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# A circle, with center O, is inscribed in square WXYZ. Point P, as

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A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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Updated on: 13 Jun 2015, 10:35
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Difficulty:

95% (hard)

Question Stats:

49% (02:18) correct 51% (01:54) wrong based on 156 sessions

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A circle, with center O, is inscribed in square WXYZ. Point P, as shown above, is where the circle and the line segment ZX intersect. If PX = 1, which of the following is closest to the value of the circumference of the circle?

A. 2π
B. 3π
C. 4π
D. 5π
E. 6π

Attachment:

Picture2-1%5B1%5D.png [ 10.55 KiB | Viewed 5074 times ]

Originally posted by vishalgc on 09 Oct 2008, 21:34.
Last edited by Bunuel on 13 Jun 2015, 10:35, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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13 Jun 2015, 11:05
5
1
petu wrote:
Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!!

A circle, with center O, is inscribed in square WXYZ. Point P, as shown above, is where the circle and the line segment ZX intersect. If PX = 1, which of the following is closest to the value of the circumference of the circle?

A. 2π
B. 3π
C. 4π
D. 5π
E. 6π

Check image below:

Notice that triangle AOX is an isosceles right triangle: AO = AX = radius.

We know that hypotenuse of an isosceles right triangle is $$side*\sqrt{2}$$, thus = $$OX=side*\sqrt{2}=radius*\sqrt{2}$$.

On the other hand, OX = OP + PX = radius + PX = radius + 1.

Therefore, $$radius + 1=radius*\sqrt{2}$$:

$$radius=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\approx{1.4+1}=2.4$$.

The circumference = $$2\pi{r}=2\pi*2.4=4.8\pi$$.

Hope it's clear.
Attachment:

Untitled.png [ 7.12 KiB | Viewed 5000 times ]

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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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09 Oct 2008, 22:32
1
5pi.

OX = OP + 1 = r + 1 (r is the radius of circle).

Now, OX^2 = r^2 +r^2

Solve for r and then 2pi4 will be approx 5pi.
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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16 Oct 2008, 17:19
scthakur wrote:
5pi.

OX = OP + 1 = r + 1 (r is the radius of circle).

Now, OX^2 = r^2 +r^2

Solve for r and then 2pi4 will be approx 5pi.

5pi

Yours is a simple way to solve the problem
I added unnecessary steps to it
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Posts: 159
Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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16 Oct 2008, 17:34
scthakur wrote:
5pi.

OX = OP + 1 = r + 1 (r is the radius of circle).

Now, OX^2 = r^2 +r^2

Solve for r and then 2pi4 will be approx 5pi.

can you pls explain why you did OX^2=r^2+r^2
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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Updated on: 16 Oct 2008, 19:44
scthakur wrote:
5pi.

OX = OP + 1 = r + 1 (r is the radius of circle).

Now, OX^2 = r^2 +r^2

Solve for r and then 2pi4 will be approx 5pi.

Ahhh!! I was not able to see this I was stumped by the question !!
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Originally posted by amitdgr on 16 Oct 2008, 19:43.
Last edited by amitdgr on 16 Oct 2008, 19:44, edited 1 time in total.
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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16 Oct 2008, 19:43
let OP=r the radius, then we know that OP+1=OX

or r+1=OX, lets look at the diagonal of the square ZX, ZX=2OX or 2(r+1)

we also know that ZX=sqrt(2)XY; XY=2R...right..with me???

2r+2=sqrt(2)2r

solve for r, you get r=1/(sqrt(2)-1) or 1/0.41

circum. =2pi*r, 2/0.41 is apprx 5...so circum=5pi
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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17 Oct 2008, 01:10
ALD wrote:
scthakur wrote:
5pi.

OX = OP + 1 = r + 1 (r is the radius of circle).

Now, OX^2 = r^2 +r^2

Solve for r and then 2pi4 will be approx 5pi.

can you pls explain why you did OX^2=r^2+r^2

The distance between mid-point of XY and O is r and distance between mid-point of XY and X is r. That is why, r^2 + r^2 = OX^2.
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GMAT 1: 730 Q49 V40
Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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13 Jun 2015, 10:27
Hi, could someone give help me with this one? I just can't understand the explanations! thanks!!!
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Joined: 30 Mar 2015
Posts: 15
GMAT 1: 730 Q49 V40
Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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13 Jun 2015, 11:32
THanks Bunuel. Clear as allways. You have few beers waiting for you if you ever come to Barcelona.
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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17 Jan 2017, 14:43
That was an excellent explanation. Thanks Brunel!
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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17 Jan 2017, 14:51
Sorry to ask what might be an obvious questions.
From your explanation, I don´t understand the part in bold.

Thanks

Check image below:

Image

Notice that triangle AOX is an isosceles right triangle: AO = AX = radius.

We know that hypotenuse of an isosceles right triangle is side∗2‾√side∗2, thus = OX=side∗2‾√=radius∗2‾√OX=side∗2=radius∗2.

On the other hand, OX = OP + PX = radius + PX = radius + 1.

12−1=2+1≈1.4+1=2.4.

The circumference = 2πr=2π∗2.4=4.8π2πr=2π∗2.4=4.8π.

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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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19 Jan 2017, 07:50
lauramo wrote:
Sorry to ask what might be an obvious questions.
From your explanation, I don´t understand the part in bold.

Thanks

Check image below:

Image

Notice that triangle AOX is an isosceles right triangle: AO = AX = radius.

We know that hypotenuse of an isosceles right triangle is side∗2‾√side∗2, thus = OX=side∗2‾√=radius∗2‾√OX=side∗2=radius∗2.

On the other hand, OX = OP + PX = radius + PX = radius + 1.

12−1=2+1≈1.4+1=2.4.

The circumference = 2πr=2π∗2.4=4.8π2πr=2π∗2.4=4.8π.

We are solving for radius (you can denote it as x for simplicity) the following equation: $$radius + 1=radius*\sqrt{2}$$ to get $$radius=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\approx{1.4+1}=2.4$$.

$$x + 1=x*\sqrt{2}$$

$$1=x*\sqrt{2}-x$$

$$1=x(\sqrt{2}-1)$$

$$x=\frac{1}{\sqrt{2}-1}$$

Multiply both numerator and denominator by $$\sqrt{2}+1$$:

$$x=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1$$.

Hope it's clear.
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Re: A circle, with center O, is inscribed in square WXYZ. Point P, as  [#permalink]

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19 Aug 2018, 18:31
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