME II Q11(Hand-draw picture)

According to the problem, we have $AB=AB'=2$ , $DC=DC'=3$ , $MB=MB'$ , $MC=MC'$ , and $B'C'=7-2-3=2$

Because $M$ is the midpoint of $BC$ , we have $BM=MC$ , so: $MB=MB'=MC'=MC.$

Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$

Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$ .

This is when we found that points $M$ , $C$ , $D$ , and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$ . This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$ .

Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (BM')^2=AB' \cdot C'D = 6$

Point $H$ is the midpoint of $B'C'$ , and $MH \perp AD$ . $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$ .

The area of this quadrilateral is the sum of areas of triangles: $S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}$ $=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2$ $=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH$ $=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}$

Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$

~DSAERF-CALMIT (https://binaryphi.site)

Solution 2

Denote by $M$ the midpoint of segment $BC$ . Let points $P$ and $Q$ be on segment $AD$ , such that $AP = AB$ and $DQ = DC$ .

Denote $\angle DAM = \alpha$ , $\angle BAD = \beta$ , $\angle BMA = \theta$ , $\angle CMD = \phi$ .

Denote $BM = x$ . Because $M$ is the midpoint of $BC$ , $CM = x$ .

Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$ , $\triangle BAM \cong \triangle PAM$ . Hence, $MP = MB$ and $\angle AMP = \theta$ . Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$ .

Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$ , $\triangle CDM \cong \triangle QDM$ . Hence, $MQ = MC$ and $\angle DMQ = \phi$ . Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$ .

Because $M$ is the midpoint of segment $BC$ , $MB = MC$ . Because $MP = MB$ and $MQ = MC$ , $MP = MQ$ . Thus, $\angle MPD = \angle MQA$ . Thus, \[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]

In $\triangle AMD$ , $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$ . In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$ . Thus, \[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]

Taking $(1) + (2)$ , we get $\alpha = \phi$ . Taking $(1) - (2)$ , we get $\beta = \theta$ .

Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$ .

Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$ . Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$ .

In $\triangle ADM$ , by applying the law of cosines, $\cos \angle AMD = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$ . Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$ . Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$ .

Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}

Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}$ .

~Steven Chen (www.professorchenedu.com)

Problem12

Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that $\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.$ Find the least possible value of $a+b.$

Solution

Denote $P = \left( x , y \right)$ .

Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$ , $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ .

Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$ .

Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$ , $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ .

Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$ .

Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses. To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ , and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ .

The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$ .

The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$ .

Hence, $2 a + 2 b = 26 + 20 = 46$ .

Therefore, $a + b = \boxed{\textbf{(023) }}.$

~Steven Chen (www.professorchenedu.com)

Problem13

There is a polynomial $P(x)$ with integer coefficients such that $P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}$ holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ .

Solution 1

Because $0 < x < 1$ , we have $\begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}$

Denote by $c_{2022}$ the coefficient of $P \left( x \right)$ . Thus, $\begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}$

Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy $105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1)$

Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$ . Hence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy $105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2)$

Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$ . Hence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy $35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3)$

Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$ . Hence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy $7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4)$

Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$ . Hence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy $b' + c' + d' + e' = 9 . \hspace{1cm} (5)$

The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = \boxed{\textbf{(220) }}$ .

~Steven Chen (www.professorchenedu.com)

Solution 2

We know that $\frac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1} a^{n-1-i}b^i$ . Applying this, we see that $P(x)=(1+x^{105}+x^{210}+...)(1+x^{70}+x^{140}+...)(1+x^{42}+x^{84}+...)(1+x^{30}+x^{60}+...)(x^{4620}-2x^{2310}+1)$ The last factor does not contribute to the $x^{2022}$ term, so we can ignore it. Thus we only have left to solve the equation $105b+70c+42d+30e=2022$ , and we can proceed from here with Solution 1.

~MathIsFun286

Problem14

For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ .

Solution 1

Notice that we must have $a = 1$ , otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$ . Using at most $c-1$ stamps of value $1$ and $b$ , it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ , every value up to $1000$ can be represented. Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ , $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ , all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.

So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$ , $b<c-1$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ . We can get the answer by solving this equation.

