reﬂections on the 69th william lowell putnam … · hilary putnam in 1970 matiyasevich solved...

Reflections on the 69thWilliam Lowell Putnam

Mathematical Competition

Armstrong Putnam Team

The Hudson Colloquium Series

Willaim Lowell Putnam II

Lawyer & BankerDied in 1923 at age around 61.wife: Elizabeth Lowell Putnam

In 1927 she establishedthe William Lowell Putnam Intercollegiate Memorial Fundin order to begin a college-level mathematics competition,

the William Lowell Putnam Mathematical Competition.The competition begun in 1935.

Hilary Putnam

In 1970 Matiyasevichsolved Hilbert’s 10th problembuilding on earlier work ofDavis, Putnam, and Robinson.

Hilbert’s 10th problem:Is there an algorithm which determines whether or not

every Diophantine equation f (x1, . . . ,xn) = 0 has an integer solution.

Hilary Putnam

In 1970 Matiyasevichsolved Hilbert’s 10th problembuilding on earlier work ofDavis, Putnam, and Robinson.

Hilbert’s 10th problem:Is there an algorithm which determines whether or not

every Diophantine equation f (x1, . . . ,xn) = 0 has an integer solution.

Ans: No, there isn’t

Putnam Mathematical Competition

When: First Saturday in December.Duration:

• 3 hrs (morning) for six problems• 2 hr for lunch• 3 hrs (afternoon) for six problems

Where: Participating universities.

The twelve problems can typically be solvedwith only basic knowledge of college mathematics

but require extensive creative thinking.


Grading:

Each problem is graded on a basis of 0 to 10 points.

All the necessary work to justify an answer and all the neces-sary steps of a proof must be shown clearly to obtain full credit.

Some partial credit may be given, but only when a contestanthas shown significant and substantial progress toward a solu-tion.


Statistics for the 69th Competition:

• 3627 participants from 545 institutions.• 1712 (47%) scored 0.• 364 (10%) scored 10.• 157 (4%) scored 20.• Putnam Fellows:

117, 110, 108, 102, 101

Each receives $2,500.

Putnam Fellows for the 69th Competition

Brian R Lawrence Cal TechSeok Hyeong Lee Stanford Univ.

Arnav Tripathy Harvard Univ.Bohua Zhan MIT

Yufei Zhao MIT

Putnam Fellowsfor the 37th Competition (1974)

Putnam Fellowsfor the 37th Competition (1974)

Grant M Roberts WaterlooJames B. Saxe Union College

Karl Rubin Princeton Univ.Phillip Strenski AASU

Thomas G. Goodwillie Harvard

Famous Putnam Fellows

Fields Medalists:Milnor, Mumford, Quillen

Nobel Prize Winners in Physics:Feynman, Wilson

Four-time Putnam Fellows

Don Coppersmith (MIT) 1968-1971Arthur Rubin (Purdue, CalTech) 1970-1973Bjorn Poonen (Harvard) 1985-1988Ravi Vakil (Toronto) 1988-1991Gabriel Carroll (UC Berk., Harvard) 2000-2003Reid Barton (MIT) 2001-2004Daniel Kane (MIT) 2003-2006

Seminar, Fall 2008

InstructorsJames Brawner, Sungkon Chang

Participating StudentsAlex Collins, Dan Holland, Scott King

Participating Faculty MembersTim McMillan, Tim Ellis

Seminar, Fall 2008

InstructorsJames Brawner, Sungkon Chang

Participating StudentsAlex Collins, Dan Holland, Scott King

Participating Faculty MembersTim McMillan, Tim Ellis

Activities• One-hour seminar per week.• We worked on seven sets of problems.• The students took one take-home (review) test.

Website: Department � Curriculum � Putnam Course

Seminar, Fall 2008

Problems of The 69th Competition

Show the test.

Solutions of Problemsof The 69th Competition

Scott King will present a solution of

Solutions of Problemsof The 69th Competition

Dan Holland will present a solution of

Solution of Problem B3

The 4-dimensional hypercube of side length 1 centered at the originis the subset

{(x1,x2,x3,x4) ∈ R4 : |xi| ≤ 1/2, i = 1,2,3,4}.


