[/quote]
,
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
,
be the incenters of triangles 
,
,
,
,
meet at a point.
satisfying
Prove
cups labeled 
,
-
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and 
,
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
,![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
to tangent 
disjoint from 
on 
and 
.
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where 
and 
,
and 
the points on sides 
and 
,
the point of intersection of segments 
and 
.
is denoted by 
.
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
be a convex pentagon such that 
,
,
.
to 
are concurrent.
Y by GoJensenOrGoHome, itslumi, Adventure10, cubres, NicoN9
For a positive integer 
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
.
should be a perfect square. Thus 
.
.![\[ \prod_{i = 2}^{n} \frac{i}{i - 1} = n \]](http://latex.artofproblemsolving.com/2/3/6/236859bfd4b55a1b1c1e8520fda5bca9170ce1c1.png)
initially, which is a perfect square. We wanted to reverse several fractions such that the final product is 
.![\[ \prod_{i = 2}^{\sqrt{n}} \frac{i}{i - 1} = \sqrt{n} \]](http://latex.artofproblemsolving.com/4/2/1/4217e7a703142d345d3f229b19996d811b6599c4.png)
is![\[ \left(\frac12\cdot \frac23\cdot \frac34\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right) = \frac{n}{k^2}=1. \]](http://latex.artofproblemsolving.com/6/2/c/62c6bdd77184c5d5534488f124bf8a969b436145.png)
Suppose that ![\[ (2\cdot 1)(3\cdot 2)\cdots (n(n-1)) = (n-1)!^2n,\]](http://latex.artofproblemsolving.com/3/9/d/39d71adbc84873a602a9a22ac0ff9f67db4e5656.png)
which means 
,
fractions. If 
,
is bad. Whenever we flip a fraction 
,
,
for some positive integer 
.
fractions.
for distinct values of 
.![\[ \left( \frac{1}{2} \right)^2 \cdot \left( \frac{2}{3} \right)^2 \cdot \dots \cdot \left( \frac{\sqrt{n}-1}{\sqrt{n}} \right)^2 = \frac1{n}\]](http://latex.artofproblemsolving.com/2/4/7/2471572cbba208df93cc24511e84518bbe6632ba.png)
to 
.
.
in both the numerator and denominator. This implies that 
fractions. The product of these fractions will then be
![\[\frac{1}{2}\cdot\frac{2}{3}\dots\frac{\sqrt{n}-1}{\sqrt{n}}\cdot\frac{\sqrt{n}+1}{\sqrt{n}}\cdot\frac{\sqrt{n}+2}{\sqrt{n}+1}\dots\frac{n}{n-1}\]](http://latex.artofproblemsolving.com/d/3/2/d32d2e1a511041ebb1a124496522d57f9e9367bb.png)
And it is easy to see that this works.![\[\frac{2}{1}\cdot\frac{3}{2}\dots\frac{n}{n-1}=n\]](http://latex.artofproblemsolving.com/6/9/f/69fe2c2adae0749a09ce9644d85148ad2e9a5a16.png)
by the construction that works to get![\[\left(\frac{2}{1}\right)^{1+\epsilon_1}\cdot\left(\frac{3}{2}\right)^{1+\epsilon_2}\dots\left(\frac{n}{n-2}\right)^{1+\epsilon_n}\]](http://latex.artofproblemsolving.com/b/3/5/b35e94d6dab521618259f2f4ac9766a3b882ac94.png)
Where 
is either 
.
must be an even number, so 
![$$\left[\frac{1}{2} \cdot \frac{2}{3} \dotsm\frac{\sqrt{n}-1}{\sqrt{n}}\right] \times \left[\frac{\sqrt{n}+1}{\sqrt{n}} \dotsm \frac{n}{n-1}\right].$$](http://latex.artofproblemsolving.com/4/f/3/4f37f626ec7b4fc891b660eb1ca6d4a8fc6f11ac.png)
All of the first bracket simplifies to 
an the second to 
and 
and the denominator 
has product 
so 
Also note that for the value to be 
.
,
since otherwise for some prime, we have a number of those primes odd and thus we cannot group them into two groups having the same cardinality.
hence 
and 
,
.
,
.
.
is a perfect square and so is 
groups of 
fractions and reciprocate every group except the first. Here's an example for 
.
.
