Problem:
(*) Let ABC be an equilateral triangle and let P be a point on its circumcircle. Let lines PA and BC intersect at D; let lines PB and CA intersect at E; and let lines PC and AB intersect at F. Prove that the area of triangle DEF is twice the area of triangle ABC.
(c) 2017, Mathematical Association of America.
Solution:
We offer several solutions. Throughout, we use bracket notation for areas: for example, [ABC] means the area of triangle ABC.
We first present three down-to-earth approaches. One of them is a coordinate geometry approach. The other two approaches utilize the fact of many pairs of similar triangles in this configuration:
BPC,FPA,FBC,APE, and BCE;
FBP and FCA;
ECP and EBA.
In these solutions, we assume the points are configured so that P is on minor arc BC of the circle, as shown in the figure.
Solution 1: (By USA(J)MO packet reviewers.) We may assume that AB=1. Then [ABC]=3/4. Set b=PB,c=PC,e=PE, and f=PF. Note that ∠FBD=∠ECD=∠BPC=120∘. Hence
Because ∠FBC=∠BPC and ∠FCB=∠PCB, triangles FCB and BCP are similar to each other, implying that
BCFC=CPCB=PBBF or 1c+f=c1=bBF
Thus, c+f=1/c and BF=b/c. Analogously, b+e=1/b and CE=c/b. It remains to show that
2=(b+e)(c+f)−BF⋅BD−CE⋅CD=bc1−cb⋅BD−bc⋅CD
Note that ∠BPD=∠CPD=60∘, so we have BD/CD=BP/CP by the Angle-Bisector theorem. Consequently, we have BD=b/(b+c) and CD=c/(b+c). Thus, we want to show that
2=bc1−cb⋅BD−bc⋅CD=bc1−c(b+c)b2−b(b+c)c2
=bc1−bc(b+c)b3+c3=bc1−b2−c2+bc
or b2+c2+bc=1, which is true by applying the Law of Cosines in triangle BPC.
Solution 2. (By USA(J)MO packet reviewers.) Note that ∠DPF=∠DPE=∠EPF=120∘. We have
[DEF]=21⋅sin120∘(PD⋅PE+PE⋅PF+PF⋅PD)
To show that [DEF]=2[ABC], it suffices to show that
PD⋅PE+PE⋅PF+PF⋅PD=2BC2
Set b=PB and c=PC. We will express the lengths of BC,PD,PE, and PF in terms of b and c. Note that ∠BPC=120∘. Applying the Law of Cosines in triangle BPC gives BC2=b2+bc+c2. Applying Ptolemy's theorem to cyclic quadrilateral ABCP yields AP⋅BC=BP⋅AC+CP⋅AB or AP=b+c. Because ∠ACB=∠ABC=∠APC=60∘, triangles ACD and APC are similar, and so
APAC=PCCD=CADA
or b2+bc+c2=AC2=AP⋅AD=(b+c)⋅AD. We conclude that
AD=b+cb2+bc+c2 and PD=AP−AD=b+c−b+cb2+bc+c2=b+cbc
Finally, because ∠FBP=180∘−∠ABP=∠ACP and ∠BPF=∠APC=60∘, triangles FBP and ACP are similar. Hence
ACFB=CPBP=PAPF
from which it follows that PF=AP⋅BP/CP=b(b+c)/c. In exactly the same way, we get PE=c(b+c)/b. It follows that
Solution 3. (By USA(J)MO packet reviewers.) Without loss of generality, we may assume that A=(0,2),B=(−3,−1), and C=(3,−1). Set P=(a,b) with a2+b2=4.
Solving for line equations y=−1 and y=a(b−2)⋅x+2 gives D=(−b−23a,−1).
Solving for line equations y=3x+2 and y=a−3(b+1)⋅(x−3)−1 gives
F=(b+4−3a3a+3b−23,b+4−3a3a+5b+2)
Solving for line equations y=−3x+2 and y=a+3(b+1)⋅(x+3)−1 gives
The next solution is by the problem authors. It uses more advanced tools that USAJMO participants are not expected to know, but offers some additional insight into the origins of the problem.
Solution 4: (By the posers.) Without loss of generality, let us assume that P lies on the arc AC, which does not contain vertex B. Because P is on the circumcircle, its isogonal conjugate, say Q, is a point at infinity. Furthermore, the intersections D′,E′,F′ of lines QA,QB,QC with lines BC,CA,AB, respectively, are the reflections of D,E,F across the midpoints of BC,CA,AB. This essentially follows from the fact that △ABC is equilateral: isogonal conjugates with respect to it are also isotomic conjugates. We are thus led to the following lemma.
Lemma 1. Let ABC be a triangle with D,E,F points lying on the lines BC,CA,AB, respectively. Let D′,E′,F′ be the reflections of D,E,F with respect to the midpoints of BC,CA,AB, respectively. Then, triangles DEF and D′E′F′ have the same area.
Proof. The statement holds regardless of the position of points D,E,F on lines BC,CA,AB, so, for convenience, in the computations below we shall assume that these all lie close enough to the midpoints of the sides so that all points D,E,F,D′,E′,F′ lie on the sides of △ABC. The proof for the other scenarios is similar.
We begin by writing
[CD′E′]=[AD′E]=[AD′C]−[CD′E]
Analogously, [AE′F′]=[BE′A]−[AE′F] and [BF′D′]=[CF′B]−[BF′D]. Adding these three together, we get [CD′E′]+[AE′F′]+[BF′D′]=[AD′C]+[BE′A]+[CF′B]−[CD′E]−[AE′F]−[BF′D]
Furthermore,
[CDE]=[BD′E]=[BEC]−[CD′E]
and similarly [AEF]=[CFA]−[AE′F] and [BFD]=[ADB]−[BF′D]. Therefore,
But D′C=DB,E′A=EC,F′B=FA, so [AD′C]=[ADB],[BE′A]=[BEC],[CF′B]=[CFA]. Using all of the above, we get
[CD′E′]+[AE′F′]+[BF′D′]=[CDE]+[AEF]+[BFD]
and so [ABC]−[D′E′F′]=[ABC]−[DEF], i.e., [DEF]=[D′E′F′], establishing the lemma.
Assuming Lemma 1, we just have to check that [D′E′F′]=2[ABC]. Because P lies on the small arc AC, points D and F lie on the extensions of segments BC and AB, respectively, and so D′ and F′ do too. Furthermore, B lies in the interior of triangle D′E′F′, therefore
[D′E′F′]=[D′BF′]+[F′BE′]+[E′BD′]
On the other hand, AD′∥CF′ implies [D′CF′]=[ACF′], which, after subtracting [BCF′] from both sides, gives [D′BF′]=[ABC]. Likewise, BE′∥CF′ gives [F′BE′]=[CBE′] and AD′∥BE′ gives [E′BD′]=[E′BA]. Hence, it follows that
[D′E′F′]=[ABC]+[CBE′]+[E′BA]=2[ABC]
as claimed.
Note: One can also establish the lemma using barycentric coordinates. Suppose points D,E,F are dividing the sides BC,CA,AB in the ratios
BD:DC=x:1−x,CE:EA=y:1−y,AF:FB=z:1−z
In terms of barycentric coordinates with respect to triangle ABC, we have
D=(1−x)B+xC,E=(1−y)C+yA,F=(1−z)A+zB
Consequently, by definition, points D′,E′,F′ satisfy
D′=xB+(1−x)C,E′=yC+(1−y)A,F′=zA+(1−z)B
Now, without loss of generality, rescale so that [ABC]=1. It can then be easily checked that
