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and 
be positive integers such that 
divides 
. Prove that 
.
where 
are nine distinct integers. Prove that there exists an integer 
such that for all integers 
the number 
is divisible by a prime number greater than 20.
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points 
in such that 
.
, there exists a balanced set consisting of 
points. for which there exists a balanced centre-free set consisting of points.
be a triangle, 
is incenter of triangle . Draw 
perpendicular to 
at 
and 
perpendicular to 
at 
, 
. Prove that: Area of triangle 
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?
with
. (It is implied that is written with a decimal representation.)
regions. Every pair of lines in the same group are parallel. Let 
and 
respectively be the number of lines in groups 1, 2, and 3. If no lines in group 3 go through the intersection of any two lines (in groups 1 and 2, of course), then the least number of lines required in order to have more than 2018 regions is ....
where 
and 
are squares with both planes being parallel. The length of the sides of and respectively are 
and 
, and the height of the frustum is 
. Points and respectively are intersections of the diagonals of and and the line 
is perpendicular to the plane . Construct the pyramids 
and 
and calculate the volume of the 3D figure which is the intersection of pyramids and .
is on the 3rd row, 4th column, and on the 6th diagonal. Meanwhile the position of the number 
is on the 3rd row, 5th column, and 7th diagonal.
based on its row, column, and diagonal.. is the set of 3-digit integers not containing the digit 
. Define a gadang number to be the element of whose digits are all distinct and the digits contained in such number are not prime, and (a gadang number leaves a remainder of 5 when divided by 7. If we pick an element of at random, what is the probability that the number we picked is a gadang number?
and satisfy 
dan 
. The number 
is a fraction in its simplest form.
. is (equal to some number) , determine all positive factors of . above (specifically in problem b), determine the probability of taking a factor who is a multiple of 4.
be given in the following graphs.
with 
for all 
where 
is the domain of 
..
so that 
. has side lengths 
cm and 
cm. All four of its vertices lie on a circle. Calculate the area of quadrilateral . and 
, with 
and 
. It is known that 
, where 
is a 3-digit number whose number in its tens place is 5. Determine the number/quantity of all possible values of .
(the original problem does not have an overline, which I fixed) is arranged from the set 
. Such number satisfies 
. Determine the quantity of different possible (such) numbers.
(i.e., the square on 
has as one of its sides and lies outside ). Show that the lines drawn from the vertices 
, 
, and 
to the centers of the opposite squares are concurrent.
. A circle O with center C has a radius of 2. Let P be a point on the circle O.
?
?, and 
are points on sides 
and 
, respectively, such that 
. If 
and 
, find the area of .
and of the equation ![\[(a^2+b)(a+b^2) = (a-b)^3.\]](http://latex.artofproblemsolving.com/b/5/7/b57940d59694d6892784e6390487010d5e036604.png)
and divide by (
) to obtain
. Therefore for some integer we must have 
.
(value 
is discarded)., we find the solutions:
? 
I found that 
became 
, yielding 
and 
. So 
should be an additional solution...
is always a solution, so henceforth we assume 
. Expanding the equation, we get![\[m^2n^2+mn+n^3=-3m^2n+3mn^2-n^3,\]](http://latex.artofproblemsolving.com/f/c/2/fc2ba10e049b732afd73265dfecd517b4439363d.png)
or![\[2n^2+n(m^2-3m)+(3m^2+m).\]](http://latex.artofproblemsolving.com/c/9/e/c9ea34f1af43b85ea869be843618975334e001bc.png)
The quadratic formula then gives![\[n=\frac{3m-m^2\pm\sqrt{(m^2-3m)^2-8(3m^3+m)}}{4}=\frac{3m-m^2\pm(m+1)\sqrt{m(m-8)}}{4}.\]](http://latex.artofproblemsolving.com/8/1/9/819c07cd91d8f0eb9207569bfe81c6560441b113.png)
Thus, 
is a perfect square, so 
. This gives the final solution set of![\[\boxed{(m,0),(-1,-1),(8,-10),(9,-6),(9,-21)}.\]](http://latex.artofproblemsolving.com/a/9/4/a94a60ded6e6fcbe8fe7b8aec56b2363f675ad73.png)
is clearly a solution so suppose ; then, expanding and dividing by we get the quadratic 
. But the discriminant is 
so 
, and thus 
, so using diff. of squares we get 
. After a bit of bashing our final answer is 
.
