[Guest]

1. You are given the four points representing the endpoints of the major and minor axes of an ellipse. Using a straight edge and compasses only, what is the simplest way to determine the positions of the the ellipse's two foci?

2. You are given a triangle. Using a straight edge and compasses only, what is the simplest way to divide the triangle into 7 equal areas?

In an earlier thread some wit asked "Would you allow me some paper and a pencil too?" Yes.

[Guest]

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Draw the lines connecting minor axis points and the major axis points.

Using compass measure the distance of one of the major axis end points to the point of intersection of the two axis.

Hinge the compass on one of the minor axis end points and keeping this 1/2 major axis length, draw arc on the line connecting the major axis end points. The two meeting points of the arc and the major axis are the two foci.

[Guest]

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Divide one of the sides of the triangle into 7 parts and connect the 6 points to the opposite side.

In order to divide one of the sides into 7 parts, do the following. Lets say we divide side AB of triangle ABC into 7 parts.

Comment on the image: The image does not actually show all parts equal coz i made it on XL using drawing tools so centres of circles may not be the actual points D1, D2 etc

Construct and line AD at any angle to AB (prefereably AD should be outside the triangle)

Now keep centre as A draw a small circle of any radius (not too large though). Let D1 be the point where the circle cuts line AD

Now keep the same compass opening, and draw another circle with point D1 as the centre. This circle cuts the line AD at D2.

Similarly draw more circles with each new intersection point as the centre and get points D1 to D7.

Now join D7 and B

Slide the straight edge and draw line parallel to D7B through D6, D5, D4, D3, D2 and D1

This gives points B1 to B6 on line AB

These points divide AB into 7 equal parts

Connect each of these to point C and this divides the triangle into 7 equal parts

Edited September 7, 2009 by DeeGee

[bonanova]

  On 9/7/2009 at 12:47 PM, DeeGee said:

  Reveal hidden contents

Divide one of the sides of the triangle into 7 parts and connect the 6 points to the opposite side.

In order to divide one of the sides into 7 parts, do the following. Lets say we divide side AB of triangle ABC into 7 parts.

Comment on the image: The image does not actually show all parts equal coz i made it on XL using drawing tools so centres of circles may not be the actual points D1, D2 etc

Construct and line AD at any angle to AB (prefereably AD should be outside the triangle)

Now keep centre as A draw a small circle of any radius (not too large though). Let D1 be the point where the circle cuts line AD

Now keep the same compass opening, and draw another circle with point D1 as the centre. This circle cuts the line AD at D2.

Similarly draw more circles with each new intersection point as the centre and get points D1 to D7.

Now join D7 and B

Slide the straight edge and draw line parallel to D7B through D6, D5, D4, D3, D2 and D1

This gives points B1 to B6 on line AB

These points divide AB into 7 equal parts

Connect each of these to point C and this divides the triangle into 7 equal parts

Nice.

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slide the straight edge and ensure no rotation as it slides?

[Guest]

I can't argue with any of that, DeeGee. In my view, definitely the simplest. The reason I posted this one was reading a book on geometric constructions in which the writer gave what, in my view, were definitely not the simplest method. Admittedly, for the second problem he added the condition that the division was to be made by parallel lines.

[Guest]

  On 9/7/2009 at 1:10 PM, jerbil said:

I can't argue with any of that, DeeGee. In my view, definitely the simplest. The reason I posted this one was reading a book on geometric constructions in which the writer gave what, in my view, were definitely not the simplest method. Admittedly, for the second problem he added the condition that the division was to be made by parallel lines.

Dividing triangles with parallel lines... hmm that would be difficult and complex too I guess!

[Guest]

  On 9/7/2009 at 1:07 PM, bonanova said:

Nice.

How does one slide the straight edge and ensure no rotation as it slides?

Well during my engineering drawing classes, we used to draw parallel lines by sliding (cough and clear throat "the drafter" clear throat again) but it should not be very difficult even with a simple stratight edge. Just got to be more careful...

[Guest]

parallel lines can be made by using a set of mariners parallel rulers. they are hinged together to allow moving a course plot line.

[Guest]

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you can use the parallel line construction to ensure you have them as well. another way that only requires one parallel line from B, use the same circle method, then use straight edge and place one end at first circle and other end at other first circle. the straight edge will go through the necessary point along AB, so mark it. repeat for each circle.

http://www.mathopenref.com/constparallel.html

http://www.mathopenref.com/constdividesegment.html

[Guest]

  On 9/7/2009 at 2:20 PM, decal-last said:

parallel lines can be made by using a set of mariners parallel rulers. they are hinged together to allow moving a course plot line.

In my view, the use of navigators' parallel rules is totally valid, since they are merely a convenience for allowed geometrical constructions. Likewise for the use of set squares (a more primitive variation of that to which you refer.) I do, in fact, have a solution to the amended problem 2#, not mentioned by the original poser (and it is fairly simple, DeeGee, though your original reply is excellent.)

Edited September 7, 2009 by jerbil

[Guest]

DeeGee, nice job. phillip1882, nice find. Another option, since you already have the circles, is to measure from point D5 to the point that the circle around D6 crosses the line going from D7 to point B and then move to D4 and mark that point on the circle around D5 and draw the line from D6 to B6, then move to D3 ... (i.e. no extra parallel line needed)

[Guest]

Here is how I did it. I'm not sure about the second one because I do not know where the foci are. I assumed that if you cut the elippsis in half that the foci woudl be in the middle of the two sides. Or that the Foci was just two circles in the middle, slightly offcenter. Cuz when you put the two circles overlapping, it creates a strange ellipsis. Like the jesus fish.

ADDENDUM: I did not read the question correctly. I did not know that it said 7 EQUAL parts. I just divided it into 7 parts.

