THE AVERAGE AREA OF A TRIANGLE
IN A PARABOLIC SECTOR
Allegheny Mountain Section Meeting of the MAA
Indiana University of Pennsylvania
April 6, 2013
Michael Woltermann
Washington and Jefferson College
THE PROBLEM
Find the average area of the triangle formed by
joining three points taken at random in (the
surface of) a parabola whose base is b and
altitude is h.
 Problem 248 proposed by Enoch Beery Seitz in
the Mathematical Visitor, 1880.
 Solution published in 1893.
 Senior MathTalk with Logan Elias (2012) at
W&J, Fall 2012.

PUBLISHED SOLUTION


Solution appears to
assume that the base
is parallel to the
directrix.
Avg =
11
𝑏ℎ
210
11
AreaP
 Avg =
140


Avg =
Avg =
      area(...)dV
dV

6       area(PQR)dV
2 
 bh 
3 
3
IS IT TRUE IN THIS CASE?



b is not parallel to the
directrix.
Is Avg =
Or
11
𝑏ℎ
210
11
AreaP ?
140
?
A PROPERTY OF PARABOLAS
𝑀𝐸 2

𝐷𝐵2

𝑣=
=
𝑏
2
𝑀𝐴
𝐷𝐴
=
𝑢
.
ℎ′
𝑢
.
ℎ′
b
v 
2
u
b
, x 
h
2
w
b
, z 
h
2

P=(v,u)
0≤u≤h′
 -v′≤v≤v′


Q=(x,w)
0≤w≤u
 -x′≤x≤x′


R=(z,y)


w≤y≤u
-z′≤z≤z′
y
h
AREA OF TRIANGLE PQR
P=(v,u)
 Q=(x,w)
 R=(z,y)
 S=(t,y)

(v  x)(u  y )
 t= v 
uw

Area(∆PQR) =
1
t  z u  wsin( )
2
AVERAGE AREA
h v  u x  u z 

Avg =
1
6       t  z u  wsin( )dzdydxdwdvdu
2
0  v 0  xw  z 
h v  h x  h z 
      dzdydxdwdvdu
0  v 0  x 0  z 


Factor out sin(ω),
And
THE AVERAGE AREA BECOMES

Avg = sin(ω)∙Seitz answer, or
11
 Avg =
bh' sin( ),
210
h
,
 Or since sin(ω) =
h
11
11
bh 
AreaP
210
140

Avg =

Link to: Excel Simulation
REFERENCES
Problems and Solutions from The Mathematical
Visitor 1877-1896, ed. By Stanley Rabinowitz,
1996, MathPro Press, Inc.
 Archimedes, What Did He Do Besides Cry
Eureka? By Sherman Stein, 1999, MAA.
 http://archive.org/details/mathematicalvis00mart
goog

THANK YOU!