Please help with the math for a pyramid box

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Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 2:38 pm

So I have an isosceles triangle traced in a square. The tips of the triangle touches the edges of the square, as you can see in the image. I have the triangle's vertex angle and the sides of the square. I wish to know the measurement of the triangle's base. I'm giving two different cases. They only differ in the vertex angle.

I can't figure out how to solve this problem :oops:. Can you please help me? But please, don't give me the answer. Instead of that, explain to me how to solve it.

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After that, I have another math problem for my pyramid box. I hope I can count with your help with that one as well, but let's start with this one ;).

PS: Am I using "in" and "on" correctly?
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 3rd, 2019, 3:10 pm

My approach to solving this would be to divide the square in half diagonally. Use some basic trigonometry to find where the creases hit the bottom edge, and then calculate reference points for that number.


It seems as though a 75deg angle would be easier to find (since it’s a multiple of 15deg).
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Re: Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 3:39 pm

No idea :oops:. Would you please show me Baltorigamist how exactly is that done?
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Re: Please help with the math for a pyramid box

Postby Euthym » September 3rd, 2019, 3:44 pm

Here you know all the angles and a length, so you can use the sinus rule multiple times, but also Pythagoreans theorem once you know another lenght. We have a triangle ABC, with A=90°, B=10°, C=80°. Using the sinus rule we can calculate that the bottom of the left triangle is 1,76 cm. So the triangle on the right must have 2 lenghts of 10-1.76cm=8.24. Using the pythagorean theorem ce can calculate that y=11.65cm
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 3rd, 2019, 4:24 pm

Gerardo -> For the 75deg example, 90 - 75 = 15--and a 15deg angle is easy to construct. You'd simply mark out 1/2 of the bottom edge and fold through the upper left corner such that the bottom corner touches the 1/2 mark.
Note that this gives you a 75deg angle from the top edge--not centered along the main diagonal--but you just need to bisect the 15deg angle and repeat in mirror image on the top edge.

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Re: Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 5:44 pm

Euthym wrote:Here you know all the angles and a length, so you can use the sinus rule multiple times, but also Pythagoreans theorem once you know another lenght. We have a triangle ABC, with A=90°, B=10°, C=80°. Using the sinus rule we can calculate that the bottom of the left triangle is 1,76 cm. So the triangle on the right must have 2 lenghts of 10-1.76cm=8.24. Using the pythagorean theorem ce can calculate that y=11.65cm

Thank you very much Euthym!

I see the triangle you're referring to the bottom left one and the top right one, which are equal. I also have no problem using the Pythagorean theorem, but could you please explain to me how exactly do I apply the sinus rule in order to arrive to the 1.76 cm result? It's been a while since I I used the rule.


Baltorigamist wrote:Gerardo -> For the 75deg example, 90 - 75 = 15--and a 15deg angle is easy to construct. You'd simply mark out 1/2 of the bottom edge and fold through the upper left corner such that the bottom corner touches the 1/2 mark.
Note that this gives you a 75deg angle from the top edge--not centered along the main diagonal--but you just need to bisect the 15deg angle and repeat in mirror image on the top edge.

Oh, now I understand what you mean Baltorigamist :D. The thing is, I'm not asking for help folding the needed creases. I'm asking for help making out a measurement purely through calculations. OK?

Thank you for helping me out.
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 3rd, 2019, 6:14 pm

All you need to do is calculate the sine of the smaller angles (in degrees) of the triangles and multiply that by the side of the square.
For the 75deg example, (90-75)/2=7.5 degrees. Sin(7.5) is about .1305, which means the bottom left triangle has a base of .1305 times the side of the square, or 1.305cm. Subtracting that from the 10cm side length leaves us with 8.695cm for the short side(s) of the bottom right triangle. Multiply that by sqrt2, or ~1.41421, for the total hypotenuse length.

The 70deg example can be done by following the same steps (but using sin(10) instead of sin(7.5)).
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Re: Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 8:57 pm

Thank you both for your help :D!

And this is my second math problem in order to make the pyramid box.

There's a big isosceles triangle, it's divided between a small isosceles triangle and an isosceles trapezoid. I have the length of the small isosceles trapezoid's base and the base angles' measurements. Aside from that, I have the result from adding the length of the trapezoids two legs to its bottom base. With that information I must figure out the length of each one of the trapezoid's legs and of the trapezoid's bottom base.

Can you please explain to me how solve this problem?

Image


Baltorigamist wrote:All you need to do is calculate the sine of the smaller angles (in degrees) of the triangles and multiply that by the side of the square.
For the 75deg example, (90-75)/2=7.5 degrees. Sin(7.5) is about .1305, which means the bottom left triangle has a base of .1305 times the side of the square, or 1.305cm. Subtracting that from the 10cm side length leaves us with 8.695cm for the short side(s) of the bottom right triangle. Multiply that by sqrt2, or ~1.41421, for the total hypotenuse length.

The 70deg example can be done by following the same steps (but using sin(10) instead of sin(7.5)).

That helped a lot Baltorigamist! Now, I'm confused about something. Someone in some other website explained the exact same process, but using the tangent instead of the sine. How is it possible that two different functions work following exactly the same process :??
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 3rd, 2019, 9:08 pm

It may very well be the tangent instead of the sine—my trig is pretty rusty.
For the example in my post, the sine and tangent of 7.5deg are close enough that the different values are nearly indistinguishable when compared with a 10cm square.
The sine, as stated above, is roughly 0.1305, while the tangent of 7.5 degrees is about 0.1316. Even when multiplied by 10, it’s only a tenth of a millimeter difference.

