igo2015 solutions english

7/25/2019 IGO2015 Solutions English

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Problems of 2nd Iranian Geometry Olympiad 2015 (Elementary)

1. We have four wooden triangles with sides 3, 4, 5 centimeters. How many convexpolygons can we make by all of these triangles?(Just draw the polygons without anyproof)A convex polygon is a polygon which all of its angles are less than 180 and thereisnt any hole in it. For example:

This polygon isnt convex This polygon is convex

Proposed by Mahdi Etesami Fard

2. Let ABCbe a triangle with A= 60. The pointsM , N , K lie on BC,AC,ABrespectively such that BK= KM =M N =N C. IfAN = 2AK, find the values ofB and C.


3. In the figure below, we know that AB = CD and BC = 2AD. Prove thatBAD = 30.

P roposed by M orteza S aghafian

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4. In rectangle ABCD, the points M , N , P, Q lie on AB,BC,CD,DA respectivelysuch that the area of triangles AQM, BMN,CNP, DPQare equal. Prove that thequadrilateral M N P Q is parallelogram.


5. Do there exist 6 circles in the plane such that every circle passes through centersof exactly 3 other circles?


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Problems of 2nd Iranian Geometry Olympiad 2015 (Medium)

1. In the figure below, the points P,A, B lie on a circle. The point Qlies inside thecircle such that P AQ= 90 andP Q= BQ. Prove that the value ofAQBP QAis equal to the arc AB.

P roposed by Davood V akili

2. In acute-angled triangle ABC, BHis the altitude of the vertex B . The points Dand Eare midpoints ofAB andACrespectively. Suppose that Fbe the reflection ofHwith respect toE D. Prove that the lineBF passes through circumcenter ofABC.


3. In triangle ABC, the points M, N, K are the midpoints ofBC,CA, AB respec-tively. Let B and C be two semicircles with diameter AC and AB respectively,

outside the triangle. Suppose that M K and M N intersect C and B at X and Yrespectively. Let the tangents at Xand Y to C and B respectively, intersect at Z.prove that AZBC.


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4. Let ABCbe an equilateral triangle with circumcircle and circumcenter O. LetPbe the point on the arc BC( the arc which A doesnt lie ). Tangent to at P in-tersects extensions ofAB andACat KandL respectively. Show that KOL >90.

P roposed by Iman M aghsoudi

5. a) Do there exist 5 circles in the plane such that every circle passes through centersof exactly 3 circles?

b) Do there exist 6 circles in the plane such that every circle passes through centersof exactly 3 circles?


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Problems of 2nd Iranian Geometry Olympiad 2015 (Advanced)

1. Two circles1and2(with centersO1andO2respectively) intersect at A and B .The point X lies on 2. Let point Ybe a point on 1 such that XBY = 90. Let

X be the second point of intersection of the line O1X and 2 and Kbe the secondpoint of intersection ofXY and 2. Prove that Xis the midpoint of arc AK.


2. Let ABCbe an equilateral triangle with circumcircle and circumcenter O. LetP be the point on the arc BC( the arc which A doesnt lie ). Tangent to at Pintersects extensions ofABand ACat Kand L respectively. Show that KOL >90.


3. Let Hbe the orthocenter of the triangle ABC. Let l1 and l2 be two lines passingthrough Hand perpendicular to each other. l1 intersects BC and extension of ABat D and Z respectively, and l2 intersects BCand extension ofAC at E and X re-spectively. Let Ybe a point such that Y D ACand Y E AB. Prove that X, Y , Z are collinear.

Proposed by Ali Golmakani

4. In triangle ABC, we draw the circle with center A and radius AB. This circleintersectsACat two points. Also we draw the circle with centerAand radiusACandthis circle intersects AB at two points. Denote these four points by A1, A2, A3, A4.Find the points B1, B2, B3, B4 and C1, C2, C3, C4 similarly. Suppose that these 12points lie on two circles. Prove that the triangle ABCis isosceles.