. Then, we have 
. This leads to the solutions 
and 
. Now, assume 
. We get 
, which we already found. Finally, assume 
, which leads to 
. Thus, all solutions of the form 
satisfy the equation. and are nonzero. Expanding the given equation gives us
Since 
, we must have 
as well. Notice that 
, so it is sufficient to find all such that 
for some integer .
Now, we have to find all such that 
is a perfect square. Let 
, so that 
. Note that if 
, then
so it is sufficient to check in the range ![$[0,7]$](http://latex.artofproblemsolving.com/2/4/0/240b93fd38140efa342f509e7759d003aa437172.png)
. We find that the only are 
, which means 
. For each of these, we can manually check what can be, and we find the solutions 
.
.
![\[m^3+m^2n^2+mn+n^3=m^3-3m^2n+3mn^2-n^3\]](http://latex.artofproblemsolving.com/d/4/c/d4c1b65cfda5d75f63134e26b3e4c1fd3c058f0c.png)
Rearranging, we get![\[n(2n^2-n(3m-m^2)+(3m^2+m))=0\]](http://latex.artofproblemsolving.com/1/4/6/14634b4e5271c58738dab247ddac70e98c4433a5.png)
The will always gives solutions, so we will assume 
from now on. Otherwise, by the quadratic formula we must have the determinant of the inner quadratic is a square,![\[\Delta^2 = (3m-m^2)^2-8(3m^2+m) = m(m+1)^2(m-8)\]](http://latex.artofproblemsolving.com/f/c/2/fc209d0d34dcd3613510cab579d7d6a83b5ae0bf.png)
Thus, 
is a square. Note that 
. and 
are squares, this happens when 
. Case 2 is when either term is 0, 
. Case 3 is when and are both of the form 
, which has no solutions.![\[(m,0), (8,-10),(-1,-1),(9,-6),(9,-21)\]](http://latex.artofproblemsolving.com/d/8/4/d84c0378e9cd8719c1998dcd1a95e2aab3a9e8c9.png)
and 
, we test until 
.
, then 
as well as 
are both perfect squares, so the perfect squares must differ by 
. The difference between two consecutive perfect squares eclipses at 
and we have 
, so we test all the way until .
, then as well as are both perfect squares, so the perfect squares must differ by . The difference between two consecutive perfect squares eclipses at and we have , so we test all the way until .
for any integer , and also the additional solutions 
![\[ m^2n^2 + 3m^2 n - 3mn^2 + mn + 2n^3 = 0 \]](http://latex.artofproblemsolving.com/b/a/1/ba119d07cdc40963e7936fa60a5bf0fbc0c08fab.png)
, then we get . Now assume . Then dividing both sides by gives ![\[m^2 n + 3m^2 - 3mn + m + 2n^2 =0\]](http://latex.artofproblemsolving.com/3/7/d/37d6343036643796aae516c616496a1ef6c2dbf9.png)
![\[2n^2 + (m^2 - 3m)n + 3m^2 + m = 0\]](http://latex.artofproblemsolving.com/8/a/e/8ae1c5d4f3ce604962f18290e1d5f2e24d3d5df7.png)
![\[(m^2 - 3m)^2 - 8(3m^2 + m) = m(m-8)(m+1)^2\]](http://latex.artofproblemsolving.com/a/6/d/a6ddac9ab9ccc2cc7c84ef4af3ecd9d9cfd432c9.png)
must be a perfect square. is a square.
, then we have 
. Assume and are nonnegative. We have 
.
, we have 
, so 
. This gives the described solutions.
.
Observe that the discriminant of this expression with respect to must be a perfect square, hence 
is a perfect square. This yields 
, which implies the desired solution set.
.Observe that the discriminant of this expression with respect to must be a perfect square, hence is a perfect square. This yields , which implies the desired solution set.?
, hence, we have 
If we use the quadratic, formula we get 
, we need, 
, to be a perfect, square, note, that 
, for all 
(integers), so we just test, out the solutions, here, to get 
, so , is not a valid solution.
