such that 
and 
are coprime if and only if 
?
with decimal representation 
,
with 
except those of this color, the resulting decimal number is rational.
array of square unit cells. Two corner cells that lie on the same longest diagonal are colored black, and the rest of the array is white. A move consists of choosing a row or a column and changing the color of every cell in the chosen row or column.
satisfying
Prove
Y by codyj, Davi-8191, Abdollahpour, itslumi, jhu08, mathematicsy, megarnie, Lamboreghini, Adventure10, Mango247, Rounak_iitr, Shadow6885, NicoN9, and 7 other users
Let 
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
Proposed by Giorgi Arabidze, Georgia.
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .Proposed by Giorgi Arabidze, Georgia.
and 
which are symmetric to 
and 
with respect to 
and find two pairs of similar triangles 
,
and 
,
.
and 
are parallel.
be the intersection of 
be points on 
such that 
and 
and 
are similar, then 
and 
are similar. Then 
and 
are similar, then 
and 
are similar. Then 
is a parallelogram. Take 
be the midpoint of 
is parallel to 
and 
.
and 
,
so 
and 
and 
so in the triangle 
we have ![\[\frac{\sin(x-A)}{\sin(x)}=\frac{AC}{AB}\]](http://latex.artofproblemsolving.com/e/0/a/e0a847df323278506b00883a5d252f2c3cc34890.png)
and by the same way ![\[\frac{\sin(y-A)}{\sin(y)}=\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/5/5/3/5538c403fa6f3417655b9b5b8b06d74da013aada.png)
![\[\sin(x-A)\sin(y-A)=\sin(x)\sin(y)\]](http://latex.artofproblemsolving.com/e/8/1/e81c6ab921fae057c0cefcd3f0bd8f39ca0c120c.png)
so 
and therefore ![\[\cos(x+y-2A)=\cos(x+y)\]](http://latex.artofproblemsolving.com/b/7/2/b72b67a7257ec66db6ec4dc99d04a8725c93b053.png)
which gives ![\[x+y=A\]](http://latex.artofproblemsolving.com/3/9/0/3909c3b8e840d629a69eb7da870cb712bb5a74ed.png)
and if ![\[<BRC+<BCR=A\]](http://latex.artofproblemsolving.com/9/9/4/99463ee90671b4a0056d7745b25ba43294524b8b.png)
.
is cycle
,
,
respectively.
is the double of 
,
is the double of 
.
has the same area as 
and 
respectively.
,
.
be the circumcenter of 
.
,
.
,
.
be the midpoints of 
.
.
lies on circle 
,
with diameter 
.
lie on 
,
and 
are collinear, i.e 
is on 
.
,
and 
are collinear, so by converse of Pascal's Theorem, 
which is precisely circle 
and 
be the midpoints of 
and 
,
be the intersection between 
and 
.
and 
.
,
and 
are similar, where we get 
(because both 
and finally:
.
is cyclic.
is cyclic as well. This ends the proof.
,
,
.
, 
and 
so the triangles 
and 
are similar hamotetic (with ratio 2) .On the other 
so we have 
lies on the circumcircle of the triangle 
.
be similar to 
.
is similar to 
and we are done!
.
.
.
.
so there exists a homothety mapping 
to 
,
we have 
,
.
.
is 
which clearly lies on the circumcircle.
From similar triangles 
and 
we have 
Combined with the fact that 
we deduce that 
implying 
Let 
We have that 
We then have 
Similarly, 
From sine rule in 
and from sine rule in 
Combining them, we see that![\[\sin^2(\theta - \angle BAC) = \sin^2 \theta \implies \cos (2(\theta - \angle BAC)) + 1 = \cos^2 \theta \\ \implies 2 \theta = \angle BAC \implies \theta = \dfrac{\angle BAC}{2}.\]](http://latex.artofproblemsolving.com/4/6/c/46c111dede65fac5a72358f01f29f3bce84b0360.png)
we have that 
so 
is cyclic, as desired. 
with 
.![\[(AM; PP_{\infty}) \overset{B}{=} (A, BM \cap (ABC); C, B) \implies K \in BM.\]](http://latex.artofproblemsolving.com/7/e/a/7eac08e0b95fc6bcf9ea8de8d3c654fffc280d21.png)
,
such that 
.
.
.
.
be the intersection of the tangents to ![\[\angle AQC=180-\angle CAQ-\angle C=180-\angle B-\angle C=\angle A=\angle BCT,\]](http://latex.artofproblemsolving.com/4/b/e/4beb5e5549419656924300328427b2dfaa8cb097.png)
so 
,![\[-1=(AX;BC)\overset{B}{=}(AX;TY)\overset{C}{=}(AN';P_{\infty, CT}Q),\]](http://latex.artofproblemsolving.com/7/6/b/76b998ba4911ceec7d11a3c6e7807600d9f9c393.png)
so 
,
,
.
and 
.
and 
,
Therefore, 
be the reflection of 
is a parallelogram. By the definitions of 
.![\[PABM \sim ACBK, \qquad QACN \sim ABCJ.\]](http://latex.artofproblemsolving.com/7/5/1/7516788eef95600fd87f2981c8bc3d5a2a2f2945.png)
Now, letting 
as desired.
,
and thus 
.
.
,
as desired. 
Consider the tangent at 
Hence, 
which finishes.
Observe that from our conditions 
But 
so by SAS 
so 
is cyclic. Hence, 
so 
the reflection of 
and 
are similar, so triangles 
and 
are similar, implying that 
.
is a parallelogram, we should have:![\[ \angle ABT + \angle ACT = \angle ABM + \angle ACN = \angle CAC’ + 180^{\circ} - \angle CAC’ = 180^{\circ} \]](http://latex.artofproblemsolving.com/7/f/5/7f55d990bd0c68210e762ac3473749724d50ab08.png)
Thus providing that 
so then its immediate that 
and similarly 
.
and 
.
by parameterizing 
whicih gives 
.
be 
.
,
,
.
and 
.
and 
.
and 
.
,
.
and 
.
.
which clearly is true so 
.
with the segment 
is the intersection of 
,
.
Therefore 
and we are done.
,
is the 
and 
are collinear, where 
which is the composition of a inversion centered at 
and a reflection over the internal angle bissector of 
.
is the intersection of the parallel through 
is the refletion of 
is the intersection of parallels through 
is cyclic, but it's immediate, since it's a isosceles trapezoid. The result follows simillarly to show the other collinearity. 
are tangent to 
.
goes through the intersection of the 
bisects 
is tangent to 
lies on ![\[ \measuredangle D_APB = \measuredangle D_APC = \measuredangle D_AAC = \measuredangle D_ABA \]](http://latex.artofproblemsolving.com/0/c/3/0c3485b72baef54123f135b58d6ccf3ba9f1c1d0.png)
as desired.
