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Difference between revisions of "2023 USAMO Problems/Problem 6"

(→‎Solution)
(→‎Solution 1)
Line 2: Line 2:
 
Let ABC be a triangle with incenter <math>I</math> and excenters <math>I_a</math>, <math>I_b</math>, <math>I_c</math> opposite <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Given an arbitrary point <math>D</math> on the circumcircle of <math>\triangle ABC</math> that does not lie on any of the lines <math>IIa</math>, <math>I_bI_c</math>, or <math>BC</math>, suppose the circumcircles of <math>\triangle DIIa</math> and <math>\triangle DI_bI_c</math> intersect at two distinct points <math>D</math> and <math>F</math>. If <math>E</math> is the intersection of lines <math>DF</math> and <math>BC</math>, prove that <math>\angle BAD = \angle EAC</math>.
 
Let ABC be a triangle with incenter <math>I</math> and excenters <math>I_a</math>, <math>I_b</math>, <math>I_c</math> opposite <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Given an arbitrary point <math>D</math> on the circumcircle of <math>\triangle ABC</math> that does not lie on any of the lines <math>IIa</math>, <math>I_bI_c</math>, or <math>BC</math>, suppose the circumcircles of <math>\triangle DIIa</math> and <math>\triangle DI_bI_c</math> intersect at two distinct points <math>D</math> and <math>F</math>. If <math>E</math> is the intersection of lines <math>DF</math> and <math>BC</math>, prove that <math>\angle BAD = \angle EAC</math>.
 
== Solution 1 ==
 
== Solution 1 ==
−
[asy]
+
<asy>
 
size(500);
 
size(500);
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q;
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q;
Line 32: Line 32:
 
draw(G--J,dashed);
 
draw(G--J,dashed);
 
draw(H--K,dashed);
 
draw(H--K,dashed);
−
dot("<math>A</math>",A,dir(A-circumcenter(A,B,C)));
+
dot("$A$",A,dir(A-circumcenter(A,B,C)));
−
dot("<math>B</math>",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C)));
+
dot("$B$",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C)));
−
dot("<math>C</math>",C,dir(C-circumcenter(A,B,C))*dir(15));
+
dot("$C$",C,dir(C-circumcenter(A,B,C))*dir(15));
−
dot("<math>D</math>",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
+
dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
−
dot("<math>E</math>",E,dir(dir(H-K)+dir(B-C)));
+
dot("$E$",E,dir(dir(H-K)+dir(B-C)));
−
dot("<math>F</math>",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F)));
+
dot("$F$",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F)));
−
dot("<math>G</math>",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
+
dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
−
dot("<math>H</math>",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C))));
+
dot("$H$",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C))));
−
dot("<math>I</math>",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
+
dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
−
dot("<math>J</math>",J,dir(J-circumcenter(A,B,C)));
+
dot("$J$",J,dir(J-circumcenter(A,B,C)));
−
dot("<math>K</math>",K,dir(K-circumcenter(A,B,C)));
+
dot("$K$",K,dir(K-circumcenter(A,B,C)));
−
dot("<math>I_A</math>",IA,dir(IA-circumcenter(D,I,IA)));
+
dot("$I_A$",IA,dir(IA-circumcenter(D,I,IA)));
−
dot("<math>I_B</math>",IB,dir(dir(IB-IC)+dir(IB-IA)));
+
dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA)));
−
dot("<math>I_C</math>",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
+
dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
−
dot("<math>P</math>",P,dir(dir(A-I)+dir(C-B)));
+
dot("$P$",P,dir(dir(A-I)+dir(C-B)));
−
dot("<math>Q</math>",Q,dir(dir(IC-IB)+dir(B-C)));
+
dot("$Q$",Q,dir(dir(IC-IB)+dir(B-C)));
−
[/asy]
+
</asy>
  
 
Consider points <math>G,H,J,K,P,</math> and <math>Q</math> such that the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DII_A</math> are <math>D</math> and <math>G</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DI_BI_C</math> are <math>D</math> and <math>H</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{II_A}</math> are <math>A</math> and <math>J</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{I_BI_C}</math> are <math>A</math> and <math>K</math>, the intersection of lines <math>\overline{II_A}</math> and <math>\overline{BC}</math> is <math>P</math>, and the intersection of lines <math>\overline{I_BI_C}</math> and <math>\overline{BC}</math> is <math>Q</math>.
 
