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C. 974. ABCD is a cyclic trapezium. Diagonal BD is extended beyond vertex D by the length of the leg and the point L is obtained. K is the point on diagonal AC whose distance from C is equal to the leg. Prove that the areas of the rectangles formed by the line segments AB and CD and the line segments AK and BL are equal.

(5 points)

Deadline expired on 16 February 2009.


Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás. Igazolandó: AB.CD=AK.BL, a szokásos jelölésekkel: ac=(e-b)(e+b). Ezt így is írhatjuk: e2=b2+ac. Az ADC háromszögben a koszinusz-tétel: e2=b2+c2-2bccos (180o-\alpha). Elegendő lenne bizonyítani, hogy c2-2bccos (180o-\alpha)=ac, azaz: c-2bcos (180o-\alpha)=a, vagyis \cos(180^{\circ}-\alpha)=\frac{\frac{c-a}{2}}{b}, illetve \cos\alpha=\frac{\frac{a-c}{2}}{b}. Ez pedig igaz.


Statistics on problem C. 974.
179 students sent a solution.
5 points:164 students.
4 points:4 students.
3 points:4 students.
2 points:2 students.
1 point:1 student.
0 point:1 student.
Unfair, not evaluated:3 solutions.


  • Problems in Mathematics of KöMaL, January 2009

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