## Seemingly simple

For the discussion of math. Duh.

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KarenRei
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### Seemingly simple

Either I'm tired, an idiot, or this problem isn't as easy as it looks (solve for X; all corners that look like right angles are right angles): Can anyone enlighten me as to which one it is?

doogly
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### Re: Seemingly simple

Similar triangles woooooooo

the lil c-x-unlabeled is similar to the (a-x)-b-unlabeled triangle

so

C/x = B/sqrt(B^2 -(A-x)^2)

and then you can finesse this a bit and get some quadratic formula goin.
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KarenRei
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### Re: Seemingly simple

C/x = B/sqrt(B^2 -(A-x)^2)
C = x * B/sqrt(B^2 -(A-x)^2)
x * B/sqrt(B^2 -(A-x)^2) - C = 0

This isn't in the form Ax^2 + Bx + C = 0... and I don't remember how to get it in that form And Wolfram Alpha last night was just looping forever, while today it says "Standard computing time exceeded"

This is one of many different formulae I arrived at with different approaches... pythagorean theorum, trig, etc... but I never got anywhere. In case anyone is curious about the real world application, I'm modeling a roof that's going to have diagonal purlins (cross supports) coming from two angles and is going to need some custom brackets; B is the spacing between rafters, C is the purlin width, and A is half the distance between repeating patterns of purlins (the other half is for the purlin pointing in the opposite direction). I could just eyeball the connection in Blender, but then I'd worry about having a bunch of brackets that come out millimeters off and don't fit Pay attention in school, kids, or you'll end up like me jaap
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### Re: Seemingly simple

KarenRei wrote:C/x = B/sqrt(B^2 +(A-x)^2)
C = x * B/sqrt(B^2 +(A-x)^2)
x * B/sqrt(B^2 +(A-x)^2) - C = 0

This isn't in the form Ax^2 + Bx + C = 0... and I don't remember how to get it in that form ;)

C = x * B/sqrt(B^2 +(A-x)^2)
C * sqrt(B^2 +(A-x)^2) = x * B
Now square both sides of the equation.

Edit: Fixed the sign errors that Cauchy pointed out.
Last edited by jaap on Tue May 10, 2016 9:08 am UTC, edited 1 time in total.

Cauchy
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### Re: Seemingly simple

doogly wrote:C/x = B/sqrt(B^2 -(A-x)^2)

This should be C/x = B/sqrt(B^2 + (A-x)^2), since sqrt(B^2 + (A-x)^2) is the length of the hypotenuse of the large right triangle. Then,

C sqrt(B^2 + (A-x)^2) = B x
C^2 (B^2 + (A-x)^2) = B^2 x^2
B^2 C^2 + A^2 C^2 - 2 A C^2 x + C^2 x^2 = B^2 x^2
0 = (B^2 - C^2) x^2 + (2 A C^2) x + (- A^2 C^2 - B^2 C^2)

This is a quadratic equation in the more familiar style.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
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KarenRei
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### Re: Seemingly simple

Cauchy wrote:
doogly wrote:C/x = B/sqrt(B^2 -(A-x)^2)

This should be C/x = B/sqrt(B^2 + (A-x)^2), since sqrt(B^2 + (A-x)^2) is the length of the hypotenuse of the large right triangle. Then,

C sqrt(B^2 + (A-x)^2) = B x
C^2 (B^2 + (A-x)^2) = B^2 x^2
B^2 C^2 + A^2 C^2 - 2 A C^2 x + C^2 x^2 = B^2 x^2
0 = (B^2 - C^2) x^2 + (2 A C^2) x + (- A^2 C^2 - B^2 C^2)

This is a quadratic equation in the more familiar style.

Now that's more like it!

( -(2A C^2) +- sqrt( (2 A C^2)^2 - 4 (B^2 - C^2) (-A^2 C^2 - B^2 C^2) ) ) / (2 (B^2 - C^2) )
( -(2A C^2) +- 2 * sqrt( A^2 C^4 - (B^2 - C^2) (-A^2 C^2 - B^2 C^2) ) ) / (2 (B^2 - C^2) )
( -(A C^2) +- sqrt( A^2 C^4 - (B^2 - C^2) (-A^2 C^2 - B^2 C^2) ) ) / (B^2 - C^2)
( -(A C^2) +- sqrt( C^2 A^2 C^2 - (B^2 - C^2)(C^2)(-A^2 - B^2) ) ) / (B^2 - C^2)
( -(A C^2) +- C sqrt( A^2 C^2 - (B^2 - C^2)(-A^2 - B^2) ) ) / (B^2 - C^2)
( -(A C^2) +- C sqrt( A^2 C^2 + B^4 + B^2 A^2 - C^2 A^2 - C^2 B^2) ) / (B^2 - C^2)
( -(A C^2) +- C sqrt(B^4 + B^2 A^2 - C^2 B^2) ) / (B^2 - C^2)
( -(A C^2) +- B C sqrt(B^2 + A^2 - C^2) ) / (B^2 - C^2)
( (A C^2) +- B C sqrt(B^2 + A^2 - C^2) ) / (C^2 - B^2)

Look right? I'll punch in some numbers to check it out this evening...

