# Problem A. 383. (November 2005)

**A. 383.** The diameter *AB* of the circle *k* is perpendicular to the line , which does not pass through either *A* or *B*. Let *C* be a point of outside *k*, and let *D* denote the other intersection of the line *AC* with the circle. Draw one of the tangents from *C* to *k *and denote the point of contact by *E*. Let *F* be the intersection of the lines *BE* and , and finally let *G* denote the intersection of the line segment *AF* with *k*. Show that the line *DF* passes through the reflection of the point *G* about *AB*.

*Based on a problem from Germany, suggested for the IMO*

(5 pont)

**Deadline expired on December 15, 2005.**

**Solution 1.** Denote the intersection of lines *AE* and *BG* by *H*. The orthocenter of triangle *ABH* is *F* and also *H* lies on *l* (Figure 1).

Figure 1 | Figure 2 |

Let *O* and *C*' be the midpoints of *AB* and *HF* and denote by *T* the intersction of *l* and *AB*. The 9-point circle of triangle *ABH* passes through points *E*, *G*, *O*, *T* and *C*'. From the right angle at *T* we know that *OC*' is a diameter of the 9-point circle and *CEO*=90^{o}. Therefore *C*'*E* is perpendicular to *OE* which is a radius of *k*. This implies that *C*'=*C* and *CE*=*CF*=*CG*=*CH*. Finally, *CG* touches *k* (Figure 2).

3. ábra

Now consider the inversion with pole *A* which maps *k* and *l* to each other. Points *C*,*D* and *F*,*G* are images of each other so these four point lie on a circle (Figure 3).

Denote by *G*' the second intersection of *DF* and *k*. Then

*G*'*GA*=*G*'*DA*=180^{o}-*CDF*=*FGC*=*CFG*.

and *GG*' is paralel to *l* and perpendicular to *AB*.

*Remark.* Since several possible configurations of the figure exist, one must use signed angles or consider each cases separately.

**Solution 2.** Denote by *H* the intersection of lines *BD* and *EG*. Apply Pascal's thorem to the degenerate hexagon *EEBDAG*. You obtain that *F*, *C* and *H* lie on a line which is *l* (Figure 4).

Figure 4

Let *G*' be the second intersection of *k* and *l*. Now apply Pascal's thorem for the degenerate hexagon *GG*'*DBBE* (Figure 5). You obtain that line *GG*' passes through the ideal intersection of *l* and the tangent line at *B*. Therefore, *GG*' is parallel to *l* and perpendicular to *AB*.

Figure 5

### Statistics:

13 students sent a solution. 5 points: Bogár 560 Péter, Jankó Zsuzsanna, Nagy 224 Csaba, Paulin Roland. 4 points: Hujter Bálint, Kisfaludi-Bak Sándor, Tomon István. 3 points: 5 students. 0 point: 1 student.

Problems in Mathematics of KöMaL, November 2005