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Contests
National and Regional Contests
Germany Contests
Bundeswettbewerb Mathematik
2017 Bundeswettbewerb Mathematik
2017 Bundeswettbewerb Mathematik
Part of
Bundeswettbewerb Mathematik
Subcontests
(4)
4
2
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Recursion And Primes
The sequence
a
0
,
a
1
,
a
2
,
…
a_0,a_1,a_2,\dots
a
0
,
a
1
,
a
2
,
…
is recursively defined by a_0 = 1 \text{and} a_n = a_{n-1} \cdot \left(4-\frac{2}{n} \right) \text{for } n \geq 1. Prove for each integer
n
≥
1
n \geq 1
n
≥
1
: (a) The number
a
n
a_n
a
n
is a positive integer. (b) Each prime
p
p
p
with
n
<
p
≤
2
n
n < p \leq 2n
n
<
p
≤
2
n
is a divisor of
a
n
a_n
a
n
. (c) If
n
n
n
is a prime, then
a
n
−
2
a_n-2
a
n
−
2
is divisible by
n
n
n
.
Consecutive Integers As Sum Of Square And Cube
We call a positive integer heinersch if it can be written as the sum of a positive square and positive cube. Prove: There are infinitely many heinersch numbers
h
h
h
, such that
h
−
1
h-1
h
−
1
and
h
+
1
h+1
h
+
1
are also heinersch.
3
2
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Tangential Quadrilateral Defines Another Quadrilateral
Let
M
M
M
be the incenter of the tangential quadrilateral
A
1
A
2
A
3
A
4
A_1A_2A_3A_4
A
1
A
2
A
3
A
4
. Let line
g
1
g_1
g
1
through
A
1
A_1
A
1
be perpendicular to
A
1
M
A_1M
A
1
M
; define
g
2
,
g
3
g_2,g_3
g
2
,
g
3
and
g
4
g_4
g
4
similarly. The lines
g
1
,
g
2
,
g
3
g_1,g_2,g_3
g
1
,
g
2
,
g
3
and
g
4
g_4
g
4
define another quadrilateral
B
1
B
2
B
3
B
4
B_1B_2B_3B_4
B
1
B
2
B
3
B
4
having
B
1
B_1
B
1
be the intersection of
g
1
g_1
g
1
and
g
2
g_2
g
2
; similarly
B
2
,
B
3
B_2,B_3
B
2
,
B
3
and
B
4
B_4
B
4
are intersections of
g
2
g_2
g
2
and
g
3
g_3
g
3
,
g
3
g_3
g
3
and
g
4
g_4
g
4
, resp.
g
4
g_4
g
4
and
g
1
g_1
g
1
. Prove that the diagonals of quadrilateral
B
1
B
2
B
3
B
4
B_1B_2B_3B_4
B
1
B
2
B
3
B
4
intersect in point
M
M
M
.[asy] import graph; size(15cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-9.773972777861085,xmax=12.231603726660566,ymin=-3.9255487671791487,ymax=7.37238601960895; pair M=(2.,2.), A_4=(-1.6391623316400197,1.2875505916864178), A_1=(3.068893183992864,-0.5728665455336459), A_2=(4.30385937824148,2.2922812065339455), A_3=(2.221541124684679,4.978916319940133), B_4=(-0.9482172571022687,-2.24176848577888), B_1=(4.5873184669543345,0.057960746374459436), B_2=(3.9796042717514277,4.848169684238838), B_3=(-2.4295496490492385,5.324816563638236); draw(circle(M,2.),linewidth(0.8)); draw(A_4--A_1,linewidth(0.8)); draw(A_1--A_2,linewidth(0.8)); draw(A_2--A_3,linewidth(0.8)); draw(A_3--A_4,linewidth(0.8)); draw(M--A_3,linewidth(0.8)+dotted); draw(M--A_2,linewidth(0.8)+dotted); draw(M--A_1,linewidth(0.8)+dotted); draw(M--A_4,linewidth(0.8)+dotted); draw((xmin,-0.07436970390935019*xmin+5.144131675605378)--(xmax,-0.07436970390935019*xmax+5.144131675605378),linewidth(0.8)); draw((xmin,-7.882338401302275*xmin+36.2167572574517)--(xmax,-7.882338401302275*xmax+36.2167572574517),linewidth(0.8)); draw((xmin,0.4154483588930812*xmin-1.847833182441644)--(xmax,0.4154483588930812*xmax-1.847833182441644),linewidth(0.8)); draw((xmin,-5.107958950031516*xmin-7.085223310768749)--(xmax,-5.107958950031516*xmax-7.085223310768749),linewidth(0.8)); dot(M,linewidth(3.pt)+ds); label("
