Tổng hợp đề thi đề nghị cho kỳ thi HSG Hình học IGO năm 2018

Indeed, we easily seen triangle OEF and T AD orthogonal because AF, DE and perpendicular bisector of BC are concurrent so the lines passing through F, E and O and are perpendicular to OD[r]

Proposed problems for Olympiad Tran Quang Hung

Let ABCD be cyclic quadrilateral with circumcenter O Let E be the point such that OE k AC and DE ⊥OA Let F be the point such that OF kBD and AF ⊥OD Assume that AB meest CD at S Prove that EF ⊥OS

O B

C

D A

S

F

E

S

A D

O

B C

F

E

T M N

X

Proof LetBM, CN be diameter of (O) Apply Pascal theorem for six points

B N D C M A

we deduce AN, M D and SO are concurrent at T Now we will show that OT ⊥EF Indeed, we easily seen triangleOEF and T AD orthogonal because AF, DE and perpendicular bisector of BC are concurrent so the lines passing through F, E and O and are perpendicular to OD, OA and AD are concurrent Since T A ⊥ OE and T D ⊥ OF we get T O ⊥ EF So we complete the proof

Problem proposal for IGO Nguyen Van Linh

23/7/2018

Problem LetABC be a triangle inscribed in (O),P be an arbitrary point onA-median of triangle ABC Circles (AP B), (AP C) meet AC, AB again at E, F, respectively EF meets BC at G Circle (AEF) meetsAP again atT Prove that the reflection ofT wrtP Glies on (O)

Proof First we introduce a lemma:

Lemma Given triangleABC inscribed in (O) Let P be an arbitrary point in the plane Circles (AP B),(AP C) intersect AC, AB again at E, F, respectively (AEF) intersects AP at K Let H be the reflection ofK wrtP ThenH ∈(O)

N

S M

L

H

K E

F

O

P A

B C

Proof LetL be the intersection ofF K with (AP C),M, S be the reflections of K, E wrtL, C We have ∠LCA=∠AF L=∠AEK thenKE kLC We getM SkKE, then∠M SA=∠KEA= ∠AF M, which follows thatAF SM is cyclic

Moreover, KA·KH = 2KA·KP = 2KF ·KL = KF ·KM, then AF M H is cyclic We get pointsA, F, S, H, M are concylic

Similarly, (AEH) passes throughN-the reflection of F wrtB Then PC/(AEH)

PC/(AF H) =−1 =

PB/(AEH) PB/(AF H) ThereforeA, B, H, C are concyclic Back to our problem

X Y U

L

K T'

T

J

G E

F

M O A

B

C P

LetX, Y be the second intersections of (AP B),(AP C) withBC, respectively, M be the midpoint ofBC,J be the center of (AEF),L be the reflection ofAwrt the perpendicular bisector of BC

We have M B·M X =M P ·M A=M C·M Y, thenM X =M Y

From this,PB/(AEF)=BA·BF =BY ·BC =CX·CB =CE·CA=PC/(AEF).ThenJ B =J C, we getL∈(AEF)

We have A(M L, BC) =−1 then (T L, F E) =−1

On the other side, letG0 be the intersection of the perpendicular bisector ofT LwithBC We have ∠T J G0 =∠T AL =∠AM B, hence J T M G0 is cyclic We get ∠J T G0 =∠J LG0 = 90◦ Then G0, E, F are collinear orG0 ≡G

LetK be the intersection ofT P with (O), U be the antipode ofK wrt (O) U T meets (O) again atT0

By the lemma above,P T =P K hence T0P =T P =KP

On the other side, ∠T T0L = ∠U KL = ∠U AL = 90◦ −∠T AL =

2∠T GL Then G is the circumcenter of triangleT T0L, which follows thatGT =GT0

ThereforeT0 is the reflection ofT wrtP Gand T0 ∈(O)

Proposed problems for Olympiad Tran Quang Hung

Let ABC be an acute-angled triangle The altitudes BE, CF meet at orthocenter H O is circumcenter of triangle ABC M is the midpoint of BC P is the point on EF such that HP ⊥HO Q is the point on HA such that P Q⊥HM Prove that QA= 3QH

