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8
8
Part of
2021 Stanford Mathematics Tournament
Problems
(1)
SMT 2021 Geometry #8
Source:
8/9/2023
In triangle
△
A
B
C
\vartriangle ABC
△
A
BC
,
A
B
=
5
AB = 5
A
B
=
5
,
B
C
=
7
BC = 7
BC
=
7
, and
C
A
=
8
CA = 8
C
A
=
8
. Let
E
E
E
and
F
F
F
be the feet of the altitudes from
B
B
B
and
C
C
C
, respectively, and let
M
M
M
be the midpoint of
B
C
BC
BC
. The area of triangle
M
E
F
MEF
MEF
can be expressed as
a
b
c
\frac{a \sqrt{b}}{c}
c
a
b
for positive integers
a
a
a
,
b
b
b
, and
c
c
c
such that the greatest common divisor of
a
a
a
and
c
c
c
is
1
1
1
and
b
b
b
is not divisible by the square of any prime. Compute
a
+
b
+
c
a + b + c
a
+
b
+
c
.
geometry