be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
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)
Y by Adventure10
Quadrilateral 
is circumscribed around a circle. Diagonals 
are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments 
and 
at points 
and 
. Given that 
is cyclic, prove that so is .
Proposed by Nikolai Beluhov (Bulgaria)
is circumscribed around a circle. Diagonals 
are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments 
and 
at points 
and 
. Given that 
is cyclic, prove that so is .Proposed by Nikolai Beluhov (Bulgaria)
This post has been edited 2 times. Last edited by parmenides51, Sep 20, 2018, 9:25 AM
Reason: Proposed
Reason: Proposed
,
and 
concur at a point (possibly at infinity). Define 
to be the quadrilateral formed by the tangency points of the incircle and sides 
respectively. A suitable application of Brianchon's and Pascal's theorems, and some harmonic bundles lead us to infer that 
concur at a point 
;
concur at 
and 
concur at 
are the 
-
,
and the other equalities hold simultaneously. The final step is just to prove that 
prevents us from having the extreme case of 
.
,
,
.
touch 
,
,
,
at 
,
,
,
.
,
is the harmonic conjugate of 
wrt.\ 
.
and 
.
,
,
concur at a point 
.![\[-1=(AB;KK^*)\stackrel{L^*}=(AC;EF).\]](http://latex.artofproblemsolving.com/7/f/0/7f036fa0f97e680bcd4109d2c14591ff72797bdc.png)
The claim readily follows by symmetry. 
![[asy] size(8cm); defaultpen(fontsize(10pt));
pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
pair I, K, L, M, NN, A, B, C, D, EE, F, G, X, Y, Kp, Lp; I=(0, 0); K=dir(120); L=dir(40); M=dir(270); NN=2*foot(I, L, foot(L, K, M))-L; A=2*(NN+K)/length(NN+K)^2; B=2*(K+L)/length(K+L)^2; C=2*(L+M)/length(L+M)^2; D=2*(M+NN)/length(M+NN)^2; EE=extension(K, M, L, NN); F=extension(K, L, M, NN); G=extension(K, NN, L, M); X=extension(A, B, C, D); Y=extension(A, D, B, C); Kp=extension(A, B, L, NN); Lp=extension(B, C, K, M);
filldraw(circumcircle(A, B, C), fil, pri); fill(A -- B -- C -- D -- cycle, fil); filldraw(circle(I, 1), sfil, sec); fill(K -- L -- M -- NN -- cycle, sfil); draw(F -- L -- M -- F, sec); draw(L -- M -- NN -- K, sec); draw(X -- B -- C -- X, pri); draw(Lp -- C -- D -- Y, pri); draw(Y -- X, tri); draw(F -- C, tri); draw(B -- D, tri); draw(Lp -- M, sec); draw(Kp -- L, sec); draw(Kp -- Lp,sec);
dot("\(A\)", A, dir(100)); dot("\(B\)", B, dir(70)); dot("\(C\)", C, SE); dot("\(D\)", D, SW); dot("\(K\)", K, dir(75)); dot("\(L\)", L, NE); dot("\(M\)", M, S); dot("\(N\)", NN, dir(240)); dot("\(X\)", X, SW); dot("\(Y\)", Y, NE); dot("\(F\)", F, NW); dot("\(E\)", EE, dir(295)); dot("\(K^*\)", Kp, dir(300)); dot("\(L^*\)", Lp, N); [/asy]](http://latex.artofproblemsolving.com/f/9/4/f942eb2a72acf495b49e2aed49442949d3cca183.png)
.
,
.
is the harmonic conjugate of 
,
lies on 
.
.![\[-1=G(K'L';F'E')=(AC;EF),\]](http://latex.artofproblemsolving.com/b/b/1/bb1b2e5fefbc2e788e142b709e2e634a20646a92.png)
where the second equality is by taking the dual wrt.\ 
.
and 
,
,
lie on 
.
,
,
.
and 
.
readily implies ![\[\angle BAD=180^\circ-\widehat{NK}=\widehat{LM}=180^\circ-\angle DCB,\]](http://latex.artofproblemsolving.com/d/2/6/d26c275d4f1a477febe0371f73bb9a89a7884835.png)
as needed.