A

B C

H

O F

E P

Q

M Figure

A

B C

H

O F

E

P Q

M

K G

T

S

Y

R

D Z

X Figure

Proof Let (O) be the circumcircle of ABC AK is the diameter of (O) D, S, T are the reflections point of H through BC, CA, AB thenD, S, T are on (O) We easily seenHCKB is a parallelogram soM lies on HK Let KH intersect (O) again at G Lines HP intersects the lines BC, DK, ST, GA atR, X, Y, Z, respectively Note that EF is the median line of triangle HST soHY = 2HP By butterfly theorem for quadrilateralBCT S then HR =HY = 2HP By butterfly theorem for quadrilateral AGDK then HZ =HX = 2HR= 2HY = 4HP Note thatP Q⊥HM soP QkAG Using Thales theorem thenHA = 4HQ, this meansQA= 3QH We complete the solution

The Problems proposed for IGO 2018

1 Problem and its proof

Problem (9th grade) Let ABC be an acute-angled triangle withAB < AC Let M, and E, F be the midpoint of BC, and the feet of the altitudes from B and C, respectively Let I and J be the incenters of the triangles ABC and AEF, respectively The lines respectively passing I perpendicular to AI and J perpendicular toM J meet at a pointK Prove that ifLis the point of intersection of KJ and EF then KJ = 2J L

Figure 1.1 Author: Le Viet An

Proof Let P, Q be the the feet of lines passing through I perpendicular to CA and AB, respectively; R be the foot of line passing through C perpendicular to BI and S be the foot of line passing through B perpendicular toCI

Firstly, since the three points A, I, J are on the interior angle bisector of BAC,[ they must be collinear

Let Dbe the foot of line passing through P perpendicular to AB (see f.1.2) Since triangle ABC is acute, points E and F are on the segments AC, AB, respectively

Since \BEC = \BF C = 90◦, BCEF is cyclic It follows AEF[ = ABC[ Hence [

AEJ = 12AEF[ = 12ABC[ =ABI.[ Combine withEAJ[ =BAI, it follows[ 4AEJ ∼ 4ABI (a.a) Hence AJAI = AEAB Note that the triangles ABE,AP D are right and

2 The Problems proposed for IGO 2018-Le Viet An

Figure 1.2

similar, andAP =AQ by properties of the tangent lines from a point to a circle, we have

AI AJ =

AE AB =

AD AP =

AD AQ

Hence, by Thales’s theorem, J D//IQ⊥AB It follows thatP, J, D are collinear and P J//IQ

Similarly, QI//P J From IP = IQ, IP J Q is a rhombus It follows that P Q is the perpendicular bisector ofIJ So, if denoteN by the intersection point ofP Q and IK then applying the mid-line theorem for triangle IJ K, N is mid-point of J K

On the orther hand, since IP C[ = 90◦ = IRC[ and IP A[ = IQA[ = 90◦, CIP R and AP IQ are cyclic It folows that

[

CP R=CIR[ =IBC[+ICB[ =

2(ABC[ +ACB) = 90[ ◦−

2BAC[ = 90

◦−[ IP Q

Hence IP R[+IP Q[ =CP R[ + 90◦+IP Q[ = 90◦+ 90◦ = 180◦, this means that the points P, Q, R are collinear It follows that R belongs toP Q

Similarly, S also belongs toP Q Thus, RS passes through N

Note that points E, F, R, S are on same circle (M) with diameter BC Hence BCRE is cyclic, it follows REC[ =RCB[ = 12ABC[ = 12AEF[ =AEJ[ This thing means that R, E, J are collinear It follows that J belongs toER

3 According to butterfly theorem with note that ERF S is a quadrilateral inscribed in circle (M) and J =ER∩F S and LN ⊥M J atJ We get J L=J N = 12J K The problem is proved

Remark A similar problem:

Problem 1’ Let ABC be an acute-angled triangle with AB < AC Let M, and E, F be the midpoint of BC, and the feet of the altitudes from B and C, respec-tively Let I and J be the incenters of the triangles ABC and AEF, respectively The lines respectively passing throughI perpendicular to M I andJ perpendicular to AJ meet at a point K Prove that if L is the point of intersection of KI and BC then KI = 2IL

