Tough Puzzle: Ratio of similar triangles

Replies

• 

  Kaustubh Katdare

  21 views so far and not a single attempt?
  
  Too tough for you, CEans?
  
  All the best! 👍
• 

  Kaustubh Katdare

  Where did all the talented people on CE go? 😒
  
  Come on its simple Geometry. Did you not learn it in school? 😲
• 

  Kaustubh Katdare

  Woo hoo.
  
  I can't believe this! CEans can't crack a simple Geometry problem? 😔 . Where are the engineers who solve the toughest problems in the world! Where are they?
  
  Bring them on CE! Come on! Oh Lord!
• 

  satheesh27887

  HELLO... ''THE BIG K''
  we don't know how to begin....
  Can u give us another hint.....😕
• 

  Kaustubh Katdare

  No, sorry. I'd have given hints if anyone had attempted it on the day it was posted.
  
  Now there are only two options - either crack the puzzle or accept the defeat! :twisted:
  
  😁
• 

  KSHIRABDHI

  i think the ratio of triangle ABC to triangle A'B'C is 2:1. as both the triangles are similar so considering AB as base and A'B' is parallel to AB then as per similarity rule the ratio is 2:1
• 

  yudi

  i think the ratio Triangle ABC/A'B'C should be eqaul to (A'C/AC)^2
  taken AB as the base of triangle ABC & A'B'||AB as ABC~A'B'C
• 

  banu

  abc:a'b'c is 2:1
• 

  satheesh27887

  The_Big_K

  No, sorry. I'd have given hints if anyone had attempted it on the day it was posted.
  
  Now there are only two options - either crack the puzzle or accept the defeat! :twisted:
  
  😁

  
  k.... i accept the defeat.... we r eager to know the answer.......
• 

  Kaustubh Katdare

  satheesh27887

  k.... i accept the defeat.... we r eager to know the answer.......

  
  No no! 😒
  
  That's not the CEan - attitude!
  
  Keep attempting, get your friends to solve, form a team - do anything - but crack it!
  
  I know you can! You must!
• 

  Prasad Ajinkya

  Answer is simple, Area A'B'C to Area ABC = 1/4
  
  Here is the solution,
  
  Construction - from A' draw a line parallel to BC to meet AB in C'
  Thus from parallel lines theory, and A-A-A theorem, all the four triangles formed as similar!!
  
  If you see triangles AA'C' and C'B'B, side A'C' is common, and since they are similar, thus both of them are congruent as well. Similarly we can prove that all the four triangles are congruent to each other.
  
  Thus, Area ABC = Area A'C'B' + Area A'B'C + Area AC'A' + Area C'BB' = 4 A'B'C
  
  Thus, 1/4.
  
  QED :-D
• 

  Prasad Ajinkya

  What confounded me, was why give AB = CM? And all the triangles are right angled triangles.
• 

  satheesh27887

  I Think That Is The Right Answer.....
  Congrats....
  Keep The Good Work.....
• 

  Kaustubh Katdare

  We DO have smartest people on the planet Earth! 😁
  
  Here's the detailed solution for those who couldn't crack the puzzle!
  
  Source: Previous monthly puzzles: July-August 2008
  

  [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
  [/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]
  
  [/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
  [/FONT]
  ​