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Kaustubh Katdare • Jul 20, 2008

# Tough Puzzle: Ratio of similar triangles

Puzzle Source: Mind and Visual Puzzles!

[FONT=Verdana, Arial, Helvetica, sans-serif]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'B’C? Hint: angle CAB is not necessarily a right angle, however triangles ABC and A'B'C are similar![/FONT]

Kaustubh Katdare • Jul 21, 2008
21 views so far and not a single attempt?

Too tough for you, CEans?

All the best! 👍
Kaustubh Katdare • Jul 22, 2008
Where did all the talented people on CE go? 😒

Come on its simple Geometry. Did you not learn it in school? 😲
Kaustubh Katdare • Jul 22, 2008
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 • Jul 28, 2008
HELLO... ''THE BIG K''
we don't know how to begin....
Can u give us another hint.....😕
Kaustubh Katdare • Jul 28, 2008
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 • Jul 28, 2008
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 • Jul 29, 2008
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 • Jul 29, 2008
abc:a'b'c is 2:1
satheesh27887 • Jul 30, 2008
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 • Jul 30, 2008
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 • Jul 31, 2008
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 • Jul 31, 2008
What confounded me, was why give AB = CM? And all the triangles are right angled triangles.
satheesh27887 • Jul 31, 2008
I Think That Is The Right Answer.....
Congrats....
Keep The Good Work.....
Kaustubh Katdare • Sep 22, 2008
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)
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]