Geometric Inequalities 1∗

Arkadii Slinko

Contents

1 Basic ideas 11.1 Elementary tools and their usage . . . . . . . . . . 11.2 Three math olympiad problems . . . . . . . . . . . 41.3 Ptolemy’s inequality . . . . . . . . . . . . . . . . . 6

2 Problems to solve 9

3 Solutions 11

This assignment presents the main tools used to prove geometric inequal-ities. We do not go far into the geometry of triangle here because this willbe the subject of a separate assignment.

We assume that our triangles, quadrilaterals and polygons are non-degenerate,which means that no three vertices are on the same line. When we say thata point is “inside” the polygon will always mean “strictly inside,” that is noton the boundary.

1 Basic ideas

1.1 Elementary tools and their usage

The first four tools will be considered as familiar to the reader.

Tool 1 (Triangle Inequality) In an arbitrary triangle the combined lengthof any two sides is greater than the length of the remaining side, i.e., in atriangle ABC we always have AB + BC > AC.

Tool 2 The area of a triangle ABC is not greater than 12AB · AC.

∗Copyright c©Arkadii Slinko. All rights reserved.

1

Tool 3 In a triangle ABC the inequality AC < BC holds if and only if� ABC < � BAC.

Tool 4 The length of the orthogonal projection of a segment s onto a line �is not greater than the length of s.

Now we will consider very useful but slightly less obvious tools. We willprove them.

Tool 5 Let ABCD be a convex quadrilateral. Then AB +CD < AC +BD,i.e., the combined length of the diagonals of a convex quadrilateral is greaterthan the combined length of any pair of its opposite sides.

Proof: Since the quadrilateral is convex, the two diagonals intersect. Let Obe the point of their intersection.

B

D

C

A

O

Then, applying the Triangle Inequality twice, we get

AB < AO + BO, CD < OC + OD.

Adding these two inequalities together, we get

AB + CD < AO + BO + OC + OD = AC + BD.

The next tool I found to be especially useful.

Tool 6 Let O be a point inside a triangle ABC (can lie on a side but mustbe different from any of the vertices). Then

AO + OC < AB + BC.

Proof: Let P be the intersection of AO with BC.

2

C

B

A

PO

We apply the Triangle Inequality twice as follows.

AO + OC < AO + OP + PC = AP + PC < AB + BP + PC = AB + AC.

We are ready for the first nontrivial (or not completely trivial) result:

Problem 1 Let O be a point inside a triangle ABC and p be a semiperimeterof ABC. Then

p < AO + BO + CO < 2p.

Proof: There are two inequalities here, and for the first one we need justthe Triangle Inequality.

C

B

A

O

We can write:

AB < OA + OB,

BC < OB + OC,

CA < OC + OA.

Adding them all together, we get 2p < 2(AO + BO + CO), which is the firstinequality that we were going to prove.

For the second inequality we need Tool 6. We get:

OA + OB < CA + CB,

OB + OC < AB + AC,

OC + OA < BC + BA.

3

Adding them all together, we get 2(AO + BO + CO) < 4p, which is thesecond inequality that we were after.

Important: The idea used here is an important idea of “symmetrizing”inequalities. Most symmetric inequalities are proved this way.

The next tool is also very important. It is used when a certain inequalitymust be proved for all values a, b, c, which are known to be the lengths of thesides of a triangle. This tool allows to replace a, b, c, which are not arbitrary,by arbitrary positive x, y, z.

Tool 7 Numbers a, b, c are the lengths of the sides of a triangle if and onlyif there exist positive numbers x, y, z such that a = x + y, b = y + z andc = z + x.

Proof: If a, b, c are the sides of a triangle, then

x =a + c − b

2> 0,

y =a + b − c

2> 0,

z =b + c − a

2> 0.

This numbers clearly satisfy a = x + y, b = y + z and c = z + x. Onthe other hand, if such x, y, z exist, then all triangle inequalities for thenumbers a = x + y, b = y + z and c = z + x are satisfied. For example,a + b = (x + y) + (y + z) = (x + z) + 2y = c + 2y > c as y > 0.

