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is given where 
are positive reals. Prove that:
be a triangle and let 
be a circle in its plane passing through 
and 
Suppose there exist circles 
such that for 
is externally tangent to 
and passes through 
and 
where 
for all 
. Prove that 
and 
have common external tangents 
and 
meeting at 
. Suppose touches at 
and touches at 
. A circle 
through and intersects again at 
and again at 
, such that quadrilateral 
is convex.
and 
meet at point 
, while lines 
and 
meet at point 
. Show that , , are collinear.
and 
be two interior points such that 
and 
. Point lies on segment . Prove that 
if and only if 
.
are obviously isogonal conjugates WRT 
Let 
cut at 
Since 
and 
the problem is equivalent to prove that there exists a point on 
such that 
are isogonal WRT 
and 
are isogonal WRT 
Assume there exists such a 
By Steiner theorem we have:
which is obvious identity. Thus, there exists such a 
are isogonal conjugates. Since the second condition is equivalent to 
, which is analogous to the first condition (i.e. are switched), it suffices to prove that the first condition implies the second. By the Law of Sines on 
in addition to the angle conditions, we have![\[\frac{BD}{\sin{APC}}=\frac{BD}{\sin(180^\circ-DPB)}=\frac{BD}{\sin{DPB}}=\frac{DP}{\sin{PBC}},\\
\frac{CD}{\sin{APB}}=\frac{CD}{\sin(180^\circ-DPC)}=\frac{CD}{\sin{DPC}}=\frac{DP}{\sin{PCB}},\]](http://latex.artofproblemsolving.com/c/a/0/ca0e2e209baf7ba9134c351c414383d16d6c43ca.png)
so![\[\frac{BD\sin{APB}}{CD\sin{APC}}=\frac{\sin{PCB}}{\sin{PBC}}\qquad(1).\]](http://latex.artofproblemsolving.com/8/2/4/824db69c2266e54bc7b31b369443caf7cee812da.png)
Now by the Law of Sines on 
and isogonal conjugates ,![\[\frac{AB}{\sin{AQB}}=\frac{AQ}{\sin{QBA}}=\frac{AQ}{\sin{PBC}}\\
\frac{AC}{\sin{AQC}}=\frac{AQ}{\sin{QCA}}=\frac{AQ}{\sin{PCB}},\]](http://latex.artofproblemsolving.com/7/5/3/7538dff7a5343a7207ac0d8680c8e2346f365775.png)
so from (1),![\[\frac{AB\sin{AQC}}{AC\sin{AQB}}=\frac{\sin{PCB}}{\sin{PBC}}=\frac{BD\sin{APB}}{CD\sin{APC}}\qquad(2).\]](http://latex.artofproblemsolving.com/a/9/e/a9e7461ec005a703b575b3aa3650505b9b0721b6.png)
But again by the Law of Sines on 
and isogonal conjugates , we have![\[\frac{BD}{\sin{DQB}}=\frac{DQ}{\sin{QBC}}=\frac{DQ}{\sin{PBA}},\\
\frac{CD}{\sin{DQC}}=\frac{DQ}{\sin{QCB}}=\frac{DQ}{\sin{PCA}},\]](http://latex.artofproblemsolving.com/f/a/8/fa8a5dbaacba57eeed5580ff73e9b3810adc944f.png)
so dividing and applying the Law of Sines on 
,![\[\frac{BD\sin{DQC}}{CD\sin{DQB}}=\frac{\sin{PCA}}{\sin{PBA}}=\frac{\frac{AP}{AC}\sin{APC}}{\frac{AP}{AB}\sin{APB}}=\frac{AB\sin{APC}}{AC\sin{APB}}.\]](http://latex.artofproblemsolving.com/4/8/1/481992bcedeeb9516fcfa246b847e9dedfebd7e6.png)
Multiplying this equation with (2), we obtain![\[\sin{DQC}\sin{AQC}=\sin{DQB}\sin{AQB},\]](http://latex.artofproblemsolving.com/c/a/0/ca0d3439f331726a7ba8b8f9b8281568775f253a.png)
or (note that 
)
Clearly angles 
are all in 
, so their differences are in 
. Thus we either have![\[DQC-AQC=DQB-AQB \implies DQC+AQB=DQB+AQC=180^\circ,\]](http://latex.artofproblemsolving.com/a/4/2/a4273a94f458e4ceddadafd2addc8ba136ca9c29.png)
as desired, or![\[DQC-AQC=-DQB+AQB \implies DQC+DQB=AQC+AQB=180^\circ,\]](http://latex.artofproblemsolving.com/a/b/9/ab9672cebca069138c8f507845fdc75ea9a4676c.png)
contradicting the fact that is inside 
.
