UMass
Math 491A: Putnam Seminar
Fall 2016
Homework 2: Solutions
Problem 1. (A1/2009) Let f be a real-valued function on the plane such that for every square
ABCD in the plane f (A) + f (B) + f (C) + f (D) = 0. Does it follows that f (P ) = 0 for all points
in the plane?
Solution. Given a point P in the plane, choose a square ABCD with center P . The midpoints of
the the sides of this square form another square A1 B1 C1 D1 . The diagonals A1 C1 and B1 D1 divide
ABCD into four smaller squares. Summing the conditions for these four squares, and subtracting
the relation for ABCD and twice the relation for A1 B1 C1 D1 gives 4f (P ) = 0, i.e., f (P ) = 0.
Problem 2. (A2/2007) Find the least possible area of a convex set in the plane that intersects
both branches of the hyperbolas xy = 1 and xy = −1.
Solution. The convex set must contain a quadrilateral with a vertex on each branch of the hyperbolas, so it suffices to minimize the area of the quadrilateral with vertices A = (a, a1 ), B = (−b, 1b ),
C = (−c, − 1c ) and D = (d, − d1 ). Note that
1
1a
b
a 1/a
[AOB] = det
=
+
≥ 1,
−b 1/b
2
2 b
a
and similarly for the other 3 triangles. Consequently, [ABCD] ≥ 4 and the equality holds for the
square (±1, ±1).
Problem 3. (A1/2012) Let d1 , d2 , . . . , d12 be real numbers in the open interval (1, 12). Show that
there exist distinct indices i, j, k ∈ {1, ..., 12} such that di , dj , dk are the side lengths of an acute
triangle.
Proof. Without loss of genereality, assume
d1 ≤ d2 ≤ ... ≤ d12 .
We want to shows that d2i+2 < d2i+1 + d2i , for some 1 ≤ i ≤ 10. Assuming the contrary, we get
d23 ≥ d22 + d21 =⇒ d23 ≥ 2d21 =⇒ d24 ≥ d23 + d22 ≥ 3d21 .
Let f1 = f2 = 1, fi+1 = fi + fi−1 be the Fibonacci sequence. If d2i ≥ fi d21 , we get by induction
d2i+1 ≥ fi d21 + fi−1 d21 = fi+1 d21 .
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UMass
Math 491A: Putnam Seminar
Fall 2016
Thus,
d12 ≥ f12 d21 = 144d21 > 144 =⇒ d12 > 12,
which is contradiction.
Problem 4. (A5/1998) Let F be a finite collection of open discs in R2 whose union contains a set
E ⊆ R2 . Show that there is a pairwise disjoint subcollection D1 , . . . , Dn in F such that
E⊆
n
[
3Dj .
j=1
Here, if D is the disc of radius r centered at P , then 3D is the disc of radius 3r centered at P .
Proof. We define the sequence Di by the following ”greedy algorithm”: Let D1 be the disc with the
largest radius, D2 - the disc with the largest radius not meeting D1 , D3 - the disc with the largest
radius not meeting D1 and D2 , etc., up to some final disc Dn (this process ends since F is finite).
For an arbitrary point X ∈ E, if X lies in one of Di , we are done. Otherwise, by our construction,
X lies in a disc D ∈ F of center O and radius r, which meets one of the Di (of center Oi ) having
radius ri ≥ r. Then by the triangle inequality
XOi ≤ OX + OO1 ≤ r + (r + ri ) ≤ 3ri ,
showing that X ∈ 3Di .
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