Problem

Source: 1974 USAMO Problem 5

Tags: rotation, geometry, geometric transformation



Consider the two triangles $ ABC$ and $ PQR$ shown below. In triangle $ ABC, \angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $ x=u+v+w$. [asy][asy]unitsize(7mm); defaultpen(linewidth(.7pt)+fontsize(10pt)); pair C=(0,0), B=4*dir(5); pair A=intersectionpoints(Circle(C,5), Circle(B,6))[0]; pair Oc=scale(sqrt(3)/3)*rotate(30)*(B-A)+A; pair Ob=scale(sqrt(3)/3)*rotate(30)*(A-C)+C; pair D=intersectionpoints(Circle(Ob,length(Ob-C)), Circle(Oc,length(Oc-B)))[1]; real s=length(A-D)+length(B-D)+length(C-D); pair P=(6,0), Q=P+(s,0), R=rotate(60)*(s,0)+P; pair M=intersectionpoints(Circle(P,length(B-C)), Circle(Q,length(A-C)))[0]; draw(A--B--C--A--D--B); draw(D--C); label("$B$",B,SE); label("$C$",C,SW); label("$A$",A,N); label("$D$",D,NE); label("$a$",midpoint(B--C),S); label("$b$",midpoint(A--C),WNW); label("$c$",midpoint(A--B),NE); label("$u$",midpoint(A--D),E); label("$v$",midpoint(B--D),N); label("$w$",midpoint(C--D),NNW); draw(P--Q--R--P--M--Q); draw(M--R); label("$P$",P,SW); label("$Q$",Q,SE); label("$R$",R,N); label("$M$",M,NW); label("$x$",midpoint(P--R),NW); label("$x$",midpoint(P--Q),S); label("$x$",midpoint(Q--R),NE); label("$c$",midpoint(R--M),ESE); label("$a$",midpoint(P--M),NW); label("$b$",midpoint(Q--M),NE);[/asy][/asy]