Inverting into a Trash Can
by agbdmrbirdyface, Aug 17, 2016, 3:20 AM
According to most nerds (like us on AoPS), we have often been subject to inversion and subsequent stuffing in trash cans.
Perhaps this inversion will not be so bad?
Solution:
Perhaps this inversion will not be so bad?
#9 in 110 Problems in Oly Geo, mild paraphrasing wrote:
Let
be a triangle with
Let
be points on sides
, respectively, such that the angles
and
are congruent. If
lies in the interior of quadrilateral
such that the circles
and
are tangent, and circles
and
are tangent, show the points
are concyclic.













Solution:
Uninverted Diagram:
![[asy]
import graph; size(200); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-3.,xmax=20.,ymin=-5.,ymax=8.;
pair A=(6.78,5.52), B=(0.04,-2.34), C=(8.2,-2.34), D=(7.699728008025581,0.4291111668443164), F=(5.412370806452737,-0.3929963195026304);
D(A--B--C--cycle, blue); D((5.127855293301196,3.5933149266094064)--D, blue); D(CR((5.969479843925995,1.6500753010771838),2.117666670289025), red); D(CR((4.12,-5.212453318604166),4.989728255883288), red); D(CR((6.969590642292293,-1.1325421485935039),1.7239088289285724), red); D(CR((1.7422895380775811,1.3483674953380538),4.062246232334359), red);
D(A); MP("$A$",(6.858681295543141,5.697058767407373),NE*lsf); D(B); MP("$B$",(0.1135230630150352,-2.147853524772011),N*lsf); D(C); MP("$C$",(8.270032338762338,-2.147853524772011),SE*lsf); D(D); MP("$D$",(7.775143011919242,0.6198608586837995),E*lsf); D((5.127855293301196,3.5933149266094064)); MP("$E$",(5.209050206066158,3.7724891630175708),N*lsf); D(F); MP("$F$",(5.483988720978989,-0.20495468605468722),NW*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/a/9/7/a97e2d24fc62ed7871c9cdc09336c6cbff34e530.png)
Inverted Diagram (about point
):
![[asy]
import graph; size(200); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.637612057179676,xmax=19.04744680047705,ymin=-6.828403687050436,ymax=8.323290524240274;
pair F=(5.412370806452737,-0.3929963195026304);
D((4.967025564951665,5.846680121804947)--(11.440094301029207,-4.603025349234083),linetype("4 4")); D((1.2991658073186674,-1.883665069011752)--(15.09166150642361,3.0858732429350657),linetype("4 4")); D(CR((2.036549626218972,2.5017918172372644),4.44701776447148),blue); D(CR((10.543985564726135,0.5340892984042673),5.214686737500074),blue); D((4.967025564951665,5.846680121804947)--(1.2991658073186674,-1.883665069011752), red); D((1.2991658073186674,-1.883665069011752)--(11.440094301029207,-4.603025349234083), red); D((4.967025564951665,5.846680121804947)--(15.09166150642361,3.0858732429350657), red); D((15.09166150642361,3.0858732429350657)--(11.440094301029207,-4.603025349234083), red);
D(F); MP("$F$",(5.522780155499318,-0.14233728173255547),SE*lsf); D((4.967025564951665,5.846680121804947)); MP("$E$'",(5.065178652473756,6.111549926283409),NE*lsf); D((11.440094301029207,-4.603025349234083)); MP("$C$'",(11.547866612002556,-4.3370177261335074),S*2); D((1.2991658073186674,-1.883665069011752)); MP("$B$'",(1.4043666282692566,-1.6168310137037991),S*2); D((15.09166150642361,3.0858732429350657)); MP("$D$'",(15.183256330483413,3.3405186024064166),NE*lsf); D((6.340610750388194,3.6202840330975685)); MP("$A$'",(6.437983161550443,3.8743870226029014),E*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](//latex.artofproblemsolving.com/9/4/d/94da1b24e70b47a271868989ee198d28d594e62b.png)
Looking at point
in our uninverted diagram, we have almost immediate motivation to invert about this point, since doing so will bring our messy tangent circles to lines. We can do so, and this is illustrated in the second diagram.
We claim that this inversion
will send circles
and
to parallel lines. We know
and
will become lines perpendicular to the line between the centers of those circles and
... but this just means the lines are perpendicular to the same line, so they must be parallel. Similarly, we conclude that
and
must also invert to a pair of parallel lines. Thus, the inversions of our four tangent circles about
go to a parallelogram.
We haven't yet shown where
is yet, but we know where
,
,
,
will go. Noting that line
does not pass through
and contains point
, we see that
inverts to a circle through
. Hence
are concyclic, and similarly,
are concyclic. Thus
is the intersection of the circumcircles of
and
.
The angle condition hasn't been dealt with yet. Let's see what this is under inversion. We can decompose this condition into:

