Download Solution 2 Let M be the midpoint of CD, and let CE intersect AF

Problem 1
Let A1A2A3 be a non-equilateral triangle with circumcentre O and incentre I, and
let T1, T2, T3 be the tangency points of the incircle of A1A2A3 with A2A3, A3A1, A1A2
respectively. Prove that the orthocentre of T1T2T3 lies on OI.
Solution
Let S1S2S3 be the orthic triangle of T 1T2T3 (that is, the triangle formed by the feet
of the altitudes of T1T2T3). It is easy to see that S1S2S3 and A1A2A3 are homothetic
and it is well-known that H is the incentre of S1S2S3.
(Notice that, in a contest, this result should be proven rigorously.)
Hence the lines A1S1, A2S2 and A3S3 are concurrent at a point P. The homothety
with centre P which maps S1S2S3 to A1A2A3 maps the incentre of S1S2S3 to the
incentre of A1A2A3, so H is mapped to I. Hence, P, H and I are collinear.
Now, since I and H are the circumcentre and orthocentre of T 1T2T3 respectively,
the points P, H and I lie on the Euler line of T 1T2T3. It is well-known that the
circumcentre W of S1S2S3, which is the centre of the nine-point circle of S1S2S3,
lies on this Euler line as well (in fact, W is the midpoint of IH). It follows that the
points P, H, I and W are all collinear.
The homothety mentioned above, mapping S 1S2S3 to A1A2A3, maps the
circumcentre of S1S2S3 to the circumcentre of A1A2A3, so W is mapped to O.
Hence P, W and O are collinear. Since P, H, I and W are collinear, we can
conclude that O, I, H, W and P lie on the Euler line of S1S2S3. Hence result.
Problem 2
Let ABC be an isosceles triangle (AB = AC) and let D be the midpoint of BC. The
point E lies on AB such that DE is perpendicular to AB. F is the midpoint of DE.
Prove that AF is perpendicular to CE.
Solution 1
Let G be the point on AB such that CG is perpendicular to AB, and let H be the
intersection of EC and AF. It is easy to see that EDA and GBC are similar. As E
and F are the midpoints of GB and ED respectively, ECB and FAD are similar.
Then, since <HCD = <ECB = <FAD = <HAD, HDCA is cyclic. It then easily
follows that <AHC = <ADC = 90 and we are done.
Solution 2
Let M be the midpoint of CD, and let CE intersect AF and AM at H and S
respectively. It is clear that ADC and AED are similar, and that ACM and ADF are
similar. Hence ADC and AFM are similar. Thus <AFM = <ADC = 90. Since FM
and CE are parallel, AF is perpendicular to CE.
Solution 3
It is sufficient to prove that AE2 + CF2 = EF2 + AC2. Notice that AE2 = AD2 – DE2,
and AC2 = AD2 + CD2. Hence we must prove that CF2 = DE2 + EF2 + CD2. Using
the formula for the length of a median we get 4CF2 = 2CE2 + 2CD2 – DE2. Hence
we must prove that 2CE2 + 2CD2 – DE2 = 4DE2 + 4EF2 + 4CD2. As 4EF2 = DE2
this reduces to proving that CE2 = CD2 + 3DE2. A simple application of the cosine
rule is sufficient to do this (notice that –2.DE.BD.cos(<EDC) = 2.DE2).
Solution 4
Use vectors.
Solution 5
Use coordinates.
Problem 3
Let ABC be an acute triangle with circumcentre O and orthocentre H. Prove that
there exist points D, E and F on [BC], [CA] and [AB] respectively such that the
lines AD, BE and CF are concurrent and DO + DH = EO + EH = FO + FH.
Solution
Let P, Q and R be the reflections of H in BC, CA and AB respectively. It is known
that P, Q and R lie on the circumcircle of ABC. Let D, E and F be the
intersections of OP, OQ and OR with BC, CA and AB respectively. It is sufficient
to prove that these points D, E and F do the job.
Notice that DO + DH = DO + DP = OP = R (where R is the radius of the
circumcircle) and similarly EO + EH = FO + FH = R.
