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The figure is made up of a series of inscribed equilateral

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The figure is made up of a series of inscribed equilateral [#permalink] New post 17 Sep 2011, 19:45
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File comment: Inscribed Traingles
InscribedTraingles.JPG
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The figure is made up of a series of inscribed equilateral triangles. If the pattern continues until the length of a side of the largest triangle (i.e. the entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded?

A. 1\4(2^0 + 2^-4 + 2^-8 + 2^-12)

B. 1\4(2^0 + 2^-2 + 2^-4 + 2^-6)

C. 3\4(2^0 + 2^-4 + 2^-8 + 2^-12)

D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6)

E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3)

PS: I am sorry for the formatting of the answer choices like this.
[Reveal] Spoiler: OA

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Re: Inscribed Equilateral Triangles. [#permalink] New post 17 Sep 2011, 20:45
The Answer is C.


To get a triangle 128 times(2^7) smaller than the main triangle, you will have to partition it 7 times.
For each Odd partition you will get 3 Shaded regions.
For each Even partition you will get 1 shaded region the will again be partitions into regions by subsequent odd partition.So, we will no consider these partitions.

Partition 1 2 3 4 5 6 7
Shaded Regions 3 1 3 1 3 1 3

Consider the area of main triangle be A.
So the shaded region will be (3)*(1/4^7)A + (3)*(1/4^5)A + (3)*(1/4^3)A + (3)*(1/4^1)A

= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A

To get the fraction of shaded region , (3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A) / A
= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 )
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Re: Inscribed Equilateral Triangles. [#permalink] New post 21 Sep 2011, 00:27
may i ask which source it is from? i dont think real test will have such kinda of question
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Re: Inscribed Equilateral Triangles. [#permalink] New post 21 Sep 2011, 00:37
IMO C.

I did not calculate it completely. Just calculated the area of first two partitions and established a pattern and then used the pattern to guess the answer. That's what I am going to do if something like this comes in real test as well :)
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Re: Inscribed Equilateral Triangles. [#permalink] New post 21 Sep 2011, 05:49
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varunmaheshwari wrote:
Attachment:
InscribedTraingles.JPG
The figure is made up of a series of inscribed equilateral triangles. If the pattern continues until the length of
a side of the largest triangle (i.e. the entire figure) is exactly 128 times that of the smallest triangle, what
fraction of the total figure will be shaded?

A. 1\4(2^0 + 2^-4 + 2^-8 + 2^-12)

B. 1\4(2^0 + 2^-2 + 2^-4 + 2^-6)

C. 3\4(2^0 + 2^-4 + 2^-8 + 2^-12)

D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6)

E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3)

PS: I am sorry for the formatting of the answer choices like this.


Area \hspace{3} of \hspace{3} equilateral \hspace{3} triangle = \frac{\sqrt{3}}{4}*(side)^2

Total \hspace{3} Area = \frac{\sqrt{3}}{4}(128)^2=\frac{\sqrt{3}}{4}(2^7)^2=\frac{\sqrt{3}}{4}*2^{14}

Shaded \hspace{3} portion \hspace{3} of \hspace{3} outer \hspace{3} most= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^7)^2 {:Side=128=2^7}
Skip the next: {:Side=64}
Shaded \hspace{3} portion \hspace{3} of 3^{rd}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^5)^2 {:Side=32=2^5}
Shaded \hspace{3} portion \hspace{3} of 5^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^3)^2 {:Side=8=2^3}
Shaded \hspace{3} portion \hspace{3} of 7^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^1)^2 {:Side=2=2^1}

Take the fraction:
\frac{\frac{3}{4}*\frac{\sqrt{3}}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{\frac{\sqrt{3}}{4}*2^{14}}

=\frac{\frac{3}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{2^{14}}

=\frac{3}{4}(\frac{2^{14}}{2^{14}}+\frac{2^{10}}{2^{14}}+\frac{2^{6}}{2^{14}}+\frac{2^{2}}{2^{14}})

=\frac{3}{4}(2^{(14-14)}+2^{(10-14)}+2^{(6-14)}+2^{(2-14)})

=\frac{3}{4}(2^0+2^{-4}+2^{-8}+2^{-12})

Ans: "C"
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Re: Inscribed Equilateral Triangles. [#permalink] New post 20 Nov 2013, 01:09
fluke wrote:
varunmaheshwari wrote:
Attachment:
InscribedTraingles.JPG
The figure is made up of a series of inscribed equilateral triangles. If the pattern continues until the length of
a side of the largest triangle (i.e. the entire figure) is exactly 128 times that of the smallest triangle, what
fraction of the total figure will be shaded?

