Problem 1878

We prove the following stronger statement: $A_1, B_1, C_1$ are midpoints of segments $HD, DE, HF$, respectively, where $H$ is the orthocenter of triangle $ABC$. Therefore, there is a dilation centered at $H$ with magnitude 2 sending triangle $A_1B_1C_1$ to $DEF$. (For readers familiar with the nine-point circle of a triangle, this stronger statement shows that $A_1, B_1, C_1$ lie on the nine-point circle of triangle $ABC$. Furthermore, the statement remains true if $D, E, F$ are arbitrarily chosen points on arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$. We leave it to the reader to show this more general result.)

We give two solutions; in each, let $X$ and $Y$ denote the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $P$ and $Q$ denote the feet of the perpendiculars from $A$ to lines $DB$ and $DC$ respectively. Then $A_1$ is the intersection of line $XY$ (or $m_a$) and line $PQ$ (or $l_a$). Let $H_a$ be the foot of the perpendicular from $A$ to line $BC$. Let $M$ be the midpoint of side $BC$. Extend segment $AH_a$ to meet the circumcircle of triangle $ABC$ at $G_a$. We also set $B = \angle ABC, C = \angle BCA$, and $A = \angle CAB$.

Solution 1.

We consider the left-hand side configuration shown below. (Our proof can be easily modified for other configurations.) Let $H$ denote the intersection of lines $DA_1$ and $AH_a$. By symmetry, it suffices to show that $A_1$ is the midpoint of segment $DH$ and $H$ is the orthocenter of triangle $ABC$.

Note that points $X, Y, M$ are collinear – they lie on the Simson line from $D$ with respect to triangle $ABC$. Indeed, because $\angle BXD = \angle BMD = 90^\circ$, $BXMD$ is cyclic, implying that
$$\angle XMB = \angle XDB = 90^\circ - \angle XBD = 90^\circ - \angle ABD = 90^\circ - \left(B + \frac{A}{2}\right) = \frac{C-B}{2}. \quad (16)$$
Because $\angle DMC = \angle DYC = 90^\circ$, $BMCY$ is cyclic, implying that
$$\angle CMY = \angle CDY = 90^\circ - \angle DCY = 90^\circ - \angle ABD = \frac{C-B}{2}, \quad (17)$$
where the third equality holds because $ABDC$ is cyclic. By (16) and (17), we know that $\angle XMB = \angle CMY$; that is, $X, M, Y$ are collinear. In exactly the same way, we know that $P, H_a, Q$ are collinear (on the Simson line from $A$ with respect to triangle $BCD$) by establishing
$$\angle BH_a P = \angle QH_a C = \frac{C - B}{2} \quad (18)$$
Furthermore, by (17) and (18), we conclude that $MA_1H_a$ is an isosceles triangle with $MA_1 = A_1H_a$. Let $N$ denote the intersection of lines $MD$ and $PA$. Then in right triangle $MNH_1$, we must have $MA_1 = NA_1 = A_1H_a$. Thus triangles $DNA_1$ and $HA_1H_a$ are congruent to each other (by SAS, since $DM \parallel HH_a$ and $NA_1 = A_1H_a$). In particular, $A_1$ is the midpoint of $AH$.

That is, $A_1H_a \parallel DG_a$. Because $A_1$ is the midpoint of side $DH$ of triangle $DHG_a$ and $A_1H_a \parallel DG_a$, $H_a$ must be the midpoint of segment $HG_a$. Consequently, line $BH$ is the reflection of line $BG_a$ across line $BC$, from which it follows that
$$\angle HBC = \angle G_a BC = \frac{\widehat{G_aC}}{2} = \angle G_a AC = \angle H_a AC = 90^\circ - C.$$
Therefore, $\angle HBC + C = 90^\circ$ or $BH \perp CA$; that is, $H$ is the orthocenter of triangle $ABC$ (as $AH \perp BC$).

Solution 2

(based on work by Thomas Swayze). Our goal will be to prove that $A_1$ is the midpoint of $HD$, where $H$ is the orthocenter of triangle $ABC$, from which the result follows immediately. Let $M$ be the midpoint of $HD$. Set up a system of complex numbers (equivalently, vectors) having the circumcircle of triangle $ABC$ as the unit circle. Then, if $A, B, C, D$ have coordinates $a, b, c, d$, then $H$ has coordinate $a+b+c$ and thus $M$ has coordinate $\frac{a+b+c+d}{2}$. The symmetry of this expression in $a, b, c$, and $d$ implies that $M$ is also the midpoint of $AH_A, BH_B$, and $CH_C$, where $H_A, H_B$, and $H_C$ are the respective orthocenters of triangles $BCD, CDA$, and $DAB$. We now disregard the condition that $D$ is the midpoint of arc $BC$ and complete the problem using two symmetric applications of the following lemma.

Lemma 1.

Let $ABDC$ be a cyclic quadrilateral and $M$ the common midpoint of $AH_A, BH_B, CH_C$, and $DH_D$, where $H_A, H_B, H_C$, and $H_D$ are the respective orthocenters of triangles $BCD, CDA, DAB$, and $ABC$. Then the Simson line of $A$ with respect to triangle $BCD$ passes through $M$.

*Proof.* By definition, the Simson line passes through the feet $P, Q, X$ of the perpendiculars from $A$ to $BD, DC$, and $CB$. But $AH_C DH_B$ is also a cyclic quadrilateral since $\angle AH_C D = 180^\circ - \angle ABD = \angle ACD = 180^\circ - \angle AH_B D$, and the Simson line of $D$ with respect to triangle $H_B AH_C$ also passes through $P$ and $Q$ as well as the foot $Y$ of the perpendicular from $D$ to $H_B H_C$. We conclude by noting that $X$ and $Y$ are the feet of corresponding altitudes of triangles $ABC$ and $H_A H_B H_C$, which are congruent through a half-turn about $M$; hence $M$ is the midpoint of $XY$ and thus lies on line $PQ$. $\square$

By the lemma, $M$ lies on both of the Simson lines $l_a, m_a$ of the problem and thus coincides with $A_1$. Applying this result to the three sides of $\triangle ABC$ yields that triangles $A_1B_1C_1$ and $DEF$ are homothetic with ratio 2 about $H$.