A Solution to Crux Mathematicorum Problem 5098

The original problem statement is intentionally not reproduced here. This article follows its notation and presents only the independently prepared solution published here and its diagram. The complete statement is available in the official issue of Crux Mathematicorum.

Solution

Cyclic quadrilateral BCAD with CB equal to CD. A₁ and B₁ are the altitude feet in triangle ABC; H bisects AA₁, and DB₁ is perpendicular to DB at D. — Figure 1. A representative configuration in which the two equivalent conditions hold.Select the diagram to enlarge it.

Proof

Work in the complex plane, and denote $A(a)$ , $B(b)$ , $C(c)$ , $D(d)$ , $A_1(a_1)$ , $B_1(b_1)$ , and $H(h)$ . Take the circumcircle as the unit circle. Let $c=1$ , and $|a|=|b|=|d|=1$ . Then $h=a+b+c=a+b+1$ .

Since $CB=CD$ and the vertices are distinct, $d=\overline b$ , with $b\ne\overline b$ and $b\ne1$ . Moreover,

\begin{aligned} a_1 &=\frac{1}{2}\left(a+b+c-\overline{a}bc\right) =\frac{1}{2}\left(a+b+1-\overline{a}b\right),\\ b_1 &=\frac{1}{2}\left(a+b+c-a\overline{b}c\right) =\frac{1}{2}\left(a+b+1-a\overline{b}\right). \end{aligned}

$H$ is the midpoint of $AA_1$ if and only if

\begin{aligned} &h=\frac{1}{2}(a+a_1)\\ \iff\;&a+b+1=\frac{1}{2}\left(a+\frac{1}{2}\left(a+b+1-\overline{a}b\right)\right)\\ \iff\;&a+3b+3=-\overline{a}b\\ \iff\;&a+3b+3=-\frac{b}{a}\\ \iff\;&a^2+3ab+3a+b=0. \end{aligned}

On the other hand, $DB_1$ is perpendicular to $DB$ if and only if

\begin{aligned} &\Re\left(\frac{b_1-d}{b-d}\right)=0\\ \iff\;&\frac{b_1-d}{b-d}=-\overline{\left(\frac{b_1-d}{b-d}\right)}\\ \iff\;&\frac{b_1-\overline{b}}{b-\overline{b}}=-\frac{\overline{b_1}-b}{\overline{b}-b}\\ \iff\;&b_1-\overline{b}=\overline{b_1}-b\\ \iff\;&\frac{1}{2}\left(a+b+1-a\overline{b}\right)-\frac{1}{b} =\frac{1}{2}\left(\overline{a}+\overline{b}+1-\overline{a}b\right)-b\\ \iff\;&(b-1)(a^2+3ab+3a+b)=0\\ \iff\;&a^2+3ab+3a+b=0. \end{aligned}

This completes the proof.

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