6:14 is the part that I would never have realized without being shown -- that the distances between the midpoints of each chord is half the distance of CD. After you showed it I was able to prove to myself that it's true, but, as I said, I wouldn't have seen it on my own -- and without seeing that I don't think the problem can be solved. Very nice! Thanks!
Nowhere in the problem statement do you say that the the green line tangent to both circles is parallel to line segment AB. This is required to show that line segment AB is perpendicular to the radius of the blue circle as well as to the diameter of the green circle, as you draw at time stamp 2:43. Without that parallelism, the circle sizes are not fixed (as I just confirmed in Geogebra). Please add that parallelism assumption to the problem--perhaps in the description text?
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Alternative approach using similar triangles: Notice, angle MPE = angle NPB = θ, Therefore, MPE and NPB are similar Right triangles. Also, angle MAE = angle MPE= θ = angle NPB = angle NBP Therefore APC and BPD are similar Right triangles. OB/AO = OD/CO = 2R/2r adding 1 to each side, we get OB/AO +1 = OD/CO+1 = (R/r)+1 (AO+OB)/AO = (CO+OD)/CO = (R/r)+1 24/AO=10/CO = (R/r)+1 Tan(θ) = CO/AO=5/12 Therefore APC is similar to a (5,12,13) Right Triangle, Right Triangle MQN is similar to Right Traingle APC, with angle MNQ = θ Therefore, MQ/MN= (R-r)/(R+r) = 5/13 R/r = 9/4 Hence 24/AO=10/CO = (R/r)+1 = 13/4 AO = 4*24/13, CO=4*10/13 From triangle APC using Pythogorous theorem, R=9 and since R/r = 9/4, r=4
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why and how we know that the cord AB is perpendicular to radius R? Different asked: how we are sure that cord AB is parallel to basic tangent of both circles?
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Very interesting question.❤️
Solution was much more interesting than the question per se.😀
Love and Prayers from India! ❤️🇮🇳
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Greetings from the USA!
6:14 is the part that I would never have realized without being shown -- that the distances between the midpoints of each chord is half the distance of CD. After you showed it I was able to prove to myself that it's true, but, as I said, I wouldn't have seen it on my own -- and without seeing that I don't think the problem can be solved. Very nice! Thanks!
تمرين جيد. رسم واضح مرتب .شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم . تحياتنا لكم من غزة غلسطين .
Nowhere in the problem statement do you say that the the green line tangent to both circles is parallel to line segment AB. This is required to show that line segment AB is perpendicular to the radius of the blue circle as well as to the diameter of the green circle, as you draw at time stamp 2:43. Without that parallelism, the circle sizes are not fixed (as I just confirmed in Geogebra). Please add that parallelism assumption to the problem--perhaps in the description text?
Right. I also mentioned that point
Awesome problem and very interesting sol sir...love from India❤️
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loved the problem, sir... it was indeed a brilliant solution. Thank you 😀😎
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Very nice problema. Thank you Sir.
Difficult problem, but you discuss it very simple method.
🥺 you are my role model , I wish I could be like you
Thanks dear!
I'm just an ordinary person. I'm very blessed and very grateful for that.
🙏
Jazakallah
😊 Excellent math problem!!!
Muchas gracias
Very very interesting. Super sir
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There's so much to learn in geometry, thanks.
Very impressive method. Not a easy problem
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Alternative approach using similar triangles:
Notice, angle MPE = angle NPB = θ,
Therefore, MPE and NPB are similar Right triangles.
Also, angle MAE = angle MPE= θ = angle NPB = angle NBP
Therefore APC and BPD are similar Right triangles.
OB/AO = OD/CO = 2R/2r
adding 1 to each side, we get
OB/AO +1 = OD/CO+1 = (R/r)+1
(AO+OB)/AO = (CO+OD)/CO = (R/r)+1
24/AO=10/CO = (R/r)+1
Tan(θ) = CO/AO=5/12
Therefore APC is similar to a (5,12,13) Right Triangle,
Right Triangle MQN is similar to Right Traingle APC, with angle MNQ = θ
Therefore, MQ/MN= (R-r)/(R+r) = 5/13
R/r = 9/4
Hence 24/AO=10/CO = (R/r)+1 = 13/4
AO = 4*24/13, CO=4*10/13
From triangle APC using Pythogorous theorem,
R=9
and since R/r = 9/4,
r=4
That was brilliant! I tried this and failed with 4 variables and not enough equations.
I also didn't even notice the green line, lol!
Great job!
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Great job at failing???
You only fail if you don't get back up again, failures are all part of lifes rich tapestry
brillant solution! Amazing!
It was very interesting sir.
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But you must mention that AB is parrelel to the common tangent. Or, the sum can't be solved. The chords could have been a little inclined.
Thank for premath
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Sir Professor, on 13:27 lenght 5 visual disturbances ( to short painted)
Very nice.
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Can u find the answer by using coordinate geometry or by using some other methods
Teach us derivation of ladder theorem
Cool 😎😎
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Very good
why and how we know that the cord AB is perpendicular to radius R? Different asked: how we are sure that cord AB is parallel to basic tangent of both circles?
Thank sir
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Let R & r be the radii of big & small circles. (R + r)sq = (R-r)sq + 144. R - r = 5, R + r = 13. R = 9, r = 4.
Areas = 81pi & 16pi = 254.47 & 50.26
When i saw 10 and 24, i immediately thought solutions would be 3,4,5 and 5,12,13 pythagoras.
50.265 and 254.469 with CAD
Interesting problem
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Right triangle
#RightTriangle
a = 5; b = 12; c = 13
#Radius
#Pythagoras #PythagoreanTheorem
Pi does NOT equal 3.14. That's approximating way too much.
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better use (22/7) as an approximation (without calculator)
@@murdock5537 better not approximate at all
A = 1 pi × 9^2 = 1 pi × 81 = 81 pi
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2r = 13 - 5 = 8
Radius
Area of green circle = 16 pi
Nice job!
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Area of blue circle = 81 pi
Well done Theo!
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Pythagorean Theorem
Square root of 169 = 13
r = 4
Radius = 4
c = 13
R = 9
R + r = 13
Pythagoras
5; 12; 13
Brainless is brainless
12; 5; 13
R + 4 = 13
9 + 4 = 13
How can we know exactly the distance EF is half of AB
How EF is equal to 12 bluff is there
R = 13-4 = 9