Small challenge

06062017, 08:05 AM
(This post was last modified: 06062017 09:44 AM by Pekis.)
Post: #1




Small challenge
Hello,
While making an Android app, I had to draw a small equilateral triangle pointing to a circle: r: Outer Circle Radius: distance OAB d: Side length of the equilateral triangle. d=a*r, where a is a fraction of the outer circle radius If we want the distance AB to be equal to 1/5 of r, what's the value of a ? Have fun & Thanks for reading 

06062017, 01:26 PM
Post: #2




RE: Small challenge
Nice one! I also have in the bookmark your "brain teaser 2" to solve.
Wikis are great, Contribute :) 

06072017, 09:53 AM
(This post was last modified: 06072017 09:57 AM by Pekis.)
Post: #3




RE: Small challenge
Hello,
Here is the solution: Distance AB = Distance AE + Distance EB Distance AE: d*cos(PI/6) =a*r*sqrt(3)/2 Distance EB: Arc height Formula r=h/2+c^2/(8*h), Here h=Distance EB and c=d => Solved for Distance EB=r*(1sqrt(4a^2)/2) Distance AB: a*r*sqrt(3)/2+r*(1sqrt(4a^2)/2), which must be equal to c*r (c=0.2 in the question) => Solve a*r*sqrt(3)/2+r*(1sqrt(4a^2)/2)=c*r => Solve a*sqrt(3)/2+1sqrt(4a^2)/2=c => Solved for a=(sqrt(3)*(c1)+sqrt((c+1)*(3c)))/2 If c=0.2 => a~=0.223694816 => d will be approx. 0.22*r while the distance AB will be 0.2*r as required If angle at B is t and (cx,cy) is the center of the outer circle with radius r: Coordinates of A: (Ax,Ay)=(cx+r*(1c)*cos t, cy+r*(1c)*sin t) Coordinates of C: (Cx,Cy)=(Ax+d*sin(PI/3t),Ay+d*cos(PI/3t)) Coordinates of D: (Dx,Dy)=(Ax+d*cos(tPI/6),Ay+d*sin(tPI/6)) Thanks for reading 

06072017, 11:35 AM
(This post was last modified: 06082017 12:54 PM by PedroLeiva.)
Post: #4




RE: Small challenge
(06072017 09:53 AM)Pekis Wrote: Hello,Can you provide a numerial example to test?. TYVM, Pedro 

06072017, 03:49 PM
(This post was last modified: 06072017 04:11 PM by Pekis.)
Post: #5




RE: Small challenge
(06072017 11:35 AM)PedroLeiva Wrote:(06072017 09:53 AM)Pekis Wrote: Hello,Can you provide a numerial exaple to test?. TYVM, Pedro Hello, Here is a numerical example: cex=0 (Outer Circle centered at (0,0)) cey=0 r=5 (Outer circle Radius) c=0.2 (Fraction of Outer circle radius left for the triangle) a=(sqrt(3)*(c1)+sqrt((c+1)*(3c)))/2=0.223694816 d=a*r=1.11847408 (Side length of equilateral triangle) t=40° (0.6981317008 rad) (Angle at B) Ax=cex+r*(1c)*cos(t)=3.064177772 Ay=cey+r*(1c)*sin(t)=2.571150439 Cx=Ax+d*sin(PI/3t)=3.446718437 Cy=Ay+d*cos(PI/3t)=3.622172279 Dx=Ax+d*cos(tPI/6)=4.165659718 Dy=Ay+d*sin(tPI/6))=2.765371425 Bx=cex+r*cos(t)=3.830222216 By=cey+r*sin(t)=3.213938048 Ex=cex+(r*(1c)+d*cos(PI/6))*cos(t)=3.806189078 Ey=cey+(r*(1c)+d*cos(PI/6))*sin(t)=3.193771851 Distance OA=sqrt((Axcex)^2+(Aycey)^2)=4 OK Distance OA=4=r*(1c) Distance AE=sqrt((AxEx)^2+(AyEy)^2)=0.9686269669 Distance EB=sqrt((ExBx)^2+(EyBy)^2)=0.03137303338 OK Distance AB=Distance AE+Distance EB=1=r*c Distance AC=sqrt((AxCx)^2+(AyCy)^2)=1.11847408 Distance CD=sqrt((CxDx)^2+(CyDy)^2)=1.11847408 Distance DA=sqrt((DxAx)^2+(DyAy)^2)=1.11847408 OK Distance AC=Distance CD=Distance DA=d => Equilateral CQFD 

06072017, 04:25 PM
Post: #6




RE: Small challenge
Or, a=(sqrt(7)2)*sqrt(3)/5


06072017, 04:31 PM
Post: #7




RE: Small challenge  
06072017, 08:42 PM
(This post was last modified: 06082017 02:24 AM by SlideRule.)
Post: #8




RE: Small challenge
I arrived at a slightly different results:
Δ ≈ 6.52° = interior angle formed by the line segments BOD (½ the central angle of the sector formed by COD) θ ≈ 23.58° = interior angle formed by line segments ADO D ≈ (0.118474, 0.993725) D ≈ (0.111847, 0.993725) typo correction verified by WolframAlpha & AnalyzeMath AD ≈ 0.22710 = d = a*R AD ≈ 0.22369 = d = a*R correction for typo [attachment=4914] [attachment=4915] The difference seems to be associated with the assumption guiding the calculation of the distance EB. I do not see how the Arc Segment S for the sector CAD with a central angle anchored to point A and the Arc Segment for the sector COD with a central angle anchored at point O can both describe the same arc segment CBD. Although the difference of the sagitta calculated this way may be small, in a manner similar to the small difference between the SIN & the TAN of a small angle, I take exception to the equality of the results, not the equivalency. Where am I going wrong? BEST! SlideRule 

