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Question of the week - 28 (In the figure above, O is the center .....)

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Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post Updated on: 27 Feb 2019, 05:36
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Question Stats:

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e-GMAT Question of the Week #28

Image

In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB?

    A. 16
    B. 18
    C. 34
    D. 58
    E. 126

Image



Attachment:
QoW28.png
QoW28.png [ 18.9 KiB | Viewed 1432 times ]

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Originally posted by EgmatQuantExpert on 21 Dec 2018, 20:21.
Last edited by EgmatQuantExpert on 27 Feb 2019, 05:36, edited 3 times in total.
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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 21 Dec 2018, 20:55
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3
EgmatQuantExpert wrote:
Image

In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB?

    A. 16
    B. 18
    C. 34
    D. 58
    E. 126

Image


It will be maximum when Pb~PC, that is PB is the tangent, and at that time it will be 24, so PB<24..
Now for Min, when PC lies on the hypotenuse itself...
So PO = √(24^2+10^2)=√676= 26, PB will be 26-10=16..
So min value is 16 and max is less than 24..

Now the property that will be useful here is (PA)^2=PB*PC...
So PB will be a factor of 24^2 but less than 24 and greater than or equal to 16..
So PB can be only 16 and 18..
Sum is 16+18=34

C
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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 25 Dec 2018, 03:30
EgmatQuantExpert
Please provide detail solution...

EgmatQuantExpert wrote:
Image

In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB?

    A. 16
    B. 18
    C. 34
    D. 58
    E. 126

Image



Attachment:
QoW28.png

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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 28 Dec 2018, 03:39

Solution


Given:
    • O is the center of the circle
    • BC is a chord
    • Radius of the circle = 10
    • PA is the tangent, whose length = 24
    • Lengths of PB and PC are integers, and PB < PC

To find:
    • The sum of all possible lengths of PB

Approach and Working:
    • Let us draw lines from A to B and from A to C

Image

Now, if we observe in triangles PAB and PAC,
    • ∠APB = ∠APC (common angle)
    • ∠PAB = ∠PCA (angles in alternate segments)

Therefore, we can say that the two triangles are similar,
    • Thus, \(\frac{PA}{PB} = \frac{PC}{PA}\)
    • \(PA^2 = PB * PC\)

Implies, PB * PC = \(24^2\) (substituting the value of PA)
    • Since, PB < PC, PB must be less than 24, and the minimum value that PB can take is when B is the intersection point of line PO and the circle.
    • In such case, \(PB = PO – OB = √(PA^2 + OA^2) – 10 = √676 – 10 = 26 – 10 = 16\)
    • Thus, 16 ≤ PB < 24, and PB must contain only 2 or 3 as its prime factors, since PB is a factor of \(24^2\) and \(24^2 = 2^6 * 3^2\)

Therefore, the only possible values of PB that satisfies the above two conditions are PB = 16 or 18.

So, the sum of all possible lengths of PB = 16 + 18 = 34

Hence the correct answer is Option C.

Answer: C

Image

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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 29 Dec 2018, 09:16
chetan2u wrote:
EgmatQuantExpert wrote:
Image

In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB?

    A. 16
    B. 18
    C. 34
    D. 58
    E. 126

Image


It will be maximum when Pb~PC, that is PB is the tangent, and at that time it will be 24, so PB<24..
Now for Min, when PC lies on the hypotenuse itself...
So PO = √(24^2+10^2)=√676= 26, PB will be 26-10=16..
So min value is 16 and max is less than 24..

Now the property that will be useful here is (PA)^2=PB*PC...
So PB will be a factor of 24^2 but less than 24 and greater than or equal to 16..
So PB can be only 16 and 18..
Sum is 16+18=34

C



That was a very comprehensive approach Chetan Sir.

:ok :ok :ok
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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 30 Dec 2018, 08:22
EgmatQuantExpert wrote:

Solution


Given:
    • O is the center of the circle
    • BC is a chord
    • Radius of the circle = 10
    • PA is the tangent, whose length = 24
    • Lengths of PB and PC are integers, and PB < PC

To find:
    • The sum of all possible lengths of PB

Approach and Working:
    • Let us draw lines from A to B and from A to C

Image

Now, if we observe in triangles PAB and PAC,
    • ∠APB = ∠APC (common angle)
    • ∠PAB = ∠PCA (angles in alternate segments)

Therefore, we can say that the two triangles are similar,
    • Thus, \(\frac{PA}{PB} = \frac{PC}{PA}\)
    • \(PA^2 = PB * PC\)

Implies, PB * PC = \(24^2\) (substituting the value of PA)
    • Since, PB < PC, PB must be less than 24, and the minimum value that PB can take is when B is the intersection point of line PO and the circle.
    • In such case, \(PB = PO – OB = √(PA^2 + OA^2) – 10 = √676 – 10 = 26 – 10 = 16\)
    • Thus, 16 ≤ PB < 24, and PB must contain only 2 or 3 as its prime factors, since PB is a factor of \(24^2\) and \(24^2 = 2^6 * 3^2\)

Therefore, the only possible values of PB that satisfies the above two conditions are PB = 16 or 18.

So, the sum of all possible lengths of PB = 16 + 18 = 34

Hence the correct answer is Option C.

Answer: C

Image


I have a doubt :)

In the figure we know:

Triangle PAO = Right Angled Triangle
PA = 24
AO = 10
=> PO = 26

Now lets look at Triangle POC:

PO = 26
OC = 10
=> PC<PO + OC
=> PC < 26+10
=> PC < 36

Since PA*PA = PB*PC

For PC = 36
=> 24*24= 36*PB => PB = 16
Hence PB != 16

That leaves us with only 1 value i.e. PB = 18

Please let me know where I went wrong.

Thanks in advance.
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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 04 Jan 2019, 05:08
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eabhgoy wrote:

I have a doubt :)

In the figure we know:

Triangle PAO = Right Angled Triangle
PA = 24
AO = 10
=> PO = 26

Now lets look at Triangle POC:

PO = 26
OC = 10
=> PC<PO + OC
=> PC < 26+10
=> PC < 36

Since PA*PA = PB*PC

For PC = 36
=> 24*24= 36*PB => PB = 16
Hence PB != 16

That leaves us with only 1 value i.e. PB = 18

Please let me know where I went wrong.

Thanks in advance.


Hi eabhgoy,

PC can be equal to OC + OP, when C is one of the end points of the diameter.

In that case PC = 36 and PB will be 36 - 20 = 16.

Regards,
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Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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New post 06 Jan 2019, 00:23
EgmatQuantExpert wrote:
eabhgoy wrote:

I have a doubt :)

In the figure we know:

Triangle PAO = Right Angled Triangle
PA = 24
AO = 10
=> PO = 26

Now lets look at Triangle POC:

PO = 26
OC = 10
=> PC<PO + OC
=> PC < 26+10
=> PC < 36

Since PA*PA = PB*PC

For PC = 36
=> 24*24= 36*PB => PB = 16
Hence PB != 16

That leaves us with only 1 value i.e. PB = 18

Please let me know where I went wrong.

Thanks in advance.


Hi eabhgoy,

PC can be equal to OC + OP, when C is one of the end points of the diameter.

In that case PC = 36 and PB will be 36 - 20 = 16.

Regards,


Thank you

So since OC is radius then it is always a part of a diameter chord.
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Re: Question of the week - 28 (In the figure above, O is the center .....)   [#permalink] 06 Jan 2019, 00:23
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