Cosine law in Cyclic quadrilateral

Question

Let $\square$ ABDC is cyclic quadrilateral , and $\triangle$ ABC is equilateral triangle of length a.

Express $ \overline {DA} ^4 + \overline {DB} ^4 + \overline {DC}^4 $ using terms of a.

Let $ \overline {DA} =x , \overline {DB} =y , \overline {DC} =z $

by Cosine law

$ x^2+ y^2 - xy = x^2 + z^2 -xz = y^2 +z^2 +yz = a^2 $

I think $ z \rightarrow 0 \ then\ y \rightarrow a ,\ x \rightarrow a $
so $x^4+y^4 +z^4 = 2a^4$ but it lacks persuation.

What is the connection with a cyclic quadriteral. Please write it in the body of the question. — user376343, Jun 05 '23 at 12:22
Is ABCD supposed to be a square, given the symbol that you used? — Calvin Lin, Jun 05 '23 at 15:21
Do you mean that $D$ is on the same circle as $A,B,C$, on the arc between $B$ and $C$? — Lutz Lehmann, Jun 06 '23 at 07:18

score 2 · Answer 1 · answered Jun 06 '23 at 08:15

By P'tolemy, $ay + az = ax$

$ y+z = x $

By Cosine Law at $\triangle BDC$,

$ y^2 + z^2 - 2yz cos \frac {2 \pi} 3 = a^2 $

$ yz = x^2 -a^2 $

$y^2 + z^2 = 2a^2 -x^2 $

$ y^4 + z^4 = ( y^2 + z^2 )^2 - 2y^2 z^2 $

$ = (2a^2 -x^2 )^2 - 2( x^2 -a^2 )^2 $

$ = 2a^4 -x^4 $

so $ x^4 + y^4 + z^4 =2a ^4 $

After thinking all night, I was finally able to reach the answer. Thank you for your interest in this problem.

1 Answers1