i find your statement at 01:00 not so obvious. Please explain. I mean: how do you know that connecting the two points will pass through the center of the circle.
The two tangents to the circle and the two radii perpendicular to the tangents form two congruent triangles. It's also like inscribing a circle in a kite. The longer diagonal goes through the center because of symmetry.
@@SyberMath Hmm. Don't see the "symmetry" part. I'd explain it as follows (wish you'd had some letters for the points on the drawing). 1) draw a line from the center point of the circle to the intersection point of the square and the big black triangle. The two radii perpendicular to the tangents and the tangents form two right angled triangles having the same hypothenuse. As those two triangles have same hypothenuse and one right handed side the same, they are congruent (case RHS). 2) Same reasoning can be followed for the two triangles formed by the line drawn from the center point of the circle to the intersection point of the equilateral triangle and the big black triangle. So also those two triangles are congruent. 3) still remains to show that the hypothenuse of the first two rectangles and the hypothenuse of the last two rectangles are colinear. This can be shown because the right hand sides of the triangles are parallel to each other because the two black tangents of the big black triangle are coming from a right angle. In that case the third side of each triangle is also parallel. And because they share one point, they are colinear. QED
Hello. Sir I have questions. How I do I solve these: a. If 1/(1+a) + 1/(1+b) + 1/(1+c) = 2, then what is the value of a/(1+a) + b/(1+b) + c/(1+c) ? b. If (a^2 - bc)/(a^2 + bc) + (b^2 - ca)/(b^2 + ca) + (c^2 - ab)/ (c^2 + ab) = 1, then find the value of a^2/(a^2 + bc) + b^2/(b^2 + ca) + c^2/(c^2 + ab). Thank you and keep safe!
Only flaw with this problem is once I saw it was a 30 60 90 right triangle it all collapsed to the inevitable. When I have time I want to try this with a 3-4-5 triangle or something.
i find your statement at 01:00 not so obvious. Please explain. I mean: how do you know that connecting the two points will pass through the center of the circle.
The two tangents to the circle and the two radii perpendicular to the tangents form two congruent triangles. It's also like inscribing a circle in a kite. The longer diagonal goes through the center because of symmetry.
@@SyberMath Hmm. Don't see the "symmetry" part. I'd explain it as follows (wish you'd had some letters for the points on the drawing). 1) draw a line from the center point of the circle to the intersection point of the square and the big black triangle. The two radii perpendicular to the tangents and the tangents form two right angled triangles having the same hypothenuse. As those two triangles have same hypothenuse and one right handed side the same, they are congruent (case RHS). 2) Same reasoning can be followed for the two triangles formed by the line drawn from the center point of the circle to the intersection point of the equilateral triangle and the big black triangle. So also those two triangles are congruent. 3) still remains to show that the hypothenuse of the first two rectangles and the hypothenuse of the last two rectangles are colinear. This can be shown because the right hand sides of the triangles are parallel to each other because the two black tangents of the big black triangle are coming from a right angle. In that case the third side of each triangle is also parallel. And because they share one point, they are colinear. QED
Hello. Sir I have questions. How I do I solve these:
a. If 1/(1+a) + 1/(1+b) + 1/(1+c) = 2, then what is the value of a/(1+a) + b/(1+b) + c/(1+c) ?
b. If (a^2 - bc)/(a^2 + bc) + (b^2 - ca)/(b^2 + ca) + (c^2 - ab)/ (c^2 + ab) = 1, then find the value of a^2/(a^2 + bc) + b^2/(b^2 + ca) + c^2/(c^2 + ab).
Thank you and keep safe!
Thank you! These are good problems! See my solutions here:
twitter.com/SyberMath/status/1365585367211012099
How do we know the right triangle in the top left corner is a 30-60-90 triangle?
Because we know that the top left angle is 60 degrees.
@@SyberMath thanks, but how was that fact known to us?
@@Happy_Abe The bottom right angle is 30 degrees. Notice the isosceles triangle next to the equilateral. It's a 30-30-120 triangle.
@@SyberMath can’t believe I missed that, oops
Thanks!
@@Happy_Abe Np. You're welcome!
The cosine of 30 degree is sqrt(3)/2 so why did you write sqrt(3)/3?
Are we using the cosine?
@@SyberMath I might be confused but it was a triangle with the hypotenuse =1 and the angle 30degree, so isn’t it the cosine of 30 degree?
Thank you very mach
You are welcome
Appreciate your efforts n please let it be in english 🎲👍
It was in English! 😁
Only flaw with this problem is once I saw it was a 30 60 90 right triangle it all collapsed to the inevitable. When I have time I want to try this with a 3-4-5 triangle or something.
The arrangement forces it to be a 30-60-90. You cannot do it with a 3-4-5.
@@SyberMath Something else would have to change of course. Sometimes these problems spur my imagination and something new arises.
どうもありがとうございました 🙏🙏🙏
😊😊😊
@@SyberMath Hi human - what language is that??? Looks like noodles with funny shapes 😊
I'm gonna, well, so, okay, actually, let me, right?
Well, actually I'm gonna have to use these words again, okay? So you'll let me use them, right?
😁😁😁
😍😍😍
算額 !!!
Idk why this video lasts 7 min, it just ended around 4
Idk either! 😁
3:32 you got all the information you actually needed
That's right!
Nyccc
Thxx