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

Question

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

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

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION: The first trick is to draw this complicated picture correctly! https://gmatclub.com/forum/download/file.php?id=27810 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. https://gmatclub.com/forum/download/file.php?id=27811 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. https://gmatclub.com/forum/download/file.php?id=27811 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. https://gmatclub.com/forum/download/file.php?id=27813 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. The correct answer is D. Screen-shot-2010-12-20-at-1.39.17-PM.png Screen-shot-2010-12-20-at-1.39.28-PM.png Screen-shot-2010-12-20-at-1.39.37-PM.png Screen-shot-2010-12-20-at-1.39.45-PM.png

kunal555 · Answer

Let's start with equilateral triangle G.
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 	riangle}  , where abc is the product of sides of a triangle and  	riangle  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{	riangle}{S}  where  	riangle  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

Answer:- D

PolarCatastrophe · Answer

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.

https://i60.tinypic.com/2e66d8o.jpg

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

https://i61.tinypic.com/vxhxqa.png

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.

ENGRTOMBA2018 · 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.

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