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 ℎ 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 ℎ 210 11 AreaP ? 140 ? A PROPERTY OF PARABOLAS 2 2 = = 2 = . ℎ′ . 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 uw Area(∆PQR) = 1 t z u wsin( ) 2 AVERAGE AREA h v u x u z Avg = 1 6 t z u wsin( )dzdydxdwdvdu 2 0 v 0 xw 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!