AoPS Academy
such that 
is the square of an integer.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 
; then the numbers on the board after step 1 are 
; after step 2 are 
; 
, prove that the number 2024 cannot appear on the board.
; 
, prove that the number 2024 can appear on the board.
(base-10) into base-, where is the smallest integer that has 8 factors. The answer will be the code to the safe.[/quote]
such that there exist positive integers 
and 
with 
satifying the condition that,
is an integer.
.
is a nonzero polynomial with integer coefficients. Prove that there exists infinitely many prime numbers 
such that for some natural number , 
.
be positive real numbers that fulfill![\[
\frac{x^2}{y^2}+\frac{4x^2-3xy-4y^2}{2xy-5y^2}=2
\]](http://latex.artofproblemsolving.com/6/c/c/6ccba85deb3e6aa205abd3bc8e9e9bb68f4df6fd.png)
Hence find the value of 
to 
on a different unit square of a 
square grid, such that consecutive integers are on adjacent squares, and is not adjacent to ? (Note that adjacent squares are squares that share a common side.)
be a prime and 
be positive integers such that 
and 
is prime. Prove that 
, 
, , and 
. The railway forms a complete loop, as shown below, and trains go in both directions. Suppose that a trip between two adjacent towns costs one ticket. Using exactly eight tickets, how many distinct ways are there of traveling from town and ending at town ? (You may pass through town in the middle).
be the solid square region with vertices at![\[
(0, 0), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0), \left(\frac{1}{2}, -\frac{1}{2}\right).
\]](http://latex.artofproblemsolving.com/a/9/8/a9898f9237f07eaef5d17a9ac689d80ca631863f.png)
consists of two copies of 
: one copy which is rotated 
counterclockwise around the origin and scaled by a factor of 
, and another copy which is also rotated counterclockwise around the origin and scaled by a factor of , and then translated by 
. and 
below.
be the limiting region of the sequence 
. can be written as 
for relatively prime positive integers and . Find 
.
shows that after each species divides and then each amoeba cell preys on exactly one bacterial cell, it turns out kill bacteria. Determine the ratio of the number of amoeba to the number of bacteria on the first day. is a positive integer. Let 
.
cm. The first nine balls are placed on the table so that cm and all the balls are arranged following the pattern of the arrangement of the balls made by Budi, calculate the height of the center of the topmost ball is measured from the table surface in the arrangement of the balls done by Bambang. whose sides are 
cm, 
cm, and 
cm. Find the maximum possible area of the rectangle can be made in the triangle .
people waiting in line to buy tickets to a show with the price of one ticket is 
Rp.. Known of them they only have 
Rp. in banknotes and the rest is only has a banknote of Rp. If the ticket seller initially only has Rp., what is the probability that the ticket seller have enough change to serve everyone according to their order in the queue?
be a root of 
. If splits in ![$\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/e/5/a/e5a21492095c9116b100e2f7626a70cc26834e82.png)
, then 
; otherwise 
. Either way, .
. We obtain the 
equality chain![\[\left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) + 1 = 0 \implies t^3 + \frac{1}{t^3} + 1 = 0 \implies t^9 = 1.\]](http://latex.artofproblemsolving.com/a/e/9/ae935843dc527f9cb9a88fca74b5a4f2cf4b7f12.png)
Thus the (multiplicative) order of divides 9, so it is one of 1, 3, or 9. If it is one of 1 or 3, then 
, meaning 
, so 
. Otherwise, the order of is 9. belongs to 
, a group with 
elements. (It is also cyclic, but we don't need that.) Then by Lagrange, 
, so 
. (If we know that the group is cyclic, we can directly deduce .) To conclude, either or .
trick has become quite standard [one easy application is to determine 
], and (2) we essentially need to deal with the field of 
elements, a concept not normally included in the olympiad canon.
are co-prime integers, and 
is a prime divisor of 
then 
. and two integers such that 
, , is not a common divisor of but 
. Take the smallest such . Obviously 
, so 
. and by smaller numbers: we show that there are some integers 
, not both zero, such that 
and 
.
and 
by their greatest common divisor. Let 
, 
and 
. notice that 
and 
, so 
is not divisible by . Then the have 
.
and 
modulo 
and 
, we can see that 
or 
. Hence there is another prime divisor 
of 
with 
.
. Then
If 
then 
, contradiction.
must be checked manually.
be a root of . If splits in , then ; otherwise . Either way, .
