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Previous Puzzles of the Month + Solutions

 
July-August 2008, Puzzle nr 118
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Puzzle # 118 Difficulty level: bulbbulbbulb, general math knowledge.

triangle

Ratio of similar triangles.
  
If the segment A'B' is tangent to the incircle of triangle ABC, and that segment AB = segment CM; then, what is the ratio of the area of the triangle ABC to the area of the small triangle A'B’C? Hint: angle CÂB is not necessarily a right angle, however triangles ABC and A'B'C are similar!

Keywords: inscribed circle, incircle, tangent.

Related puzzles:
- Soccer ball problem.
- Prof. Gibbus' angle.


Source of the puzzle:
©G. Sarcone, "Focus Braintrainer Magazine #16", page 61.


solution

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
solution 1

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


solution 2

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
solution by Fu Su

 

cup winnerThe Winners of the Puzzle of the Month are:
Alex Quenon, USA flag usa
Patrick John Reidy, Australia italian flag
Su Fu, ?
Abhilash PP, India indian flag

Congratulations!


© 2005 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!


More Math Facts behind the puzzle

incircle triangle

Let the circle (O, OM) be the incircle of the triangle ABC above.

Consider that:
AB = c
BC = a
AC = b
AL = AN = x
BL = BM = y
CM = CN = z
Semiperimeter p = 1/2(a + b + c)

Then:
2x = (x + y) + (x + z) - (y + z)
2x = AB + AC - BC
and
x = 1/2(AB + AC - BC)
x = 1/2(AB + AC + BC) - BC
x = p - a

Furthermore:
y = p - b
z = p - c

Formulae
Area A of the triangle ABC:
A = square root(px · py · pz) = square root[p(p - a)(p - b)(p - c)] = r · p
A
= 1/2(bc · sinα) = 1/2(ac · sinβ) = 1/2(ab · sinγ)



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