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October 3rd, 2019, 01:20 AM  #1 
Newbie Joined: Sep 2019 From: Fier, Albania Posts: 3 Thanks: 0  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. 
October 3rd, 2019, 02:58 AM  #2 
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151 
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. 
October 3rd, 2019, 03:31 AM  #3 
Newbie Joined: Sep 2019 From: Fier, Albania Posts: 3 Thanks: 0 
How do I prove that then? That they are not in a halfcircle.

October 3rd, 2019, 03:43 AM  #4 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 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.

October 3rd, 2019, 03:44 AM  #5 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
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 √(BC²+CA²) = √(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 
October 3rd, 2019, 03:50 AM  #6 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle 
The formula for a circle is $x^2+y^2+ax+by+c=0$ $x=10, \; y=12 \Rightarrow 100+14410a12b+c=0 \Rightarrow 10a+12bc=244$ $x=6, \; y=8 \Rightarrow 36+64+6a+8b+c=0 \Rightarrow 6a+8b+c=100$ $x=2, \; y=14 \Rightarrow 4+1962a14b+c=0 \Rightarrow 2a+14bc=200$ 
October 3rd, 2019, 03:58 AM  #7 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
^ That could work. Seems like a lot more effort, though.

October 3rd, 2019, 02:41 PM  #8  
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271  Quote:
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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