It remains to prove that AD, BE and CF are concurrent. Let AO meet BC and D’
and the circumcircle of ABC at A’. Define points E’ and F’ similarly. Let PO meet
the circumcircle again at P’. It is clear that APA’P’ is a rectangle and A’P // BC.
Hence A’ is the reflection of P in the perpendicular bisector of BC and D’ is the
reflection of D in the same line. Similarly, E’ and F’ are the reflections of E and F
in the perpendicular bisectors of CA and AB respectively. Since AD’, BE’, CF’ are
concurrent at O, AD, BE and CF are concurrent at the isotomic conjugate of O.
Problem 4
Let ABCD be a convex quadrilateral for which there exists a point K on [AB]
satisfying <KDA = <BCD and a point L on AC such that KL and BC are parallel.
Prove that <KDB = <LDC.
Solution
Let M be the point on AD such that LM // CD. Since AK/KB = AL/LC = AM/MD,
we have KM // CD. Hence <KLM = <BCD = <KDA = <KDM, so KLDM is a cyclic
quadrilateral. Hence <MKD = <MLD. Since <MKD = <KDB and <MLD = <LDC we
then get <KDB = <LDC, so we are done.
Problem 5
Let ABC be a triangle with circumcentre O, orthocentre H and circumradius R.
Let D, E and F be the reflections of A, B and C in BC, CA and AB respectively.
Prove that D, E and F are collinear if and only if OH = 2R.
Solution
Construct points A’, B’ and C’ so that A, B and C are the midpoints of B’C’, C’A’
and A’B’ respectively. Let G be the centroid of ABC. Consider the homothety h
with centre G and ratio –1/2. It is obvious that h(A’) = A, h(B’) = B and h(C’) = C.
Let D’, E’ and F’ be the feet of the perpendiculars from O to B’C’, C’A’ and A’B’
respectively. We will prove that h(D) = D’, h(E) = E’ and h(F) = F’.
Let M be the midpoint of BC. Since OD’ is perpendicular to B’C’, OM is
perpendicular to BC and B’C’ // BC, the points O, D’ and M are collinear. AD is
also perpendicular to BC, and hence 2MD’ = AD. Consider GMD’ and GAD.
Since MD’ // AD, <GMD’ = <GAD, and also GM/GA = MD’/AD = 1/2. It follows
that GMD’ and GAD are similar and that D, G, D’ are collinear with D’G/DG = 1/2.
Hence we get h(D) = D’, as claimed, and similarly h(E) = E’ and h(F) = F’. It
follows that D, E and F are collinear if and only if D’, E’ and F’ are collinear. By
Simson’s theorem, D’, E’ and F’ are collinear if and only if O lies on the
circumcircle of A’B’C’. Since A’B’C’ has circumcentre H and circumradius 2R (this
follows for example from the Euler line result), O lies on the circumcircle of A’B’C’
if and only if OH = 2R, and so we are done.
Problem 6
Let ABCD be a convex quadrilateral such that <DAB = <ABC = <BCD. Let ABC
have circumcentre O and orthocentre H. Prove that H, O and D are collinear.
Solution
Let AD and BC intersect at P, and let AB and CD intersect at Q. Let M and N be
the midpoints of AB and BC respectively. One can see that PAB and QBC are
isosceles (and similar) triangles. Hence PM is perpendicular to AB, and QN is
perpendicular to BC. It follows that the intersection of PM and QN is O (the
circumcentre of ABC). Let G be the centroid of ABC: AN and CM intersect at G.
By the Euler line result, O, G and H are collinear. Now notice that O, G and D are
collinear by Pappus’ theorem applied for lines AMQ and CNP (since O is the
intersection of PM and QN, D is the intersection of AP and CQ and G is the
intersection of AN and CM). Hence O, G, H and D are all collinear. Hence result.
Problem 7
Let A, B and C be fixed points on a line (in this order). Let  be a circle passing
through A and C (AC is not a diameter of ). Suppose that the tangents to  at A
and C meet at P, and that  meets the segment PB at Q. Prove that the angle
bisector of <AQC passes through a fixed point (independent of ).