A hyperplane in R4 is a subset{(x1,x2,x3,x4) ∈ R4 : ax1 +bx2 + cx3 +dx4 = e},

which is 3-dimensional.

A plane in R4 isthe intersection of two hyperplanes,


A line in R4 isthe intersection of three hyperplanes,



Parametric Vector Equations

A hyperplane in R4 is given by{sa+ tb+uc+d : s, t,u ∈ R},

which is 3-dimensional,

A plane in R4 is given by{sa+ tb+ c : s, t ∈ R},


A line in R4 is given by{sa+b : s ∈ R},



Parametric Vector Equations

A hypersurface in R4 is given by{(

f (s, t,u),g(s, t,u),h(s, t,u), j(s, t,u))∈ R4 : s, t,u ∈ R},

which is 3-dimensional,

A surface in R4 is given by{(

f (s, t),g(s, t),h(s, t), j(s, t))∈ R4 : s, t ∈ R},which is 2-dimensional.

A curve in R4 is given by{(

f (s),g(s),h(s), j(s))∈ R4 : s ∈ R},



A circle in R3


A circle in R4 is the intersection of

a (3-dimensional) hyper-sphere:

(x1− c1)2 + · · ·+(x4− c4)2 = r2,

and a plane H.


The 4-dimensional hypercube of side length 1 centered at the originis the subset

{(x1,x2,x3,x4) ∈ R4 : |xi| ≤ 1/2, i = 1,2,3,4}.

A circle centered at the origin in R4 is{(x1,x2,x3,x4) ∈ H : x2

1 + · · ·+ x24 = r2}.

where H is a (2-dimensional) plane.


Lemma: Any circle C within the unit hypercube can be translated sothat it is centered on the origin and contained in the hypercube.



proof: Let C be contained in the hyper-sphere:

(x1− c1)2 + · · ·+(x4− c4)2 = r2.

Let (a1,a2,a3,a4) be a point in C. Then, the point at the opposite side is

(2c1−a1, . . . ,2c4−a4).

Thus, −.5≤ ak ≤ .5 and −.5≤ 2ck−ak ≤ .5. This implies that

−.5≤ ak− ck ≤ .5.



Cor: It is sufficient to consider circles centered at the origin.


Prop: All circles centered at the origin in R4 are given by{a cosθ +b sinθ : θ real}where a and b are orthog. to e.o., and |a|= |b|.



proof: Let C be a circle centered at the origin in R4.Then, by definition, C is the intersection of a plane and a

hyper-sphere: C = {sa+ t b : |sa+ t b|= r}. So, |a|= |b|= r.



proof: Let C be a circle centered at the origin in R4.Then, by definition, C is the intersection of a plane and a

hyper-sphere: C = {sa+ t b : |sa+ t b|= r}. So, |a|= |b|= r.We can choose the two vectors such that a ·b = 0. Then,

r2 = |sa+ t b|2 = (sa+ t b) · (sa+ t b) = r2(s2 + t2).

Thus, s2 + t2 = 1 and hence,s = cosθ and t = sinθ for some θ .

Therefore, C is given by such a parametric vector equation.


Lemma: A circle C contained within the unit hypercube cannot havea radius more than

√2/2.



√2/2.

proof: Recall the prop: C = {a cosθ +b sinθ} for some a and b.The coordinates of the points on this circle can, by the prop., begiven by

ak cosθ +bk sinθ = 〈ak,bk〉 · 〈cosθ ,sinθ〉.



√2/2.

proof: Recall the prop: C = {a cosθ +b sinθ} for some a and b.The coordinates of the points on this circle can, by the prop., begiven by

ak cosθ +bk sinθ = 〈ak,bk〉 · 〈cosθ ,sinθ〉.

Hence, |ak cosθ +bk sinθ | ≤√

a2k +b2

k.Moreover, the equality holds for some θ .

