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

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...

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)

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 can do it in much simple way. Angle BOC is 60 degree. and BOC is equiletaral triangle. and Hence R= 6

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

Apply two times Laws of Cosine, square to eliminate the angle (Pythagorean Theorem and co-function identity) and you get a nice quartic

Better off keeping the final result in the form R = 3 * sqrt((5 + 2 * sqrt(2))/2),

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

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...

2 r^2 = a^2 + b^2 + ab root(2)

6( AOC=30 grade, COB= 60grade, rezulta COB triunghi echilateral, adica raza =6)

ببساطه bcg=,7 على قوس الدائره وr=5.6m والباقي أصبح سهلا

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)

You applied very critical rules and not define them briefly. First apply easy possible method then apply tough in alternate methods.

Cosine rule

Very ok

Love 🎉it