[/quote]
,
.
.
is the intersection of the parallel line from 
.
be the point where the external bisector of 
intersects with 
.
have a common point.
dwarves, whose weigths are 
,
each time without sinking. For each journey of the boat a non-empty subset of the dwarves to be taken to the other side is chosen and one of these dwarves is chosen as the 
for that journey. Since return journeys are counter-current, no dwarf is able to do the rower for more than one return journey. Is it possible for all the dwarves to reach the other side of the river?
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
,
and that line 
is not parallel to line 
.
and 
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
,
intersects the extension of 
at point 
.
is the midpoint of 
,
intersects the circumcircle of 
.
.
be positive real numbers such that 
.![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
,
.
be reals such that 
P
,
,
is __
satisfying 
Y by mathematicsy, CoolJupiter, Adventure10, Mango247
Let the incircle of a nonisosceles triangle 
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
be the incircle of 
.
,
is the point diametrically opposite 
is 
.
,
and 
be the line through 
parallel to 
median then the rest is trivial
refers to directed angles.
be the incentre and ![[asy]
import geometry; import olympiad;
unitsize(4cm);pair A = dir(135), B = dir(210), C = -1/B;
pair I = incenter(A, B, C);
pair D = foot(I, B, C), E = foot(I, C, A), F = foot(I, A, B);
pair T_ = 2I - D, N = B + C - D, T = intersectionpoints(incircle(A,B,C), A -- N)[1];
pair K = extension(E, F, D, T);
pair L = extension(D, I, A, K);
draw(circumcircle(A,E,F)^^circumcircle(D, T, A), linetype("0 2")); draw(D--L, linetype("2 4"));
draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--N); draw(D--K--F); draw(A--K, dashed);
draw(rightanglemark(A, L, D, 1.5)^^rightanglemark(I, D, B, 1.5));
draw(rightanglemark(A, F, I, 1.5)^^rightanglemark(A, E, I, 1.5));
draw(rightanglemark(A, T, D, 1.5));
draw(I--F^^I--E, linetype("2 5"));
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(T); dot(T_); dot(N); dot(K); dot(I); dot(L);
label("A", A, dir(90));
label("B", B, dir(-90));
label("C", C, dir(-90));
label("D", D, dir(-90));
label("E", E, dir(90));
label("F", F, -dir(30));
label("K", K, dir(90));
label("T$'$", T_, dir(-135));
label("T", T, dir(-7));
label("I", I, dir(-45));
label("N", N, dir(-90));
label("L", L, dir(90));[/asy]](http://latex.artofproblemsolving.com/1/7/6/17683b1851b292373700bf0479d24b6665b21dbd.png)
are concyclic with diameter 
.![\[\measuredangle AEI = 90^{\circ} = \measuredangle AFI\]](http://latex.artofproblemsolving.com/8/7/7/877120eed02179c1a9c7530890b852803c843be4.png)
![\[\measuredangle ATD = 90^{\circ}\]](http://latex.artofproblemsolving.com/9/b/3/9b3bf32dba5f84cd72b1d481b6e00e45c5f75712.png)
,
.
is a diameter of 
.
are collinear and 
.
and 
,
is the radical axis of 
is the radical axis of 
and 
.
and 
,
.
meets 
and ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.176743075134857, xmax = 11.428752031839883, ymin = -3.7969421960496423, ymax = 7.034290346656071; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (0.6570484595545034,0.), X = (2.3429513665770023,0.);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw((0.6570484595545034,3.597818826016786)--D, linewidth(1.));
draw(X--A, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", A,dir(90));
dot(B,dotstyle);
label("$B$", B,dir(180));
dot(C,dotstyle);
label("$C$", C,dir(0));
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", D,dir(-90));
dot((0.6570484595545034,3.597818826016786),linewidth(4.pt) + dotstyle);
label("$E$", (0.6570484595545034,3.597818826016786),dir(-135));
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", X,dir(-90));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/b/d/5bd0babd042ab4b6bb32a68205f401ed7e5f4d37.png)
and 
respectively. As 
.
