be reals such that 
P
,
,
is __
satisfying 
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
,
is obtuse
is acute
,
is semi perimeter
,
Y by Adventure10, Mango247, Rounak_iitr
Let 
be two circles with centers 
and 
, respectively. These two circles intersect each other at points 
and 
. Line 
intersects 
for the second time at point 
, and line 
intersects 
for the second time at point 
. Let 
be the second intersection of 
and . Also 
is the second intersection point of 
and . Prove that 
.
Proposed by Alireza Dadgarnia
be two circles with centers 
and 
, respectively. These two circles intersect each other at points 
and 
. Line 
intersects 
for the second time at point 
, and line 
intersects 
for the second time at point 
. Let 
be the second intersection of 
and . Also 
is the second intersection point of 
and . Prove that 
.Proposed by Alireza Dadgarnia
This post has been edited 1 time. Last edited by parmenides51, Sep 20, 2018, 9:23 AM
Reason: Proposed
Reason: Proposed
and 
.
.
![[asy]
import graph; size(15cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps);
pen dotstyle = black; /* point style */
real xmin = -22.067805141047277, xmax = 20.294112477713487, ymin = -13.369930481706609, ymax = 10.644325415162337; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen qqzzff = rgb(0,0.6,1);
/* draw figures */
draw(circle((2.0839894493890574,0.4939698298878338), 5), linewidth(2) + wrwrwr);
draw(circle((-7.54,0.01), 7), linewidth(2) + wrwrwr);
draw((-4.483683057485903,6.307533385929867)--(-2.8360298327762967,1.384705628607698), linewidth(2) + qqzzff);
draw((-0.5583658847362666,0.5167396576010096)--(1.0892873399733358,-4.406088099721154), linewidth(2) + qqzzff);
draw((-2.8360298327762967,1.384705628607698)--(-1.3085861196788156,-3.178962378602244), linewidth(2) + wrwrwr);
draw((-1.659946616111183,3.8080221435133406)--(-0.5583658847362666,0.5167396576010096), linewidth(2) + wrwrwr);
draw((-4.483683057485903,6.307533385929867)--(2.0839894493890574,0.4939698298878338), linewidth(2) + wrwrwr);
draw((-7.54,0.01)--(1.0892873399733358,-4.406088099721154), linewidth(2) + wrwrwr);
/* dots and labels */
dot((2.0839894493890574,0.4939698298878338),dotstyle);
label("$O_2$", (2.1940204042429814,0.7690472170226442), NE * labelscalefactor);
dot((-7.54,0.01),dotstyle);
label("$O_1$", (-7.433688145475375,0.2739079201799855), NE * labelscalefactor);
dot((-1.659946616111183,3.8080221435133406),linewidth(4pt) + dotstyle);
label("$A$", (-1.5470320607904369,4.014960385213406), NE * labelscalefactor);
dot((-1.3085861196788156,-3.178962378602244),linewidth(4pt) + dotstyle);
label("$B$", (-1.1894314575151836,-2.972005248010777), NE * labelscalefactor);
dot((1.0892873399733358,-4.406088099721154),linewidth(4pt) + dotstyle);
label("$C$", (1.2037418105576647,-4.182345751403942), NE * labelscalefactor);
dot((-4.483683057485903,6.307533385929867),linewidth(4pt) + dotstyle);
label("$D$", (-4.380329148278982,6.518164608140181), NE * labelscalefactor);
dot((-0.5583658847362666,0.5167396576010096),linewidth(4pt) + dotstyle);
label("$X$", (-0.44672251225119625,0.7415394783091631), NE * labelscalefactor);
dot((-2.8360298327762967,1.384705628607698),linewidth(4pt) + dotstyle);
label("$Y$", (-2.7298648254701208,1.5942793784270752), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/6/d/0/6d018cc864792b2acc3fd134548edbb9186094a2.png)
and 
be the antipodes of 
.
and 
it follows that 
implying that 
and 
.
is similar to 
.
by cyclicity
.
.
is a parallelogram which implies 
.
.
.
.
so 
and 
.
and 
hence 
.
,
.
hence 
.
Hence, by the last two we conclude that 
then 
is diameter of 
then 
is diameter of 
and 
but since 
we have that 
thus 
as desired.
and 
We have 
so 
![\[\frac{CA}{CK}=\frac{sin \angle 90-\alpha-\beta}{sin \angle 90-\beta}\]](http://latex.artofproblemsolving.com/4/3/1/4316795513ad7e8af9d61130fa146260b3bcce0c.png)
![\[\frac{DA}{DB}=\frac{sin \angle \beta}{sin \angle 90- \alpha}\]](http://latex.artofproblemsolving.com/d/2/c/d2c3190ab6e1b9e97a7b72597541df82050c0a50.png)
![\[\frac{BC}{DL}=\frac{\frac{QC}{sin \angle \alpha}}{\frac{QL}{sin \angle 90-\alpha-\beta}}=\frac{QC}{QL}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DA}{DL}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DP}{DB}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DP}{KD}.\frac{KD}{DB}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}\]](http://latex.artofproblemsolving.com/b/6/9/b699de83eeb372d041f2c0fc885708f86160ae4b.png)
![\[=\frac{sin \angle \alpha}{\sin \angle 90-\alpha}.\frac{sin \angle \beta}{sin \angle 90-\beta}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{sin \angle \beta}{sin \angle 90-\alpha }.\frac{sin \angle 90-\alpha-\beta}{sin \angle 90-\beta}=\frac{CA}{CK}.\frac{DA}{DB}\]](http://latex.artofproblemsolving.com/6/8/b/68b89a6ece08b3be876e62362cf1d199f272df7e.png)
Which means 
as desired.
