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2023 USAMO Problems/Problem 6

Revision as of 22:36, 10 March 2024 by Mathisfun286 (talk | contribs)

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 $II_{a}$, $I_{b}I_{c}$, 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 $\angle BAD = \angle EAC$.

Video Solution by mop 2024

https://youtube.com/watch?v=LAuyU2OuVzE

~r00tsOfUnity

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 $\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AC}=\tfrac{AB}{AG}$. 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 $\triangle{}E'I_BI_C$ 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

Solution 2

Set $\triangle ABC$ as the reference triangle, and let $D=(x_0,y_0,z_0)$ with homogenized coordinates. To find the equation of circle $DII_A$, we note that $I=(a:b:c)$ (not homogenized) and $I_A=(-a:b:c)$. Thus, for this circle with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_1x+v_1y+w_1z)$, we compute that \[u_1=bc,\] \[v_1=\frac{bc^2x_0}{bz_0-cy_0},\] \[w_1=\frac{b^2cx_0}{cy_0-bz_0}.\] For circle $DI_BI_C$ with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_2x+v_2y+w_2z)$,, we find that \[u_2=-bc,\] \[v_2=\frac{bc^2x_0}{bz_0+cy_0},\] \[w_2=\frac{b^2cx_0}{cy_0+bz_0}.\] The equation of the radical axis is $ux+vy+wz=0$ with $u=u_1-u_2$, $v=v_1-v_2$, and $w=w_1-w_2$. We want to consider the intersection of this line with line $\overline{BC}$, so set $x=0$. The equation reduces to $vy+wz=0$. We see that $v=\frac{2bc^3x_0y_0}{b^2z_0^2-c^2y_0^2}$ and $w=\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}$, so \[\frac{v}{w}=\frac{b^2z_0}{c^2y_0},\] which is the required condition for $\overline{AD}$ and $\overline{AE}$ to be isogonal.

See Also

2023 USAMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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