, it has no root either in 
or 
, otherwise 
which is obviously false. Then, there is no such that 
.
be a root of . If splits in , then ; otherwise . Either way, ., it has no root either in or , otherwise which is obviously false. Then, there is no such that .
modulo .
does not represent the "field" of integers modulo 
(which can't be a field!); instead, it is the unique (up to isomorphism) degree-two extension of 
. has no root in , this polynomial is irreducible, and so the quotient ![$\tfrac{\mathbb F_5[x]}{(x^2 - 4x + 1)}$](http://latex.artofproblemsolving.com/4/b/5/4b593657c53a83056fec776871e1e8baa5778c8b.png)
is a field. We can describe this field symbolically as follows: if 
(formally!) satisfies 
, then the elements of this field are the elements of the form 
, where 
. elements up to isomorphism; we denote this field by . This construction generalizes to finite fields of arbitrary prime power.
for some 
.
has order .
This implies 
, so has order dividing . However, 
since 
. Therefore, has order exactly . 
so we're done.
for some . has order .This implies , so has order dividing . However, since . Therefore, has order exactly . so we're done.
is defined as follows: to be a root of an irreducible quadratic (say 
) in . Then is the set
with 
, which means that we can assume is the root of 
for some 
, i.e. 
for some quadratic non-residue 
mod .
in . If its discriminant 
is a quadratic residue mod , then the quadratic formula gives us two roots in 
. Otherwise, 
is a quadratic residue mod , say 
. Then notice that
.
since we divide by 
a couple of times, but we can just check it manually then.
for some . has order .This implies , so has order dividing . However, since . Therefore, has order exactly . so we're done.?
is (related to the field in question) implies in divides the order of the field. Is that what you want?
to 
. By the quadratic formula, if 
is a QR then in fact 
, otherwise is an NQR and we still have . Thus we may substitute 
, whence
where all equalities are in . This implies that the order of in is , since 
. This then implies , so we're done. 
, so that the equation becomes 
This requires the order of mod to be precisely equal to . On the other hand, 
is an element of 
, so and we have the result.
has one root in 
if , three roots in if 
, and no roots in otherwise. When , the only root is 
, from now on, assume . Throughout, let 
be a root of 
and 
be the Frobenius automorphism sending to 
. Solving the cubic shows has roots 
, 
, and 
.
, the degree of the field extension 
is 
if 
, if 
, and or 
otherwise.
if 
, which is enough since 
is Galois. We have![\[3 \ge [\mathbb F_p(\omega) : \mathbb F_p] = [\mathbb F_p(\omega) : \mathbb F_p(\omega + \omega^{-1})][\mathbb F_p(\omega + \omega^{-1}) : \mathbb F_p] \ge 2[\mathbb F_p(\omega + \omega^{-1}) : \mathbb F_p],\]](http://latex.artofproblemsolving.com/c/c/a/cca28ef38bcb80aa99a43e4a873d1490c5f366b9.png)
so 
is nontrivial and .
if , which is enough since is Galois. If , 
and we are done. If , the minimal polynomial of is![\[(x - \omega)(x - \sigma(\omega)) = x^2 - (\omega + \omega^p)x + \omega^{p + 1} = x^2 - (\omega + \omega^{-1})x + 1.\]](http://latex.artofproblemsolving.com/a/e/1/ae18d5065aea4203367a95e87869fff68fa9ceec.png)
But the minimal polynomial has coefficients in , so 
is in .
in ![$\mathbb{F}_{p^2}=\mathbb{F}_{p}[\sqrt{a^2-4}]$](http://latex.artofproblemsolving.com/1/7/a/17a67895dc1552660c66270d5cc0faa583a128d1.png)
(which has order 
)and so we have (since ) ![\[x^6+x^3+1 \equiv 0 \pmod p \iff x^9 \equiv 1 \pmod p \iff 9 \mid p^2-1\]](http://latex.artofproblemsolving.com/c/c/b/ccba0a6fb10b7c08d92175fe02972c5896d5ec94.png)
And done.
such that 
(we can't do this in modulo since 
could be a qnr). We get 
Therefore, is the order of in 
Since the order of is exactly 
it follows 
which implies the statement.
,
or 
,
is integer.
is a quadratic equation, and every quadratic with coefficients in 
for 
so 
or 
.
and 
then 
.
we have 
or 
