## 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.

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]

2. Let the distance between the parallel lines
(i)EF and BC 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

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.

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

5. name l1 distance from EF to a, l2 from EF to b => l/l1 = (2x+a+b)/(x+b) (1)
Draw CKL //AB => l/l1 = (b-a)/(x-a) (2)
From (1) and (2) a²+b² = 2x²