Problem 2 (IMO, 1983) let a, b, c be the lengths of the sides of a triangle.Prove that

a2b(a − b) + b2c(b − c) + c2a(c − a) ≥ 0.

Determine when equality occurs.

Solution: This was the most difficult problem of that IMO but Tool 7makes it trivial in a sense that it reduces it to some very well-known ordinaryinequalities. Substituting a = x+y, b = y + z and c = z +x and simplifying,we get

x3y + y3z + z3x ≥ x2yz + xy2z + xyz2,

which is a partial case of the Muirhead’s inequality. For those who are unfa-miliar with it we may suggest the following argument. Since the inequality

4

is homogeneous, we may prove it only for those x, y, z for which xyz = 1.Then our inequality takes form

x2

z+

y2

x+

z2

y≥ x + y + z,

which follows directly from the rearrangement inequality. Indeed, withoutloss of generality, we may assume x ≥ y ≥ z and hence x2 ≥ y2 ≥ zz, while

we have1

x≤ 1

y≤ 1

z. Equality occurs when x = y = z, i.e., a = b = c.

1.2 Three math olympiad problems

Problem 3 In the plane there are n red points and n blue points. Provethat it is possible to draw n segments, each joining one red point and oneblue point, so that no two of them intersect or have a common endpoint.

Solution: This problem does not seem like a problem about inequalitiesbut this impression is premature. There are n! ways of joining red and bluepoints by segments. How to choose the right one? Our solution will in factgive an algorithm how to do it. We can observe that, if two of the segmentsintersected, say AB and CD with the points A and C being red and thepoints B and D being blue (see Figure),

D

A

B

C

then we can choose the segments AD and BC, instead of AB and CD, and,by using Tool 5, reduce the combined length of segments. Eventually, wewill get the right configuration. In a less constructive way, we may considerthe configuration with a minimal total combined length of segments, it existsbecause we have only finite number of possibilities (it might not be unique butit does not matter). We can be certain that the segments of this configurationdo not intersect.

Problem 4 (IMO, Jury) The area of a quadrilateral with the lengths ofthe sides a, b, c, d is equal to S. Prove that

S ≤ a + c

2· b + d

2.

5

Solution: We have to be attentive and note that the quadrilateral is notnecessary convex. If the quadrilateral ABCD is not convex, then one ofits diagonals, say BD (see Figure) does not have common points with theinterior of the quadrilateral.

C'

B

C

D

A

Reflecting C about BD we will get a convex quadrilateral ABC ′D with thesame sides but of greater area. Now, having said that, we can assume thatthe quadrilateral is convex. Let us divide it by one of the diagonals into twotriangles and use Tool 2. These triangles will have the areas, which are notgreater than ab/2 and cd/2. Thus 2S ≤ ab + cd. Similarly, 2S ≤ bc + da.Adding these two inequalities (we are symmetrizing again!) we get

4S ≤ ab + cd + bc + da = (a + c)(b + d),

which is equivalent to what we have to prove.

Problem 5 (IMO, 64) Let a, b, c be the sides of a triangle. Prove that

a2(b + c − a) + b2(c + a − b) + c2(a + b − c) ≤ 3abc.

Solution: For such problems Tool 7 is just magic and we use it again.Indeed, substituting a = x + y, b = y + z and c = z + x we will have a newinequality

2(x + y)2z + 2(y + z)2x + 2(z + x)2y ≤ 3(x + y)(y + z)(z + x),

which has to be proved for x ≥ 0, y ≥ 0, z ≥ 0. This is a great advantage,these constraints on indeterminates are much easier to handle. Actually atthis stage we can forget about the triangles and treat this inequality as wenormally do. Simplifying, we get

x2y + y2z + z2x + xy2 + yz2 + zx2 ≥ 6xyz,

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which splits into inequalities (x2 + y2)z ≥ 2xyz, (y2 + z2)x ≥ 2xyz, (z2 +x2)y ≥ 2xyz. See also a solution by means of rearrangement inequality inthe paper “Rearrangement Inequality” by K.Wu and A.Liu1

1.3 Ptolemy’s inequality

One of the most important tools is

Ptolemy’s Inequality. Let A, B, C, D be arbitrary points in the plane,not on the same line. Then

AB · CD + BC · AD ≥ AC · BD.

