Geometry/Trigonometry/Math Question

[Jim Koepke]

To start, my apologies for not having a better image for this.

A simple way to draw a 45º angle from a point on a line was shown in the book By Hand & Eye by Jim Tolpin & George R. Walker.

Simple 45º Angle Construction.jpg

The only tools required are; a straight edge, compass and pencil.

From the point of origin “O” a radius is struck to produce point P1 and P2 on the line. a dashed circle was drawn to indicate this radius.

From P1 and P2 using a larger radius, two arcs are drawn to intersect above point O. This is how to drop a line from R1 to O creating a right angle (90º). In this case a full circle was drawn. Using the same radius an arc is drawn from R1 to R2. The arc at R3 was drawn from R2 to show a center line from P2. As is commonly known continuing around the circle with its radius produces the layout for a hexagon, I am sure this is part of the reason this works.

A line drawn from point O to the point at R2 creates a 45º angle from point O on the line formed from P1 to P2.

Since varying the radiuses results in the same orientation of the hexagon in relation to line R1 - O and P1 - P2 it seems this plays a part in why a 45º angle results when a line is drawn from O to R2.

I am curious as to why this works.

Anyone know the reason?

jtk

[Mike Cutler]

A right triangle was created by the bisection of the lines.
The opposite and the adjacent were determined by the circumference/ radii of the circles, and are equal. Therefore the hypotenuse has to be a 45 degree angle.

[Jim Koepke]

A right triangle was created by the bisection of the lines.
The opposite and the adjacent were determined by the circumference/ radii of the circles, and are equal. Therefore the hypotenuse has to be a 45 degree angle.

The right triangle was created after the 45º line was drawn using a (30-60-90) triangle as a straight edge and square against line P1-P2. The line where P1-P2 was given a second going over by my pencil in this process is a bit darker.

The 45º angle was derived from the intersection of arcs at R2. R2 is one radius of the larger circle from the intersection of arcs at R1.

A triangle drawn from P2-R2-R3 would be an equilateral triangle, all corners would be 60º.

Any line perpendicular to the base line along the 45º vector will be an isosceles, right triangle.

Sorry for the confusion, I should have not included the extra lines.

jtk

[Steve Demuth]

I don't believe you have the construction right. You are intended to actually draw the perpendicular from O through R1 until it intercepts the first circle you drew (solid blue line). From that intercept, using the larger radius, you draw an arc that intercepts the second circle (dashed blue line is the radius, purple arc the actual arc). Then the red dashed triangle (which would be P2 - R1 - R2 if we followed your labeling) is equilateral, and the green line by construction bisects the angle at R2, and thus the red dashed line opposite, and thus, because triangle P2 - O - R1 is isosceles, angle P2 - O - R1, which is right, so angle P2 - O - R2 is half that, or 45^o.

Note that your construction does not yield a 45^o angle in general. You can easily see that if you shorten the second radius, it will move your R1 down on the diagram, and your R2 to the anti-clockwise on the circle, to the point where in the limit that your second radius approaches the first, you get a 60^o angle.


Capture.jpg

(Note: corrected a mistake in first post. The limiting case I mentioned was wrong).

[Jim Koepke]

I don't believe you have the construction right. You are intended to actually draw the perpendicular from O through R1 until it intercepts the first circle you drew (solid blue line). From that intercept, using the larger radius, you draw an arc that intercepts the second circle (dashed blue line is the radius, purple arc the actual arc). Then the red dashed triangle (which would be P2 - R1 - R2 if we followed your labeling) is equilateral, and the green line by construction bisects the angle at R2, and thus the red dashed line opposite, and thus, because triangle P2 - O - R1 is isosceles, angle P2 - O - R1, which is right, so angle P2 - O - R2 is half that, or 45^o.

Note that your construction does not yield a 45^o angle in general. You can easily see that if you shorten the second radius, it will move your R1 down on the diagram, and your R2 to the anti-clockwise on the circle, to the point where in the limit that your second radius approaches the first, you get a 60^o angle.


Capture.jpg

(Note: corrected a mistake in first post. The limiting case I mentioned was wrong).

Mike, Steve thanks to both of you for offering a response.

Steve, the first circle was only drawn to show how P1 & P2 came to be. In the construction, that circle is not present.

