Tough Puzzle: Ratio of similar triangles
Puzzle Source: #LinkSnipped#
[FONT=Verdana, Arial, Helvetica, sansserif]If the segment A'B' is tangent to the inscribed circle 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'Bâ€™C? Hint: angle CAB is not necessarily a right angle, however triangles ABC and A'B'C are similar![/FONT]
[FONT=Verdana, Arial, Helvetica, sansserif]If the segment A'B' is tangent to the inscribed circle 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'Bâ€™C? Hint: angle CAB is not necessarily a right angle, however triangles ABC and A'B'C are similar![/FONT]
Replies

Kaustubh Katdare21 views so far and not a single attempt?
Too tough for you, CEans?
All the best! 👍 
Kaustubh KatdareWhere did all the talented people on CE go? 😒
Come on its simple Geometry. Did you not learn it in school? 😲 
Kaustubh KatdareWoo 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! 
satheesh27887HELLO... ''THE BIG K''
we don't know how to begin....
Can u give us another hint.....😕 
Kaustubh KatdareNo, 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:
😁 
KSHIRABDHIi 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

yudii 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 
banuabc:a'b'c is 2:1

satheesh27887The_Big_KNo, 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 Katdaresatheesh27887k.... 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 AjinkyaAnswer 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 AAA 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 AjinkyaWhat confounded me, was why give AB = CM? And all the triangles are right angled triangles.

satheesh27887I Think That Is The Right Answer.....
Congrats....
Keep The Good Work..... 
Kaustubh KatdareWe 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: JulyAugust 2008
[FONT=Verdana, Arial, Helvetica, sansserif]a) It is given:
CM = AB = m;
triangles ABC and Aâ€™Bâ€™C are similar; so, tangent Aâ€™B is parallel to the side AB.[/FONT]
[FONT=Verdana, Arial, Helvetica, sansserif]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, sansserif]Small triangle Aâ€™Bâ€™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:
Aâ€™N = Aâ€™Q e Bâ€™Q = Bâ€™M (fig. 2)
f) Thus the perimeter of the small triangle Aâ€™Bâ€™C is:
(CM  Bâ€™M) + (CN  Aâ€™N) + (Aâ€™Q + Bâ€™Q)
that is:
2CM  Bâ€™M  Aâ€™N + Aâ€™Q + Bâ€™Q = 2CM = 2m[/FONT]
[FONT=Verdana, Arial, Helvetica, sansserif]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 Aâ€™Bâ€™C is:
(4m/2m)2 = 4[/FONT][FONT=Verdana, Arial, Helvetica, sansserif]
[/FONT]
[FONT=Verdana, Arial, Helvetica, sansserif]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, sansserif]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, sansserif]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, sansserif]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, sansserif]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]
You are reading an archived discussion.
Related Posts
Q. Which leadership guru coined the term 'transformational leadership'?
1. Pradep Sindhu
2. Phillip Kotlar
3. Samuel A Dipiazza
4. James MacGregor Burns
Guess the answer and also post your...
well can any one please explain why the input impedance of any electronic circuit should be high and why the output impedance should be low
Alright.
I just came to know that Mr. Narayana Murthy, founder of Infosys is an Electrical Engineer.
😁
*is proud to be a crazy electrical engineer*
😁
CEans,
For those who need a serious motivation in their lives; go read the book by Donald Trump:
Think Big & Kick Ass!
I recommend the book to everyone!
CEans,
I know  there is an entrepreneur in each one of us. The world is full of wannabe entrepreneurs. The engineers have been the greatest entrepreneurs: Mr. Edison, to...