Consider points <math>G,H,J,K,P,</math> and <math>Q</math> such that the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DII_A</math> are <math>D</math> and <math>G</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with the circumcircle of <math>\triangle{}DI_BI_C</math> are <math>D</math> and <math>H</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{II_A}</math> are <math>A</math> and <math>J</math>, the intersections of the circumcircle of <math>\triangle{}ABC</math> with line <math>\overline{I_BI_C}</math> are <math>A</math> and <math>K</math>, the intersection of lines <math>\overline{II_A}</math> and <math>\overline{BC}</math> is <math>P</math>, and the intersection of lines <math>\overline{I_BI_C}</math> and <math>\overline{BC}</math> is <math>Q</math>.
Line 54: Line 54:
 
Since <math>IBI_AC</math> is cyclic, the pairwise radical axes of the circumcircles of <math>\triangle{}DII_A,\triangle{}ABC,</math> and <math>IBI_AC</math> concur. The pairwise radical axes of these circles are <math>\overline{GD},\overline{II_A},</math> and <math>\overline{BC}</math>, so <math>G,P,</math> and <math>D</math> are collinear. Similarly, since <math>BCI_BI_C</math> is cyclic, the pairwise radical axes of the cirucmcircles of <math>\triangle{}DI_BI_C,\triangle{}ABC,</math> and <math>BCI_BI_C</math> concur. The pairwise radical axes of these circles are <math>\overline{HD},\overline{I_BI_C},</math> and <math>\overline{BC}</math>, so <math>H,Q,</math> and <math>D</math> are collinear. This means that <math>-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)</math>, so the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>G</math> and <math>H</math> intersect on <math>\overline{BC}</math>. Let this intersection be <math>X</math>. Also, let the intersection of the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>K</math> and <math>J</math> be a point at infinity on <math>\overline{BC}</math> called <math>Y</math> and let the intersection of lines <math>\overline{KG}</math> and <math>\overline{}HJ</math> be <math>Z</math>. Then, let the intersection of lines <math>\overline{GJ}</math> and <math>\overline{HK}</math> be <math>E'</math>. By Pascal's Theorem on <math>GGJHHK</math> and <math>GJJHKK</math>, we get that <math>X,E',</math> and <math>Z</math> are collinear and that <math>E',Y,</math> and <math>Z</math> are collinear, so <math>E',X,</math> and <math>Y</math> are collinear, meaning that <math>E</math> lies on <math>\overline{BC}</math> since both <math>X</math> and <math>Y</math> lie on <math>\overline{BC}</math>.
 
Since <math>IBI_AC</math> is cyclic, the pairwise radical axes of the circumcircles of <math>\triangle{}DII_A,\triangle{}ABC,</math> and <math>IBI_AC</math> concur. The pairwise radical axes of these circles are <math>\overline{GD},\overline{II_A},</math> and <math>\overline{BC}</math>, so <math>G,P,</math> and <math>D</math> are collinear. Similarly, since <math>BCI_BI_C</math> is cyclic, the pairwise radical axes of the cirucmcircles of <math>\triangle{}DI_BI_C,\triangle{}ABC,</math> and <math>BCI_BI_C</math> concur. The pairwise radical axes of these circles are <math>\overline{HD},\overline{I_BI_C},</math> and <math>\overline{BC}</math>, so <math>H,Q,</math> and <math>D</math> are collinear. This means that <math>-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)</math>, so the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>G</math> and <math>H</math> intersect on <math>\overline{BC}</math>. Let this intersection be <math>X</math>. Also, let the intersection of the tangents to the circumcircle of <math>\triangle{}ABC</math> at <math>K</math> and <math>J</math> be a point at infinity on <math>\overline{BC}</math> called <math>Y</math> and let the intersection of lines <math>\overline{KG}</math> and <math>\overline{}HJ</math> be <math>Z</math>. Then, let the intersection of lines <math>\overline{GJ}</math> and <math>\overline{HK}</math> be <math>E'</math>. By Pascal's Theorem on <math>GGJHHK</math> and <math>GJJHKK</math>, we get that <math>X,E',</math> and <math>Z</math> are collinear and that <math>E',Y,</math> and <math>Z</math> are collinear, so <math>E',X,</math> and <math>Y</math> are collinear, meaning that <math>E</math> lies on <math>\overline{BC}</math> since both <math>X</math> and <math>Y</math> lie on <math>\overline{BC}</math>.
  