This inequality becomes equality if and only if the points A, B, C, D areconcyclic and each of the two arcs determined by the points A, C containsone of the two remaining points.

Proof: Two simple ones can be found on http://MathOlymp.com in theGeometry section of Tutorials.

Problem 6 Let ABC be an equilateral triangle and P be an arbitrary pointin the plane. Prove that the lenghts of PA, PB, PC are always the lengthsof the sides of a triangle. Determine, when this triangle is degenerate.

Solution: By Ptolemy’s inequality

PA · BC + PC · AB ≥ PB · AC,

and since AB = AC = BC, this becomes

PA + PC ≥ PB

with equality, when P is on the circumcircle of ABC. Similarly, we can get

PA + PB ≥ PC.

Hence, the triangle becomes degenerate, only when P is on the circumcircleof ABC.

1reproduced on http://MathOlymp.com in the Algebra section of Tutorials with apermission of the authors.

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Problem 7 Let M, A1, A2, . . . , An (n ≥ 3) be distinct points in the planesuch that

A1A2 = A2A3 = · · · = An−1An = AnA1.

Prove the inequality

1

MA1 · MA2

+1

MA2 · MA3

+ · · · + 1

MAn−1 · MAn

≥ 1

MA1 · MAn

,

and determine all cases when equality occurs.

Solution 1: Applying Ptolemy’s inequality to MA1AkAk+1 we get

MA1 · AkAk+1 + A1Ak · MAk+1 ≥ A1Ak+1 · MAk,

(which for k = 1 degenerates into an equality). We divide this expression byMA1 · MAk · MAk+1 to get

AkAk+1

MAk · MAk+1

≥ A1Ak+1

MA1 · MAk+1

− A1Ak

MA1 · MAk

.

Summing these inequalities for k = 1, 2, . . . , n−1 we obtain

A1A2

MA1 · MA2

+A2A3

MA2 · MA3

+ · · · + An−1An

MAn−1 · MAn

≥

A1An

MA1 · MAn

− A1A1

MA1 · MA1

=A1An

MA1 · MAn

.

As A1A2 = A2A3 = · · · = AnA1 we get the desired inequality.

Equality holds iff it holds for each inequality that was added, that is forall k = 1, 2, . . . , n−1. It happens when the points A1, Ak, Ak+1, M lie on acircle and in this particular order. This means that the points M, A1, . . . , An

lie on the circle in this order. In this case A1, A2, . . . , An is a regular n-gonand M belongs to the shortest arc A1An.

Solution 2: Consider an inversion i with pole M and any coefficient r.Let A′

1, A′2, . . . , A

′n be the images of A1, A2, . . . , An, respectively, under this

inversion. Applying the Triangle inequality (in fact, several times) for thepoints A′

1, A′2, . . . , A

′n, we get

A′1A

′2 + A′

2A′3 + · · · + A′

n−1A′n ≥ A′

1A′n. (1)

It is well-known (or easy to prove) how distances between images of pointsunder inversion can be expressed. In our case, if X, Y are any two pointsdifferent from M , and if X ′, Y ′ are their images under i then

X ′Y ′ =r2 · XY

MX · MY.

8

This formula can be applied to any pair of points A1, A2, . . . , An because theyare all different from M . So we rewrite (1) in the form

r2 · A1A2

MA1 · MA2

+r2 · A2A3

MA2 · MA3

+ · · · + r2 · An−1An

MAn−1 · MAn

≥ r2 · A1An

MA1 · MAn

.

