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805+ (Hard)|   Geometry|      
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gmatt1476
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The figure above shows 2 circles. The larger circle has center A, radi 21 Sep 2019, 06:57
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95% (hard)

Question Stats:

49% (03:01) correct 51% (03:08) wrong based on 1354 sessions

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The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}\)

B. \(\frac{2}{\sqrt{2} - 1}\)

C. \(\frac{2}{2\sqrt{2} + 1}\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\)


PS75571.01


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D
Every tricky question should trigger a follow up quiz. Build that habit with GMAT Club Forum Quiz →.
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The figure above shows 2 circles. The larger circle has center A, radi 28 Sep 2019, 05:14
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side of the square = 2R
diagonal of the square = 2√2R
AE = half of the diagonal of the square (A, B, C, and E are collinear and lies on the diagonal of the square)
AE = √2R
AE = AB+BC+CE
√2R = R+r+√2r (BC = r--radius of small circle CE = diagonal of small square)

R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{√2+1}{√2-1}\)

D is the answer
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The figure above shows 2 circles. The larger circle has center A, radi 21 Sep 2019, 11:09
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)


PS75571.01
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The big and the small squares are similar to each other, so we can set up an equation between the ratio of their side lengths and the ratio of their diagonals.

\(\frac{2R}{r}=\frac{2(R+r+r\sqrt{2})}{r\sqrt{2}}\)

\(R\sqrt{2}-R=r+r\sqrt{2}\)

\(R(\sqrt{2}-1)=r(1+\sqrt{2})\)

\(\frac{R}{r}=\frac{1+\sqrt{2}}{\sqrt{2}-1}\)

Answer: D
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The figure above shows 2 circles. The larger circle has center A, radi 21 Sep 2019, 07:24
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

PS75571.01


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Definitely NOT a 2 min Question!

From the figure, AC = R + r
Draw a horizontal line from A and vertical line from C. Let both the lines meet at point H

Note that, triangle ACH is an isosceles right angled triangle, with AH = CH = R - r
--> \(AC^2 = AH^2 + CH^2\)
--> \((R + r)^2 = (R - r)^2 + (R - r)^2\)
--> \(R^2 + 2Rr + r^2 = 2R^2 - 4Rr + 2r^2\)
--> \(R^2 - 6Rr + r^2 = 0\)
--> \((R/r)^2 - 6R/r + 1 = 0\)
Let R/r = x
--> \(x^2 - 6x + 1 = 0\)
--> \(x = (6 ± \sqrt{36 - 4*1*1})/(2*1)\)
--> \(x = (6 ± \sqrt{32})/2\)
--> \(x = (6 ± 4\sqrt{2})/2\)
--> \(x = (3 ± 2\sqrt{2}\)
--> \(x = 3 + 2\sqrt{2}\) [Since R>r --> x > 1]

\(x = (3 + 2\sqrt{2})\) is equivalent to Option D
--> \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\) (Multiply numerator & Denominator by \(\sqrt{2} + 1\) )
--> \((\sqrt{2} + 1)^2 / (\sqrt{2}^2 - 1^2)\)
--> \((2 + 1 + 2\sqrt{2})/1\)
--> \(3 + 2\sqrt{2}\)

IMO Option D

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The figure above shows 2 circles. The larger circle has center A, radi 25 Sep 2019, 00:27
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logically we can tell that
AE = AB+ BC + CE
AE = root(2) R
AB =R
Bc =r
Ce= root(2)r
equat the following values

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The figure above shows 2 circles. The larger circle has center A, radi 28 Sep 2019, 02:16
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Let's see..

As soon as i read "If points A, B, C, and E are collinear" i had a feeling the question will probably revolve around the length of AE. Further on seeing the fig it's quite obvious we can form a right triangle. S, with these things in mind, let's proceed.

What's AE ?
AE = R + r + rRoot2 (i.e the diagonal of CFDE)

Sorry for not providing any visual representation but just try to visualise with me.

Now, if we take the right triangle with AE as the hypotenuse.
R^2 + R^2 = (R+r+rRoot2)^2
2R^2 = (R+r+rRoot2)^2
Taking root (since dis can't be -ve)
(Root2)R = R+r+rRoot2
Solve this and you'll get D

If you still have problem solving it. Let me know, i'll post the detailed soln
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The figure above shows 2 circles. The larger circle has center A, radi 08 Jul 2020, 03:49
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[quote="gmatt1476"]
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 05:20
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)
Show more


Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 06:48
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)


Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks
Show more


The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 07:07
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OmotayoH
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)


Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks


The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.
Show more


Thanks a lot for responding.

I am still unclear about this ' Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

When i calculate the square root of \(\sqrt{8R^2}\) i get 2R\(\sqrt{2}\)

Do you mind clarifying please?

Also is there a reason why the 45-45-90 triangle rule doesn't work here?
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 07:17
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MikeScarnScott
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MikeScarnScott
gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)


Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks


The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.


Thanks a lot for responding.

I am still unclear about this ' Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

When i calculate the square root of \(\sqrt{8R^2}\) i get 2R\(\sqrt{2}\)

Do you mind clarifying please?

Also is there a reason why the 45-45-90 triangle rule doesn't work here?
Show more

R is outside the square-root and 2R\(\sqrt{2}\) is same as 2\(\sqrt{2}\)R
45-45-90 will also work
1:1:\(\sqrt{2}\) = 2R:2R:2R\(\sqrt{2}\)
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 07:21
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OmotayoH
Your 90-45-45 right angle Isosceles triangle is nothing but two halves of square. We have considered side of bigger square as 2R
That's why the diagonal is 2√2R
You can also write it like 2R, 2R and 2√2R (45°, 45°, 90° triangle)
No such difference. Both are two sides of same coin.

