I did some calculations but could not come up with the right answer.
Since the radius length is 2
s^2 +t^2 = 4
and (s+root3)^2 + (t-1)^2 = 8
but was not able to resolve this two equations further.
I have attached the problem as a image.
Gmat Prep - Triangle Co ordinates
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Last edited by subha_sri8 on Wed Jun 11, 2008 9:37 pm, edited 4 times in total.
- VerbalAttack
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you can simplify them to t = root3s using the two equations you have given...
can we see the complete question?
Cheers
can we see the complete question?
Cheers
- VerbalAttack
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First of all, if 2 lines are perpendicular then the product of their slopes is -1.
Since both lines cross thru centre (0,0);
Slope 1 = (1 - 0) / (-root3 - 0) = - 1 / root3
Slope 2 = t - 0 / s - 0 = t/s
product of slopes = t/s * -1/root3 = -1 ==> t = root3 s
Now lets use the triangles;
r^2 = 1^2 + (-root3)^2 = 1 + 3 = 4
radius r = 2
apply this the other triangle;
r^2 = s^2 + t^2
4 = s^2 + (root3 s)^2 = s^2 + 3 s^2 = 4 s^2
2 = 2s ==> s = 1
Cheers
Since both lines cross thru centre (0,0);
Slope 1 = (1 - 0) / (-root3 - 0) = - 1 / root3
Slope 2 = t - 0 / s - 0 = t/s
product of slopes = t/s * -1/root3 = -1 ==> t = root3 s
Now lets use the triangles;
r^2 = 1^2 + (-root3)^2 = 1 + 3 = 4
radius r = 2
apply this the other triangle;
r^2 = s^2 + t^2
4 = s^2 + (root3 s)^2 = s^2 + 3 s^2 = 4 s^2
2 = 2s ==> s = 1
Cheers
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- AleksandrM
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There is a little-known shortcut for this kind of a problem. Whenever you have a set-up such as this, the opposite side will simply be flipped. In other words, since the coordinates of one side are x = -sqroot3, and y = 1, then the other side will be x = 1 and y = sqroot3.
It's similar to the "testing points" approach on a line. If you know that points less than a given coordinate are negative, then the testing points will be in the form negative, positive, negative.
It's similar to the "testing points" approach on a line. If you know that points less than a given coordinate are negative, then the testing points will be in the form negative, positive, negative.
Looking at the co-ordinates of P, we can deduce that OP=2 (Draw a line perpendicular to the x-axis from P to point B and use the pythogoras theorem).
If OP=2, PB=1 and OB=|root 3| we can deduce triangle OPB to be a 30-60-90 triangle with angle POB = 30.
Draw a perpendicular line from Q to the x-axis to Point A. So Angle QOA = 180 - (90+30)= 60.
Hence Triangle QOA is also 30-60-90 with hypotenuse=radius=2
Hence s= side opp to angle 30 degrees = 1/2 hypotenuse = 1
s=1, t= root3
If OP=2, PB=1 and OB=|root 3| we can deduce triangle OPB to be a 30-60-90 triangle with angle POB = 30.
Draw a perpendicular line from Q to the x-axis to Point A. So Angle QOA = 180 - (90+30)= 60.
Hence Triangle QOA is also 30-60-90 with hypotenuse=radius=2
Hence s= side opp to angle 30 degrees = 1/2 hypotenuse = 1
s=1, t= root3
- AleksandrM
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Sorry. I shouldn't have included that. I was just making a comparison to make a point of how shortcuts can sometimes be useful.smohsin wrote:AleksandrM - Could you please explain the "touch point" formula please with an example?
Testing points is from geometry. You apply it to inequalities. Don't worry about it, has nothing to do with this problem.