and 
share circumcircle 
and incircle 
so that points 
and 
occur in this order along 
be the triangle formed by lines 
and 
and define triangles 
similarly. Furthermore, let 
and 
be the circumcircle and incircle of triangle 
similarly.
and the two common external tangents to 
are either concurrent or pairwise parallel.
,
and 
similarly. Prove that points 
,
Prove that
be an acute-angled triangle with 
.
be the orthocenter of triangle 
be the midpoint of the side 
and the points 
is perpendicular to the common chord of the circumscribed circles of triangle 
.
as the set of all numbers of the form 
for some nonnegative 
and 
.
such that 
and 
.
over 
,
.
children, each of whom selects
tables, with 
Prove that
and 
.![$$ a^2-ab+b^2 \geq \sqrt[3]{2} $$](http://latex.artofproblemsolving.com/7/9/8/798484c6b985ebbace2a3dde816f11a46635feba.png)
Let 
.![$$ 3\geq a^2+ab+b^2 \geq \sqrt[3]{4} $$](http://latex.artofproblemsolving.com/d/7/7/d77fbdc50a88395259e58d31413b9ac453e5dfd0.png)
of odd degree with real coefficients such that 
.
be non-negative real numbers such that 
Prove that
and 
.
intersects the side 
and the circumcircle of triangle 
intersects the side 
.
lie on the same side of 
and 
.
and 
are tangent.
Define a cell with 
property if after remove that cell from the board The board can be tile with 
tetromino.
Note: 
Y by Davi-8191, nguyendangkhoa17112003, TurtleKing123, HWenslawski, Adventure10, centslordm, megarnie, proxima1681, Mahmood.sy, Mango247, Rounak_iitr, and 2 other users
Let 
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from 
to the lines 
, 
, 
, respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with 
.
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from 
to the lines 
, 
, 
, respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with 
.Attachments:
solved in about 30 minutes (which is extremely few for me)
and 
be the altitudes from 
,
intersect 
is a right angle.
,
is cyclic. Then 
.
.
,
is cyclic. Thus 
.
are collinear.
.
.
,
.
.
.
.
.
.
if and only if 
.
and 
.
.
.
be a triangle, 
an arbitrary point and 
,
,
the pedal points wrt the sides 
,
,
.
,
denotes the circrumradius of 
.
and 
.
.
denote the intersection of the angle bisectors of 
and 
.
,
.
,
,
,
and 
denote the intersections of the angle bisectors of angles 
.
intersect 
,
,
and 
be points diametrically opposite on the circumcircle of 
such that 
is the perpendicular bisector of 
.
be the point diametrically opposite to 
is cyclic, 
is a rectangle (cyclic), so 
.
,
.
are collinear and are the Simson line of 
,
and 
,
is directly similar to 
.
to 
is the midpoint of 
so we are done.
and 
we have 
.
is similar to 
and therefore 
.
in triangle 
we obtain:
.
(because of (1)). The conclusion is now obvious.
,
the projections of 
.
i
and 
are inscribed in circles with diamete
,
.
a
(
,
we hav
.
,
,
and 
similar. (This follows by a very simple angle chasing, as in the above posts). This also gives us the fact that triangles 
and 
are similar, therefore 
are collinear (Simson's line)
. Points 
are collinear and lie on the lines 
respectively. Applying Menelaus' Theorem, we have:
,
by relation (1). The conclusion follows immediately.
is the Simson line of 
,
.
which implies the result.
![\[\frac{PQ}{QD} = \frac{\sin \angle PDQ}{\sin \angle QPD} = \frac{\sin \angle BAC}{\sin \angle DAC},\]](http://latex.artofproblemsolving.com/4/7/b/47b4bfe991ad452d3070f27a123061593e412f6f.png)
and ![\[\frac{QR}{QD} = \frac{\sin \angle QDR}{\sin \angle DRQ} = \frac{\sin \angle ACB}{\sin \angle DBA}. \]](http://latex.artofproblemsolving.com/9/f/c/9fc3398e2050595949a50e32917faeadd49a9347.png)
Hence, ![\[\frac{PQ}{QR} = \frac{\sin \angle BAC}{\sin \angle DAC} \cdot \frac{\sin \angle DBA}{\sin \angle ACB} = \frac{BC}{BA} \cdot \frac{DA}{DC} = 1 \iff ABCD \text{ is harmonic,}\]](http://latex.artofproblemsolving.com/9/3/e/93e04784e3cf59eecc6df7e9d2b22a55101a8343.png)
as desired. 
.
,
,
.
we get 
(we can do this since P, Q, R lie on one line aka Simpson line) and 
.
