Problem:
Let ABCD be a quadrilateral, and let E and F be points on sides AD and BC, respectively, such that AE/ED=BF/FC. Ray FE meets rays BA and CD at S and T, respectively. Prove that the circumcircles of triangles SAE,SBF,TCF, and TDE pass through a common point.
Solution:
First Solution. Let P be the second intersection of the circumcircles of triangles TCF and TDE. Because the quadrilateral PEDT is cyclic, ∠PET=∠PDT, or
∠PEF=∠PDC(*)
Because the quadrilateral PFCT is cyclic,
∠PFE=∠PFT=∠PCT=∠PCD(**)
By equations (∗) and (∗∗), it follows that triangle PEF is similar to triangle PDC. Hence ∠FPE=∠CPD and PF/PE=PC/PD. Note also that ∠FPC=∠FPE+∠EPC= ∠CPD+∠EPC=∠EPD. Thus, triangle EPD is similar to triangle FPC. Another way to say this is that there is a spiral similarity centered at P that sends triangle PFE to triangle PCD, which implies that there is also a spiral similarity, centered at P, that sends triangle PFC to triangle PED, and vice versa. In terms of complex numbers, this amounts to saying that
E−PD−P=F−PC−P⟹F−PE−P=C−PD−P
Because AE/ED=BF/FC, points A and B are obtained by extending corresponding segments of two similar triangles PED and PFC, namely, DE and CF, by the identical proportion. We conclude that triangle PDA is similar to triangle PCB, implying that triangle PAE is similar to triangle PBF. Therefore, as shown before, we can establish the similarity between triangles PBA and PFE, implying that
∠PBS=∠PBA=∠PFE=∠PFS and ∠PAB=∠PEF
The first equation above shows that PBFS is cyclic. The second equation shows that ∠PAS=180∘−∠BAP=180∘−∠FEP=∠PES; that is, PAES is cyclic. We conclude that the circumcircles of triangles SAE,SBF,TCF, and TDE pass through point P.
Note. There are two spiral similarities that send segment EF to segment CD. One of them sends E and F to D and C, respectively; the point P is the center of this spiral similarity. The other sends E and F to C and D, respectively; the center of this spiral similarity is the second intersection (other than T ) of the circumcircles of triangles TFD and TEC.
Second Solution. We will give a solution using complex coordinates. The first step is the following lemma.
Lemma. Suppose s and t are real numbers and x,y and z are complex. The circle in the complex plane passing through x,x+ty and x+(s+t)z also passes through the point x+syz/(y−z), independent of t.
Proof. Four points z1,z2,z3 and z4 in the complex plane lie on a circle if and only if the\
cross-ratio
cr(z1,z2,z3,z4)=(z1−z4)(z2−z3)(z1−z3)(z2−z4)
is real. Since we compute
cr(x,x+ty,x+(s+t)z,x+syz/(y−z))=ss+t
the given points are on a circle.
Lay down complex coordinates with S=0 and E and F on the positive real axis. Then there are real r1,r2 and R with B=r1A,F=r2E and D=E+R(A−E) and hence AE/ED=BF/FC gives
C=F+R(B−F)=r2(1−R)E+r1RA
The line CD consists of all points of the form sC+(1−s)D for real s. Since T lies on this line and has zero imaginary part, we see from Im(sC+(1−s)D)=(sr1R+(1−s)R)Im(A) that it corresponds to s=−1/(r1−1). Thus
T=r1−1r1D−C=r1−1(r2−r1)(R−1)E
Apply the lemma with x=E,y=A−E,z=(r2−r1)E/(r1−1), and s=(r2−1)(r1−r2). Setting t=1 gives
(x,x+y,x+(s+1)z)=(E,A,S=0)
and setting t=R gives
(x,x+Ry,x+(s+R)z)=(E,D,T)
Therefore the circumcircles to SAE and TDE meet at
x+y−zsyz=(1−r1)E−(1−r2)AAE(r1−r2)=A+F−B−EAF−BE
This last expression is invariant under simultaneously interchanging A and B and interchanging E and F. Therefore it is also the intersection of the circumcircles of SBF and TCF.
This problem was proposed by Zuming Feng and Zhonghao Ye.
The problems on this page are the property of the MAA's American Mathematics Competitions