5. Rectangles ABA1B2, BC B1C2, CAC1A2 lie otside triangle ABC. Let C be a

point such that CA1 A1C2 and CB2 B2C1. Points A and B are defined simi-larly. Prove that lines AA, BB , CC concur.

P roposed by Alexey Zaslavsky(Russia)

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Solutions of 2nd Iranian Geometry Olympiad 2015 (Elementary)

1. We have four wooden triangles with sides 3, 4, 5 centimeters. How many convexpolygons can we make by all of these triangles?(Just draw the polygons without anyproof)A convex polygon is a polygon which all of its angles are less than 180 and thereisnt any hole in it. For example:

This polygon isnt convex This polygon is convex


Solution.

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2. Let ABCbe a triangle with A= 60. The pointsM , N , K lie on BC,AC,ABrespectively such that BK= KM =M N =N C. IfAN = 2AK, find the values ofB and C.


Solution.Suppose the point P be the midpoint ofAN. Therefore AK = AP = AN and

so we can sayAP K is the equilateral triangle. So ANK = KPA2

= 30 LetACB = NM C=. Therfore ABC= KM B= 120 . So KM N= 60.ThereforeKM N is the equilateral triangle. Now we know that M N A = 90.Therefore = 45. So we have C= 45 and B= 75.

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3. In the figure below, we know that AB = CD and BC = 2AD. Prove thatBAD = 30.


Solution 1.

Let two points E and F on BC and AB respectively such that DFBC andDEAB. We can say DF = DC

2 = AB

2 .(because ofBC D= 30 and DF C= 90)

Also we know that DF =B E, therfore DE is the perpendicular bisector ofAB. SoBD = AD.

Let Hbe a point on CD such that BHCD. therefore BH= BC2

=B D, so wecan say D H and BDC= 90. Therefore ABD = BAD = 30.

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Solution 2.Suppose thatPis the point such that triangle DCPis Equilateral. We know that

P CBCand P C=C D= AB, therfore quadrilateral ABCP is Rectangular.

AP D= AP C

DP C= 90

60 = 30

In other hand, DP = DC and AP = BC. SoADP andBDCare congruent.Therfore AD= BD.

Let the point H on CD such that BHCD. therefore BH = BC2

= BD, so we

can say D H and BDC= 90

. Therefore ABD = BAD = 30

.

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4. In rectangle ABCD, the points M , N , P, Q lie on AB,BC,CD,DA respectivelysuch that the area of triangles AQM, BMN,CNP, DPQare equal. Prove that thequadrilateral M N P Q is parallelogram.


Solution.Let AB = CD = a,AD = BC = b and AM = x, AQ= z, P C = y , N C = t. If

x =y, we can assume that x > y. We know that:

y < x a x < a y (1)

SAQM

=SCNP

zx = y t

z < t

b

t < b

z (2)

According to inequality 1, 2:

(a x)(b t)< (a y)(b z) SBMN < SDPQits a contradiction. Therfore x = y , soz=t Now we can say two triangles AM Q

and CP N are congruent. Therefore M Q = NP and similarly M N = P Q. So thequadrilateral M N P Q is parallelogram.

Comment.If quadrilateralABCD be the parallelogram, similarly we can show that quadri-

lateral M N P Q is parallelogram.

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5. Do there exist 6 circles in the plane such that every circle passes through centersof exactly 3 other circles?


Solution.In the picture below, we have 6 points in the plane such that for every point there

exists exactly 3 other points on a circle with radius 1 centimeter.

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Solutions of 2nd Iranian Geometry Olympiad 2015 (Medium)

1. In the figure below, the points P,A, B lie on a circle. The point Qlies inside thecircle such that P AQ= 90 andP Q= BQ. Prove that the value ofAQBP QAis equal to the arc AB.


Solution 1.Let point Mbe the midpoint ofP B. So we can say P M Q= 90 and we know

that P AQ= 90

, therefore quadrilateral PAMQ is cyclic. Therefore:

AP M= AQM

In the other hand:

AQB AQP = P QM+ AQMAQP = 2AQM

So we can say that the subtract AQB from P QAis equal to arc AB.