Therefore,√

a2k +b2

k ≤ 1/2.



√2/2.

proof: So,√

a2k +b2

k ≤ 1/2. Since r = |a|= |b|,

2r2 = |a|2 + |b|2

= (a21 + · · ·+a2

4)+(b21 + · · ·+b2

4)

=4

∑k=1

(a2k +b2

k)≤ 4 · (1/4) = 1.


Lemma: Let a = 〈.5, .5,0,0〉 and b = 〈0,0, .5, .5〉.Then, the following circle has radius

√2/2:

{a cosθ +b sinθ : θ real}.


Lemma: Let a = 〈.5, .5,0,0〉 and b = 〈0,0, .5, .5〉.Then, the following circle has radius

√2/2:

{a cosθ +b sinθ : θ real}.

Ans: The largest circle has radius√

2/2.

Problem A5

Problem A5:

Let n ≥ 3 be an integer. Let f (x) and g(x) be polynomials withreal coefficients such that the points(

f (1),g(1)),(

f (2),g(2)), . . . ,

(f (n),g(n)

)are the vertices of a regular n-gon in counterclockwise order.Prove that at least one of f (x) and g(x) has degree greater thanor equal to n−1.

Example for Problem A5

Let n = 3, and consider three points(cos(2πk/3),sin(2πk/3)

)for k = 0,1,2.



)for k = 0,1,2.

The x-coordinates: (1,1), (2,−1/2), (3,−1/2).The y-coordinates: (1,0), (2,

√3/2), (3,−

√3/2).

f (x) 6= ax+b;

g(x) 6= cx+d.



)for k = 0,1,2.


√3/2), (3,−

√3/2).

F(x) = 1 · (x−2)(x−3)(1−2)(1−3)

+(−1/2) · (x−1)(x−3)(2−1)(2−3)

+(−1/2) · (x−1)(x−2)(3−1)(3−2)

.



)for k = 0,1,2.


√3/2), (3,−

√3/2).

G(x) = 0 · (x−2)(x−3)(1−2)(1−3)

+(√

3/2) · (x−1)(x−3)(2−1)(2−3)

+(−√

3/2) · (x−1)(x−2)(3−1)(3−2)

.



)for k = 0,1,2.


√3/2), (3,−

√3/2).

F(x) and G(x) are quadratic,and are called Lagrange Interpolations.



)for k = 0,1,2.


√3/2), (3,−

√3/2).

Lemma: If f (x) and g(x) are polynomials which interpolate those x-and y-coordinates, resp., then

f (x) = (x−1)(x−2)(x−3)Q1(x)+F(x);

g(x) = (x−1)(x−2)(x−3)Q2(x)+G(x)

proof: f (x)−F(x) and g(x)−G(x) are divisible by(x−1)(x−2)(x−3).

Solutions for Problem A5

Let (k,ak) for k = 1, . . . ,n be n points.

Lemma: If f (x) and F(x) are two polynomials which interpolate then points, then

f (x) = (x−1) · · ·(x−n)Q(x)+F(x).

Cor: There is at most one polynomial of degree ≤ n−1 which inter-polates the n points.


Let (k,ak) for k = 1, . . . ,n be n points.

Lemma: If f (x) and F(x) are two polynomials which interpolate then points, then

f (x) = (x−1) · · ·(x−n)Q(x)+F(x).

Cor: There is at most one polynomial of degree ≤ n−1 which inter-polates the n points.

Cor: The Lagrange interpolation is the polynomial of the smallestdegree that interpolates the n points.


Let n = 5, and consider the n points(cos(2πk/n),sin(2πk/n)

)for k = 0,1, . . . ,(n−1).

Let F(x) and G(x) be the Lagrange interpolationsfor the x- and y-coordinates, resp.

Let ak = cos(2πk/n) and bk = sin(2πk/n). Then,

F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)

where Pk(x) is the polynomial (x−1) · · ·(x−5)/(x− k).



