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point not on AC

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by sanju09 » Wed May 19, 2010 4:06 am
The points A, B, C lie on a line in that order with AB = 9, BC = 21. Let O be a point not on AC such that AO = CO and the distances AO and BO are integral. What is the sum of all possible perimeters of triangle ACO?
(A) 320
(B) 350
(C) 380
(D) 410
(E) 420


[spoiler]My answer: 128, choices...? source: BTG PS Forum[/spoiler]
Last edited by sanju09 on Wed May 19, 2010 6:28 am, edited 1 time in total.
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by liferocks » Wed May 19, 2010 5:24 am
@sanju09..great question..as of now I am able to deduce only this

the point O has to be on perpendicular bisector line of AC at distance of 15

let it intersect AC at P

so AP=15 and BP=6

6 can form only one right angle triangle when its all three sides are integer 6,8,10
so point on the bisector is 8 unit away from P and 17 unit away from A or C...perimeter= 2*(17+17+30) or 128

but this is not in ans..am I in right track or..its time for a break..
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by liferocks » Wed May 19, 2010 9:13 am
ok..here is the complete solution,
let distance from point be x and that from point B be y...
so x^2-15^2=y^2-6^2
or (x+y)(x-y)=189

now 189 can be expressed as multiple of 2 integers in the following ways (1,189),(3,63),(7,27),(9,21) since x>y we get
(x+y)----(x-y)----2x
189-------1-------190
63---------3-------66
27---------7-------34
21---------9-------30......this will be the case when the point in on AC

So sum of all possible perimeters is 190+66+34+3*30=380

ans option C
Last edited by liferocks on Thu May 20, 2010 12:49 am, edited 1 time in total.
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by sanju09 » Thu May 20, 2010 12:39 am
liferocks wrote:ok..here is the complete solution,
let distance from point be x and that from point B be y...
so x^2-15^2=y^2-6^2
or (x+y)(x-y)=189

now 189 can be expressed as multiple of 2 integers in the following ways (1,189),(3,63),(7,27),(9,21) since x>y we get
(x+y)----(x-y)----2x
189-------1-------190
63---------3-------66
27---------7-------34
21---------9-------30......this will be the case when the point in on AC

So sum of all possible perimeters is 190+66+34+3*30=320

ans option A
I must say that you are a real fighter, and hence you finally proved one of the choices as correct. Good part about your work is that all possibilities shown by you maintain the most important feature of triangle ACO, which is, AO > 15. Everything is perfect EXCEPT the [spoiler]answer, which in my humble opinion, should be (C) instead of (A).

Nonetheless, you have been included to the hit list of my Quant Commandos on BTG, which reads

Many old members followed by...gmatmachoman, shashank.ism, thephoenix, ajith, harshavardhanc, rockeyb,...and with few present names definitely missing, and finally it's you liferocks. You certainly deserve my thanks too![/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by liferocks » Thu May 20, 2010 12:45 am
sanju09 wrote:
liferocks wrote:ok..here is the complete solution,
let distance from point be x and that from point B be y...
so x^2-15^2=y^2-6^2
or (x+y)(x-y)=189

now 189 can be expressed as multiple of 2 integers in the following ways (1,189),(3,63),(7,27),(9,21) since x>y we get
(x+y)----(x-y)----2x
189-------1-------190
63---------3-------66
27---------7-------34
21---------9-------30......this will be the case when the point in on AC

So sum of all possible perimeters is 190+66+34+3*30=320

ans option A
I must say that you are a real fighter, and hence you finally proved one of the choices as correct. Good part about your work is that all possibilities shown by you maintain the most important feature of triangle ACO, which is, AO > 15. Everything is perfect EXCEPT the [spoiler]answer, which in my humble opinion, should be (C) instead of (A).