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Solution 2.Let the point Kbe the reflection ofP to AQ. We have to show:

2AP B= AQB AQP

Now we know thatAQ is the perpendicular bisector ofP K. So AQP = AQKand P Q = KQ = BQ, therefore the point Q is the circumcenter of triangle P KB.We know that:

2AP B= KQB= AQB AQK= AQB AQPTherefore the subtract AQB from P QAis equal to arc AB.

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2. In acute-angled triangle ABC, BHis the altitude of the vertex B . The points Dand Eare midpoints ofAB andACrespectively. Suppose that Fbe the reflection ofHwith respect toE D. Prove that the lineBF passes through circumcenter ofABC.


Solution 1.The circumcenter ofABC denote by O. We know that OBA = 90 C,

therfore we have to show that F BA = 90C. We know that AD= BD= DH,also DH=DF.

Therfore quadrilateralAHF B is cyclic (with circumcenter D)

F BA = F HE= 90 DEH , DE BC DEH= C

F BA = 90 C

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Solution 2.The circumcenter ofABCdenote by O. We know that quadrilateral ADOE is

cyclic. Also we know that AD= H D= DB, therefore:

A= DH A= 180 DHE= 180 DF E ADFE:cyclic

So we can say ADFOEis cyclic, therefore quadrilateral DFOEis cyclic.

C= DEA= DEF = DOF

In the other hand: C= DOB so DOF = DOB , thereforeB , F, Oare collinear.

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3. In triangle ABC, the points M, N, K are the midpoints ofBC,CA, AB respec-tively. Let B and C be two semicircles with diameter AC and AB respectively,outside the triangle. Suppose that M K and M N intersect C and B at X and Yrespectively. Let the tangents at Xand Y to C and B respectively, intersect at Z.

prove that AZBC.


Solution 1.

Let point H on BC such that AHBC. Therefore quadrilaterals AXBH andAY CHare cyclic. We know that KM and M Nare parallel to AC and AB respec-tively. So we can say AKX= ANY = A, therefore ABX= ACY = A

2 and

XAB= Y AC= 90 A2

. So X,A, Yare collinear.

AHX= ABX=A

2 , AHY = ACY =

A

2 XHY = XM Y = A

Therefore quadrilateral XHMYis cyclic. Also we know that M XZ= M Y Z=90, therefore quadrilateralMXZYis cyclic. So we can sayZXHMY is cyclic. ther-fore quadrilateralHXZY is cyclic.

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In the other hand: ZY X= ACY = A2

ZHX= ZY X=A

2 , AHX=

A

2 ZHX= AHX

So the points Z,A, Hare collinear, therefore AZBC.

Solution 2.Let pointHonBCsuch that AHBC. We know thatK MandM Nare parallel

to AC and AB respectively. So we can say AKX = ANY = A, thereforeABX= ACY = A

2 and XAB= Y AC= 90 A

2 . So X,A, Yare collinear.

ZXY = ZY X= A2 ZX=Z Y

So the point Z lie on the radical axis of two these semicirculars. Also we know thatthe lineAHis the radical axis of two these semicirculars. Therefore the pointsZ,A, Hare collinear, therefore AZBC.

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Solution 1.Suppose that M and Nbe the midpoints ofAB and AC respectively. We know

that quadrilateral BMNC is cyclic. Also BP C = 120 > 90, so we can saythe point P is in the circumcircle of quadrilateral BMNC. Therefore: M P N >M BN= 30

In the other hand, quadrilaterals KMOP and NOPLare cyclic. Therefore:

M KO= M P O , N LO= NP O AKO + ALO= M P N >30

KOL= A + AKO+ ALO >90

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Solution 2.Suppose that KOL 90, therfore KL2 OK2 +OL2. Assume that R is the

radius of a circumcircle ABC. LetBK=x and LC=y andAB =AC=BC=a.According to law of cosines in triangle AKL, we have:

KL2 =AK2 + AL2AK.AL.cos(A) KL2 = (a + x)2 + (a + y)2 (a + x)(a + y)

In the other hand:

KB.KA= OK2 R2 OK2 =R2 + x(a + x)

LC.LA= OL2 R2 OL2 =R2 + y(a + y)We know that KL2 OK2 + OL2 and a= R3, therfore:

(a + x)2 + (a + y)2 (a + x)(a + y) 2R2 + x(a + x) + y(a + y)

R2 xy (1)KLis tangent to circumcircle ofABCat P. So we have:

KP2 =KB.KA= x(a + x)> x2 KP > x (2)

LP2 =LC.LA= y(a + y)> y2 LP > y (3)According to inequality 2, 3 we can say: xy < KP.LP (4)

Now According to inequality 1, 4 we have: R2 < KP.LP (5)

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We know that KOL 90, therefore K OL is acute-triangle. Suppose that H isorthocenter ofKOL. So the point H lies on OPand we can say HP OP.

Inother hand, HKP = P OL and KHP = OLP, therefore two trianglesT HP and OP Lare similar. So we have:

KP

HP =

OP

LP KP.LP=HP.OP OP2 =R2

But according to inequality 5, we have R2 < KP.LPand its a contradiction.

Therfore KOL >90.

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5. a) Do there exist 5 circles in the plane such that every circle passes through centersof exactly 3 circles?

b) Do there exist 6 circles in the plane such that every circle passes through centersof exactly 3 circles?


a)Solution.There arent such 5 circles. Suppose that these circles exists, therefore their centers

are 5 points that each point has same distance from 3 other points and has diffrentdistance from the remaining point. We draw an arrow from each point to its diffrentdistance point.

- lemma 1. We dont have two points such Oi, Oj that each one is the diffrentdistance point of the other one.

proof. If we have such thing then Oi and Oj both have same distance to theremaining points, therefore both of them are circumcenter of the remaining points,which is wrong.- lemma 2. We dont have 4 points such Oi, Oj , Ok, Ol that Oi, Oj put their arrowin Ok and OKputs its arrow in Ol.

proof. If we name the remaining point Om then the distances ofOi from Oj, Ol,Om are equal and the distances ofOj from Oi, Ol, Om are equal. Therefore each ofOl, Om is the diffrent distance point of another which is wrong (according to lemma

1).so each point sends an arrow and recives an arrow. Because of lemma 1 we donthave 3 or 4 points cycles. Therefore we only have one 5 points cycle. So each pair ofthese 5 points should have equal distance. which is impossible.

b)Solution.in the picture below, we have 6 points in the plane such that for every point there

exists exactly 3 other points on a circle with radius 1 centimeter.

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Solutions of 2nd Iranian Geometry Olympiad 2015 (Advanced)

1. Two circles1and 2(with centers O1and O2 respectively) intersect at A and B .

The point X lies on 2. Let point Ybe a point on 1 such that

XBY = 90

. LetX be the second point of intersection of the line O1X and 2 and Kbe the secondpoint of intersection ofXY and 2. Prove that Xis the midpoint of arc AK.


Solution.Suppose that the point Zbe the intersection ofB Xand circle 1. We know that

Y BZ= 90

, therefore the points Y, O1, Zare collinear.

O1Y A= ABX= AXX Y AXO1: cyclic

In the other hand, we know that AO1 = Y O1 so AXX = Y XO1 = XX

K.Therefore the point Xlies on the midpoint of arc AK.

