Calculate the Radius of the semicircle | Radii of Green circles are 2 and 3 | Fun Geometry Olympiad

[@predator1702]

Beautiful question and more beautiful explanation 👍. Thank you teacher 🌹🙏.

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@ghhdcdvv5069]

تمرين جميل جيد. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين

[@ghhdcdvv5069]

تمرين جميل جيد .رسم واصح مرتب. شرح واصح مرتب .شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة غلسطين

[@PreMath]

شكرًا عزيزي

[@theoyanto]

This was difficult, I never could have done this without the video.. I nearly always find it difficult to get started. Ah well, onward and downward 🤓 But thoroughly enjoyed your solution, thanks again 👍🏻

[@PreMath]

Thanks for your feedback! Cheers! You are amazing, Ian. Keep smiling👍 Love and prayers from the USA! 😀

[@fauxcube]

What a fantastic solution, you have taught me so much from this. I give all my thanks to you.

[@PreMath]

You are most welcome😀

[@dannymeslier6658]

Brilliant! Loved it, thanks

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Danny. Keep smiling👍 Love and prayers from the USA! 😀

[@luigipirandello5919]

Beautiful question. Thank you Sir.

[@PreMath]

Glad you think so! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@alster724]

Another technique in solving 96r²-500r-625= 0 Is by using the _Edge Test_ factoring method 24 25 4 -25 Then by cross multiplying and adding, we get the value of b (24)(-25)+(4)(25) -600+100 -500 So our factors are (24r+25)(4r-25) r= -25/24, 25/4 Since this is a geometric figure, we take the positive value, we have 6.25 units or 25/4

[@PreMath]

Excellent! Thanks for your feedback! Cheers! You are awesome, Kevin. Keep it up 👍 Love and prayers from the USA! 😀

[@ghhdcdvv5069]

شكرا جزيلا لكم والله يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين ​@@PreMath

[@PreMath]

@@ghhdcdvv5069 Thanks dear

[@HappyFamilyOnline]

Great 👍 Thanks for sharing😊

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@VictorLonmo]

I was not able to get this on my own. I like a good challenge. Great video PreMath!

[@murdock5537]

Many thanks, Sir, this is a real challenge - and a masterpiece solving it. btw: sin⁡(φ) = 2/(r - 2) = 8/17 → sin⁡(δ) = 3/(r - 3) = 12/13 → φ + δ > 90°

[@bertblankenstein3738]

Nice. I did not see an obviously solution that could be guessed at, or found by inspection. At always, the building blocks like the quadratic and Pythagoras formulas were there, and the solution lies in putting things together the right way which makes it interesting.

[@alexandermorozov2248]

Very nice!

[@PreMath]

Glad you think so! You are awesome, Alexander. Keep smiling👍

[@wackojacko3962]

I absolutely love "Famous Identities" !🙂

[@PreMath]

Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@odayazeez777]

well very useful

[@bigm383]

Thanks Professor, another gnarly little problem!🥂❤

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@bigm383]

@@PreMath 🍺😃👌

[@neillawrence4198]

Finally, a problem that someone didn't claim to solve in 35 seconds in their head.

[@voltalimwabbit5492]

Sir can we possibly work out the length of the vertical line M?

[@PreMath]

Good question! Never thought about that...

[@ybodoN]

When the green circles are in the ratio 3:2, the vertical line is equal to the diameter of the larger of the two circles 😉

[@timeonly1401]

I poked around a found the length of the vertical line segment MC (If we call the point C where that vertical intersects the semicircle): From the video, we have r=25/4, and that x=√(r²-6r) = √{(25/4)[(25/4)-6]} = √(25/16) = 5/4. In right triangle AOP, the long leg AO = 5 - x = 5 - (5/4) = 15/4. Since AM = 2, and AO = 15/4, we have MO = AO - AM = 15/4 - 2 = 7/4. Draw radius OC. Then the 3 segments MO, MC & OC form a right triangle. By Pythag's Thm: MC² = OC² - MO² = r² - MO² = (25/4)² - (7/4)² = 576/16, and MC = √(576/16) = 24/4 = 6. Done!! [I'm STUNNED that that vertical turned out to be a "nice number" 6. 😅 ]

[@gelbkehlchen]

Solution: R = radius of semicircle. From the center of the semicircle to the centers of the two small circles it is R-2 and R-3. These are the hypotenuses of two right-angled triangles whose centers are 2+3 = 5 apart. This gives the equation: √[(R-2)²-2²]+√[(R-3)²-3²] = 5 ⟹ √[R²-4R+4-4]+√[(R²-6R+9-9] = 5 ⟹ √[R²-4R]+√[R²-6R] = 5 |-√[R²-6R] ⟹ √[R²-4R] = 5-√[R²-6R] |()² ⟹ R²-4R = 25-10*√[R²-6R]+R²-6R |-R²+4R+10*√[R²-6R] ⟹ 10*√[R²-6R] = 25-2R |/10 ⟹ √[R²-6R] = 2,5-0,2R |()² ⟹ R²-6R = 6,25-R+0,04R² |-0,04R²+R-6,25 ⟹ 0,96R²-5R-6,25 = 0 |/0,96 ⟹ R²-125/24*R-625/96 = 0 |p-q-formula ⟹ R1/2 = 125/48±√[(125/48)²+625/96] = 125/48±√[15625/2304+15000/2304] = 125/48±√[30625/2304] = 125/48±175/48 ⟹ R1 = 125/48+175/48 = 300/48 = 6,25 and R2 = 125/48-175/48 = -50/48 = -25/24 [invalid in geometry]