A. 1\4(2^0 + 2^-4 + 2^-8 + 2^-12)

B. 1\4(2^0 + 2^-2 + 2^-4 + 2^-6)

C. 3\4(2^0 + 2^-4 + 2^-8 + 2^-12)

D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6)

E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3)

PS: I am sorry for the formatting of the answer choices like this.


Area \hspace{3} of \hspace{3} equilateral \hspace{3} triangle = \frac{\sqrt{3}}{4}*(side)^2

Total \hspace{3} Area = \frac{\sqrt{3}}{4}(128)^2=\frac{\sqrt{3}}{4}(2^7)^2=\frac{\sqrt{3}}{4}*2^{14}

Shaded \hspace{3} portion \hspace{3} of \hspace{3} outer \hspace{3} most= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^7)^2 {:Side=128=2^7}
Skip the next: {:Side=64}
Shaded \hspace{3} portion \hspace{3} of 3^{rd}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^5)^2 {:Side=32=2^5}
Shaded \hspace{3} portion \hspace{3} of 5^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^3)^2 {:Side=8=2^3}
Shaded \hspace{3} portion \hspace{3} of 7^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^1)^2 {:Side=2=2^1}

Take the fraction:
\frac{\frac{3}{4}*\frac{\sqrt{3}}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{\frac{\sqrt{3}}{4}*2^{14}}

=\frac{\frac{3}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{2^{14}}

=\frac{3}{4}(\frac{2^{14}}{2^{14}}+\frac{2^{10}}{2^{14}}+\frac{2^{6}}{2^{14}}+\frac{2^{2}}{2^{14}})

=\frac{3}{4}(2^{(14-14)}+2^{(10-14)}+2^{(6-14)}+2^{(2-14)})

=\frac{3}{4}(2^0+2^{-4}+2^{-8}+2^{-12})

Ans: "C"



Can you please explain why u skipped certain sides?
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Re: Inscribed Equilateral Triangles. [#permalink] New post 02 Jan 2014, 09:06
vinodmallapu wrote:
The Answer is C.


To get a triangle 128 times(2^7) smaller than the main triangle, you will have to partition it 7 times.
For each Odd partition you will get 3 Shaded regions.
For each Even partition you will get 1 shaded region the will again be partitions into regions by subsequent odd partition.So, we will no consider these partitions.

Partition 1 2 3 4 5 6 7
Shaded Regions 3 1 3 1 3 1 3

Consider the area of main triangle be A.
So the shaded region will be (3)*(1/4^7)A + (3)*(1/4^5)A + (3)*(1/4^3)A + (3)*(1/4^1)A

= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A

To get the fraction of shaded region , (3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A) / A
= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 )


Excuse me but would you explain why you skipped the even partitions?

Cheers!
J :)
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Re: The figure is made up of a series of inscribed equilateral [#permalink] New post 10 Feb 2014, 05:34
I believe the even portions are skipped because those are considered shaded in the next odd portions.
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Re: The figure is made up of a series of inscribed equilateral [#permalink] New post 10 Feb 2014, 22:14
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varunmaheshwari wrote:
Attachment:
InscribedTraingles.JPG
The figure is made up of a series of inscribed equilateral triangles. If the pattern continues until the length of a side of the largest triangle (i.e. the entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded?

A. 1\4(2^0 + 2^-4 + 2^-8 + 2^-12)

B. 1\4(2^0 + 2^-2 + 2^-4 + 2^-6)

C. 3\4(2^0 + 2^-4 + 2^-8 + 2^-12)

D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6)

E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3)

PS: I am sorry for the formatting of the answer choices like this.


We can arrive at the answer quickly by estimating. The outermost triangle is split into 4 equal triangles 3 of which are completely shaded. Hence the fraction shaded is more than 3/4. So we are probably looking at options (C), (D) and (E).

We are left with the center triangle which occupies 1/4th of the area. 3/4th of this 1/4th is not shaded while a part of 1/4th of 1/4th is shaded. How much of this is completely shaded? 3/4th since the 1/4th of 1/4th triangle is also split into 4 equal areas and 3 of those areas are shaded. Hence the area of the shaded region would be something like this:

3/4 + (3/4)*(1/4)*(1/4) + ... = (3/4)(1 + 1/16 + ...)

We see that this matches option (C) which is 3/4(1 + 1/16 + ...)

Answer (C)

Note that none of the other options can give an area close to this one.
D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6) = (3\4)(1 + 1/4 + ...) Too much
E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3) = (3\4)(1 + 1/2 + ...) Too much again
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Re: The figure is made up of a series of inscribed equilateral   [#permalink] 10 Feb 2014, 22:14
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