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# The figure above shows 2 circles. The larger circle has center A, radi

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The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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21 Sep 2019, 05:57
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Question Stats:

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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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2019-09-21_1753.png [ 27.97 KiB | Viewed 5414 times ]
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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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28 Sep 2019, 04:14
13
7
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)

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

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##### General Discussion
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The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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21 Sep 2019, 06:24
1
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gmatt1476 wrote:

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

Attachment:
2019-09-21_1753.png

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

Pls Hit kudos if you like the solution
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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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21 Sep 2019, 10:09
7
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gmatt1476 wrote:

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

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.

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

$$R\sqrt{2}-R=r+r\sqrt{2}$$

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

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

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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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24 Sep 2019, 23:27
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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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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28 Sep 2019, 01:16
1
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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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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08 Jul 2020, 02:49
[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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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 04:20
MikeScarnScott wrote:
gmatt1476 wrote:

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
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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 05:48
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:

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.
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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 06:07
MikeScarnScott wrote:
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:

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$$\sqrt{2}$$

Also is there a reason why the 45-45-90 triangle rule doesn't work here?
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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 06:17
1
OmotayoH wrote:
MikeScarnScott wrote:
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:

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$$\sqrt{2}$$

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

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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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 06:21
1
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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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 07:32

This is how I have done.

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Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

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10 Jul 2020, 10:10
gmatt1476 wrote:

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

Attachment:
2019-09-21_1753.png

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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Re: The figure above shows 2 circles. The larger circle has center A, radi   [#permalink] 10 Jul 2020, 10:10