Geometric inequalities marathon the first 100 problems and solutions

• Let P be any point inside 4ABC, and let Q be an arbitrary point in the plane.. • Let the internal angle bisectors issued from A, B, C be la, lb, lcrespectively, which meet at the incen

Geometric Inequalities Marathon1

The First 100 Problems and Solutions

Contributors Typesetting and Editing Members of Mathlinks Samer Seraj (BigSams)

On Wednesday, April 20, 2011, at 8:00 PM, I was inspired by the existing Mathlinks marathons to create

a marathon on Geometric Inequalities - the fusion of the beautiful worlds of Geometry and MultivariableInequalities It was the result of the need for expository material on GI techniques, such as the crucial Rrs,which were well-explored by only a small fraction of the community Four months later, the thread has over

100 problems with full solutions, and not a single pending problem On Friday, August 26, 2011, at 5:30

PM, I locked the thread indefinitely with the following post:

The reason is that most of the known techniques have been displayed, which was my goal Recent problemsare tending to to be similar to old ones or they require methods that few are capable of utilizing at this time.Until the community is ready for a new wave of more diffcult GI, and until more of these new generation GIhave been distributed to the public (through journals, articles, books, internet, etc.), this topic will remainlocked

This collection is a tribute to our hard work over the last few months, but, more importantly, it is a source

of creative problems for future students of GI My own abilities have increased at least several fold since theexposure to the ideas behind these problems, and all those who strive to find proofs independently will findthemselves ready to tackle nearly any geometric inequality on an olympiad or competition

The following document is dedicated to my friends Constantin Mateescu and R´eda Afare (Thalesmaster),and the pioneers Panagiote Ligouras and Virgil Nicula, all four of whom have contributed much to theevolution of GI through the collection and creation of GI on Mathlinks

The file may be distributed physically or electronically, in whole or in part, but for and only for commercial purposes References to problems or solutions should credit the corresponding authors

non-To report errors, a Mathlinks PM can be sent BigSams, or an email to samer_seraj@hotmail.com

Samer SerajSeptember 4, 2011

1 The original thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=403006/

2 Notation

For a 4ABC:

• Let AB = c, BC = a, CA = b be the sides of 4ABC

• Let A = m (∠BAC), B = m (∠ABC), C = m (∠BCA) be measures of the angles of 4ABC

• Let ∆ be the area of 4ABC

• Let P be any point inside 4ABC, and let Q be an arbitrary point in the plane Let the ceviansthrough P and A, B, C intersect a, b, c at Pa, Pb, Pc respectively

• Let the distance from P to a, b, c, extended if necessary, be da, db, dc respectively

• Let arbitrary cevians issued from A, B, C be d, e, f respectively

• Let the semiperimeter, inradius, and circumradius be s, r, R respectively

• Let the heights issued from A, B, C be ha, hb, hc respectively, which meet at the orthocenter H

• Let the feet of the perpendiculars from H to BC, CA, AB be Ha, Hb, Hc respectively

• Let the medians issued from A, B, C be ma, mb, mc respectively, which meet at the centroid G

• Let the midpoints of A, B, C be Ma, Mb, Mc respectively

• Let the internal angle bisectors issued from A, B, C be la, lb, lcrespectively, which meet at the incenter

I, and intersect their corresponding opposite sides at La, Lb, Lc respectively

• Let the feet of the perpendiculars from I to BC, CA, AB be Γa, Γb, Γc respectively

• Let the centers of the excircles tangent to BC, CA, AB be Ia, Ib, Ic respectively, and the excircles betangent to BC, CA, AB at Ea, Eb, Ec

• Let the radii of the excircles tangent to BC, CA, AB be ra, rb, rc respectively

• Let the symmedians issued from A, B, C be sa, sb, sc respectively, which meet at the Lemoine Point

S, and intersect their corresponding opposite sides at Sa, Sb, Sc respectively

• Let Γ be the Gergonne Point, and the Gergonne cevians through A, B, C be ga, gb, gc respectively

• Let N be the Nagel Point, and the Nagel cevians through A, B, C be na, nb, nc respectively

Let [X] denote the area of polygon X

AllXandYsymbols without indices are cyclic

 denotes the end of a proof, either for a lemma or the original problem

3 Problems

1 For 4ABC, prove that R ≥ 2r (Euler’s Inequality)

2 For 4ABC, prove thatXAB >XP A

3 For 4ABC, prove that ab + bc + ca

R.

