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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3219
Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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12 00:00

Difficulty:   95% (hard)

Question Stats: 25% (02:37) correct 75% (03:24) wrong based on 71 sessions

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Question of the Week #28 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

Attachment: QoW28.png [ 18.9 KiB | Viewed 1714 times ]

_________________

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.
Math Expert V
Joined: 02 Aug 2009
Posts: 8335
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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1
3
EgmatQuantExpert wrote: 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

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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EgmatQuantExpert

EgmatQuantExpert wrote: 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

Attachment:
QoW28.png
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3219
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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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 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.

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Senior Manager  P
Joined: 03 Mar 2017
Posts: 369
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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chetan2u wrote:
EgmatQuantExpert wrote: 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

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.   _________________
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Manager  G
Joined: 12 Apr 2011
Posts: 149
Location: United Arab Emirates
Concentration: Strategy, Marketing
Schools: CBS '21, Yale '21, INSEAD
GMAT 1: 670 Q50 V31
GMAT 2: 720 Q50 V37 WE: Marketing (Telecommunications)
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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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 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.

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.

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3219
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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1
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.

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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Manager  G
Joined: 12 Apr 2011
Posts: 149
Location: United Arab Emirates
Concentration: Strategy, Marketing
Schools: CBS '21, Yale '21, INSEAD
GMAT 1: 670 Q50 V31
GMAT 2: 720 Q50 V37 WE: Marketing (Telecommunications)
Re: Question of the week - 28 (In the figure above, O is the center .....)  [#permalink]

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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.

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. Re: Question of the week - 28 (In the figure above, O is the center .....)   [#permalink] 06 Jan 2019, 00:23
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