Edited September 8, 2009 by William Ryan Kus

[Guest]

I'll post my answer to the "parallel lines" version of problem 2# in a couple of days, if no-one has any objection. I still think DeeGee deserves the prize for his original answer.

[Guest]

I don't know if anyone is even still checking this thread but I believe I got the answer to the parallel line problem.

Holy Moly this ain't simple but I think it will work:

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1st off we need to figure out what we need, so take a triangle and define the height as H and the base as B, then

A=1/2 B * H,

so we need the top triangle to be

1/7 A = 1/2 b * h where b and h are the base and height of that upper most triangle.

Now if f=b/B which will also =h/H then

1/7 A = 1/2 (f*B) * (f*H) = f^2 * 1/2 B * H = f^2 * A

so f=sqrt(1/7)

Now if we create a triangle that is 2 times the area of the first so that when you subtract the area of the first triangle you are left with an area = 1/7 the total triangle you have:

2/7 A = f'^2 * A where f' is the fraction for this second triangle and

f' = sqrt(2/7)

and so on so f(n) = sqrt (n/7) where n is 1-7 for the 7 triangles and f(n) is the fraction of the height from the apex that each new base is drawn.

Now we need to get sqrt(1/7):

the rt. triangle 1/7,sqrt(1/7),2*sqrt(2)/7 will work

Start with height and divide it into 7 (see the above solutions for how to do that).

Next draw lines parallel to the base at 1/7 and 2/7.

Copy 2/7 length onto the horizontal line at 2/7 to make a 45deg triangle

Measure the hypotenuse (which will be 2sqrt(2)/7) and sweep to the horizontal line at 1/7

Now measure the length from the height line to the new point (this is sqrt(1/7))

Now mark that length from the apex on the height line and call this point A.

Draw a line parallel to the base at this point A and you have triangle #1

****

Next put compass point on A and sweep from apex and mark on the base of triangle #1.

Now put compass on apex and sweep from the mark you just made (sqrt(2/7)) down onto the height line and call this point B.

Draw a line parallel to the base at this point B and you have triangle #2

****

Next put compass point on A and sweep from apex around to height line and call this point D (=2*sqrt(1/7))

Draw a line parallel to the base at this point D and you have triangle #4

****

Now put compass on apex and sweep from point D up to the base of triange #1 (create X,X sqrt(3), 2X triangle)

Measure the length on the base of triangle #1 from the height line to the new point you created (sqrt(3/7))

And copy that length from the apex down on the height line and call this point C

Draw a line parallel to the base at this point C and you have triangle #3

****

Now measure the length from the apex to point D

And copy that length onto the base of triangle #1 from the height (create X,2X, Xsqrt(5) triangle)

Now put compass on apex and sweep from the mark you just made (sqrt(5/7)) down onto the height line and call this point E.

Draw a line parallel to the base at this point E and you have triangle #5

****

Finally put compass on point C and sweep to base of triangle #3 (45deg triangle)

Now put compass on apex and sweep from the mark you just made (sqrt(6/7)) down onto the height line and call this point F.

Draw a line parallel to the base at this point F and you have triangle #6

****

And you're DONE

I attached a figure where I did it all, and it all looks good until you get to the end, in which the distance between the last 2 lines is too big, that is a problem will all the compounding errors (and my sloppy work), but theoretically I believe the method correct.

7 areas(2).pdf

[Guest]

Coo, Doc! I use Turbocad myself. I fully agree that a graphical method for finding square roots is at the heart of the matter. The use of the number 7 is of course irrelevant. 19 or 23 would do as well.

[Guest]

And, as promised, I attach my solution.

Explanation.doc

[Guest]

Seems that trying to solve it visually is harder than trying to figure it out abstractly with numberz. I'm not even in Math 115 yet so give me a break if I'm like a caveman compared to you guys.

Edited September 13, 2009 by William Ryan Kus

[Guest]

Dang that is a lot easier than mine!!! Very cool.

[Guest]

I like to see the proof so just in case anyone else is interested:

As I showed in my earlier answer (the long and complicated one) the ratio of the distance from the top to the total height must be equal to the fraction of the area desired (i.e. the height of the top triangle in this sol'n would be sqrt(1/7), ...), also, since we are talking ratio of the height to the total height, the same equation applies if you use the length of one side of the triangle, so as an equation it is

h (length from top)= sqrt( a/H ) * H = sqrt(aH), where H is the total length and a/H is the fraction of the area you are trying to get.

Now with the origin at the left/bottom of the half circle and using the diameter of the circle (H) as the x axis, the equation of the 1/2 circle is:

(x-H/2)^2 + y^2 = (H/2)^2 OR y^2 = xH - x^2

so if this method is right then the hypotenuse of the triangle made by the perpendicular bisectors (base a and height y) should be our h so:

h^2 = a^2 + y^2 = a^2 + xH - x^2, and x = H-a, so

h^2 = a^2 + (H^2 - aH) - (H^2 - 2aH + a^2) = aH

very cool

[Guest]

Do I detect a kindred spirit, Doc? I hate being told of a simple mathematical result without being able to work out how to prove it for myself! And welcome to the club, Caveman (alias William Ryan Krus.)

[Guest]

Did you guys solve the ellipsis problem also?

The problem with solving things visually is that the eyes create optical illusions. Such as seeing motion when there is none. Or seeing a curve in a line when there is none. Or seeing objects in the distance as smaller even though they are the same size.

Math takes out a lot of the miscalculations that the eye can cause.

[Guest]

Yes, Caveman (excuse me for calling you that Kus, but I think it a superb sobriquet - in reality we are all cavemen), but DeeGee caught that immediately.

If our moderators allow, I will describe why DeeGee's description is so good, in a further post which I will immediately prepare.