Instead of saying that taking a point in the minor arc would be no good, you could have just stated why. It's a basic explanation and you just subtract both angles by 360°. Btw I had this in my curriculum 5 years ago and I still haven't forgotten this beautiful theorem wow good job bprp
Akshat K Agarwal can u elaborate your statement? I didn't understand subtracting 360 from the 2 angles (I also don't know which two angles you are talking about)
Varun Murali sure. Say you're given that the angle of the minor arc AC is 60° and the point B is on the minor arc instead of the major arc. Then then angle ABC(the angle inside the circle) would be 150° since you can consider the central angle as convex. So it'll be 300° (because 360°-60°). And using the same theorem, the angle ABC becomes 150°
The proof I like most is the one that starts with A&B on opposite sides of a diameter. Once you prove that, you can prove the other two cases you have quickly.
Picking a point on the minor arc AC is still good. In which case, the central angle AOC is a "reflex angle", and the inscribed angle ABC is an obtuse angle and is half of the reflex angle AOC.
Stuff from the 8:40 point is unnessecary. Angle AOB equals (180-2y+2x) from AOB triangle. Angle COB equals (180-2y) from COB triangle. Angle AOC equals angle AOB minus angle COB and that's (180-2y+2x)-(180-2y)=180-2y+2x-180+2y=2x, Q.E.D.
On that figure, OB=OA(radius) => OBA=OAB=x(let) 180°=AOB+OBA+OCA(angle sum property) =>180°=2x+AOB 180°=AOB+AOC(linear pair) On equating, 2x+AOB=AOB+AOC =>2x=AOC =>2AOB=AOC
What is the integral of sqroot(sinx) from 0 to pi? I found it 2√(2/π)(gamma(3/4))^2 in wolfram alpha which seems very interesting.can you provide me the magical steps?
Debraj Banerjee what no. Differentiate the substitution. You'll get (1/2)sec^2(x/2)dx=du Then change the sec to tan if you want. I'm sure you can figure out what to do next if you talk about gamma in your integrations
Debraj Banerjee also if you graph the function, you'll notice that it is symmetric at x=π/2 so computer the integral to that point maybe and then multiply by 2 because at π/2,you can use a property of definite integrals and change u from u to π/2 - u This is a higher level integral and I don't think he'd be able to cover it anytime soon. If you need help, you can still ask or Google it. All the best
I think until 5:45 was the essential part of the proof, since if we join A and C we get a chord AC of a circle and angles subtended by a chord of a circle on it are all equal. But I really like that you did the other case as well for completeness :)
In the second case, why didn't you use the triangles OCB and AOC. The OCB one gives you the angle on the right side (180-2y) and the AOC one gives you the whole angle (180-2*(y-x)), so the angle we are looking for is: 180-2*(y-x) - (180-2y) = 180-2y+2x-180+2y = 2x Therefore the angle in the middle is twice as big as the one on the outside. Yet I liked your way, too, but it was a bit too complicated, I guess
Hey can you tell me why that angle is y - x? EDIT: Nevermind, I understand now. It looked confusing because of the dimensions of the diagram, I forgot that it was isosceles.
If point B coincided with point C wouldn't the angle ABC be zero. In that case angle ABC must approach zero as B approaches C and hence cannot be constant?
Point C is kinda like the border between the big AC arc and the small AC arc. Point B, as we know, must be on the big AC arc. The theorem doesn't work if point B is on the small arc or on its borders. If point B and point C are on the same place, the angle ABC is 0, but the theorem is not appliable for this case. Even if point B is very close to point C, as long as it is on the big arc, angle ABC is two times smaller than its corresponding central angle.
minus Y, plus Y also say bye bye
+
My Number Theory professor disagreed with the theorem after I told him what it was. LOL
I subscribed for blackpenredpen, not for blackpenredpenbluepen. I am very dissapointed.
I think redpenblackpen is _way_ better than blackpenredpen.
Alitrux
i am sorry........ : )
In 2019, it may be that his name will be blackpenredpenbluepenyellowpengreenpenorangepenpurplepen.
I never understood that blackpenredpen thing, why not just use blackredpen^2? It just doesn’t compute to me..
Is it commutative though?
night before maths exam. this man is amazing
Fantastic.....especially the second part of the proof!
Thanks!!
Precisely my thinking, I like to show precisely these three variations. Thank you for the video.
We can use it for writing equation of angle bisector but it will be a little bit longer than comparing distances of point from both rays of angle
Yay...I just learned a new trick..Thanks BPRP
Love geometry theorems and concepts!
I wish I was you brodi. Imma fail this Damn test
Instead of saying that taking a point in the minor arc would be no good, you could have just stated why. It's a basic explanation and you just subtract both angles by 360°.
Btw I had this in my curriculum 5 years ago and I still haven't forgotten this beautiful theorem wow good job bprp
Akshat K Agarwal can u elaborate your statement?
I didn't understand subtracting 360 from the 2 angles (I also don't know which two angles you are talking about)
Varun Murali sure.
Say you're given that the angle of the minor arc AC is 60° and the point B is on the minor arc instead of the major arc. Then then angle ABC(the angle inside the circle) would be 150° since you can consider the central angle as convex. So it'll be 300° (because 360°-60°). And using the same theorem, the angle ABC becomes 150°
Why are you keep this name blackpenredpen
The geometry is incredible
The proof I like most is the one that starts with A&B on opposite sides of a diameter. Once you prove that, you can prove the other two cases you have quickly.