Since A1A2 = A2A3 = · · · = An−1An = AnA1, the latter yields

1

MA1 · MA2

+1

MA2 · MA3

+ · · · + 1

MAn−1 · MAn

≥ 1

MA1 · MAn

, (2)

as desired.

It is clear from the above argument that (1) and (2) are simultaneouslystrict inequalities or equalities. So it remains to establish all cases when (1)is equality. This happens if and only if A′

1, A′2, . . . , A

′n are points on a line

� arranged in this order. Now, the preimage of � under i can be either

a) the same line, if it passes through the pole M ,

b) a circle C through M , if � does not contain M .

In case a) the disposition of A′1, A

′2, . . . , A

′n on � in this order is equivalent

to M, A1, A2, . . . , An being arranged in the order

Ak, Ak−1, . . . , A1, M, An, . . . , Ak+1

on �. This is, however, impossible, given the condition A1A2 = A2A3 = · · · =An−1An = AnA1.

In case b), the condition is satisfied if and only if M, A1, A2, . . . , An areon a circle, the points A1, A2, . . . , An are arranged in this order in one of thetwo possible directions (clockwise or anticlockwise) with M being on the arc

A1An of C that does not contain any of the points A2, A3, . . . , An−1.Summing up, we conclude that equality occurs if and only if the points

M, A1, A2, . . . , An lie on a circle, and arranged in this order with respect toone of the two possible directions.

Comment 1: The argument is independent of whether or not the givenpoints lie in a plane. Nothing changes if they are in three-dimensional space.

Comment 2: It follows from the above solution that arbitrary distinctpoints M, A1, A2, . . . , An satisfy the inequality

A1A2

MA1 · MA2

+A2A3

MA2 · MA3

+ · · · + An−1An

MAn−1 · MAn

≥ A1An

MA1 · MAn

,

and that the only possible cases of equality are the ones listed above.

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2 Problems to solve

1. (USSR, 84) Four points on a straight line are situated in the orderA, B, C, D. Prove that for every point E, which is not on the line,

AE + ED + |AB − CD| > BE + EC.

2. (Czecho-Slovakia,76) One convex polygon is situated inside anotherconvex polygon. Prove that the perimeter of the inner polygon issmaller than the perimeter of the outer one.

3. How many sides can have a convex polygon if all its diagonals haveequal length?

4. A point M is given inside a triangle ABC. Prove that

4Area (ABC) ≤ AM · BC + BM · AC + CM · AB.

5. (Yugoslavia, 75) The midpoints of adjacent sides of a convex polygonare joined by segments which become the sides of a new polygon. Showthat its area is not less than half the area of the original polygon.

6. A parallelogram is inside the triangle. Prove that the area of the par-allelogram is not greater than half the area of the triangle.

7. (IMO, 66) Points M, K, L are on the sides AB, BC, CA of a triangleABC. Prove that, of the triangles MAL, KBM , LCK, the area of atleast one triangle is not greater than a quarter of the area of ABC.

8. (a) Prove that every convex polygon of area S and perimeter P con-tains a circle of radius greater than S/P .

(b) A convex polygon of area S1 and perimeter P1 contains a convexpolygon of the S2 and perimeter P2. Prove that 2S1/P1 > S2/P2.

9. Prove that if a, b, c are the lengths of the sides of a triangle, then

(a + b − c)(a − b + c)(−a + b + c) ≤ abc.

10. (Shortlist IMO, 97) Let ABCDEF be a convex hexagon such thatAB = BC, CD = DE, EF = FA. Prove that

BC

BE+

DE

DA+

FA

FC≥ 3

2.

10

11. Let A1A2A3A4 be a square and P be an arbitrary point in the plane.Of the distances PA1, PA2, PA3, PA4, let M be the maximal and mbe the minimal. Prove that

PA1 + PA2 + PA3 + PA4 ≥ (1 +√

2)M + m,

with equality, when P is on the circumcircle of the square and strictinequality otherwise.

