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3
2017 Geometry Tiebreaker #3
2017 Geometry Tiebreaker #3
Source:
November 19, 2022
2017
Geometry Tiebreaker
Problem Statement
Triangle
A
B
C
ABC
A
BC
has
A
B
=
4
,
B
C
=
6
,
C
A
=
5
AB=4,BC=6,CA=5
A
B
=
4
,
BC
=
6
,
C
A
=
5
. Let
M
M
M
be the midpoint of
B
C
‾
\overline{BC}
BC
and
P
P
P
the point on the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
such that
∠
M
P
A
=
9
0
∘
\angle MPA=90^\circ
∠
MP
A
=
9
0
∘
. Let points
D
D
D
and
E
E
E
lie on
A
C
‾
\overline{AC}
A
C
and
A
B
‾
\overline{AB}
A
B
respectively such that
B
D
‾
⊥
A
C
‾
\overline{BD}\perp\overline{AC}
B
D
⊥
A
C
and
C
E
‾
⊥
A
B
‾
\overline{CE}\perp\overline{AB}
CE
⊥
A
B
. Find
P
D
P
E
\tfrac{PD}{PE}
PE
P
D
.
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