−
[asy]
+
<asy>
 
size(500);
 
size(500);
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP;
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP;
Line 81: Line 81:
 
draw(C--GP);
 
draw(C--GP);
 
draw(circumcircle(A,E,J),dashed);
 
draw(circumcircle(A,E,J),dashed);
−
dot("<math>A</math>",A,dir(A-circumcenter(A,B,C)));
+
dot("$A$",A,dir(A-circumcenter(A,B,C)));
−
dot("<math>B</math>",B,dir(B-circumcenter(A,B,C)));
+
dot("$B$",B,dir(B-circumcenter(A,B,C)));
−
dot("<math>C</math>",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
+
dot("$C$",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
−
dot("<math>D</math>",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
+
dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
−
dot("<math>E'</math>",E,dir(dir(J-G)+dir(B-C)));
+
dot("$E'$",E,dir(dir(J-G)+dir(B-C)));
−
dot("<math>G</math>",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
+
dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
−
dot("<math>I</math>",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
+
dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
−
dot("<math>J</math>",J,dir(J-circumcenter(A,B,C)));
+
dot("$J$",J,dir(J-circumcenter(A,B,C)));
−
dot("<math>P</math>",P,dir(dir(A-I)+dir(C-B)));
+
dot("$P$",P,dir(dir(A-I)+dir(C-B)));
−
dot("<math>G'</math>",GP,dir(GP-B));
+
dot("$G'$",GP,dir(GP-B));
−
[/asy]
+
</asy>
  
 
Consider the transformation which is the composition of an inversion centered at <math>A</math> and a reflection over the angle bisector of <math>\angle{}CAB</math> that sends <math>B</math> to <math>C</math> and <math>C</math> to <math>B</math>. We claim that this sends <math>D</math> to <math>E'</math> and <math>E'</math> to <math>D</math>. It is sufficient to prove that if the transformation sends <math>G</math> to <math>G'</math>, then <math>AE'JG'</math> is cyclic. Notice that <math>\triangle{}AGB\sim\triangle{}ACG'</math> since <math>\angle{}GAB=\angle{}G'AC</math> and <math>\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AB}=\tfrac{AC}{AG}</math>. Therefore, we get that <math>\angle{}AG'E'=\angle{}ABG=\angle{}AJE'</math>, so <math>AE'JG'</math> is cyclic, proving the claim. This means that <math>\angle{}BAE'=\angle{}CAD</math>.
 
Consider the transformation which is the composition of an inversion centered at <math>A</math> and a reflection over the angle bisector of <math>\angle{}CAB</math> that sends <math>B</math> to <math>C</math> and <math>C</math> to <math>B</math>. We claim that this sends <math>D</math> to <math>E'</math> and <math>E'</math> to <math>D</math>. It is sufficient to prove that if the transformation sends <math>G</math> to <math>G'</math>, then <math>AE'JG'</math> is cyclic. Notice that <math>\triangle{}AGB\sim\triangle{}ACG'</math> since <math>\angle{}GAB=\angle{}G'AC</math> and <math>\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AB}=\tfrac{AC}{AG}</math>. Therefore, we get that <math>\angle{}AG'E'=\angle{}ABG=\angle{}AJE'</math>, so <math>AE'JG'</math> is cyclic, proving the claim. This means that <math>\angle{}BAE'=\angle{}CAD</math>.
  
