Find w

I have solved it but the proof of my solution is too dank to fit in the margin of this shitpost

>If you can't solve that in under 5 minutes
I hate sniveling competition, isn't it good enough that I can solve it?

there are so many things wrong with this, it has to be bait

prove it

It's about 6cm

There is no "w".

ayy

Say we call the height of the left triangle l, and that of the right triangle, r. Since we have the lengths of both hypotenuses, we can solve for l in terms of r, or r in terms of l.

Let's express the longer height L in terms of the shorter S and flip the picture so that the vertical leg of the triangle of shorter height is on the left. Lets call the bottom left corner the origin, so that the point where the 5m touches the x-axis is called (x, 0). So, now we have two linear equations in x. We'll call the height of this perpendicular line in the middle h.

1. h = L/w * x
2. h = -S/w * x + S

=> L/w * x = -S/w * x + S
=> (L + S)/w * x = S
=> x = w * S/(L + S)

Because L is expressed in terms of S, S/(L + S) reduces to some value a, 0 < a < 1.

We now know the position of the vertical leg as a percentage of w, and we know the lengths of the hypotenuses of the two big right triangles, we can now calculate the the values of the hypotenuses of the two little right triangles. This gives us the lengths of both the height and the hypotenuse for both of the little triangles. We simply have to calculate the bottom of both triangles and sum the lengths of the bottoms together to get w.

Here's a picture to illustrate what I mean.

what's the answer then

there is no possible solution
die and be damned

>2^3^3^3^3^3
>about 60

user, I...

Wrong

arccos(w/10) = arctan(5/x)
arccos(w/12) = arctan(5/(w-x))

solve

got a 4th degree polynomial

...

Simple Geometry
t. Hanzo

was it w^4 - 244w^2 + 13775?

Yes.

w can't exist.

such a configuration isn't possible.

any positive height < 5m is possible though.

Substitute w' = w2 and you'll have a nice quadratic to solve.

heh.

>7' 9 3/16''
My god, what is this mess?

Well done OP, I was about to call you a Nigger and a Jew, because I didn't think your pic was possible because you obviously shooped the 5m in.

w = 19.5256m

I've solved it numerically since I don't think there is an analytical solution, but I'm not sure since my math is a little rusty. You are still a faggot though.

w < 10

er, maybe..
are the full line lengths 10 and 12, or just the top parts?

Blue circle is 10.
Green circle is 12.
(0,0) is the center of both circles.
Red line is 5.

With grid. This seems to be the only solution for h=5 btw.

[eqn]w \approx 4.297328005 [/eqn]

workable

forgot puzzle lol

[eqn] 12^2 = a^2 + w^2 \\
10^2 = b^2 + w^2 \\
\frac{5}{a} + \frac{5}{b} = 1 [/eqn]

Solving this system gives

Forgot picture

This solution involves taking a new measurement that is conducive to calculating w

how'd you get the 5/a + 5/b = 1

[eqn] -w^8 + 388w^6 - 51736w^4 + 2610400w^2 = 32890000 [/eqn]

en.wikipedia.org/wiki/Intercept_theorem

Once used for both of the triangles.

I thought the upper segments were 12 and 10m, not the full lines

>doing someone's homework

nice

I don't think it can be correct with 5m there though just looking at it

ah nvm, it's probably ok

so we get wa/wb = sa/sb = xa/xb = a/b = 12/10

5/a = wb / w = sb / sb+12 = 10 / 10+sa

[eqn] \frac{5}{a} = \frac{w_a}{w} [/eqn]
[eqn] \frac{5}{b} = \frac{w_b}{w} = \frac{w - w_a}{w} [/eqn]
Add them together
[eqn] \frac{5}{a} + \frac{5}{b} = 1 [/eqn]

Do the measurements "12m" and "10m" refer to the larger line segments, or the smaller line segments?

If it refers to the smaller line segments, I think w has length 21m (approx). I spent way too long on calculating that answer. Label the upper-left acute angle x, and the upper-right acute angle y. By drawing a horizontal line at the top of the 5m segment, we can create some additional triangles, which allows us to find:

You can draw a few more triangles to get:
w = 5 tan y + 5 tan x

With all of that, you can also get:
w = 12 sin x + 10 sin y
12 sin x = 5 tan y
10 sin y = 5 tan x

Through a miserable amount of pain-in-the-ass basic trig and basic algebra, one eventually gets:
720(sin^4)(x) + (-551)(sin^2)(x) = 0

The relevant solution is:
(sin^2) x = 0.76527777777 (approx)

⇒ x = arcsin(sqrt(0.76527777777)) = 61.0214996 degree (approx)
⇒ y = arctan(12 / 5 * sin(x)) = 64.5316102 degree (approx)
⇒ w = 12 sin x + 10 sin y = 19.5258454256 (approx)

Double checking result:
⇒ y = arcsin(5 / 10 * tan(x)) = 64.5316102 degree (approx)

Double checking result:
⇒ w = 5 tan y + 5 tan x = 19.5258453784 (approx)

Looks right to me. I re-checked my work enough, so I think I didn't make a mistake, but goddamnit I haven't done this sort of basic math in years.

If you assume the "12m" and "10m" are the lengths of the larger line segments, then:
w = 12 sin x
w = 10 sin y

Draw a few more triangles, and you still get:
w = 5 tan y + 5 tan x

I know that narrows it down to an exact solution, but I don't know if you can solve that algebriacally. I'm getting a non-trivial 6th degree polynomial, which cannot be solved algebraically. I don't know if I'm missing some other way to solve that. As for numerical solutions, using Wolfram-Alpha, I get:
x = 0.366244 radians (approx)
y = 0.444197 radians (approx)

w = 12 sin x = 4.29733281528 (approx)

double checking work:
w = 10 sin y = 4.29732945233 (approx)
w = 5 tan y + 5 tan x = 4.29733145046 (approx)

>length 21m (approx).
Please ignore typo. I provide correct answer later in same post:
> 19.5258454256 (approx)

Did you make 5m 5 foot?

...

lel

Here he is!

Why does 5m look about the same size as 10m?

You don't expect these things to be drawn to scale, but come on, make it look somewhat realistic.

I couldn't do it.
What's the bad news?

found the communist

i cant stop laughing at this

What the hell are you talking about?

There's no unique solution. We need to know at least two angles subtending w and/or at least two other lengths or something. Correct me if I'm wrong

Kek it's impossible for both bottom line segment to be the same length as their top counterpart, unless the thick black lines are not perfectly vertical.

lol

Are you laughing at pic related or some error you think I made? I'm 100% sure this isn't solvable without more information

Well, you're 100% wrong. Look at my solutions above.

God you people are fucking brainlets.

It's trivial that following an analytic continuation, w is -1/12 units in length.

You're wrong though. Just try to manipulate the line segments in your head. Maintain the known measurements but increase the length of each line segment whose length is unknown. You will see that w changes as a result, and so we must conclude that without more information there is no solution

Don't know what you're talking about. The description is somewhat ambiguous whether the lengths refer to the larger or smaller line segments, but both problems are solvable.

Except he's right by your own admission. Why would you ask if they are the smallest or biggest line segments?

?
Because it's unclear if the "12" and "10" are labels on the smaller line segments, or the larger line segments.

I got approx. 19.6 but idk since I have autism.

Which would change the answer. Although not sure why he said angles.

This is what I mean

Except in w' the lines aren't bounded by the vertical thick black lines.

...

Damn, you're right. Guess I'm gonna shoot myself now ;_;

Neat