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

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.

Answer: C

That was a very comprehensive approach Chetan Sir.

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.

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.

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.

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

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.

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?

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 PB<PC.
We use Pythagorean to get PO=26. Let the intersect between the circle and the line PO be called X. So PX+XO=26.
We know XO=10, so PX=16. Because for a triangle, we have that the sum of two sides is always greater than the third side, and because we have the lengths of PO, PB, and BO, we can take a look at the triangle PBO to and say that:

PB+BO>PO -> PB+10>26 -> PB >16
In total we have 16<PB<24.
Now we know from the Power of Point Theorem that PA^2=(PB)(PC). Because PA=24 it follows
24^2=(2)^6(3)^2
All factors of 24^2 can be depicted by possible combinations of 2^6 and 3^2. But 16<PB<24, and so, the only factor we can depict is 18, which is 3^2*2.

Where did I go wrong?

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.

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

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.

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