If I can find some ink and get a chance to redo this, it will be done in ink without any unnecessary lines or full circles to make the construction clearer.

jtk

[Jim Koepke]

It has been over forty years since I've done an inked drawing so this is a little fat in places.

Simpiler 45º Angle Construction.jpg

From the point of origin “O” a radius is struck to produce point P1 and P2 on the line.

From P1 and P2 using a larger radius, two arcs are drawn to intersect above point O. Using this same radius an arc is drawn from R1 to R2.

A line drawn from point O to the point at R2 creates a 45º angle from point O on the line formed from P1 to P2.

My attempt at providing more information may have been confusing the construction.

jtk

[Mel Fulks]

Jim , the drawing is fine , especially since I now see that it IS a drawing , and not a picture of a broken kite on a cloths -line.

[Jim Koepke]

LOL!

jtk

[Charlie Velasquez]

It has been over forty years since I've done an inked drawing so this is a little fat in places.

Simpiler 45º Angle Construction.jpg

From the point of origin “O” a radius is struck to produce point P1 and P2 on the line.

From P1 and P2 using a larger radius, two arcs are drawn to intersect above point O. Using this same radius an arc is drawn from R1 to R2.

A line drawn from point O to the point at R2 creates a 45º angle from point O on the line formed from P1 to P2.

My attempt at providing more information may have been confusing the construction.

jtk

I don’t think it does.
IMG_1399.jpg


OR1. is a perpendicular bisector by definition.
Assume segment OP2 = 3, OP1 =5. Then…

OP2R1 is a right triangle with sides 3,4,5
and angles 90°, 53.13°, and 36.87° (within 0.01), 53.13° is our angle of interest.

Triangle R1R2P2 is equilateral by definition, then angle R1P2R2 is 60°
Angle OP2R2 = 53.13°
The sum of these two is 113.13°

The two sides of triangle OR2P2 are 3 and 5 as given.

Using the side-angle-side theorem we can compute the other side and the angles.
Rather than do the calculations I just used an online calculator.

IMG_1400.jpg

42.808°

edit to add: If you were to scribe distance OP2 on line OR1, then scribe your arc from that point, bisecting the right angle, you would indeed have a 45° angle.

[Howard Garner]

Too complicated for me.
Using the previous definitions:

Draw a complete half circle from point O to P1 and P2
Using P1 and P2 draw arcs that intersect above the first arc.
From the intersection to point O is perpendicular to the P1-P2 line
A line from the perpendicular crossing the first arc to either P1 or P2 will be 45 degrees.

Howard Garner

[Steve Demuth]

It has been over forty years since I've done an inked drawing so this is a little fat in places.

Simpiler 45º Angle Construction.jpg

From the point of origin “O” a radius is struck to produce point P1 and P2 on the line.

From P1 and P2 using a larger radius, two arcs are drawn to intersect above point O. Using this same radius an arc is drawn from R1 to R2.

A line drawn from point O to the point at R2 creates a 45º angle from point O on the line formed from P1 to P2.

My attempt at providing more information may have been confusing the construction.

jtk

I understood your intent. But it doesn't work the way you describe it - it won't create a 45^o angle, except in the case where you choose the second, longer, radius to be the square root of 2 times the first, shorter radius (which is equivalent to doing what I suggested - extending the perpendicular out to intersect the circle you'd drawn and working from there).

Here's a little demonstration:

Choosing the second radius to spot on at 1.414 times the first, you get this (note the measured angle):On the Money.PNG

Choosing the second radius larger than 1.414 times the first, you get this (measured angle is under 45^o):Over.PNG

Choosing the second radius small than 1.414 times the first, but still longer than the first, you get this (measured angle over 45^o):Under.PNG


Sorry, but you can't figure out why it works, because it doesn't work.

[Jim Koepke]

I understood your intent. But it doesn't work the way you describe it - it won't create a 45o angle, except in the case where you choose the second, longer, radius to be the square root of 2 times the first

Steve, my understanding of this construction is the size of the radius striking P1 & P2 from O doesn't matter. Neither does the radius from P1to R1 as long long as it is the same as the radius from P2 to R1. From P2 a circle can be drawn and then, starting at R1 be stepped off to construct a hexagon using the radius of P1 - R2.

