[/quote]
,
be the set of all positive integers. Determine all subsets 
of 
such that
and 
intersect at 
and 
.
be the circumcircle of the triangle 
;
and 
,
and 
intersects 
.
on 
.
and 
meet at 
,
.
.
?
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
,
,
,
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 
,
such that in each Japanese triangle there is a ninja path containing at least 
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
and 
.
passes through the midpoint of 
.
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
for 
.
,
,
divides 
.
,
.
,
.
,
,
.
,
,
.
and permutations.
works. Clearly it satisfies the conditions given. By the Euclidean Algorithm 
.
is not divisible by 5 or 7. However, note that 
,
so that 
.
,
so that 
.
and 
are coprime.
Adding all of the inequalities gives us the desired result. Equality holds only when 
.
Y by Adventure10, Mango247
The 30th Irish Mathematical Olympiad took place on Saturday the 6th of May 2017
This Thread contains all 10 problems asked over 2 papers. Solutions (particularly clear solutions) are missing for some of these problems so feel free to post your solutions!
Answer list :
By LateralUnits
By PepsiCola
by jasonhu4
By a1267ab
This Thread contains all 10 problems asked over 2 papers. Solutions (particularly clear solutions) are missing for some of these problems so feel free to post your solutions!
Answer list :
By LateralUnits
By PepsiCola
by jasonhu4
By a1267ab
This post has been edited 5 times. Last edited by LateralUnits, May 14, 2017, 3:56 PM
for 
and 
as diameters. The two circles passing through 
,
, two circles at 
at 
and the two circles at 
at 
.
and 
is similar to 
,
.
and permutations.
.
denote the number of distinct colorings satisfying this condition.
.
is defined by 
and
Prove that 
is divisible by 
is divisible by 
and the two numbers 
and 
are co-prime?
draw both teams gain 
point, and for other draws 
e
is divided into 
and 
is a right angle. Prove that :
,
When is equality attained?
.
into 
,
.
are quadratic residues mod 
,
for some nonzero 
in the denominator of the expression. 
and 
are both integers mod 
t
mod 
,
is called 
for all positive integers 
-
is just casework and for 
,
,
be the smallest positive integer (if exists) that 
.
.
)
and 
must be zero. We can prove by induction that 
for all 
.
for all 
.
)
be real numbers and 
be positive real numbers. Then
Equality holds only when 
,
for some nonzero 
.
s.t. 
(by symmetry: 
)
with 
)
We know from Fermat's Little Theorem that 
and we know from the definition of the Quadratic Residue and Euler's Criterion that 
- but I don't see how to put it all together.
,
![\[a^3 -\frac{a}{2} - abc = 0 \]](http://latex.artofproblemsolving.com/c/c/8/cc8ab8489238acdbba833a57af560c258e230fd2.png)
![\[b^3 -\frac{b}{2} - abc = 0 \]](http://latex.artofproblemsolving.com/d/2/f/d2f90a02a2bc854feee71bc40dff7a8a754cb79d.png)
![\[c^3 -\frac{c}{2} - abc = 0 \]](http://latex.artofproblemsolving.com/0/6/e/06ee2af858df3e618cbe64314ed4d2a280ce63bc.png)
Adding these equations together (and taking into account that 
)![\[a^3+b^3+c^3-3abc=0\]](http://latex.artofproblemsolving.com/8/7/8/878081ca8ad1b20901fbf3c03153c03d5bfb0d9d.png)
Substituting 
above, 
,