be an interior point of 
, and let , 
, and 
be the projections of onto , 
, and 
, respectively. Let 
be the circumcenter of 
, and let 
be the midpoint of 
. Then 
is a common chord of the circumcircles of 
and , so 
. Hence,![\[ \angle MOF \&= \frac{\angle EOF}{2} \&= \angle EDF = \angle EDQ + \angle QDF \&= \angle ACQ + \angle QBA. \]](http://latex.artofproblemsolving.com/b/b/4/bb4dfa94b757c863bca4b562875d272d2cf7937e.png)
be the reflection of across , and let be the isogonal conjugate of . Then 
. and 
be the midpoints of 
and 
, respectively, and let be the projection from onto . By the lemma,
with factor 2 reveals that 
. Similarly, we find that 
and 
. Hence, lies on the intersection of the circumcircles of 
, 
, and 
. But these circumcircles intersect only at the point , so we have 
, as desired.
and 
be the reflections of and across , respectively, and let 
. Because 
lies on the axis of symmetry of isosceles trapezoid 
, must lie on as well. As varies from to , 
increases and 
decreases, so there exists at most one point such that 
and 
. We claim that 
satisfies both of these conditions.
. Let be the the midpoint of 
, let be the midpoint of 
, and let be the midpoint of 
. By the corollary to our lemma, is the circumcenter of the -pedal triangle of , so by the lemma, 
. A homothety centered at with factor 2 sends to , to , and to , so 
, which completes our proof.
with respect to 
is the reflection 
of in line . The condition 
is therefore equivalent to , and being collinear. Similarly, the condition 
is equivalent to , and being collienear where is the reflection of in . Now note that quadrilaterals 
and 
are reflections in . Therefore 
and 
divide in the same ratio and concur on . This implies the result.
and 
and 
and 
.After Steiner's theorem , this problem is equivalent to 
.By sine law to the triangles 
and 
, 
is equivalent to 
, which is true by Ceva's theorem.
be the isogonal conjugate of 
.
be the isogonal conjugate of 
of 
.
be the isogonal conjugate of 
of 
.
are symmetry WRT 
.
be the intersection of 
and 
.
lie on satisfy 
are isogonal conjugate of 
.
are isogonal conjugate of 
,
are isogonal conjugate of 
.
are isogonal conjugate of .
are isogonal conjugate of .
(Easy angle chasing) ,
are symmetry WRT .
are isogonal conjugate of 
are isogonal conjugate of 
be the pedal triangle of WRT and let 
be the triangle obtained by reflecting in 
Since 
are isogonal conjugates, it is well-known (Corollary to Fact 3) that is the circumcenter of 
Therefore 
is the perpendicular bisector of 
, implying that ![\[\measuredangle BQ_aP = -\measuredangle Q_aPB - \measuredangle PBQ_a = -\measuredangle Q_aPB - \measuredangle ABC = -\measuredangle Q_aQ_bQ_c - \measuredangle ABC.\]](http://latex.artofproblemsolving.com/9/f/0/9f08e1e334f8902d31fa198aa6eafef034153deb.png)
Furthermore, since 
are inscribed in the circle of diameter 
, with similar relations holding for the two other sets of vertices, we have ![\[\measuredangle Q_aQ_bQ_c = \measuredangle A'B'C' = \measuredangle A'B'Q + \measuredangle QB'C' = \measuredangle A'CQ + \measuredangle QAC' = \measuredangle BCQ + \measuredangle QAB = -\measuredangle CQA - \measuredangle ABC,\]](http://latex.artofproblemsolving.com/1/9/c/19c50d31020a7999089a22029432a33dcd8583b1.png)
where the last step follows from examining quadrilateral 
Consequently, if 
cuts at , then ![\[\measuredangle BQD' = -\measuredangle BQ_aD' = -\measuredangle BQ_aP = -\measuredangle CQA,\]](http://latex.artofproblemsolving.com/c/4/b/c4b7fb66ef1433748ce6c4d71157070030c699bd.png)
implying that 
Thus, if 
is the reflection of in , then by symmetry in trapezoid 
, we have 
, and analagous arguments show that 
The desired result follows. 
