GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Dec 2018, 14:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

December 16, 2018

December 16, 2018

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# Within equilateral triangle A, circle B is inscribed; within circle B,

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51218
Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

07 Sep 2015, 03:40
2
17
00:00

Difficulty:

95% (hard)

Question Stats:

49% (02:33) correct 51% (02:49) wrong based on 154 sessions

### HideShow timer Statistics

Within equilateral triangle A, circle B is inscribed; within circle B, square C is inscribed; within square C, circle D is inscribed; within circle D, square E is inscribed; within square E, circle F is inscribed; finally, within circle F, equilateral triangle G is inscribed.

What is the ratio of the area of triangle A to the area of triangle G?

(A) 4 √2 : 1
(B) 8 : 1
(C) 8 √2 : 1
(D) 16 : 1
(E) 16 √2 : 1

_________________
Manager
Joined: 29 Jul 2015
Posts: 158
Re: Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

07 Sep 2015, 11:18
1
Bunuel wrote:
Within equilateral triangle A, circle B is inscribed; within circle B, square C is inscribed; within square C, circle D is inscribed; within circle D, square E is inscribed; within square E, circle F is inscribed; finally, within circle F, equilateral triangle G is inscribed.

What is the ratio of the area of triangle A to the area of triangle G?

(A) 4 √2 : 1
(B) 8 : 1
(C) 8 √2 : 1
(D) 16 : 1
(E) 16 √2 : 1

Triangle G is inscribed within circle F or circle F is the circumcircle of triangle G.
circumradius of a circumcircle of a triangle is given by $$\frac{abc}{4 \triangle}$$ , where abc is the product of sides of a triangle and $$\triangle$$ is the area of triangle.
let the side of triangle G be $$x$$
then the area of triangle G will be $$\frac{\sqrt3}{4}$$ * $$x^2$$
then the circumradius of triangle G or radius of circle F will be $$\frac{x^3}{4* \frac{\sqrt3}{4} * x^2}$$ = $$\frac{x}{\sqrt3}$$ = $$\frac{\sqrt3x}{3}$$

Circle F is inscribed within square E, so the side of square will be equal to diameter of circle F = $$2\frac{\sqrt3x}{3}$$
The diagonal of a square is given by $$\sqrt2$$ * side of square. Diagonal of square E will be $$2*\sqrt{2}*\frac{\sqrt3x}{3}$$ = $$2\frac{\sqrt6x}{3}$$
This diagonal will be equal to diameter of circle D which circumscribes square E and the diameter of circle D will be equal to the side of square C in which circle D is inscribed.
So the side of square C will be $$2\frac{\sqrt6x}{3}$$
Diagonal of square C will be $$\sqrt2$$*$$2*\frac{\sqrt6x}{3}$$ = $$4\frac{\sqrt3x}{3}$$
Diagonal of square C will be equal to diameter of Circle B
The radius of circle circle B will be $$2\frac{\sqrt3x}{3}$$ ....(1)

Circle B is the incircle of triangle A.
Inradius of a circle is given by $$\frac{\triangle}{S}$$ where $$\triangle$$ is the area of triangle and S is semi perimeter = 1/2(sum of sides of triangle.)
Let y be the side of traingle A,
the area of triangle A is $$\frac{\sqrt3}{4}$$ * $$y^2$$
and semi perimeter is $$3y/2$$
Inradius or radius of circle B will be $$\frac{\frac{\sqrt3}{4}y^2}{\frac{3y}{2}}$$ = $$\frac{\sqrt3}{6}y$$ ....(2)

(1) = (2) =radius of circle B
or $$2\frac{\sqrt3x}{3}$$ = $$\frac{\sqrt3}{6}y$$
or 4x = y
or $$\frac{y}{x}=4$$

Ratio of area of triangle A and area of triangle G will be equal to ratio of square of their sides.
or $$\frac{y^2}{x^2}=4^2=16$$
So the ratio will be $$16:1$$

Intern
Joined: 02 Jun 2015
Posts: 33
Location: United States
Concentration: Operations, Technology
GMAT Date: 08-22-2015
GPA: 3.92
WE: Science (Other)
Re: Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

07 Sep 2015, 15:06
2
Alright, let's be honest, even if you have all the geometric relationships memorized, it is going to be extremely tough to punch out this problem in less than 2 minutes. When I first saw this problem I thought "there is no way I'm finishing this one in 2 minutes". I opted for guesswork, realizing that the answer choices are relatively spaced. ~6:1, 8:1, ~12:1, 16:1, ~24:1.

I just drew the figure as accurately as possible in a minute and then tried to cut up the areas to get a decent comparison.

Breaking down the triangle it appears that the ratio is 16:1 just by insepction

It was pretty obvious that the answer wasn't 6:1, 8:1 or 24:1. 16:1 felt the best, so that is what I guessed.

I'm sure there is a quick way to do this mathematically, but I'm not sure that come test time I'd be able to invest time into that route. Hopefully someone else can provide a more savvy "real" answer.
CEO
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

07 Sep 2015, 15:37
1
PolarCatastrophe wrote:
Alright, let's be honest, even if you have all the geometric relationships memorized, it is going to be extremely tough to punch out this problem in less than 2 minutes. When I first saw this problem I thought "there is no way I'm finishing this one in 2 minutes". I opted for guesswork, realizing that the answer choices are relatively spaced. ~6:1, 8:1, ~12:1, 16:1, ~24:1.

I just drew the figure as accurately as possible in a minute and then tried to cut up the areas to get a decent comparison.

Breaking down the triangle it appears that the ratio is 16:1 just by insepction

It was pretty obvious that the answer wasn't 6:1, 8:1 or 24:1. 16:1 felt the best, so that is what I guessed.

I'm sure there is a quick way to do this mathematically, but I'm not sure that come test time I'd be able to invest time into that route. Hopefully someone else can provide a more savvy "real" answer.

I will give it a shot with the "savvy real answer", using the figure that you attached.

Once you draw the collection of figures as you have done, the symmetry of the figure will give you the following observations:

1. 'O' the incenter of triangle 'G', will also be the incenter of triangle 'A'.
2. $$OH$$ = $$\frac{OD}{4}$$ .. this is easy to prove and also evident from the figure you have drawn.

(2) will lead to the final answer as follows: Let 2a and 2a' be the sides of triangles A and G respectively.

In triangle ODC, right angled at D, $$OD = \frac{CD}{\sqrt{3}}$$ = $$\frac{a}{\sqrt{3}}$$

Similarly, in triangle OEH, right angled at H,

$$OH = \frac{EH}{\sqrt{3}}$$ =$$\frac{a'}{\sqrt{3}}$$

As per (2) above, $$\frac{a'}{\sqrt{3}}$$ = $$\frac{a}{4*\sqrt{3}}$$ ---->$$a' = \frac{a}{4}$$

As the ratio of areas of the triangles A and G are proportional to the ratio of square of the sides,

$$\frac{Area of A}{Area of G}$$ = $$\frac{(4a')^2}{(a')^2}$$ = 16:1

Does look like a long way but once you figure out the relation as mentioned in (2), it will be straightforward.

Hope this helps.
Attachments

2015-09-07_19-16-50.jpg [ 35.22 KiB | Viewed 2874 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

13 Sep 2015, 09:07
Bunuel wrote:
Within equilateral triangle A, circle B is inscribed; within circle B, square C is inscribed; within square C, circle D is inscribed; within circle D, square E is inscribed; within square E, circle F is inscribed; finally, within circle F, equilateral triangle G is inscribed.

What is the ratio of the area of triangle A to the area of triangle G?

(A) 4 √2 : 1
(B) 8 : 1
(C) 8 √2 : 1
(D) 16 : 1
(E) 16 √2 : 1

MANHATTAN GMAT OFFICIAL SOLUTION:

The first trick is to draw this complicated picture correctly!

Next, work your way from the inside out. Pick the radius of circle F as 1. Notice that this is the distance from the center of triangle G to a vertex of G. For now, don’t worry about finding the area of triangle G—we’ll see a shortcut later.

Now, what is the radius of the next circle out, circle D? 1 is the distance from the center of square E to the center of any side of that square. So we can draw a 45-45-90 triangle and find that the “half-diagonal” of square E is √2. This is also the radius of circle D.

Applying the same reasoning, we can see that √2 is the distance from the center of square C to the center of any side of that square. We can draw another 45-45-90 triangle and find that the “half-diagonal” of square C is 2. This is also the radius of circle B.

Finally, we can draw a 30-60-90 triangle within equilateral triangle A and see that the distance from the center to any vertex of triangle A is 4.

Now, rather than calculate each area and divide, we can use a huge shortcut. The distance from center to vertex for triangle G was 1; the same distance for triangle A is 4. Since these two triangles are similar, this means that every “distance” ratio for the two triangles will be 4 : 1. For instance, their side lengths will be in a 4 : 1 ratio. And since areas are squares of distances, the ratio of areas will be 4^2 to 1^2, or 16 : 1.

Attachment:

Screen-shot-2010-12-20-at-1.39.17-PM.png [ 15.65 KiB | Viewed 3067 times ]

Attachment:

Screen-shot-2010-12-20-at-1.39.28-PM.png [ 9.07 KiB | Viewed 3065 times ]

Attachment:

Screen-shot-2010-12-20-at-1.39.37-PM.png [ 9.38 KiB | Viewed 2725 times ]

Attachment:

Screen-shot-2010-12-20-at-1.39.45-PM.png [ 9.46 KiB | Viewed 3064 times ]

_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9166
Re: Within equilateral triangle A, circle B is inscribed; within circle B,  [#permalink]

### Show Tags

01 Mar 2018, 09:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Within equilateral triangle A, circle B is inscribed; within circle B, &nbs [#permalink] 01 Mar 2018, 09:05
Display posts from previous: Sort by