Oops. Sorry about that. In my previous comment, I said that this problem represented a regular hexagon inscribed into a circle, but on closer inspection... it isn't! The shorter line segment of three units if extended would pass right out of the area of the circle. It is at the wrong angle to be another side of a hexagon. That means that the radius must be slightly shorter than six and trigonometry actually is the best way to solve for it. I was much too hasty on my assumption that the figure described part of a hexagon in a circle. Thanks for the kind comments from those who supported me, but I was wrong.
You can easily construct this on graph paper, by slightly rotating the figure clockwise, such that the side BC with length six is oriented vertically: * draw BC = 6 cm with midpoint D * draw BA in 135° degrees with 3 cm. * construct the midpoint M of the circle with two perpendicular bisectors of BC and BA. Now it's easily seen that MD = 3 + 1.5*sqrt(2) ~ 5.1213... And the radius MC = sqrt((3 + 1.5*sqrt(2) )^2 + 3^2) ~ 5.935...
This is not correct. Chord length is not linearly proportional to central angle size. So just because the chords are 6 and 3, that doesn't mean that the angles will be 60 and 30 degrees, adding up to 90. Consider a circle with radius R. A 60 degree central angle yields a chord of length R. However, a 90 degree central angle yields a chord of length R^(1/2). The angle is 1.5 times bigger, but the chord is only about 1.414 times bigger.
I enjoy your videos and I enjoy your math because it is very well presented, and very detailed. I appreciate the detail. Sometimes I like to jump ahead in math problems and skip the math if I don't think it's necessary. I did that with this problem. I came up with an answer of 6 for the radius because looking at the construction it looked to be part of a regular hexagon inscribed within a circle and if the sides of the hexagon are six then the radius of the circle is six. So when I saw this problem today I skipped the math and just went to six as my answer. However, your mathematics points out that 6 is only approximate. That surprised me because I thought that since the pentagram is subscribed within a circle, 6 would have been a little more precise. Anyway, I appreciate the detail that you go into and it certainly adds a lot of insight. Thanks for everything you are doing.
i solved this with a repeated calculation considering the proportions of the graphics. it can also be done solving 2 equations with 2 unknown numbers r and angle of r pointing from 0 to point c: 10 print "mathbooster-a nice geometry problem from germany":dim x(1,2),y(1,2) 20 l1=3:l2=6:rg=1:sw=rg^2/(rg+l1+l2):xc=sw:xa=0:ya=rg:xb=rg:yb=0:goto 50:rem ein gewaehlter radius 30 yc=sqr(rg^2-xc^2):l1r=sqr((xa-xc)^2+(ya-yc)^2):l2r=sqr((xc-xb)^2+(yc-yb)^2) 40 dgu1=l1r/l2r:dgu2=l1/l2:dg=dgu1-dgu2:return 50 gosub 30 60 dg1=dg:xc1=xc:xc=xc+sw:xc2=xc:gosub 30:if dg1*dg>0 then 60 70 xc=(xc1+xc2)/2:gosub 30:if dg1*dg>0 then xc1=xc else xc2=xc 80 if abs(dg)>1E-10 then 70 90 r=l2/l2r:xc=xc*l2/l2r:yc=yc*l2/l2r:ya=r:xb=r 100 