Ken's POTW


Equal Areas
Here's an easier bit of math than the recent puzzles. Make a figure with two squares: ABCD and EFGC, with DCG perpendicular to ECB. (The squares are not [necessarily] the same size.) Add lines DE, DF and AE. DF crosses AE at point H. AE crosses DC at point I. DF crosses EC at point J. Show that the areas of DHE and HJCI are equal.

Source: Submitted by Jason Telgren, citing Sarah Everding. I'd be interested in similar puzzles others might have.

Solutions were received from Radu Ionescu and Phillipe Fondanaiche:
Radu Ionescu sent: 
Let AB=a,CG=b, IC=x and JC=y
ABE and ICE are similar==>x=ab/(a+b)
DGF and DCJ are similar==>y=ab/(a+b)
Aria(ICE)=abb/2(a+b) and Aria(DEJ)=(a/2)(b-ab/(a+b))=abb/2(a+b)==>
Aria(ICE)=Aria(DEJ)==>Aria(ICE)-Aria(HJE)=Aria(DEJ)-Aria(HJE)==>
Aria(HICJ)=Aria(DHE)
Phillipe Fondanaiche sent: 
Let a=side of ABCD and b=side of EFGC.

As ADI and CEI are similar triangles,CI/b = (a-CI)/a => CI = ab/(a+b) and DI =
a^2/(a+b)
As CDJ and EFG are similar triangles,CJ/a =(b-CJ)/b => CJ = ab/(a+b) = CI 

Area(HJCI) + area(DIH) = area(DCJ) = DC x CJ /2 = a^2xb/2(a+b)
Area(DHE) + area(DIH) = area(DEI) = DI x CE /2 = a^2xb/2(a+b)

So, area (HJCI) = area(DHE)
My (Ken's) solution is a little more complicated: 
Add a point X at the intersection of AD and EF.
Now AE divides rectangle AXEB in half.
And DF divides rectangle DXFG in half.

So, (all combinations of letters describe areas):

DFG - DXF  +  ABE - AXE  = 0

Expanding (areas EJH and DIH have cancelled out.):
(ABCI - ADI) + (CJFG - EJF) -2DXE + 2(HICJ - DHE)  = 0

Let EF=a, AD=b.  Then:
ADI = (b/2)(b/a+b)(b) =  (b^3)/[2(a+b)]
ABCI = (b^2) - ADI
EJF = (a^3)/[2(a+b)]
CJFG = (a^2) = EJF
DXE = ab/2

Substituting:
(a^2 + b^2) - (a^3 + b^3)/(a+b) - ab + 2(HICJ - DHE)  = 0

(a^2 + b^2)(a+b) - (a^3 + b^3) - ab(a+b) + 2(a+b)(HICJ - DHE) = 0

a^3 + ab^2 + ba^2 + b^3 - a^3 - b^3 - ba^2 - ab^2 + 2(a+b)(HICJ - DHE) = 0

0 + 2(a+b)(HICJ - DHE) = 0

So DHE = HICJ.
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