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| − | <asy>
| + | == Problem == |
| − | size(500);
| + | 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>. |
| − | pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q;
| + | == Solution == |
| − | B=(0,0);
| + | {{Solution}} |
| − | 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 <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>.
| + | {{MAA Notice}} |
| − | | |
| − | 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>
| |
| − | 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 <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>
| |
| − | 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));
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| − | 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 <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>.
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| − | | |
| − | <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);
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| − | A=intersectionpoint(Circle(B,6),Circle(C,9));
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| − | I=incenter(A,B,C);
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| − | path c=circumcircle(A,B,C);
| |
| − | J=intersectionpoint(I--(4*I-3*A),c);
| |
| − | IA=2*J-I;
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| − | 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);
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| − | D=intersectionpoint(I--(I+(10,-12)),c);
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| − | path c1=circumcircle(D,I,IA),c2=circumcircle(D,IB,IC);
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| − | F=intersectionpoints(c1,c2)[1];
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| − | E=extension(B,C,D,F);
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| − | G=intersectionpoint(c1,c);
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| − | H=intersectionpoint(c2,c);
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| − | P=extension(A,I,B,C);
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| − | Q=extension(IB,IC,B,C);
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| − | JP=2*J-E;
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| − | KP=2*K-E;
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| − | draw(A--B--C--A);
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| − | draw(c);
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| − | draw(A--J);
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| − | draw(circumcircle(D,I,IA));
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| − | draw(circumcircle(D,IB,IC));
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| − | draw(D--F,dashed);
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| − | draw(B--Q--IB);
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| − | draw(G--JP);
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| − | draw(H--KP);
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| − | 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)));
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| − | 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))));
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| − | dot("$E'$",E,dir(dir(H-K)+dir(B-C)));
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| − | dot("$F$",F,dir(dir(90)*dir(F-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(D,IB,IC)-F)));
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| − | dot("$G$",G,dir(dir(90)*dir(G-circumcenter(D,I,IA))+dir(90)*dir(circumcenter(A,B,C)-G)));
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| − | dot("$H$",H,dir(dir(90)*dir(circumcenter(D,IB,IC)-H)+dir(90)*dir(H-circumcenter(A,B,C))));
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| − | dot("$I$",I,dir(dir(90)*dir(circumcenter(D,I,IA)-I)+dir(A-I)));
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| − | dot("$J$",J,dir(dir(circumcenter(A,B,C)-J)*dir(90)+dir(J-G)));
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| − | dot("$K$",K,dir(dir(K-circumcenter(A,B,C))*dir(90)+dir(K-H)));
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| − | dot("$I_A$",IA,dir(IA-circumcenter(D,I,IA)));
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| − | dot("$I_B$",IB,dir(dir(IB-IC)+dir(IB-IA)));
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| − | dot("$I_C$",IC,dir(dir(90)*dir(circumcenter(D,IB,IC)-IC)+dir(IC-IB)));
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| − | dot("$P$",P,dir(dir(A-I)+dir(C-B)));
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| − | dot("$Q$",Q,dir(dir(IC-IB)+dir(B-C)));
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| − | dot("$J'$",JP,dir(JP-circumcenter(D,I,IA)));
| |
| − | dot("$K'$",KP,dir(KP-circumcenter(D,IB,IC)));
| |
| − | </asy>
| |
| − | | |
| − | Let line <math>\overline{E'J}</math> intersect the circumcircle of <math>\triangle{}DII_A</math> at <math>G</math> and <math>J'</math>. Notice that <math>J</math> is the midpoint of <math>\overline{II_A}</math> and <math>\angle{}IE'I_A=\angle{}IDI_A=\angle{}IJ'I_A</math>, so <math>IE'I_AJ'</math> is a parallelogram with center <math>J</math>, so <math>\tfrac{}{EJ}{EJ'}=\tfrac{1}{2}</math>. Similarly, we get that if line <math>\overline{E'K}</math> intersects the circumcircle of <math>\triangle{}DI_BI_C</math> at <math>H</math> and <math>K'</math>, we have that <math>\tfrac{EK}{EK'}=\tfrac{1}{2}</math>, so <math>\overline{KJ}\parallel\overline{K'J'}</math>, so <math>\angle{}HGJ'=\angle{}HGJ=\angle{}HKJ=\angle{}HK'J'</math>, so <math>G,H,J',K'</math> are concyclic. Then, the pairwise radical axes of the circumcircles of <math>\triangle{}DII_A,\triangle{}DI_BI_C,</math> and <math>GHJ'K'</math> are <math>\overline{DF},\overline{HK'},</math> and <math>\overline{GJ'}</math>, so <math>\overline{DF},\overline{HK'},</math> and <math>\overline{GJ'}</math> concur, so <math>\overline{DF},\overline{HK},</math> and <math>\overline{GJ}</math> concur, so <math>E=E'</math>. We are then done since <math>\angle{}BAE'=\angle{}CAD</math>.
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| − | | |
| − | ~Zhaom
| |
and excenters 
, 
, 
opposite 
, 
, and 
, respectively. Given an arbitrary point 
on the circumcircle of 
that does not lie on any of the lines 
, 
, or 
, suppose the circumcircles of 
and 
intersect at two distinct points 
. If 
is the intersection of lines 
and 
.
