Tuesday, March 5, 2013

Problem 862: Trapezoid, Parallel, Equal Areas, RMS, Root Mean Square

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 862.

Online Geometry Problem 862: Trapezoid, Parallel, Equal Areas, RMS, Root Mean Square

4 comments:

  1. Let EF=x, BE:EA=m:n

    From similar triangle, we have
    m/n = (x-a)/(b-x)
    x = (na + mb)/(m+n)

    From same area, we have
    m(x+a) = n(x+b)
    x = -(ma - nb)/(m-n)

    Hence,
    (m+n)(ma - nb) = (n-m)(na + mb)
    (m^2 + 2mn - n^2)a + (m^2 - 2mn - n^2)b = 0
    (a+b)m^2 + 2(a-b)mn - (a+b)n^2 = 0
    m/n = [(b-a) + √[2(a^2 + b^2)]] / (a+b)

    x = [a + (m/n)b] / (1 + m/n)
    = [a^2 + b^2 + b√[2(a^2 + b^2)]] / [2b + √[2(a^2 + b^2)]]
    = √[(a^2 + b^2) / 2]

    ReplyDelete
  2. Let the distance between the parallel lines
    (i)EF and BC be h'
    (ii)EF and AD be h"
    Let EF = x
    (AEFD) = (BECF) = (1/2)(ABCD)
    So (x + a)h' = (x + b)h" = (1/2)(a + b)(h' + h")
    Each = k say
    So h' = k/(x + a), h" = k/(x + b)
    and h' + h" = 2k/(a + b)
    Follows 1/(x + a) + 1/(x + b) = 2/(a +b)
    (a + b)(2x + a + b)= 2(x + a)(x + b)
    (a + b)^2 = 2x^2 + 2ab
    2x^2 = a^2 + b^2 etc

    ReplyDelete
  3. Call S the equal areas, and call X the intersection of AB and CD. Call L the area of triangle XBC. By similarity we have

    1) XBC ~ XEF gives L/(S+L) = (a/EF)²
    2) XAD ~ XEF gives (L+2S)/(S+L) = (b/EF)²

    Adding these two, we have: 2(L+S)/(L+S) = (a²+b²)/EF², so that EF²=(a²+b²)/2

    The result follows.

    ReplyDelete
  4. prof Radu Ion,Sc Bozioru,BuzauMarch 14, 2013 at 4:49 AM

    Notam cu x=EF,m inaltimea BCFE,n inaltimea EEFDA si avem(a+x).m=(x+b).n.Ducem paralela prin C la AB sinotam cu G ,H intersectia acestei paralele cu EF,AD=>TFA
    (x-a).(m+n)=m(b-a)<=>n(x-a)=m(b-x)=>(x-a)(x+a)=(b-x)(b+x)=> x²=(a²+b²)/2

    ReplyDelete