Campamento_2012

Geometric Inequalities Anthony Erb Lugo April 2012

Geometric inequalities combine the elegance and cleverness evident in both algebra and geometry to create beautiful expressions in the form of inequalities. We’ll look into these expressions only briefly, so I recommend further reading on your own.

1

Notation & Identities • Given a triangle ABC we denote its sides by a = BC, b = AC and c = BC. • The angles of triangle ABC are denoted as A = ∠BAC, B = ∠ABC and C = ∠BCA. • The altitudes from vertices A, B, C will be denoted as ha , hb and hc , respectively.

• The area of the triangle ABC will be denoted as [ABC].

• The orthocenter, incenter, centroid and circumcenter are denoted as H, I, G and O, respectively.

• The semiperimeter, inradius and circumradius will be denoted as s, r and R, respectively.

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

• Ravi substitution uses the fact that we can write the sides of a triangle in the form of a = y + z, b = x + z and z = x + y where x, y and z are positive real numbers.

• Area formulas include: [ABC] =

p ab sin C a · ha abc = = rs = s(s − a)(s − b)(s − c) = 2 4R 2

Our first theorem follows intuitively Theorem 1.1 (Triangle Inequality): Let ABC be a triangle, then AB + BC > AC follows and so do its cyclic equivalents.

OMPR 2012

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Geometric Inequalities

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Examples

Example 2.1: Let P be a point inside of triangle ABC. Prove that 2(P A + P B + P C) > a + b + c Proof. It follows from adding the following results of the Triangle Inequality P A + P B > AB = c P A + P C > AC = b P B + P C > BC = a

Example 2.2: Let a, b and c be the sides of a triangle. Prove that abc ≥ (a + b − c)(b + c − a)(c + a − b) Proof. Here we’ll use Ravi substitution and note that the inequality is then equivalent with (y + z)(x + z)(x + y) ≥ 8xyz which follows from the Arithmetic-Mean and Geometric-Mean.

Example 2.3: Let ABC be a triangle with altitudes ha , hb and hc . Prove that ha + hb + hc ≥ 9r Proof. Let’s first note an important identity: 1 s s a b c 1 1 1 = = = + + = + + . r rs [ABC] 2[ABC] 2[ABC] 2[ABC] ha hb hc Which implies that our inequality is equivalent to

1 1 1 1 ≥ 9. + + = (ha + hb + hc ) (ha + hb + hc ) r ha hb hc Since this follows from the AM-GM Inequality, we’re done. Note, by AM-GM we have p ha + hb + hc ≥ 3 3 ha hb hc and

1 1 1 3 + + ≥ √ 3 ha hb hc ha hb hc

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Geometric Inequalities

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Now, any geometric inequalities lecture would be incomplete if this classic problem were not included.

Example 2.4: (Euler’s Inequality) Given any triangle ABC. Prove that R ≥ 2r.

Proof. The classic proof explores the relation between the distance IO and both R and r, but I’ll leave this proof to be found by the reader. Here we’ll use the following identities [ABC] = rs =

abc p = s(s − a)(s − b)(s − c).

4R

Note that we can rewrite the inequality as 1 r 1 [ABC] 4[ABC] ≥ ⇔ ≥ · 2 R 2 s abc which rearranges into proving that abc · s ≥ 8[ABC]2 = 8s(s − a)(s − b)(s − c) now divide both sides by s and the inequality is then abc ≥ 8(s − a)(s − b)(s − c) which, by distributing the factor of 2 to each term, is equivalent to abc ≥ (b + c − a)(a + c − b)(a + b − c) In other words, the first example problem.

Now, let’s take a look at one of the first IMO problems and how we may use our tools to quickly solve it.

Example 2.5: (IMO 1961) Let a, b, c be the sides of a triangle, and S its area. Prove √ a2 + b2 + c2 ≥ 4S 3 In what case does equality hold?

Proof. Let’s take advantage of Ravi Substitution once again and note y+z x+z x+y x+y+z p s(s − a)(s − b)(s − c) p S = (x + y + z)xyz

a b c s S

= = = = =

Furthermore, our inequality is then equivalent to

p (y + z)2 + (x + z)2 + (x + y)2 ≥ 4 3(x + y + z)xyz.

Which, after expanding and simplifying, can written as p x2 + y 2 + z 2 + xy + yz + xz ≥ 2 3(x + y + z)xyz.

