Distance From Center Of Circle To Intersection Of Diagonals Of A Cyclic Quadrilateral

Question

Given a cyclic quadrilateral $ABCD$ with length $AB=w$, $BC=x$, $CD=y$, and $DA=z$, compute the distance from the center of the circumscribing circle to the intersection point of the diagonals $AC$ and $BD$ in terms of $w,x,y,z$.

Answer 1

As in @dan_fulea's answer, I'll take the sides of the cyclic quadrilateral $\square ABCD$ to be $a$, $b$, $c$, $d$ (in order). Let the diagonals be $p$ and $q$ (either order). Let the diagonals meet at $L$, let their midpoints be $M$ and $N$ (either order), and let the circumcenter be $O$.

Each diagonal is a chord of the circle; therefore, it's perpendicular to the segment joining its midpoint to $O$. Consequently, $\square OMLN$ has an opposite pair of right angles, making it cyclic; moreover, its circumdiameter is exactly $|OL|$, the distance we seek. By the (Extended) Law of Sines and the area formula $|\square ABCD|=\tfrac12pq\sin\angle MLN$ (valid for any quadrilateral), we can write

$$|OL| = \frac{|MN|}{\sin\angle MLN} = \frac{p q \;|MN|}{2\;|\square ABCD|} \tag{$\star$}$$

From here, we proceed as @dan did, leveraging existing results to get expressions for substitution into $(\star)$.

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Euler's (not-necessarily-cyclic) Quadrilateral Theorem states that $$a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4 |MN|^2 \tag{1}$$ and it is "known" that the diagonals of a cyclic quadrilateral satisfy $$p^2 + q^2 = \frac{(ac+bd)(ad+bc)}{ab+cd}+\frac{(ac+bd)(ab+cd)}{ad+bc} \qquad\qquad pq = ac+bd \tag{2}$$ (the latter of which is Ptolemy's Theorem). Finally, Brahmagupta's formula for the area of a cyclic quadrilateral gives $$|\square ABCD|^2 = (s-a)(s-b)(s-c)(s-d) \tag{3}$$ where $s:=\tfrac12(a+b+c+d)$ is the semiperimeter of $\square ABCD$.

Substituting $p^2+q^2$ from $(2)$ into $(1)$, and solving, we get $$|MN|^2 = \frac{ac(b^2-d^2)^2+bd(a^2-c^2)^2}{4(ab+cd)(ad+bc)} \tag{4}$$

Therefore, squaring to avoid radicals, $(\star)$ becomes ...

$$|OL|^2 = \frac{1}{16}\;\frac{ac(b^2-d^2)^2+bd(a^2-c^2)^2}{(s-a)(s-b)(s-c)(s-d)}\;\frac{(ac+bd)^2}{(ab+cd)(ad+bc)} \tag{$\star\star$}$$

... which agrees with @dan's solution. $\square$

Answer 2

I will use picture and the notations from

https://en.wikipedia.org/wiki/Cyclic_quadrilateral

so the cyclic quadrilateral under study is $ABCD$, it sides are $AB=a$, $BC=b$, $CD=c$, $DA=d$ (instead of $w,x,y,z$, which i cannot type in this context without error.)

The intersection of the diagonals is $P$, and the center of the circle $(ABCD)$ is $O$, we denote by $R$ the radius $R=OA=OB=OC=OD$. When two letters are used in connection with metric relation, we mean always the corresponding lengths.

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We will use the following facts and solve the issue:

• The power of the interior point $P$ is $$PA\cdot PC=PB\cdot PD=R^2-OP^2\ .$$

• The relations of Brahmagupta and Parameshvara, and similar relations for the trigonometric functions of the angles in $A,B,C,D$, and of the angle $\theta$ between the diagonals.

