[/quote]
,
(use in a problem for junior) to practice. Can anyone share to me?
be an acute angled scalene triangle with circumcircle 
.
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and 
Prove that 
and 
intersect on the circumcircle of 
.
grid with 
rows and 
columns, and each square of the grid is filled with a non-negative integer. Let 
be the average of all the numbers in the grid. Prove that if 
the there exist two identical rows (meaning same numbers in the same order).
is called feirense if it is possible to write on a sheet of paper some integers such that every positive divisor of 
be a square with 
side length and with center 
.
is drawn where point 
.
,
where 
is the number of times the number 
appears in the list. So, for example, for the list 
,
.
numbers. Then he asked him to apply the program to the resulting list, and so on, until a number greater than or equal to 
appears in the list. Find the largest value of 
,
table is colored black or white. Every cell on table border is black. It is known, that in every 
square there are cells of two colors. Prove, that exist 
be midpoints of 
and 
be reflections of 
.
have the same centroid.
and 
are acute and 
.
,
.
.
,
,
.
at 
.
be the midpoint of 
,
meet 
.
at 
is tangent to 
Y by AlastorMoody, integrated_JRC, saksham1729, nguyendangkhoa17112003, RudraRockstar, Purple_Planet, mathematicsy, michaelwenquan, ImSh95, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with 
. Let 
be a point on the segment 
and 
be a point on line 
such that 
is tangent to the circumcircle of triangle 
at 
and 
is perpendicular to . Given that 
and 
. Determine 
in degrees.
be a triangle with 
. Let 
be a point on the segment 
and 
be a point on line 
such that 
is tangent to the circumcircle of triangle 
at 
and 
is perpendicular to . Given that 
and 
. Determine 
in degrees.This post has been edited 2 times. Last edited by div5252, Jan 20, 2019, 5:54 PM
and Let 
,
and we already have 
is isosceles, hence, 
Simple trig 
):
.
.
.
to 
.
is a rhombus. Therefore 
.
is cyclic.
.
.
.
,
is the center of 
.
.
Now we will choose a variable Point 
Such 
Now there are 
cases
Which is only possible if 
or 
Hence 
in this case.
And here also it is only possible if 
and 
be the foot of perpendicular from 
.
is cyclic.
and 
is midpoint of 
is cyclic.
.
so 
.
parallel with 
meet 
so 
is cyclic so 
so 
so 
.
.
.
such that 
to be a rhombus
a, since 
we have 
, also we have 
we have 
is a cyclic quadrilateral since we have 
.
.
,
and also 
.![\[\frac{XD}{DE} = \frac{XE}{AE}\implies \frac{AD}{2\cdot DE} = \frac{AD+2\cdot DE}{2\cdot (AD+DE)} \implies AD = \sqrt{2}\cdot DE\]](http://latex.artofproblemsolving.com/7/c/0/7c091e6d1d499d96f080e86e8612ac81e58e4724.png)
and we also have![\[\frac{AB}{AE}=\frac{AB}{CE} = \frac{CX}{BE} = \frac{XD}{DE} \implies AB = \sqrt{2}AE\]](http://latex.artofproblemsolving.com/1/7/e/17e28b63d4908cf97ca359b23cfe27c09e2e0f58.png)
Thus, 
.
and 
and let 
.![\[\alpha=\angle ADC=\angle DAC=\angle ACE,\]](http://latex.artofproblemsolving.com/5/e/5/5e552cd848e8325dbfa64a069edac565cab1ccfe.png)
so![\[\angle BAC+\angle ACE=\angle BAD+2\alpha=\angle ACD+\angle DAC+\angle ADC=180^{\circ},\]](http://latex.artofproblemsolving.com/7/5/f/75ff5236aa3136ec0e72d039e4585047423fe042.png)
so 
and 
.
.
by AA similarity.
centered at 
.
.
since they are parallel. Since 
,
.
,
.
.
,
.
