@@imetroangola17 Fico muito feliz que você tenha gostado deste vídeo. Espero que continue me apoiando assim, isso me inspira a criar vídeos ainda melhores. Obrigado!
@@lesliederrar9718 Thank you so much for your kind words! I'm glad you enjoyed the video. Your support means a lot to me and motivates me to create more smart and helpful content. Stay tuned for more exciting videos!
@@powerofSanatandharma Thank you so much for your kind words! Your support means the world to me, and it's because of wonderful viewers like you that I’m inspired to keep creating. I’m so grateful that the video resonated with you. Wishing you lots of learning and joy ahead! 🙏💖
As we are given the ratio of BH to HA as 10:3, we know that the actual length of BH and HA will be 10k and 3k, with k being an as-yet unknown constant. As BH and FB are tangents to circle O that intersect at B, then by the two tangents theorem, FB = BH = 10k. Similarly, as HA and AD are tangents to circle O that intersect at A so AD = HA = 3k. As AC and CB are tangent to circle O at D and F respectively, then ∠ODC = ∠CFO = 90°, as OD and OF are radii of circle O. Since ∠DCF = 90° as well, then ∠FOD must equal 360°-3(90°) = 90°, and as OF and OD are equal in length and adjacent, quadrilateral ODCF is a square with side length of 2. The side lengths of triangle ∆ACB are thus as follows: BA = 10k+3k = 13k, AC = 3k+2, CB = 2+10k. Triangle ∆ACB: AC² + CB² = BA² (3k+2)² + (10k+2)² = (13k)² 9k² + 12k + 4 + 100k² + 40k + 4 = 169k² 109k² + 52k + 8 = 169k² 60k² - 52k - 8 = 0 15k² - 13k - 2 = 0 15k² - 15k + 2k - 2 = 0 15k(k-1) + 2(k-1) = 0 (k-1)(15k+2) = 0 k - 1 = 0 | 15k + 2 = 0 k = 1 | k = -2/15 ❌ k ≥ 0 BA = 13(1) = 13 AC = 3(1) + 2 = 5 CB = 2 + 10(1) = 12 As CB is tangent to circle Q at G, then ∠QGB = 90°, as QG is a radius of circle Q. As AC is tangent to circle P at E, then ∠AEP = 90°, as PE is a radius of circle P. As PE and OD are both perpendicular to AC and thus parallel to each other, then ∠EPA and ∠DOA are corresponding angles and thus congruent. As ∠PAE = ∠OAD, then ∆AEP and ∆ADO are similar triangles by angle-angle similarity. As the point of tangency between two circles is collinear with the two centers, then OP = 2+r₁. Triangle ∆ADO: AD² + OD² = OA² 3² + 2² = OA² OA² = 9 + 4 = 13 OA = √13 OD/OA = PE/PA 2/√13 = r₁/(√13-(2+r₁)) √13r₁ = 2(√13-2-r₁) √13r₁ = 2√13 - 4 - 2r₁ √13r₁ + 2r₁ = 2√13 - 4 r₁(√13+2) = 2√13 - 4 r₁ = (2√13-4)/(√13+2) r₁ = (2√13-4)(√13-2)/(√13+2)(√13-2) r₁ = (26-4√13-4√13+8)/(13-4) [ r₁ = (34-8√13)/9 ≈ 0.573cm ] As OF and QG are both perpendicular to CB and thus parallel to each other, then ∠BQG and ∠BOF are corresponding angles and thus congruent. As ∠GBQ = ∠FBO, then ∆QGB and ∆OFB are similar triangles by angle-angle similarity. As the point of tangency between two circles is collinear with the two centers, then OQ = 2+r₂. Triangle ∆OFB: OF² + FB² = OB² 2² + 10² = OB² OB² = 4 + 100 = 104 OB = √104 = 2√26 OF/OB = QG/QB 2/2√26 = r₂/(2√26-(2+r₂)) 1/√26 = r₂/(2√26-2-r₂) √26r₂ = 2√26 - 2 - r₂ √26r₂ + r₂ = 2√26 - 2 r₂(√26+1) = 2√26 - 2 r₂ = (2√26-2)/(√26+1) r₂ = (2√26-2)(√26-1)/(√26+1)(√26-1) r₂ = (52-2√26-2√26+2)/(26-1) [ r₂ = (54-4√26)/25 ≈ 1.344cm ]
@@quigonkenny Thank you so much for sharing your solution! I really appreciate the effort you've put in, and it's always interesting to see different approaches to solving a problem. It's great that we both arrived at the correct answer through different methods-this shows the beauty of math, where there can be multiple ways to reach the same conclusion. Keep up the great work, and feel free to share more of your ideas!