−
[asy]
+
<asy>
 
size(500);
 
size(500);
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP,BP,CP,DP;
 
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP,BP,CP,DP;
Line 126: Line 126:
 
draw(circumcircle(B,IB,IC));
 
draw(circumcircle(B,IB,IC));
 
draw(circumcircle(E,IB,IC));
 
draw(circumcircle(E,IB,IC));
−
dot("<math>A</math>",A,2*dir(dir(IB-A)+dir(C-A)));
+
dot("$A$",A,2*dir(dir(IB-A)+dir(C-A)));
−
dot("<math>B</math>",B,dir(B-circumcenter(A,B,C)));
+
dot("$B$",B,dir(B-circumcenter(A,B,C)));
−
dot("<math>C</math>",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
+
dot("$C$",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B)));
−
dot("<math>D</math>",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
+
dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC))));
−
dot("<math>E'</math>",E,dir(B-C)*dir(90));
+
dot("$E'$",E,dir(B-C)*dir(90));
−
dot("<math>I_B</math>",IB,dir(dir(IB-IC)+dir(IB-IA)));
+
dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA)));
−
dot("<math>I_C</math>",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
+
dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
−
dot("<math>B'</math>",BP,dir(BP-circumcenter(B,IB,IC)));
+
dot("$B'$",BP,dir(BP-circumcenter(B,IB,IC)));
−
dot("<math>C'</math>",CP,dir(CP-circumcenter(B,IB,IC)));
+
dot("$C'$",CP,dir(CP-circumcenter(B,IB,IC)));
−
dot("<math>D'</math>",DP,dir(DP-E));
+
dot("$D'$",DP,dir(DP-E));
−
[/asy]
+
</asy>
  
 
We claim that <math>\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ</math>. Construct <math>D'</math> to be the intersection of line <math>\overline{AE}</math> and the circumcircle of <math>\triangle{}EI_BI_C</math> and let <math>B'</math> and <math>C'</math> be the intersections of lines <math>\overline{AC}</math> and <math>\overline{AB}</math> with the circumcircle of <math>\triangle{}BI_BI_C</math>. Since <math>B'</math> and <math>C'</math> are the reflections of <math>B</math> and <math>C</math> over <math>\overline{I_BI_C}</math>, it is sufficient to prove that <math>A,B',C',D'</math> are concyclic. Since <math>\overline{B'C},\overline{D'E'},</math> and <math>\overline{I_BI_C}</math> concur and <math>D',E',I_B,I_C</math> and <math>I_B,I_C,B',C</math> are concyclic, we have that <math>B',C,D',E'</math> are concyclic, so <math>\angle{}B'D'A=\angle{}ACE'=\angle{}AC'B'</math>, so <math>A,B',C',D'</math> are concyclic, proving the claim. We can similarly get that <math>\angle{}IE'I_A=\angle{}IDI_A</math>.
 
We claim that <math>\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ</math>. Construct <math>D'</math> to be the intersection of line <math>\overline{AE}</math> and the circumcircle of <math>\triangle{}EI_BI_C</math> and let <math>B'</math> and <math>C'</math> be the intersections of lines <math>\overline{AC}</math> and <math>\overline{AB}</math> with the circumcircle of <math>\triangle{}BI_BI_C</math>. Since <math>B'</math> and <math>C'</math> are the reflections of <math>B</math> and <math>C</math> over <math>\overline{I_BI_C}</math>, it is sufficient to prove that <math>A,B',C',D'</math> are concyclic. Since <math>\overline{B'C},\overline{D'E'},</math> and <math>\overline{I_BI_C}</math> concur and <math>D',E',I_B,I_C</math> and <math>I_B,I_C,B',C</math> are concyclic, we have that <math>B',C,D',E'</math> are concyclic, so <math>\angle{}B'D'A=\angle{}ACE'=\angle{}AC'B'</math>, so <math>A,B',C',D'</math> are concyclic, proving the claim. We can similarly get that <math>\angle{}IE'I_A=\angle{}IDI_A</math>.

Revision as of 17:05, 6 August 2023

Problem

Let ABC be a triangle with incenter $I$ and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively. Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$ that does not lie on any of the lines $IIa$, $I_bI_c$, or $BC$, suppose the circumcircles of $\triangle DIIa$ and $\triangle DI_bI_c$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\angle BAD = \angle EAC$.