F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)




F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)

where Pk(x) is the polynomial (x−1) · · ·(x−5)/(x− k).The coefficients of x4 in F(x) and G(x) are

A =5

∑k=1

ak

Pk(k)and B =

5

∑k=1

bk

Pk(k).



F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)


A =5

∑k=1

ak · ck and B =5

∑k=1

bk · ck

where ck = 1/Pk(k). Then,



F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)


A =5

∑k=1

ak · ck and B =5

∑k=1

bk · ck

where ck = 1/Pk(k). Then,A+ iB = c1(a1 + ib1)+ · · ·+ c5(a5 + ib5) = c1ζ 0

5 + · · ·+ c5ζ 45 .


Let ζ5 = cos(2π/5)+ isin(2π/5),Pk(x) is the polynomial (x−1) · · ·(x−5)/(x− k),

and ck = 1/Pk(k).Then, A+ iB = c1ζ 0

5 + · · ·+ c5ζ 45 .

c1 = 1/(−1)(−2)(−3)(−4)

c2 = 1/(1)(−1)(−2)(−3)

c3 = 1/(2)(1)(−1)(−2)

c4 = 1/(3)(2)(1)(−1)

c5 = 1/(4)(3)(2)(1)




5 + · · ·+ c5ζ 45 .

c1 = 1/(−1)(−2)(−3)(−4)

c2 = c1 · (−4)/(1)

c3 = 1/(2)(1)(−1)(−2)

c4 = 1/(3)(2)(1)(−1)

c5 = 1/(4)(3)(2)(1)




5 + · · ·+ c5ζ 45 .

c1 = 1/(−1)(−2)(−3)(−4)

c2 = c1 · (−4)/(1)

c3 = c1 · (−4)(−3)/(1)(2)

c4 = 1/(3)(2)(1)(−1)

c5 = 1/(4)(3)(2)(1)




5 + · · ·+ c5ζ 45 .

c1 = 1/(−1)(−2)(−3)(−4)

c2 = c1 · (−4)/(1)

c3 = c1 · (−4)(−3)/(1)(2)

c4 = c1 · (−4)(−3)(−2)/(1)(2)(3)

c5 = c1 · (−4)(−3)(−2)(−1)/(1)(2)(3)(4)


c1 = 1/(−1)(−2)(−3)(−4)

c2 = c1 · (−4)/(1)

c3 = c1 · (−4)(−3)/(1)(2)

c4 = c1 · (−4)(−3)(−2)/(1)(2)(3)

c5 = c1 · (−4)(−3)(−2)(−1)/(1)(2)(3)(4)

Then,

A+ iB = = c1ζ05 + · · ·+ c5ζ

45

= c1ζ05 − c1

(41

)ζ5 + c1

(42

)ζ

25 − c1

(43

)ζ

35 + c1

(44

)ζ

45 .

= c1(1−ζ5)4 ⇒ AB 6= 0.



F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)


We showed thatF(x) and G(x) have degree 4.



F(x) = a1P1(x)P1(1)

+ · · ·+a5P5(x)P5(5)

,

G(x) = b1P1(x)P1(1)

+ · · ·+b5P5(x)P5(5)


Lemma: If f (x) and g(x) are polynomials which interpolate ak’s andbk’s, resp., then their degress are greater than or equal to 4.


Lemma: Let f (x) and g(x) be polynomials which interpolate the x-and the y-coordinates of a regular pentagon not necessarily centeredat the origin. Then, at least one has degree at least 4.


Lemma: Let f (x) and g(x) be polynomials which interpolate the x-and the y-coordinates of a regular pentagon not necessarily centeredat the origin. Then, at least one has degree at least 4.

proof: Suppose that their degrees are both < 4. Via a shift, arotation, and a dialation, we have(

fg

)→ t

(( f +m)cosθ − (g+n)sinθ

( f +m)sinθ +(g+n)cosθ

)

The 70th William Lowell PutnamMathematical Competition,

December 5, 2009

reﬂections on the 69th william lowell putnam … · hilary putnam in 1970 matiyasevich solved...

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