No matter what the length of O to P1 or P2, an arc with a larger radius drawn from P1 & P2 will in such a way as to be 90º from O. If multiple intersections of different radiuses were drawn above to create R1a & R2a, R1b & R2b and so on, the different radius would make concentric arcs. The circles & hexagons formed, if one were to draw them out, would all be concentric. The lengths of the lines nor the radiuses used to plot points have nothing to do with a solution since they are infinitely variable.

The relationship of the 90º line and the (undrawn) hexagon may be the key to this. It has been many years since I have done much of this kind of calculations and I never took Trigonometry, I'm kind of amazed I can spell it right.

jtk

[Steve Demuth]

You are right, the size of the radius that creates P1 and P2 on the line, equidistant from O, doesn't matter. I chose it to be 1 in the pictures because it has to be something, but it's completely arbitrary.

In your description of the construction, the second radius, used to construct R1 and R2, can be any length, provided it's longer than the first. But if you follow your instructions, the constructed angle will not in most cases be 45º. The procedure you describe simply doesn't work. I drew the sketches with calculated measurements using different arc lengths to show that it doesn't work. You can easily do the same on paper with a compass and ruler. Follow your procedure, but deliberately choose the second arc length to be roughly twice the first (by setting your compass to the distance from P1 to P2). Measure the angle you construct. It'll be right around 40º. I can easly prove it doesn't work using trignometry, but it involves a lot of algebra and application of the law of cosines and law of sines - pretty ugly stuff, and really not very convincing compared to just working a counter example. It does work if the second arc is precisely the arc required to make triangle P2 - O - R1 an isoceles right triangle, which is sqrt(2) times the first arc.

The construction is attempting to create a 45º angle by bisecting the right angle initially constructed by creating the line from O to the point you label as R1. This is a classic Eulcidean compass and ruler construction, but it requires R1 and P2 to be equal distances from O to work. That's what I described in my first post - extend the line O - R1 until you can create a new point on it that is the same distance fro mO as P1 and P2 are. Use this as the basis for constructing R2, and you'll always get a 45º angle. BTW, no trignometry is necessary to get to this - it's pure, measurement-free (trigonometry means "triangle measurement"), geometric construction straight out of Euclid.

[Jim McCue]

I can maybe draw diagrams and take pictures later but hard to find the time. I don't have the book but I remember doing exercises in math class using just a straight edge and compass.

I think there is something missing or confusing in your description.

You can bisect any angle with the same technique, you are just starting with a straight line (or 180* angle) first. When you make the P1 and P2 marks, those are just spots for you to rest the sharp compass point for your next arcs. The radius for the arcs has to be longer than the initial radius, otherwise they would intersect at only one point, the origin. When the radius is longer, there will be an intersection above and below the origin. You can draw a line through an intersection and the origin or through the top and bottom intersections to bisect your original line segment, making a 90* angle. After this you can erase or forget about P1 and P2 and the tick marks you just made and used.

Now you have a 90* angle, and you can bisect it the same way. Put the sharp compass point at the right angle vertex, then make tick marks at an arbitrary equal distance, just like you originally did for P1 and P2. You can call these P3 and P4? Moving your sharp compass point to these spots, you can make arcs (or entire circles) to see where they intersect. If you don't adjust the radius from when you marked P3 and P4, one of the intersections will be at the origin and one will be out in the distance. Even if you do adjust the radius the intersections should line up bisecting the angle. So you can draw a line through your intersections to bisect the angle.

You can bisect any angle with this technique. Just draw a random acute or obtuse angle and draw the arcs and connect the dots.

As far as trigonometry or geometry you are really taxing my memory. I remember there were theorems and proofs where if you know some sides or angles, you can make assumptions about the remaining sides and angles. But I think the main point is you are making two symmetric triangles. Two of their legs are arbitrary lengths from your compass, but their hypotenuse is shared. And since two of the angles are butted up at your origin, their angles will be 1/2 of the original angle.

[Jim Koepke]

This may be a case of "don't believe everything you read." The book By Hand & Eye gives very few details on this method.

For the last few minutes I've been drawing this out using the same points for P1 & P2 with different radii for deriving R1 & R2. My drawing may be inaccurate, but the R2 points do not clearly align. This was done on a clipboard using a pencil. Maybe later I will try again with my drawing board set up on my bench.

Currently I'm starting to believe what others have said above. Maybe the easiest method to construct a 45º angle is to bisect a previously drawn right angle.

jtk