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

Question

e-GMAT Question of the Week #28 https://gmatclub.com/forum/download/file.php?id=45648 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 https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg QoW28.png

chetan2u · Accepted Answer

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

Archit3110 · Answer

EgmatQuantExpert 
Please provide detail solution...

EgmatQuantExpert · Answer

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

https://e-gmat.com/blogs/wp-content/uploads/2018/12/QoW28-2.png

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

https://e-gmat.com/blogs/wp-content/uploads/2018/07/FT-GC-FB-event-e1532524398949.jpg

warrior1991 · Answer

That was a very comprehensive approach Chetan Sir.

eabhgoy · Answer

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 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.

EgmatQuantExpert · Answer

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,

eabhgoy · Answer

Thank you

So since OC is radius then it is always a part of a diameter chord.

Balkrishna · Answer

Hi  EgmatQuantExpert ,  Bunuel ,  VeritasKarishma

Lets say line segment PO intersect the circle at point X. PX=16.

It is clear cut from the diagram that PB>PX. Now, there has been shown a possibility that PC might pass from the center of the circle. However, if PC had passed from O, then line segment PC would have been the extension of the line segment PO. But that is not the case here.

How can PB be 16? PB has to be greater than 16.

Would you please help clear this doubt?

Thank you.

KarishmaB · Answer

Note that PB > PX is a not a constraint given in our question. The line PBC shown in the diagram is just one example of various cases (and hence the question asks you for all possible values of PB)

The actual constraints on PB are just these two:
- the lengths of PB and PC are distinct positive integers, 
- PB < PC

So PX = PB is a valid case.

hadimadi · Answer

chetan2u  
one thing I don't get is:

The sum of two sides of a triangle has to be greater than the third side.

If PO=26, PB= 16, and we know that BO=10=radius of circle, then

BO+PB=26 !> 26.

How can this be?

Thanks

chetan2u · Answer

Hi, 
Which triangle are you looking for?
POB need not be a triangle, here PBO is a straight line. We are just looking at different values of PB.

hadimadi · Answer

chetan2u

Are you sure? POB looks like a triangle to me, or am I looking at point B wrongly? According to the drawing it isn't on a line?

hadimadi · Answer

chetan2u POB doesn't look like a straight line, rather, B is below P and O according to the drawing? I somehow get (B) as a result: We already know that PB < 24 and PBPO -> PB+10>26 -> PB >16 In total we have 16

chetan2u · Answer

Hi

The question nowhere mentions that POB is a triangle. What it tells us is that PB and PC are distinct, B and C cannot be the same point.

The question wants us to draw different possible lines PBC, so the situation you are talking of PB=16, the radius is 10, so when you extend PBO further to meet the circumference at C, you get line PBC, where PB = 16 and PC = PB+BO+OC=16+10+10 = 36
Next such point is when PB is 18, then PC= PB+BC
24^2=18*PC or PC=32

What they have shown in the sketch is just one of the possible layout of line PBC. It can be anywhere on the sketch. Nothing to do with triangle POB. There may or may not be a triangle POB.

hadimadi · Answer

chetan2u 
I think I understood it:

My answer is correct were the line PC drawn as is on the picture and the line was fixed. But the line isn't fixed, and could also just be straight (like POB).

That explains why PO=PB=16 is possible.

Now I got it. But to be quite frank, I find it irritating the way it's drawn ...

Thanks a lot for your time chetan

mpobisetty · Answer

Does GMAT expect us to know this property? Can you confirm if you have seen this property being used in any official question?

chetan2u  and other experts.

bumpbot · Answer

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

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