Brain fart - "easy" geometry problem

[DrPizza]

Okay, this one is bugging me. (shame on me). Can't upload a pic from my iPad, but it's easy to describe. In a circle, draw 2 perpendicular diagonals. From the endpoint of one of the diagonals, draw a chord that's divided in a 3:1 ratio by the other diagonal. What is the angle formed between the first diagonal and the chord?

 

[allisolm]

Do your own homework. Oh, wait...

 

[IronWing]

nm. Misunderstood, back to the drawing board.

 

[yh125d]

This?

 

[disappoint]

30,60,90 right triangle?

 

[Smoblikat]

This?

Thats the way I saw it in my mind. It would all depend on where the chord is.

 

[Red Squirrel]

I don't think that would be solvable unless you can determine a few things such as the distance between the two lines at the far point and the length of at least one line or the circumference (which would be used to get the length of the line assuming it passes dead center). Math was not my strong point, so I'll shut up now. In a real life situation I'd just draw it out in autocad with the distances I do know and then measure the distances of the rest.

 

[Einz]

Okay, this one is bugging me. (shame on me). Can't upload a pic from my iPad, but it's easy to describe. In a circle, draw 2 perpendicular diagonals. From the endpoint of one of the diagonals, draw a chord that's divided in a 3:1 ratio by the other diagonal. What is the angle formed between the first diagonal and the chord?

If the above diagrams are right, then the key is to draw an additional chord from the end if the chord to the other end of the first diagonal, making another right triangle that is similar to the one in question.

Now with solutiony goodness below

Triangle BDE is similar to ABC, therefore,
BE/BD = BC/AB, with BC=3/4*BD
Rearranging that, you end up with the ratio of AB to BC is 2/sqrt(6). The angle is ~35 degrees per the posts below

 

[yh125d]

I know enough to know that it is solvable, the only info that matters is the two diagonals are perpendicular, and the 3:1 ratio of the chord (I mistakenly drew it more 2:1). I just have no clue how to mathematically do it

I'd also try to do it in autocad, but I'm not sure how I'd manage to get the 3:1 ratio right though. It is pretty awesome how you can use cad to solve a lot of geometry stuff if you understand how to make it do what you want it to

 

[Gunslinger08]

Thinking.

 

[Northern Lawn]

I can pretty evenly cut up a pumpkin pie into 6 or even 8 pieces.

 

[Ken g6]

Well, let's look at it algebraically. Take the two diagonals, and make them an x/y coordinate system. Make the circle have a unit radius.

x^2+y^2=1

Take the top half of the circle:
y(x) = sqrt(1-x^2)

Now, let's see. The length of the chord is proportional to its projection onto the x axis. The ratio of the lengths is supposed to be 1/3:3/3. So, given that the chord ends at (1,0), we can say that the chord begins on the left at point (x=-1/3, y(x)).
y(-1/3) = sqrt(8/9) = 2/3*sqrt(2) ~= .943

Now it's just trigonometry to find the angle.
tan(a) = opposite/adjacent = 2/3*sqrt(2)/(4/3) = sqrt(2)/2
tan(a) ~= .7071

Google says arctan(.7071) = 0.615475188 rad ~= 35.264 degrees.

 

[Mr. Pedantic]

Very confusing. Which is the "first" diagonal?

So, if x is the 1 in the 3:1 ratio, r is the radius of the circle, and y is the angle, then:

4x/2r = cos(y) \\from the big triangle formed with the diameter as the hypotenuse
r/3x = cos(y) \\from the small triangle formed from both diameters and the chord
12x^2 = 2r^2
6x^2 = r^2
sqrt(6)x = r

Then you just substitute this for one of the equations up top, and you get a ratio of sqrt(6)/3 for cos(y), which gives y = 35.26°. Or 0.6155 radians.

 

[Einz]

Well, let's look at it algebraically. Take the two diagonals, and make them an x/y coordinate system. Make the circle have a unit radius.

x^2+y^2=1

Take the top half of the circle:
y(x) = sqrt(1-x^2)

Now, let's see. The length of the chord is proportional to its projection onto the x axis. The ratio of the lengths is supposed to be 1/3:3/3. So, given that the chord ends at (1,0), we can say that the chord begins on the left at point (x=-1/3, y(x)).
y(-1/3) = sqrt(8/9) = 2/3*sqrt(2) ~= .943

Now it's just trigonometry to find the angle.
tan(a) = opposite/adjacent = 2/3*sqrt(2)/(4/3) = sqrt(2)/2
tan(a) ~= .7071

Google says arctan(.7071) = 0.615475188 rad ~= 35.264 degrees.

