Putnam Marathon
by djmathman, Dec 31, 2020, 5:20 PM
Differential geometry destroyed me last semester, but at least I still had time to do some vintage Putnam.
Joy of Mathematics. Let
. Prove that
![\[
2\sin\alpha + \tan\alpha \geq 3\alpha.
\]](//latex.artofproblemsolving.com/3/1/2/312a08e810c59b737f37cf4922ae0472385da15e.png)
Putnam 1965 B3. Find all twice differentiable functions
satisfying
![\[
f(x)^2 - f(y)^2 = f(x-y)f(x+y).
\]](//latex.artofproblemsolving.com/0/6/7/0679da38da7c678fac5782ad6f983c616fa36645.png)
Putnam 1968 A6. Determine all polynomials of the form
with 
(
, 
) such that each has only real zeros.
Putnam 1968 B6. A set of real numbers is compact if it is closed and bounded. Show that there does not exist a sequence
of compact sets of rational numbers such that each compact set of rationals is contained in at least one 
.
Putnam 1969 A6. Let a sequence
be given, and let 
, 
. Suppose that the sequence 
converges. Prove that the sequence also converges.
Putnam 1972 A6. Let
be an integrable function in 
and suppose 
, 
, 
, 
and 
. Show that 
in a set of positive measure.
Putnam 1972 B4. Let
be an integer greater than 
. Show that there exists a polynomial 
with integral coefficients such that 
.
Putnam 1972 B6. Let
be a set of positive integers. Prove that the polynomial 
has no roots inside the circle 
.
Putnam 1976 A5. In the
plane, if 
is the set of points inside and on a convex polygon, let 
be the distance from to the nearest point of . (a) Show that there exist constants 
, 
, and 
, independent of , such that
![\[
\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-D(x,y)}\,dx\,dy = a + bL + cA,
\]](//latex.artofproblemsolving.com/f/d/8/fd8c84800620fec871b09a515fed29f3d9627aac.png)
where 
is the perimeter of and 
is the area of . (b) Find the values of , , and .
Putnam 1976 B6. As usual, let
denote the sum of all the (positive integral) divisors of 
. For example, if 
is a prime, then 
. Motivated by notion of a "perfect" number, a positive integer is called "quasiperfect" if 
. Prove that every quasiperfect number is the square of an odd integer.
Steinhaus. Let be a Lebesgue measurable set in 
with 
. Then the difference set
![\[
A - A \coloneqq \{x - y: x, y\in A\}
\]](//latex.artofproblemsolving.com/1/b/7/1b74b5ace07134e09db61ea6a5a33b356e1ea122.png)
contains a nonzero neighborhood of the origin.
Joy of Mathematics. Let
. Prove that![\[
2\sin\alpha + \tan\alpha \geq 3\alpha.
\]](http://latex.artofproblemsolving.com/3/1/2/312a08e810c59b737f37cf4922ae0472385da15e.png)
.
and![\[
f'(\alpha) = 2\cos\alpha + \sec^2\alpha = \cos \alpha + \cos\alpha + \frac{1}{\cos^2\alpha} \geq 3,
\]](http://latex.artofproblemsolving.com/b/b/3/bb341dc6e5cd4aa9071eade2dac8b87004e00b87.png)
where the last inequality is by AM-GM. Thus 
by the Fundamental Theorem of Calculus.Putnam 1965 B3. Find all twice differentiable functions
satisfying![\[
f(x)^2 - f(y)^2 = f(x-y)f(x+y).
\]](http://latex.artofproblemsolving.com/0/6/7/0679da38da7c678fac5782ad6f983c616fa36645.png)
to get![\[
2f(x)f'(x) = f(x+y)f'(x-y) + f'(x+y)f(x-y).
\]](http://latex.artofproblemsolving.com/5/3/b/53b9a4a47eef0abc46c6269da28e5af63ba2b2a3.png)
Now observe that the RHS is independent of 
since the LHS is by default. That is, we may reparametrize to get the existence of a differentiable function 
satisfying![\[
f(x)f'(s-x) + f'(x)f(s-x) =: g(s).
\]](http://latex.artofproblemsolving.com/d/5/b/d5bef9360dd3155df714e14e897aa6f398eeec3c.png)
Differentiate wrt 
.
or 
for some constant 
.
you get linear, if 
you get exponential (hyperbolic trig), and if 
you get trigonometric.Putnam 1968 A6. Determine all polynomials of the form
with 
(
, 
) such that each has only real zeros.
without loss. We claim that the only polynomials of this form are 
,
,
,
,
,
(whoops missed these last two possibilities at first). These polynomials trivially work, so it remains to show they are the only ones.