Ed: Seems like the most natural form for this problem is:

(B*C*sqrt(A² + B² - C²) - A*C²) / (B² - C²)

f5r5e5d
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### Re: Seemingly simple

don't need to create equations, solve by hand when you can just draw in GeoGebra http://www.geogebra.org/ all done with the construction tools

gmalivuk
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### Re: Seemingly simple

f5r5e5d wrote:don't need to create equations, solve by hand when you can just draw in GeoGebra http://www.geogebra.org/
Can that give you exact values in terms of the other sides, though?
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f5r5e5d
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### Re: Seemingly simple

for engineering values of "exact", you can set numerical entry/display number of digits up to 15 digits

besides just dragging segment endpoints and watching the dynamically updated display of length on the right;
you can enter numerical pair defined points or vectors
or segments can be created with numerically entered lengths, drag and snap may work for axis aligned segments
or the center, radius circle takes a numerical entry for radius, intersecting the circles, base line, constructing lines through the intersection points letting you construct SSS triangles...

the construction protocol defines geometric relations, you can drag the construction points and the whole diagram updates, you can add sliders to control numeric inputs dynamically

the CAS (Computer Algebra System) Symbolic Math "view" lets you do the equations, solve quadratics - uses a weak, free Symbolic Math package but can get that far

gmalivuk
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### Re: Seemingly simple

So "no", in other words.

(Even if it had infinite precision, we also want it in terms of the known values.)
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f5r5e5d
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### Re: Seemingly simple

who is "we"? - for that matter the Euclidian metric is "wrong" too for something constructed on the Earth's surface...

I'm modeling a roof that's going to have diagonal purlins (cross supports) coming from two angles and is going to need some custom brackets; B is the spacing between rafters...

I think GeoGebra is worth pointing out to people wanting to play with middle/high school math again

doogly
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### Re: Seemingly simple

f5r5e5d wrote:who is "we"?

Anyone who has ever drawn a diagram with the sides looking like the one in the original post, ever. Especially when such a solution is possible.
LE4dGOLEM: What's a Doug?
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f5r5e5d
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### Re: Seemingly simple

good job encouraging contributions to this forum

gmalivuk
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### Re: Seemingly simple

It's a cool site, but it doesn't really answer the OP's question (which had variables rather than explicit values for the given side lengths), and that fact should be made clearer.
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Cauchy
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### Re: Seemingly simple

We found out in Post 3 that another question (maybe even the more pertinent one) was "I'm building a roof, and I'm afraid of being off a few millimeters if I try to eyeball the answer to this problem. Help?". That's a question which GeoGebra is perfectly suited to answering, so the dismissal of it for not producing an exact answer seems unjustified. You can pretty easily set it up so that it takes the inputs A, B, and C as two lengths of segments and the radius of a circle, and then has a final answer X for you at the end.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

Flumble
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### Re: Seemingly simple

Exact solution: douze points
GeoGebra: dix points

Sure it's not an analytic solver, but it's a nifty experimentation/construction tool that deserves mentioning. We already have the analytic solution anyway.
While toying with it, I found another construction that is parametrized in a, b and c (in the form of the distances AC, AB and BD), after observing that the (continuation of) the outer diagonal must also have a distance c to the top of the triangle (where b and x meet in the original image).

Or put differently, the outer diagonal is a tangent line of the circle around B passing through C. And then you're free to draw a line through B either parallel to the horizontal axis or parallel to the diagonal and get your distance from the newly appeared intersection point. On the top it neatly shows a secant with hypotenuse c, but judging by the earlier solutions, the corresponding adjacent side is nothing nice.

 adjusted capitalization to tell points A, B, C, D in this post's image apart from lengths a, b, c, x in the first post.
Last edited by Flumble on Thu May 12, 2016 1:32 pm UTC, edited 1 time in total.

PsiCubed
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### Re: Seemingly simple

A different way to reach the same equation:

The big shape is a right-angle trapezoid, which can be divided into a big right triangle and a parallelogram which rests on that triangle's hypotenuse.

The area of the trapezoid is:

[height] * [the average of the bases] = B*(A+x)/2 = BA/2 + Bx/2

The area of the parallelogram is:

[height] * [base] = C*sqrt[B2+(A-x)2]

And the area of the big triangle is:

[height] * [base] / 2 = B*(A-x)/2 = BA-xB/2

So:

BA/2 + Bx/2 = BA/2 - Bx/2 + C*sqrt[B2+(A-x)2]

Bx/2 = - Bx/2 + C*sqrt[B2+(A-x)2]

Bx = C*sqrt[B2+(A-x)2]

C = Bx/sqrt[B2+(A-x)2] Return to “Mathematics”

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