Is the answer 4-(2*sqrt(2)). ?
I will explain if i am correct.
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Circles and Square.....pls help
Source: Beat The GMAT — Problem Solving |
I will try my best to explain.
Name the square as ABCD where AC represents the diagonal of the square.
Name the centers of the circles as O1 (of the circle near to vertex A) and O2 (of the circle near to vertex C)
So the diagonal line is AO1O2C ( the length = 4sqrt(2) ). Let the radius of the circles be r.
To determine r, we should know the distance between A and O1 and the distance between O2 and C. The distance between O1 and O2 is 2r.
To determine distance between A and O1
------------------------------------------------
Draw a line between A and O1 and a line between O1 and the tangential point P in AB. Drawing a line between O1 and the tangential point P in AB creates a right angle ( If i remember correctly ). The distance of the line O1P is again r.
Apply hypotenuse theorem here. AO1^2 = r^2 + r^2 = 2*r^2
=> AO1 = r*sqrt(2)
Similarly O2C = r*sqrt(2)
AC = AO1 + O1O2 + O2C
= r*sqrt(2) + 2r + r*sqrt(2)
4sqrt(2) = 2r*sqrt(2) + 2r
2sqrt(2) = r(1+ sqrt(2))
Solving this, we get r = 4-(2*sqrt(2))
HTH.
Name the square as ABCD where AC represents the diagonal of the square.
Name the centers of the circles as O1 (of the circle near to vertex A) and O2 (of the circle near to vertex C)
So the diagonal line is AO1O2C ( the length = 4sqrt(2) ). Let the radius of the circles be r.
To determine r, we should know the distance between A and O1 and the distance between O2 and C. The distance between O1 and O2 is 2r.
To determine distance between A and O1
------------------------------------------------
Draw a line between A and O1 and a line between O1 and the tangential point P in AB. Drawing a line between O1 and the tangential point P in AB creates a right angle ( If i remember correctly ). The distance of the line O1P is again r.
Apply hypotenuse theorem here. AO1^2 = r^2 + r^2 = 2*r^2
=> AO1 = r*sqrt(2)
Similarly O2C = r*sqrt(2)
AC = AO1 + O1O2 + O2C
= r*sqrt(2) + 2r + r*sqrt(2)
4sqrt(2) = 2r*sqrt(2) + 2r
2sqrt(2) = r(1+ sqrt(2))
Solving this, we get r = 4-(2*sqrt(2))
HTH.
Good approach to be able to select the ans from amongst the choices, but this is not exactly the radius, and the actual radius would be slightly smaller than this, cos the diagonal of the square is not exactly equal to 4r, but extends a bit on both sides beyond the circles.stop@800 wrote:My attempt!!
Square side = 4
diagonal = 4sqrt(2)
this squares diagonal is sum to two diagonals of circle
so 2d = 4sqrt(2)
d = 2sqrt(2)
and radius = sqrt(2)
However, given that thr would be 5 choiced to select from, this should be pretty good to estimate and guess the right answer!
-TheGenius
I am stuck with this thingMani_mba wrote: 2sqrt(2) = r(1+ sqrt(2))
Solving this, we get r = 4-(2*sqrt(2))
Anybody. please elaborate
2sqrt(2) = r(1+sqrt(2))
=> r = 2*sqrt(2) / (1+sqrt(2))
Multiply (1-sqrt(2)) on both Numerator and Denominator
=> r = 2sqrt(2) * (1-sqrt(2)) / (1+sqrt(2)) (1-sqrt(2))
= (2sqrt(2) - 4) / (-1)
= 4-2sqrt(2)
HTH.
=> r = 2*sqrt(2) / (1+sqrt(2))
Multiply (1-sqrt(2)) on both Numerator and Denominator
=> r = 2sqrt(2) * (1-sqrt(2)) / (1+sqrt(2)) (1-sqrt(2))
= (2sqrt(2) - 4) / (-1)
= 4-2sqrt(2)
HTH.
Hi Mani_mba,Mani_mba wrote:I will try my best to explain.
Name the square as ABCD where AC represents the diagonal of the square.
Name the centers of the circles as O1 (of the circle near to vertex A) and O2 (of the circle near to vertex C)
So the diagonal line is AO1O2C ( the length = 4sqrt(2) ). Let the radius of the circles be r.
To determine r, we should know the distance between A and O1 and the distance between O2 and C. The distance between O1 and O2 is 2r.
To determine distance between A and O1
------------------------------------------------
Draw a line between A and O1 and a line between O1 and the tangential point P in AB. Drawing a line between O1 and the tangential point P in AB creates a right angle ( If i remember correctly ). The distance of the line O1P is again r.
Apply hypotenuse theorem here. AO1^2 = r^2 + r^2 = 2*r^2
=> AO1 = r*sqrt(2)
Similarly O2C = r*sqrt(2)
AC = AO1 + O1O2 + O2C
= r*sqrt(2) + 2r + r*sqrt(2)
4sqrt(2) = 2r*sqrt(2) + 2r
2sqrt(2) = r(1+ sqrt(2))
Solving this, we get r = 4-(2*sqrt(2))
HTH.
I dint quite understand how the length of AP = r ??
Can u pls elaborate??
thanks
-V
I just knew AP should be r by intuition. I will try to explain mathematically:
Let us name the tangential point in the side AD as P1.
As i said already, Angle O1PA = 90 and Angle A = 90. Hence the lines DA and O1P must be parallel.
Applying same for the tangential point P1, we get
Angle O1P1A=90 and it is known that Angle A = 90. Hence the lines O1P1 and PA must be parallel.
=> APO1P1 looks like a square and all the sides measure the same. Hence AP = PO1 = O1P1 = P1A = r.
HTH.
Let us name the tangential point in the side AD as P1.
As i said already, Angle O1PA = 90 and Angle A = 90. Hence the lines DA and O1P must be parallel.
Applying same for the tangential point P1, we get
Angle O1P1A=90 and it is known that Angle A = 90. Hence the lines O1P1 and PA must be parallel.
=> APO1P1 looks like a square and all the sides measure the same. Hence AP = PO1 = O1P1 = P1A = r.
HTH.
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I received a PM asking me to reply. This problem is incomplete - it's either missing statements (if it's DS) or answer choices (if it's PS). Can the original poster please post the problem in its entirety?
Also, this one seems more complicated than what we would see on the GMAT - just FYI. To people who are wondering whether to study this, I would spend time on things that are more likely to appear on the real test, especially because this one is incomplete. On a complicated geometry problem, estimation is often the best approach.
Also, this one seems more complicated than what we would see on the GMAT - just FYI. To people who are wondering whether to study this, I would spend time on things that are more likely to appear on the real test, especially because this one is incomplete. On a complicated geometry problem, estimation is often the best approach.
Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!
Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
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Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