Picking a point on the minor arc AC is still good. In which case, the central angle AOC is a "reflex angle", and the inscribed angle ABC is an obtuse angle and is half of the reflex angle AOC.
our school requires us to write angle at centre twice angle at circumference
Beautiful.
steamroller82 yay!!!
Stuff from the 8:40 point is unnessecary. Angle AOB equals (180-2y+2x) from AOB triangle. Angle COB equals (180-2y) from COB triangle. Angle AOC equals angle AOB minus angle COB and that's (180-2y+2x)-(180-2y)=180-2y+2x-180+2y=2x, Q.E.D.
I use central angle theorem for construction regular 2n gon after construction n gon with given side length
Best explanation
Thanks for the daily doze of math
You forgot the prove the case when the segment AB colinear to the segment AO!!
Thanks! lol
On that figure,
OB=OA(radius)
=> OBA=OAB=x(let)
180°=AOB+OBA+OCA(angle sum property)
=>180°=2x+AOB
180°=AOB+AOC(linear pair)
On equating,
2x+AOB=AOB+AOC
=>2x=AOC
=>2AOB=AOC
@Kiritsu Insert x=0 in the formula's on the left halve of the board and they describe the situation you suggest.
The day blackpenredpen did simple things.
へええ これはみんな覚えとくと便利だねえ😄
What is the integral of sqroot(sinx) from 0 to pi?
I found it 2√(2/π)(gamma(3/4))^2 in wolfram alpha which seems very interesting.can you provide me the magical steps?
Substitute tan(x/2) = u and you'll get everything after that
Akshat K Agarwal after this substitution the integral would be 2sqrt(2u/(1+u^2)^3) which would not work
Debraj Banerjee what no. Differentiate the substitution. You'll get (1/2)sec^2(x/2)dx=du
Then change the sec to tan if you want.
I'm sure you can figure out what to do next if you talk about gamma in your integrations
Debraj Banerjee also if you graph the function, you'll notice that it is symmetric at x=π/2 so computer the integral to that point maybe and then multiply by 2 because at π/2,you can use a property of definite integrals and change u from u to π/2 - u
This is a higher level integral and I don't think he'd be able to cover it anytime soon. If you need help, you can still ask or Google it. All the best
Hilbert Black yesss
Nice explanation sir.....
crap, this is obvious -.-
I feel embarrassed now
Yeah, it actually makes way too much sense...
for the real life, what the situation we can explaint about this theorem?
Now it makes sense how Thales thm works, cause the diameter's central angle is 180 degrees.
When i was 13yo just solved it in like 20 sec, but now with 24 i couldn't. Is that bad?(don't know what to feel)
Hopefully you like this? I LOVE THIS.
Is there a way to prove this without splitting into multiple cases?
What about interior angle
Cool bro
let us peep the shoe game
I think until 5:45 was the essential part of the proof, since if we join A and C we get a chord AC of a circle and angles subtended by a chord of a circle on it are all equal. But I really like that you did the other case as well for completeness :)
I’m pretty sure that that sentence derived from the central angle theorem and not vice versa
If I squint my eyes and tilt my head, your diagram kinda looks like a pentagram.
I FINALLY UNDERSTAND THIS!!!!!!!!!!!
Good
In the second case, why didn't you use the triangles OCB and AOC.
The OCB one gives you the angle on the right side (180-2y) and the AOC one gives you the whole angle (180-2*(y-x)), so the angle we are looking for is:
180-2*(y-x) - (180-2y) =
180-2y+2x-180+2y =
2x
Therefore the angle in the middle is twice as big as the one on the outside.
Yet I liked your way, too, but it was a bit too complicated, I guess
Hey can you tell me why that angle is y - x?
EDIT: Nevermind, I understand now. It looked confusing because of the dimensions of the diagram, I forgot that it was isosceles.
@Anurag MFT OC = OB so if OCB = y then OBC = y cause it's an isosceles triangle. Then it's easy to see that OBA = y - x
If point B coincided with point C wouldn't the angle ABC be zero. In that case angle ABC must approach zero as B approaches C and hence cannot be constant?
Point C is kinda like the border between the big AC arc and the small AC arc. Point B, as we know, must be on the big AC arc. The theorem doesn't work if point B is on the small arc or on its borders. If point B and point C are on the same place, the angle ABC is 0, but the theorem is not appliable for this case. Even if point B is very close to point C, as long as it is on the big arc, angle ABC is two times smaller than its corresponding central angle.
nice rhyme in 10:19
Would you need a third case when A, O and B are co-linear?
Akai Nishin hi
I should but I forgot. The proof is similar.
That's same as the first case but with x=0.
That one has a trivial proof.
David Alexander
Great video, as always.
I really love mathematics, isn't it! :)
I was taught that Central Angle in a Circle was always 90°.
'semi circle...
Hi, can you figured out the derivative of n ! ?
Yes
Yaaaayyyyyyy
7:40 why is this angle y-x?
Because it is an isosceles triangle, the angles OCB and OBC has to be the same
What?
Why don't you just write ∠AOC = 2∠ABC instead of m∠AOC = 2m∠ABC ? m is confusing.
how did the y-x come to be
i guess you can say that it's a central theorem for angles in a circle
but what if B is on the left side instead of the right side :^)
You can get around that by slapping a WLOG onto the second scenario ::^^))))
Copy his proof on the inside of a window, then look at it from outside.
That's the Star Trek symbol...
K
666 likes
nonsense