[/quote]
,
and 
Prove that
of the pizza, where 
is a positive integer. Find 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
.
is stylish because the set 
can be partitioned as 
,
,
.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
that verify the following equation for all 
:
is a prime number
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
,
and 
although its supposed to be a quick comparison.
is a function for which the expression 
for all 
is always a perfect square. Prove that 
for all 
.
such that 
.
be a quadrilateral inscribed in a circle 
,
be a point on the extension of 
such that 
and 
are tangent to 
and the line 
at 
.
be the second point of intersection between 
and 
,
be a triangle with incenter 
,
and 
.
and 
the reflections of 
with 
,
be the midpoint of 
and 
are perpendicular.
.
be the feet of the altitudes from 
to 
respectively. Let 
be the circle with diameter 
be the intersection of 
and 
be the intersection of 
be the lines through 
respectively. Let 
and 
.
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
is constant.
be tangent to the side 
.
on 
on 
lies on the circumcircle of triangle 
Y by Adventure10, Mango247, Mango247, Mango247
Let 
be a diameter of a circle and let 
be a point on the segement such that 
. Let 
be a point on the circle such that 
is perpendicular to . Let 
be the diameter through . If ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
denotes the area of the triangle 
, find ![$[ABD]/[CDE]$](//latex.artofproblemsolving.com/2/5/d/25d664cba9fde10261c491a2465bfd758de0ea39.png)
to the nearest integer.
be a diameter of a circle and let 
be a point on the segement such that 
. Let 
be a point on the circle such that 
is perpendicular to . Let 
be the diameter through . If ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
denotes the area of the triangle 
, find ![$[ABD]/[CDE]$](http://latex.artofproblemsolving.com/2/5/d/25d664cba9fde10261c491a2465bfd758de0ea39.png)
to the nearest integer.
- reflection of 
we have ![$[CDE]=[CDC']$](http://latex.artofproblemsolving.com/5/a/8/5a81dd07fb30a11d3311ccdf577055387ca7116a.png)
, so ![$[ABD]/[CDE]=AB/CC'=13$](http://latex.artofproblemsolving.com/a/4/7/a4763009911713b682c67ddd8eeec6cadb3b1a25.png)
.
,![$[CDE]=2[CDO]=CD\cdot OC$](http://latex.artofproblemsolving.com/3/0/c/30ccbf813067b4a75ef4dd8c6e86e5876703b77d.png)
and ![$[ABD]=2[BOD]=OB\cdot CD$](http://latex.artofproblemsolving.com/a/2/e/a2effc04f63d6ee03da77cbc7552d1e43c1d8cfb.png)
so ![\[\frac{[ABD]}{[CDE]}=\frac{r\cdot CD}{CD\cdot OC}=\frac{r}{OC}=\frac{r}{r-6x}=\frac{r}{\frac{13r-12r}{13}}=\boxed{13}\]](http://latex.artofproblemsolving.com/b/1/d/b1db56e99c1374b416084432da593aadf93af3c3.png)