satisfying both conditions at once. To construct it, take the ellipse with foci and tangent to the sides , , at , , . We claim this works.
bisects 
. Similarly, 
bisects 
and 
bisects 
. Thus 
follows, and in the same way .
are symmetric to the points 
with respect to the line , then 
and 
are isogonal conjugates with respect to 
and also 
and are isogonal conjugates with respect to 
![[asy]
import olympiad;
size(12cm);
defaultpen(fontsize(10pt));
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair X = (7A+2C)/(7+2);
pair Y = (5A+3B)/(5+3);
pair P = extension(B,X,C,Y);
pair P0 = foot(P,B,C);
pair Ps = 2*P0 - P;
pair X0 = foot(P,A,C);
pair S = 2*X0 - P;
pair Y0 = foot(P,A,B);
pair K = 2*Y0 - P;
pair Q = circumcenter(Ps,S,K);
pair Q0 = foot(Q,B,C);
pair Qs = 2*Q0 - Q;
pair D = extension(P,Qs,Q,Ps);
pair T = (8P-5A)/(8-5);
pair Z = (5Q-1.5*A)/(5-1.5);
draw(A--T, dotted);
draw(A--Z, dotted);
draw(A--B--C--A,red);
draw(B--P--C, black);
draw(B--Q--C, black);
draw(Q--Ps, dashed);
draw(P--Qs, dashed);
draw(B--Ps, black);
draw(C--Ps, black);
draw(B--Qs, black);
draw(C--Qs, black);
draw(anglemark(Qs,B,C), blue);
draw(anglemark(D,B,Q), blue);
draw(anglemark(P,B,A), blue);
draw(anglemark(A,C,P), darkgreen);
draw(anglemark(Q,C,D), darkgreen);
draw(anglemark(D,C,Qs), darkgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$Q$", Q, dir(120));
dot("$P$", P, dir(120));
dot("$P^\ast$", Ps, dir(Ps));
dot("$Q^\ast$", Qs, dir(Qs));
dot("$D$", D, dir(380));
[/asy]](http://latex.artofproblemsolving.com/a/6/d/a6da7db34808fc1050659a85af94246ac1427bf1.png)
therefore 
and are isogonal conjugates with respect to 
Similarly, 
and are isogonal conjugates with respect to 
and 
are isogonal conjugates relative to 
, as well as the lines 
and 
are isogonal conjugates relative to 
whence 
Then it suffices to prove that 
lies on the ray if and only if lies on the ray 
which is obvious, since and 
intersect on 
such that 
must be such that 
and 
are isogonal conjugates relative to . Let intersect line at , then this result implies![\[\frac{BX}{XC}\cdot\frac{BD_1}{D_1C}=\frac{BP^2}{PC^2}.\]](http://latex.artofproblemsolving.com/2/a/b/2abc4eb8f4d7c43809d15dcbb31cf916f3595676.png)
Then if we define as the intersection of lines and , it suffices to check![\[\frac{BP^2}{PC^2}\cdot\frac{XC}{XB}=\frac{BQ^2}{QC^2}\cdot\frac{YC}{YB}\iff \frac{BP^2}{PC^2}\cdot\frac{XC^2}{XB^2}=\frac{BQ^2}{QC^2}\cdot\frac{YC}{YB}\cdot\frac{XC}{XB}=\frac{BQ^2}{QC^2}\cdot\frac{AC^2}{AB^2}.\]](http://latex.artofproblemsolving.com/f/d/8/fd806b9db1cebc7c5f5f32ab5df6732ebbbf4913.png)
So it is equivalent to check![\[1=\frac{BP\cdot XC\cdot QC\cdot AB}{BQ\cdot AC\cdot PC\cdot XB}=\frac{AP\cdot \frac{\sin\angle BAP}{\sin\angle ABP}\cdot AC\cdot\frac{\sin\angle PAC}{\sin\angle AXC}\cdot AQ\cdot\frac{\sin\angle QAC}{\sin\angle QCA}\cdot AB}{AQ\cdot\frac{\sin\angle BAQ}{\sin\angle ABQ}\cdot AC\cdot AP\cdot\frac{\sin\angle PAC}{\sin\angle ACP}\cdot AB\cdot\frac{\sin\angle PAB}{\sin\angle AXB}}=\]](http://latex.artofproblemsolving.com/1/5/b/15b8611ee16b724091ee05d66e4143d8dd48f770.png)
