IMO-Auswahlklausur

Geometry Difficulty 9.0 Shortlist Prove it Germany

Problem:

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ intersect at the point $E$ and whose sides $AD$ and $BC$ lie on lines that intersect at the point $F$ . Let the midpoints of the segments $AB$ and $CD$ be denoted by $G$ and $H$ , respectively. Prove that the line $EF$ is tangent at $E$ to the circle through $E$ , $G$ and $H$ .

Solution

Solution:

A homothety with center $E$ and scale factor $2$ maps $G$ to $G'$ and $H$ to $H'$ . Then

(1) $\text{GHE = G'H'E}$ .

Since $\text{ABF = 180 - CBA = ADC}$ and $\text{FAB = 180 - BAD = DCB}$ (cyclic quadrilateral $ABCD$ ), the triangles $FBA$ and $FDC$ are similar and are mapped onto each other by a glide reflection with center $F$ along the bisector of $\text{BFA}$ . Under this map, in particular $A \rightarrow C$ , $B \rightarrow D$ , $G \rightarrow H$ (midpoints!).

By the definition of $H'$ , the diagonals $EH'$ and $CD$ bisect each other; hence $DECH'$ is a parallelogram and we have $\text{H'CD = EDC = BDC = BAC = BAE}$ . Analogously, $\text{CDH' = EBA}$ . Thus the triangles $ABE$ and $CDH'$

are similar. Because of the similarity of $FBA$ and $FDC$ , the quadrilaterals $FBEA$ and $FDH'C$ are also similar, so that under the glide reflection under consideration $E$ is mapped to $H'$ . Hence the triangles $FGE$ and $FHH'$ are similar, and we have

(2) $\text{GEF = HH'F = EH'F}$ .

Let $C'$ and $D'$ be the preimages of $A$ and $B$ , respectively, under the glide reflection under consideration. Then, because of the similarity of the triangles $FC'D'$ and $FAB$ and because $|FB| / |FD'| = |FD| / |FB|$ , the quadrilateral $D'C'BA$ is a cyclic quadrilateral similar to $BADC$ . Let its diagonal intersection point be $G''$ . Then $\text{G''AB = C'AB = C'D'B = ABD = ABE}$ holds, and it follows that $G''A \parallel BE$ and analogously $G''B \parallel AE$ . Hence $AG''BE$ is a parallelogram, and by the definition of $G'$ we have $G' = G''$ . Consequently, $G'$ is mapped to $E$ under the glide reflection under consideration. But since $E$ is mapped to $H'$ , under the two-fold application of the map $G'$ is mapped to $H'$ . In this two-fold application the two reflections cancel each other out, leaving a homothety centered at $F$ . Therefore $F$ , $G'$ and $H'$ are collinear. From this it follows that

(3) $\text{FH'E = G'H'E}$ .

From (1), (2) and (3) it now follows that $\text{GHE = G'H'E = FH'E = FEG}$ .

From the converse of the tangent-chord angle theorem it follows that $FE$ is tangent to the circle through $E$ , $G$ and $H$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from de; metadata (topic, difficulty) added by this project.