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Geometric Inequalities

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Moreover, we can note the well-known result (a + b + c)2 ≥ 3(ab + bc + ac) to conclude that (xy + yz + xz)2 ≥ 3(x + y + z)xyz =⇒ xy + yz + xz ≥ Note that we now have

Thus, if

p 3(x + y + z)xyz.

p 2(xy + yz + xz) ≥ 2 3(x + y + z)xyz. x2 + y 2 + z 2 + xy + yz + xz ≥ 2(xy + yz + xz)

holds then we’re done. But this is equivalent to

(x − y)2 + (y − z)2 + (z − x)2 ≥ 0 and so the problem is solved. Equality then holds for when x = y = z, i.e. when a = b = c.

3

Practice Problems 1. Let r and R be the inradius and circumradius of triangle ABC, respectively. Prove that 3 r 1+ ≤ R 2 2. Let ma , mb and mc be the medians of triangle ABC. Prove that 3 ma + mb + mc > a+b+c 4 3. Let a, b and c be the sides of a triangle. Prove that 2>

a2 + b2 + c2 ≥1 ab + bc + ac

4. Let r, R and s be the inradius, circumradius and semiperimeter of triangle ABC, respectively. Prove that 2s2 R≤ r 5. Let a, b and c be the sides of a triangle. Prove that 2(ab + bc + ac) >

6abc + a2 + b2 + c2 a+b+c

6. Prove the stronger inequality for a triangle with area S √ a2 + b2 + c2 ≥ 4S 3 + (a − b)2 + (b − c)2 + (c − a)2 7. (IMO, 1964) Prove that if a, b and c are the sides of a triangle, then a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc 4

Geometric Inequalities Anthony Erb Lugo April 2012 Geometric inequalities combine the elegance and cleverness evident in both algebra and geometry to create beautiful expressions in the form of inequalities. We’ll look into these expressions only briefly, so I recommend further reading on your own.

1

Notation & Identities • Given a triangle ABC we denote its sides by a = BC, b = AC and c = BC. • The angles of triangle ABC are denoted as A = ∠BAC, B = ∠ABC and C = ∠BCA. • The altitudes from vertices A, B, C will be denoted as ha , hb and hc , respectively.

• The area of the triangle ABC will be denoted as [ABC].

• The orthocenter, incenter, centroid and circumcenter are denoted as H, I, G and O, respectively.

• The semiperimeter, inradius and circumradius will be denoted as s, r and R, respectively.

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

• Ravi substitution uses the fact that we can write the sides of a triangle in the form of a = y + z, b = x + z and z = x + y where x, y and z are positive real numbers.

• Area formulas include: [ABC] =

2

p abc a · ha ab sin C = = rs = s(s − a)(s − b)(s − c) = 2 4R 2

Examples

Example 2.1: Let a, b and c be the sides of a triangle. Prove that abc ≥ (a + b − c)(b + c − a)(c + a − b) Proof. Here we’ll use Ravi substitution and note that the inequality is then equivalent with (y + z)(x + z)(x + y) ≥ 8xyz which follows from the Arithmetic-Mean and Geometric-Mean.

Geometric Inequalities

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Example 2.2: Let ABC be a triangle. Suppose we keep points B and C fixed but we let A vary such that the perimeter, 2s, stays constant. What is the largest possible area in terms of s and BC? Proof. We let a = BC and note that the points that satisfy the conditions define a ellipse whose foci are B and C and the point A lies on the graph of the ellipse. Furthermore, we maximize the area by maximizing the distance from the point A to the segment BC (altitude in the triangle) since the base remains constant. Clearly, the point with the largest distance is the intersection of the perpendicular bisector of BC with the graph of the parabola. This point is then equidistant from the foci, so our triangle is isosceles. If we let the second/third side be denoted as b, then the area of the triangle is p

s(s − a)(s − b)(s − c) =

which follows from noting that

p ap s(s − a)(s − b)2 = s(s − a) 2

a a + 2b = 2s =⇒ s − b = .

2

Now, any geometric inequalities lecture would be incomplete if this classic problem were not included.

Example 2.3: (Euler’s Inequality) Given any triangle ABC. Prove that R ≥ 2r.

Proof. The classic proof explores the relation between the distance IO and both R and r, but I’ll leave this proof to be found by the reader. Here we’ll use the following identities [ABC] = rs =

abc p = s(s − a)(s − b)(s − c).