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Proposed by Iman Maghsoudi

Solution 1.Suppose that M and Nbe the midpoints ofAB and AC respectively. We know

that quadrilateral BMNC is cyclic. Also BP C = 120 > 90, so we can saythe point P is in the circumcircle of quadrilateral BMNC. Therefore: M P N >M BN= 30

In the other hand, quadrilaterals KMOP and NOPLare cyclic. Therefore:

M KO= M P O , N LO= NP O AKO + ALO= M P N >30

KOL= A + AKO+ ALO >90

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Solution 2.Suppose that KOL 90, therfore KL2 OK2 +OL2. Assume that R is the

radius of a circumcircle ABC. LetBK=x and LC=y andAB =AC=BC=a.According to law of cosines in triangle AKL, we have:

KL2 =AK2 + AL2AK.AL.cos(A) KL2 = (a + x)2 + (a + y)2 (a + x)(a + y)

In the other hand:

KB.KA= OK2 R2 OK2 =R2 + x(a + x)

LC.LA= OL2 R2 OL2 =R2 + y(a + y)We know that KL2 OK2 + OL2 and a= R3, therfore:

(a + x)2 + (a + y)2 (a + x)(a + y) 2R2 + x(a + x) + y(a + y)

R2 xy (1)KLis tangent to circumcircle ofABCat P. So we have:

KP2 =KB.KA= x(a + x)> x2 KP > x (2)

LP2 =LC.LA= y(a + y)> y2 LP > y (3)According to inequality 2, 3 we can say: xy < KP.LP (4)

Now According to inequality 1, 4 we have: R2 < KP.LP (5)

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We know that KOL 90, therefore K OL is acute-triangle. Suppose that H isorthocenter ofKOL. So the point H lies on OPand we can say HP OP.

Inother hand, HKP = P OL and KHP = OLP, therefore two trianglesT HP and OP Lare similar. So we have:

KP

HP =

OP

LP KP.LP=HP.OP OP2 =R2

But according to inequality 5, we have R2 < KP.LPand its a contradiction.

Therfore KOL >90.

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3. Let Hbe the orthocenter of the triangle ABC. Let l1 and l2 be two lines passingthrough Hand perpendicular to each other. l1 intersects BC and extension of ABat D and Z respectively, and l2 intersects BCand extension ofAC at E and X re-spectively. Let Ybe a point such that Y D ACand Y E AB. Prove that X, Y , Z are collinear.

Proposed by Ali Golmakani

Solution.

Suppose that HZ intersects ACat P and HX intersects AB at Q. According toMenelauss theorem in two triangles AQX and AP Zwe can say:

CX

AC

.AB

BQ

.QE

EX

= 1 (1) and BZ

AB

.AC

P C

.P D

DZ

= 1 (2)

In the other hand, His the orthocenter ofABC. So BHACand we know thatDH E = 90, therefore HXA = BH Z = . Similarly we can say HZA =CHX=.

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According to law of sines inHP C,HCXandHP X:

sin(90 )P C

= sin(HCP)

HP ,

sin()

CX =

sin(HCX)

HX ,

HP

HX =

sin()

sin(90

)

P CCX

= tan()

tan()

Similarly, according to law of sines in HBQ, HBZand HQZ, we can show:

BZBQ

= tan()

tan() BZ

BQ=

P C

CX P C

BZ =

CX

BQ (3)

According to equality 1, 2 and 3, we can say:

XE

EQ = P D

ZD (4)

Suppose that the line which passes through Eand parallel to AB, intersects ZX atY1 and the line which passes through D and parallel to AC, intersects ZX at Y2.According to Thaless theorem we can say:

Y1X

ZY1=

XE

EQ ,

Y2X

ZY2=

P D

ZD

According to equality 4, we show that Y1 Y2, therefore the point Y lies on ZX.

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4. In triangle ABC, we draw the circle with center A and radius AB. This circleintersectsACat two points. Also we draw the circle with centerAand radiusACandthis circle intersects AB at two points. Denote these four points by A1, A2, A3, A4.Find the points B1, B2, B3, B4 and C1, C2, C3, C4 similarly. Suppose that these 12

points lie on two circles. Prove that the triangle ABCis isosceles.


Solution 1.Suppose that triangle ABC isnt isosceles and a > b > c. In this case, there

are four points (from these 12 points) on each side ofABC. Suppose that these12 points lie on two circles 1 and 2. Therefore each one of the circles 1 and 2

intersects each side ofABCexactly at two points. Suppose that P(A, 1), P(A, 2)are power of the point A with respect to circles 1, 2 respectively. Now we knowthat:

P(A, 1).P(A, 2) =b.b.(a c).(a + c) =c.c.(a b)(a + b) b2(a2 c2) =c2(a2 b2) a2(b2 c2) = 0 b= c

But we know that b > c and its a contradiction. Therefore the triangle ABC isisosceles.

