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Three identical circles are inscribed within an equilateral

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Three identical circles are inscribed within an equilateral [#permalink]

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Three identical circles are inscribed within an equilateral triangle, as shown above. If each of the radii of the circles is 3, what is the perimeter of the triangle?

A. 18(1+√3)
B. 9(1+2√3)
C. 18√3
D. 9(1+√3)
E. 6(1+√3)

I was able to solve it by guesstimating the answer, however I would love to see a detailed explanation one. (Question source - Master GMAT)

Thanks.

Originally posted by eladshush on 28 Sep 2010, 06:40.
Last edited by Bunuel on 22 Nov 2013, 01:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Circles in a triangle [#permalink]

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New post 28 Sep 2010, 07:33
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Three identical circles are inscribed within an equilateral triangle, as shown above. If each of the radii of the circles is 3, what is the perimeter of the triangle?
A. 18(1+√3)
B. 9(1+2√3)
C. 18√3
D. 9(1+√3)
E. 6(1+√3)

See attached file:
Attachment:
Triangle.PNG
Triangle.PNG [ 19.58 KiB | Viewed 26168 times ]


Side of this equilateral triangle equals to 2r (middle part) + 2AB. Now, AB is a leg opposite 60 degrees in 30-60-90 right triangle. Sides in 30-60-90 right triangle are in the ration \(1:\sqrt{3}:2\), so \(AB=r\sqrt{3}\)

So you can see that: side equals to \(2r+2*(r\sqrt{3})=6+6\sqrt{3}\), so \(P=3*(6+6\sqrt{3})=18(1+\sqrt{3})\).

Answer: A.
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Re: Circles in a triangle [#permalink]

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New post 28 Sep 2010, 08:15
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Thank you Bunuel. This was very useful as always.
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 02:12
I guessed this by using an approximate value of each side of the triangle...If u see the figure then it can be visualized that each side of the equilateral triangle have a length slightly greater than the sum of the 4 radius=4*3 =12. Therefore perimeter must be greater than 12+12+12=36 approx...

which is choice A.
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 03:21
Just.... How do you prove that angle BAE (with E the center of the circle on the left hand side) is equal to 30 degree ???
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 03:27
utin wrote:
I guessed this by using an approximate value of each side of the triangle...If u see the figure then it can be visualized that each side of the equilateral triangle have a length slightly greater than the sum of the 4 radius=4*3 =12. Therefore perimeter must be greater than 12+12+12=36 approx...

which is choice A.


This approach would give 100% accurate answer if only one option were more than 36, but in our case both A and B are more than 36. So chances are 50/50 to pick the right answer. Still good way of thinking for educated guess.

alexn49 wrote:
Just.... How do you prove that angle BAE (with E the center of the circle on the left hand side) is equal to 30 degree ???


The big triangle is equilateral, so the angels are 60 degrees each and BAE is half of it, so 30 degrees.
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 03:41
I reformulate : How do you know BAE is half ?
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 04:00
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alexn49 wrote:
I reformulate : How do you know BAE is half ?


How else? Look at the attachment below:
Attachment:
Triangle2.PNG
Triangle2.PNG [ 19.7 KiB | Viewed 26008 times ]
Right triangles ACE and ABE are congruent (they share the same hypotenuse, CE=BE=radius, hence the third sides are also equal). So <CAE=<BAE=<CAB/2=30.

Hope it's clear.
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Re: Circles in a triangle [#permalink]

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New post 29 Sep 2010, 04:42
Of course' ! Thx... Kudos !!!
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Re: Circles in a triangle [#permalink]

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New post 30 Sep 2010, 11:41
I have a more straightforward way to solve it

If we take the three centers of the inner circles we can draw an isoceles triangle with a total perimeter of 27 (9 X 3).

There is a relationship between increase in in the side of an isoceles triangle and the perimeter of it. In that case we are increasing the above mentioned triangle in three. as it is the length of the ratio the inner circles. In that case we will increase the perimeter from 27 to (9 + 3 ) = 12 X 3 = 36. The only answer that has an aproximate value to 36 is A.

Please comment.
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Re: Circles in a triangle [#permalink]

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New post 30 Sep 2010, 11:48
Bull78 wrote:
I have a more straightforward way to solve it

If we take the three centers of the inner circles we can draw an isoceles triangle with a total perimeter of 27 (9 X 3).

There is a relationship between increase in in the side of an isoceles triangle and the perimeter of it. In that case we are increasing the above mentioned triangle in three. as it is the length of the ratio the inner circles. In that case we will increase the perimeter from 27 to (9 + 3 ) = 12 X 3 = 36. The only answer that has an aproximate value to 36 is A.

Please comment.


I dont think I understand what you just did there.
The triangle formed by joining the centers, is not isoceles, it is equilateral and also its perimeter is 6*3=18 and not 27. And finally not sure where you get that 36 from. The answers above are the exact answers not approximations
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Re: Circles in a triangle [#permalink]

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New post 30 Sep 2010, 12:01
shrouded1 wrote:
Bull78 wrote:
I have a more straightforward way to solve it

If we take the three centers of the inner circles we can draw an isoceles triangle with a total perimeter of 27 (9 X 3).

There is a relationship between increase in in the side of an isoceles triangle and the perimeter of it. In that case we are increasing the above mentioned triangle in three. as it is the length of the ratio the inner circles. In that case we will increase the perimeter from 27 to (9 + 3 ) = 12 X 3 = 36. The only answer that has an aproximate value to 36 is A.