x(0,0)=0:y(0,0)=0:x(0,1)=r:y(0,1)=0:x(0,2)=xc:y(0,2)=yc 110 x(1,0)=0:y(1,0)=0:x(1,1)=xc:y(1,1)=yc:x(1,2)=0:y(1,2)=r 120 print "der radius= ";r:mass=8E2/r:goto 140 130 xbu=x*mass:ybu=y*mass:return 140 for a=0 to 1:gcol8+a:x=x(a,0):y=y(a,0):gosub 130:xba=xbu:yba=ybu:for b=1 to 3 150 ib=b:if ib=3 then ib=0 160 x=x(a,ib):y=y(a,ib):gosub 130:xbn=xbu:ybn=ybu:goto 180 170 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 180 gosub 170:next b:next a:gcol8:xba=0:yba=0:gosub 130:circle xba,yba,r*mass mathbooster-a nice geometry problem from germany der radius= 5.93531145 > run in bbc basic sdl and hit ctrl tab to copy from the results window
Nice Geometry provlem , but you use Trigonometry . You can avoid Trigonometry by using Generalisation of Pythagoras Theorem. AB²=AC²+BC²+2AC⋅CD (*) (R√2)²=3²+6²+ 2⋅3 ⋅ 3√2 …………….. R=√(22.5+9⋅√2) (*) Cause Δ BCD is orthogonal and Equilateral triangle. So CD²+BD²=BC ² => 2CD²=6² => CD=3√2
I also found a non trigonometric solution, that isn't better, in this way: drop perpendiculars from C to OB and to OA , call them CD and CE setting CD = x, CE = y, radius = R we can write 3 pythagorean identities: 1) OC² = CD² + CE² R² = x² + y² 2) BC² = CD² + DB² 6² = x² + (R - y)² 3) AC² = CE² + AE² 3² = y² + (R - x)²
Solution without trigonometry:
Using coordinate geometry, with O at origin and quarter circle in first quadrant:
O = (0, 0) A = (0, r) B = (r, 0) C = (x, y)
AC = 3 → (x−0)² + (y−r)² = 3² → x² + y² − 2ry + r² = 9 . . . . (1)
BC = 6 → (x−r)² + (y−0)² = 3² → x² − 2rx + r² + y² = 36 . . . (2)
OC = r → (x−0)² + (y−0)² = r² → x² + y² = r² . . . . . . . . . . . . (3)
(3)−(1) 2ry − r² = r² − 9 → 2ry = 2r² − 9 → y = (2r² − 9) / (2r)
(3)−(2) 2rx − r² = r² − 36 → 2rx = 2r² − 36 → x = (2r² − 36) / (2r)
Plug these values of x and y into (3)
[(2r² − 36) / (2r)]² + [(2r² − 9) / (2r)]² = r²
(4r⁴ − 144r² + 1296) / (4r²) + (4r⁴ − 36r² + 81) / (4r²) = r²
4r⁴ − 144r² + 1296 + 4r⁴ − 36r² + 81 = 4r⁴
4r⁴ − 180r² + 1377 = 0
r² = (180 ± √(180² − 4·4·1377)) / (2·4)
r² = (180 ± 72√2) / (2·4)
r² = (45 ± 18√2) / 2
Note: C(x,y) is in first quadrant, therefore x = (2r² − 36) / (2r) > 0 → r² > 18
r² = (45 + 18√2) / 2 = 9 (5 + 2√2) / 2
*r = 3 √[(5+2√2)/2] ≈ 5.9353*
6:34 As you quoted Pythagoras you should quote *Al Kashi* for this theorem
Oops. Sorry about that. In my previous comment, I said that this problem represented a regular hexagon inscribed into a circle, but on closer inspection... it isn't! The shorter line segment of three units if extended would pass right out of the area of the circle. It is at the wrong angle to be another side of a hexagon.
That means that the radius must be slightly shorter than six and trigonometry actually is the best way to solve for it.
I was much too hasty on my assumption that the figure described part of a hexagon in a circle. Thanks for the kind comments from those who supported me, but I was wrong.
Another possibility:
In triangle AOC AC^2 = OA^2 + OC^2 -2.OA.OC.cos(angle AOC), so 9 = 2.R^2 - 2.R^2.cos(t) = 2.R^2. (1 -cos(t)), with t = angleAOC
In triangle BOC, by the same way: 36 = 2.R^2.( 1- cos(90° -t) , so 36 = 2.R^2.(1 -sin(t)).