The sine of an angle gives the y value of the angle’s endpoint on the unit circle, whereas the tangent gives the ratio of the x and y values.

EDIT: It is the tangent function that's needed, not the sine. The sine gives the y value on the unit circle, not the y value on x=1. My bad.
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Re: Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 9:51 pm

Thank you for clarifying that Baltorigamist :). Would you please help me with my second math problem when you have a moment?
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 3rd, 2019, 10:00 pm

Regarding your second question:
I'm using the first example.
Imagine a line down the center of the large triangle. The height of the resulting right triangle is ~1.428 times the length of the base (since tan(55deg)= ~1.428). This means that the small isosceles triangle (on top) is (1.428(2.1))cm tall--around 2.99cm.
Using this information, we can find the length of the top portion of the right triangle's hypotenuse by the Pythagorean Theorem: ((2.99)^2)+((2.1)^2)=~13.35. The hypotenuse of the smaller right triangle is (sqrt(13.35))cm, or ~3.65cm.

Now, you stated that the sides of the trapezoid plus its bottom base are 10cm in total. This means that one side plus half the bottom base is 5cm. (We can subtract 2.1cm from that 5. By drawing a vertical line down from the corner of the small right triangle, we can isolate the portion that we need to find and use proportions to solve for the rest.)
We already know that the smaller right triangle has proportions 1:1.428. After subtracting the 2.1cm from the base of the right triangle, we're left with a smaller (yet similar) right triangle whose base and hypotenuse need to be 2.9cm in total.
The base:hypotenuse ratio of the triangles is 2.1:3.65 (or about 1:1.74), so we need to divide 2.9 into that proportion. To do that, we add the two parts together and divide 2.9 by the result: 2.9/2.74=~1.058. Multiplying that by 1.74 gives the hypotenuse of the smaller triangle (a in your example) as 1.841cm. The value of b, then, is (2(1.058+2.1)), or a little less than 6.32cm.

It's a complicated problem, but I hope I explained the steps well enough. Let me know if you need any clarification.
Also, note that the final values might be off by a couple percent due to the way I rounded, but they should be close enough to work for what you need.
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Re: Please help with the math for a pyramid box

Postby Gerardo » September 3rd, 2019, 11:23 pm

You've been really helpful Baltorigamist. You really enjoy this kind of math problems, don't you :)?

I was able to understand everything up to here:

Baltorigamist wrote:The base:hypotenuse ratio of the triangles is 2.1:3.65 (or about 1:1.74), so we need to divide 2.9 into that proportion. To do that, we add the two parts together and divide 2.9 by the result: 2.9/2.74=~1.058. Multiplying that by 1.74 gives the hypotenuse of the smaller triangle (a in your example) as 1.841cm. The value of b, then, is (2(1.058+2.1)), or a little less than 6.32cm.

OK, I got that we are now concentrated on a small right triangle on the bottom right. After that, I don't understand at all what's going on :oops:. Can you please explain it a bit further?
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 4th, 2019, 12:08 am

Yes, I do. :)

The numbers I used in that step came from the first steps (using the Pythagorean Theorem along with tan(55)). Essentially, I'm finding two numbers in that proportion which sum to 2.9.
In order to do that, I first need to have the ratio in simplest terms (1:x), so I divide both sides by 2.1. Having the desired ratio in simplest terms means that dividing 2.9 by the sum of the proportions (in this case, 2.74) will give me one side of the proportion directly. I just need to multiply the result (1.058) by the rest of the proportion (subtracting it from 2.9 would also work) to find the second part of the ratio.

To check my work:
- 1.841+1.058=2.899, which can be rounded to 2.9
- 2(1.841) = 3.682; 3.682+6.32= 10.002, which can be rounded down to 10cm.

I hope this is a clear enough explanation.
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Re: Please help with the math for a pyramid box

Postby Gerardo » September 4th, 2019, 3:01 am

OK, I understood everything and was able to translate all this into a set of equations. Now, I don't get why is the following true:
Baltorigamist wrote:Having the desired ratio in simplest terms means that dividing 2.9 by the sum of the proportions (in this case, 2.74) will give me one side of the proportion directly


And also...
Baltorigamist wrote:I just need to multiply the result (1.058) by the rest of the proportion to find the second part of the ratio [the hypotenuse of the small right triangle on the bottom right].


Can you please argue these statements?
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Re: Please help with the math for a pyramid box

Postby Baltorigamist » September 4th, 2019, 9:35 pm

If the ratio is in simplest terms, I can more easily find the second part in terms of the first.
Say you wanted to divide a number (5, for instance) into a 3:4 proportion. You know that there has to be a total of 7 parts; three of them will be grouped into one part of the ratio, and the other four will make up the second part.
Dividing 5 by 7 directly means an extra step is involved: namely combining the groups of (5/7)s into 15/7 and 20/7. But if the ratio is in simplest terms (1:1 1/3), you just need to divide 5 by 2 1/3, which gives you 15/7 directly. Then just multiply that 15/7 by 1 1/3 to get 20/7.

In the end, it’s just another way of getting to the same answer. But it always helps to have things in simplest terms.
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