Solutions
Annie says:
While this was sort of a contrived problem, there were some very good answers. Eighteen people got it right; 6 people got it right but didn't provide a general solution, and 35 people got it wrong.
A number of people correctly said that there were a whole bunch of seats, and described how you might find them (check out Rosina Pannone's solution, which is excellent), but didn't describe what the set of solutions looked like, namely a circle (or two circles, as it were). So folks with credit for the general solution included the circle, while those who are only right didn't get part. Some folks were right in saying that there is more than one seat, but didn't provide much explanation beyond that, so they didn't get any credit.
I like Amy's solution in that she pointed out what she learned by experimenting with the angles and the circles. She also described the resulting figure as a 'figure 8', which is really nice.
Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are also available.
Rosina Pannone
Grade 10
Cheshire High School, Cheshire, CT
The man who said that there is more than one seat where the lines
of sight between home plate and first base meet at a 25 degree
angle is correct. In order to figure this out, I drew a diagram of a
baseball diamond. I used the line between home plate and first base
as the base of a triangle. Then, since the third angle had to equal
25 degrees, that left 155 degrees to split up between the two base
angles. There are many different ways to split up 155 degrees
between two angles, and each produces different vertices where a
good seat is. I did five triangles, and arrived at five different seats,
and there are even more. Therefore, there is more than one best
seat.
Amy Forster
Age 11, Grade 7, Home schooling
Wilkins/Forster family
Crooked Tree Point
Cygnet, Tasmania, Australia.
The second person, who said you could sit in more than one seat, is
correct.
Solution: I drew two points, Home plate and first base,on a piece of
paper. I cut a piece of paper so that it was at a 25 degree angle.
I lined the 25 degree angled piece of paper with the home plate and
first base, and put points whereever it was 25 degrees.The points I
drew formed two circles that intersected at home plate and first
base, so you got a sort of figure of 8 of possible places to sit with a
25 degree angle between home plate and first base. Since spectators
can't sit in the field the possible seats that give a 25 degree view are
two arcs on opposite sides of the players within the sitting area
(assuming spectators can sit 360 degrees round the players;
otherwise there is only one arc of possible seating.
From doing this problem I discovered there must be a relationship
for circles where lines from each end of a chord to a point on the
circumference of the circle make the same angle for any point on
the circumference.
Mike Sue
Grade: 10
School: Granada
The second guy is right. There are many places where you get
this 25 degrees and they are all on 2 circles with the radius
of approximately 106.5 ft. The circles are centered on the
perpendicular bisector of the line segment between home and
first approx. 96.5 ft. toward the pitcher mound, and the other
96.5 feet away from the pitcher mound.
Let Home to First be a line segment from H to F. Select a
point V such that angle HVF = 25 degrees. Then construct a
circumscribed circle which can be called O about triangle HVF.
the measure of arc HF = 2 times the measure of angle HVF = 50.
Let X be any point on arc HVF, such that X can't be H or F.
The measurement of angle HXF = one half of the measurement
of arc HF = 25.
Thus the locus of vertices X forms the measurement of angle HXF
= 25 = the arcHXF where H and F aren't included.
To find the center of circle O:
Assume the baseball diamond is a 90 ft square. Therefore the
line segment HF = 90 ft. Construct a perpendicular bisector to HF
intersecting H at M. This line passes through the center of circle o.
The measurement of angle HOF = measurement of arc HF = 50.
The measurement of angle HOM = one half the angle HOF = 25.
HM = one half of HF = 45.
R = 45/(sin 45) = 106.5 ft.
OM = 45/(tan 45) = 96.5 ft.
Arc HXF can be on either side of segment HF.
Ben Warfield
Garfield High
Home of the 1995 AL West Champion Seattle Mariners
I hope you don't tell the deal about the good seats to the Mariners
management: they're always looking for ways to make the extra
buck (or 8 million), and since there are a very large number of
such seats (defined by the set of vertices of a triangle with a fixed
opposite side of 90 feet and a vertex angle of 25 degrees), it might
get a little rough on those of us who actually buy tickets there
regularly if they decided to put a premium on them. I can also say
that your basic assumption, at least for the Kingdome, is a crock:
you are much better off with outfield seats out in the open than
with perfectly angled seats under the roof on the 200 level. If
you're planning on seeing one of the Cleveland games, you'd better
call ahead for tickets, they're selling out.
Boston plays at Seattle on the 30 and 31st of May and the 1st and
2nd of June. I'll be in section 116, myself.
The seats described are on the circumference of any circle
constructed such that the circumference passes through home plate
and first base and the central angle to the bases is 50 degrees.
There are probably three sets of relevant circles, each consisting of
a large number of cirles very close to each other (to adjust for the
pitch of the bleachers), one for each deck, each circle having a
radius of just over 106 feet (I'm not sure how far up the 300 level
is). Having thought about it some, I am willing to concede that that
is a pretty good measure of a seat, but I maintain that the circles
that extend to the 200 level, while the angles to the bases may be
great, do not provide a worthwhile baseball experience.
I hope you're not a Twins fan: Randy Johnson should be pitching.