MathDB
Problems
Contests
National and Regional Contests
Brazil Contests
Brazil National Olympiad
1979 Brazil National Olympiad
1979 Brazil National Olympiad
Part of
Brazil National Olympiad
Subcontests
(5)
5
1
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Quadrllaterals, areas and midpoints
[*] ABCD is a square with side 1. M is the midpoint of AB, and N is the midpoint of BC. The lines CM and DN meet at I. Find the area of the triangle CIN. [*] The midpoints of the sides AB, BC, CD, DA of the parallelogram ABCD are M, N, P, Q respectively. Each midpoint is joined to the two vertices not on its side. Show that the area outside the resulting 8-pointed star is
2
5
\frac{2}{5}
5
2
the area of the parallelogram. [*] ABC is a triangle with CA = CB and centroid G. Show that the area of AGB is
1
3
\frac{1}{3}
3
1
of the area of ABC. [*] Is (ii) true for all convex quadrilaterals ABCD?
4
1
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Diophantine Equations with the same number of solutions
Show that the number of positive integer solutions to
x
1
+
2
3
x
2
+
3
3
x
3
+
…
+
1
0
3
x
10
=
3025
x_1 + 2^3x_2 + 3^3x_3 + \ldots + 10^3x_{10} = 3025
x
1
+
2
3
x
2
+
3
3
x
3
+
…
+
1
0
3
x
10
=
3025
(*) equals the number of non-negative integer solutions to the equation
y
1
+
2
3
y
2
+
3
3
y
3
+
…
+
1
0
3
y
10
=
0
y_1 + 2^3y_2 + 3^3y_3 + \ldots + 10^3y_{10} = 0
y
1
+
2
3
y
2
+
3
3
y
3
+
…
+
1
0
3
y
10
=
0
. Hence show that (*) has a unique solution in positive integers and find it.
3
1
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Locus of the ortocenter
The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter.
2
1
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Remainder in the divison by a quadratic polynomial
The remainder on dividing the polynomial
p
(
x
)
p(x)
p
(
x
)
by
x
2
−
(
a
+
b
)
x
+
a
b
x^2 - (a+b)x + ab
x
2
−
(
a
+
b
)
x
+
ab
(where
a
≠
b
a \not = b
a
=
b
) is
m
x
+
n
mx + n
m
x
+
n
. Find the coefficients
m
,
n
m, n
m
,
n
in terms of
a
,
b
a, b
a
,
b
. Find
m
,
n
m, n
m
,
n
for the case
p
(
x
)
=
x
200
p(x) = x^{200}
p
(
x
)
=
x
200
divided by
x
2
−
x
−
2
x^2 - x - 2
x
2
−
x
−
2
and show that they are integral.
1
1
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Sine functions and intervals
Show that if
a
<
b
a < b
a
<
b
are in the interval
[
0
,
π
2
]
\left[0, \frac{\pi}{2}\right]
[
0
,
2
π
]
then
a
−
sin
a
<
b
−
sin
b
a - \sin a < b - \sin b
a
−
sin
a
<
b
−
sin
b
. Is this true for
a
<
b
a < b
a
<
b
in the interval
[
π
,
3
π
2
]
\left[\pi,\frac{3\pi}{2}\right]
[
π
,
2
3
π
]
?