The Problems proposed for IGO 2018 Problem and its proof

Problem (10th grade) LetABCD be a cyclic quadrilateral A circle passing through A, B and tangent to the segment CD at E, and a circle passing through

C, D and tangent to the segment AB at F Let G and H be the points of inter-section for two lines of each pair of (AE, DF)and (BE, CF), respectively Prove that the incenters of the triangles AGF, BHF, CHE, DGE lie on a circle

Figure 3.1 Author: Le Viet An

Proof Define thatI, J, K, L be the incenters of the triangles AGF, BHF,CHE,

DGE respectively In case of AB//CD then ABCD an isosceles trapezoid and the configuration has an axis of symmetry that is the perpendicular bisector of

AB We easily see that IJ and KL have same axis of symmetry It follows that

IJ KLis an isosceles trapezoid and is cyclic This means that the problem is true in this case

Suppose that the rays BA and CD meet at point M (see f.3.2) SinceABCDis cyclic,M Eis tangent toJ

(ABE)andM F is tangent toJ

(CDF),

M E2 =PM/J

(ABE) =M A.M B =PM/J

(ABCD) =M B.M C =PM/J

(CDE) =M F

It follows that M E =M F

From M E = M F and M E is tangent to J

(ABE), we get ∠M EF = ∠M F E

and ∠M EA=∠M BE Hence

∠AEF =∠M EF −∠M EA=∠M F E−∠M BE =∠BEF

2 The Problems proposed for IGO 2018-Le Viet An

Figure 3.2

This thing means that EF is the interior angle bisector of angle AEB Similarly, F E is the interior angle bisector of angle CF D

Since HJ, HK are all the angle bisector of angle BHF, three points H, J, K are collinear and ∠F J H = 90◦+∠F BH2

Thus

∠F J K = 90◦+∠M BE

2 = 90

◦+ ∠M EA = 90◦+180

◦ −

∠AEC

2 = 180

◦− ∠AEC = 180◦− ∠AEB+∠BEC

2 = 180

◦−

(∠F EB+∠BEK) = 180◦−∠F EK

This thing means that EF J K is cyclic Similarly, EF IL is also cyclic

SinceEF is the interior angle bisector of angleGEH andGF H, we easily see that the triangles GEF and HEF are equal It follows EG = EH and F G = F H Hence GEGF = HEHF So, applying the properties of angle bisector of a triangle, the exterior angle bisector from vertex G of 4GEF, from vertex H of 4HEF and the lineEF are concurrent or the pair of these lines are parallel

•If the exterior angle bisector from vertexGof4GEF, from vertex Hof4HEF

and the line EF are pairwise parallel, then we have that isosceles trapezoids

3 points of L, K with respect to the perpendicular bisector of EF Hence IJ KLis cyclic

•If the exterior angle bisector from vertexGof4GEF, from vertex Hof4HEF

and the lineEF meet at a point, denote by this pointP ThenP =J K∩IL∩EF

and

P J P K =PP /J

(EF J K) =P E.P F =PP /J

(EF IL) =P I.P L This means that IJ KLis cyclic

Thus, in any case, we always have thatIJ KLis cyclic The problem is proved Remark The problem can be extended as follows:

Problem 3’ LetABCDbe a convex quadrilateral satisfiyingAB.CD =BC.DA Supposes that the tangent line at A of the circumcircle of triangle ABD meets the rays CB and CD at E and H respectively; and the tangent line at C of the circumcircle of triangleCBDmeets the raysAB andADatF andGrespectively Prove that the incenters the trianglesABE,BCF, CDGandDAH lie on a circle Problem 3’ was proposed the author that can be constructed from a nice result as follows:

Nice result Let ABC be a triangle with circumcircle (O) Supposes that the tangent line at B of (O)meets CA at E; and tangent line at C at (O) meetsAB

at F Let I and J be the incenters of the triangles ABE and ACF respectively Prove that four points B, C, I and J are concyclic (see f.3.3)