Under inversion, we can easily see that:

Also, in our inverted diagram, since we have a parallelogram, we can easily claim that
.
With this information, we claim
are collinear. The proof is a simple angle chase, denoting
as the angle going from
to
, and
as the angle going the other direction:



Hence we can easily see that
, which proves our claim.
Noting this, we can show that
are also collinear. This is another angle chase, splitting
.
Then, with help from a couple of cyclic quads, and our parallel lines:




Hence
is a line, and we can note that
is a circle through
,
,
, and
. Therefore, since a line not passing through the center of inversion goes to a circle through it, we know that
are concyclic, completing the problem.
.
![[asy]
import graph; size(200); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-3.,xmax=20.,ymin=-5.,ymax=8.;
pair A=(6.78,5.52), B=(0.04,-2.34), C=(8.2,-2.34), D=(7.699728008025581,0.4291111668443164), F=(5.412370806452737,-0.3929963195026304);
D(A--B--C--cycle, blue); D((5.127855293301196,3.5933149266094064)--D, blue); D(CR((5.969479843925995,1.6500753010771838),2.117666670289025), red); D(CR((4.12,-5.212453318604166),4.989728255883288), red); D(CR((6.969590642292293,-1.1325421485935039),1.7239088289285724), red); D(CR((1.7422895380775811,1.3483674953380538),4.062246232334359), red);
D(A); MP("$A$",(6.858681295543141,5.697058767407373),NE*lsf); D(B); MP("$B$",(0.1135230630150352,-2.147853524772011),N*lsf); D(C); MP("$C$",(8.270032338762338,-2.147853524772011),SE*lsf); D(D); MP("$D$",(7.775143011919242,0.6198608586837995),E*lsf); D((5.127855293301196,3.5933149266094064)); MP("$E$",(5.209050206066158,3.7724891630175708),N*lsf); D(F); MP("$F$",(5.483988720978989,-0.20495468605468722),NW*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/a/9/7/a97e2d24fc62ed7871c9cdc09336c6cbff34e530.png)
Inverted Diagram (about point

![[asy]
import graph; size(200); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.637612057179676,xmax=19.04744680047705,ymin=-6.828403687050436,ymax=8.323290524240274;
pair F=(5.412370806452737,-0.3929963195026304);
D((4.967025564951665,5.846680121804947)--(11.440094301029207,-4.603025349234083),linetype("4 4")); D((1.2991658073186674,-1.883665069011752)--(15.09166150642361,3.0858732429350657),linetype("4 4")); D(CR((2.036549626218972,2.5017918172372644),4.44701776447148),blue); D(CR((10.543985564726135,0.5340892984042673),5.214686737500074),blue); D((4.967025564951665,5.846680121804947)--(1.2991658073186674,-1.883665069011752), red); D((1.2991658073186674,-1.883665069011752)--(11.440094301029207,-4.603025349234083), red); D((4.967025564951665,5.846680121804947)--(15.09166150642361,3.0858732429350657), red); D((15.09166150642361,3.0858732429350657)--(11.440094301029207,-4.603025349234083), red);
D(F); MP("$F$",(5.522780155499318,-0.14233728173255547),SE*lsf); D((4.967025564951665,5.846680121804947)); MP("$E$'",(5.065178652473756,6.111549926283409),NE*lsf); D((11.440094301029207,-4.603025349234083)); MP("$C$'",(11.547866612002556,-4.3370177261335074),S*2); D((1.2991658073186674,-1.883665069011752)); MP("$B$'",(1.4043666282692566,-1.6168310137037991),S*2); D((15.09166150642361,3.0858732429350657)); MP("$D$'",(15.183256330483413,3.3405186024064166),NE*lsf); D((6.340610750388194,3.6202840330975685)); MP("$A$'",(6.437983161550443,3.8743870226029014),E*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/9/4/d/94da1b24e70b47a271868989ee198d28d594e62b.png)
Looking at point

We claim that this inversion









We haven't yet shown where















The angle condition hasn't been dealt with yet. Let's see what this is under inversion. We can decompose this condition into:

Under inversion, we can easily see that:

Also, in our inverted diagram, since we have a parallelogram, we can easily claim that

With this information, we claim








Hence we can easily see that

Noting this, we can show that


Then, with help from a couple of cyclic quads, and our parallel lines:




Hence