12. (IMO, 1995) Let ABCDEF be a hexagon with AB = BC = CD andDE = EF = FA, and � BCD = � EFA = 60◦. Let also G and H betwo arbitrary points. Then

AG + BG + GH + DH + EH ≥ CF.

13. A square with the length of its side 1 is given. A broken line of lengthL is drawn inside it in such a way that the distance from every pointof the interior of the square to a certain point of the broken line is lessthan ε. Prove that

L ≥ 1

2ε− π

2ε.

14. A 7-gon A1A2 . . . A7 is inscribed in a circle. Prove that, if the centre ofthe circle is inside the 7-gon, then

� A1 + � A3 + � A5 < 450◦.

15. (Shortlist IMO, 96) Let ABCD be a convex quadrilateral, and let RA,RB, RC and RD denote the circumradii of the triangles DAB, ABC,BCD and CDA respectively. Prove that RA + RC > RB + RD if andonly if � A + � C > � B + � D.

3 Solutions

1. Suppose that CD > AB. We choose a point D′ on CD such thatAB = CD′.

11

F

E

DA

D'CB

M

Now D′D = |AB −CD| and ED + |AB −CD| > ED′ by the Triangleinequality. Therefore it is sufficient to prove that

AE + ED′ > BE + EC.

Note that the midpoints of AD′ and BC coincide. Let us constructa point F which is centrally symmetric to E about M . Then bothquadrilaterals AED′F and BECF are parallelograms. Now we useTool 7 to get

AE + ED′ = AE + AF > BE + BF = BE + EC.

2. Solution 1: For every vertex of the inner polygon let us erect in theexterior of it two rays which are perpendicular to the two sides of thepolygon which meet at this vertex as shown below. Since the polygonis convex each angle of it is less than 180◦ and the shaded areas inthe figure do not intersect. Let us consider the side AB of the innerpolygon and two rays r and q which are perpendicular to it.

12

r

q

ED

CB

A

The two rays r and q cut a peace of the boundary of the outer polygonwhich is a broken line situated between these rays. The side AB will bethe projection of this broken line and will be shorter than this brokenline (Tool 4). Since we can find such a broken line for every side of theinner polygon and these broken lines do not have points in common,the statement is proved. Note that we did not use the convexity of theouter polygon.

L

K

AB

Solution 2: Let us extend the side AB of the inner polygon to thepoints K and L of intersection of AB with the sides of the outer polygonP . Then, due to the convexity of P , the segment KL divides it intoconvex polygons P1 and P2. Since the inner polygon is also convex itlies inside either P1 or P2. Both P1 and P2 have perimeters smaller thanP since in each case we replace a broken line by the straight segmentKL. If we repeat this procedure with another side of the inner polygonwe will get yet another outer polygon whose perimeter is again smallerthan the perimeter of the previous outer polygon. When we repeatthis procedure for all sides of the inner polygon the outer and the inner

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polygons will coincide. Since the perimeters of all outer polygons weredecreasing, the statement is proved.

3. It can have five sides as in the case of a regular pentagon. But it cannothave more. If it had six or more sides, then we can choose two non-intersecting diagonals AB and CD with vertices A, B, C, D all differentfrom each other. But then AC and BD are also diagonals. All theyhave equal length which contradicts to our Tool 6.

4. Let P, Q,R be the points of intersection of AM , BM , CM with BC,CA, AB, respectively. Let α = � APB, β = � BQC, γ = � CRA.

B

Q CA

M

PR

We have

S =1

2AB(CM + MR) sin γ,

S =1

2BC(AM + MP ) sin α,

S =1

2CA(BM + MQ) sin β.

Adding these together, we get

3S =1

2(AB · CM sin γ + BC · AM sin α + CA · BM sin β) + S

since

1

2(AB · MR sin γ + BC · MP sin α + CA · MQ sin β) =

Area (ABM) + Area (BCM) + Area (CAM) = S.

