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# Triagle ABC is right isosceles and PQR is circle with center O. OQ is

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Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 12:33
1
3
00:00

Difficulty:

65% (hard)

Question Stats:

61% (02:25) correct 39% (02:46) wrong based on 41 sessions

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Triagle ABC is right isosceles and PQR is circle with center O. OQ is 1cm, BQ is variable x. what is value of x^2?

a) 2
b) 3 + sqrt(2)
c) 1 + 2[sqrt(2)]
d) 3 + 2[sqrt(2)]
e) 1 + sqrt(2)
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 12:46
1
I once got bogged down in the solution to a problem similar to this. Maybe it's not worth "solving"...use the answer choices instead.

Hint: 3 of the answers are totally unreasonable, and of the 2 that remain, I think one is much more in the ballpark than the other.
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 12:52
esledge wrote:

Even to make an educated guess, what is the remotest possibility? Do you believe these kind of questions could be experimental?
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 13:01
3
1
Hey,

This problem just seems difficult but it isn't.

From the figure (refer to the one I have attached), we can calculate OB = $$\sqrt{2}$$

Since we know that OQ = 1

x = BQ = OQ + OB = 1 + $$\sqrt{2}$$

Therefore, $$x^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}$$

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incircle.png [ 27.42 KiB | Viewed 2507 times ]

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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 13:10
1
sriharimurthy wrote:
Hey,

This problem just seems difficult but it isn't.

How can you prove that OR || PB, and similarly OP || BR? Why should OR be exactly perpendicular to BC?
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 13:44
2
SensibleGuy wrote:
sriharimurthy wrote:
Hey,

This problem just seems difficult but it isn't.

How can you prove that OR || PB, and similarly OP || BR? Why should OR be exactly perpendicular to BC?

Circle Property : Any line touching the circle is a tangent and makes a 90 degree angle with radius.

Since both AB and BC touch the circle at just one point, that is, points P and R respectively, they can be considered tangents to the circle.

From the circle property mentioned above, any line drawn from O to AB and from O to BC will form an angle of 90 degrees since OR and OP will be the radius of the circle.

Now, since AB and BC are perpendicular to each other, any lines drawn perpendicular to them (OP and OR) will also be perpendicular to each other.

Thus, OPBR forms a square since all angles are 90 degrees and the length of two adjacent is equal (OP = OR = radius of the circle).

Hence we can conclude the following :

PB = BR = OP = OR = radius of the circle = 1 cm.

Hope this helps. If you still have difficulty in understanding any of the above steps, I'll be more than happy to try and explain further.

Cheers.
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1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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Updated on: 18 Nov 2009, 15:19
1
SensibleGuy wrote:
Even to make an educated guess, what is the remotest possibility? Do you believe these kind of questions could be experimental?

Yeah, any question could be experimental. I doubt there are throw-away or stickier/trickier questions deliberately included in the experimental pool. After all, the GMAC tests the experimental questions with the intent of using them as official questions...

Here's my guessing rationale. You know the diameter of the circle is 2, so x is more than 2 and x^2 must be more than 4. That eliminates ACE.

Now the guessing issue is just how much more than 2 is x? With knowledge of the 45-45-90 triangle/diagonal stuff, a scale drawing, and a critical eye, it's not too much of a stretch to just approximate OB as root(2) = 1.4.

Therefore, x is at least around 1 + 1.4 = 2.4, and x^2 is 5.76-ish, or (D).

Also, (B) is just too close to 4 to be as reasonable a guess as (D).

Originally posted by esledge on 18 Nov 2009, 13:44.
Last edited by esledge on 18 Nov 2009, 15:19, edited 1 time in total.
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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18 Nov 2009, 15:11
srihari you are outstanding dude!!!

esledge, your guessing technique is impressive as well. Learning to guess properly is in itself an art and can save some questions for sure!!!!
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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13 May 2011, 07:39
BO^2 = 1 + 1 = 2

BO = root(2)

OQ = 1

BO + OQ = BQ = {root(2) + 1}

BQ^2 = 2 + 1 + 2*root(2)*1 = 3 + 2root(2)

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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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13 May 2011, 10:57
I also guessed the answer as D.

First of all the x > 2

Leave with 2 options:
b) 3 + sqrt(2) = 4.4
d) 3 + 2[sqrt(2)] =6.8

I chose D over B cause there is substantial about of x outside diameter of circle which is 4. After squaring x it will sure be more then 4.4
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Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is  [#permalink]

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19 Feb 2020, 07:44
BarneyStinson wrote:
Attachment:
incircle.JPG

Triagle ABC is right isosceles and PQR is circle with center O. OQ is 1cm, BQ is variable x. what is value of x^2?

a) 2
b) 3 + sqrt(2)
c) 1 + 2[sqrt(2)]
d) 3 + 2[sqrt(2)]
e) 1 + sqrt(2)

circle_r=1
imagine a square with side r = 1
this square will have diagonal = 1√2
BQ=x=1√2+r=1+√2
x^2=(1+√2)^2=1^2+√2^2+2(1)(√2)=3+2√2

Ans (D)
Re: Triagle ABC is right isosceles and PQR is circle with center O. OQ is   [#permalink] 19 Feb 2020, 07:44
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