Perpendicular geometric objects

[Campbell1982]

Trying to solve a drawing task. The following has to be achieved:

- Produce a Square (vertices of ABCD)
- Cut line BC into two equal parts, label the point in between as M.
- Draw a line between point M and A.
- Divide angle of BMA into two equal parts. Label the location where the bisector touches the line AB.
- Build a line that is named PM. This line is perpendicular to line NM, and touches AD. (The point the line PM touches AD is named P).
- Draw a circumcircle around APM.

I have attached a rough drawing that I produced in paint (Square 2) . I think I have gone wrong somewhere, but I'm not sure where. Can anyone help please?

Matt.

 

[DEvens]

You should check some things, calculate some things, and mark some things on the diagram.

You have the distance AB as the side of a square and the distance BM as half the side of a square. So, what is the size of angle AMB?

So then you should be able to work out the angle MP makes with the horizontal. And then you should be able to check that the point P is in fact on the segment AD and inside the square. You should be able to calculate how far down from the corner D the point P is.

To draw the circum-circle of three points I presume you have a standard procedure. Maybe something like, you need to work out where the centre is, what the radius is, and so mark off a radial line segment.

 

[Campbell1982]

Hello DEvens, firstly thank you for helping me. I have redrawn the diagram using constructions but it has come out the exact same. I found the size of angle AMB to be 60 degrees, I'm not sure if that is correct though. The line of MP is perpendicular to NM and touches AD, but for line MP to be perpendicular (right angles) to line MN it causes point P to be outside of the square.. Have I made a mistake somewhere that has caused this?

 

[Campbell1982]

Just to add to that previous statement, Point P is on the line AD (not line segment), but it is above point D.. which makes me think something has gone wrong.

 

[DEvens]

Just to add to that previous statement, Point P is on the line AD (not line segment), but it is above point D.. which makes me think something has gone wrong.

The point P being above the point D is what I was trying to hint at without actually saying it. That means you need to re-do your diagram a little bit.

Also, check your calculation of the angle.

 

[LCKurtz]

I agree D (corrected: P) is above the square and you have the angle wrong. My question is, where are you headed with this construction? Once you have the circumcircle, then what?

 

[Campbell1982]

Thank you both for your help. I have attached a new version of the drawing (It's rough). I have calculated AMB as 30 degrees now, instead of 60 degrees. I'm glad point P is above the point d. I'm trying to eventually investigate the triangle of MPA and the circumcircle.

 

[LCKurtz]

Angle AMB is close to 60 degrees but it isn't 60 degrees and half of it isn't 30 degrees. Tangent(AngleAMB) = 2 and the inverse tangent of that is about 63.4 degrees. Also, for what it's worth, I drew a pretty accurate sketch and it looks to me like, if the angle bisector from M hits AB at N, that point N may be on the circumcircle. Not sure if that's exactly true, or even relevant.

[Edit:] After some calculations I don't believe N is on the circumcircle. It's just close.

 

[Campbell1982]

LCKurtz, I have attached a more accurate drawing done by hand. Point N is not on the circumcircle on my drawing. The angle AMB is 31.7, the other two angles in that triangle at 90 degrees and 58.3. Looking at anything of interest within the circumcircle, I can only find that triangle AMP is isosceles.

 

[LCKurtz]

Triangle AMP is not isosceles. It couldn't be unless P = D.
[Edit:] I thought you meant AM = MP. You are correct that AM = AP.

And it turns out that N is exactly on the circumcircle. I have attached a picture created with Maple:

 

[Campbell1982]

Triangle AMP is not isosceles. It couldn't be unless P = D.
[Edit:] I thought you meant AM = MP. You are correct that AM = AP.

And it turns out that N is exactly on the circumcircle. I have attached a picture created with Maple:
View attachment 79226

I have been steering at this all day and I can't find anything special about the circle. The circle's centre isn't the centre of the square or triangle. The only thing I can think is there is a special relationship between the chords..

 

[Campbell1982]

I have been steering at this all day and I can't find anything special about the circle. The circle's centre isn't the centre of the square or triangle. The only thing I can think is there is a special relationship between the chords..

That Maple software looks pretty good as well, I think I may have to look at investing in that.

 

[LCKurtz]

That Maple software looks pretty good as well, I think I may have to look at investing in that.

You might want to think about that some. There is a pretty good learning curve involved to get Maple to do what needed to be done to produce that picture.

 

[Campbell1982]

You might want to think about that some. There is a pretty good learning curve involved to get Maple to do what needed to be done to produce that picture.

Yey, I've checked Maple out. That is some serious software.. You made it look easy.

The square is not a cyclic quadrilteral, but A is touching the circumference. Points N and M touches the side of the square and circumference.
Point P touches the circumference but not the square. Line CB is tangent to the circle.
Angles PAM and APM are both 30 degrees, resulting in PMA being 120 degrees. Diameter of circumcircle can be calculated by measuring the triangles sides
and using the following formula: abc / sqr(root)(a + b + c)(b + c - a)(c + a - b)(a + b - c).
BC is a tangent to the the circumcircle, allowing alternative segment theorem to be used, therefore BMA = APM and CMP = PMA, which reinforces
previous angle calculations.
PM and MA are both segments. Is there anything notable to state about the centres of the triangle with regards to the centre of the circle? I have found
that Euler's line lies crosses through the midpoint of AP. It seem also that the centre of the circle is not the midpoint of the triangle AMP.