Tong Huu Nhan Proposal Problems for IGO 2018

Problem Let triangle ABC be with circumcircle (O), orthocenter H and altitude AD Lines that pass throughH and parallel toAB, AC, will cuts AC, AB at P, Q, respectively P Q meets BC at X Points Y, Z are defined similarly Prove that

(a) X, Y, Z lie on a line that is perpendicular to OH at S (b) A, O, D, S lie on a circle

O A

B C

H

D Q

P

X Y

Z S

E

M Ha

Ta T

A0

Solution (a) Let M be midpoint of BC and AA0 be diameter of (O) It is known that

H, M, A0 are collinear and HBA0C is parallelogram So, if ray M H meets (O) atT then

AT ⊥HM

From AH ⊥ BC, BH ⊥ AP, CH ⊥ P H, we have 4HBC ∼ 4P AH Notice that

P AQH, HBA0C are parallelograms, we get4HBA0 ∼ 4P AQ Then

∠P QA=∠HA0B =∠T AB,

so P QkAT, and HM ⊥P Qat Ta (1)

Let(E)be Euler circle of 4ABC andHa be midpoint ofP Q AsAP HQis parallelogram,

then Ha is also midpoint of AH, which implies that HaM is diameter of (E) (2)

From (1) and (2), we get Ta ∈(E), which implies that H lies on the polar of X with(E)

And by La Hire’s Theorem, similarly, we get that X, Y, Z lie on the polar of H with (E), which is perpendicular to EH (or OH)

Proposal Problems for IGO 2018 Tong Huu Nhan

(b) Let R be radius of(O), then R/2is radius of (E) As X, Y, Z is the polar ofH with

(E), we have

EH·ES = R

4 ⇔EH·(EH+HS) = R2

4

⇔HE·HS=HE2− R

2

4 =PH/(E) =HHa·HD

⇔HO·HS=HA·HD (HO = 2·HE, HA= 2·HHa),

which implies that A, O, D, S lie on a circle

IGO 2018 SUGGESTION PROBLEMS

(Le Phuc Lu, Viet Nam)

Let ABC be a triangle with ∠A = 60◦, AB < AC and I, O, G are the incenter, the circumcenter, the intersection of A-median and A-Apollonius circle (differs from A) Prove that the midpoint of the segment that connects two orthocenters of triangleABC and IGO lies on the nine point circle of triangle IGO

Solution Let AM, AD be the median, angle bisector of triangle ABC and H, K be the orthocenter of triangle ABC, IGO It is easy to check that ∠BHC = ∠BIC =

∠BOC= 120◦ thenB, C, H, I, O belong to the circle of centerT, which is the midpoint of minor arc of (O)

DenoteS as the second intersection of Apollonius circle with (O) then SB SC =

AB

AC which implies that ABSC is harmonic quadrilateral Then AS is the symmedian of triangle

ABC, SBC so D is the incenter of triangle AM S Hence, M A, M S are symmetric respect toBC then

∠BGC =∠BSC =∠180◦−∠A= 120◦ Thus, G ∈ (T) Consider the homothety center K and ratio

2 maps circle (IGO) to the nine point cirlce of triangle IGO But H ∈ (IGO) then the midpoint of HK lies on the nine-point circle of triangleIGO

IGO 2018 SUGGESTION PROBLEMS

(Le Phuc Lu, Viet Nam)

In the space, let (P) be a plane and the line d lies on (P), the line d0 does not lie on (P), not perpendicular to (P) such that d, d0 meet at O Consider arbitrary, distinct points A, B on d0 such that the distance from A, B to(P) are equal Suppose that C, D are ond such that the acute angles form by lines AC, BD with (P) are equal (C, D lie on the same side with respect to O) Prove that the ratio CD

AB is fixed

Solution The distances from A, B to (P) are equal then OA = OB Since C, D ∈ d and the angle between AC, BD with (P) are equal then

∠ACO=∠(AC,(P)) = ∠(BD,(P)) =∠BDO

Denote B0 as the reflection of B respect to d then AB0 k d since the midpoint of AB lies ond

