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# Tough Puzzle: Ratio of similar triangles

Puzzle Source: Mind and Visual Puzzles!

[FONT=Verdana, Arial, Helvetica, sans-serif]If the segment

[FONT=Verdana, Arial, Helvetica, sans-serif]If the segment

**is tangent to the inscribed circle of triangle***A'B'***, and that***ABC**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]
21 views so far and not a single attempt?

Too tough for you, CEans?

All the best! 👍

Too tough for you, CEans?

All the best! 👍

Where did all the talented people on CE go? 😒

Come on its simple Geometry. Did you not learn it in school? 😲

Come on its simple Geometry. Did you not learn it in school? 😲

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!

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!

HELLO... ''THE BIG K''

we don't know how to begin....

Can u give us another hint.....😕

we don't know how to begin....

Can u give us another hint.....😕

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:

😁

Now there are only two options - either crack the puzzle or accept the defeat! :twisted:

😁

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

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

taken AB as the base of triangle ABC & A'B'||AB as ABC~A'B'C

abc:a'b'c is 2:1

The_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.......

satheesh27887k.... 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!

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

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

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

I Think That Is The Right Answer.....

Congrats....

Keep The Good Work.....

Congrats....

Keep The Good Work.....

We DO have smartest people on the planet Earth! 😁

Here's the detailed solution for those who couldn't crack the puzzle!

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]

triangles

[FONT=Verdana, Arial, Helvetica, sans-serif]Triangle

2

2(

[/FONT]

[FONT=Verdana, Arial, Helvetica, sans-serif]Small triangle

(

that is:

2

[FONT=Verdana, Arial, Helvetica, sans-serif]In conclusion

Then, the ratio of the area of triangle

(4

[/FONT]

[FONT=Verdana, Arial, Helvetica, sans-serif]Another solution

And here is an algebrical solution submitted by Fu Su:

[FONT=Verdana, Arial, Helvetica, sans-serif]

a + b + (c + e) = 2(a + b) [since c + e = a + b]

[FONT=Verdana, Arial, Helvetica, sans-serif]

triangle

=

[FONT=Verdana, Arial, Helvetica, sans-serif]

= [(a + b) - (c + d)] / 2(a + b) =

= 1/2 - 1/2[(c + d) / (a + b)]

= (

[FONT=Verdana, Arial, Helvetica, sans-serif]

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**) 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*=2

*CM*+ (*BM*+*AN*) +*AB*=2(

*CM*+*AB*) = 4*CM*= 4*m*[/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:

2

*CM*-*B’M*-*A’N*+*A’Q*+*B’Q*= 2*CM*= 2*m*[/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:(4

*m*/2*m*)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*: 2*r*(a + b)[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]

**e**) Area of triangle*A'B'C*:triangle

*ABC*- quadrilateral*ABB'A'*= 2*r*(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)] / 2*r*(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]