![\[m^2n^2+mn+n^3=-n^3+3mn^2-3m^2n \iff 2n^2+3m^2+m^2n-3mn+m=0 \text{ if } n \neq 0.\]](http://latex.artofproblemsolving.com/a/2/5/a254165aa858e1589da35a7e66a8c74347c0057f.png)
Grouping the terms in terms of , we find that![\[(n+3)m^2+(-3n+1)m+2n^2=0,\]](http://latex.artofproblemsolving.com/3/8/5/385cd3921a750096ff8982fedfa5f5f80effce37.png)
and by the quadratic formula, we get that![\[m=\frac{(3n-1)\pm \sqrt{-8n^3-15n^2-6n+1}}{2(n+3)}=\frac{(3n-1)\pm \sqrt{-(8n-1)(n+1)^2}}{2(n+3)},\]](http://latex.artofproblemsolving.com/9/9/6/9969c117af3704a66f187681fd8933f5fe92af5c.png)
meaning that 
is a perfect square. Letting 
, rewriting in terms of gives us that![\[m=\frac{\frac{-5-3k^2}{8}\pm \sqrt{k^2\left(\frac{9-k^2}{8}\right)^2}}{2\left(\frac{25-k^2}{8}\right)}=\frac{(-5-3k^2)\pm k(9-k^2)}{2(5+k)(5-k)}.\]](http://latex.artofproblemsolving.com/a/1/7/a170e5d67238809a789b91b23e34eb825f36aa34.png)
We now take this into cases.
. In this case, note that would then be 
, giving us that![\[(m^2-3)(m+9)=m^3+9m^2-3m-27=(m+3)^3=m^3+9m^2+27m+27,\]](http://latex.artofproblemsolving.com/8/8/b/88bb6c8cfec29b6e078d2a3e7706266e6e86ec80.png)
which only has a solution at 
, which is not an integer. Therefore this case has no solutions.![\[m=\frac{(-5-3k^2)+k(9-k^2)}{2(5+k)(5-k)},\]](http://latex.artofproblemsolving.com/a/5/7/a574157c9b9d3e123fcfb73994c389254bbbb10b.png)
for 
.
, meaning that is actually equal to![\[\frac{k^2-2k+1}{2(k-5)},\]](http://latex.artofproblemsolving.com/8/5/7/85752aacc40a9e138259c7da73d7ca68753cbfc3.png)
which means that 
. Note that since 
has a remainder of when divided by 
, we actually have that 
. Therefore, the possibilities for in this case are 
, , 
, 
, 
, 
, 
, 
, 
, and 
. Checking all of these, we find that the only valid are 
, , , and , giving us the solutions , 
, 
, and 
.![\[m=\frac{(-5-3k^2)-k(9-k^2)}{2(5+k)(5-k)},\]](http://latex.artofproblemsolving.com/3/0/6/3069fef17186d7f2f16d952784f7b8060fce25c8.png)
for . Note that since 
, we have that is actually equal to![\[\frac{k^2+2k+1}{2(k+5)},\]](http://latex.artofproblemsolving.com/c/a/f/caf1c3ab91b4cf4409bb24a92d25409821565594.png)
which means that 
, which has a remainder of when divided by 
, meaning that 
. However, this is symmetric to the second case, meaning that all solutions that could be found here have already been found., , , and , and we are done.
![\[m^3 + n^3 + m^2n^2 +mn = m^3 - 3m^2n + 3mn^2 - n^3\]](http://latex.artofproblemsolving.com/d/4/9/d4983bbb3b7b5249a0332216246b31cd0efc77e6.png)
![\[2n^3 + (m^2 - 3m)(n^2) + (3m^2 + m)(n) = 0 \]](http://latex.artofproblemsolving.com/e/7/a/e7a919671dc8cebac17a41fde54b800275d51953.png)
, and then using the Quadratic Formula gets us:![\[ \frac{-m^2 + 3m \pm \sqrt{m^4 - 6m^3 - 15m^2 - 8m}}{4}\]](http://latex.artofproblemsolving.com/7/8/8/788bd5bc5eb732ce7770f8616dbdd27a118b61a5.png)
![\[\frac{-m^2 + 3m \pm \sqrt{m(m - 8)(m + 1)^2}}{4}\]](http://latex.artofproblemsolving.com/2/0/9/209d31bed3b141d8e5a60a0653b0db20a25f776b.png)
must be a perfect square. If 
, then this still holds true for , as 
This also holds true for 
.![\[m(m - 8) = k^2\]](http://latex.artofproblemsolving.com/3/8/6/386761592a403b46ccf4c6fa6ad41721e06e75d2.png)
![\[(m - 4)^2 - k^2 = 16\]](http://latex.artofproblemsolving.com/9/c/c/9cc3e4183e06d795a8d697255d3a977921a5a026.png)
![\[(m - 4 + k)(m - 4 - k) = 16\]](http://latex.artofproblemsolving.com/8/e/7/8e79112b7681f80f0392c7497778ff5348b6f8fc.png)
.
.
This simplifies to 
This rearranges to 
Now quadratic formula gives 
where the discriminant factors as . Then for some we have 
. This then becomes,
This then gives the claimed solution set.