"Great video! The way you explained the geometric concepts was so clear and easy to understand. The visual aids and examples really helped solidify my understanding. Keep up the fantastic work!"
@@DhanushJoshi-u1r Thank you so much for your kind words and for liking the video! I'm really glad you enjoyed it. Your support means a lot to me and motivates me to keep creating helpful content. Stay tuned for more...
@@VirendraKumar-z4u Thank you for your support, which I received in the form of a comment. I hope it continues to inspire me to create even better videos.
yes it is tricky but can be solved. the repeated calculation changes the centerpoint of the circle with its known radius until the quotient "fl=" as written in line 20, is reached: 10 print "quant circle-how to solve this tricky geometry problem with 3 circles" 20 dim x(1,2),y(1,2),r(2):r(0)=2:sw=r(0)/10:fl=10/3:y(1,0)=r(0):ym=r(0):y(1,0)=ym:nr=30 30 nr=30 :goto 150 40 ys=y(1,0)+sqr(abs(r(0)^2-(xs-xm)^2)):dgu1=(xs^2+ys^2)/r(0)^2:dgu2=xs*xm/r(0)^2 50 dgu3=ys*ym/r(0)^2:dg=dgu1-dgu2-dgu3:return 60 xm=r(0)+sw:gosub 40 70 dg1=dg:xm1=xm:xm=xm+sw:if xm>20*r(0) then stop 80 xm2=xm:gosub 40:if dg1*dg>0 then 70 90 xm=(xm1+xm2)/2:gosub 40:if dg1*dg>0 then xm1=xm else xm2=xm 100 if abs(dg)>1E-10 then 90 110 return 120 gosub 60:ye=ys/xs*(xm+r(0)):lbh=sqr(xs^2+ys^2):lah=sqr((xm+r(0)-xs)^2+(ye-ys)^2) 130 lac=ye:df=lbh/lah: df=df-fl:return 140 return 150 xs=sw:gosub 120 160 xs1=xs:df1=df:xs=xs+sw:if xs>nr*r(0) then stop 170 xs2=xs:gosub 120:if df1*df>0 then 160 180 xs=(xs1+xs2)/2:gosub 120:if df1*df>0 then xs1=xs else xs2=xs 190 if abs(df)>1E-10 then 180 200 lbc=sqr((lbh+lah)^2-lac^2):lab=lbh+lah:print "lbh=";lbh:print"lah=";lah:print "lab=";lab 210 print "lbc=";lbc;"lac";lac: print xs,"%",ys 220 x(0,0)=0:y(0,0)=0:x(0,1)=lbc:x(0,2)=x(0,1):y(0,2)=ye:x(1,0)=xm 230 xmu=sw:goto 260:rem den berechneten mittelpunkt verwenden *** 240 ymu=xmu/xm*ym:ru=ymu:dgu1=(xm-xmu)^2/r(0)^2:dgu2=(ym-ymu)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2 250 dg=dgu1+dgu2-dgu3 :return 260 gosub 240 270 dg1=dg:xmu1=xmu:xmu=xmu+sw:xmu2=xmu:gosub 240:if dg1*dg>0 then 270 280 xmu=(xmu1+xmu2)/2:gosub 240:if dg1*dg>0 then xmu1=xmu else xmu2=xmu 290 if abs(dg)>1E-10 then 280 300 x(1,1)=xmu:y(1,1)=ymu:r(1)=ru:dx=xm-lbc:dy=ym-ye:k=sw:goto 330 310 dxk=k*dx:dyk=k*dy:xmu=lbc+dxk:ymu=ye+dyk:ru=lbc-xmu:dgu1=(xmu-xm)^2/r(0)^2 320 dgu2=(ymu-ym)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2:dg=dgu1+dgu2-dgu3:return 330 gosub 310 340 dg1=dg:k1=k:k=k+sw:k2=k:gosub 310:if dg1*dg>0 then 340 350 k=(k1+k2)/2:gosub 310:if dg1*dg>0 then k1=k else k2=k 360 if abs(dg)>1E-10 then 350 370 x(1,2)=xmu:y(1,2)=ymu:r(2)=ru:masx=1200/x(0,1):masy=850/y(0,2) 380 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the start.
