weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers 
.
and 
be positive integers with 
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
and 
.
![$ a,b,c\in [0,1] $](http://latex.artofproblemsolving.com/2/1/b/21b329d0a92c18c6a2861742177969d5977d4558.png)
. Prove that
.
.
.
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
Y by Davi-8191, tenplusten, Western_Spadefoot_Toad, Adventure10, Mango247, GoldenLisa, Rounak_iitr, HWenslawski, and 3 other users
Trapezoid 
, with 
, is inscribed in circle 
and point 
lies inside triangle 
. Rays 
and 
meet again at points 
and 
, respectively. Let the line through parallel to 
intersects 
and 
at points 
and 
, respectively. Prove that quadrilateral 
is cyclic if and only if 
bisects 
.
, with 
, is inscribed in circle 
and point 
lies inside triangle 
. Rays 
and 
meet again at points 
and 
, respectively. Let the line through parallel to 
intersects 
and 
at points 
and 
, respectively. Prove that quadrilateral 
is cyclic if and only if 
bisects 
.
takes 
to 
takes their circumcircles to each other (and therefore their centers), their tangents at 
.
.
is the intersection of the common tangent at 
.
,
are cyclic.
,
.
,
,
be the radical center of the circles of 
.
are collinear, 
,
is cyclic. Hence, 
.
and 
,
,
.
,
.
,
is cyclic and 
.
is cyclic and 
,
.
is cyclic.
be on 
such that 
,
be the intersection of 
with 
,
.
.
.
.
,
is cyclic. But we know that 
is cyclic. This is a contradiction, so our original assumption was false and 
and 
)
are parallel, the circumcircle of 
,
,
exists since it is the radical center of 
,
are collinear so 
.
so those lines intersect at infinity so 
.
bisects 
.
,
.
so 
is cyclic. Let 
.
,
are collinear. But the angle bisector implies that 
so 
so 
,
so 
to meet 
to meet 
.
,
,
,
lie on the same line. Call 
.
must be on the line through 
,
,
.
,
.
if and only if 
and 
intersect on the perpendicular bisector of 
above 
.
and 
,
,
,
is tangent to line 
.
is parallel to line 
,
on line 
,
on 
.
is the point on line 
,
is a trapezoid, it's cyclic if and only if it's isosceles.
to meet circle 
.
,
.
parallel to 
are the same. If they are, 
is a reflection of 
across that line, and 
meets the circle at the same point as the perpendicular bisector of 
with arbitrary radius, denoting 
is the inverse of 
We see 
are points on 
such that 
Also, 
is on 
and 
Since 
is a trapezoid, it suffices that 
Notice 
so we can let 
so 
is equivalent to 
and 
Since 
we have 
. NOOO, why is no one Inverting at 
in any of the earlier posts and Inverting with 
at 
.
,
through 
and 
at 
and 
repectively and also let 
,
,
lies on the tangent through 
.
.
is the angle bisector.
and 
.
and similarly we also get 
.
which gives that 
is cyclic and similarly 
is also cyclic. Thus we have the following final POP relations 
.
and radius 
which swaps 
,
and 
.
,
.
also get swapped. We also deduce that 
also get swapped under this Inversion. Thus now it just suffices to prove that 
,
is cyclic which simply follows from 
and we are done by Inverting back 
.
are collinear. So now Reim's on the the lines 
and 
with the circles 
and 
is parallel to 
.
,
from the initial relation we deduced at the beginning of the solution. This clearly gives 
which immediately finishes as then 
and we are finally done 
.
be the circumcircle of 
and 
.
and 
such that 
.
and tangent at 
other than 
.
is the bisector of 
if and only if 
is cyclic.
.![\[ f(X)=P(X, \gamma)-P(X, \omega) \]](http://latex.artofproblemsolving.com/c/0/f/c0f1c9e8a8145910f51b7867758299fd50d5e205.png)
It is known that 
is linear. So by linearity we have ![\[ \frac{ST}{SR}=\frac{f(S)-f(T)}{f(S)-f(R)}=\frac{TA^2}{P(R, \omega)} \implies P(R, \omega)=\frac{TA^2\cdot SR}{ST} \]](http://latex.artofproblemsolving.com/f/0/3/f035ade1e9fc5fbe901ebde2878406077366f48a.png)
Let 
be the center of 
therefore is sufficient to show 
or equivalently 
.
which is easy angle chase.
is tangent to 
and 
;
,
.
.
is an angle bisector. Then 
hence 
,
by same computation, hence 
.
and let 
to 
so 
are tangent so 
Next, let 
intersect 
at 
We get 
and 
so 
and letting 
again at 
we get that 
We find that 
so this is true iff 
,
is tangent to 
,
be the line through 
and 
be the intersections of 
and 
with 
is cyclic, then by Reim's 
.
we then find 
passes through 
.
is on the perpendicular bisector, i.e., 
.
(points are defined the same way). For the most part, everything here is identical to the previous. By Pascal's on 
and 
being given as symmetric, we instead have 
symmetric. Thus 
and 
are clearly symmetric, so 
and reflect 
.
which isn't hard if you notice the radical center of 
(I didn't oops)
intersect 
.
intersect 
.
and line 
lie on the same line, which is equivalent to tangent at 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/8/2/2/8228d1790927481d50930c5e1ca5ba1d3d067cec.png)
centered at 
We know that 
and 
and let 
.
Hence, 
is cyclic, so 
and 
,
Now 
is cyclic. However, since 
,
the perpendicular bisector of segment 
.
and since 
,
and 
are symmetric with respect to 
.
and we are done.
Since 
and 
are also symmetric wrt to 
lies on 
are tangent.
and let 
Additionally by sine law in 
Similarly by applying the extended sine law on 
With 
we conclude
So
This implies 
and hence 
.
Hence 
Hence 
is tangent to 
.
Hence 
.
be a triangle, 
such that 
is tangent to 
and 
lie on 
such that 
is tangent to 
Finally, 
Show that 
Note that 
is isosceles, so it suffices to prove that 
is isosceles. However, 
is isosceles.![[asy]
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be intersections of 
with 
.
,
,
,
.
.
are symmetric and 
and 
,
