Difference between revisions of "2010 USAMO Problems/Problem 1"
Danielguo94 (talk | contribs) |
m (Fixed USAMO box. Also changed British "anti-clockwise" to American "counterclockwise".) |
||
Line 84: | Line 84: | ||
Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>. | Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>. | ||
− | Now <math>SY</math> is perpendicular to <math>AZ</math> so the direction <math>SY</math> is <math>\alpha</math> | + | Now <math>SY</math> is perpendicular to <math>AZ</math> so the direction <math>SY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YSR = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical. |
− | Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> | + | Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical. |
Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma | Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma | ||
Line 98: | Line 98: | ||
below the point <math>Y</math>. | below the point <math>Y</math>. | ||
− | Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> | + | Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise |
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math> | from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math> | ||
horizontally to the right of <math>Y</math>. | horizontally to the right of <math>Y</math>. | ||
Line 115: | Line 115: | ||
meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>. | meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>. | ||
− | {{ | + | {{USAMO newbox|year=2010|beforetext=|before=First Problem|num-a=2}} |
Revision as of 19:46, 5 April 2011
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.
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.
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
.
2010 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |