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Comments:#13 Motivation for assuming r = 4 stems from requirement that the area must be an integer, so 15 * r is an integer and r = 4 is a great guess.correct formula are tan( 2*arctan(1/2) ) = 4/3 and arctan(3/4) + arctan(4/3) = pi/2
9:38 the f(x) function is copied down incorrectly , the function given in the question is f(x) = x{x} not f(x) = x[x]
13:11 What's the NAME of the theorem that angle CBT= angle A? I can easily prove it.
@@mikono2022 Thanks
@@xiaodongjin8051 Also known as the alternate segment theorem in this particular case.
Just double application of inscribed angle theorem. I don't know if there is an exact name for that theorem.
Two formulae in 8:00 - 8:14 on Q13 were wrong.
13:23 how do u know XM is perpendicular to AC? may seem like a stupid question
since XMYT is a parallelogram, XM is parallel to TY. By corresponding angles, since angle Y is right, XM and AC are perpendicular.
Comments:
#13 Motivation for assuming r = 4 stems from requirement that the area must be an integer, so 15 * r is an integer and r = 4 is a great guess.
correct formula are tan( 2*arctan(1/2) ) = 4/3 and arctan(3/4) + arctan(4/3) = pi/2
9:38 the f(x) function is copied down incorrectly , the function given in the question is f(x) = x{x} not f(x) = x[x]
13:11 What's the NAME of the theorem that angle CBT= angle A? I can easily prove it.
@@mikono2022 Thanks
@@xiaodongjin8051 Also known as the alternate segment theorem in this particular case.
Just double application of inscribed angle theorem. I don't know if there is an exact name for that theorem.
Two formulae in 8:00 - 8:14 on Q13 were wrong.
13:23 how do u know XM is perpendicular to AC? may seem like a stupid question
since XMYT is a parallelogram, XM is parallel to TY. By corresponding angles, since angle Y is right, XM and AC are perpendicular.