.
both sides from the equation to obtain ![\[ab + a^2b^2 + b^3 = -3a^2b + 3ab^2 - b^3.\]](http://latex.artofproblemsolving.com/8/2/a/82a608de49a7158a1831cc8dcdf99e432bfa526e.png)
Dividing by on both sides (
) as well as rearranging, we obtain the quadratic in ![\[2b^2 + b(a^2 - 3a) + (3b^2 + b) = 0.\]](http://latex.artofproblemsolving.com/c/5/c/c5c0480a8a0bd2b0bd62c3ea58e7366db3cc499e.png)
Now we must have the discriminant 
, as must be an integer. Now upon factoring we find ![\[\Delta = a(a + 1)^2(a - 8) \implies a + 1 = 0, \text{ or } a^2 - 8a = (a - p)^2, \text{ for } p \in \mathbb{N}.\]](http://latex.artofproblemsolving.com/c/7/f/c7f3bc6f9f3206f3e8fd90a97b29fb0f6e413746.png)
). Then 
.
Then ![\[a = \frac{1}{2} \left(\frac{p^2}{p - 4} \right) \implies p = -12, -4, 2, 6, 8, 12, 20 \implies \boxed{a = -1, 9, 8}.\]](http://latex.artofproblemsolving.com/f/b/2/fb2cd401f064fe1788d194cbd99ca4fc94a2df0f.png)
Now 
was covered in the first case, so that 
yields 
and 
, as claimed. 
. Note that produces a valid solution for any and implies . Henceforth, assume 
.![\begin{align*}
(m^2+n)(m+n^2) - (m-n)^3 &= [m^3+m^2n^2+mn+n^3] - [m^3-3m^2n+3mn^2-n^3] \\
&= 2n^3 + n^2(m^2-3m) + n(3m^2+n) = 0 \\
&\iff 2n^2+(m^2-3m)n+(3m^2+n) = 0.
\end{align*}](http://latex.artofproblemsolving.com/2/2/a/22aa488243b041a090c65dca1d05bf373d407fa1.png)
, it suffices to find all values of such that![\[\Delta_n = (m^2-3m)^2-4 \cdot 2 \cdot (3m^2+n) = m(m-8)(m+1)^2\]](http://latex.artofproblemsolving.com/8/e/a/8ead751a17e8324bf4904b44c3002c648fd5a664.png)
produces a solution, but otherwise, we must have![\[m(m-8)=k^2 \iff (m-4)^2-k^2 = 16.\]](http://latex.artofproblemsolving.com/2/2/2/2226e72bc88623fd70693bc3542f3b8e88be1388.png)
or 
; the desired solution sets follow.
, , and allow for zero solutions.![\[n\left(2n^2+(m^2-3m)n+(3m^2+m)\right) = 0.\]](http://latex.artofproblemsolving.com/7/6/1/761f091fc7f16885e92f8c97a42c5bd8f88ff076.png)
, and the other two are of the form![\[n = \frac{-(m^2-3m) \pm (m+1) \sqrt{m(m-8)}}{4}.\]](http://latex.artofproblemsolving.com/f/0/4/f043dd10ff91e33e801f0dbe351cf83711497d42.png)
are 16 and 25. Substituting back in, we get our solutions![\[\boxed{(m,0), (-1,-1), (8,-10), (9,-21), (9,-6)}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/9/c/f/9cfda85e55a31afc6f1b28d2b96e74cca6d9fc25.png)
must be a perfect square. This simplifies to . This means that either , , 
, or is a perfect square. to be a perfect square, notice that 
is away and also a perfect square, so then in that case .
.
Simplifying as a quadratic in terms of yields 
so is a perfect square. This gives the solutions , , , and , along with the family of solutions.
If 
we can simplify and rewrite a quadratic in terms of 
where we can factorize the LHS as 
By solving 
we see that 
We get that 
Clearly, the answer is all integers.
or 
,
,
,
,
.
.
side length of 
area of 
is a right angle, and 
,
.
is a right angle, 
,
.
and 
both contain a right angle and 
through AA postulate. Because they are similar, 
,
,
,
.
.
,
.
In Farenhajt algebra, 
.
.
.
.
.
,
.
.
work, so the solution set is 
.
. We can discard the 
case, as the problem states 
.
.
. Looking at the expression in the root, we see that 
.
becomes 
.
must be a perfect square as 
for some integer 
,
.
is also a perfect square, and the only case for this is where 
.
and 
.
and 
,
.
,
.
,
,
,
must be a square for 
which immediately gives the solutions 