Let's divide these two equations. We get: 1/4 = (1 -cos(t))/(1 -sint), and so 4 - 4.cos(t) = 1 -sin(t) or cos(t) = (3 + sin(t))/4
Knowing cos(t)^2 + sin(t)^2 = 1, we get: (9 + 6.sin(t) +sin(t)^2)/16 + sin(t)^2 = 1, or 17.sin(t)^2 +6.sin(t) -7 = 0
Deltaprime = (3)^2 -17.(-7) = 128 = (8.sqrt(2))^2, and so sin(t) = (-3 - 8.sqrt(2))/17 which is rejected as negative,
or sin(t) =(-3 -8.sqrt(2))/17. Then we get: 36 = 2.R^2.(1- (-3 +8.sqrt(2))/17) = 2.R^2.(20 -8.sqrt(2))/17
Then: R^2 = (36.17)/ 2.(20 -8.sqrt(2)) = (9.17)/(10 -4.sqrt(2)) = (9.17)(10+4.sqrt(2)))/68 = 9.(5 +2.sqrt(2))/2 = (45/2) +9.sqrt(2)
You can easily construct this on graph paper, by slightly rotating the figure clockwise, such that the side BC with length six is oriented vertically:
* draw BC = 6 cm with midpoint D
* draw BA in 135° degrees with 3 cm.
* construct the midpoint M of the circle with two perpendicular bisectors of BC and BA.
Now it's easily seen that MD = 3 + 1.5*sqrt(2) ~ 5.1213...
And the radius MC = sqrt((3 + 1.5*sqrt(2) )^2 + 3^2) ~ 5.935...
you can do it in much simple way. Angle BOC is 60 degree. and BOC is equiletaral triangle. and Hence R= 6
Equal chords subtend equal angles at centre , so chord with 3 will subtend half the angle at centre than chord with 6 .
This is not correct. Chord length is not linearly proportional to central angle size. So just because the chords are 6 and 3, that doesn't mean that the angles will be 60 and 30 degrees, adding up to 90.
Consider a circle with radius R. A 60 degree central angle yields a chord of length R. However, a 90 degree central angle yields a chord of length R^(1/2). The angle is 1.5 times bigger, but the chord is only about 1.414 times bigger.
@@subtlethingsinlife It's true that equal chords subtend equal angles at center, but chord length is not linearly proportional to central angle size.
I enjoy your videos and I enjoy your math because it is very well presented, and very detailed. I appreciate the detail. Sometimes I like to jump ahead in math problems and skip the math if I don't think it's necessary.
I did that with this problem. I came up with an answer of 6 for the radius because looking at the construction it looked to be part of a regular hexagon inscribed within a circle and if the sides of the hexagon are six then the radius of the circle is six. So when I saw this problem today I skipped the math and just went to six as my answer.
However, your mathematics points out that 6 is only approximate. That surprised me because I thought that since the pentagram is subscribed within a circle, 6 would have been a little more precise.
Anyway, I appreciate the detail that you go into and it certainly adds a lot of insight.
Thanks for everything you are doing.
You're very kind. Of course it is 6 units.
@@philippeganty Yeah... I just didn't want to be a jerk about it.