,
,
denote the roots of the given polynomial. By Vieta,![\[
\sum_{i=1}^nr_i^2 = a_1^2 - 2a_2 = 1 - 2a_2.
\]](http://latex.artofproblemsolving.com/d/b/d/dbd3d4d3e6d7b2927429edcb4ebf3d3610dac67a.png)
This expression is negative when 
,
,
.![\[
\frac 3n = \frac 1n\sum_{i=1}^nr_i^2 \geq \left(\prod_{i=1}^nr_i^2\right)^{1/n} = 1;\quad(*)
\]](http://latex.artofproblemsolving.com/a/b/e/abec31da813f73a60654bb2a778132d6f6ed2dcb.png)
ergo, 
.
is easy since both possible polynomials are linear. The case 
is easy by manually checking the two possibilities. Finally, in the case 
,
,
for all 
;Putnam 1968 B6. A set of real numbers is compact if it is closed and bounded. Show that there does not exist a sequence
of compact sets of rational numbers such that each compact set of rationals is contained in at least one 
.
of compact sets. For each positive integer 
,
.
be any irrational number in the interval 
.
such that 
,
(contradicting the fact that 
is closed). As a result, we may choose 
.
.
is a bounded set of rational numbers whose sole limit point is 
-- hence 
for each Putnam 1969 A6. Let a sequence
be given, and let 
, 
. Suppose that the sequence 
converges. Prove that the sequence also converges.
;
from each 
we may assume 
.
as 
;
be arbitrary. Observe that, since 
as 
such that 
for all 
.![\[
|x_n| = \frac{|y_n - x_{n-1}|}2 \leq \frac{|y_n| + |x_{n-1}|}2 < \frac{\tfrac \varepsilon 2 + |x_{n-1}|}2.
\]](http://latex.artofproblemsolving.com/2/3/9/239da847f7cd8faebf52d40b853d3d9bca3d59f3.png)
Thus 
,![\[
|x_n| < \frac\varepsilon 2 + \frac{|x_N|}{2^{n-N}} < \varepsilon
\]](http://latex.artofproblemsolving.com/b/4/3/b43806d0337812b0008945d55e19c064f7e9fd9d.png)
for sufficiently large Putnam 1972 A6. Let
be an integrable function in 
and suppose 
, 
, 
, 
and 
. Show that 
in a set of positive measure.
this took forever >.<Putnam 1972 B4. Let
be an integer greater than 
. Show that there exists a polynomial 
with integral coefficients such that 
.
with integer coefficients such that 
where 
.![\[
x + x^c - (x+x^c)^c = x - cx^{2c + 1}\cdots.
\]](http://latex.artofproblemsolving.com/d/8/d/d8d72bd1fc878a5f577aed49024a43707cfbbd00.png)
Inductively push the minimum degree of the nonlinear terms higher and higher.
,
,
for nonnegative integers 
and 
.Putnam 1972 B6. Let
be a set of positive integers. Prove that the polynomial 
has no roots inside the circle 
.
.
then we are trivially done, so assume 
.
,
be a complex number satisfying 
.
This means 
,
.
,![\[
(1-z)P(z) = (1-z)\left(1 + z + \sum_{j=2}^kz^{n_j}\right) = 1 - z^2 + \sum_{m=3}^{n_k + 1}\varepsilon_mz^m,
\]](http://latex.artofproblemsolving.com/c/b/6/cb64d993024880cea9bf8304202271cdcaab9c53.png)
where each 
is either 
,![\[
1 = \left|z^2 - \sum_{m=3}^{n_k+1}\varepsilon_mz^m\right| \leq |z|^2 + |z|^3 + \cdots = \frac{|z|^2}{1 - |z|},
\]](http://latex.artofproblemsolving.com/d/5/8/d58620da71f6d7ed996e5e8a72a44a7bbd9154b9.png)
leading to the same bound as before.Putnam 1976 A5. In the
plane, if 
is the set of points inside and on a convex polygon, let 
be the distance from to the nearest point of . (a) Show that there exist constants 
, 
, and 
, independent of , such that![\[
\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-D(x,y)}\,dx\,dy = a + bL + cA,
\]](http://latex.artofproblemsolving.com/f/d/8/fd8c84800620fec871b09a515fed29f3d9627aac.png)
where 
is the perimeter of and 
is the area of . (b) Find the values of , , and .
and vertices 
,
.
and 
through the endpoints of 
denote the region bounded by 
into several unbounded regions 
,
into several parts by writing
There are three cases to consider.
.
,
.
to 
.
,![\[
\int_{P_j}e^{-D(x,y)}\,dx\,dy = \int_0^{\text{length}(s_j)}\int_0^\infty e^{-x}\,dx = \text{length}(s_j).
