Problem:
Let ABC be an acute, scalene triangle, and let M,N, and P be the midpoints of BC,CA, and AB, respectively. Let the perpendicular bisectors of AB and AC intersect ray AM in points D and E respectively, and let lines BD and CE intersect in point F, inside of triangle ABC. Prove that points A,N,F, and P all lie on one circle.
Solution:
First solution: Let O be the circumcenter of triangle ABC. We prove that
∠APO=∠ANO=∠AFO=90∘(1)
It will then follow that A,P,O,F,N lie on the circle with diameter AO. Indeed, the fact that the first two angles in (1) are right is immediate because OP and ON are the perpendicular bisectors of AB and AC, respectively. Thus we need only prove that ∠AFO=90∘.
We may assume, without loss of generality, that AB>AC. This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because PO is the perpendicular bisector of AB, it follows that triangle ADB is an isosceles triangle with AD=BD. Likewise, triangle AEC is isosceles with AE=CE. Let x=∠ABD=∠BAD and y=∠CAE=∠ACE, so x+y=∠BAC.
Applying the Law of Sines to triangles ABM and ACM gives
sinxBM=sin∠BMAAB and sinyCM=sin∠CMAAC
Taking the quotient of the two equations and noting that sin∠BMA=sin∠CMA we find
CMBMsinxsiny=ACABsin∠BMAsin∠CMA=ACAB
Because BM=MC, we have
sinysinx=ABAC(2)
Applying the law of sines to triangles ABF and ACF we find
sinxAF=sin∠AFBAB and sinyAF=sin∠AFCAC
Taking the quotient of the two equations yields
sinysinx=ABACsin∠AFCsin∠AFB, so by (2), sin∠AFB=sin∠AFC(3)
Because ∠ADF is an exterior angle to triangle ADB, we have ∠EDF=2x. Similarly, ∠DEF=2y. Hence
∠EFD=180∘−2x−2y=180∘−2∠BAC
Thus ∠BFC=2∠BAC=∠BOC, so BOFC is cyclic. In addition,
∠AFB+∠AFC=360∘−2∠BAC>180∘
and hence, from (3), ∠AFB=∠AFC=180∘−∠BAC. Because BOFC is cyclic and △BOC is isosceles with vertex angle ∠BOC=2∠BAC, we have ∠OFB=∠OCB= 90∘−∠BAC. Therefore,
∠AFO=∠AFB−∠OFB=(180∘−∠BAC)−(90∘−∠BAC)=90∘
This completes the proof.
Second solution: Invert the figure about a circle centered at A, and let X′ denote the image of the point X under this inversion. Find point F1′ so that AB′F1′C′ is a parallelogram and let Z′ denote the center of this parallelogram. Note that △BAC∼ △C′AB′ and △BAD∼△D′AB′. Because M is the midpoint of BC and Z′ is the midpoint of B′C′, we also have △BAM∼△C′AZ′. Thus
∠AF1′B′=∠F1′AC′=∠Z′AC′=∠MAB=∠DAB=∠DBA=∠AD′B′
Hence quadrilateral AB′D′F1′ is cyclic and, by a similar argument, quadrilateral AC′E′F1′ is also cyclic. Because the images under the inversion of lines BDF and CFE are circles that intersect in A and F′, it follows that F1′=F′.
Next note that B′,Z′, and C′ are collinear and are the images of P′,F′, and N′, respectively, under a homothety centered at A and with ratio 1/2. It follows that P′,F′ and N′ are collinear, and then that the points A,P,F and N lie on a circle.
This problem was suggested by Zuming Feng. The second solution was contributed by Gabriel Carroll.
The problems on this page are the property of the MAA's American Mathematics Competitions