TriangleDBB0 is isosceles then ∠B0DO =∠BDO=∠ACO which implies that AC k

B0D Hence, ACDB0 is a parallelogram

ThusAB0 =CD and note that AB0B is a right triangle then we have CD

AB = AB0

AB = cosB

AB= cosα

with α is the acute angle betweend, d0 This value clearly is a fixed number

The Problems proposed for IGO 2018

1 Problem and its proof

Problem (8th grade) Given a convex quadrilateral ABCD andX be a point lying inside it that satisfies the conditions

XA.XC2 =XB.XD2 and AXD\+BXC\ =CXD\ Prove that XAD\+XCD\ =XBC\+XDC\

H`ınh 2.1 Author: Le Viet An

Proof Since CXD\ = AXD\ +BXC >\ AXD\, we can contruct a triangle CXY

such that

4CXY ∼ 4DXA,

(1)

and rayXY lies between two rays XC v`aXD (see f.2.2) From this, we get XYXC = XA

XD It follows XY =

XA.XC XD

From XA.XC2 =XB.XD2, we get XA.XC = XB.XD2

XC , we deduce that

XY XD =

XA.XC XD

XD =

XA.XC XD2 =

XB.XD2 XC

XD2 = XB XC

Since Y XD\ =CXD\ −Y XC\= (AXD\+BXC\)−AXD\ =BXC\,

4DXY ∼ 4CXB (s.a.s)

(2)

2 The Problems proposed for IGO 2018-Le Viet An

Figure 2.2 From (1) and (2), we get

CY DA =

CX DX =

CB DY ,

and

\

BCY =BCX\+XCY\=Y DX\+XDA\ =\ADY

It follows that 4CBY ∼ 4DAY (s.a.s) Hence

\

CBY =\DAY

(3)

Again, from (1) and (2), we get

XY XA =

XC XD =

XB XY ,

and

\

AXY =AXD\+DXY\ =Y XC\+DXY\ =DXC\

=DXY\ +Y XC\ =CXB\+Y XC\=XY B.\

It follows that 4XAY ∼ 4XDC ∼XY B (s.a.s) Hence

\

XAY =XDC\ =XY B\ and XY A\=XCD\ =XBY \

(4)

From (3) and (4), we get

\

XAD+XCD\ =\XAY +\Y AD+XCD\

3

The problem is proved

Remark The conditionAXD\+BXC\ =CXD\ is equivalent to 12AXB\+CXD\ = 180◦

IGO 2018 SUGGESTION PROBLEMS

(Le Phuc Lu, Viet Nam)

Let ABC be a triangle with M, N, P are midpoints of BC, CA, AB respectively De-note D, E, F as arbitrary points on segments BC, CA, AB such that AD, BE, CF are concurrent at O (D 6= M, E 6= N, F 6=P) Suppose A0, B0, C0 lie on the perimeter of triangleABC such that each of DA0, EB0, F C0 bisects the area of triangleABC Prove that DA0, EB0, F C0 are concurrent at some point O0 and the length of OO0 does not exceed length of the longest median of triangleABC

Solution Suppose that AB≤ CA≤AB Then A0 such that DA0 bisects the area of triangleABC will belong to segment CA.Note that

SAM C =

1

2SABC =SDA0C thusSDA0M =SAA0M which implies thatAD kM A0 Similarly, BE kN B0, CF kP C0

DenoteG as the centroid of triangleABC and O1 is the intersection ofOG withM A0 It is easy to check thatGlies betweenO, O1and

GO GO1

= GA

GM = by Thales’s theorem Similarly defineO2, O3 then we can see that they also satisfy the same condition which meansO1, O2, O3 coincide

From this, we conclude that M A0, N B0, P C0 are concurrent at O0 and OO0 = 2OG Continue, we have the following lemma

Lemma LetT be a point lie inside triangle XY Z then XT <max{XY, XZ} SinceO lies strictly inside one of GAB, GBC, GCAthen

OG <max{GA, GB, GC}=

3 length of longest median of triangle ABC

This finishes our proof

IGO 2018 SUGGESTION PROBLEMS

(Le Phuc Lu, Viet Nam)

Consider a set X of points {A, B, C, D, E, F, O} such that ABCDEF is a regular hexagon with the center O How many angles equal to 60◦ such that the vertex in X

and two sides pass through another two points in X? (Note that for any three point