Ótimo questão, parabéns!!!🎉🎉🎉
@@imetroangola17
Fico muito feliz que você tenha gostado deste vídeo. Espero que continue me apoiando assim, isso me inspira a criar vídeos ainda melhores. Obrigado!
very smart thank you
@@lesliederrar9718
Thank you so much for your kind words! I'm glad you enjoyed the video. Your support means a lot to me and motivates me to create more smart and helpful content. Stay tuned for more exciting videos!
Nice sir
@@powerofSanatandharma
Thank you so much for your kind words! Your support means the world to me, and it's because of wonderful viewers like you that I’m inspired to keep creating. I’m so grateful that the video resonated with you. Wishing you lots of learning and joy ahead! 🙏💖
As we are given the ratio of BH to HA as 10:3, we know that the actual length of BH and HA will be 10k and 3k, with k being an as-yet unknown constant.
As BH and FB are tangents to circle O that intersect at B, then by the two tangents theorem, FB = BH = 10k. Similarly, as HA and AD are tangents to circle O that intersect at A so AD = HA = 3k.
As AC and CB are tangent to circle O at D and F respectively, then ∠ODC = ∠CFO = 90°, as OD and OF are radii of circle O. Since ∠DCF = 90° as well, then ∠FOD must equal 360°-3(90°) = 90°, and as OF and OD are equal in length and adjacent, quadrilateral ODCF is a square with side length of 2.
The side lengths of triangle ∆ACB are thus as follows: BA = 10k+3k = 13k, AC = 3k+2, CB = 2+10k.
Triangle ∆ACB:
AC² + CB² = BA²
(3k+2)² + (10k+2)² = (13k)²
9k² + 12k + 4 + 100k² + 40k + 4 = 169k²
109k² + 52k + 8 = 169k²
60k² - 52k - 8 = 0
15k² - 13k - 2 = 0
15k² - 15k + 2k - 2 = 0
15k(k-1) + 2(k-1) = 0
(k-1)(15k+2) = 0
k - 1 = 0 | 15k + 2 = 0
k = 1 | k = -2/15 ❌ k ≥ 0
BA = 13(1) = 13
AC = 3(1) + 2 = 5
CB = 2 + 10(1) = 12
As CB is tangent to circle Q at G, then ∠QGB = 90°, as QG is a radius of circle Q. As AC is tangent to circle P at E, then ∠AEP = 90°, as PE is a radius of circle P.
As PE and OD are both perpendicular to AC and thus parallel to each other, then ∠EPA and ∠DOA are corresponding angles and thus congruent. As ∠PAE = ∠OAD, then ∆AEP and ∆ADO are similar triangles by angle-angle similarity.
As the point of tangency between two circles is collinear with the two centers, then OP = 2+r₁.
Triangle ∆ADO:
AD² + OD² = OA²
3² + 2² = OA²
OA² = 9 + 4 = 13
OA = √13
OD/OA = PE/PA
2/√13 = r₁/(√13-(2+r₁))
√13r₁ = 2(√13-2-r₁)
√13r₁ = 2√13 - 4 - 2r₁
√13r₁ + 2r₁ = 2√13 - 4
r₁(√13+2) = 2√13 - 4
r₁ = (2√13-4)/(√13+2)
r₁ = (2√13-4)(√13-2)/(√13+2)(√13-2)
r₁ = (26-4√13-4√13+8)/(13-4)
[ r₁ = (34-8√13)/9 ≈ 0.573cm ]
As OF and QG are both perpendicular to CB and thus parallel to each other, then ∠BQG and ∠BOF are corresponding angles and thus congruent. As ∠GBQ = ∠FBO, then ∆QGB and ∆OFB are similar triangles by angle-angle similarity.
As the point of tangency between two circles is collinear with the two centers, then OQ = 2+r₂.
Triangle ∆OFB:
OF² + FB² = OB²
2² + 10² = OB²
OB² = 4 + 100 = 104
OB = √104 = 2√26
OF/OB = QG/QB
2/2√26 = r₂/(2√26-(2+r₂))
1/√26 = r₂/(2√26-2-r₂)
√26r₂ = 2√26 - 2 - r₂
√26r₂ + r₂ = 2√26 - 2
r₂(√26+1) = 2√26 - 2
r₂ = (2√26-2)/(√26+1)
r₂ = (2√26-2)(√26-1)/(√26+1)(√26-1)
r₂ = (52-2√26-2√26+2)/(26-1)
[ r₂ = (54-4√26)/25 ≈ 1.344cm ]
@@quigonkenny
Thank you so much for sharing your solution! I really appreciate the effort you've put in, and it's always interesting to see different approaches to solving a problem. It's great that we both arrived at the correct answer through different methods-this shows the beauty of math, where there can be multiple ways to reach the same conclusion. Keep up the great work, and feel free to share more of your ideas!
