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October 3rd, 2019, 01:20 AM   #1
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Prove that these 3 points are in the same half circle.

Prove that A(-10;-12) B(6;8) C(-2;-14) are in the same half circle.

Please help; been torturing myself all morning.

Last edited by skipjack; October 3rd, 2019 at 02:03 PM.
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October 3rd, 2019, 02:58 AM   #2
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The coordinates of point $\displaystyle B$ are obscured by an emoticon.

If the coordinates of $\displaystyle B$ are ($\displaystyle 6,8$) then the answer is that the three points are not on the same half circle. Clearly they are on a circle.
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October 3rd, 2019, 03:31 AM   #3
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How do I prove that then? That they are not in a half-circle.
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October 3rd, 2019, 03:43 AM   #4
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Math Focus: Area of Circle
Calculate the center of the circle. Draw line from one of a given point to center. Check if two other points lie on the same side of the line.
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October 3rd, 2019, 03:44 AM   #5
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Plotting the points, it is too close to tell decisively.

From geometry: If three vertices of a triangle are on the same semicircle, there will be one angle of 90° or more.

We could use trig, but we can get away with Pythagoras.
AB = (16,20), |AB| = √(16²+20²) = √656 = 25.6124
BC = (-8,-22), |BC| = √(8²+22²) = √548 = 23.4094
CA = (-8,2), |CA| = √(8²+2²) = √68 = 8.2462

If this were a right triangle, then the hypotenuse |AB| would be √(|BC|²+|CA|²) = √(548+68) = √616 = 24.8193.

|AB| is longer than this, so <ACB is greater than 90°. Therefore the angle of the arc clockwise from A to B is greater than 180°, so the angle of the arc containing the points B, C, and A is less than 180°. The three points lie on a semicircle. QED
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October 3rd, 2019, 03:50 AM   #6
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Math Focus: Area of Circle
The formula for a circle is $x^2+y^2+ax+by+c=0$

$x=-10, \; y=-12 \Rightarrow 100+144-10a-12b+c=0 \Rightarrow 10a+12b-c=244$

$x=6, \; y=8 \Rightarrow 36+64+6a+8b+c=0 \Rightarrow 6a+8b+c=-100$

$x=-2, \; y=-14 \Rightarrow 4+196-2a-14b+c=0 \Rightarrow 2a+14b-c=200$
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October 3rd, 2019, 03:58 AM   #7
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^ That could work. Seems like a lot more effort, though.
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October 3rd, 2019, 02:41 PM   #8
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Quote:
Originally Posted by mrtwhs View Post
. . . the answer is that the three points are not on the same half circle.
That's incorrect - they do lie on the same semicircle.

The points are the vertices of a triangle. It suffices to show that the triangle has an obtuse angle (or right angle), as already shown above.
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