Mathematical and Physical Journal
for High Schools
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# 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 solutions.

Problems in Mathematics of KöMaL, October 2011