6 For 4ABC, prove thatp12(R2− Rr + r2) ≥XAI ≥ 6r

7 A circle with center I is inscribed inside quadrilateral ABCD Prove that XAB ≥√

2 ·XAI

8 For 4ABC, prove that 9R2≥Xa2 (Leibniz’s Inequality)

9 Prove that for any non-degenerate quadrilateral with sides a, b, c, d, it is true that a

2+ b2+ c2

d2 ≥ 1

3.

10 For 4ABC, prove that 3 ·Xa sin A ≥Xa·Xsin A≥ 3(a sin C + b sin B + c sin A)

11 For acute 4ABC, prove thatXcot3A + 6 ·Ycot A ≥Xcot A

12 For 4ABC, prove that

XcosA2



·

XcscA2

4 · a ≥ ha+ max{b, c}

18 For 4ABC, prove that s ·Xha≥ 9∆ with equality holding if and only if 4ABC is equilateral

19 Prove that the semiperimeter of a triangle is greater than or equal to the perimeter of its orthic triangle

20 Prove that of all triangles with same base and area, the isosceles triangle has the least perimeter

21 ABCD is a convex quadrilateral with area 1 Prove that AC + BD +XAB ≥ 4 + 2√

2

22 For 4ABC, prove thatXcscA

2 ≥ 4

rR

r.

23 For 4ABC, prove thatXsin2A

2 ≥3

4.

24 Of all triangles with a fixed perimeter, dtermine the triangle with the greatest area.

25 Let ABCD be a parallelogram such that ∠A ≤ 90 Altitudes from A meet BC, CD at E, F respectively.Let r be the inradius of 4CEF Prove that AC ≥ 4r Determine when equality holds

26 For 4ABC, the feet of the altitudes from B, C to AC, AB respectively, are E, D respectively Let thefeet of the altitudes from D, E to BC be G, H respectively Prove that DG + EH ≤ BC Determinewhen equality holds

27 For 4ABC, a line l intersects AB, CA at M, N respectively K is a point inside 4ABC such that itlies on l Prove that ∆ ≥ 8 ·p[BM K] + [CN K]

28 For 4ABC, prove that

r15

4 +

Xcos(A − B) ≥Xsin A

29 Let pI be the perimeter of the Intouch/Contact Triangle of 4ABC Prove that pI ≥ 6r s

4R

1

30 In addition to 4ABC, let 4A0B0C0 be an arbitrary triangle Prove that 1 + R

32 For 4ABC, prove thatXha≥ 9r

33 For 4ABC, prove thatXcosA − B

37 For 4ABC, prove thatXa2b(a − b) ≥ 0

38 Show that for all 0 < a, b < π

2 we have

sin3asin b +

cos3acos b ≥ sec(a − b)

39 For all parallelograms with a given perimeter, explicitly define those with the maximum area

40 Show that the sum of the lengths of the diagonals of a parallelogram is less than or equal to theperimeter of the parallelogram

41 For 4ABC, the parallels through P to AB, BC, CA meet BC, CA, AB respectively, at L, M, Nrespectively Prove that 1

43 For 4ABC, it is true that BC = CA and BC ⊥ CA P is a point on AB, and Q, R are the feet

of the perpendiculars from P to BC, CA respectively Prove that regardless of the location of P ,max{[AP R], [BP Q], [P QCR]} ≥ 4

9∆ (Generalization of Canada 1969)

44 For 4ABC, prove thatXa2+√abc

3R≥ 4(abc)2

45 For 4ABC, prove that 6R ≥Xa

2+ b2

m2 c

2 Determine when equality holds.

47 For 4ABC, prove that s√

49 For 4ABC, prove thatXa2≥ 4√3∆ · max ma

50 A1A2B1B2C1C2is a hexagon with A1B2∩C1A2= A, B1C2∩A1B2= B, C1A2∩B1C2= C and AA1=

AA2= BC, BB1= BB2= CA, CC1= CC2= AB Prove that [A1A2B1B2C1C2] ≥ 13 · [ABC]

51 For 4ABC, let r1, r2 denote the inradii of 4ABMa, 4ACMa Prove that 1



52 For 4ABC, prove thatXcsc2A

59 For 4ABC, prove that 3 ·Xd2a ≥XP A2sin2A

60 For 4ABC, if CA + AB > 2 · BC, then prove that ∠ABC + ∠ACB > ∠BAC (Euclid Contest 2010)

61 For 4ABC, prove that p7 · P a2+ 2 ·P ab

2 ≥Xma (Dorin Andrica)

62 For 4ABC, prove thatXcosA

2 ≥

√2

2 +

r1

2 + (3

√

3 − 2√2) · s2R.