O1, O2, O3, two angles ∠O1O2O3 and ∠O3O2O1 are considered as the same)

Solution Consider the segment AC, we can see that there are satisfied angles are

∠ADC =∠AEC =∠AF C= 60◦

Similarly with segments BD, CE, DF, EA, F B then if we not consider vertex O, there are 3×6 = 18 angles in total

TakeO into account, we have more angles are

∠AOB =∠BOC =∠COD =∠DOE =∠EOF =∠F OA= 60◦

Therefore, we have 18 + = 24 angles

Tong Huu Nhan Proposal Problems for IGO 2018

Problem Let triangle ABC be with circumcircle(O), orthocenterH and M is midpoint ofBC Let J be a point on the opposite ray of ray M O such that M J = 12M O Circle

(J, J B) cuts Euler circle of 4ABC at P, Q Prove that H, P, Q are collinear

O A B C H F E S M P Q J E G

Solution Let (E) and BE, CF be Euler circle and altitudes of 4ABC, respectively.EF meets BC atS Since BF EC is cyclic, we get

PS/(E)=SE·SF =SB·SC =PS/(J),

so S lies on radical axis P Q of (E)and (J) (1)

LetG be centroid of 4ABC, we have OG OE =

1 ·OH ·OH

= =

OM OJ ,

then M G k JE In addition, P Q is common chord of (E) and (J), then JE ⊥ P Q, so P Q⊥AM (2)

Applying Brocard’s Theorem for complete quadrilateral BF EC.AS, then SH ⊥ AM Combining with (1) and (2), we get that H, P, Q are collinear

Proposal Problems for IGO 2018 Minh Ngoc Tran

Problem (Grade 10) Let acute triangle ABC be with centroid G Let AM AN, be two tangents of the circle with diameter BC Let BP BQ, be two tangents of the circle with diameter CA Let CR CS, be two tangents of the circle with diameter AB Prove that

, , , , ,

M N P Q R S lie on a circle with center G

First, we prove that MN PQ RS, , are concurrent at H

Let AD BE CF, , be altitudes of ABC, O be midpoint of BC and H be orthocenter of ABC



Since BDHF is cyclic and AM tangents to  O , we get  

AH AD AF AB AM

So AMH ＳADM  AMH  ADM In addition, A D M N O, , , , lie on a circle and

AMN is a isosceles triangle , then ADM  ANM  AMN So AMH AMN , ,

M N H is collinear Analogously, P Q H, ,  , R S H, ,  are collinear D

K H

N

M

E

F

O A

We haveC F M N, , ,  , C F P Q, , ,  are concyclic So HM HN HC HF HP HQ , , ,

M N P Q

 are concyclic Analogously, P Q R S, , , are concyclic

Let I J, be midpoint of AB AC, , respectively, we have AO BI CJ, , are perpendicular bisector of MN PQ RS, , , respectively and they are concurrent at G So

GM GN GPGQGRGS, then M N P Q R S, , , , , lie on a circle with center G N

M

O F

E

D H

S

G R

J P

Q

I A

Problem proposal for IGO Nguyen Van Linh

23/7/2018

Problem LetABC be a triangle inscribed in (O) LetE, F be two points on sidesAC, AB such that AE = AF =BC (AEF) meets (O) again at P Let K, L be the midpoints of arc AC, AB of (O) P K, P L meet (AEF) again atX, Y, respectively A-excircle of triangle ABC is tangent to BC

atD Prove thatAD bisects XY

T R N I D M L X K Y P F E C A B

Proof LetI be the incenter of triangle ABC

We have∠Y AF = 180◦−∠Y P F = 180◦−∠LP A−∠AP F =∠AEF −∠ACI = 90◦−∠IAC−

∠ACI =∠IBC

∠AY F =∠AP F = 180◦−∠AEF = 90◦+1

2∠BAC =∠BIC Then 4AY F =4BIC (a.s.a)

We get AY =BI, thenAY BI is a parallelogram Similarly,AXCI is also a parallelogram Then 4AXY =4IBC

Let M, N be the midpoints of XY, BC,respectively Let T be the tangency of (I) with BC,R be the reflection ofT wrtI