centered at 
to 
which is the incircle to the excircle of ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.98, xmax = 16.42, ymin = -4.14, ymax = 9.34; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (-1.0131969613578624,2.4670075966053435), F = (0.6570484595545034,0.), T = (2.03941567143766,0.6477638648687565);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw(A--(2.3429515404454966,0.), linewidth(1.));
draw(A--(11.327350454273194,5.), linewidth(1.));
draw(F--(11.327350454273194,5.), linewidth(1.));
draw((11.327350454273194,5.)--D, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", (0.08,5.2), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (-2.3,-0.38), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (5.08,0.2), NE * labelscalefactor);
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", (-1.46,2.3), NE * labelscalefactor);
dot((1.9290695423130608,3.070930457686939),linewidth(4.pt) + dotstyle);
label("$E$", (2.,3.24), NE * labelscalefactor);
dot(F,linewidth(4.pt) + dotstyle);
label("$F$", (0.6,-0.38), NE * labelscalefactor);
dot((2.3429515404454966,0.),linewidth(4.pt) + dotstyle);
label("$N$", (2.34,-0.42), NE * labelscalefactor);
dot(T,linewidth(4.pt) + dotstyle);
label("$T$", (2.2,0.42), NE * labelscalefactor);
dot((11.327350454273194,5.),linewidth(4.pt) + dotstyle);
label("$K'$", (11.4,5.16), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/c/4/9c453ff80361494d0162a7a20827c6cbfaf8c915.png)
.
.
and 
.
and using the lemma, we'll be done.
,![\[\frac{s-b}{b}=\frac{\sin(\alpha - C)}{\sin \alpha}=\cos C - \sin C\cot \alpha.\]](http://latex.artofproblemsolving.com/8/0/1/801e922637cb8d60401a8e05f86e97ec011838bb.png)
So, 
.
,![\[\frac{AK'}{AD}=\frac{\sin\left(\frac{B+C}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}\cdot \frac{\cos\left(\frac{B+C}{2}\right)}{\cos\left(\frac{B+C}{2}\right)}=\frac{\sin(B+C)}{\sin B-\sin C}=\frac{\sin A}{\sin B-\sin C}.\]](http://latex.artofproblemsolving.com/c/1/c/c1cf85916facfe3707b0fdf468dff3e358395cd2.png)
So, 
.
at 
,![\[AX=\frac{AC\cdot AK}{FC+AK}=\frac{b\cdot AK}{FC+AK}.\]](http://latex.artofproblemsolving.com/d/9/7/d97debfb0c900fdc9251f96dca83cbf5f4e50050.png)
.![\[\cot \beta = \frac{\frac{AK'}{AX}-\cos C}{\sin C}=\frac{FC+AK}{b\sin C}-\cot C = \frac{s-c+\frac{(s-a)a}{b-c}}{b\sin C}-\cot C.\]](http://latex.artofproblemsolving.com/4/e/4/4e4c308c54c51f5c748ebaaa9739e7e85b64344a.png)
or 
.
or ![\[b^2=ab\cos C - (s-b)(s-c)+\left(\frac{(s-a)a}{b-c}\right)\left(b\cos C - (s-b)\right).\]](http://latex.artofproblemsolving.com/e/5/e/e5ec295ee4a8ffe7761bf7f332f926f84293bad8.png)
,![\[0=\frac{a^2-b^2-c^2}{2}-\frac{(a+c-b)(a+b-c)}{4}+\left(\frac{a(s-a)}{b-c}\right)\left(\frac{a^2+b^2-c^2}{2a}-(s-b)\right).\]](http://latex.artofproblemsolving.com/6/0/0/600f7ae3623af9273b49851f7df96457f4449190.png)
.
or 
.
or 
which is clearly true.
be the point diametrically opposite 
.
colinear.
Also:
So 
as desired.
so 
and project from A we get thd desired result.
meet on the incircle at the antipode of 
are collinear. Now, let 
meet the line parallel to 
,
.
.
is cyclic from 
.
from 
is cyclic and 
,
are concurrent. Hence 
,
.
is diametrically op. to 
is a harmonic quad. Since 
collinear
.Where 
.
are harmonic pencils ! where 
is parallel to 
with the incircle of 
at 
,
are harmonic quadrilaterals, we have: 
= 
,
= 
= 
or 
meet the incircle again at a point 
,
and 
meet at a point 
is collinear, so 
,
,
.
collinear. Since 
is a diameter, 
.
,
at 
respectively.
be the line through 
be the antipode of 
.
,
are concurrent.
passes through 
by parallel lines.
and 
are cyclic.
are collinear, 
proving the first claim.
which proves our second claim. 