63 For 4ABC, prove that pP a2b2



64 For 4ABC, prove the following and determine which is stronger: (Samer Seraj)

(a) ∆ ≥ r ·

r1

3 ·Xmamb+1

2 ·Xab

(b) ∆ ≥ r ·

r2

66 For 4ABC, prove that s2≥Xl2

67 ABCD is a quadrilateral inscribed in a circle with center O P is the intersection of its diagonals and

R is the intersection of the segments joining the midpoints of the opposite sides Prove that OP ≥ OR

68 For 4ABC, prove that 5

S2, where S1= [AM L] and S2= [BN L]

70 For 4ABC, prove thatX(b + c)P A ≥ 8∆

71 Right 4ABC has hypotenuse AB The arbitrary point P is on segment CA, but different from thevertices A, C Prove that AB − BP

74 Let P be a point inside a regular n-gon, with side length s, situated at the distances x1, x2, , xn

from the sides, which are extended if necessary Prove that

76 For 4ABC, find the smallest constant k such that it always holds that k ·Xab >Xa2

77 For 4ABC, prove thatXabdadb ≤4

3∆

2, and determine when equality holds

78 For 4ABC, let AI, BI, CI extended intersect the circumcircle of 4ABC again at X, Y, Z respectively.Prove thatYIX ≥YAI

79 Let {a, b, c} ⊂ R+ such thatXa

2+ b2− c2

ab > 2 Prove that a, b, c are sides of triangle.

80 Let AP be the internal angle bisector of ∠BAC and suppose Q is the point on segment BC such that

BQ = P C Prove that AQ ≥ AP

81 For 4ABC, prove that ∆2≥ r ·Yla

82 For 4ABC, prove that 9R ≥ a ≥ 18Rr.

83 For 4ABC, prove thatX(P A · P B · c) ≥ abc

84 For 4ABC, prove that 8R3≥YIEa

85 For 4ABC, prove that

XsinA2



·

XtanA2



≥3

√3

2 .

86 For 4ABC, prove thatXa3+ 6abc ≥Xa·Xab>Xa3+ 5abc

87 D and E are points on congruent sides AB and AC, respectively, of isosceles 4ABC such that AD =

CE Prove that 2EF ≥ BC Determine when the equality holds

88 For 4ABC, prove thatXAH

90 For 4ABC, determine minnXQA2o

91 For 4ABC, prove that

q

8 ·P a2+ 4√

3∆

3 ≥XGA

92 For 4ABC, prove that a2+ b2+ R2≥ c2, and determine when equality holds

93 For 4ABC, prove thatXab· (s2+ r2) ≥ 4abcs + 36R2r2

94 For 4ABC, prove that P a2

A 2

1 − sinA2 .

96 In 4ABC, the internal angle bisectors of angles A, B, C intersect the circumcircle of 4ABC again at

X, Y, Z respectively Prove that AX + BY + CZ > a + b + c (Australia 1982)

97 An arbitrary line ` through the incenter I of 4ABC cuts AB and AC at M and N Show that

100 Let m ∈ R+ and φ ∈ (0, π) For 4ABC, prove that

(1 − m cos φ) · a2+ m (m − cos φ) · b2+ m cos φ · c2 ≥ 4m sin φ · ∆

Equality holds if and only if m = a

b and φ = C.

For m = 1 and φ = 60◦obtain Weitzenb¨ock’s Inequality (Virgil Nicula)

R = cos A + cos B + cos C.