Solution
Let the bisector of <AQC meet AC at R. We will show that R is fixed. It is known
that the line QC is a symmedian of AQC. Hence (using the Steiner theorem, or
otherwise) we get AB/BC = (AQ/QC)2. Also, AR/RC = AQ/QC by the angle
bisector theorem. Hence AB/BC = (AR/RC)2. Since B is fixed, the ratio AB/BC is
fixed, and hence the ratio AR/RC is also constant. Hence R is a fixed point.
Problem 8
Let ABC be an acute triangle. Let D be the foot of the altitude from A to BC, and
let E and F be the feet of the perpendiculars from D to AB and AC respectively.
Let tA be the line passing through A and perpendicular to EF. Define t B and tC
similarly. Prove that tA, tB and tC are concurrent.
Solution
Let S be the intersection of tA and EF. Since AEDF is a cyclic quadrilateral we
have <ADE = <AFE = <AFS. We also get <AED = <ASF = 90. It follows that AED
and ASF are similar. Hence <EAD = <SAF or <EAS = <DAF, so t A and AD are
isogonal conjugates with respect to the angle bisector of <BAC.
Similarly, tB and tC are the isogonal conjugates of the other two altitudes of ABC.
The altitudes of ABC are concurrent at the orthocentre H of ABC, so the lines t A,
tB and tC are concurrent at the isogonal conjugate point of H (which is the
circumcentre O of ABC).
Problem 9
Let ABCD be a convex quadrilateral such that <ADB + <ACB = 90, <DAB is
acute and <DBC + 2<DBA = 180. Prove that (DB + BC)2 = AD2 + AC2.
Solution
Let D’ be the reflection of D in AB. It is easy to see that D’, B and C are collinear
in this order (since <DBC + 2<DBA = 180). Since <ADB + <ACB = 90, we also
get <AD’C + <ACD’ = 90, so CAD’ is a right angle. An application of the
Pythagoras theorem in ACD’ then gives
(DB + BC)2 = (D’B + BC)2 = D’C2 = D’A2 + AC2 = DA2 + AC2.
Problem 10
Let ABC be a triangle such that AC = 2AB and let  be its circumcircle. The
tangents to  at A and C meet at P. Prove that the intersection of BP and the
perpendicular bisector of BC lies on .
Solution
Let BP intersect  again at D. It is sufficient to show that BD = CD. Let M be the
midpoint of AC. It is well-known that BP is a symmedian in ABC. Consequently,
we have <ABM = <DBC. Since ABCD is cyclic, we also have <BAM = <BDC, so
ABM and DBC are similar. Since AM = AC/2 = AB, it follows that BD = CD.
Problem 11
Let ABCD be a quadrilateral such that <BCD = <CDA, and let the angle bisector
of <ABC meet CD at E. Prove that <AEB = 90 if and only if AB = AD + BC.
Solution
Suppose first that <AEB = 90. Let F be the reflection of C in BE (which is a point
on AB, since BE bisects <ABC). We get BC = BF and <BFE = <BCE = <EDA.
Hence A, D, E and F are concyclic. Since <AEB = 90 and <CEB = <BEF we also
get <FEA = <AED, so that <FDA = <FEA = <AED = <ADF, and AF = AD. Finally,
we get AB = AF + FB = AD + BC.
Conversely, if AB = AD + BC, there exists a point F on AB satisfying BF = BC
and AF = AD. Then BCE and BFE are congruent, A, D, E and F are concyclic
and <FDA = <AFD. Hence <FEA = <FDA = <AFD = <AED, which proves that AE
bisects <DEF. It follows immediately that <AEB = 90.
Problem 12
Let ABC be a scalene triangle. The circle with diameter BC and centre O
intersects AB and AC at M and N respectively. Suppose that the angle bisectors
of <BAC and <MON intersect at R. Prove that the circumcircles of BMR and CNR
have a common point which lies on BC.
Solution
Notice that the angle bisector of <MON is also the perpendicular bisector of MN
(since OM = ON), and hence the point R, being the intersection of the angle
bisector of <MAN and the perpendicular bisector of MN, lies on the circumcircle
of triangle AMN. It follows that AMRN is cyclic.