Solution 1

[asy] size(500); pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q; B=(0,0); C=(8,0); A=intersectionpoint(Circle(B,6),Circle(C,9)); I=incenter(A,B,C); path c=circumcircle(A,B,C); J=intersectionpoint(I--(4*I-3*A),c); IA=2*J-I; IB=2*intersectionpoint(I--(4*I-3*B),c)-I; IC=2*intersectionpoint(I--(4*I-3*C),c)-I; K=intersectionpoint(IB--IC,c); D=intersectionpoint(I--(I+(10,-12)),c); path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC); F=intersectionpoints(c1,c2)[1]; E=extension(B,C,D,F); G=intersectionpoint(c1,c); H=intersectionpoint(c2,c); P=extension(A,I,B,C); Q=extension(IB,IC,B,C); draw(A--B--C--A); draw(c); draw(A--J); draw(circumcircle(D,I,IA)); draw(circumcircle(D,IB,IC)); draw(D--F); draw(B--Q--IB); draw(G--J,dashed); draw(H--K,dashed); dot("$A$",A,dir(A-circumcenter(A,B,C))); dot("$B$",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C))); dot("$C$",C,dir(C-circumcenter(A,B,C))*dir(15)); dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC)))); dot("$E$",E,dir(dir(H-K)+dir(B-C))); dot("$F$",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F))); dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G))); dot("$H$",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C)))); dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I))); dot("$J$",J,dir(J-circumcenter(A,B,C))); dot("$K$",K,dir(K-circumcenter(A,B,C))); dot("$I_A$",IA,dir(IA-circumcenter(D,I,IA))); dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA))); dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB))); dot("$P$",P,dir(dir(A-I)+dir(C-B))); dot("$Q$",Q,dir(dir(IC-IB)+dir(B-C))); [/asy]

Consider points $G,H,J,K,P,$ and $Q$ such that the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DII_A$ are $D$ and $G$, the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DI_BI_C$ are $D$ and $H$, the intersections of the circumcircle of $\triangle{}ABC$ with line $\overline{II_A}$ are $A$ and $J$, the intersections of the circumcircle of $\triangle{}ABC$ with line $\overline{I_BI_C}$ are $A$ and $K$, the intersection of lines $\overline{II_A}$ and $\overline{BC}$ is $P$, and the intersection of lines $\overline{I_BI_C}$ and $\overline{BC}$ is $Q$.

Since $IBI_AC$ is cyclic, the pairwise radical axes of the circumcircles of $\triangle{}DII_A,\triangle{}ABC,$ and $IBI_AC$ concur. The pairwise radical axes of these circles are $\overline{GD},\overline{II_A},$ and $\overline{BC}$, so $G,P,$ and $D$ are collinear. Similarly, since $BCI_BI_C$ is cyclic, the pairwise radical axes of the cirucmcircles of $\triangle{}DI_BI_C,\triangle{}ABC,$ and $BCI_BI_C$ concur. The pairwise radical axes of these circles are $\overline{HD},\overline{I_BI_C},$ and $\overline{BC}$, so $H,Q,$ and $D$ are collinear. This means that $-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)$, so the tangents to the circumcircle of $\triangle{}ABC$ at $G$ and $H$ intersect on $\overline{BC}$. Let this intersection be $X$. Also, let the intersection of the tangents to the circumcircle of $\triangle{}ABC$ at $K$ and $J$ be a point at infinity on $\overline{BC}$ called $Y$ and let the intersection of lines $\overline{KG}$ and $\overline{}HJ$ be $Z$. Then, let the intersection of lines $\overline{GJ}$ and $\overline{HK}$ be $E'$. By Pascal's Theorem on $GGJHHK$ and $GJJHKK$, we get that $X,E',$ and $Z$ are collinear and that $E',Y,$ and $Z$ are collinear, so $E',X,$ and $Y$ are collinear, meaning that $E$ lies on $\overline{BC}$ since both $X$ and $Y$ lie on $\overline{BC}$.