So it is sufficient to prove that 2 cos A + 2 cos B + 2 cos C ≤ 3

It’s easy to verify that this inequality is equivalent to (1 − (cos B + cos C))2+ (sin B − sin C)2 ≥ 0,which is true by the Trivial Inequality 

1 Author: BigSams

For positive reals x, y, z it is true that (x + y)(y + z)(z + x) ≥ 8xyz by AM-GM:Yx + y

2 ≥Y√xy =xyz By Ravi Substitution, let a, b, c be side lengths of a triangle such that a = x+y, b = y+z, c = z +x.The inequality becomes abc ≥ 8(s − a)(s − b)(s − c) By Heron’s Theorem, the inequality issabc ≥ 8S2 ⇐⇒ abc

4∆ ≥2∆

s Using the fact that ∆ =

abc4R = sr, R ≥ 2r 

This lemma implies that

3 Author: Goutham

Let x = s − a, y = s − b, z = s − c all greater than 0, and s = x + y + z, ∆2= xyzs

We have Xx2≥Xxy =⇒ X(x2+ 3xy) ≥ 4Xxy

ButX(x2+ 3xy) =X(x + y)(x + z) =Xab

And so, P ab

4xyzs ≥P xy

xyzs Therefore, we have

P ab4∆ ≥X 1

s(s − a) 

4 Author: Mateescu Constantin

Using the well-known formula for area i.e ∆ = sr, the inequality rewrites as: s√

cosA

2 =

r(s − b)(s − c)bc

sinB

2 =

rs(s − a)bc

R = abc4∆

Using Ravi’s substitution:

, the inequality is equivalent to: (2x + y + z)2 ≥ 8x(y + z),

which is true according to AM-GM Inequality 

a + b + c from Heron’s Formula.

Hence, it suffices to proveX

r

a2+ abc

a + b + c ≥ 6

rabc

a + b + c ⇐⇒ X pa(a + b)(a + c) ≥ 6√

abc.However, using AM-GM Inequality twice givesX pa(a + b)(a + c) ≥ 3p6 abc(a + b)2(b + c)2(c + a)2≥

abc · 64(abc)2≥ 6√abc, as desired 

Left Side

Lemma AI + BI + CI ≤ 2(R + r) (Author: Mateescu Constantin)

Proof Show easily that AI = bc

s2 ·X √bc ·ps(s − a)

2 C.B.S.

≤ 1

s2 · (ab + bc + ca) ·Xs(s − a) = ab + bc + ca ≤ 4(R + r)2.The last inequality is due to Gerretsen i.e s2 ≤ 4R2+ 4Rr + 3r2 Therefore, we have shown that:

AI + BI + CI ≤ 2(R + r) 

As a direct consequence of the lemma, it suffices to prove 2(R + r) ≤ 2p3(R2− Rr + r2) ⇐⇒2R2− 5Rr + 2r2≥ 0

However, ths is equivalent to (2R − r)(R − 2r) ≥ 0, which is indeed true 

For both inequalities, equality holds for 4ABC equilateral

3(AI2+ BI2+ CI2) ≥ (AI + BI + CI)2 ⇐⇒ p3(s2+ r2− 8Rr) ≥ AI + BI + CI

So it suffices to show that

Applying this cyclically to BI and CI, the left hand side is equivalent to

6r ≤ r

sinA2 +

rsinB2 +

rsinC2 ⇐⇒ 2 ≤

1sinA2 +

1sinB2 +

1sinC23

which follows from Jensen’s Inequality since cscx

BC =

rsin ∠BIC.

Combining, AI · DI

AD =

rsin ∠AID =

rsin ∠BIC =

BI · CIBC

2 =

r(s − b)(s − c)bc

XcotA

2 =

sr

,

the inequality is equivalent to X

rs(s − a)bc

!

· X

s

bc(s − b)(s − c)

!

· X

s

bc(s − a)(s − b)

!

≥ X 4

ss(s − a)(s − b)(s − c)

!2

=

X

√

s − a

√r

2 cot

C

2 ≥ 3√3Which is true according to AM-GM and Mitrinovic’s Inequality:

Let sa+ sb+ sc= n Then the number of lines is 2010 ≤ n

3 + n + 1 Thus, n ≥ 77 Indeed, plugging

in sa= sb= 26, sc = 25 works, so our answer is 77 

14 Author: mcrasher

SinceXsin A = s

R, it suffices to show that

Xsin A ≤ 3

√3

2 , which is true by Jensen’s Inequality 

⇐⇒ 4Xcos A + 2Ycos A ≥ 4 + 3Xcos A · cos B

⇐⇒ X(2 − cos A) · (2 − cos B) ≥ 2Y(2 − cos A) ⇐⇒ X 1

2 − cos A ≥ 2 Right Side

16 Author: Mateescu Constantin

Using the relation: YsinA

2 =

r4R, the inequality reduces to 2r ≤ R, which is due to Euler 

17 Author: ftong

Let θ = ∠C, and assume without loss of generality that 0◦≤ θ ≤ 45◦, or equivalently, b ≥ c

Now hA= b sin θ, and a = b

cos θ, so we wish to prove that cos θ(sin θ + 1) ≤

3√34

It seems now that we must use resort calculus to find the maximum of f (θ) = cos θ(sin θ + 1) over thegiven interval

Taking the derivative, we have f0(θ) = 1 − sin θ − 2 sin2θ, so that f takes extremal values at sin θ = 1

2and sin θ = −1

We discard the latter because sin θ is positive in our interval, so the maximum occurs at θ = π

6, atwhich point f (θ) = 3

√3

4 as desired 

of M0M00with AB and AC.

Triangles AM M0 and AM M00 are isosceles, hence ∠M0AM00 = 2∠A = const, thus M0M00, i.e therequired perimeter, is minimal when AM0= AM00= AM is minimal, which is obviously attained if M

is the foot of the perpendicular from A to BC (∗)

Now we note that the orthic triangle has the property that, when one of its vertices is reflected aboutthe remaining two sides of the initial triangle, the two reflections are collinear with the two remainingvertices of the orthic triangle - which is easy to prove: ∠M P N = π − 2∠C ∧ ∠M P B = ∠C

Therefore the triangle obtained by the argument (∗) is indeed the orthic triangle, as claimed 

Using the lemma, the orthic triangle does not have a greater perimeter than the medial triangle, whichhas a perimeter equal to the semiperimeter of the original triangle 

20 Author: BigSams

Let 4ABC be an arbitrary triangle with a constant area ∆ and constant base a Since the areaand a base are constant, then the height ha with foot on a is also constant since it can be expressed interms of constants: a · ha

2 = ∆ =⇒ ha=

2X

a .Let AB = c, CA = b Let ha intersect BC = a (extended if necessary) at P Let P C = a1, P B = a2.Note that the perimeter is minimized when b + c is minimized, since a is a constant

21 Author: r1234

Let O be the point of intersection of the two diagonals Now [ABCD] = 1

2 · AC · BD · sin ∠ACD So[ABCD] ≤ AC · BD

Now again [ABCD] = 1

2 · AB · BC · sin B ≤ 1

2· AB · BC similarly we get [ABCD] ≤ 1

2· CD · DA onthe other hand we get other two inequalities [ABCD] ≤ 1

2 · AB · CD and [ABCD] ≤ 1

2· BC · AD.Adding the last four inequalities we get(AB + CD)(BC + DA) ≥ 4 This implies that (AB + BC +

CD + DA)2≥ 4(AB + CD)(BC + AD) ≥ 16 or AB + BC + CD + DA ≥ 4

On the other hand we get AC · BD ≥ 2 or (AC + BD)2≥ 8 or AC + BD ≥ 2√2

bc =

r yz(x + y)(x + z).

So the inequality is equivalent toX

sinB

2 · sinC2



≥ 2 ·

rYsinA

3, which occurs when a = b = c 

25 Author: math explorer

Since ∠AEC and ∠AF C are both right, the points AECF are cyclic and AC is a diameter Therefore

AC is twice the circumradius of 4CEF

By Euler’s inequality of a triangle in 4CEF the circumradius is at least twice the inradius, so

AC ≥ 4r1, with equality iff 4CEF is equilateral iff ∠C = 60◦ and A lies on the angle bisector of

∠ECF iff ABCD is a rhombus and ∠C = 60◦ 

26 Author: truongtansang89

Note that DG · BC = DB · DC ⇒ DG · BC = BC2· cos B sin B ⇒ DG = 1

2BC sin 2B.

Similarly, EH = 1

2BC sin 2C ⇒ DG + EH = BC · sin A · cos(B − C) ≤ BC.