"Great video! The way you explained the geometric concepts was so clear and easy to understand. The visual aids and examples really helped solidify my understanding. Keep up the fantastic work!"
@@DhanushJoshi-u1r
Thank you so much for your kind words and for liking the video! I'm really glad you enjoyed it. Your support means a lot to me and motivates me to keep creating helpful content. Stay tuned for more...
Nice animation
@@VirendraKumar-z4u
Thank you for your support, which I received in the form of a comment. I hope it continues to inspire me to create even better videos.
r1 = 1, r2=1
(2)^2=4 180ABC/4=4.20 4.2^10 4.2^2^5 2^2.1^1^1 1^2 .1 2(ABC ➖ 2ABC+1).
yes it is tricky but can be solved. the repeated calculation changes the centerpoint of the circle with its known radius until
the quotient "fl=" as written in line 20, is reached:
10 print "quant circle-how to solve this tricky geometry problem with 3 circles"
20 dim x(1,2),y(1,2),r(2):r(0)=2:sw=r(0)/10:fl=10/3:y(1,0)=r(0):ym=r(0):y(1,0)=ym:nr=30
30 nr=30 :goto 150
40 ys=y(1,0)+sqr(abs(r(0)^2-(xs-xm)^2)):dgu1=(xs^2+ys^2)/r(0)^2:dgu2=xs*xm/r(0)^2
50 dgu3=ys*ym/r(0)^2:dg=dgu1-dgu2-dgu3:return
60 xm=r(0)+sw:gosub 40
70 dg1=dg:xm1=xm:xm=xm+sw:if xm>20*r(0) then stop
80 xm2=xm:gosub 40:if dg1*dg>0 then 70
90 xm=(xm1+xm2)/2:gosub 40:if dg1*dg>0 then xm1=xm else xm2=xm
100 if abs(dg)>1E-10 then 90
110 return
120 gosub 60:ye=ys/xs*(xm+r(0)):lbh=sqr(xs^2+ys^2):lah=sqr((xm+r(0)-xs)^2+(ye-ys)^2)
130 lac=ye:df=lbh/lah: df=df-fl:return
140 return
150 xs=sw:gosub 120
160 xs1=xs:df1=df:xs=xs+sw:if xs>nr*r(0) then stop
170 xs2=xs:gosub 120:if df1*df>0 then 160
180 xs=(xs1+xs2)/2:gosub 120:if df1*df>0 then xs1=xs else xs2=xs
190 if abs(df)>1E-10 then 180
200 lbc=sqr((lbh+lah)^2-lac^2):lab=lbh+lah:print "lbh=";lbh:print"lah=";lah:print "lab=";lab
210 print "lbc=";lbc;"lac";lac: print xs,"%",ys
220 x(0,0)=0:y(0,0)=0:x(0,1)=lbc:x(0,2)=x(0,1):y(0,2)=ye:x(1,0)=xm
230 xmu=sw:goto 260:rem den berechneten mittelpunkt verwenden ***
240 ymu=xmu/xm*ym:ru=ymu:dgu1=(xm-xmu)^2/r(0)^2:dgu2=(ym-ymu)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2
250 dg=dgu1+dgu2-dgu3 :return
260 gosub 240
270 dg1=dg:xmu1=xmu:xmu=xmu+sw:xmu2=xmu:gosub 240:if dg1*dg>0 then 270
280 xmu=(xmu1+xmu2)/2:gosub 240:if dg1*dg>0 then xmu1=xmu else xmu2=xmu
290 if abs(dg)>1E-10 then 280
300 x(1,1)=xmu:y(1,1)=ymu:r(1)=ru:dx=xm-lbc:dy=ym-ye:k=sw:goto 330
310 dxk=k*dx:dyk=k*dy:xmu=lbc+dxk:ymu=ye+dyk:ru=lbc-xmu:dgu1=(xmu-xm)^2/r(0)^2
320 dgu2=(ymu-ym)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2:dg=dgu1+dgu2-dgu3:return
330 gosub 310
340 dg1=dg:k1=k:k=k+sw:k2=k:gosub 310:if dg1*dg>0 then 340
350 k=(k1+k2)/2:gosub 310:if dg1*dg>0 then k1=k else k2=k
360 if abs(dg)>1E-10 then 350
370 x(1,2)=xmu:y(1,2)=ymu:r(2)=ru:masx=1200/x(0,1):masy=850/y(0,2)
380 if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the start.
@@zdrastvutye amazing! 👏