be positive real numbers such that 
.![\[\frac{1}{a^2+155}+\frac{1}{b^2+155}+\frac{1}{c^2+155}\le \frac{11}{780}\]](http://latex.artofproblemsolving.com/4/2/7/427e12cbfdbbba6f26407dbb5c2430aee26c15eb.png)
When does the equality hold ?
and 
Prove that
where 
,
,
be non-negative real numbers, no two of which are zero. Prove that :
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 
.
![$ 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 nguyendangkhoa17112003, samrocksnature, Adventure10, ehuseyinyigit, Rounak_iitr
Let 
be a triangle with circumcircle 
, incenter 
, and 
-excenter 
. Let the incircle and the -excircle hit 
at 
and 
, respectively, and let 
be the midpoint of arc without . Consider the circle tangent to at and arc 
at 
. If 
intersects again at 
, prove that 
and 
meet on .
Amol Aggarwal.
be a triangle with circumcircle 
, incenter 
, and 
-excenter 
. Let the incircle and the -excircle hit 
at 
and 
, respectively, and let 
be the midpoint of arc without . Consider the circle tangent to at and arc 
at 
. If 
intersects again at 
, prove that 
and 
meet on .Amol Aggarwal.
the circle internally tangent to 
cut 
Since 
then 
arcs 
and 
of 
bisects 
and 
cuts 
It's well known that 
lie on the circle 
with center 
thus inversion WRT 
into 
and 
and 
are orthogonal to 
Lines 
.
are collinear .
,
is the reflection of 
in the bisector of 
.
,
.
,
,
we get 
,
we get 
are collinear .
a chord of the bigger circle tangent to the smaller circle in a point 
.
is the internal bisector of 
.
are collinear. Also is well known that 
,
,
.
,
,
.
and 
,
,
.
,
,
,
.
and 
are similar, thus 
.
,
,
.
being trivial.)
and let 
meet 
.
and so 
are concyclic.
inversion we get that 
is the point of intersection of 
where 
is the touch-point of the 
.
meet 
.
are col-linear.
.
.
.
and 
intersect on 
be the intersection of 
,
,
with 
be the intersection of 
with 
,
,
and we're done.
.![\[D=(0:s-c:s-b), E=(0:s-b:s-c),I=(a:b:c),I_a=(-a:b:c),M=(-a^2:b^2+bc:c^2+bc).\]](http://latex.artofproblemsolving.com/f/b/a/fba8f9a89cdb32a9ee35870de0df2602790d200d.png)
All fine and good, but now how do we proceed? We need to think very carefully about how to make this barybash feasible. In particular, the definition of 
are collinear by homothety. This makes 
.
and 
by intersecting this line with the circle, and finally check that 
.
is![\[\begin{vmatrix}
x & y & z \\
-a^2 & b^2+bc & c^2+bc \\
0 & s-c & s-b
\end{vmatrix}=0\iff\]](http://latex.artofproblemsolving.com/a/c/a/acab8f7852ff2a975ed8d420773d5c23cf393bd8.png)
![\[\iff x(b+c)(b(s-b)-c(s-c))+y(a^2)(s-b)-z(a^2)(s-c)=0\]](http://latex.artofproblemsolving.com/8/1/f/81f029d04b8e06f5871b760d542102e413cf0df9.png)
![\[\iff x(c^2-b^2)(s-a)+y(a^2)(s-b)-z(a^2)(s-c)=0\]](http://latex.artofproblemsolving.com/d/9/5/d95076b7907a44cc81034d1a65dc7711ffd062d0.png)
![\[\iff z=x\cdot \frac{(c^2-b^2)(s-a)}{a^2(s-c)}+y\cdot \frac{(s-b)}{(s-c)}.\]](http://latex.artofproblemsolving.com/e/6/1/e616fda636cb85a0561314c343a384720d5a9d9b.png)
The circumcircle has equation 
,
into the equation: ![\[y^2\cdot \frac{a^2(s-b)}{(s-c)}+x^2\cdot \frac{b^2(c^2-b^2)(s-a)}{a^2(s-c)}+Qxy=0,\]](http://latex.artofproblemsolving.com/a/4/4/a4498b7c486c7ec1359b9e561ecb18b2e5e2e811.png)