PQ = QP. In the other direction is the same because the pair of similar triangles is there, so we get that 
which is equivalent to the fact that the bisectors meet on AC. We are ready.
be the second intersection of line 
with the circle 
.![\[ \angle BXD = \angle BAD = \angle RAD = \angle RQD, \]](http://latex.artofproblemsolving.com/f/3/e/f3e9b2d72dcb4df3034ed7954adaf2be5f4ba27c.png)
and thus, 
is parallel to the Simson line.![\[ (A,C;B,D) \stackrel{X}{=} (A,C;\overline{BX} \cap \overline{AC}, Q) \stackrel{B}{=} (R,P;P_\infty,Q), \]](http://latex.artofproblemsolving.com/4/0/0/400be725e10384210ef4d845c4c12978e09d40bd.png)
and since 
,
.
be the arc midpoints of 
meet a point whose polar is parallel to 
are concurrent (or 
)
being the intersection of 
.
,
,
.
.
.
.
,
.
.
.
,
,
,
![\[\frac{AB}{CB}=\frac{AD}{CD} \iff AB*CD=AD*BC \iff (AC;BD)=-1,\]](http://latex.artofproblemsolving.com/2/d/e/2de0e2a0c8f03a52ef8ba89ce41c3910b08dd7a5.png)
and by the uniqueness of the harmonic conjugate, this means that 
is the unique point on 
such that 
.
.![\[PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD.\]](http://latex.artofproblemsolving.com/a/d/7/ad7323de5d9cc36d5e5a1093e101fc3ad51d0d76.png)
to 
.
,
,![\[\angle RPD=\angle QPD=\angle QCD=\angle ACD,\]](http://latex.artofproblemsolving.com/1/b/3/1b3b8311fd02ed0d79219642bf7ec6e2a6ed2032.png)
and since 
,![\[\angle PRD=\angle QRD=\angle QAD=\angle CAD,\]](http://latex.artofproblemsolving.com/2/6/3/263845ccc7d451492d5b1435866d733f4084290a.png)
and these two combined gives us that 
,![\[\angle Q\rightarrow M \iff \angle MDQ=\angle CDP \iff 90-\angle MDQ=90-\angle CDP \iff \angle DMQ=\angle DCP,\]](http://latex.artofproblemsolving.com/5/8/8/588a478726b3d4a37d4cde027c30a63c3c07588d.png)
and since 
by cyclic properties, and 
,![\[\iff \angle BAD=\angle DMC.\]](http://latex.artofproblemsolving.com/c/5/b/c5b1f34c7f2e1820ec5bbe6ff0d9ea0536564320.png)
,
must be the unique point on 
is fixed, 
must also be fixed, meaning that there is a unique point on ![\[\angle ADM=180-\angle AMD-\angle MAD=\angle CMD-\angle MAD=\angle BAD-\angle MAD=\angle BAC=\angle BDC,\]](http://latex.artofproblemsolving.com/f/e/a/feab2154e02d110e097b4bcce4144d1216ccc1a6.png)
which means that 
.
,![\[PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD,\]](http://latex.artofproblemsolving.com/8/4/7/847c470413e4e9625c97efd6415b784adaaeae61.png)
as desired. Since![\[BD\text{ is the } D\text{-symmedian of }\triangle ACD\iff (AC;BD)=-1\iff \text{the angle bisector of }\angle ABC \text{ and } \angle ADC \text{ meet on } AC,\]](http://latex.artofproblemsolving.com/4/b/f/4bf6101cc874721164e4aa00192c408b368557da.png)
we have that the angle bisectors meet on 
![\[PQ = QR \iff AD \sin \angle A = CD \sin \angle C \iff AD \cdot BC = CD \cdot AB. \quad \blacksquare\]](http://latex.artofproblemsolving.com/f/6/e/f6e45c06f0081e424e81d1f0b8614847f3c46808.png)
again at 
.
are collinear, and 
so 
.
which concludes by angle bisector theorem.
.
,
with transversal line 
,
,
which gives the proportion.
is a cyclic quadrilateral as 
Therefore by the Law of Sines 
Similarly 
Dividing these equations, it follows that 
by the Law of Sines. QED
,
,![\[2ac+2bd=ab+bc+cd+da\]](http://latex.artofproblemsolving.com/c/a/2/ca2c9e927f7c2511ea59b3ec27944d334843f298.png)
in complex. The second condition is equivalent to ![\[(a-b)(c-d)=-(c-b)(a-d)\Longleftrightarrow 2ac+2bc=ab+bc+cd+da.\]](http://latex.artofproblemsolving.com/f/7/1/f71c36105481dba7acdf536bfe710fd9ca9f780e.png)
Wow. 
be on 
such that 
,
.![\[
-1=(P,R;Q,\infty_{PR})
\stackrel{D}{=}(\infty_{\perp BA},\infty_{\perp BC};\infty_{\perp AC},\infty_{PR})
\stackrel{\text{rotate }90^\circ}{=}(\infty_{BA},\infty_{BC};\infty_{AC},\infty_{\perp PR})
\stackrel{B}{=}(A,C;B',B\infty_{\perp PR}\cap(ABCD)).
\]](http://latex.artofproblemsolving.com/d/f/6/df6f367297c652ec12aba95e682f794fa1859642.png)
But we claim that the unique such point 
is exactly 
(since 
,
and the angle bisector thing follows by the angle bisector theorem. Everything in this proof was reversible so we get the other direction too. 
