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Intern  Joined: 01 Sep 2010
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Three identical circles are inscribed within an equilateral  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 69% (02:42) correct 31% (02:34) wrong based on 169 sessions

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Attachment: Triangle.jpg [ 9.38 KiB | Viewed 41640 times ]

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, 05:40.
Last edited by Bunuel on 22 Nov 2013, 00:33, 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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3
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 [ 19.58 KiB | Viewed 41465 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 $$P=3*(6+6\sqrt{3})=18(1+\sqrt{3})$$.

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Intern  Joined: 01 Sep 2010
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Re: Circles in a triangle  [#permalink]

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Thank you Bunuel. This was very useful as always.
Manager  Joined: 26 Mar 2010
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Re: Circles in a triangle  [#permalink]

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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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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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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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I reformulate : How do you know BAE is half ?
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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 [ 19.7 KiB | Viewed 41142 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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Of course' ! Thx... Kudos !!!
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Re: Circles in a triangle  [#permalink]

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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.

Attachments triangle.ppt [30 KiB] triangle.ppt [30 KiB]

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Re: Circles in a triangle  [#permalink]

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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.

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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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.

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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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+

Most of the ppl guessing this one will end up with B .......... it helps to know the correct way to solve question Thanks Bunuel.
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Re: Circles in a triangle  [#permalink]

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HI all,

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
Cheers!
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Re: Three identical circles are inscribed within an equilateral  [#permalink]

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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

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

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

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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

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

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

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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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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] 07 Oct 2019, 16:44

# Three identical circles are inscribed within an equilateral  