,
gives![\[(\sqrt2+1)a^2+b^2=(\sqrt2+1)^2 a^2\Longrightarrow b=\sqrt{2+\sqrt2} a\Longrightarrow DE=\sqrt{4+2\sqrt2}\cdot\frac{a}{2}.\]](http://latex.artofproblemsolving.com/e/2/9/e29efa2b7a123d4242b2a509f8b4b59de2db90f5.png)
Thus![\[\cos\alpha=\cos\angle FDE=\frac{a/2}{DE}=\frac{1}{\sqrt{4+2\sqrt2}}\Longrightarrow
\cos(2\alpha)=\frac{2}{4+2\sqrt2}-1=\frac{1}{2+\sqrt2}-1=\frac{2-\sqrt2}{2}-1=\frac{-\sqrt2}{2}\]](http://latex.artofproblemsolving.com/d/8/f/d8fe5a72d25db17120b83d85103fb7ed792d0c58.png)
![\[\Longrightarrow
\cos(\angle ACB)=\frac{\sqrt2}{2}\Longrightarrow
\boxed{\angle ACB=45^{\circ}}.\]](http://latex.artofproblemsolving.com/5/3/a/53a9bbd994ec44f14649934ef45e1503c7091894.png)
such that 
to meet 
at 
.
is isosceles.)
.
.
.
is cyclic.
(vertically opposite).
is isosceles 
.
.
is isosceles 
.
.
.
.
at ![\[\angle BAE=\angle ACD=2a.\]](http://latex.artofproblemsolving.com/0/0/d/00d1c7cbda9855e04ffe6fa9410deaf1a753436c.png)
Additionally, since ![\[\angle AEC=180-2\angle EAC=180-2\angle DAC=\angle ACD=2a,\]](http://latex.artofproblemsolving.com/1/f/6/1f6fa8997fac987b84081cccc22bab5c7b774925.png)
since 
,
.![\[\angle ABD=\angle ECD,\]](http://latex.artofproblemsolving.com/3/2/e/32eee0dcfdd6701ec7edf625a01e94c20c69f53c.png)
and![\[\angle BAD=\angle DEC,\]](http://latex.artofproblemsolving.com/b/2/2/b22abb481e436858b21f4c2b56a45fe647509132.png)
meaning that 
be on line 
.
.
and 
,
is isosceles with 
.
is also isosceles with 
.
.![\[\frac{BD}{DC}=\frac{AD}{DE}.\]](http://latex.artofproblemsolving.com/c/5/b/c5b743185cca4df0445b84d934ed879e36da7d9d.png)
Additionally, by ![\[\frac{BD}{DC}=\frac{XD}{AD}=\frac{2*DE}{AD}.\]](http://latex.artofproblemsolving.com/3/0/b/30b046916626cf6c346ddeced3b97230bf831820.png)
Equating the two equations gives that![\[2*\frac{DE}{AD}=\frac{AD}{DE},\]](http://latex.artofproblemsolving.com/f/d/d/fdd0392d6eb153fca33d2f275ac8bb0ae659acbb.png)
implying that 
.![\[\frac{EA}{AB}=\frac{EC}{AB}=\frac{ED}{AD}=\frac{\sqrt{2}}{2},\]](http://latex.artofproblemsolving.com/0/8/7/087bb71fb4b546bb958438edd4aad26946fe4272.png)
by 
,![\[\angle BCA=\angle ACD=2a=45,\]](http://latex.artofproblemsolving.com/b/f/8/bf8d780557c7fcffe6088a9026d8189b5b315bbd.png)
finishing the problem.
.
,
is sent to the foot of 
is cyclic. We reverse reconstruct 
.![\[\measuredangle DAB'=\measuredangle BCA=\measuredangle AEC=\measuredangle AB'C\]](http://latex.artofproblemsolving.com/2/a/c/2ac21ec9c9a54f9e5e0924890821ecf8a650b883.png)
Letting 
,
is then a symmedian of 
(Isos triangle). This means that 
is tangent to 
,
and 
, now we express 
Now we equate both of them to get