[asy] size(500); pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP; B=(0,0); C=(8,0); A=intersectionpoint(Circle(B,6),Circle(C,9)); I=incenter(A,B,C); path c=circumcircle(A,B,C); J=intersectionpoint(I--(4*I-3*A),c); IA=2*J-I; IB=2*intersectionpoint(I--(4*I-3*B),c)-I; IC=2*intersectionpoint(I--(4*I-3*C),c)-I; K=intersectionpoint(IB--IC,c); D=intersectionpoint(I--(I+(10,-12)),c); path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC); F=intersectionpoints(c1,c2)[1]; E=extension(B,C,D,F); G=intersectionpoint(c1,c); H=intersectionpoint(c2,c); P=extension(A,I,B,C); Q=extension(IB,IC,B,C); GP=extension(A,2*foot(G,A,I)-G,B,C); draw(A--B--C--A); draw(c); draw(A--J--G--D); draw(C--GP); draw(circumcircle(A,E,J),dashed); dot("$A$",A,dir(A-circumcenter(A,B,C))); dot("$B$",B,dir(B-circumcenter(A,B,C))); dot("$C$",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B))); dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC)))); dot("$E'$",E,dir(dir(J-G)+dir(B-C))); dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G))); dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I))); dot("$J$",J,dir(J-circumcenter(A,B,C))); dot("$P$",P,dir(dir(A-I)+dir(C-B))); dot("$G'$",GP,dir(GP-B)); [/asy]

Consider the transformation which is the composition of an inversion centered at $A$ and a reflection over the angle bisector of $\angle{}CAB$ that sends $B$ to $C$ and $C$ to $B$. We claim that this sends $D$ to $E'$ and $E'$ to $D$. It is sufficient to prove that if the transformation sends $G$ to $G'$, then $AE'JG'$ is cyclic. Notice that $\triangle{}AGB\sim\triangle{}ACG'$ since $\angle{}GAB=\angle{}G'AC$ and $\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AB}=\tfrac{AC}{AG}$. Therefore, we get that $\angle{}AG'E'=\angle{}ABG=\angle{}AJE'$, so $AE'JG'$ is cyclic, proving the claim. This means that $\angle{}BAE'=\angle{}CAD$.

[asy] size(500); pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,GP,BP,CP,DP; B=(0,0); C=(8,0); A=intersectionpoint(Circle(B,6),Circle(C,9)); I=incenter(A,B,C); path c=circumcircle(A,B,C); J=intersectionpoint(I--(4*I-3*A),c); IA=2*J-I; IB=2*intersectionpoint(I--(4*I-3*B),c)-I; IC=2*intersectionpoint(I--(4*I-3*C),c)-I; K=intersectionpoint(IB--IC,c); D=intersectionpoint(I--(I+(10,-12)),c); path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC); F=intersectionpoints(c1,c2)[1]; E=extension(B,C,D,F); G=intersectionpoint(c1,c); H=intersectionpoint(c2,c); P=extension(A,I,B,C); Q=extension(IB,IC,B,C); BP=2*foot(B,IB,IC)-B; CP=2*foot(C,IB,IC)-C; DP=2*foot(D,IB,IC)-D; draw(A--B--C--A); draw(E--DP); draw(BP--A--CP); draw(IB--IC); draw(c); draw(circumcircle(B,IB,IC)); draw(circumcircle(E,IB,IC)); dot("$A$",A,2*dir(dir(IB-A)+dir(C-A))); dot("$B$",B,dir(B-circumcenter(A,B,C))); dot("$C$",C,dir(dir(90)*dir(circumcenter(A,B,C)-C)+dir(C-B))); dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC)))); dot("$E'$",E,dir(B-C)*dir(90)); dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA))); dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB))); dot("$B'$",BP,dir(BP-circumcenter(B,IB,IC))); dot("$C'$",CP,dir(CP-circumcenter(B,IB,IC))); dot("$D'$",DP,dir(DP-E)); [/asy]

We claim that $\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ$. Construct $D'$ to be the intersection of line $\overline{AE}$ and the circumcircle of $\triangle{}EI_BI_C$ and let $B'$ and $C'$ be the intersections of lines $\overline{AC}$ and $\overline{AB}$ with the circumcircle of $\triangle{}BI_BI_C$. Since $B'$ and $C'$ are the reflections of $B$ and $C$ over $\overline{I_BI_C}$, it is sufficient to prove that $A,B',C',D'$ are concyclic. Since $\overline{B'C},\overline{D'E'},$ and $\overline{I_BI_C}$ concur and $D',E',I_B,I_C$ and $I_B,I_C,B',C$ are concyclic, we have that $B',C,D',E'$ are concyclic, so $\angle{}B'D'A=\angle{}ACE'=\angle{}AC'B'$, so $A,B',C',D'$ are concyclic, proving the claim. We can similarly get that $\angle{}IE'I_A=\angle{}IDI_A$.