$c > \lfloor \frac{c-1}{b} \rfloor + b-1$

$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

$c^2 - 97c + 999 > 0$ , $c > 85.3$ or $c < 11.7$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$

$97 > \frac{999}{c}$ , $c>10.3$

 For , , 
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The $3$ least values of $c$ is $11$ , $88$ , $89$ . $11 + 88+ 89 = \boxed{\textbf{188}}$

~isabelchen

Problem15

Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. $[asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]$

Solution 1

First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $A'B'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $B'A'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2B'$ and $A'O_1C$ are congruent, so $B'D = A'C$ and quadrilateral $B'A'CD$ is an isosceles trapezoid. $[asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Bp = dir(105), Ap = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Bp--Ap--O1--C--D--O2--cycle); label("$B'$",Bp,dir(origin--Bp)); label("$A'$",Ap,dir(origin--Ap)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Bp--D^^Ap--C,linetype("2 2")); [/asy]$ Next, remark that $A'O_1 = DO_2$ , so quadrilateral $A'O_1DO_2$ is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$ , and similarly $B'C = 15$ . Thus, Ptolmey's theorem on $B'A'CD$ yields $B'D\cdot A'C + 2\cdot 16 = 15^2$ , whence $B'D = A'C = \sqrt{193}$ . Let $\alpha = \angle B'A'D$ . The Law of Cosines on triangle $B'A'D$ yields $\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,$ and hence $\sin\alpha = \tfrac 45$ . Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$ .

Now let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$ ; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$ , which implies the measure of angle $B'O_2D$ is $180^\circ - \alpha$ . Therefore, the Law of Cosines applied to triangle $\triangle B'O_2D$ yields $\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}$

Thus $r_1r_2 = 40$ , and so the area of triangle $B'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$ .

Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$ .

~djmathman

Solution 2

Denote by $O$ the center of $\Omega$ . Denote by $r$ the radius of $\Omega$ .

We have $O_1$ , $O_2$ , $A$ , $B$ , $C$ , $D$ are all on circle $\Omega$ .

Denote $\angle O_1 O O_2 = 2 \theta$ . Denote $\angle O_1 O B = \alpha$ . Denote $\angle O_2 O A = \beta$ .

Because $B$ and $C$ are on circles $\omega_1$ and $\Omega$ , $BC$ is a perpendicular bisector of $O_1 O$ . Hence, $\angle O_1 O C = \alpha$ .

Because $A$ and $D$ are on circles $\omega_2$ and $\Omega$ , $AD$ is a perpendicular bisector of $O_2 O$ . Hence, $\angle O_2 O D = \beta$ .

In $\triangle O O_1 O_2$ , $O_1 O_2 = 2 r \sin \theta .$

Hence, $2 r \sin \theta = 15 .$

In $\triangle O AB$ , $\begin{align*} AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}$

Hence, $15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 2 . \hspace{1cm} (1)$

In $\triangle O CD$ , $\begin{align*} CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}$

Hence, $15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 16 . \hspace{1cm} (2)$

Taking $\frac{(1) + (2)}{30}$ , we get $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ . Thus, $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ .

Taking these into (1), we get $2 r \cos \theta = \frac{35}{4}$ . Hence, $\begin{align*} 2 r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\ & = \frac{5}{4} \sqrt{193} . \end{align*}$

Hence, $\cos \theta = \frac{7}{\sqrt{193}}$ .

In $\triangle O O_1 B$ , $O_1 B = 2 r \sin \frac{\alpha}{2} .$

In $\triangle O O_2 A$ , by applying the law of sines, we get $O_2 A = 2 r \sin \frac{\beta}{2} .$

Because circles $\omega_1$ and $\omega_2$ are externally tangent, $B$ is on circle $\omega_1$ , $A$ is on circle $\omega_2$ , $\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\ & = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\ & = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . \end{align*}$

Thus, $\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}$ .

Now, we compute $\sin \alpha$ and $\sin \beta$ .

Recall $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ and $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ . Thus, $e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}$ .

We also have $\begin{align*} \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right) \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right) . \end{align*}$

Thus, $\begin{align*} \sin \alpha + \sin \beta & = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha} + e^{i \beta} - e^{-i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( e^{i \alpha} + e^{i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right) \left( \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2} {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 } + 1 \right) \\ & = \frac{167 \cdot 8}{193 \cdot 5 } . \end{align*}$

Therefore, $\begin{align*} {\rm Area} \ ABO_1CDO_2 & = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1 + {\rm Area} \ \triangle O_3 O_1 C \\ & \quad + {\rm Area} \ \triangle O_3 C D + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right) + \sin \beta + \sin \beta \right) \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) - \sin \left( 2 \theta + \alpha + \beta \right) + 2 \sin \alpha + 2 \sin \beta \right) \\ & = r^2 \left( - \cos 2 \theta \sin \left( \alpha + \beta \right) + \sin \alpha + \sin \beta \right) \\ & = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} + \sin \alpha + \sin \beta \right) \\ & = \boxed{\textbf{(140) }} . \end{align*}$