[/quote]
,
be a triangle. Let 
be a circle passing through 
intersecting 
at 
and 
at 
.
be the intersection of 
and 
and 
be the intersections of 
and 
with the tangent to 
at 
.
be the second intersection of 
and 
and 
,
is the midpoint of 
.
is an arbitrary point on 
.
be the intersection of 
and 
.
intersects with the circumcircle of 
,
.
is the circumcenter of triangle 
.
.
where 
are integers such that 
for all 
.
![\[
r_1 \ge r_a,\quad r_2 \ge r_b,\quad r_3 \ge r_c
\]](http://latex.artofproblemsolving.com/8/2/1/8214c052ab9ad2c979414f92b6707fa6de33664f.png)
because if any of the radii were smaller than these, the circles would not be able to reach and touch the incircle.![\[
\frac{r_1 + r_2 + r_3}{r} \ge \frac{r_a + r_b + r_c}{r}
\]](http://latex.artofproblemsolving.com/b/b/a/bbadb40e6aa197207abfa907981e0a4a88926969.png)
![\[
\frac{r_a + r_b + r_c}{r} = \frac{h_a - 2r}{h_a} + \frac{h_b - 2r}{h_b} + \frac{h_c - 2r}{h_c}
\]](http://latex.artofproblemsolving.com/5/3/2/5329fec1f02a728b5f34817e0ed43ded663516ae.png)
![\[
= \left(1 - \frac{2r}{h_a}\right) + \left(1 - \frac{2r}{h_b}\right) + \left(1 - \frac{2r}{h_c}\right)
= 3 - 2r \left( \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} \right)
\]](http://latex.artofproblemsolving.com/1/e/0/1e087ea87566fb4cf8cd2ae0cda3389247095eea.png)
![\[
\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} = \frac{a + b + c}{2S}
\]](http://latex.artofproblemsolving.com/a/4/a/a4ab49131584cf2028270e7248fcbb5d2c6a7adb.png)
![\[
3 - 2r \cdot \frac{a + b + c}{2S} = 3 - \frac{4S}{a + b + c} \cdot \frac{a + b + c}{2S} = 3 - 2 = 1
\]](http://latex.artofproblemsolving.com/0/f/3/0f3ada5ca22deeecc1644d79138942caf04d5134.png)
![\[
\frac{r_1 + r_2 + r_3}{r} \ge 1
\]](http://latex.artofproblemsolving.com/7/1/e/71ea14c8fb8e3860a1db427a0afcdb3b2fd6254b.png)
a prime number with 
and 
when 
.
else 
.
be real numbers such that 
.
be positive integers and 
be prime. We know that 
and 
.
is a perfect square.
be a triangle with circumcentre 
.
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
.
be the circle passing through 
with 
such that ![\[ (a-1)(b-1)(c-1) \]](http://latex.artofproblemsolving.com/f/8/c/f8c33b59bdba1fdd12dec51f16a0b1b93e60754f.png)
is a divisor of 
is called good if there exists a permutation 
of the numbers 
,
and 
have different parities for every 
is a quadratic residue modulo 
for every 
is considered a quadratic residue modulo 
such that 
.
| 
with 
are positive integers and 
)
be a real numbers such that 
P
Where 
.
be non-negative real numbers. Prove that
Looking for an algebraic solution!
Y by Adventure10, megarnie, Mango247
A box whose shape is a parallelepiped can be completely filled with cubes of side 
If we put in it the maximum possible number of cubes, each of volume 
, with the sides parallel to those of the box, then exactly 
percent of the volume of the box is occupied. Determine the possible dimensions of the box.
If we put in it the maximum possible number of cubes, each of volume 
, with the sides parallel to those of the box, then exactly 
percent of the volume of the box is occupied. Determine the possible dimensions of the box.