This implies

4S = AB · CM sin γ + BC · AM sin α + CA · BM sin β

≤ AB · CM + BC · AM + CA · BM,

as required.

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5. Let A1, A2, . . . , An be the vertices of the n-gon, S be its area, andB1, B2, . . . , Bn be the midpoints of the sides A1A2, A2A3, . . . , AnA1,respectively. For convenience of notation we will also set An+1 = A1

and An+2 = A2. Then each of the triangles

AiAi+1Ai+2 i = 1, 2, . . . , n

intersects only with triangles Ai−1AiAi+1 and Ai+1Ai+2Ai+3 (only pointsof the interior of the polygon are considered, not just common vertex)and no three of these triangles intersect.

Ai+2

Ai+1

Ai

Bi

Bi+1

Hence

2S ≥n∑

i=1

Area (AiAi+1Ai+2).

We also notice that

Area (BiAi+1Bi+1) =1

4Area (AiAi+1Ai+2).

Therefore

Area (B1B2 . . . Bn) = S −n∑

i=1

Area (BiAi+1Bi+1) =

S − 1

4

n∑

i=1

Area (AiAi+1Ai+2) ≥ S − 1

2S =

1

2S.

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6. Suppose that a parallelogram PQRS is situated inside a triangle ABC.We may assume that at least three vertices of the parallelogram are onthe sides of the triangle. Indeed, if we had only two such vertices, thenone of the sides of the triangle does not contain a vertex or else oneof the vertices of the parallelogram is simultaneously a vertex of thetriangle. In both cases we can find a triangle of smaller area whichcontain the same parallelogram as shown below.

Of course, if we have less than two vertices of the parallelogram on thesides of the triangle this argument also works.

If three vertices of the parallelogram PQRS are on the sides of thetriangle ABC, we can always find a parallelogram PQR′S ′, which is ofthe same area as PQRS, has three vertices on the boundary of ABC,and whose one side is lying completely on a certain side of the triangle.As shown in the Figure we have to extend the side of the parallelogramwhich has only one vertex on the boundary of the triangle. In thisparticular case, we extended SR and replace PQRS with PQR′S ′,where S ′ is the point of intersection of RS with AC.

B

A C

R

P

Q

S

R'

S'

Let us consider now this latter case. We can assume from the verybeginning that one side of the parallelogram, namely PS, is on a sideof the triangle, in this case AC (see Figure below).

16

B

A CS

Q R

TP

Since the parallelogram AQRT is of the same area as PQRT , by re-placing PQRT by AQRT , this case in turn may be reduced to the case,when one of the angles of the parallelogram coincides with one of theangles of the triangle.

Let us consider this remaining latter case (see Figure).

B

A=P C

RQ

S

Let us denote U = Area (PQRS), V = Area (BQR), U = Area (RSC).Then Area (PQR) = Area (PRS) = U/2. We have

2V

U=

Area (BQR)

Area (PQR)=

BQ

QA=

BR

RC=

AS

SC=

Area (PRS)

Area (SRC)=

U

2W.

or 4V W = U2. Now AM-GM inequality gives us that (V + W )2 ≥4V W = U2, or V + W ≥ U , and hence U = Area (PQRS) is less thanhalf the area of the triangle ABC.

7. Since Area (MBK) = MB ·BK · sin � B and Area (ABC) = AB ·BC ·sin � B, we have

Area (MBK)

Area (ABC)=

MB · BK

AB · BC

17

B

A C

KM

L

Similarly

Area (MAL)

Area (ABC)=

AM · AL

AB · AC,

Area (KLC)

Area (ABC)=

KC · LC

BC · AC.

Suppose that all these three ratios are greater than 1/4. Then multi-plying all three of them together we get

BM · BK · AM · AL · CK · CL

AB · BC · AB · AC · BC · AC>

1

64

orAM · BM

AB2· BK · CK

BC2· AL · CL

AC2· >

1

64.