![\[\frac{\frac{\sin\angle BAP}{\sin\angle ABP}\cdot \frac{1}{\sin\angle PCB}}{\frac{\sin\angle PAC}{\sin\angle PBC}\cdot \frac{1}{\sin\angle ACP}}=\frac{\sin\angle BAP}{\sin\angle ABP}\cdot\frac{\sin\angle PBC}{\sin\angle PCB}\cdot\frac{\sin\angle ACP}{\sin\angle PAC}=\frac{BP}{AP}\cdot\frac{PC}{PB}\cdot\frac{AP}{CP},\]](http://latex.artofproblemsolving.com/8/1/a/81abaaf80050841a9dcc688fc9d39c607d0ab803.png)
which is trivial.
and 
are isogonal with respect to 
and 
, then 
and are isogonal conjugates. Let and be the intersections of and with respectively. By our theorem, we have the two equations: 
and we wish to prove 
and 
so 
and 
. Therefore, we have 
by Trig Ceva's, as desired. 
. In addition, we define 
and 
. It's clear that and are isogonal conjugates wrt 
.
which implies 
and 
are isogonal in . Thus, applying Steiner's Ratio Theorem twice yields 
and 
Now, the LoS and Ratio Lemma give 
Hence, the Converse of Steiner's implies 
, which clearly finishes. 
and are isogonal conjugates. So if we denote 
and 
, we have that 
.![\[\left(\frac{QB}{QC}\right)^2\cdot \frac{YC}{YB}=\left(\frac{BP}{PC}\right)^2\cdot \frac{XC}{XB}\]](http://latex.artofproblemsolving.com/a/1/6/a16f727557ed3d40f8a2491f25926aa34d9ed783.png)
which reduces, after using the sine law in the triangles 
, 
, 
and 
, to ![\[\frac{\sin{\angle QCA}}{\sin{\angle QCB}}\cdot \frac{\sin{\angle QBC}}{\sin{\angle{QBA}}}\cdot \frac{\sin{\angle BAQ}}{\sin{\angle{CAQ}}}=1\]](http://latex.artofproblemsolving.com/d/5/0/d50bccd0852968d88381807504b2d181b95315c4.png)
which is true, by Ceva.
inside quadrilateral has an isogonal conjugate iff 
.
kills the problem.
and let 
. It suffices to show that![\[\left(\frac{PB}{PC}\right)^2\div \frac{XB}{XC}=\left(\frac{QB}{QC}\right)^2\div \frac{YB}{YC}\]](http://latex.artofproblemsolving.com/5/e/f/5efa7157708284b7a2531977931e200b603007ef.png)
or that![\[\frac{PB}{PC}\div \frac{QB}{QC}=\sqrt{\frac{XB}{XC}\div \frac{YB}{YC}}\]](http://latex.artofproblemsolving.com/2/5/0/250af5053bd50373d96150c1ad46b82b84e59ebc.png)
with , as cevians. After some annoying calculations we can find![\[\frac{XB}{XC}\div \frac{YB}{YC}=\frac{\sin^2 \angle BAP}{\sin^2 \angle CAP}\]](http://latex.artofproblemsolving.com/d/9/b/d9b94fc0d675a9e572e1cc74677c631c2cb31808.png)
and LoS on and 
gives![\[\sin BAP=\frac{PB\cdot \sin \angle ABP}{PA}\]](http://latex.artofproblemsolving.com/1/6/d/16d0ccf37704be866c5a44ee354ef6997b646b89.png)
and dividing gives![\[\frac{XB}{XC}\div \frac{YB}{YC}=\left(\frac{PB\cdot QC}{PC\cdot QB}\right)^2\]](http://latex.artofproblemsolving.com/6/0/8/6087db18722477d9f4fe2e062aa855f0fc31f46e.png)
and we're done.