![\[ x = \left\lfloor\frac{a}{\sqrt[3]{2}}\right\rfloor \ \ \ \
y = \left\lfloor\frac{b}{\sqrt[3]{2}}\right\rfloor \ \ \ \
z = \left\lfloor\frac{c}{\sqrt[3]{2}}\right\rfloor \ \ \ \]](http://latex.artofproblemsolving.com/0/a/d/0ad67bde4ee74c09e63db306324de702f21d2751.png)
is the maximal number of cubes with sides of length ![$ \sqrt[3]{2}$](http://latex.artofproblemsolving.com/5/f/3/5f31adac765e62322c233a3f9f0da7a5157aee33.png)
that
.
,![\[ \frac ax\cdot \frac by\cdot \frac cz=5\ \ \ \ \ \ \ \ (1)\]](http://latex.artofproblemsolving.com/9/6/0/960d40c904c4d01ec487308bed476d8e01fabc06.png)
and 
for 
.
and 
.![\[ \begin{tabular}{|c|c|c|}
\hline
a & x & a/x \\ \hline
1 & 0 & - \\ \hline
2 & 1 & 2 \\ \hline
3 & 2 & 3/2 \\ \hline
4 & 3 & 4/3 \\ \hline
5 & 3 & 5/3 \\ \hline
6 & 4 & 3/2 \\ \hline
7 & 5 & 7/5 \\ \hline
8 & 6 & 4/3 \\
\hline
\end{tabular}\]](http://latex.artofproblemsolving.com/b/a/4/ba4d41cec5c40b5e625794c8194b423ecf38cde8.png)
:
and 
.
.![$ \frac ax\ge \sqrt[3]{5}$](http://latex.artofproblemsolving.com/1/c/c/1cc7a26e9415f483cfed8aab4c7a64a939985711.png)
.![\[ x< a/\sqrt[3]2 < x+1,\]](http://latex.artofproblemsolving.com/2/2/2/22282bcfa2c315e02a88fab0d24555b83e83d433.png)
![\[ \sqrt[3]2< a/x < \sqrt[3]2(1+\frac 1x)\]](http://latex.artofproblemsolving.com/2/d/a/2dab5f5b847e7944412e1ad54855f9993daeaeea.png)
then 
and ![$ \frac ax<\sqrt[3]2(1+\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$](http://latex.artofproblemsolving.com/9/6/c/96c16643e10510c44a1a8428da091c24eff8865c.png)
.
and 
.![$ a/x=3/2<\sqrt[3]5$](http://latex.artofproblemsolving.com/0/2/1/021973fca4530cac7cc9e5f647841612d0fde603.png)
and so necessarily 
.![\[ \frac by \cdot \frac cz =\frac 52 \ \ \ \ \ \ (2)\]](http://latex.artofproblemsolving.com/2/1/4/2144675a9ba5255ed07b89ca0c2aae0f9fb14e3a.png)
imply ![\[ y< b/\sqrt[3]2 < y+1,\ \
\textrm{and} \ \ z< c/\sqrt[3]2 < z+1.\ \ \ \ (3)\]](http://latex.artofproblemsolving.com/2/8/3/2833fc55bd61aa71f73cfcaff46a3ab0d620a8d3.png)
that![\[ \frac by \ge \sqrt{5/2}\ \ \ \ \ (4)\]](http://latex.artofproblemsolving.com/0/b/8/0b8c820d050436ebf792122ddb8cf401229082b1.png)
then 
and we would have ![$ c/z=5/4<\sqrt[3]2$](http://latex.artofproblemsolving.com/4/6/5/465b7d23bed449d6d8981a5371915488cfdf4815.png)
,
.
then 
and ![$ \frac by<\sqrt[3]2(1+\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$](http://latex.artofproblemsolving.com/3/c/c/3cc39ccf0d1acfcd6d40aba780140161e34c7cbb.png)
as 
.
.
we have 
and for 
we have
and so necessarily 
and 
)
and 
.
and ![$ \frac cz<\sqrt[3]2(1+\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$](http://latex.artofproblemsolving.com/7/1/c/71c6369719964ab482b413d7676206c831beee3c.png)
.
there are the only two possible values
and 
which yield the known solutions.