Let S be the intersection of AR and BC. It is obviously sufficient to prove that
quadrilaterals BMRS and CNRS are cyclic.
To prove that BMRS is cyclic, notice that, since AMRN is cyclic, <ARM = <ANM,
and, since BCNM is cyclic, <ANM = <ABC. Hence <ARM = <ABC = <MBS, which
is sufficient to conclude that BMRS is cyclic, and similarly, that CNRS is cyclic.
Problem 13
Let ABC a right-angled triangle with AB = AC, and let D be the point on BC
satisfying BD = 2CD. Let E be the foot of the perpendicular from B to AD.
Determine <CED.
Solution 1
Construct point F such that ABFC is a square and let M be the intersection of the
diagonals of the square. In ACF, CM is a median. Since the point D divides the
segment CM in the ratio 2 : 1, D must be the centroid of ACF. Let BE and AD
meet AC and CF at S and T respectively. Notice that T is the midpoint of CF
(since D is the centroid of ACF). It follows that S is the midpoint of AC (since ABS
and CAT are congruent). Since <SET = <SCT = 90, SETC is cyclic. It then
follows that <CED = <CET = <CST, but since S and T are the midpoints of AC
and CF respectively, <CST = <CAF = 45. Hence <CED = 45.
Solution 2
Let P be the foot of the perpendicular from C to AD. It is easy to see that BED
and CPD are similar, and BE = 2PC. One can also see that ABE and CAP are
congruent, and hence PC = AE = AP – PE = BE – PE = 2PC – PE. It easily
follows that PE = PC, and hence <CED = <CEP = 45.
Solution 3
Use the cosine rule 1000 times.
Problem 14
Let ABC be a triangle with circumcentre O and incentre I. Given that <ABC = 90,
AC = 2 and OI = 2 – 1, prove that the triangle is isosceles.
Solution
Let BI intersect the circumcircle again at M. It is well-known that MA = MC = MI. It
is obvious that OM = 1 (O is, in fact, the midpoint of AC) and that MI = MA = 2.
Hence, by the triangle inequality, MI  OM + OI, so that OI  2 – 1. For equality
to hold, M, O and I must be collinear. In that case, the angle bisector of <ABC
and the perpendicular bisector of AC coincide, so AB = BC.
Problem 15
The angle bisectors of ABC intersect the circumcircle of ABC again at A’, B’, C’.
Let [P] denote the area of a polygon P. Prove that [ABC]  [A’B’C’].
Solution
Let I be the incentre of ABC. An easy angle chase shows that AA’ is
perpendicular to B’C’, and similarly, BB’ and CC’ are also altitudes of A’B’C’.
Consequently I is the orthocentre of A’B’C’ and A, B and C are the reflections of I
in B’C’, C’A’ and A’B’ respectively (because the reflection of the orthocentre of a
triangle in one of the sides lies on the circumcircle). Hence [AB’C’] = [IB’C’], and
similarly, [BC’A’] = [IC’A’] and [CA’B’] = [IA’B’]. Thus, [AB’CA’BC’] = 2[A’B’C’].
Let H be the orthocentre of ABC. Let A”, B”, C” be the reflections of H in BC, CA
and AB respectively (as before, these points lie on the circumcircle). It is easy to
see that [AB’C]  [AB”C] (this follows from AB’ AB”). Similarly [BC’A]  [BC”A]
and [CA’B]  [CA”B]. Hence [AB’C] + [BC’A] + [CA’B]  [ABC]. From this it follows
that [AB’CA’BC’] 2[ABC]. Since [AB’CA’BC’] = 2[A’B’C’], we are done.
Problem 16
Three congruent circles have a common point O. The circles (taken in pairs) also
meet at A, B and C. Let A’B’C’ be the triangle containing these three circles, such
that every side of the triangle is tangent to exactly two of the given circles. Prove
that [A’B’C’] 9[ABC] (where the square brackets denote the area of a triangle).