06072017, 09:54 PM
Post: #9




RE: Small challenge
(06072017 08:42 PM)SlideRule Wrote: I arrived at a slightly different results: Hello, Don't forget that the points C and D belong to the outer circle with radius r. So, although they also belong to the triangle ACD, the formula I used to calculate the arc height seems perfectly valid. You could imagine a OCD triangle to apply the formula. What do you think ? Thanks 

06072017, 10:53 PM
(This post was last modified: 06082017 02:26 AM by SlideRule.)
Post: #10




RE: Small challenge
Pekis
I used the LineCircle intersect equations to calculate the coordinates of point D & since the other points can all be derived by rotating the line segment thru points O & B to align with the y axis, the rest is simple trigonometry / geometry using the distance formula, etc. I've attached a grahpmatica screen capture to illustrate the deviation in the two Arcs thru pts CD created by the two segments CAD (with central angle @ A) & COD (with central angle @ O). The approximate delta is 0.0235. [attachment=4916] BEST! SlideRule edit: found and corrected numeric typo  omitted a digit! 

06082017, 05:10 AM
(This post was last modified: 06082017 05:12 AM by Pekis.)
Post: #11




RE: Small challenge
(06072017 10:53 PM)SlideRule Wrote: Pekis But there is no arc with central angle at A ! The only existing arc is created by outer circle with radius r and points C and D, with central angle at O, and for which I used the arc height formula. Another completely different thing is the equilateral triangle ACD ... But I understand it can be confusing 

06082017, 12:12 PM
(This post was last modified: 06082017 12:13 PM by SlideRule.)
Post: #12




RE: Small challenge
Pekis
Talk about one mistake leading to another  I shouldn't do these challenges after midnight. I was looking for the reason underlying the difference in the magnitude of my solution and made a hasty & incorrect assumption with respect to the calculation of the sagitta, very embarrassing senior moment! I wonder, is there a third approach to the solution of the problem? BEST! SlideRule 

06082017, 12:59 PM
Post: #13




RE: Small challenge  
06092017, 07:08 AM
Post: #14




RE: Small challenge
Hello,
I also tried to complete the drawing of the equilateral triangle with the surface below the arc (transforming the equilateral triangle into a pie slice). It's then good to know that: Arc length=r*2*arcsin(a/2) Arc height=r*(1sqrt(4a^2)/2) Arc angle span=Arc lengh/r=2*arcsin(a/2) Arc starting angle (at C)=t+arcsin(a/2) Arc ending angle (at D)=tarcsin(a/2) Thanks to all ! 

06092017, 01:14 PM
(This post was last modified: 06092017 01:35 PM by Vtile.)
Post: #15




RE: Small challenge
(06082017 12:12 PM)SlideRule Wrote: PekisAt least it is nice to know that I'm not the only one writing gibberish at times while surfing at the midnight on MoHPC! I'll throw in another small and easy challenge ( especially since I give a picture. ). You are making cones out of cardboard, what are the measures of the cut out piece when it is then bend to cone and the dimensions for the cone are as follows. angle Alpha: 30 deg. R: 30 Hx: 5 The given values are random. 

06112017, 10:59 AM
Post: #16




RE: Small challenge
(06092017 01:14 PM)Vtile Wrote: I'll throw in another small and easy challenge ( especially since I give a picture. ). You are making cones out of cardboard, what are the measures of the cut out piece when it is then bend to cone and the dimensions for the cone are as follows. From these equations (\(\beta\) in radian): \(R_2 · \beta=2 · r · \pi\), and \((R_2+s) · \beta=2 · R · \pi\) we get: \(R_2=\frac{\frac{h_x}{cos \alpha}}{\frac{R}{Rh_x}1}\), and \(\beta=\frac{2 · (Rh_x) · \pi}{R_2}\) All the other simplification is your task Csaba 

06112017, 09:58 PM
(This post was last modified: 06122017 08:12 AM by Pekis.)
Post: #17




RE: Small challenge
Hello,
I just don't fully understand the figure in your cone challenge ... Anyway, I wanted to give an epilogue to my challenge with a generalized formula with a isosceles triangle (with one base and two equal sides) instead of an equilateral one: let c=Fraction of Outer circle radius left for the isosceles triangle let p=1c=Fraction of Outer circle radius for the figure inside the inner circle let b=Base length of the isosceles triangle let e=Fraction of the outer circle radius for base length => b=e*r let d=Side length of the isosceles triangle let a=Fraction of the outer circle radius for side length => d=a*r let k=a/e=Ratio between Side and Base of the isosceles triangle => a=(sqrt(4*k²p²)p*sqrt(4*k²1))/(2*k) It's good looking ... Arc length: r*2*arcsin(a/(2*k)) Arc height: r*sqrt(4(a/k)²)/2 Arc Angle span: 2*arcsin(a/(2*k)) Arc Start angle: t+arcsin(a/(2*k)) Arc End angle: tarcsin(a/(2*k)) And instead of the PI/6 angle in ACE, we now have Angle ACE=arcsin(1/(2*k)) For an equilateral triangle, k=1 and it leads to a=((sqrt(4p²)p*sqrt(3))/2 (same as previous formula a=(sqrt(3)*(c1)+sqrt((c+1)*(3c)))/2)) Thanks 

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