A Geometry Problem from IMO 2016

(1)

Azim Premji University At Right Angles, July 2022 79

Student Corner

I

n this article, we solve a geometry problem from the International Mathematics Olympiad (IMO) 2016 (Hong Kong). It was proposed by Art Waeterschoot from Belgium, who received an honourable mention during the IMO 2015 (Thailand). The problem was given to the problem-solving group of our school.

Figure 1

1

A Geometry Problem from IMO 2016

SHREYAS ADIGA

Keywords: Triangle, parallelogram, concurrent, radical axis, concyclic, IMO, INMO, USAMO

(2)

80 Azim Premji University At Right Angles, July 2022

Problem 1.TriangleBCFhas a right angle atB. LetAbe the point on lineCFsuch thatFA=FBandF lies betweenAandC. PointDis chosen such thatDA=DCandACis the bisector of∡DAB. PointEis chosen such thatEA=EDandADis the bisector of∡EAC.

LetMbe the midpoint ofCF. LetXbe the point such thatAMXEis a parallelogram (whereAM∥EXand AE∥MX).

Prove that linesBD,FXandMEare concurrent.

Solution

We find three circles such thatBD,FX,MEare the respective radical axes associated with pairs of these circles. Then we apply the radical axis theorem.

From the data of the problem, we can denote∡CAB=∡FAD= ∡DAE= ∡EDA= θ.

The main claims in the proof are the following:

• PointsD,F,B,C,Xare concyclic.

• PointsB,A,E,D,Mare concyclic.

• PointsE,F,M,Xare concyclic.

• Claim 1: PointsE,D,Xare collinear.

Proof: SinceADis the angle bisector of∡EAC, we have∡CAD= ∡EAD.

SinceED=AE,△AEDis isosceles and hence∡EAD=∡EDA. Thus∡CAD=∡EDA.

This implies thatED ∥AC.

Now, asAMXEis a parallelogram,EX∥AC. It follows thatE,D,Xare collinear (Figure 1).

• Claim 2: PointsD,F,B,Care concyclic.

Proof: From the observation that△ABFis similar to△ACD, we have, AB

AC = AF AD.

Next, observe that∡BAF=∡FAD. It therefore follows that△ABCis similar to△AFD.

We further note that∡AFD=∡ABC=90^◦+θ. Since∡DCF=θand∡AFDis the exterior angle of△CFD, we conclude that∡FDC=90^◦and henceB,C,D,Fare concyclic withCFas diameter. Denote this circle byΓ1(Figure 2).

• Claim 3: B,E,Fare collinear.

Proof: Observe that△CDAis isosceles, with∡DCA=∡DAC=θ.

From the previous claim, we know thatB,C,D,Fare concyclic. Hence∡FBD=∡FCD=θ(as these are angles on the same segmentFD).

(3)

Azim Premji University At Right Angles, July 2022 81 Figure 2

From the observation that△ABFis similar to△ADEwe have:

AB AD = AF

AE.

Since∡BAD=∡FAE, it follows that△ABDis similar to△AFE.

Now∡AFE=∡ABD= θ+∡FBD=θ+∡FCD=θ+θ=2θ.

But in△ABF,∡AFE=2θ=180−∡BFA. This proves the claim (Figure 1).

• Claim 4: A,B,M,Dare concyclic.

Proof: SinceB,C,D,Fare concyclic (from Claim 2) withMas centre of the circumcircle of the

△FBC,∡AMD=∡FMD=2∡ACD=2θ.

Again,B,C,D,Fbeing concyclic implies that∡EBD=θ, and hence ABD=∡ABE+∡EBD=θ+θ=2θ.

We conclude thatA,B,M,Dare concyclic. Denote this circle byΓ2(Figure 2).

• Claim 5: E lies onΓ2.

Proof: From Claim 2, we know thatB,C,D,Fare concyclic, so∡FBD=∡FCD= θ.

SinceE,F,Bare collinear,∡EBD= ∡FBD=θ.

But∡EAD= θ. ThusE,A,B,Dare concyclic.

Now, from Claim 4,A,B,M,Dare concyclic and henceElies onΓ2.

• Claim 6: Xlies onΓ1.

Proof: From the observation thatMDEAis cyclic, it follows that∡MDX=∡EAM.SinceAMXE is a parallelogram,∡EAM= ∡DXM.

Therefore,△MDXis isosceles and soMD= MX, thus proving the claim (Figure 2).

(4)

82 Azim Premji University At Right Angles, July 2022

• Claim 7: M,F,E,Xare concyclic.

Proof: We observe that∡AFE=2θ(this is the exterior angle for the triangle AFB). SoEF=EA.

SinceAEXMis a parallelogram,EA=MX. ThereforeEF=MXand soMFEXis an isosceles trapezium. It follows thatM,F,E,Xare concyclic (an isosceles trapezium is always cyclic). Let us denote circleMFEXbyΓ3.

Now observe thatBD,FX,MEare the radical axes of the circle pairs(Γ1,Γ2),(Γ3,Γ1),(Γ2,Γ3). By applying the radical axis theorem, we get the desired result. (The Radical Axis Theorem states:

The three pairwise radical axes of three circles concur at a point. The point where the lines meet is called the ‘radical centre’ of the three circles.)

Two problems for the reader

Problem 2(USAMO 1997).LetABCbe a triangle. Take pointsD,E,Fon the perpendicular bisectors of BC,CA,ABrespectively. Show that the lines throughA,B,Cperpendicular toEF,FD,DErespectively are concurrent.

(Here, USAMO refers to the Mathematics Olympiad held in USA. )

Problem 3(IMO 1995).LetA,B,C,Dbe four distinct points on a line, in that order. The circles with diametersACandBDintersect atXandY. LineXYmeetsBCatZ. LetPbe a point on lineXYother than Z. LineCPintersects the circle with diameterACatCandM, and lineBPintersects the circle with diameterBDatBandN. Prove that the linesAM,DN,XYare concurrent.

References

Evan Chen,Euclidean Geometry in Mathematical Olympiads, Mathematical Association of America (MAA), 2016.

SHREYAS S ADIGA is a student of class 12 at the Learning Centre PU College, Mangalore, Karnataka. He has a deep love for problem solving in Mathematics, especially Geometry and Number theory. He received a Winner diploma in 16th I F Sharygin Geometry Olympiad. He also got HM in APMO 2022. He cleared the Indian National Mathematics Olympiad (INMO) in 2021 and 2022. He has a keen interest in chess. He may be contacted at [email protected].