geometry query

[halfeye]

This is puzzling me, and I clearly don't understand it:

Starting with a right angle triangle, sides of 1, 2, and square root of 5, with angles of 90, 60 and 30 degrees. Then bisect the right angle. Now we have two triangles, the one which we are interested in having angles 45, 60 and 75 degrees. We know two of the sides of this triangle, the side opposite the 45 degree angle is half the square root of five (i.e. one and a fraction) whereas the side opposite the 75 degree angle is exactly one. Surely this is wrong? I am sure it's a requirement that the longer sides are opposite the widest angles?

[factotum]

the side opposite the 45 degree angle is half the square root of five (i.e. one and a fraction) whereas the side opposite the 75 degree angle is exactly one. Surely this is wrong?

I'm pretty sure it is wrong, yes. Bisecting the right angle splits the original hypotenuse unevenly, so neither piece is going to be exactly root 5 over two. A quick scribble on a piece of paper suggests you'd have to split the right angle into 60 and 30 degree chunks rather than 45 degree ones to exactly split the hypotenuse in half.

[gomipile]

This is puzzling me, and I clearly don't understand it:

Starting with a right angle triangle, sides of 1, 2, and square root of 5, with angles of 90, 60 and 30 degrees. Then bisect the right angle. Now we have two triangles, the one which we are interested in having angles 45, 60 and 75 degrees. We know two of the sides of this triangle, the side opposite the 45 degree angle is half the square root of five (i.e. one and a fraction) whereas the side opposite the 75 degree angle is exactly one. Surely this is wrong? I am sure it's a requirement that the longer sides are opposite the widest angles?

A 1,2,root 5 triangle is not a 30°, 60°, 90° triangle. A 30°, 60°, 90° triangle has sides proportional to 1, root 3, and 2.

[halfeye]

A 1,2,root 5 triangle is not a 30°, 60°, 90° triangle. A 30°, 60°, 90° triangle has sides proportional to 1, root 3, and 2.

That's a bit better but it doesn't fix things. Now we have sides of one and one opposite different angles. Now I think about it, I don't know where I got the 90, 60 and 30 degree triangle having those sides from, are you sure the sides you quote are right? The bisection of the right angle bisecting the hypotenuse being right has to be the case I think, and with angles of 45 (from bisecting the right angle) and 60 (preexisting) degrees the other angle is 75 degrees.

[Knaight]

Now I think about it, I don't know where I got the 90 60 30 degree triangle having those sides from, are you sure the sides you quote are right?

Consider the unit circle, then pick either 30 or 60 degrees on the unit circle (pi/6 or pi/3) respectively. Take the cosine and sine of that and you get sqrt(3)/2 and 1/2, with 1 being the hypotenuse because it's a unit circle. Scale up by a factor of 2, and you get sqrt(3), 1, and 2. Those are correct.

The bisection of the right angle bisecting the hypotenuse being right has to be the case I think

Now for the bisection line: Bisecting the 90 degree angle does create two new triangles, with angles 60, 75, 45 and 30, 105, 45. That 75-105 pair adds up to 180 as it should, and what this tells you is that your assumption that the initial hypotenuse is split in half with 90 degree angles with the intersection is wrong.

We can also check this another way. Going with the 30 degrees from the unit circle, we know the initial angle of the hypotenuse is 30 degrees. Flipping the triangle such that it rests on axis gets you an initial angle of -30 degrees. The intersection line is known to be 45 degrees, and the difference between them tells you the angle, which gets that 75 back. Similarly if you use the 60 degree perspective you recreate that 105.

This also means that the assumption that the side lengths of the hypotenuse are split evenly between the two new triangles is flatly wrong.

[Lord Torath]

If the lengths of the legs were identical, then when you bisect the 90^o angle, you also bisect the hypotenuse. But that's the only case this is true for a Right triangle.

This also holds true for any isosceles triangle. If you bisect the angle between the congruent sides, you also bisect the side opposite the angle. A CAD program or graph paper and a protractor are the easiest ways to verify this yourself.

[halfeye]

Consider the unit circle, then pick either 30 or 60 degrees on the unit circle (pi/6 or pi/3) respectively. Take the cosine and sine of that and you get sqrt(3)/2 and 1/2, with 1 being the hypotenuse because it's a unit circle. Scale up by a factor of 2, and you get sqrt(3), 1, and 2. Those are correct.