for some function 
of 
and observe that one solution for 
is when 
,![\[\left( \frac{-b(b+c)}{a^2}\right) \left(\frac{a^2(s-b)}{(s-c)}\right)y=\frac{(c^2-b^2)(s-a)}{a^2(s-c)}. \]](http://latex.artofproblemsolving.com/0/0/5/005ed2064ada4bfaf51f374656214866b40059d9.png)
This reduces to ![\[y=\frac{b(b-c)(s-a)}{a^2(s-b)}.\]](http://latex.artofproblemsolving.com/f/4/6/f466966de24176fbbb676cb19debbff6a111c28a.png)
Now since ![\[z=\frac{c(c-b)(s-a)}{a^2(s-c)},\]](http://latex.artofproblemsolving.com/e/b/c/ebc74f56207d1b2e3fcdfbad906ee5cb4f5b3d5d.png)
which is acceptable since it has to agree (symmetrically) with ![\[T=\left(a^2(s-b)(s-c):b(b-c)(s-a)(s-c):c(c-b)(s-a)(s-b)\right).\]](http://latex.artofproblemsolving.com/b/f/c/bfc926442360eddf246c0d807f2f6d8e4cf7f738.png)
The computation of ![\[X=\left( a^2(s-b)(s-c):b(c-b)s(s-b):c(b-c)s(s-c) \right).\]](http://latex.artofproblemsolving.com/a/1/1/a11ea57053bc08081b91646cd6ebd95ab178b199.png)
(Note to reader: We can also synthetically observe that 
have the same x-coordinate [when normalized, although in the above forms they still do.] This could be used to find 
.
has equation![\[\begin{vmatrix}
x & y & z \\
a & b & c \\
a^2(s-b)(s-c) & b(b-c)(s-a)(s-c) & c(c-b)(s-a)(s-b)
\end{vmatrix}=0\]](http://latex.artofproblemsolving.com/0/f/8/0f8a9f2147bdb3d71371196c2c78918aca50bb0b.png)
![\[\iff x(bc)(s-a)(c-b)a+y(ac)(s-b)(a(s-c)+(b-c)(s-a))-z(ab)(s-c)(a(s-b)+(c-b)(s-a))=0.\]](http://latex.artofproblemsolving.com/7/7/8/778df41edd696e0966bd61003c9def4338ec30b0.png)
Now, we divide through by ![\[a(s-c)+(b-c)(s-a)=S_c, a(s-b)+(c-b)(s-a)=S_b.\]](http://latex.artofproblemsolving.com/2/3/f/23f0472d17b17130490e4a787732c7312f207975.png)
Thus line ![\[x(bc)(s-a)(c-b)+y(c)(s-b)S_c-z(b)(s-c)S_b=0.\]](http://latex.artofproblemsolving.com/4/7/5/4756c96f04edd00ffd82b94e8cdc70adcc4bb04b.png)
Again computing 
is done in nearly the same way; somewhat surprisingly, we again find a nice simplification, this time![\[a(s-b)+(b-c)s=S_c, a(s-c)+(c-b)s=S_b,\]](http://latex.artofproblemsolving.com/d/3/4/d342258fc4a65b0b26011ca3a79b86deefc58c13.png)
which allows us to reduce the equation of ![\[x(bcs)(b-c)+y(c)(s-c)S_c-z(b)(s-b)S_b=0.\]](http://latex.artofproblemsolving.com/e/d/8/ed85606dddd9200d0433395cc26db6c1203001c9.png)
Now we will compute ![\[z=x\cdot \frac{cs(b-c)}{(s-b)S_b}+y\cdot \frac{c(s-c)S_c}{b(s-b)S_b}.\]](http://latex.artofproblemsolving.com/c/0/1/c01e0544502ba7fe71949223f1ef2108cd591572.png)
Plugging this into the circumcircle equation we find![\[y^2\cdot \left(\frac{a^2c(s-c)S_c}{b(s-b)S_b}\right)+x^2\cdot \left( \frac{b^2cs(b-c)}{(s-b)S_b}\right)+Q\cdot xy=0,\]](http://latex.artofproblemsolving.com/6/0/0/6002dcad3ad2e159752827274ee5bb632f653b2a.png)
for some function ![\[y=\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}.\]](http://latex.artofproblemsolving.com/9/3/e/93e822368e67de4b5906144c2bed55ea50edcf37.png)
Therefore the one we care about satisfies![\[\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}\cdot \frac{a^2c(s-c)S_c}{b(s-b)S_b}\cdot y=\frac{b^2cs(b-c)}{(s-b)S_b,}\]](http://latex.artofproblemsolving.com/a/3/b/a3bdfc20ad198b28e91a0102110dedc4606019e3.png)
which miraculously simplifies to 
.