\]](http://latex.artofproblemsolving.com/c/d/2/cd25fb167155fe5f80bcf9aa1f5eb9ca53bfae4d.png)
Summing over all ![\[
\sum_{j=1}^k\int_{P_j}e^{-D(x,y)}\,dx\,dy = \sum_{j=1}^k\text{length}(s_j) = L.
\]](http://latex.artofproblemsolving.com/f/b/8/fb8f1710cb010d4fef6ef669e67672641a7be8d8.png)
,
.
is actually the entire plane 
.![\[
\sum_{j=1}^k\int_{Q_j}e^{-D(x,y)}\,dx\,dy = \int_{\mathbb R^2}e^{-\sqrt{x^2+y^2}}\,dx\,dy = 2\pi\int_0^\infty re^{-r}\,dr = 2\pi.
\]](http://latex.artofproblemsolving.com/c/c/8/cc82ac99425a3709d70c88e82e7221ec131241a3.png)
,
,
in the expression above.Putnam 1976 B6. As usual, let
denote the sum of all the (positive integral) divisors of 
. For example, if 
is a prime, then 
. Motivated by notion of a "perfect" number, a positive integer is called "quasiperfect" if 
. Prove that every quasiperfect number is the square of an odd integer.
,![\[
\sigma(N) = \sigma(p_1^{e_1})\cdots\sigma(p_k^{e_k}) = (1+p_1 + \cdots + p_1^{e_1})\cdots (1+p_k + \cdots + p_k^{e_k}).
\]](http://latex.artofproblemsolving.com/4/1/f/41f70b6b14cd8e6f321774eb1fa47abcb972c092.png)
Now each factor 
is odd if and only if either 
or (in the case where 
)
is even. Thus, since 
for some odd positive integer ![\[
(2^{a+1} - 1)\sigma(m^2) = 2^{a+1}m^2 + 1.
\]](http://latex.artofproblemsolving.com/f/e/5/fe53f78e761d384dcbb197d7fad4afb7327b9eca.png)
Now suppose 
.
,
with 
.![\[
2^{a+1}m^2 + 1 \equiv m^2 + 1 \equiv 0\pmod p,
\]](http://latex.artofproblemsolving.com/f/3/3/f33c963a6f8469ca9862f6dfba7a5c5e1f2999db.png)
contradiction since 
and 
is the square of an odd integer.Steinhaus. Let
be a Lebesgue measurable set in 
with 
. Then the difference set![\[
A - A \coloneqq \{x - y: x, y\in A\}
\]](http://latex.artofproblemsolving.com/1/b/7/1b74b5ace07134e09db61ea6a5a33b356e1ea122.png)
contains a nonzero neighborhood of the origin.
.![\[
m(A\cap I)\leq \lambda m(I)
\]](http://latex.artofproblemsolving.com/6/b/e/6be24b9bbd098a8851eff4670863c82985347e96.png)
for every interval 
.
.
satisfying 
,![\[
m(A) \leq \sum_{n\geq 1}m(A\cap I_n)\leq \sum_{n\geq 1}\lambda m(I_n) = \lambda\sum_{n\geq 1}m(I_n).
\]](http://latex.artofproblemsolving.com/0/f/4/0f47c6bb313747d570b4b77fdd5a526740bf7755.png)
Taking the infimum of the right hand side over all such collections of intervals yields 
,
we deduce 
.![$I = [0,1]$](http://latex.artofproblemsolving.com/1/f/c/1fce074fdcba0b6f4deeb98871c80f60d51fb443.png)
,
and 
.
,![\begin{align*}
m(B\cap (B+\delta)) &= m(B) + m(B+\delta) - m(B\cup(B+\delta)) \\
&\geq 2(\varepsilon + \tfrac12) - m([1,1+\varepsilon]) = \varepsilon.
\end{align*}](http://latex.artofproblemsolving.com/9/3/f/93f54d82f86762dcc1fb1afa02db47e17fcb0d8f.png)
Hence there exists some 
,
.
.![\[
m(A) = \lim_{n\to\infty}m(A\cap[-n,n]),
\]](http://latex.artofproblemsolving.com/a/8/d/a8db06a8c73e36d540bc7b573b26fbe8e0a95e26.png)
so there exists 
such that ![$m(A\cap[-n,n]) > 0$](http://latex.artofproblemsolving.com/8/c/c/8ccec2c3fe62b4c3e148ab76a96d13a893a229cf.png)
.![$A\cap[-n,n]$](http://latex.artofproblemsolving.com/a/6/8/a682a3e9e6519935a556edf7fec755d0aee193d8.png)
.
such that 
.
.This post has been edited 1 time. Last edited by djmathman, Dec 31, 2020, 5:36 PM
April 2025
November 2024