Embarrassing trigonometry problem

[FreshDK]

Hi Guys,

There maybe a specific forum for math problems somewhere, but I figure there must be enough brain power in here to solve this rather embarrassing case for me.

I actually got a engineering degree and consider myself skilled in the mathematical world, but 10 years of project management work is staring to leave its marks :D I spend more hours on this already that I would care to admit.

So, I got a circle with a known diameter. Inside that circle there is a smaller 6-sided polygon with same center as the circle. I need a equation to determinate the side length of the 6-sided polygon as a function of the circle diameter and "perpendicular corner offset" between the circle diameter and the 6-sided polygon.

I need S as a function of h and r referring to below (poor) illustration. Any who can solve this?

 

[6StringJazzer]

Is the hexagon a regular hexagon? It isn't in your drawing but maybe you intend it to be.

 

[FreshDK]

It is regular yes. All sides and angels are equal.

 

[6StringJazzer]

"perpendicular corner offset" between the circle diameter and the 6-sided polygon.

The circle diameter does not seem to be in play in any way here. It appears that h is a segment from a vertex drawn to the circle at right angles to a side.

I don't want to raise false hopes because this is not my specialty but it's interesting and I'll give it a try.

 

[FreshDK]

The circle diameter does not seem to be in play in any way here. It appears that h is a segment from a vertex drawn to the circle at right angles to a side.

I don't want to raise false hopes because this is not my specialty but it's interesting and I'll give it a try.

Wrong phrase by me, sorry! English is not my native language

I meant circle circumference. I hope the illustration shows the problem when you set aside the inaccuracies from drawing by hand in mspaint

 

[6StringJazzer]

I gave this my best shot and I'm just not going to get it. Not embarrassing at all. This is a tough one.

 

[Anthony47]

Good, old trigonometry…
A test workboog can be downloaded from here: ForFreshDK.xlsm

The coordinates of the starting point of H are 0,S
The equation of the line that your segment belongs to is
Y=2*x + S

The equation of the circle is
X^2 + y^2 = R^2

To get the intersection(s) of the line with the circle we need to resolve the following system of equations:
X^2+Y^2 = R^2
Y = 2*X +S

The next step is evolving to X^2 + (2*x + S)^2 = R^2
And thus 5*X^2 + 4S*X + (S^2 - R^2) = 0

Since this is a second degree equation, X is resolved as:
X= [(-b +/- (b^2 - 4ac)^(0.5)]/2a

That is:
Xi1= {-4S + [(16 S^2 - 20 (S^2 - R^2)]^0.5}/10 and
Xi2 = {-4S - [(16 S^2 - 20 (S^2 - R^2)]^0.5}/10

For the given problem we only need to consider the solution where X>=0, and this can only be Xi1

At that point we may calculate Yi:
Yi = 2*Xi + S

So the coordinates of the intersection are X1, Yi
The coordinates of the starting point of h are 0,S

Now the distance H between Xi,Yi and 0,S
H = [(Xi-0)^2 + (Yi-S)^2]^0.5

In formulas:
Radius is in B2
S is in B3

In B6, the first Xi using

=(-4*B3 + (16*(B3)^2 -20*(B3^2-B2^2))^0.5)/10

In B7 the second Xi using

=(-4*B3 - (16*(B3)^2 -20*(B3^2-B2^2))^0.5)/10

In B10 the valid Xi using

=MAX(B6:B7)

In B12 the correspondig Yi using

=2*B10+B3

Finally in B14 the Distance (h) using

=(B10^2+(B12-B3)^2)^0.5

But I didn't forget you was looking for S, as a function of R and H, and not H as a function of R and S

After having tryed several times to reverse the formulas to get S as a function of R and h, having obtained in each of the exercise always a wrong result I decided to do this job using a macro, that evolves S (ie range B3) and compare the calculated Distance (h) against the set value .
The macro that I used:

