MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1951 Poland - Second Round
1951 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
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circle construction
The given points are
A
A
A
and
B
B
B
and the circle
k
k
k
. Draw a circle passing through the points
A
A
A
and
B
B
B
and defining, at the intersection with the circle
k
k
k
, a common chord of a given length
d
d
d
.
5
1
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(a^2 + b^2) \sin (A - B) = (a^2 - b^2) \sin (A + B)
Prove that if the relationship between the sides and opposite angles
A
A
A
and
B
B
B
of the triangle
A
B
C
ABC
A
BC
is
(
a
2
+
b
2
)
sin
(
A
−
B
)
=
(
a
2
−
b
2
)
sin
(
A
+
B
)
(a^2 + b^2) \sin (A - B) = (a^2 - b^2) \sin (A + B)
(
a
2
+
b
2
)
sin
(
A
−
B
)
=
(
a
2
−
b
2
)
sin
(
A
+
B
)
then such a triangle is right-angled or isosceles.
4
1
Hide problems
(n - q)^2 - (m - p) (np - mq) = 0
Prove that if equations
x
2
+
m
x
+
n
=
0
a
n
d
x
2
+
p
x
+
q
=
0
x^2 + mx + n = 0 \,\,\,\, and\,\, \,\, x^2 + px + q = 0
x
2
+
m
x
+
n
=
0
an
d
x
2
+
p
x
+
q
=
0
have a common root, there is a relationship between the coefficients of these equations
(
n
−
q
)
2
−
(
m
−
p
)
(
n
p
−
m
q
)
=
0.
(n - q)^2 - (m - p) (np - mq) = 0.
(
n
−
q
)
2
−
(
m
−
p
)
(
n
p
−
m
q
)
=
0.
3
1
Hide problems
\frac{m^2}{a-x} + \frac{n^2}{b-x} = 1
Prove that the equation
m
2
a
−
x
+
n
2
b
−
x
=
1
,
\frac{m^2}{a-x} + \frac{n^2}{b-x} = 1,
a
−
x
m
2
+
b
−
x
n
2
=
1
,
where
m
≠
0
m \ne 0
m
=
0
,
n
≠
0
n \ne 0
n
=
0
,
a
≠
b
a \ne b
a
=
b
, has real roots (
m
m
m
,
n
n
n
,
a
a
a
,
b
b
b
denote real numbers).
2
1
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trinagle inscirbed in triangle, equal ratio
In the triangle
A
B
C
ABC
A
BC
on the sides
B
C
BC
BC
,
C
A
CA
C
A
,
A
B
AB
A
B
, the points
D
D
D
,
E
E
E
,
F
F
F
are chosen respectively in such a way that
B
D
:
D
C
=
C
E
:
E
A
=
A
F
:
F
B
=
k
,
BD \colon DC = CE \colon EA = AF \colon FB = k,
B
D
:
D
C
=
CE
:
E
A
=
A
F
:
FB
=
k
,
where
k
k
k
is a given positive number. Given the area
S
S
S
of the triangle
A
B
C
ABC
A
BC
, calculate the area of the triangle
D
E
F
DEF
D
EF
1
1
Hide problems
sum of inradii = altitude
In a right triangle
A
B
C
ABC
A
BC
, the altitude
C
D
CD
C
D
is drawn from the vertex of the right angle
C
C
C
and a circle is inscribed in each of the triangles
A
B
C
ABC
A
BC
,
A
C
D
ACD
A
C
D
and
B
C
D
BCD
BC
D
. Prove that the sum of the radii of these circles equals the height
C
D
CD
C
D
.