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m21 #20

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Re: m21 #20 [#permalink] New post 13 May 2013, 08:09
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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Re: m21 #20 [#permalink] New post 13 May 2013, 09:11
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] New post 13 May 2013, 21:43
How we did know that the line is diameter?
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Re: m21 #20 [#permalink] New post 13 May 2013, 22:08
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jrzayev001 wrote:
How we did know that the line is diameter?


The ratio of the sides of triangle ABC is 1:1:\sqrt{2}, so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2}).

So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20}.

Hope it's clear.

Complete solution is here: m21-76117.html#p1224302
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Re: m21 #20 [#permalink] New post 15 May 2013, 01:34
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] New post 29 Apr 2014, 15:04
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)

Q.E.D.
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Re: m21 #20 [#permalink] New post 04 May 2014, 12:21
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we can solve this problem in many ways
first by finding the equation of circle whose center's co-ordinate is (a,b) and radius is r
(x-a)^2+(y-b)^2=r^2
here withe the given co-ordinates, we can find 3 equation and we have three variable.
upon solving we get the co-ordinate of center= (3,3)
Now we can use distance formula to solve the radius
sqrt(a1-a2)^2 +(b1-b2)^2 = distance between tow numbers, where (a1, a2) and (b1, b2) are co-ordinates.
Method:2
Any line which passes through the mid point and center of a circle is perpendicular . Here with the given co-ordinates, we can find the equations of chords. So we find the equation of perpendicular bisector of chord. by solving two perpendicular bisector's equation we can find the center of co-ordinate. then by using distance formula we can find the distance between center and any given point. this found distance is the radius
2* radius = diameter
Re: m21 #20   [#permalink] 04 May 2014, 12:21
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