M
M
M
",(2.0593440948136896,2.0872038897020024),NE*lsf); dot(A_4,linewidth(3.pt)+ds); label("
A
4
A_4
A
4
",(-2.6355449660387147,1.085078446888477),NE*lsf); dot(A_1,linewidth(3.pt)+ds); label("
A
1
A_1
A
1
",(3.1575637581709772,-1.2486383377457595),NE*lsf); dot(A_2,linewidth(3.pt)+ds); label("
A
2
A_2
A
2
",(4.502882845783654,2.30684782237346),NE*lsf); dot(A_3,linewidth(3.pt)+ds); label("
A
3
A_3
A
3
",(2.169166061149418,5.203402184478307),NE*lsf); label("
g
3
g_3
g
3
",(-9.691606303109287,5.354407388189934),NE*lsf); label("
g
2
g_2
g
2
",(3.0889250292111465,6.727181967386543),NE*lsf); label("
g
1
g_1
g
1
",(-4.763345563793459,-3.4725331560442676),NE*lsf); label("
g
4
g_4
g
4
",(-2.663000457622647,6.878187171098171),NE*lsf); dot(B_4,linewidth(3.pt)+ds); label("
B
4
B_4
B
4
",(-1.5647807942653595,-3.0332452907013523),NE*lsf); dot(B_1,linewidth(3.pt)+ds); label("
B
1
B_1
B
1
",(4.955898456918535,-0.6583452686912173),NE*lsf); dot(B_2,linewidth(3.pt)+ds); label("
B
2
B_2
B
2
",(4.104778217816637,5.0661247265586455),NE*lsf); dot(B_3,linewidth(3.pt)+ds); label("
B
3
B_3
B
3
",(-3.4454819677647146,5.656417795613188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
Condition a+b=3c Implies Interesting Perpendicularity
Given is a triangle with side lengths
a
,
b
a,b
a
,
b
and
c
c
c
, incenter
I
I
I
and centroid
S
S
S
. Prove: If
a
+
b
=
3
c
a+b=3c
a
+
b
=
3
c
, then
S
≠
I
S \neq I
S
=
I
and line
S
I
SI
S
I
is perpendicular to one of the sides of the triangle.
2
2
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Maximum Number Of Acute Angles In 2017-Gon
What is the maximum number of acute interior angles a non-overlapping planar
2017
2017
2017
-gon can have?
Isosceles Triangle In Colored 35-Gon
In a convex regular
35
35
35
-gon
15
15
15
vertices are colored in red. Are there always three red vertices that make an isosceles triangle?
1
2
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Removing Numbers On A Blackboard
The numbers
1
,
2
,
3
,
…
,
2017
1,2,3,\dots,2017
1
,
2
,
3
,
…
,
2017
are on the blackboard. Amelie and Boris take turns removing one of those until only two numbers remain on the board. Amelie starts. If the sum of the last two numbers is divisible by
8
8
8
, then Amelie wins. Else Boris wins. Who can force a victory?
Pairwise Distinct Differences
For which integers
n
≥
4
n \geq 4
n
≥
4
is the following procedure possible? Remove one number of the integers
1
,
2
,
3
,
…
,
n
+
1
1,2,3,\dots,n+1
1
,
2
,
3
,
…
,
n
+
1
and arrange them in a sequence
a
1
,
a
2
,
…
,
a
n
a_1,a_2,\dots,a_n
a
1
,
a
2
,
…
,
a
n
such that of the
n
n
n
numbers
∣
a
1
−
a
2
∣
,
∣
a
2
−
a
3
∣
,
…
,
∣
a
n
−
1
−
a
n
∣
,
∣
a
n
−
a
1
∣
|a_1-a_2|,|a_2-a_3|,\dots,|a_{n-1}-a_n|,|a_n-a_1|
∣
a
1
−
a
2
∣
,
∣
a
2
−
a
3
∣
,
…
,
∣
a
n
−
1
−
a
n
∣
,
∣
a
n
−
a
1
∣
no two are equal.