Sub SearchS()
Dim dI As Double, Radius As Long, lS As Double, I As Long
Dim Loops As Integer, S As Double, sDist As Double
'
Radius = Range("B2").Value
sDist = Abs(Range("B17").Value)
If sDist > Radius Then Beep: Exit Sub
Loops = 15
'
dI = Radius / 10
For I = 1 To Loops
    For S = 0 To 20
        lS = lS + dI
        Range("b3").Value = lS
        If Abs(Range("B14") - sDist) < 0.000000001 Then
            Exit Sub
        End If
        If I Mod 2 = 1 Then
            If Range("B14") < sDist Then
                dI = dI / 10 * (-1)
                Exit For
            End If
        Else
            If Range("B14") > sDist Then
                dI = dI / 10 * (-1)
                Exit For
            End If
        End If
    Next S
DoEvents
Next I
End Sub

I guess that the Excel internal Solver could also be used instead of the macro, but I preferred the risky path

The cells in yellow have to be filled with Radius and Distance H
The cells in Orage will show the searched S and the corresponding H

Hope this make some sense

Bye

 

[Pilgrim_Paul]

S=0.5*(SQRT(4*r^2-h^2) - SQRT(3)*h)

 

[johnnyL]

This thread is making my head hurt.

 

[BSALV]

driehoeken.xlsm
	A	B	C	D	E	F	G	H	I	J	K	L	M
1	graden	radialen	x	y				1		0,60
2	0	0	1	0	B	edges (degrees)	Edges (Rad)	X (1)	Y (1)	X (0,6)	Y (0,6)
3	1	0,017453	0,999848	0,017452		30	0,52	0,87	0,50	0,52	0,30
4	2	0,034907	0,999391	0,034899		90	1,57	0,00	1,00	0,00	0,60
5	3	0,05236	0,99863	0,052336		150	2,62	-0,87	0,50	-0,52	0,30
6	4	0,069813	0,997564	0,069756		210	3,67	-0,87	-0,50	-0,52	-0,30
7	5	0,087266	0,996195	0,087156		270	4,71	0,00	-1,00	0,00	-0,60
8	6	0,10472	0,994522	0,104528		330	5,76	0,87	-0,50	0,52	-0,30
9	7	0,122173	0,992546	0,121869		390	6,81	0,87	0,50	0,52	0,30
10	8	0,139626	0,990268	0,139173
11	9	0,15708	0,987688	0,156434
12	10	0,174533	0,984808	0,173648
13	11	0,191986	0,981627	0,190809
14	12	0,20944	0,978148	0,207912		A	0	0
15	13	0,226893	0,97437	0,224951		B	0,5196	0,3
16	14	0,244346	0,970296	0,241922		C	0,9578	0,3
17	15	0,261799	0,965926	0,258819		D	0,9578	0
18	16	0,279253	0,961262	0,275637		A	0	0
19	17	0,296706	0,956305	0,292372		B0	0,5196	0
20	18	0,314159	0,951057	0,309017		C	0,9578	0,3
21	19	0,331613	0,945519	0,325568		A	0	0
22	20	0,349066	0,939693	0,34202
23	21	0,366519	0,93358	0,358368
Blad1

Cell Formulas
Range	Formula
A2:A362	A2	=SEQUENCE(361,,0)
B2:B23,G3:G9	B2	=PI()*A2/180
C2:C23	C2	=1*COS(B2)
D2:D23	D2	=1*SIN(B2)
J2	J2	="X (" &J1 & ")"
K2	K2	="Y (" &J1 & ")"
F3:F9	F3	=SEQUENCE(7,,30,60)
H3:H9	H3	=$H$1*COS(G3)
I3:I9	I3	=$H$1*SIN(G3)
J3:K9	J3	=+H3*$J$1
G15:H15	H15	=+K3
H16,G17	H16	=+H15
G16	G16	=COS(ATAN(H16))
G19,G20:H20	G19	=+G15
Dynamic array formulas.