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Contests
National and Regional Contests
France Contests
French Mathematical Olympiad
2000 French Mathematical Olympiad
Problem
Problem
Part of
2000 French Mathematical Olympiad
Problems
(1)
cartesian triangles
Source: French MO 2000 Problem
4/9/2021
In this problem we consider so-called cartesian triangles, that is, triangles
A
B
C
ABC
A
BC
with integer sides
B
C
=
a
,
C
A
=
b
,
A
B
=
c
BC=a,CA=b,AB=c
BC
=
a
,
C
A
=
b
,
A
B
=
c
and
∠
A
=
2
π
3
\angle A=\frac{2\pi}3
∠
A
=
3
2
π
. Unless noted otherwise,
△
A
B
C
\triangle ABC
△
A
BC
is assumed to be cartesian.(a) If
U
,
V
,
W
U,V,W
U
,
V
,
W
are the projections of the orthocenter
H
H
H
to
B
C
,
C
A
,
A
B
BC,CA,AB
BC
,
C
A
,
A
B
, respectively, specify which of the segments
A
U
AU
A
U
,
B
V
BV
B
V
,
C
W
CW
C
W
,
H
A
HA
H
A
,
H
B
HB
H
B
,
H
C
HC
H
C
,
H
U
HU
H
U
,
H
V
HV
H
V
,
H
W
HW
H
W
,
A
W
AW
A
W
,
A
V
AV
A
V
,
B
U
BU
B
U
,
B
W
BW
B
W
,
C
V
CV
C
V
,
C
U
CU
C
U
have rational length. (b) If
I
I
I
is the incenter,
J
J
J
the excenter across
A
A
A
, and
P
,
Q
P,Q
P
,
Q
the intersection points of the two bisectors at
A
A
A
with the line
B
C
BC
BC
, specify those of the segments
P
B
PB
PB
,
P
C
PC
PC
,
Q
B
QB
QB
,
Q
C
QC
QC
,
A
I
AI
A
I
,
A
J
AJ
A
J
,
A
P
AP
A
P
,
A
Q
AQ
A
Q
having rational length. (c) Assume that
b
b
b
and
c
c
c
are prime. Prove that exactly one of the numbers
a
+
b
−
c
a+b-c
a
+
b
−
c
and
a
−
b
+
c
a-b+c
a
−
b
+
c
is a multiple of
3
3
3
. (d) Assume that
a
+
b
−
c
3
c
=
p
q
\frac{a+b-c}{3c}=\frac pq
3
c
a
+
b
−
c
=
q
p
, where
p
p
p
and
q
q
q
are coprime, and denote by
d
d
d
the
gcd
\gcd
g
cd
of
p
(
3
p
+
2
q
)
p(3p+2q)
p
(
3
p
+
2
q
)
and
q
(
2
p
+
q
)
q(2p+q)
q
(
2
p
+
q
)
. Compute
a
,
b
,
c
a,b,c
a
,
b
,
c
in terms of
p
,
q
,
d
p,q,d
p
,
q
,
d
. (e) Prove that if
q
q
q
is not a multiple of
3
3
3
, then
d
=
1
d=1
d
=
1
. (f) Deduce a necessary and sufficient condition for a triangle to be cartesian with coprime integer sides, and by geometrical observations derive an analogous characterization of triangles
A
B
C
ABC
A
BC
with coprime sides
B
C
=
a
BC=a
BC
=
a
,
C
A
=
b
CA=b
C
A
=
b
,
A
B
=
c
AB=c
A
B
=
c
and
∠
A
=
π
3
\angle A=\frac\pi3
∠
A
=
3
π
.
geometry
Triangle