Nonetheless, you have been included to the hit list of my Quant Commandos on BTG, which reads

Many old members followed by...gmatmachoman, shashank.ism, thephoenix, ajith, harshavardhanc, rockeyb,...and with few present names definitely missing, and finally it's you liferocks. You certainly deserve my thanks too![/spoiler]
finally an error in summation...this is miserable :(..still correcting the original post
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by Haldiram Bhujiawala » Thu May 20, 2010 12:56 am
sanju09 wrote:
liferocks wrote:ok..here is the complete solution,
let distance from point be x and that from point B be y...
so x^2-15^2=y^2-6^2
or (x+y)(x-y)=189

now 189 can be expressed as multiple of 2 integers in the following ways (1,189),(3,63),(7,27),(9,21) since x>y we get
(x+y)----(x-y)----2x
189-------1-------190
63---------3-------66
27---------7-------34
21---------9-------30......this will be the case when the point in on AC

So sum of all possible perimeters is 190+66+34+3*30=320

ans option A
I must say that you are a real fighter, and hence you finally proved one of the choices as correct. Good part about your work is that all possibilities shown by you maintain the most important feature of triangle ACO, which is, AO > 15. Everything is perfect EXCEPT the [spoiler]answer, which in my humble opinion, should be (C) instead of (A).

Nonetheless, you have been included to the hit list of my Quant Commandos on BTG, which reads

Many old members followed by...gmatmachoman, shashank.ism, thephoenix, ajith, harshavardhanc, rockeyb,...and with few present names definitely missing, and finally it's you liferocks. You certainly deserve my thanks too![/spoiler]
Good Quotation of Edward Richards ; I hope you follow those words in letter and spirit !!!!
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by sanju09 » Thu May 20, 2010 1:06 am
Good Quotation of Edward Richards ; I hope you follow those words in letter and spirit !!!!
I only wish to be WISE, neither OLD nor OWL
The mind is everything. What you think you become. -Lord Buddha



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by Haldiram Bhujiawala » Thu May 20, 2010 1:29 am
sanju09 wrote:
Good Quotation of Edward Richards ; I hope you follow those words in letter and spirit !!!!
I only wish to be WISE, neither OLD nor OWL
" THE LESS HE SPOKE , THE MORE HE HEARD " That's the fundamental message , by Edward Richards !!!
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by sanju09 » Thu May 20, 2010 1:41 am
Haldiram Bhujiawala wrote:
sanju09 wrote:
Good Quotation of Edward Richards ; I hope you follow those words in letter and spirit !!!!
I only wish to be WISE, neither OLD nor OWL
" THE LESS HE SPOKE , THE MORE HE HEARD " That's the fundamental message , by Edward Richards !!!
No one likes to hear an owl speaking, that's why he's got to speak less, and the curse turns to boon when he gets a fairer chance to hear (learn) than the others. Isn't that the fundamental message, by Edward H. Richards?
The mind is everything. What you think you become. -Lord Buddha



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by gmatmachoman » Thu May 20, 2010 6:15 am
Sanju bhai!

Thx for the kind words...getting a "comment" that too positive one is welcome....

U forgot to mention fibbonacci .......She is one real fighter like Liferocks!!

I wish we all should finish off GMAT in a smoother succesful fashion!!
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by Haldiram Bhujiawala » Thu May 20, 2010 8:52 am
sanju09 wrote:
No one likes to hear an owl speaking, that's why he's got to speak less, and the curse turns to boon when he gets a fairer chance to hear (learn) than the others. Isn't that the fundamental message, by Edward H. Richards?
Some weird interpretaion indeed !!!
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by sanju09 » Thu May 20, 2010 11:38 pm
gmatmachoman wrote:Sanju bhai!

Thx for the kind words...getting a "comment" that too positive one is welcome....

U forgot to mention fibbonacci .......She is one real fighter like Liferocks!!

I wish we all should finish off GMAT in a smoother succesful fashion!!
please wait till I release the original list of my Quant Commandos on BTG, then tell who really went unnoticed from my inner sight, Govindacharya ji
The mind is everything. What you think you become. -Lord Buddha



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by sanju09 » Thu May 20, 2010 11:39 pm
Haldiram Bhujiawala wrote:
sanju09 wrote:
No one likes to hear an owl speaking, that's why he's got to speak less, and the curse turns to boon when he gets a fairer chance to hear (learn) than the others. Isn't that the fundamental message, by Edward H. Richards?
Some weird interpretaion indeed !!!
sorry if it didn't hurt an owl
The mind is everything. What you think you become. -Lord Buddha



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