We have IN k AM On the other side, A, R, D are collinear and IN is parallel to RD, then

A, M, R, Dare collinear

ThereforeAD passes through the midpoint ofXY

IGO 2018 SUGGESTION PROBLEMS

(Le Phuc Lu, Viet Nam)

Let ABC be a triangle with ∠A = 45◦, AB < AC with O is its circumcenter and

AD, AH are angle bisector, altitude, (D, H ∈ BC) Suppose that OD cuts AH at E

and K is the circumcenter of triangle EBC Prove that HK kAD

Solution We will use calculation, the beautiful solution is currently unknown

Denote P, N as the projections ofO, K on AE and M, T as the midpoints of BC, the arc BC of (O)

PutAH =h and R, R0 are radius of circles (O),(K) By Thales’s theorem, we have HE

HA = M O M T =

M O OT −OM =

√

2 +

1

then HE = (p(2) + 1)h By applying Pythagorean’s theorem, we have BK2−BO2 =KM2−OM2

⇒R02−R2 =KM2−OM2

⇒KM2 =R02− R

2 Similarly, we also have

AO2−(AH−OM)2 =OP2 =KN2 =KE2−(EH−M K)2

⇔R2 −

h−√R

2

2

=R02−(√2 + 1)h−M K2

⇔M K2+hR√2 = R02− R

2 −(

√

2 + 1)2h2+ (2 + 2√2)h·M K

⇔R√2 = (2 + 2√2)h+ (2 + 2√2)M K

⇔M K =h+ R

√

2

2 + 2√2 =h+M T

Therefore,KT =M K−M T =h=AH which implies thatAHKT is a parallelogram,

henceAD is parallel to HK

Proposed problems for Olympiad

Tran Quang Hung

Let ABC be a triangle inscribed in the circle (O) The A-mixtilinear incircle of ABC is tangent to (O)at D and is tangent to CA, AB at E, F, respectively AD cuts EF atG Prove that midpoints of the segments BC, EF, AG are collinear

A

B C

O

D F

E

G N

I

M

A

B C

O

D F

E

M I

G N

S T

P

Proof LetIbe the midpoints ofEF thenIalso is the incenter ofABC LetM, Nare midpoints

of BC, AG Let AI intersects (O) again at S ST is diamter of (O) We known that D, I, T

are collinear SD intersects BC at P We see that SI2 = SB2 = SD.SP, this shows that

∠SIP = 90◦soP lies onEF Note that triangleAIGis right atI so∠N AI =∠N IA Now the quadrilateralsP IM SandP T M Dare cyclic so∠M IS =∠M P S =∠DT S =∠DAS =∠N IA This means N, I, M are collinear We are done

Tong Huu Nhan Proposal Problems for IGO 2018 Problem Let triangle ABC be with circumcircle (O) An arbitrary line that passes throughOmeetsAC, AB atX, Y LetM, N, I be midpoints ofBX, CY, XY, respectively Prove that

(a) O, I, M, N lie on a circle

(b) The Simson line of O wrt triangle IM N always passes through a fixed point as XY moves around O

O A B C Y X M N B0 C0 S I H K

P J Q

Solution (a) Let BB0, CC0 be diameters of (O) As X, Y, O are collinear, by Pascal’s Theorem, we get that B0X meets C0Y at a point S that lies on (O)

From the property of midsegment, notice that BCB0C0 is rectangle, we get (OM, ON)≡(SB0, SC0)≡(AB, AC)≡(IM, IN) (mod π),

which implies that O, I, M, N lie on a circle

(b) LetH, K be prọjections ofOonIM, IN, thenHK is the Simson line of Owrt4IM N From the property of midsegment, we get IM kAB, so OH ⊥AB at midpoint P ofAB Similar, OK ⊥AC at midpoint Q of AC

LetJ be midpoint ofP Q By Thales’s Theorem, notice that I is midpoint ofXY, we have

HP HO · KO KQ = IY IO · IO

IX =−1 =

J P J Q,

and by Menelaus’s Theorem, we get that H, K, J are collinear

In conclusion, the Simson line of O wrt 4IM N always passes through the fixed point J