Solution
Let O1, O2 and O3 be the centres of the given circles 1, 2 and 3 (such that 1
and 2 intersect at C and O, 2 and 3 intersect at A and O, and 3 and 1 intersect
at B and O) and let R be the radius of the given circles. It is easy to see that
AO2OO3 is a rhombus (AO2 = AO3 = OO2 = OO3 = R), and similarly for BO1OO3
and CO1OO2. Hence AO2 and O3O are parallel and equal in length, and similarly
for BO1 and O3O. Hence AO2 and BO1 are parallel and equal in length. It follows
that ABO1O2 is a parallelogram and AB = O1O2.
Similarly one can show that BC = O2O3 and CA = O3O1. Hence ABC and O1O2O3
are congruent and it is sufficient to show that [A’B’C’]  9[O1O2O3]. Since the
corresponding sides of these two triangles are parallel, they are homothetic, and
hence the lines AO1, BO2 and CO3 are concurrent at a point I, which is the
incentre of both A’B’C’ and O1O2O3. Let r be the inradius of O1O2O3. It is easy to
see that point I is a distance r from the sides of O 1O2O3, and a distance R + r
from the sides of A’B’C’. However, O1O2O3 has inradius r and circumradius R. By
the Euler inequality we get R 2r, so R + r 3r. Hence the inradius of A’B’C’ is at
least 3r. It follows that the ratio of the homothety mentioned above is at least 3.
Hence the ratio of the areas of triangles O1O2O3 and A’B’C’ is at least 9.
Problem 17
Let ABC be a triangle and let M be the midpoint of BC. Let the line AM meet the
circumcircle of ABC again at A1 and let A2 be the reflection of A1 in M. Define the
points B2 and C2 similarly. Prove that the orthocentre of ABC lies on the
circumcircle of A2B2C2.
Solution
A stronger result is true. In fact, the points A2, B2 and C2.lie on the circle with
diameter GH, where G and H are the centroid and orthocentre of ABC
respectively. It is therefore sufficient to prove that <GA2H = 90. Let AH meet BC
at P. Since A2 lies on AG, and <APM = 90, it is sufficient to prove that PMA2H is
cyclic, which is equivalent to AA2.AM = AH.AP (*). Hence we focus on proving (*).
Notice that AA2.AM = (AM – A1M).AM = AM2 – A1M.AM = AM2 – BM.CM (by
power of a point) which equals ¼(2b2 + 2c2 – a2) – ¼a2 (by the formula for the
length of a median), which is equal to ½(b2 + c2 – a2) = bc.cos A (by the cosine
rule). (As usual we write a = BC, b = CA, c = AB). Let BH meet AC at Q. It is easy
to see that PHQC is cyclic and hence AH.AP = AQ.AC = (c.cos A) .b = bc.cos A. It
follows that AH.AP = AA2.AM and we are done.
Problem 18
Let ABC be an acute-angled triangle and let P and Q be points inside ABC such
that <ACP = <BCQ and <CAP = <BAQ. Let D, E and F be the orthogonal
projections of P onto BC, CA and AB respectively. Prove that <DEF = 90 if and
only if Q is the orthocentre of BDF.
Solution
It is obvious that P and Q are isogonal conjugates with respect to ABC. Let D’, E’
and F’ be the projections of Q on BC, CA and AB respectively. It is a well-known
property of isogonal conjugates that D, E, F, D’, E’ and F’ are concyclic.
Suppose that Q is the orthocentre of BDF. It follows that FQ and BD are
orthogonal. Since D’Q is perpendicular to BD, F, Q and D’ are collinear. It follows
that <DD’F = 90. Since E lies on the circumcircle of DD’F, we get <DEF = 90.
Suppose that <DEF = 90, then <DD’F = <DF’F = 90. As before, we see that F, Q
and D’ are collinear (and so are D, Q and F’) and that FD’ and DF’ are altitudes
of BDF. Hence their intersection, which is Q, is the orthocentre of BDF.
Problem 19
Let ABC be a triangle with <C = 90 and let the altitude CH meet the angle
bisectors AM and BN at P and Q respectively. Prove that the line connecting the
midpoints of PM and QN is parallel to AB.