4R

Note that we can rewrite the inequality as

r 1 [ABC] 4[ABC] 1 ≥ ⇔ ≥ · 2 R 2 s abc which rearranges into proving that abc · s ≥ 8[ABC]2 = 8s(s − a)(s − b)(s − c) now divide both sides by s and the inequality is then abc ≥ 8(s − a)(s − b)(s − c) which, by distributing the factor of 2 to each term, is equivalent to abc ≥ (b + c − a)(a + c − b)(a + b − c) In other words, the first example problem.

Now, let’s take a look at one of the first IMO problems and how we may use our tools to quickly solve it.

2

Geometric Inequalities

OMPR 2012

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Example 2.4: (IMO 1961) Let a, b, c be the sides of a triangle, and S its area. Prove √ a2 + b2 + c2 ≥ 4S 3 In what case does equality hold?

Proof. Let’s take advantage of Ravi Substitution once again and note y+z x+z x+y x+y+z p s(s − a)(s − b)(s − c) p (x + y + z)xyz S = a b c s S

= = = = =

Furthermore, our inequality is then equivalent to

p (y + z)2 + (x + z)2 + (x + y)2 ≥ 4 3(x + y + z)xyz.

Which, after expanding and simplifying, can written as p x2 + y 2 + z 2 + xy + yz + xz ≥ 2 3(x + y + z)xyz.

Moreover, we can note the well-known result

(a + b + c)2 ≥ 3(ab + bc + ac) to conclude that (xy + yz + xz)2 ≥ 3(x + y + z)xyz =⇒ xy + yz + xz ≥ Note that we now have

Thus, if

p 3(x + y + z)xyz.

p 2(xy + yz + xz) ≥ 2 3(x + y + z)xyz. x2 + y 2 + z 2 + xy + yz + xz ≥ 2(xy + yz + xz)

holds then we’re done. But this is equivalent to (x − y)2 + (y − z)2 + (z − x)2 ≥ 0 and so the problem is solved. Equality then holds for when x = y = z, i.e. when a = b = c.

Example 2.5: (Germany TST, 2002) Let x, y and z be the angle bisectors of a triangle with perimeter 6, then we have: 1 1 1 + 2 + 2 ≥1 2 x y z 3

Geometric Inequalities

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Proof. We can’t do much with how the inequality stands right now. So let’s try to rewrite this in a better way. We’ll use Stewart’s theorem to find another expression for x. Let the angle bisector from A intersect the side BC at A′ . It’s easy to verify that ac ab and A′ C = .

BA′ = b+c b+c Therefore, from Stewart’s Theorem, we have that

ab ac ac ab 2 2 = b + c2 . a x + b+c b+c b+c b+c Rearranging gives

2

x = bc 1 −

a b+c

2 !

.

Moreover, our inequality is now equivalent with X 1 2 ! ≥ 1.

a cyc (bc) 1 − b+c Furthermore, H¨older’s Inequality tells us that !

2 ! X

X X a bc 1− b+c cyc cyc cyc

(bc) 1 −

1

a b+c

3 2 ! ≥ 3

Thus, it is sufficient to prove that 33 P

bc

27 ≥

X

cyc

or

Now note that

P

cyc

cyc

bc

!

1−

X cyc

a b+c

1−

2 ! ≥ 1

a b+c

2 !

.

(a + b + c)2 ≥ 3(ab + bc + ac) =⇒ 12 ≥ ab + bc + ac.

So it is then also sufficient to prove that

2 a 9 X 1− ≥ 4 b+c cyc

which is equivalent with

X a 2 3 9 ≥3− = . b+c 4 4 cyc

It follows from the Cauchy-Schwarz Inequality used as follows ! !2 X a 2 X a 9 3 ≥ ≥ b+c b+c 4 cyc cyc P a ≥ 32 . and dividing by 3. Note the use of Nesbitt’s Inequality: cyc b+c 4

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Practice Problems 1. Let a, b and c be the sides of a triangle. Prove that a2 + b 2 + c 2 2> ≥1 ab + bc + ac 2. Let a, b and c be the sides of a triangle. Prove that 2(ab + bc + ac) >

