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Harvard-MIT Mathematics Tournament
2014 HMIC
2014 HMIC
Part of
Harvard-MIT Mathematics Tournament
Subcontests
(5)
5
1
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2014 HMIC #5
Let
n
n
n
be a positive integer, and let
A
A
A
and
B
B
B
be
n
×
n
n\times n
n
×
n
matrices with complex entries such that
A
2
=
B
2
A^2=B^2
A
2
=
B
2
. Show that there exists an
n
×
n
n\times n
n
×
n
invertible matrix
S
S
S
with complex entries that satisfies
S
(
A
B
−
B
A
)
=
(
B
A
−
A
B
)
S
S(AB-BA)=(BA-AB)S
S
(
A
B
−
B
A
)
=
(
B
A
−
A
B
)
S
.
4
1
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2014 HMIC #4
Let
ω
\omega
ω
be a root of unity and
f
f
f
be a polynomial with integer coefficients. Show that if
∣
f
(
ω
)
∣
=
1
|f(\omega)|=1
∣
f
(
ω
)
∣
=
1
, then
f
(
ω
)
f(\omega)
f
(
ω
)
is also a root of unity.
3
1
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2014 HMIC #3
Fix positive integers
m
m
m
and
n
n
n
. Suppose that
a
1
,
a
2
,
…
,
a
m
a_1, a_2, \dots, a_m
a
1
,
a
2
,
…
,
a
m
are reals, and that pairwise distinct vectors
v
1
,
…
,
v
m
∈
R
n
v_1, \dots, v_m\in \mathbb{R}^n
v
1
,
…
,
v
m
∈
R
n
satisfy
∑
j
≠
i
a
j
v
j
−
v
i
∣
∣
v
j
−
v
i
∣
∣
3
=
0
\sum_{j\neq i} a_j \frac{v_j-v_i}{||v_j-v_i||^3}=0
j
=
i
∑
a
j
∣∣
v
j
−
v
i
∣
∣
3
v
j
−
v
i
=
0
for
i
=
1
,
2
,
…
,
m
i=1,2,\dots,m
i
=
1
,
2
,
…
,
m
. Prove that
∑
1
≤
i
<
j
≤
m
a
i
a
j
∣
∣
v
j
−
v
i
∣
∣
=
0.
\sum_{1\le i<j\le m} \frac{a_ia_j}{||v_j-v_i||}=0.
1
≤
i
<
j
≤
m
∑
∣∣
v
j
−
v
i
∣∣
a
i
a
j
=
0.
2
1
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2014 HMIC #2
2014
2014
2014
triangles have non-overlapping interiors contained in a circle of radius
1
1
1
. What is the largest possible value of the sum of their areas?
1
1
Hide problems
2014 HMIC #1
Consider a regular
n
n
n
-gon with
n
>
3
n>3
n
>
3
, call a line acceptable if it passes through the interior of this
n
n
n
-gon. Draw
m
m
m
different acceptable lines, so that the
n
n
n
-gon is divided into several smaller polygons.(a) Prove that there exists an
m
m
m
, depending only on
n
n
n
, such that any collection of
m
m
m
acceptable lines results in one of the smaller polygons having
3
3
3
or
4
4
4
sides.(b) Find the smallest possible
m
m
m
which guarantees that at least one of the smaller polygons will have
3
3
3
or
4
4
4
sides.