Solution 2.Suppose that triangle ABC isnt isosceles. In this case, there are four points

(from these 12 points) on each side ofABC. Suppose that these 12 points lie ontwo circles1and 2. Therefore each one of the circles 1 and2 intersects each sideofABCexactly at two points (and each one of the circles 1 and 2 doesnt passthrough A,B,C). We know that the intersections of1 and the sides ofABC iseven number. Also the intersections of2 and the sides ofABC is even number.But Among the these 12 points, just 3 points lie on the sides ofABCand this isodd number. So its a contradiction. Therefore the triangle ABCis isosceles.

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5. Rectangles ABA1B2, BC B1C2, CAC1A2 lie otside triangle ABC. Let C be a

point such that CA1 A1C2 and CB2 B2C1. Points A and B are defined simi-larly. Prove that lines AA, BB , CC concur.

P roposed by Alexey Zaslavsky(Russia)

Solution.Suppose that lA is the line which passes through A and perpendicular to B2C1.

Let lB and lC similarly. Suppose that CB1 = BC2 = x and BA1 = AB2 = y andAC1=C A2=z. According to angles equality, we can say:

sin(A1)

sin(A2)

=y

z

, sin(B1)

sin(B2)

= x

y

, sin(C1)

sin(C2)

= z

x

According to sine form of Cevas theorem in ABC,lA, lB, lCare concur. Supposethat lA, lB, lCpass through the point P. We know thatP BC andAC2B1 areequal. ( because ofBP AC2, CP AB1, BC B1C2 and BC = B1C2 ). So wehave:

P A =x , P C =y , P B =z P ABC , P BAC , P C AB

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Suppose that P A, P B, P C intersects BC,AC,AB at D, E , F respectively and:P D= m , P E =n , P F =t. According to before figure, we have:

sin(A1)

sin(A2)=

n

t =

y

z ,

sin(B1)

sin(B2)=

t

m=

x

y ,

sin(C1)

sin(C2)=

m

n =

z

x

Ifn= ky, then: t= kz , m= kyzx

.

Nowdraw the line from A such that be parallel to BC. The intersection of thisline and extension AB and ACdenote by B3 and C3 respectively. Let the pointA

be the intersection ofAA and BC. According to Thaless theorem, we have:

BA

CA

=B3A

C3A

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Let B3P A = and C3P A

= . We know that the quadrilaterals P F B3A

and P EC3A are cyclic. Therefore B3F A

= and C3EA =.

According to law of sines inP B3A andP C3A andP C3B3:B3A

C3A=

tan()

tan()

Also according to law of sines inP F A:t

x=

sin(B+ 90)cos()

= cos(B+ )

cos() =cos(B) tan().sin(B)

tan() =

cos(B) tx

sin(B)

Similarly we can say:

tan() =cos(C) n

x

sin(C) B3A

C3A=

BA

CA=

x.cos(B) tx.cos(C) n .

sin(C)

sin(B)

Similarly, two other fractions can be calculated.

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According to Cevas theorem inABC, we have to that:

x.cos(B) tx.cos(C)

n

.sin(C)

sin(B).

z.cos(C)mz.cos(A)

t

.sin(A)

sin(C).y.cos(A) ny.cos(B)

m

.sin(B)

sin(A)= 1

x.cos(B) tx.cos(C) n .

z.cos(C)mz.cos(A) t .

y.cos(A) ny.cos(B)m = 1

In other hand, we know that:

n= ky , t= kz , m=kyz

x

x.cos(B) kzx.cos(C) ky .x.cos(

C) ky

x.cos(A) kx . x.cos(

A) kxx.cos(B) kz = 1

Therfore, we show that AA, BB , CC are concur.

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igo2015 solutions english

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