[asy] size(500); pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q,JP,KP; B=(0,0); C=(8,0); A=intersectionpoint(Circle(B,6),Circle(C,9)); I=incenter(A,B,C); path c=circumcircle(A,B,C); J=intersectionpoint(I--(4*I-3*A),c); IA=2*J-I; IB=2*intersectionpoint(I--(4*I-3*B),c)-I; IC=2*intersectionpoint(I--(4*I-3*C),c)-I; K=intersectionpoint(IB--IC,c); D=intersectionpoint(I--(I+(10,-12)),c); path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC); F=intersectionpoints(c1,c2)[1]; E=extension(B,C,D,F); G=intersectionpoint(c1,c); H=intersectionpoint(c2,c); P=extension(A,I,B,C); Q=extension(IB,IC,B,C); JP=2*J-E; KP=2*K-E; draw(A--B--C--A); draw(c); draw(A--J); draw(circumcircle(D,I,IA)); draw(circumcircle(D,IB,IC)); draw(D--F,dashed); draw(B--Q--IB); draw(G--JP); draw(H--KP); dot("$A$",A,dir(A-circumcenter(A,B,C))); dot("$B$",B,1/2*dir(B-dir(circumcenter(A,B,C))*dir(90)+dir(B-C))); dot("$C$",C,dir(C-circumcenter(A,B,C))*dir(15)); dot("$D$",D,dir(dir(90)*dir(circumcenter(D,I,IA)-D)+dir(90)*dir(D-circumcenter(D,IB,IC)))); dot("$E'$",E,dir(dir(H-K)+dir(B-C))); dot("$F$",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F))); dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G))); dot("$H$",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C)))); dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I))); dot("$J$",J,dir(dir(circumcenter(A,B,C)-J)*dir(90)+dir(J-G))); dot("$K$",K,dir(dir(K-circumcenter(A,B,C))*dir(90)+dir(K-H))); dot("$I_A$",IA,dir(IA-circumcenter(D,I,IA))); dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA))); dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB))); dot("$P$",P,dir(dir(A-I)+dir(C-B))); dot("$Q$",Q,dir(dir(IC-IB)+dir(B-C))); dot("$J'$",JP,dir(JP-circumcenter(D,I,IA))); dot("$K'$",KP,dir(KP-circumcenter(D,IB,IC))); [/asy]

Let line $\overline{E'J}$ intersect the circumcircle of $\triangle{}DII_A$ at $G$ and $J'$. Notice that $J$ is the midpoint of $\overline{II_A}$ and $\angle{}IE'I_A=\angle{}IDI_A=\angle{}IJ'I_A$, so $IE'I_AJ'$ is a parallelogram with center $J$, so $\tfrac{}{EJ}{EJ'}=\tfrac{1}{2}$. Similarly, we get that if line $\overline{E'K}$ intersects the circumcircle of $\triangle{}DI_BI_C$ at $H$ and $K'$, we have that $\tfrac{EK}{EK'}=\tfrac{1}{2}$, so $\overline{KJ}\parallel\overline{K'J'}$, so $\angle{}HGJ'=\angle{}HGJ=\angle{}HKJ=\angle{}HK'J'$, so $G,H,J',K'$ are concyclic. Then, the pairwise radical axes of the circumcircles of $\triangle{}DII_A,\triangle{}DI_BI_C,$ and $GHJ'K'$ are $\overline{DF},\overline{HK'},$ and $\overline{GJ'}$, so $\overline{DF},\overline{HK'},$ and $\overline{GJ'}$ concur, so $\overline{DF},\overline{HK},$ and $\overline{GJ}$ concur, so $E=E'$. We are then done since $\angle{}BAE'=\angle{}CAD$.

~Zhaom

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