2010 USAMO Problems/Problem 1
Problem
Let be a convex pentagon inscribed in a semicircle of diameter
. Denote by
the feet of the perpendiculars from
onto
lines
, respectively. Prove that the acute angle
formed by lines
and
is half the size of
, where
is the midpoint of segment
.
Solution
Let ,
.
Since
is a chord of the circle with diameter
,
. From the chord
,
we conclude
.
Triangles and
are both right-triangles, and share the
angle
, therefore they are similar, and so the ratio
. Now by Thales' theorem the angles
are all right-angles. Also,
,
being the fourth angle in a quadrilateral with 3 right-angles is
again a right-angle. Therefore
and
.
Similarly,
, and so
.
Now is perpendicular to
so the direction
is
counterclockwise from the vertical, and since
we see that
is
clockwise from the vertical. (Draw an actual vertical line segment if necessary.)
Similarly, is perpendicular to
so the direction
is
clockwise from the vertical, and since
is
we see that
is
counterclockwise from the vertical.
Therefore the lines and
intersect at an angle
. Now by the central angle theorem
and
, and so
,
and we are done.
Note that is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?
Footnote
We can prove a bit more. Namely, the extensions of the segments
and
meet at a point on the diameter
that is vertically
below the point
.
Since and is inclined
counterclockwise
from the vertical, the point
is
horizontally to the right of
.
Now , so
is
vertically above the diameter
. Also,
the segment
is inclined
clockwise from the vertical,
so if we extend it down from
towards the diameter
it will
meet the diameter at a point which is
horizontally to the left of
. This places the intersection point
of
and
vertically below
.
Similarly, and by symmetry the intersection point of and
is directly below
on
, so the lines through
and
meet at a point
on the diameter that is vertically below
.
Footnote to the Footnote
The Footnote's claim is more easily proved as follows.
Note that because and
are both complementary to
, they must be equal. Now, let
intersect diameter
at
. Then
is cyclic and so
. Hence
is cyclic as well, and so we deduce that
Hence
are collinear and so
. This proves the Footnote.
See Also
2010 USAMO (Problems • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
2010 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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