Please comment.


I dont think I understand what you just did there.
The triangle formed by joining the centers, is not isoceles, it is equilateral and also its perimeter is 6*3=18 and not 27. And finally not sure where you get that 36 from. The answers above are the exact answers not approximations


Also I'd add that the answer choices A and B are both more than 36: \(A\approx{49}\) and \(B\approx{40}\).

As for 36, we can think in the following way: the side of the triangle must be more than \(4r=12\), so the perimeter must be more than \(3*12=36\) (see utin's post above). This approach would give 100% accurate answer if only one option were more than 36, but in our case both A and B are more than 36. So chances are 50/50 to pick the right answer. Still good way of thinking for educated guess.

Hope it's clear.
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Re: Circles in a triangle [#permalink]

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New post 30 Sep 2010, 12:19
Makes sense. Educated guess can be really helpful if one has no idea of how to proceed with the problem.

In this problem, by guessing one is looking for something which is 36+

A is about 49
B is about 40

Most of the ppl guessing this one will end up with B :P .......... it helps to know the correct way to solve question :-D

Thanks Bunuel.
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Re: Circles in a triangle [#permalink]

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New post 21 Nov 2013, 14:59
eladshush wrote:
HI all,

Please consider the following problem:

Attachment:
Triangle.jpg


Three identical circles are inscribed within an equilateral triangle, as shown above. If each of the radii of the circles is 3, what is the perimeter of the triangle?

a. 18(1+√3)
b. 9(1+2√3)
c. 18√3
d. 9(1+√3)
e. 6(1+√3)

I was able to solve it by guesstimating the answer, however I would love to see a detailed explanation one. (Question source - Master GMAT)

Thanks.


Nice explanation Bunuel. Honestly, I could'nt find an easy way so I just ballparked.

So we have that each side is more than 2(2r) >12 so three sides >36. The only answer choice that makes sense is A.

Hope it helps
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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New post 01 Mar 2014, 14:22
This question can be solved in 3 easy step
1. calculate the area of 3 inscribed circle
2. calculate the approximate side of equilateral triangle from answers and compute area of equilateral triangle
3. compare the answer correctly.

Solution
Area of 3 inscribed circle = 3*3.14*9 =84. So area of triangle must be greater than 84 as circle are inscribed in triangle

a. 18(1+√3) =>6(2.7) =>16 A=(256*√3)/4=64√3 (approximately greater than 100 ) = 108
b. 9(1+2√3) =>3(4.46) => 13.3 =13 A=(169*√3)/4= 42√3(approximately 70) =74
c. 18√3 => 6√3 => 10.3 =10 . No need of calculation
d. 9(1+√3) => 3(2.7) => 8 .No need of calculation
e. 6(1+√3) => 2(2.7) => 6 .No need of calculation

SO answer is A.
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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New post 03 Mar 2014, 01:20
vikrantgulia wrote:
This question can be solved in 3 easy step
1. calculate the area of 3 inscribed circle
2. calculate the approximate side of equilateral triangle from answers and compute area of equilateral triangle
3. compare the answer correctly.

Solution
Area of 3 inscribed circle = 3*3.14*9 =84. So area of triangle must be greater than 84 as circle are inscribed in triangle

a. 18(1+√3) =>6(2.7) =>16 A=(256*√3)/4=64√3 (approximately greater than 100 ) = 108
b. 9(1+2√3) =>3(4.46) => 13.3 =13 A=(169*√3)/4= 42√3(approximately 70) =74
c. 18√3 => 6√3 => 10.3 =10 . No need of calculation
d. 9(1+√3) => 3(2.7) => 8 .No need of calculation
e. 6(1+√3) => 2(2.7) => 6 .No need of calculation

SO answer is A.



They are asking perimeter of the big triangle. Would this method still hold true. ?
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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New post 02 Jul 2014, 16:48
great explanation as always, thanks Bunuel
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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New post 26 Oct 2015, 15:38
picked B, but purely guess.
I could eliminate C, D, and E.
Suppose we draw a diagonal for each circle so that to form 3 equilateral triangles with sides 6
now we have 6 equal sides of these small equilateral triangles, which is 36. But it has to be slightly more than 36, since we have 3 equal parts that are not covered between these 3 small triangles.

C,D,and E are <36. Between A and B...picked B.

bunnuel's explanation is good...have to analyze it more...
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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New post 03 Jan 2016, 12:52
bumped into it another time...still had difficulties in understanding the process to solve it...nevertheless, I could eliminate C, D, and E, and after some more thinking, B as well..

so we have 3 circles, diameter of each is 6. thus, the length of each side should be more than 12. now we have 12+ * 3 = 36+.

A. 18(1+√3) = 18+18√3 = 18+slightly less than 36, or overall slightly less than 54. of these two, picked A.
B. 9(1+2√3) = 9+ 18√3, so 9+slightly less than 36, or overall -> slightly less than 45.
C. 18√3 = less than 36, so out.
D. 9(1+√3) = 9+9√3 = 8+18- => slightly less than 26, so out.
E. 6(1+√3) = 6 + 6√3 = that's slightly less than 18, so out.
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Re: Three identical circles are inscribed within an equilateral [#permalink]

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Re: Three identical circles are inscribed within an equilateral   [#permalink] 12 Jun 2018, 08:10
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