That's not completely true, since the chord traces an arch. I think that invalidates the next statement?

 

[Ken g6]

That's not completely true, since the chord traces an arch. I think that invalidates the next statement?

Sure it's true. It's just a line, and we've ensured that it meets the circle at both ends. Consider similar triangles. The triangle of (0,0), (1,0), (0,f(0)) is similar to (-1/3,0), (1,0), (-1/3,f(-1/3)=y(-1/3)), even though I don't know what f(0) is.

If you want to go ahead and calculate what f(x) is from the two points given, I'm sure my statement will hold up.

 

[Einz]

Sure it's true. It's just a line, and we've ensured that it meets the circle at both ends. Consider similar triangles. The triangle of (0,0), (1,0), (0,f(0)) is similar to (-1/3,0), (1,0), (-1/3,f(-1/3)=y(-1/3)), even though I don't know what f(0) is.

If you want to go ahead and calculate what f(x) is from the two points given, I'm sure my statement will hold up.

False. Take your y=sqrt(1-x^2) formula for example. At (0,0), (1,0), and (0,1), it's a simple 45/45/90 right triangle. At (-1/3,0), (1,0), and (-1/3,0.94), it's no longer a simple 45/45/90 right triangle. The lengths of your two right legs are now 0.94, and 4/3

 

[Ken g6]

I said (0,0), (1,0), (0,f(0)). I should have specified that I meant f(x) to describe the line between (1,0), and (-1/3,0.943). I don't know what f(0) is Scratch that, yes, I do. Because of similar triangles, f(0) ~= 3/4*0.943 = .707

Edit: So f(x) = .707-.707x Satisfied?

 

[Einz]

I said (0,0), (1,0), (0,f(0)). I should have specified that I meant f(x) to describe the line between (1,0), and (-1/3,0.943). I don't know what f(0) is Scratch that, yes, I do. Because of similar triangles, f(0) ~= 3/4*0.943 = .707

Edit: So f(x) = .707-.707x Satisfied?

Nope. That point of -1/3, 0.943 doesn't meet the condition that the distance from (0,1) to (0,0.707) is 3:1 the distance from (0,0.707) to (-1/3,0.943)

*EDIT* Humbly corrected, per below posts.

 

[Engineer]

I think DrPizza is trolling with the "easy" in the title. What do you guys think? :hmm:

 

[Ken g6]

Fine, I'll calculate it.

I assert: 1*sqrt((1-0)^2+(.7071-0)^2) = 3*sqrt((0-1/3)^2+(.943-.7071)^2)
Simplify: sqrt(1+.7071^2) = 3*sqrt((1/3)^2+(.943-.7071)^2)
Calculator: 1.224740956 = 3*0.408362487 = 1.225087462
It's within the three significant digits used. By the way, .7071 is an approximation of sqrt(2)/2.

Oh, and did you know that Google is a graphing calculator now? So if anyone's having trouble visualizing:

 

[Einz]

Fine, I'll calculate it.

I assert: 1*sqrt((1-0)^2+(.7071-0)^2) = 3*sqrt((0-1/3)^2+(.943-.7071)^2)
Simplify: sqrt(1+.7071^2) = 3*sqrt((1/3)^2+(.943-.7071)^2)
Calculator: 1.224740956 = 3*0.408362487 = 1.225087462
It's within the three significant digits used. By the way, .7071 is an approximation of sqrt(2)/2.

Oh, and did you know that Google is a graphing calculator now? So if anyone's having trouble visualizing:

I humbly stand corrected.

 

[BladeVenom]

42

 

[Maximilian]

... idk

 

[DrPizza]

If the above diagrams are right, then the key is to draw an additional chord from the end if the chord to the other end of the first diagonal, making another right triangle that is similar to the one in question.

Now with solutiony goodness below

Triangle BDE is similar to ABC, therefore,
BE/BD = BC/AB, with BC=3/4*BD
Rearranging that, you end up with the ratio of AB to BC is 2/sqrt(6). The angle is ~35 degrees per the posts below

Thanks. That extra chord, with a neatly drawn picture made all the difference in the world. I was drawing it in my head and overlooked the right angle. From that point on, the rest follows easily. I knew it would be easy; I just kept making harder and harder by drawing all sorts of extra segments.

 

[piasabird]

A lot depends on where the perpendiculars intersecet? You didnt specifically say it was the center of the circle.