and are isogonal conjugates. Then we may consider the inellipse with foci and , inscribed in 
and denote by , and 
its tangency points to , 
and 
. is the unique point along satisfying 
.
proving the claim. proves the claim and we're done.
and are isogonal conjugates in . The condition implies that has an isogonal conjugate wrt degenerate quadrilateral , which is clearly as 
and 
. Clearly the reverse holds true as well, so we are done. and is tangent to sides and 
of deg. quadrilateral , so and are isogonal conjugates in .
and 
are isogonal conjugates with respect to 
, we know that 
and 
are isogonal lines, meaning the angles they form with the triangle's sides are equal when measured from corresponding vertices. If 
is the intersection of with 
and 
is the intersection of with , we aim to prove the existence of a point 
on such that 
and 
are isogonal with respect to 
and 
and 
are isogonal with respect to 
., , and segments on , and specifically using the points , , and , we get the following relationships:![\[
\frac{BN}{NC} \cdot \frac{BD}{DC} = \frac{BP^2}{CP^2}
\]](http://latex.artofproblemsolving.com/b/9/2/b92f6904342c20fa10448cef8cca882b380640f5.png)
![\[
\frac{BM}{MC} \cdot \frac{BD}{DC} = \frac{BQ^2}{CQ^2}
\]](http://latex.artofproblemsolving.com/a/7/5/a754fa555df6c31a2ee440f618639001198b896b.png)
![\[
\frac{BQ^2}{CQ^2} \cdot \frac{MC}{BM} = \frac{BP^2}{CP^2} \cdot \frac{NC}{BN}
\]](http://latex.artofproblemsolving.com/b/8/0/b8008180febd49834aba2b64ae55814a550e8d95.png)
, 
, 
, and 
, we can express these ratios in terms of sines of angles, giving us:![\[
\frac{BQ}{CQ} \cdot \frac{\sin \angle AQC}{\sin \angle AQB} = \frac{BP}{CP} \cdot \frac{\sin \angle APC}{\sin \angle APB}
\]](http://latex.artofproblemsolving.com/7/8/4/784b122a11d30c9f6d58fb7bc9a07fa39150f5b2.png)
![\[
\frac{\sin \angle AQC}{\sin \angle QCA} \cdot \frac{\sin \angle QBA}{\sin \angle AQB} = \frac{\sin \angle APC}{\sin \angle PCA} \cdot \frac{\sin \angle PBA}{\sin \angle APB}
\]](http://latex.artofproblemsolving.com/7/d/7/7d7de3b8b6c95502e5308c9c2aa807dd98f041a8.png)
and are isogonal conjugates, we obtain:![\[
\frac{AC}{AQ} \cdot \frac{AQ}{AB} = \frac{AC}{AP} \cdot \frac{AP}{AB}
\]](http://latex.artofproblemsolving.com/b/8/7/b87fa0d72a07f2e80b4152d470b7c2be71f5635d.png)
on 
that satisfies the required conditions. By verifying the conditions, we conclude that and are isogonal with respect to , and and are isogonal with respect to . Therefore, the existence of such a point is guaranteed by the properties of isogonal conjugates and the applications of trigonometric Ceva's theorem.
and are isogonal conjugates wrt 
. So 
and are isogonal conjugates in 
.
and are isogonal conjugates in . Then 
implies has an isogonal conjugate in the degenerate quadrilateral so it must in fact be so the desired result follows. Same goes for the other direction as they are both symmetric.
it is sufficient to show, by continuity 
etc
(1)
(2)
(3)