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30

Let $$f(a,b) := \frac{\sqrt{a^2+3ab+b^2-2a-2b+4}}{ab+4}.$$ John's procedure is now to repeatedly replace the leftmost two values $a,b$ on the blackboard by the single value $f(a,b)$. I claim that John never writes a negative value onto the blackboard. Proof: the blackboard always begins with positive values. Suppose $a$ and $b$ are positive values John ...


23

Alain Remillard has given the mathematician's answer. Here's the physicist's one: Step 1: Obviously, in such a universe, regardless of their speed, the cannonballs will travel in a straight line and hit each other in the middle. Step 2: Assume "Step 1" does not exist. Therefore


21

The answer is I imagine the line of reasoning the author wants is as follows:


20

The answer is yes, this is possible. Fix $t = 100$. A big key is found in this excellent puzzle:


20

Vladimir starts with daughter 1 on the motorbike. After 48 km (that's one hour and 12 min at 40 kph), he stops and daughter 1 continues on foot. She reaches the dacha after exactly three hours. Vladimir drives back to meet daughter 2, who has already covered 12 km. They meet 36 min later (one hour and 48 min after the start), 18 km from the starting point (...


19

Shkeil got all of the nontrivial answers, but he forgot the simplest, so I'll add it here: For completeness, I'll add the rest of my thought process.


18

Okay so I think the answer is Google cannot find me, reflected I'm lemon and lime, googling 'me' is when you fall, me and meme will answer all.


16

It is Reasoning:


15

To illustrate the puzzle I made the following image: All angles I entered can be simply calculated using the fact any n-gon has a total of $180 + (n-3) \cdot 180 °$. Now let's call the intersection of line $b$ and $a$ $X_1$ and the intersection of $b$ and $c$ $X_2$. To prove that $X_1 = X_2$ I'm going to show that $P_1X_1 = P_1X_2$ First, let's look at $...


14

Assume that a bumble bee costs $x$ cents and that a honey bee costs $y$ cents, where $x$ and $y$ are integers. Then the problem statement gives $125x<175y$, which yields $5x<7y$ and hence $5x+1\le7y$ $175y<126x$, which yields $25y<18x$ and hence $25y+1\le18x$ We multiply the first inequality $5x+1\le7y$ by $18$ and the second inequality $25y+1\...


14

There is Reasonning.


14

First of all, let's see why your brute-forcing fails. (This is the puzzle part, the rest is plain old math.) No matter which you chose, the number at the bottom right would have to be both odd and even at the same time, so there's no integer solution. However, there are four equations and four unknowns, so we should have at least one solution (unless the ...


13

(I'm assuming throughout that the concentration of wine at the end needs to be 50%.) Unless I'm missing something: There are $n$ pints of wine to begin with. After the first 3 pints are taken, you have an $(n-3)/n$ fraction of wine after the water is put back. After the second 3 pints are taken and replaced, you just multiply by the same fraction again, so ...


12

And here is the number you are probably thinking of: It works only for $ab$ where $a \le b$. I suppose that it is a mistake in the problem statement. Others have proven that as it is, the problem is unsolvable. And here is how I came to that number. PS: I have been playing with this problem. You can extend it to $ab$ with $a > b$ with the ...


12

The number is because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively: Adding the first two gives Also from the first equation, So the number must be


11

As for why it is the only solution:


11

By brute-force search, yes. I started by searching over all tuples $(r,s,t)$ less than $1000$, stopping at the first example I found: $$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$ In order to find the smallest example, I used the following strategy to search all examples smaller than the previous best (which I'll call $x$). Since the ...


11

The minimum greater than 1000 is: Because:


11

@Tony Ruth's answer provides this alternate form of the equation: Adding 2 to both sides and factoring, Then Substituting this back in, So Add 2 to both sides again: So And For every $y>1$, this is true for some integer $n$, so we can choose the $y$ that gives the biggest value of $m$ less than 1 million. This is: So the answer is


11

They will I did it mathematically Suppose the horizontal distance between both cannons is $d$ and the up angle from right cannon is $\theta$. Then, the left cannon is at a height of $d\tan\theta$ and aim down at an angle of $\theta$. Since the horizontal speed of the cannonballs are the same, there is à time when they are both at the same ...


10

Note: 51.5625 = 825/16 To calculate the sum for a given $n$ we can do $\frac{n \cdot (n+1)}{2}$. Then we need to have that $\frac{n \cdot (n+1)}{2} - \frac{825 \cdot (n - 4)}{16}$ is a number that can be expressed as four even consecutive numbers. Let $x$ be the lowest of the four consecutive numbers. The four even consecutive numbers are of the form $x+x+2+...


10

This picture shows the best strategy to get to the Dacha.


10

Inscribe the original $n$-gon in a circle of radius 1. The apothem of the large $n$-gon is $\cos(\frac{\pi}{n})$ and the apothem of the small $n$-gon is $\sin(\frac{\pi}{2n})$. Therefore the ratio of their areas is $\left(\frac{\cos(\frac{\pi}{n})}{\sin(\frac{\pi}{2n})}\right)^2$. For $n=99$ this is about $3968.53$. Demonstration on a heptagon: $OA=1$, $\...


10

The doesn't belong, because


9

Following up on the suggestion by Dennis Meng, I'm posting my suggestion as an answer instead of as part of the question. So here are my thoughts on the problem. (1) As Joe Z comments, the cattle owner needs to increase prices by one eighth in order to make the stated profit. (2) For some reason, the price is given as 63 daler for 3 oxen, rather than the ...


9

The smallest possible value of $n$ is Claim: We can get every non-negative integer $n\leq 2016$ on the board. Proof: By induction. We start with $n=0$ on the board. We can get $1$ using Lord of the Dark's method: $2016!x+2016!=0$ has $-1$ as a root, $-x^2+2016!=0$ has $\pm\sqrt{2016!}$ as roots, and $\sqrt{2016!}x-\sqrt{2016!}$ has $1$ as a root. Now ...


9

Edit:


9

Is the answer Googling it returns me and meme will answer all. Can't make the middle two lines fit though


9

This is probably similar to msh210's idea but if we write


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