Problem B. 4383. (October 2011)
B. 4383. ABCD is a convex quadrilateral, such that AB>BC>CD>DA. The inscribed circles of the triangles ABD and BCD touch the diagonal BD of the quadrilateral at the points E and F, respectively. Show that EF=GH.
(Suggested by Sz. Miklós, Herceghalom)
(3 pont)
Deadline expired on November 10, 2011.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Az érintőszakaszok egyenlősége alapján
\(\displaystyle BE=\frac{BD+AB-DA}{2}\qquad\hbox{\rm és}\qquad BF=\frac{BD+BC-CD}{2},\)
ahonnan
\(\displaystyle BE-BF=\frac{(AB-BC)+(CD-DA)}{2}>0,\)
tehát
\(\displaystyle EF=BE-BF=\frac{(AB+CD)-(BC+DA)}{2}.\)
Hasonló módon kapjuk, hogy
\(\displaystyle GH=AH-AG=\frac{AC+AB-BC}{2}-\frac{AC+DA-CD}{2}=\frac{(AB+CD)-(BC+DA)}{2}.\)
Statistics:
98 students sent a solution. 3 points: 79 students. 2 points: 17 students. Unfair, not evaluated: 2 solutionss.
Problems in Mathematics of KöMaL, October 2011