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m21 #20

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Re: m21 #20 [#permalink]

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New post 09 May 2011, 04:39
The distance between points is sqrt10, sqrt10, sqrt20. This is 45, 45, 90 triangle. By circle properties if any inscribed triangle has 90 degree angle its hypotenuse passes through its center. So diameter is hypotenuse of triangle =sqrt20.
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New post 14 May 2012, 07:09
newdawn wrote:
My approach would be:
Let x,y is the center of the circle, then distance between center and any point on the circle is same equal to R.
Given that (1,2), (2,5) and (5,4) lie on circle, we can have following equations:

\((x-1)^2 + (y-2)^2 = (x-5)^2 +(y-4)^2\)
solve it for

Next take (2,5) and (5,4)
\((x-2)^2 + (y-5)^2 = (x-5)^2 +(y-4)^2\)
and you will get

Solve for x and y center is (3,3) and distance i.e. Radius is \(\sqrt{(5-3)^2+(4-3)^2} = \sqrt{5}\)

So Diameter is \(\sqrt{20}\)

I did the same. However it's way too long and complex calculation, gmat suggests easier solutions to be within 2 minutes limit
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New post 13 May 2013, 09:09
In case the triangle ABC is not a right angled triangle, we can find the diameter using the following general formula:

Diameter = abc/(2*Area(ABC))

a,b,c are the sides opposite to angles A, B and C respectively.

So if we have the coordinates of all the vertices of the triangle we can easily find the diameter using the above formula
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New post 13 May 2013, 10:11
Just want to note that I've seen a lot of people throwing around the "rule" that if a triangle is inscribed in a circle, then it either a) must be a right triangle or b) have it's hypotenuse pass through the circle's center. Neither of these are true...

The rule is that IF a right triangle is inscribed in a circle, then it's hypotenuse passes through the center of the circle (and is thus a diameter). You can inscribe a virtually infinite number of triangles in a circle which would not be right triangles and would thus not pass through the circle's center.

So to do this problem you must first determine that the triangle formed by the points is indeed a right triangle. Using (Y1-Y2)/(X1-X2) to find the slopes of the three line segments, you will see that segments intersecting at (2,5) have inverse and opposite slopes (3, -1/3) and are thus perpendicular to each other and meet at a right angle. From there it is as simple as applying the pythagorean theorm to find the long side, in this case x^2 = 4^2 + 2^2.

It seems like some people may have gotten "lucky" assuming that the long side was the hypotenuse of a right triangle, and thus a diameter (and the answer) - Just wanted to note that wouldn't always be the case
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Re: m21 #20 [#permalink]

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New post 13 May 2013, 22:43
How we did know that the line is diameter?
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New post 15 May 2013, 02:34
Good Question. For a sec I thought it is a tough question till I worked out the length of the sides.

The length of the sides work out to \sqrt{10} \sqrt{10} and \sqrt{20}

It implies that it is an isosceles triangle with a right angle.
All triangles with base as diameter and the vertex touching a point on circumference of the circle - the vertex angle is 90 degree.

So it works out. The base with side \sqrt{20} is the diameter of the circle as well.
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Re: m21 #20 [#permalink]

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New post 29 Apr 2014, 16:04
deepak4mba wrote:
Guys, correct me if I am wrong in my logic.
Ignore the point (2,5). Considering the fact that the circle touches all the points, the line joining the (1,2)--say point A and (5,4)---say point B is the diameter(AB).
Now, draw perpendiculars from A and B across the coordinates which will join at Point C---(5,2).
Triangle, thus, formed is right angled triangle- triangle ABC.
Using simple coordinate numbers, the length of AC = 4 and length of BC = 2.
Applying Pythogoras Theorem, the diameter i.e. the hypotenuse of the triangle ABC = sqrt (20).

draw the above figure with simple coordinates to follow the above logic.

this is exactly how i did it.

Its much easier to visual if you have either a horizontal and/or vertical line in your triangle.

I drew dotted lines from points (5,4) and (1,2) to point (5,2) creating a right triangle angle at that new point with a shared hypotenuse with the original triangle. The length of the shared hypotenuse is what we are looking for.

side x which goes from point (1,2) to (5,2) has length 4
side y which goes from point (5,4) to (5,2) has length 2

a^2 + B^2 = c^2
c is the value we are looking for as length c is equal to side z the shared hypotenuse.

4^2 + 2^2 = c^2 16+4 = sqrt (20)

Re: m21 #20   [#permalink] 29 Apr 2014, 16:04

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m21 #20

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