: <a href="https://www.archimedes-lab.org/monthly_puzzles_73.html" target="_blank" rel="noopener noreferrer">Previous monthly puzzles: July-August 2008</a>
[FONT=Verdana, Arial, Helvetica, sans-serif]
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.[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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
![[IMG]](proxy.php?image=http%3A%2F%2Fwww.archimedes-lab.org%2Fim_maths%2Fmon_p_july_2k8_sol1.gif&hash=0ba82779ad13a3655e4ae1de170c96db)
[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]
![[IMG]](proxy.php?image=http%3A%2F%2Fwww.archimedes-lab.org%2Fim_maths%2Fmon_p_july_2k8_sol2.gif&hash=decb3dd1713d4ea318f3d04aaad1f1a8)
[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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)[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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)[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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)][/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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[/FONT]
[FONT=Verdana, Arial, Helvetica, sans-serif]
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
![[IMG]](proxy.php?image=http%3A%2F%2Fwww.archimedes-lab.org%2Fim_maths%2Fmon_p_july_2k8_sol4.gif&hash=5e266cd013932fd02df8e0cadf71fee7)
[/FONT]
![[IMG]](proxy.php?image=http%3A%2F%2Fwww.archimedes-lab.org%2Fim_maths%2Fmon_p_july_2k8.gif&hash=974a0a2b438be43cc193cff916f57ff2)