• The sine theorem, for instance $$ \begin{aligned} \frac{PA}{DA} &= \frac {\sin \widehat{PDA}} {\sin \widehat{APD}} = \frac {\sin \widehat{BDA}} {\sin \theta} = \frac {\sin \frac 12\widehat{BOA}} {\sin \theta} = \frac {\frac{AB/2}R} {\sin \theta}\ , \\[2mm] &\qquad \text{ which implies} \\[2mm] PA &= \frac{AD\cdot AB}{2R\sin\theta} = \frac{da}{2R\sin\theta} \ ,\qquad\text{ and similarly} \\ PC &= \frac{CD\cdot CB}{2R\sin\theta} = \frac{cb}{2R\sin\theta} \ . \end{aligned} $$

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We are now in position to join the following relations: $$ \begin{aligned} R^2-OP^2 &= PA\cdot PC =\frac {abcd}{4R^2\sin^2\theta}\ , \\ 4R^2 &= \frac 14\cdot \frac {(ab+cd)(ac+bd)(ad+bc)} {(s-a)(s-b)(s-c)(s-d)}\text{ (Parameshvara) with } \\ s&=\frac 12(a+b+c+d)\ , \\[2mm] &\qquad\text{ and we use now} \\ \tan^2\frac\theta 2 &= \frac{(s-b)(s-d)}{(s-a)(s-c)}\text{ to compute} \\ \frac1{\cos^2\frac\theta 2} &= 1+ \tan^2\frac\theta 2 = \frac{(s-a)(s-c)+(s-b)(s-d)}{(s-a)(s-c)} \sin^2\theta \\ \cos^2\frac\theta 2 &= \frac {(s-a)(s-c)} {(s-a)(s-c)+(s-b)(s-d)} \\ \sin^2\frac\theta 2 &= \frac {(s-b)(s-d)} {(s-a)(s-c)+(s-b)(s-d)} \\ \sin^2\theta &= 4 \sin^2\frac\theta 2 \cos^2\frac\theta 2 \\ &= 4\cdot \frac {(s-a)(s-b)(s-c)(s-d)} {(\; (s-a)(s-c)+(s-b)(s-d)\;)^2}\ , \\ 4R^2\sin^2\theta &= \frac {(ab+cd)(ac+bd)(ad+bc)} {(\;(s-a)(s-c)+(s-b)(s-d)\;)^2}\ , \\ PA\cdot PC &=\frac {abcd}{4R^2\sin^2\theta} = \frac {abcd\;(\;(s-a)(s-c)+(s-b)(s-d)\;)^2} {(ab+cd)(ac+bd)(ad+bc)} \ ,\\[2mm] &\qquad\text{ and finally} \\ \color{blue} {OP^2} &= R^2-PA\cdot PC \\ &= \color{blue} { \frac 1{16} \cdot \frac 1{(s-a)(s-b)(s-c)(s-d)} \cdot \frac {(ac+bd)^2} {(ab+cd)(ad+bc)} \cdot \Big(\ bd(a^2-c^2)^2+ac(b^2-d^2)^2 \ \Big) }\ . \end{aligned} $$

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Note: At the last step we have used sage to factorize. Code and results:

sage: S.<a,b,c,d> = PolynomialRing(QQ)
sage: RR = 1/16 * (a*c+b*d)*(a*b+c*d)*(a*d+b*c) / (s-a) / (s-b) / (s-c) / (s-d)
sage: PAPC = a*b*c*d * ((s-a)*(s-c)+(s-b)*(s-d))^2 / (  (a*c+b*d)*(a*b+c*d)*(a*d+b*c) )
sage: factor( RR - PAPC )
-(a*b^4*c + a^4*b*d - 2*a^2*b*c^2*d + b*c^4*d 
                    - 2*a*b^2*c*d^2 + a*c*d^4)
* (a*c + b*d)^2
/ ((a*b + c*d)*(b*c + a*d)
  *(a + b + c - d)
  *(a + b - c + d)
  *(a - b + c + d)
  *(a - b - c - d))

(Lines were manually broken.)