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1964 MMPC
1964 MMPC
Part of
Michigan Mathematics Prize Competition
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1964 MMPC , Part 2 = Michigan Mathematics Prize Competition
p1. The edges of a tetrahedron are all tangent to a sphere. Prove that the sum of the lengths of any pair of opposite edges equals the sum of the lengths of any other pair of opposite edges. (Two edges of a tetrahedron are said to be opposite if they do not have a vertex in common.) p2. Find the simplest formula possible for the product of the following
2
n
−
2
2n - 2
2
n
−
2
factors:
(
1
+
1
2
)
,
(
1
−
1
2
)
,
(
1
+
1
3
)
,
(
1
−
1
3
)
,
.
.
.
,
(
1
+
1
n
)
,
(
1
−
1
n
)
\left(1+\frac12 \right),\left(1-\frac12 \right), \left(1+\frac13 \right) , \left(1-\frac13 \right),...,\left(1+\frac{1}{n} \right), \left(1-\frac{1}{n} \right)
(
1
+
2
1
)
,
(
1
−
2
1
)
,
(
1
+
3
1
)
,
(
1
−
3
1
)
,
...
,
(
1
+
n
1
)
,
(
1
−
n
1
)
. Prove that your formula is correct.p3. Solve
(
x
+
1
)
2
+
1
x
+
1
+
(
x
+
4
)
2
+
4
x
+
4
=
(
x
+
2
)
2
+
2
x
+
2
+
(
x
+
3
)
2
+
3
x
+
3
\frac{(x + 1)^2+1}{x + 1} + \frac{(x + 4)^2+4}{x + 4}=\frac{(x + 2)^2+2}{x + 2}+\frac{(x + 3)^2+3}{x + 3}
x
+
1
(
x
+
1
)
2
+
1
+
x
+
4
(
x
+
4
)
2
+
4
=
x
+
2
(
x
+
2
)
2
+
2
+
x
+
3
(
x
+
3
)
2
+
3
p4. Triangle
A
B
C
ABC
A
BC
is inscribed in a circle,
B
D
BD
B
D
is tangent to this circle and
C
D
CD
C
D
is perpendicular to
B
D
BD
B
D
.
B
H
BH
B
H
is the altitude from
B
B
B
to
A
C
AC
A
C
. Prove that the line
D
H
DH
DH
is parallel to
A
B
AB
A
B
. https://cdn.artofproblemsolving.com/attachments/e/9/4d0b136dca4a9b68104f00300951837adef84c.png p5. Consider the picture below as a section of a city street map. There are several paths from
A
A
A
to
B
B
B
, and if one always walks along the street, the shortest paths are
15
15
15
blocks in length. Find the number of paths of this length between
A
A
A
and
B
B
B
. https://cdn.artofproblemsolving.com/attachments/8/d/60c426ea71db98775399cfa5ea80e94d2ea9d2.pngp6. A finite graph is a set of points, called vertices, together with a set of arcs, called edges. Each edge connects two of the vertices (it is not necessary that every pair of vertices be connected by an edge). The order of a vertex in a finite graph is the number of edges attached to that vertex. Example The figure at the right is a finite graph with
4
4
4
vertices and
7
7
7
edges. https://cdn.artofproblemsolving.com/attachments/5/9/84d479c5dbd0a6f61a66970e46ab15830d8fba.png One vertex has order
5
5
5
and the other vertices order
3
3
3
.Define a finite graph to be heterogeneous if no two vertices have the same order. Prove that no graph with two or more vertices is heterogeneous.PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.