6abc + a2 + b 2 + c 2 a+b+c

3. Prove the stronger inequality for a triangle with area S √ a2 + b2 + c2 ≥ 4S 3 + (a − b)2 + (b − c)2 + (c − a)2 4. (IMO, 1964) Prove that if a, b and c are the sides of a triangle, then a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc 5. (Russian Geometry Olympiad, 2008) Prove that the triangle having sides a, b, c and area S satisfies the inequality √ 1 a2 + b2 + c2 − (|a − b| + |b − c| + |c − a|)2 ≥ 4 3S 2 6. (JBMO, 1997) Let ABC be a triangle and let I be the incenter. Let N, M be the midpoints of the sides AB and AC, respectively. The lines BI and CI meet M N at K and L, respectively. Prove that AI + BI + CI > BC + KL. 7. (IMO, 1983) Let a, b and c be the lengths of the sides of a triangle. Prove that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0 Determine when equality occurs.

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Processes and Operations Roman Kvasov

Introduction In this lecture we will consider problems of the following type – given the initial configuration and some set of operations, is it possible to obtain some particular configuration, starting from the initial configuration and applying given operations. In a variety of these problems it is either extremely difficult or impossible to clearly point out a desired configuration among the possible ones. Usually the solution of this sort of problems is divided into two important parts. The first part consists of proposing a process of changing the configuration (some rule to apply the operations). The second part usually contains a proof of the fact that this process is finite and eventually leads to a desired configuration.

Problem 1 At the second round of OMPR every student has at most 3 other students as friends on facebook. Prove that it is possible to divide all students into 2 groups where each student has at most one student as his friend on facebook.

Problem 2 Gabriel wrote a number in each entry of some table mĂ—n. He is allowed to change the signs of all the numbers in some row or in some column. Can Gabriel always make all the sums of the numbers in each row and each column nonnegative?

Problem 3 Enthusiastic Jack wants to construct a regular polygon with the number of sides greater than 6 in such way that all its vertices have two integer coordinates. Help him.

Problem 4 There are 2012 gas stations built along some circular road. It is known that the total amount of gas in these gas stations is enough to drive the complete circle. Anthony claims that he can drive the complete circle starting with an empty tank starting at some particular gas station. Can he really do that?

Problem 5 Marcos picked up a number n such that its last digit is not a zero. Show that he can always find such integer k that kn has no zeroes.

Problem 6 There is a very long binary number written on the board. Roberto and Francis substitute every fragment “10” by either “0001” or “0000001”. Can they do it forever?

Problem 7 Every member of some parliament slapped in the face exactly one member of the parliament. Show that it is possible to divide all the members into 3 political parties such that none of the members slapped any other member of the same party.

Problem 8 There are 2012 points and 2012 lines given on the plane. Show that it is possible to assign them numbers from 1 to 2012, such that the perpendiculars drawn from the points to the lines with the same numbers do not intersect.

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Pigeonhole Principle

April 13, 2012

Pigeonhole Principle Jiacheng Feng I. Definition: If n pigeons are put into m pigeonholes with n > m, then at least one pigeonhole must contain more than one pigeon.

II. Uses: The Pigeonhole principle is used to help solve existential problems with finite (and sometimes infinite) sets. This means that this principle can show something exists but now where it exists.

The main challenge of problems that require the pigeonhole principles is to identify the pigeons and the pigeonholes.

III. Example 1. A tennis star, preparing for a tournament, decides to play at least one match a day, but no more than 20 matches altogether, over a period of two weeks. Show that during some set of consecutive days the star must play exactly 7 matches.

Solution: Let the number games he played up until day k be Sk. A={S1, S2, …, S14}; B={ S1+7, S2+7, …, S14+7} 1 S1<S2<…< S14 20; 8 S1+7<S2+7<…< S14+7 27 There are 28 numbers in A and B, but can only take on 27 values. By pigeonhole principle, 2 of these numbers are equal. However, since each element in A are distinct, and likewise in B, A and B have one element in common, or Si=Sj+7. By definition, this means on the ith day, the tennis player has played 7 more games than on the jth day.

IV. Warm-up 1. Among three people, two have the same gender.

2. Among 13 people, two are born in the same month.

Feng 1

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3. There are 10 black socks and 17 white socks in a drawer. How many socks must be taken out to find a matching pair?

4. Prove that at a party, there exist two people who shook hands with the same number of people.

5. Let S be a set of 55 distinct positive integers, from 1 to 100 inclusive. Show that there must be two elements of S that differ by 9.