But on the other hand,

√AM · BM ≤ AM + BM

2=

AB

2

andAM · BM

AB2≤ 1

4.

SimilarlyBK · CK

BC2≤ 1

4,

AL · CL

AC2≤ 1

4.

ThusAM · BM

AB2· BK · CK

BC2· AL · CL

AC2· ≤ 1

64,

the contradiction.

8. (a) This is an application of Pigeonhole principle in geometry. Forevery side of the polygon we construct a rectangle, whose one side

18

coincides with the given side of the polygon, whose another sideis S/P , and which is oriented so that it intersects the interior ofthe polygon. The total area of all such rectangles is S and anytwo neighboring rectangles intersect, hence the area, which theyjointly cover, is less than S. Hence there exist an uncovered area,which is of course also a polygon. Taking any interior point ofthis polygon as the centre, we may draw a circle with the centreat this point of radius R > S/P , which lies inside the polygon.

(b) Let us prove first that, if a circle or radius R is situated inside apolygon of area S and perimeter P , then R ≤ 2S/P . To prove thiswe join the centre of the circle with the vertices of the polygon,which will then be divided into n nonintersecting triangles. Leth1, h2, . . . , hn be the altitudes of these triangles drawn from thecentre of the circle to the sides of the polygon. Then, as hi ≥ Rfor all i = 1, 2, . . . , n

S =P (h1 + h2 + . . . + hn)

2≥ PR

2,

which proves the statement. Part (b) can be deduced from part(a) and the argument above. Indeed, (a) guarantees that a circleof radius R > S2/P2 can be inscribed into the smaller polygon.Then by the argument above,

S2

P2

< R ≤ 2S1

P1

,

and the statement (b) is proved.

9. Using Tool 9, we substitute a = x + y, b = y + z and c = z + x, wherex ≥ 0, y ≥ 0, z ≥ 0. The new inequality will be

8xyz ≤ (x + y)(y + z)(z + x),

which reduces to

6xyz ≤ x2y + y2z + z2x + xy2 + yz2 + zx2,

which splits into three inequalities (x2+y2)z ≥ 2xyz, (y2+z2)x ≥ 2xyz,(z2 + x2)y ≥ 2xyz, which are AM-GM inequalities. Or else it can besplitted into two inequalities

3xyz ≤ x2y + y2z + z2x, 3xyz ≤ xy2 + yz2 + zx2,

which follow from the Rearrangement inequality (see Algebra Tutorial 1on http://MathOlymp.com).

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10. Applying the Ptolemy’s inequality for the quadrilateral ACEF we get

AC · EF + CE · AF ≥ AE · CF.

Since EF = AF , it impliesFA

FC≥ c

a + b. Similarly,

DE

DA≥ b

c + aand

BC

BE≥ a

b + c. It follows now that

BC

BE+

DE

DA+

FA

FC≥ a

b + c+

b

c + a+

c

a + b≥ 3

2(1)

by the well-known inequality.

For the equality to occur we need (1) to be an equality and also weneed an equality each time when the Ptolemy’s inequality was used.The latter happens when the quadrilaterals ACEF , ABCE, ACDEare cyclic, that is, ABCDEF is a cyclic hexagon. Also for the equalityin (1) we need a = b = c. Hence the equality occurs if and only if thehexagon is regular.

To prove the inequality (1) let us make the the substitution x = a + b,y = c + a, z = b + c after which the inequality takes the form

1

2

(x + z − y

y+

x + y − z

z+

y + z − x

x

)

=

1

2

(x

y+

y

x+

x

z+

z

x+

y

z+

z

y− 3

)

≥ 3

2,

and it is now obvious.

11. Let M = PA4, then it is easy to see that m = PA2. By the Ptolemy’sinequality

PA1 · A3A4 + PA3 · A1A4 ≥ PA4 · A1A3

with equality occuring if and only if A1, P, A3, A4 are concyclic. SinceA1A4 = A3A4 = A1A3/

√2, we get

PA1 + PA3 ≥√

2PA4.