Hence, equality holds when A = π

(q + 1)(r + 1)· [ABC](∗) Moreover, we can write the following relations:

qrt, which is clearly true by AM-GM inequality Equality occurs if and only

2

≥sR

RX

cos A = 1 + r

R, the above inequality becomes

(P cos A)2+ (P sin A)2

2Note that sin2A + cos2A = 1 =⇒ Xsin2A +Xcos2A = 3

Note that cos(A − B) = cos A cos B + sin A sin B

=⇒ 2Xcos(A − B) = 2X(cos A cos B) + 2X(sin A sin B)

Adding these gives 3 + 2Xcos(A − B)

=Xsin2A +Xcos2A + 2X(cos A cos B) + 2X(sin A sin B)

4+

(P cos A)2+ (P sin A)2

2becomes ⇐⇒ Xsin A ≤

r15

4 +

Xcos(A − B), as desired 

YsinA

2 =

r4R

By AM-GM, s =Xa0 ≥ 3 ·Ya0

1

= 3 ·

Y2(s − a) sinA

2

1

= 6r s4R

1 

30 Author: Thalesmaster

Let x, y, z be positive real numbers

Klamkin’s Inequality states that x sin A0+ y sin B0+ z sin C0≤ 1

2(xy + yz + zx)

r x + y + zxyz .For x = 1

sin A, y =

1sin B, z =

1sin C, we obtain

Xsin A0sin A ≤ 1

2

P sin A

Q sin A

qXsin B sin C

By Euler’s Inequality, R ≥ 2r ⇐⇒ (2R + r)(R − 2r) ≥ 0 ⇐⇒ 16Rr − 5r ≥ 22Rr − 4R − r

By Gerretsen’s Inequality, s2≥ 16Rr − 5r2

Combining, s2≥ 22Rr − 4R2− r2 ⇐⇒ 3 + 1 +

r R

R

Xcos A = 1 + r

R

YsinA

2 =

r4R

⇐⇒ 24 ·YsinA

2 ≤3 + (P cos A)2+ (P sin A)2

4

= 1

4 ·3 +Xcos2A + 2Xcos A · cos B +Xsin2A + 2Xsin A · sin B

Note the identities:

a +1

b +1 c

2 cos

C

2 + 12 ·

YsinA2

≥ 12 ·



X

sinA2



·

XsinB

2 sin

C2

+ 3 ·XsinA

Z = π − C2, and the identities:

RX

cos Y cos Z =s

2+ r2− 4R2

4R2

Ycos X = s

2− (2R + r)2

4R2

Xsin Y sin Z = s

2+ r2+ 4Rr4R2

where s, R, r respectively denote the semiperimeter, circumradius and inradius of 4XY Z

We find that the previous inequality is equivalent to:

4 ·Xcos X

3

+Xcos Y cos Z +Xsin Y sin Z + 12 ·Ycos X

≥ 12 ·Xcos X·Xcos Y · cos Z+ 3 ·Xcos X

2 the inequality reduces toY

cosB − C

2 ≥ 8 ·YsinA

2.Using cosB − C

2 =

(ra+ r)4R sinA2 and r = 4R

YsinA

2 the inequality reduces to

Y(ra+ r) ≥ 32Rr2

(b + c) ≥ 8abc which trivially comes from AM-GM inequality 

Now let BD : DC = x : y, CE : EA = y : z and AF : F B = z : x Now using Menelaus’s theorem

we get OD : OA = (x + y + z) : (y + z) and similar for others Hence the inequality reduces to(x + y + z) ·

3 ≤ 36Rrs Since 4∆R = 4Rrs = abc, we have 4(2s)∆√

Use Ravi Substitution a = x + y, b = x + z, c = y + z

Then it becomes X(x2+ y2+ 2xy)(xy + yz − xz − z2) ≥ 0

After expanding and simplifying Xx3y − 2xyzXx ≥ 0 ⇐⇒ Xx3y ≥ 2xyzXx

We have Xa2b(a − b) ≥ 0 ⇐⇒ Xc(a + b − c)(a − b)(a − c) ≥ 0, which is true according to thelemma, since c(a + b − c), b(c + a − b) and a(b + c − a) are the side lengths of a triangle 

38 Author: BigSams

By CS, (sin a · sin b + cos a · cos b) · sin3

asin b +

cos3acos b

sin a · sin b + cos a · cos b = sec(a − b)