[/quote]
,
and 
Prove that
be a trapezoid with parallel sides 
.
and 
lie on the line segments 
and 
,
.
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in 
(or 
)
positive integers 
such that 
for any 
,
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
.
,
is the smallest number different from 
which doesn't divide 
.
such that 
.
and 
meet the sides 
and 
respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
is a point on the segment 
.
and 
lie on the circles 
and 
,
and 
.
and 
are concyclic.
and the recursive relation
for all natural number 
.
let 
be the circumcircle of 
and let 
be its radius. Point 
such taht 
and point 
is the foot of the perpendicular from 
.
mets 
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
and 
.
It is known that 
whenever 
Prove that 
for all points 
such that 
for all 
,
,
with 
or 
for each 
,
,
,
,
are constructed by the following rules: ![\[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\]](http://latex.artofproblemsolving.com/d/f/d/dfd49833078b6b676cd261cc358e8114b2c4b7f4.png)
![\[\begin{array}{lll}
a_{i+1} =
\begin{cases}
2a_{i-1} + 3a_i, \\
3a_{i-1} + a_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_i = 0, \\
\text{if } \varepsilon_i = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1, \\[15pt]
b_{i+1}=
\begin{cases}
2b_{i-1} + 3b_i, \\
3b_{i-1} + b_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_{n-i} = 0, \\
\text{if } \varepsilon_{n-i} = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1.
\end{array}\]](http://latex.artofproblemsolving.com/8/9/0/8908021c60992e39acdb4fcf6b587f96b88539e0.png)
Prove that 
.
Y by Adventure10, Mango247
Let 
be the midpoint of side 
of triangle 
(
), and let 
be the bisector of the angle 
. The line passing through perpendicular to intersects the side 
at the point 
. Prove that 
is equal to half the perimeter of triangle .
be the midpoint of side 
of triangle 
(
), and let 
be the bisector of the angle 
. The line passing through perpendicular to intersects the side 
at the point 
. Prove that 
is equal to half the perimeter of triangle .
so we need to prove that 
.
as the point on side 
(and hence 
)
i.e. that 
.
.
.
.
so 
hence 
is isosceles.
as required.
cut resp. 
at 
it s clear that 
are isoceles and 
where 
pass through the midpoint of 
are the midpoints of 
hence 
Then, by LoS and 
isosceles, 
Hence, 