Hence

PA1 + PA2 + PA3 + PA4 ≥ (1 +√

2)PA4 + PA2 = (1 +√

2)M + m.

Equality holds, when P lies on the circumcircle of the square.

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12. As BCD and AEF are equilateral triangles, ABDE is a kite.

G

H

C’

C

F’

F

A

D E

B

Reflect C and F about BE to get points C ′ and F ′. Then CF = C ′F ′

and we obtain equilateral triangles DEF ′ and ABC ′. Let X be anypoint on the segment GH.

By Ptolemy’s Inequality, we have F ′H · DE ≤ F ′E · DH + F ′D · EHor F ′H ≤ DH + EH, and C ′G · AB ≤ C ′A · BG + C ′B · AG orC ′G ≤ BG + AG. Thus AG + BG + GH + DH + EH ≥ C ′G + GH +HF ′ ≥ C ′F ′ = CF .

This beautiful problem was submitted to the IMO by New Zealandand the author of it is Alastair McNaughton. This problem can also besolved using Fermat Point inequality (see Chapter “Geometric Trans-formations.”)

13. Let us consider a segment of length � and all points which are withindistance ε from this segment. This points belong to the figure shownbelow

21

which is of area πε2 + 2ε�. Suppose that the broken line consists of nstraight segments of lengths �1, �2, . . . , �n. Let us construct such figuresfor all segments of the broken line. We see that the intersection of anytwo such neighboring figures will contain a disc of radius ε.

Hence the total area covered will be not greater than

nπε2 + 2ε(�1 + �2 + . . . + �n) − (n − 1)πε2 = 2εL + πε2.

Since these figures cover the whole square, we have 1 ≤ 2εL + πε2, or

L ≥ 1

2ε− π

2ε.

14. We shall count the arcs which the angles subtend. We have

2� A1 = arcA2A7, 2� A3 = arcA4A2, 2� A5 = arcA6A4,

where denote arcs using the agreement that the arcPQ denotes the arc,which P must travel clockwise in order to meet Q.

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A3

A1

A2A4

A5

A6A7

Thus2(� A1 + � A3 + � A5) = 720◦ + arcA6A7.

Now as the center of the circle is inside the 7-gon we have arcA6A7 <180◦, which together with the equation above gives

2(� A1 + � A3 + � A5) < 900◦.

or� A1 + � A3 + � A5 < 450◦.

15. Let the diagonals AC and BD meet at X. Of the two angles � AXD and� AXB, at least one must be greater than or equal to 90◦. Withoutloss of generality, we assume that � AXD ≥ 90◦. Let � 1 = � CAD,� 2 = � CBD, � 3 = � ADB, � 4 = � ACB. These four angles are allacute and � 1 + � 2 = � 3 + � 4.

1

3

2

4

D

A

B

C

X

We note that

RA =AB

2 sin � 3, RB =

AB

2 sin � 4, RC =

CD

2 sin � 2, RD =

CD

2 sin � 1,

23

whenceRA

RB

=sin � 4

sin � 3.

We now consider three different cases:

• � A+ � C = � B+ � D. Then � B+ � D = 180◦, and the quadrilateralis cyclic. This implies � 1 = � 2, � 3 = � 4, and hence RA = RB,RC = RD. Thus, in this case RA + RC = RB + RD.

• � A + � C < � B + � D. Then � B + � D > 180◦, and the vertexB lies inside the circle circumscribed about ACD. This implies� 1 < � 2, � 3 > � 4. This further translates to and hence RA < RB

and RC < RD. Thus, in this case RA + RC < RB + RD.

• � A + � C > � B + � D. Then � B + � D < 180◦, and the vertexB lies outside the circle circumscribed about ACD. This implies� 1 > � 2, � 3 < � 4. This further translates to and hence RA > RB

and RC > RD. Thus, in this case RA + RC > RB + RD.

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