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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 08:32
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Please refer to the pic uploaded,

This is how I have done.

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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2020, 11:10
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)



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Along a diagonal;
\(2(R + r + r\sqrt{2}) = 2R\sqrt{2}\)
\(R (\sqrt{2} - 1) = r(\sqrt{2}+1)\)
\(\frac{R}{r} = \frac{\sqrt{2}+1}{\sqrt{2} - 1}\)

IMO D
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The figure above shows 2 circles. The larger circle has center A, radi 13 Sep 2020, 00:27
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shridhar786
side of the square = 2R
diagonal of the square = 2√2R
AE = half of the diagonal of the square (A, B, C, and E are collinear and lies on the diagonal of the square)
AE = √2R
AE = AB+BC+CE
√2R = R+r+√2r (BC = r--radius of small circle CE = diagonal of small square)

R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{√2+1}{√2-1}\)

D is the answer
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Hi @Bunual

How can we say that A, B, C and E are collinear?
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The figure above shows 2 circles. The larger circle has center A, radi 06 Oct 2020, 12:29
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I am unable to view the answer options. All of them appear as thumbnail icons. I have refreshed the page multiple times and opened in different browsers but still not able to read the options Bunuel
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The figure above shows 2 circles. The larger circle has center A, radi 06 Oct 2020, 12:51
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I am unable to view the answer options. All of them appear as thumbnail icons. I have refreshed the page multiple times and opened in different browsers but still not able to read the options Bunuel
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Fixed the options. Thank you for notifying!
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The figure above shows 2 circles. The larger circle has center A, radi 09 Dec 2020, 11:44
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The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}\)

B. \(\frac{2}{\sqrt{2} - 1}\)

C. \(\frac{2}{2\sqrt{2} + 1}\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\)


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Sharing an alternate method to solve this question

1. Let's assume the side of the square to be \(10\). You can assume any number here. The idea is to make the calculations simpler

2. The diameter of the larger circle i.e. \(R\) will become \(5\)

3. \(AE\) = \(R\) \(+\) \(r\) \(+\) \(\sqrt{2}r\) (\(\sqrt{2}r\) = \(CE\) by Pythagorean theorem as \(CD\) and \(DE\) are each \(r\) as both are radii of the smaller circle)

4. Now, \(R = 5\) hence \(AE\) \(=\) \(5 + R + \sqrt{2}r\)

5. \(AE\) \(=\) \(5\sqrt{2}\) again using Pythagorean theorem as the larger circle is within a square and the perpendiculars from \(A\) to the right and to the bottom are radiis of the larger circle

6. \(5\sqrt{2}\) \(=\) \(5 + r + \sqrt{2}r\)

7. \(5\sqrt{2} - 5\) \(=\) \(r + \sqrt{2}r\)

8. \(5\)(\(\sqrt{2} - 1\)) \(=\) \(r\) (\(\sqrt{2} + 1\))

9. \(r\) \(=\) 5\(\frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1)}\)

10. Now \(\frac{R}{r}\) \(=\) \(5 / \frac{5(\sqrt{2} - 1)}{(\sqrt{2} + 1)}\) = \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

Ans. D
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The figure above shows 2 circles. The larger circle has center A, radi 10 Jul 2021, 07:50
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}\)

B. \(\frac{2}{\sqrt{2} - 1}\)

C. \(\frac{2}{2\sqrt{2} + 1}\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\)


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Answer: Option D
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The figure above shows 2 circles. The larger circle has center A, radi 18 Apr 2022, 11:03
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gmatt1476

The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}\)

B. \(\frac{2}{\sqrt{2} - 1}\)

C. \(\frac{2}{2\sqrt{2} + 1}\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\)

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Other users have already posted solid solutions to this question, so I won't waste any time re-solving it.

However, if you're unsure how to solve the question, or if you're quite far behind timing-wise, you can apply the following property:
The diagrams in GMAT Problem Solving questions are DRAWN TO SCALE unless stated otherwise.
So, we may be able to use this fact to solve the question (or at least eliminate some answer choices before guessing) by simply "eyeballing" the diagram.

First, let's (mentally) add the two radii (R and r) as horizontal line segments:


Then (mentally) move the line segment for r close to the line segment for R:


At this point, we can see that the length of R is 4 to 6 times the length of r.
In other words, the value of R/r is somewhere in the 4 to 6 range.

ASIDE: Before test day, all students should have the following approximations memorized: √2 ≈ 1.4, √3 ≈ 1.7 and √5 ≈ 2.2
Now substitute 1.4 for √2 in the answer choices to get:

A. \(\frac{2}{1.4 + 1} = \frac{2}{2.4}\) = some number less than 1. ELIMINATE.

B. \(\frac{2}{1.4 - 1} = \frac{2}{0.4}\) = approximately 5. KEEP.

C. \(\frac{2}{2(1.4) + 1} = \frac{2}{1.8}\) = some number a little greater than 1. ELIMINATE.

D. \(\frac{1.4+ 1}{1.4 - 1}= \frac{2.4}{0.4}\) = approximately 6. KEEP.

E. \(\frac{2(1.4) + 1}{2(1.4) - 1}= \frac{3.8}{1.8}\) = approximately 2. ELIMINATE.

So, guess either B and D.
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