2 r^2 = a^2 + b^2 + ab root(2)
Better off keeping the final result in the form R = 3 * sqrt((5 + 2 * sqrt(2))/2),
Apply two times Laws of Cosine, square to eliminate the angle (Pythagorean Theorem and co-function identity) and you get a nice quartic
i solved this with a repeated calculation considering the proportions of the graphics. it can also be done solving 2 equations with 2 unknown numbers r and angle of r pointing from 0 to point c:
10 print "mathbooster-a nice geometry problem from germany":dim x(1,2),y(1,2)
20 l1=3:l2=6:rg=1:sw=rg^2/(rg+l1+l2):xc=sw:xa=0:ya=rg:xb=rg:yb=0:goto 50:rem ein gewaehlter radius
30 yc=sqr(rg^2-xc^2):l1r=sqr((xa-xc)^2+(ya-yc)^2):l2r=sqr((xc-xb)^2+(yc-yb)^2)
40 dgu1=l1r/l2r:dgu2=l1/l2:dg=dgu1-dgu2:return
50 gosub 30
60 dg1=dg:xc1=xc:xc=xc+sw:xc2=xc:gosub 30:if dg1*dg>0 then 60
70 xc=(xc1+xc2)/2:gosub 30:if dg1*dg>0 then xc1=xc else xc2=xc
80 if abs(dg)>1E-10 then 70
90 r=l2/l2r:xc=xc*l2/l2r:yc=yc*l2/l2r:ya=r:xb=r
100 x(0,0)=0:y(0,0)=0:x(0,1)=r:y(0,1)=0:x(0,2)=xc:y(0,2)=yc
110 x(1,0)=0:y(1,0)=0:x(1,1)=xc:y(1,1)=yc:x(1,2)=0:y(1,2)=r
120 print "der radius= ";r:mass=8E2/r:goto 140
130 xbu=x*mass:ybu=y*mass:return
140 for a=0 to 1:gcol8+a:x=x(a,0):y=y(a,0):gosub 130:xba=xbu:yba=ybu:for b=1 to 3
150 ib=b:if ib=3 then ib=0
160 x=x(a,ib):y=y(a,ib):gosub 130:xbn=xbu:ybn=ybu:goto 180
170 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
180 gosub 170:next b:next a:gcol8:xba=0:yba=0:gosub 130:circle xba,yba,r*mass
mathbooster-a nice geometry problem from germany
der radius= 5.93531145
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
Nice Geometry provlem , but you use Trigonometry . You can avoid Trigonometry by using Generalisation of Pythagoras Theorem.
AB²=AC²+BC²+2AC⋅CD (*)
(R√2)²=3²+6²+ 2⋅3 ⋅ 3√2
…………….. R=√(22.5+9⋅√2)
(*) Cause Δ BCD is orthogonal and Equilateral triangle. So CD²+BD²=BC ²
=> 2CD²=6² => CD=3√2
ببساطه bcg=,7 على قوس الدائره وr=5.6m والباقي أصبح سهلا
6( AOC=30 grade, COB= 60grade, rezulta COB triunghi echilateral, adica raza =6)
I also found a non trigonometric solution, that isn't better, in this way:
drop perpendiculars from C to OB and to OA , call them CD and CE
setting CD = x, CE = y, radius = R
we can write 3 pythagorean identities:
1) OC² = CD² + CE²
R² = x² + y²
2) BC² = CD² + DB²
6² = x² + (R - y)²
3) AC² = CE² + AE²
3² = y² + (R - x)²
r = (3√2/2)√(5 + 2√2); AOC = ϑ → cos(ϑ) = (2/17)(6 + √2)
Another easy method
Split 2 triangles of base 6 & 3
Apex angles be c and 90 -c for base 6. & 3
2R*sin((90-c)/2)/2R*sin(c/2)= 3/6
2R cancelled
sin(45-c/2)/sin(c/2)= 1/2
sin 45 * cos(c/2) - cos 45*sin(c/2)/ sin(c/2) = 1/2
1/√2 (cos(c/2)- sin(c/2))/sin(c/2) =1/2
cot(c/2) -1= 1/√2
cot(c/2) = 1/√2+1 = (1+√2)/√2
Making right angled triangle with base (1+√2), height √2, hypotenuse= √[(1+√2)^2+(√2)^2] =√(5+2√2)
sin(c/2) = √2/√(5+2√2)
2R sin(c/2)= 6
R= 6/(2 * sin(c/2))
R= 3/(√2/√(5+2√2))= 3√(5+2√2)/√2
R=5.935...
Cosine rule
Very ok
You applied very critical rules and not define them briefly. First apply easy possible method then apply tough in alternate methods.
Love 🎉it