,![\[S=(-S_bS_c:b^2S_b:c^2S_c).\]](http://latex.artofproblemsolving.com/a/a/8/aa8812c25750aaaff692b26ea473334e9be75808.png)
We're almost there! We just need to check that this lies on ![\[-S_bS_cbc(s-a)(c-b)+b^2c(s-b)S_bS_c-c^2b(s-c)S_bS_c=0.\]](http://latex.artofproblemsolving.com/b/c/b/bcbcf493f1ef074f8a8e6819ef2cbf7b53df7f9d.png)
We can cancel the 
t
from everything, leaving only![\[-(s-a)(c-b)+b(s-b)-c(s-c)=0.\]](http://latex.artofproblemsolving.com/8/f/1/8f1a62e61b377fbeec0276eccc43fccd0e26a724.png)
Now factoring 
from this, it becomes![\[(b-c)(2s-a-b-c)=0,\]](http://latex.artofproblemsolving.com/6/1/2/61238c2ce30e612c7689f50da86f0bc791bd52a3.png)
which is clearly true. Hooray!
)
as the intersection of 
are collinear.
so 
so 
so 
is tangent to the circumcircle of 
so 
.
with center 
.
with 
,
is cyclic so 
and thus 
is also cyclic so 
we obtain that 
is cyclic.
.
are collinear and we are done.
,
.
touch 
in 
.
and so 
,
.Let 
.
but as 
.
and so by angle bisector theore
we have that 
as 
.
are collinear.
is diameter of 
,
.
.
,
and 
are tangent to each other. Let 
be line passing through 
are concurrent, so 
are concurrent. Let 
.
.
are collinear. Since 
,
.
to 
.
are symmetric about perpendicular bisector of 
,
.
,
to 
.
.
,
and since 
,
,
, also I am so out of touch that I didn't even recognize the sharky devil point 
because of Curvilinear circle lemma from EGMO (or homothety or inversion).
and let the image of 
which would be obvious if I had recognized the point 
). Due to the inversion we also obtain that 
as 
and hence we obtain that 
is cyclic. By inverting back we get that this concyclicity directly implies the question as the intersection point is the inverted image of 
,
,
.![\[ME \cdot MK = MC^2 = MI_A^2 \implies \triangle MEI_A \sim \triangle MI_AK,\]](http://latex.artofproblemsolving.com/2/6/6/266d3f2c1f072152a4c158256ef607257f95d261.png)
which indeed gives![\[\measuredangle AKI_A = \measuredangle AKM + \measuredangle MKI_A = \measuredangle (AM, BC) + \measuredangle EI_AM = 90. \quad \blacksquare\]](http://latex.artofproblemsolving.com/7/7/f/77f31e43b80dda4d85728975707692826e4ebc9b.png)
,
.
are collinear.
inversion followed by a reflection across the bisector. The image of 
is an isosceles trapezoid, 
is isogonal to 
,
,
,
.
as well, 
is cyclic, so we are done.
by the tangent circles configuration, after which angle chasing suffices as we can show 
.
and we immediately know what to do from here.
.
are concyclic. But 
,
.
and note,
,
and 
concur on 
meet 
as 
cyclic, we find 
.
which gives our first desired concylicity. The other follows as 
.
are collinear but since 
which we name 
are concyclic
since 
since 
The problem reduces to showing that 
is cyclic(due to inversion).
So 
hence we are done as 
.
and 
,
be the antipode of 
be the reflection of 
collinear. Reflecting this over the perpendicular bisector of 
collinear. So it suffices to show that 
collinear. Then we want![\[-1=(TT';NM)\stackrel{S}{=}(II_A;(NS\cap AM)M).\]](http://latex.artofproblemsolving.com/4/1/8/4185ac6a11d488bd86a125c78d1a47ec1fe49400.png)
Thus we want 
.![\[MI^2=MB^2=(MD)(MT)\Longrightarrow \angle ITM=\angle DIM=\angle AMN=\angle ATN.\]](http://latex.artofproblemsolving.com/e/5/6/e56c18afdf20542810db77f1a88292acc838f000.png)
So 
too. Yay. 
.
s
,
swap with 
and 
as desired
collinear, so 
.
,
,
.