6. Among n+1 integers from {1,2,…,2n} there are two which one divides the other.

7. Among n+1 integers from {1,2,…,2n} there are two which are relatively prime to each other.

V. Problems 1. Five points are chosen in a 2 x 2 square. Show that two points are within a distance of ¥2 of each other.

2. Every point in the plane is colored red, green, or blue. Prove that there exists a rectangle in the plane such that all four of its vertices are the same color.

Is this if each point is colored in one of N colors?

3. From a set of 10 distinct 2 digit numbers, there exist two disjoint subsets such that their elements have the same sum.

4. Find the greatest positive integer n for which there exist n nonnegative integers X1, X2, …, Xn, not all zero, such that for any sequence Y1, Y2, …, Yn composed of elements of {-1,0,1} and not all zero, n3 does not divide X1Y1+ X2Y2+…+ XnYn 5 Each of m cards is labeled by one of the numbers 1, 2, …, m. Prove that if the sum of the labels of any subset of cards is not a multiple of m+1, then each card is labeled by the same number.

Feng 2

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VI. Hints 1. Divide the square into areas.

2. Choose a plane of a specific size. Use pigeonhole twice (for vertical and horizontal lines) 3. How many subset sums are there? How many sums are possible?

4. Number of subsets of {X1, X2, …, Xn } greater than n3 5. Consider mod m+1 of the sums of cards from 1 to n, 1 n m

Bonus:

Feng 3

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▼ ② ◆ !♦♥ ♣✉♥&♦! ♠❡❞✐♦! ❡♥ ❧♦! ❧❛❞♦! ❞❡❧ ❝✉❛❞.❛❞♦ ❆❇❈❉ ✱ ❞♦♥❞❡ ❚ ❡! ✉♥ ♣✉♥&♦ ❞❡ &❛♥❣❡♥❝✐❛✳ ❙✐ ❈❚ = ❦ ✱ ❡♥❝♦♥&.❛. ❡❧ 3.❡❛ ❞❡❧ ❝✉❛❞.❛❞♦ ❆❇❈❉ ✳

!♦❜❧❡♠❛ ✸

❉♦" ❝✐%❝✉♥❢❡%❡♥❝✐❛" Γ✶ ② Γ✷ "❡ ✐♥,❡%"❡❝,❛♥ ❡♥ ❧♦" ♣✉♥,♦" ❆ ② ❇ ✳ ❈♦♥"✐❞❡%❡ ✉♥❛ ❝✐%❝✉♥❢❡%❡♥❝✐❛ Γ ❝♦♥,❡♥✐❞❛ ❡♥ Γ✶ ② Γ✷ ✱ ,❛♥❣❡♥,❡ ❛ ❡❧❧❛" %❡"♣❡❝,✐✈❛♠❡♥,❡ ❡♥ ❉ ② ❊ ✳ ❙❡❛♥ ❈ ✉♥♦ ❞❡ ❧♦" ♣✉♥,♦" ❞❡ ✐♥,❡%"❡❝❝✐7♥ ❞❡ ❧❛ %❡❝,❛ ❆❇ ❝♦♥ Γ✱ ❋ ❧❛ ✐♥,❡%"❡❝❝✐7♥ ❞❡ ❧❛ %❡❝,❛ ❊❈ ❝♦♥ Γ✷ ✱ ② ● ❧❛ ✐♥,❡%"❡❝❝✐7♥ ❞❡ ❧❛ %❡❝,❛ ❉❈ ❝♦♥ Γ✶ ✳ ❙❡❛♥ ❍ ❡ ■ ❧♦" ♣✉♥,♦" ❞❡ ✐♥,❡%"❡❝❝✐7♥ ❞❡ ❧❛ %❡❝,❛ ❊❉ ❝♦♥ Γ✶ ② Γ✷ %❡"♣❡❝,✐✈❛♠❡♥,❡✳ ❉❡♠✉❡",%❡ 8✉❡ ❋ ✱ ● ✱ ❍ ❡ ■ ❡",9♥ "♦❜%❡ ✉♥❛ ♠✐"♠❛ ❝✐%❝✉♥❢❡%❡♥❝✐❛✳

!♦❜❧❡♠❛ ✷✱ ❈❊◆❚❘❖ ✷✵✵✾