,
,
,
.
circles are collinear, then the 
be a point such that 
so 
lies on the polar of 
.
so we are done.
concur ,at 
,
let 
and let 
It is well-known that 
Notice 
so 
is the point at infinity along 
,
,
then its known that if u let 
are colinear, now taking polars w.r.t. 
concurrent and that means 
and the line through 
where 
Projecting (with perspective at 
it suffices to show that 
where 
However, it is well-known that 
where 
where 
and 
.
.
which along with the fact that 
and as 
is tangent to the incircle, we have that 
is tangent to the incircle.
and 
.
lies on the polar of 
,
is the line 
.
and we are done.
-
such that 
.
are collinear, so 
and 
can easily be seen to be cyclic, and by radical center on the incircle, 
,
,
,
,
concur, hence 
.
passes through 
It is well-known that 
is just the midpoint of 
which implies the desired result. 
and 
and the incircle. Then from EGMO chapter 4, we know that 
.
.
is tangent to the incircle, this is enough to imply that line 
But since 
,
,
,
.
.
onto 
and 
are concurrent. Let 
so that 
.
.
concur at some point 
be the other intersection point of 
is a diameter of the incircle.![\[ (DE;ST) \overset{F}= (DE;PK) \overset{A}= (B,C;M,\overline{AK} \cap \overline{BC}) = -1 \]](http://latex.artofproblemsolving.com/0/9/e/09ec0b588794b3e7e9863549622b8de0845af24c.png)
which implies that 
intersect at 
,
denote the incircle. Define 
.
is a diameter of 
.
.
Now projecting through 
so we indeed have 
through Diameter of the Incircle lemma.
is the pole of 
.
,
concur by an incircle concurrence lemma.
,
.
,
,
with 
,
.
.
is the polar of 
is tangent to 
,
is a right triangle.
are collinear.
,
,
,
.![\[(-a+b+c)x+(-a+b-c)y+(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/d/f/f/dfff949c49cae5f346f134ca66b3e5c21d134c08.png)
line ![\[(a-b-c)(c-b)x+a(a-b+c)y+a(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/c/a/6/ca674a37cc0b1d9d6c655bb8418be5358d7d485f.png)
and line ![\[(0)x+(1)y+(-1)z=0.\]](http://latex.artofproblemsolving.com/c/f/a/cfacd23ed3ee2b980cfeeb8bca9a79df33839a60.png)
Now we plug everything into the concurrence lemma:
as desired.![\[
-1=(Z,Y;P,T)\stackrel{X}{=}(Z,Y;Q,K).
\]](http://latex.artofproblemsolving.com/9/6/6/966390ec9e48635548387e793d367b9078d44b00.png)
However, since![\[
-1=(B,C;M,\infty_{BC})\stackrel{A}{=}(Z,Y;Q,(A\infty_{BC}\cap ZY)),
\]](http://latex.artofproblemsolving.com/f/f/8/ff89b604fe7859cb57c73454a1f4bb57a32bd84d.png)
we indeed have 
,
and 
.
and also 
.
.
.
and since FM = MN, then MT = MF and as MF is tangent to the incircle, we get that MT is also tangent to the incircle. Now its obvious that the polar of A is DE 
lies on the polar of A. Now by La Hire, A should lie on the polar of 
and since D and E lie on a circle, we have that X lies on the polar of 
is the line 
so 
,
so 
is cyclic. Thus 
is also cyclic, so 
(follows by taking inverses of lines 
concur.
Since (where 
and 
and 
we know 
so 
are collinear, and we are done.
denote a harmonic bundle.
.
.
be a point on 
and 
.
On the other hand, using the ratio property of symmedians, we obtain 
We have therefore proven 
and thus 
is the antipode of 
.
.
is harmonic, we have![\[-1 = (Z,Y;T',T) \overset{X}{=} (Z,Y;P,K) \overset{A}{=} (B,C;M,Q).\]](http://latex.artofproblemsolving.com/6/f/e/6fea4d27d814f01286214cd0efe86b1d0dab7b26.png)
,
,
,
,
is the midpoint of 
and 
be 
.
as per the OTIS problem. Let 
are collinear (proof by a homothety sending the incircle to the 
be the 
implies 
.
implies 
collinear.
.
says 
are collinear.
Clearly, 
We have that 
so 
so 
so 
so 
lie on the polar of 
Thus, 
which finishes
,
intersects 
at 
intersects 
and 
implies that 
collinear. It is well-known that 
passes through 
implies that 
and since 
