Auswahlklausur /, 2014

Solution:

Without loss of generality, let $|AC| > |AB|$. Since $D$ and $E$ are the midpoints of sides $\overline{AB}$ and $\overline{AC}$, by the intercept theorem $DE$ is parallel to $BC$. In the case $E = P$ or $D = Q$, $CEDB$ would be a cyclic quadrilateral with parallel sides $BC$ and $DE$, hence an isosceles trapezoid with $|BD| = |EC|$, so $|AB| = |AC|$, which is excluded. If one can show that the (non-degenerate) triangles $APC$ and $BQA$ are similar, the claim follows, since due to $\angle PQA = \measuredangle PDA = 180^{\circ} - \measuredangle BDP = \measuredangle PCB = \measuredangle PCA + \gamma = \measuredangle BAQ + \measuredangle AED = \measuredangle DPQ + \measuredangle APD = \measuredangle APQ$ (using the inscribed angle theorem) the triangle $APQ$ is isosceles with $|AP| = |AQ|$.

Proof that triangles $APC$ and $BQA$ are similar: First, by the inscribed angle theorem we have
$$\measuredangle DQB = 360^{\circ} - \measuredangle EQD - \measuredangle BQE = \measuredangle DAE + \measuredangle ECB = \alpha + \gamma = 180^{\circ} - \beta = 180^{\circ} - \measuredangle EDA = \measuredangle APE,$$
$\measuredangle EPC = \measuredangle DPC - \measuredangle DPE = (180^{\circ} - \beta) - \alpha = \gamma = \measuredangle AED = \measuredangle AQD$.

From this follows the similarity of triangles $APC$ and $BQA$:

1st proof: Let $Q'$ be the uniquely determined point on the same side of $AC$ as $P$, such that triangle $AQ'C$ is similar to triangle $BQA$. Since $D$ and $E$ are the midpoints of sides $\overline{AB}$ and $\overline{AC}$, the sub-triangles $AQ'E$ and $BQD$, as well as $CEQ'$ and $ADQ$, are also similar. Thus, because of (1) and (2), i.e. $\measuredangle APE = \measuredangle BQD = \measuredangle AQ'E$ and $\measuredangle EPC = \measuredangle DQA = \measuredangle EQ'C$, $Q'$ lies on the circumcircles of $AEP$ and $CEP$, which intersect at $E$ and $P$. $Q' = E$ is excluded since $Q \neq D$. Thus $Q' = P$, and the triangles $APC$ and $BQA$ are similar.

2nd proof: By the law of sines in $ADQ$ and $BQD$ we have $|AQ| : |AD| = \sin \measuredangle QDA : \sin \measuredangle AQD$, $|BQ| : |BD| = \sin \measuredangle BDQ : \sin \measuredangle DQB$. With $\measuredangle BDQ = 180^{\circ} - \measuredangle QDA$ it follows that $|AQ| : |BQ| = \sin \measuredangle DQB : \sin \measuredangle AQD$. Analogously, $|CP| : |PA| = \sin \measuredangle APE : \sin \measuredangle EPC$. Because of (1) and (2), $APC$ and $BQA$ are similar by the SAS similarity criterion.

Solution:

Using inversion at a circle: Invert with respect to a circle centered at $A$ with radius $1$. Denote the image point of a point $X$ by $X'$. Since $D$ and $E$ are the midpoints of sides $\overline{AB}$ and $\overline{AC}$, $B'$ and $C'$ are the midpoints of sides $\overline{AD'}$ and $\overline{AE'}$. The inversion maps the circumcircle of triangle $ADE$ to the line $D'E'$, so $P' \neq D'$ and $Q' \neq E'$ are the second intersection points of the circumcircles of triangles $B'C'D'$ and $B'C'E'$ with the line $D'E'$. Since the lines $D'E'$ and $B'C'$ are parallel, the circumcircles of triangles $B'C'D'$ and $B'C'E'$ are symmetric with respect to the perpendicular bisector of segment $\overline{B'C'}$; in particular, the points $Q'$, $B'$, $P'$ map to $E'$, $C'$, $D'$, respectively. Thus $|P'C'| = |B'D'| = |AB'|$ (since $B'$ is the midpoint of $\overline{AD'}$) and $|B'Q'| = |C'E'| = |AC'|$ (since $C'$ is the midpoint of $\overline{AE'}$) and $\measuredangle Q'B'D' = \measuredangle P'C'E'$. From this follows $\measuredangle AB'Q' = 180^{\circ} - \measuredangle Q'B'D' = 180^{\circ} - \measuredangle P'C'E' = \measuredangle AC'P'$. Thus the triangles $AP'C'$ and $AQ'B'$ are congruent by the SAS congruence criterion. Hence $|AP'| = |AQ'|$ and $|AP| = 1 / |AP'| = 1 / |AQ'| = |AQ|$.