[@alinayfeh4961]

we draw Auxiliary lines ON , OT for center and point of contact According Phythagorean thereom Radius rof semicircle r (r-2)²=4+x² x=√(r²-6) (5-x)²+2²=(r-2)² 25-10x+x²+4=r²-4r+4 r=6.25

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@tehribo]

How do we know that ON and OT pass through P and Q

[@rchandos]

We don't know that in advance,. He assumes that they do, contingent on locating the center of the semicircle such that ON equals OT.

[@PreMath]

Dear TB, I believe I covered this topic in some of videos. Very simple explanation. I'll cover again pretty soon. No worries. Keep watching😀

[@alexandermorozov2248]

Кстати, 2-е решение (когда радиус отрицательный) тоже имеет место быть, оно вполне реальное, и его можно даже построить на рисунке. Задание: подумайте, как это изобразить 😜 === By the way, the 2nd solution (when the radius is negative) also takes place, it is quite real, and it can even be built in the figure. Task: think about how to portray it 😜

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@photonomist6345]

I was wondering and thinking it must mean something, but I am not clever enough to visualise it!

[@alexandermorozov2248]

@@photonomist6345 :) I can build a drawing with visualization especially for you.

[@alexandermorozov2248]

@@photonomist6345 A negative radius can be represented as a radius laid in the opposite direction (hint).

[@photonomist6345]

@@alexandermorozov2248 Thank you! I have it!

[@NLGeebee]

Made a sketch in CAD and eyeballed the semi circle. Measured it and is was about 6,3 so it must be 2pi ;)

[@kennethstevenson976]

Tough problem involving two variables and superior insight to get the problem set up and produce the final answer. Again you gave each step required to produce the answer. I could not do this problem!

[@PreMath]

Thanks for your feedback! Cheers! You are awesome, Kenneth. Keep smiling👍 Love and prayers from the USA! 😀

[@alfonsorodriguez2739]

I would like to get in contact with you premath. How can I do so?

[@PreMath]

You may email us: [e-mail] Thanks for asking. Cheers

[@zdrastvutye]

it took longer for me than expected because once the root must be substracted and once it must be added, which confused me: 10 r1=2:r2=3:sw=.01:ru=sw:ym1=r1:ym2=r2:ng=r1^2+r2^2:goto 60 20 dis1=(ru-r1)^2-ym1^2:dis2=(ru-r2)^2-ym2^2 30 if dis1

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@zdrastvutye]

@@PreMath thanks for your attention, there was a problem with the result inside the sqr( ). now i have worked this out and have edited and updated the comment

[@vara1499]

👏

[@PreMath]

Thanks dear for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@4iriska]

Why are we sure that OT and ON go through the centers of green circles?

[@rchandos]

They have to so that both the semicircle and the smaller circles are tangent at T and N. His assumption turns out to be justified because he manages to locate the center of the semicircle such that OT=ON.

[@PreMath]

Hello dear, I believe I covered this topic in some of videos. Very simple explanation. I'll cover again pretty soon. No worries. Keep watching😀

[@4iriska]

@@rchandos Thank you

[@vaibhavgupta5359]

Sir what is your age

[@PreMath]

Dear Vaibhav, I'd talk about myself pretty soon. I can understand your curiosity. Thanks for your continued love and support! You are awesome. Keep smiling👍

[@wackojacko3962]

I absolutely love "Famous Identities"!🙂

[@SohailKhan-xx8wi]

35years old.

[@vaibhavgupta5359]

@@PreMath sir you have a sharp brain I want to know how do u relate all the concepts of maths so easily

[@maturaczynicuda]

@@vaibhavgupta5359 true, the possibility of learning from such a great teacher is a great privilege

[@peterkrauliz5400]

Where is the proof that these respective points are collinear? It looks obvious, but how can one be sure!

[@Songwriter376]

Ok, I knew that. Yep. So, what's for dinner?

[@mohamadtaufik5770]

(R^2-6R)^0.5+(R^2-4R)^0.5=2V6 --> R=6.223

[@ybodoN]

If the two circles were tangent, this would be the correct answer...

[@mohamadtaufik5770]

I thought there were tangent, thanks for the correction

[@محمدالنجار-و2ف]

Difficult

[@wajiraabayasinghe7718]

First go to a school and lean how to pronounce English Language. Your Math skills are good but pronunciation is to be improved

[@johnplong3644]

Ok I need help with this one

[@amitabhshrivastava3121]

You can still solve even if the green circles were tangential

[@claudeabraham2347]

Very good!