Solution
Let S be the midpoint of PM. An easy angle chase reveals that PCM is isosceles.
It follows that <CSA = 90. Since <CHA = 90, AHSC is cyclic. Some more angle
chasing gives <HCS = <HAS = <CAS = <CHS. Hence S lies on the perpendicular
bisector of CH. Similarly one can show that the midpoint of QN belongs to the
perpendicular bisector of CH as well, and hence the line connecting the
midpoints of PM and QN is the perpendicular bisector of CH. It is now obvious
that this line is parallel to AB.
Problem 20
Let ABC be a triangle with semiperimeter s and inradius r, such that
(cot A/2)2 + (2 cot B/2) 2 + (3 cot C/2)2 = (6s/7r)2.
Prove ABC is similar to a triangle T, the sides of which have coprime positive
integers as lengths, and find these integers.
Solution
It is obvious that cot A/2 = (s –a)/r. Together with the similar relations for B and C
we get that
49[(s – a)2 + 4(s – b)2 + 9(s – c)2] = 36s2.
From the Cauchy–Schwarz inequality we get
[36 + 9 + 4][(s – a)2 + 4(s – b)2 + 9(s – c)2] 36[(s – a) + (s – b) + (s – c)]2 = 36s2.
Since equality holds, we must have
(s – a)/6 = 2(s – b)/3 = 3(s – c)/2.
Hence there exists a real number t such that
s – a = 36t, s – b = 9t and s – c = 4t.
Adding these equations we get s = 49t and hence
a = 13t, b = 40t and c = 45t.
We conclude that all triangles ABC, satisfying the given equality, are similar to
the triangle T with sides 13, 40 and 45.
METRIC RELATIONS IN A TRIANGLE
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Area of a triangle
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Distances
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Lengths
S = bh/2 = (ab sin C)/2
S = abc/(4R) = 2R2 sin A sin B sin C
S2 = s(s – a)(s – b)(s – c)
S = sr = (s – a)r[a] = (s – b)r[b] = (s – c)r[c]
OI2 = R2 – 2Rr
OH2 = 9R2 – (a2 + b2 + c2)
OH = 3OG = 2OW
IW = R/2 – r (Feuerbach)
4m[a]2 = 2b2 + 2c2 – a2 (and G trisects the medians)
l[a] = (2bc cos A/2)/(b + c)
l[a]2 = AL2 = AB.AC – LB.LC = 4bcs(s – a)/(b + c)2
Stewart’s theorem
Angle bisector theorem
Tangents to the incircle/excircles (Ravi substitution)
Ceva (usual form, trigonometric form) and Menelaos
cos A + cos B + cos C = 1 + r/R, r = 4R sin A/2 sin B/2 sin C/2
ISOGONAL CONJUGATES
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Definition – existence
The pedal triangles of two isogonal conjugate points have the same circumcircle.
Points O and H are isogonal conjugates
Steiner: if D, E are points on BC such that <BAD = <CAE then (BD BE)/(CD CE) = (AB/AC) 2.
Symmedians, and the Lemoine point K
The tangents to the circumcircle of a triangle at two of the vertices intersect on the symmedian
passing through the third vertex of the triangle.
Existence and definition of isotomic conjugates
NINE POINT–CIRCLE
MORE SPECIAL POINTS
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Excentres of a triangle
Existence of Brocard points, the Brocard angle (cot W = cot A + cot B + cot C)
Fermat points
Gergonne and Nagel points (N) (G, I, N are collinear and NG = 2IG)
The intersection of an angle bisector (interior or exterior) and the perpendicular bisector of the
opposite side lies on the circumcircle.
The reflection of H in one of the sides lies on the circumcircle.
H is the incentre of the orthic triangle.
If M is the second intersection of AI with the circumcircle then MB = MC = MI = MI[a]
GOOD TO KNOW
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Simson line (existence; the Simson line of P bisects PH; …)
The Feuerbach theorem
The Taylor circle
The Poncelet porism
Trigonometric substitution to prove inequalities
Trigonometric inequalities and convexity
Weitzenböck inequality, Erdös–Mordell and other triangle inequalities