a)
It is given:
CM = AB = m;
triangles ABC and A’B’C are
similar; so, tangent A’B is parallel
to the side AB.
Triangle ABC
b) According to the tangent property:
the lengths of intersecting tangents from their intersecting
points to their points of contact with the enclosed
circle are always equal, we have:
CM = CN (fig 1), and...
c) BM + AN = AP + BP = AB
d) Thus, the perimeter of the triangle ABC is:
CM + BM + CN + AN + AB =
2CM + (BM + AN) + AB =
2(CM + AB) = 4CM = 4m
Small
triangle A’B’C
e) As above, the lengths of intersecting
tangents from their intersecting points to their
points of contact with the enclosed circle are always
equal, therefore:
A’N = A’Q e B’Q = B’M (fig.
2)
f) Thus the perimeter of the small
triangle A’B’C is:
(CM - B’M) + (CN - A’N)
+ (A’Q + B’Q)
that is:
2CM - B’M - A’N + A’Q + B’Q =
2CM = 2m
In
conclusion
g) According to Euclid, if two triangles
are similar, then the ratio of their areas is the
square of the ratio of any two corresponding sides.
Then, the ratio of the area of triangle ABC /
area triangle A’B’C is:
(4m/2m)2 = 4
Another
solution
And here is an algebrical solution submitted by Fu
Su:
a) Semiperimeter p' of
quadrilateral ABB'A':
p' = a + b + c + d (see drawing below)
b) Area of quadrilateral ABB'A':
r(a + b + c + d)
c)
Semiperimeter p of triangle ABC:
a + b + (c + e) = 2(a + b) [since c + e = a + b]
d) Area of triangle ABC:
2r(a + b)
e)
Area of triangle A'B'C:
triangle ABC - quadrilateral ABB'A' =
2r(a + b) - r(a + b + c + d) =
= r[(a + b) - (c + d)]
f)
Then, Area of triangle A'B'C / Area of
triangle ABC:
r[(a + b) - (c + d)] / 2r(a + b)
=
= [(a + b) - (c + d)] / 2(a + b) =
= 1/2 - 1/2[(c + d) / (a + b)]
g) Also Area triangle A'B'C /
Area triangle ABC =
= (A'B'/AB)2 = [(c +
d) / (a + b)]2
h)
Let, (c + d) / (a + b) = x,
then x2 = 1/2 - x/2 (see paragraph 'f')
or 2x2 + x - 1 = 0
hence (2x - 1)(x + 1) = 0 and x = 1/2 or -1 (cannot
be)
Area triangle A'B'C/Area triangle ABC =
x2 = 1/4