,
)
,
,
,
,
,
over and our product becomes ![\[\underbrace{\frac{1}{a-1}}_{\text{first } a-2\text{ terms}}\cdot\underbrace{\frac{1}{a+1}}_{\text{terms }a+1\text{ to }a^2-1}\cdot\frac{a-1}{a}\cdot\frac{a+1}{a}\cdot\frac{a^2}{a^2-1}=1\]](http://latex.artofproblemsolving.com/a/e/9/ae9e06bb37e39a92944e9c0f59c0b6fc8b96cc0e.png)
![\[\frac{1}{2}*\frac{2}{3}*\cdots{}*\frac{k-1}{k}=\frac{1}{k}\]](http://latex.artofproblemsolving.com/8/d/7/8d7b98d599001044af590ba0af55f9aa718f84f5.png)
and![\[\frac{k+1}{k}*\frac{k+2}{k+1}*\cdots{}*\frac{k^2}{k^2-1}=k\]](http://latex.artofproblemsolving.com/2/4/4/244cb4144bea2434fc89e9158e6a5d7aec6bdf4f.png)
which we claim to be ![\[x=\frac{1}{\sqrt{n}}.\]](http://latex.artofproblemsolving.com/9/6/a/96a5531d5f1dc37ee45a528f5213b868ef560007.png)
We also know that 
.![\[\frac21, \frac32, \cdots, \frac{a^2}{a^2-1},\]](http://latex.artofproblemsolving.com/c/b/e/cbe57c2292a17355eec7ff3ea8a8fbf821829e2d.png)
and we want a product of the reciprocal of some of these to be 
.
,![\[\frac{1}{2}\cdot \frac{2}{3}\cdot \dots \frac{a-1}{a}\]](http://latex.artofproblemsolving.com/2/0/d/20d7a5bf10c1b6b3245eb07da3670bcb1c2817fd.png)
![\[\frac{1}{a},\]](http://latex.artofproblemsolving.com/9/1/1/911adf4e5042e448721f34035540ccdd5eb2e29e.png)
which means that 
.
Let 
be a subset of 
such that all the numbers of the form 
are reciprocated.
.![\[\frac{n}{\left(\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1}\right)^2}=1\]](http://latex.artofproblemsolving.com/c/6/2/c62b3134e90ce0ff0bde358510c800ef3a5477df.png)
implying that we require![\[\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1} =\sqrt{n}.\]](http://latex.artofproblemsolving.com/8/a/8/8a8fe0daf964ba643018f72dbb24642151bc6d22.png)
Since the LHS product is rational, we require the RHS to be rational as well. But 
Let ![\[\frac{k^2}{k^2-1}\cdot\frac{k+1}{k}\cdot\frac{k-1}{k-2}\cdot\frac{k-2}{k-3}\dots\frac{3}{2}\cdot\frac{2}{1}=k.\]](http://latex.artofproblemsolving.com/2/6/a/26a4000da8494c0cb732a230dd8dcb450c0c8280.png)
So the answer is 
.
,
,
.
did work, and then I wondered if the composite numbers worked, so I tried 
,
,
.
Notice the product of the reciprocated fractions is 
we multiply by 
so each prime factor gets changed by a multiple of 
![\[ \left(\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right). \]](http://latex.artofproblemsolving.com/6/9/2/6920428a850470fcda59431e7f3e817a95e2fc2c.png)
![\[n!(n-1)!=n((n-1)!)^2\]](http://latex.artofproblemsolving.com/7/4/d/74d724e8c64e2d772edb677359308fdc7e5367b0.png)
is a square, so 
and win. 
must be a perfect square, so 
where 
is an odd number it is impossible to divide it by two (such that 
of the numerator is equal to the denominator - so that the final product is 1). Now, notice that when 
for odd 
Thus, if ![\[\frac{1}{2}\cdot \frac{2}{3} \cdots \frac{k-1}{k}\cdot \frac{k+1}{k} \cdot \frac{k+2}{k+1} \cdots \frac{k^2}{k^2-1}\]](http://latex.artofproblemsolving.com/2/b/2/2b2977d9d34270b3deb7cd4f70bdd90330f1bfe0.png)
,
.
.
.
,
the product of all the numerators and denominators of the fractions must be a perfect square. Note that the product of the numerators is 
Therefore, since this quantity must be a perfect square, we know that 
as desired, concluding the proof. 
I claim 
work. We use induction. 
case is obvious. If the result is true for 
,
look at the final 
terms because the first 
.
, done.
work.
Dividing equations yields
since 
terms also multiply to 
Reciprocate either the first 
which should obviously be distinct. Then we have that 
Thus 
Proof of necessity:
Assume you reciprocate some of the fractions among the fractions to get 1 .
as 
.
,
where 
and 
but we know that 
.
.
as desired.
be the ``prime factorization" of the product of (possibly reciprocated) fractions, where each 
is a not necessarily positive integer. Clearly, the parity of each 
,
.
.
where 
,
.
we must have 
and thus 
and keep the rest the same. Then we have 
are valid. Let 
Thus, 
Now, I claim that if 
will just multiply the current value by the square number 
Thus, since 
