Solution to Problem #13



Solutions were found by Nick Baxter, Robin Stokes of the University of New England (Australia), Philippe Fondanaiche of Paris (France), and Dr. Peter-M. Schmidt of Stuttgart (Germany). Philippe Fondanaiche also found several generalizations. First, here is Nick Baxter's solution:

I will first find the ratio of the area of the small triangle (K) to
the area of triangle XYZ.  Since a(XYZ)/a(ABC) is easily 
detemined to be 7/16 (1- 3*3/16), the solution will then follow.

Draw XZ, intersecting PY at T.  The goal is to find the ratio of
ZT to TX.  Draw perpendicular lines from points Z,T,Y all to line BC, 
intersecting BC at L,M,N respectively.  Assuming the edges of the
large triangle are 1, we now look for the length of PM.

BL=1/8, LP=3/8, PC=1/2, NC=3/8, ZL=sqrt(3)/16, NY=3*sqrt(3)/8, and
PN = PC-NC = 1/8

Let PM=x and TM=y.  MX=PX-PM = 1/4 - x.
Triangles ZLX and TMX are similar, giving sqrt(3)/5 = y/(1/4 -x).
Triangles YNP and TMP are similar, giving 3*sqrt(3) = y/x.
Solving gives x=1/16.

TX/ZT = MX/ML = (1/4 - x)/(3/8 + x) = 15/64 / 25/64 = 3/5

So, in triangle XYZ, the cevian YT divides XZ in the ratio 3/5.
By symmetry, the other two cevians that form the smaller triangle
also divide the opposite edge of XYZ in the ratio 3/5.

Nick Baxter's Cevian Theorem: if the cevians of a triangle divide
the opposite sides in the ratios m,n,p, then the ratio of the area of
the triangle formed by the cevians to the area of the larger triangle
is equal to
	(mnp-1)^2/((m^2+mn+1)*(n^2+np+1)*(p^2+pm+1))

When the cevians divide the sides with the same ratio, then the 
formula reduced to (p-1)^3/(p^3-1).

Let p=3/5.  We get K/a(XYZ)= (2/5)^3 /(1 - (3/5)^3)
  = 8/98 = 4/49.

Back to the original problem, K/a(ABC) = K/a(XYZ) * a(XYZ)/a(ABC)
	= 4/49 * 7/16 = 1/28

Here are Philippe Fondanaiche's generalizations:

For Triangles 
Given any P on BC such as BP/BC = p, Q on AC such as CQ/CA = p and R on AB
such as AR/AB = p. Any X on PC such as PX/PC = q, Y on QA such as QY/QA = q
and Z on RB such as RZ/RB = q with p and q included between 0 and 1, let
k = the ratio of the area of the shaded triangle DEF to the area of the
triangle ABC.

Let r = p +(1-p)*q
We have the following identity:
 area(DEF) = area(ABC) -area(AZQ) - area(BRX) - area(CPY) + area(DPX)         
                   +area(EQY) + area(FRZ).

The areas of AZQ, BRX and CPY are equal to p*(1-p)*S.

Through an analytical approach using the equations of the lines PY and RX, we
determine the coordinates of D,then the area of DPX and in the same way the
areas of EQY and FRZ.

At the end, we get :
 k = [ r * (3*p - 2) + 1 - p ] ^ 2  / [r ^ 2 + (1- p ) ^ 2 - (1 - p ) * r ]

We check that for p = 1/2 and r = 3/4, we find again k = 1/28.

Note that k can be equal to zero, when p< or = 1/2 and r =(1-p) / (2-3p).

For Squares 
We consider a square ABCD . P is the midpoint of AB, Q is the midpoint of
BC, R is the midpoint of CD and T is the midpoint of DA. X is the midpoint
of PB, Y is the midpoint of QC, Z is the midpoint of RD and W is the 
midpoint of TA.

Two configurations are possible:

  1 : we consider PY,QZ,RW and TX which intersect in 4 points D,E,F and G
which delimit a square DEFG. The ratio of the area of this square to the area
S of the original square ABCD is  such as:
 area(DEFG) = DE^2 = (PY - PD - EY)^2 =  S * [sqr(13) - 1/(2*sqr(13))
-3/(4*sqr(13)]^2 =  4*S/13 ==> k=4/13

  2 : we consider PZ,QW,RX and TY which intersect in 4 points D', E', F',
G'
which delimit a square D'E'F'G'. It's easy to demonstrate that in this case
k=1/17

It's easy to generalize these formulas with the ratios p and r analogous
to those in the triangular case.

  1 :  k = (1 - p - r + 2*p*r )^2 / [ (1-p)^2 + r^2 ]
 
  2 :  k = (r-p)^2 / [ 1 + (p+r-1)^2 ]
Here are a few more questions: