AoPS Academy
be a quadrilateral with 
. Diagonals 
and 
meet at 
. Let 
and 
denote the circumcircle and the circumcenter of triangle 
. Let 
and 
denote the circumcircle and circumcenter of triangle 
. Segment 
meets and again at 
and 
(other than 
and 
), respectively. Let 
and 
be the midpoints of minor arcs 
(not including ) and 
(not including ). Prove that 
.
such that 
. Find the minimum value of:
be positive real numbers such that 
and 
, prove that
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.
P![$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$](http://latex.artofproblemsolving.com/6/b/6/6b6011a32b103fe94652974a521eeae479cba1d8.png)
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 
,
.
,
.
,
,
,
and then get that 
must be a square for 
which immediately gives the solutions 