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]
Problem
Source: 1974 USAMO Problem 5
Tags: rotation, geometry, geometric transformation
13.03.2010 08:27
This is actually true due to the Fermat Point (Toricelli point ) iof triangle $ ABC$ and by Pompeiu's theorem 's construction in $ PQR$, whereby you rotate with respect to $ Q$ by $ 60$ degrees.
15.06.2014 19:27
13.04.2023 18:42
Are there any shorter solutions?
13.04.2023 19:29
huashiliao2020 wrote: Are there any shorter solutions? I believe algebra bashing is the only viable option. Although it could be made shorter by the usage of a nice formula. By the sine area formula, we have that $[ABC]=[ADC]+[BDC]+[BDA]=\frac{1}{2} \sin{120^{\circ}}(uv + vw + wu) = \frac{(uv + vw + wu) \sqrt{3}}{4}$. Now by LoC, we have that $a^2=v^2+vw+w^2,b^2=w^2+wu+u^2,c^2=u^2+uv+v^2$. Now we apply the rarely useful formula ( I probably wouldn't do this in contest ) but here goes: In general, if you have an equilateral triangle of side $x$ and the distances of a point inside to the three vertices are $a,b,c$ , then $x^2=\frac{1}{2}(a^2+b^2+c^2)+2\sqrt{3} \Delta$ where $\Delta$ is the area of a triangle with side $a,b,c$. More information can be found here. This gives $x^2=\frac{1}{2}(v^2+vw+w^2+w^2+wu+u^2+u^2+uv+v^2)+2\sqrt{3} \frac{(uv + vw + wu) \sqrt{3}}{4}$, which simplifies to $x^2=\frac{1}{2}(2u^2+2v^2+2w^2+uv+vw+wu)+\frac{3(uv + vw + wu)}{2}$. Now we split the large fraction into $x^2=\frac{1}{2}(uv+vw+wu)+u^2+v^2+w^2+\frac{3(uv + vw + wu)}{2}$, which then simplifies into $x^2=u^2+v^2+w^2+2(uv+vw+wu)$. However, we notice that the right hand side is simply $(u+v+w)^2$, so thus we have $\boxed{x=u+v+w}$. .