I knew sin and cos were going to show up, I never liked those. Thanks for the confirmation of the sides.

Now for the bisection line: Bisecting the 90 degree angle does create two new triangles, with angles 60, 75, 45 and 30, 105, 45. That 75-105 pair adds up to 180 as it should, and what this tells you is that your assumption that the initial hypotenuse is split in half with 90 degree angles with the intersection is wrong.

I never assumed that the the split had a 90 degree angle, it obviously doesn't, I'm still not sure that's required for a bisection of the line length.

We can also check this another way. Going with the 30 degrees from the unit circle, we know the initial angle of the hypotenuse is 30 degrees. Flipping the triangle such that it rests on axis gets you an initial angle of -30 degrees. The intersection line is known to be 45 degrees, and the difference between them tells you the angle, which gets that 75 back. Similarly if you use the 60 degree perspective you recreate that 105.

I am in no sense disputing that the angles are 75 and 105 degrees or that they make 180 degrees total.

This also means that the assumption that the side lengths of the hypotenuse are split evenly between the two new triangles is flatly wrong.

There are an infinite number of lines lines that intersect the hypotenuse at 90 degrees, most of these do not go anywhere near the right angle, and all but one don't divide it equally, and we agree I think that that one doesn't go through the right angle.

[Peelee]

If the lengths of the legs were identical, then when you bisect the 90^o angle, you also bisect the hypotenuse. But that's the only case this is true for a Right triangle.

I was about to say this. for a 30-60-90 triangle, you can draw a line to make the 90º angle into two 45º angles, or you can draw a line the make the hypotenuse half the hypotenuse. But you cannot do both with the same line.

Bytheway, Lord Torath, instead of superscripting a o, if you hold the Alt key ahd nit 1 6 7 on the numpad, you get a nice little degree symbol. It's totally worth knowing the alt ASCII codes, IMO. ¿No?

[halfeye]

If the lengths of the legs were identical, then when you bisect the 90^o angle, you also bisect the hypotenuse. But that's the only case this is true for a Right triangle.

This also holds true for any isosceles triangle. If you bisect the angle between the congruent sides, you also bisect the side opposite the angle. A CAD program or graph paper and a protractor are the easiest ways to verify this yourself.

That's something I understand to be true, but I feel I remember that if you bisect an angle in a triangle, you also bisect the opposite side, no matter the angle of the intersection.

[Peelee]

That's something I understand to be true, but I feel I remember that if you bisect an angle in a triangle, you also bisect the opposite side, no matter the angle of the intersection.

Nah. Imagine a right triangle with two sides set as 1, and hypotenuse √2. Then imagine stretching one of the sides until the hypotenuse equals 1,000,000*. The 90º angle has not changed, and if you draw a line bisecting the 90º angle, it still ends in the same place, but it's no longer bisecting the hypotenuse, since the side that was stretched is now much longer.

*I'm a fan of reduction to absurdity in math.

[halfeye]

Nah. Imagine a right triangle with two sides set as 1, and hypotenuse √2. Then imagine stretching one of the sides until the hypotenuse equals 1,000,000*. The 90º angle has not changed, and if you draw a line bisecting the 90º angle, it still ends in the same place, but it's no longer bisecting the hypotenuse, since the side that was stretched is now much longer.

*I'm a fan of reduction to absurdity in math.

That's a good example, thanks for the explanation. I'll need to let that settle for a week or two, if I don't forget it entirely, but many thanks.

[Peelee]

That's a good example, thanks for the explanation. I'll need to let that settle for a week or two, if I don't forget it entirely, but many thanks.

No poblem! Unreasonably big numbers in examples really help, I've learned. Like with the Monty Hall problem. The solution is way easier to understand with a billion doors than with three doors.

[Knaight]

No poblem! Unreasonably big numbers in examples really help, I've learned. Like with the Monty Hall problem. The solution is way easier to understand with a billion doors than with three doors.

This is generally what I like to do when trying to get my head around or sense check an equation. Stick some big numbers in, drop values to zero, and just look at how it behaves when you do that.

[Leewei]

No poblem! Unreasonably big numbers in examples really help, I've learned. Like with the Monty Hall problem. The solution is way easier to understand with a billion doors than with three doors.

Huh. Funny thing is, I find Monty Hall easier to understand with three, since you can easily list the possibilities.