A Solution to Crux Mathematicorum Problem 5098

Problem 5098

Proposed by Mihaela Berindeanu; modified by the Editorial Board

Given a cyclic quadrilateral $BCAD$ (with $A$ and $B$ separating $C$ from $D$ ) such that $CB=CD$ , define $A_1$ and $B_1$ to be the feet of the altitudes from $A$ and $B$ in triangle $ABC$ . Prove that the orthocenter $H$ of $\triangle ABC$ is the midpoint of $AA_1$ if and only if $DB_1$ is perpendicular to $DB$ .

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.

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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Solution

Diagrams and complete source