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Problem B. 4517. (February 2013)

B. 4517. A\neB are interior points of the quadrant XY centred at O. The parallels drawn to the line OX through the points A, B intersect the radius OY at the points AY and BY. The lines drawn parallel to line OY intersect the radius OX at AX and BX. Determine the total area of the quadrilaterals AAXBXB and AAYBYB as a function of the length AB.

Suggested by Gy. Károlyi, Budapest, Brisbane

(4 pont)

Deadline expired on March 11, 2013.


Sorry, the solution is available only in Hungarian. Google translation

Megoldási ötlet: Használjunk szögfüggvényeket.

 

Megoldás. Legyen AOA_X\sphericalangle=\alpha, BOB_X\sphericalangle=\beta. Helyezzük el az ábrát a derékszögű koordinátarendszerben az ábra szerint, vizsgáljuk azt az esetet, amikor \beta\ge\alpha.

Az AAXBXB és a AAYBYB négyszög is trapéz. A területük összege


t_{AA_XB_XB} + t_{AA_YB_YB} =
\tfrac12 (AA_X+BB_X)\cdot A_XB_X + \tfrac12 (AA_Y+BB_Y)\cdot A_YB_Y =


= \tfrac12 (\sin\alpha+\sin\beta) \cdot (\cos\alpha-\cos\beta)
+ \tfrac12 (\cos\alpha+\cos\beta) \cdot (\sin\beta-\sin\alpha) =

=cos \alphasin \beta-sin \alphacos \beta=sin (\beta-\alpha).

Az OAB egyenlő szárú háromszögből látható, hogy


AB = 2 \sin\frac{\beta-\alpha}{2},

amiből


\sin(\beta-\alpha) =
2 \sin\frac{\beta-\alpha}2 \cos\frac{\beta-\alpha}2 =
2 \sin\frac{\beta-\alpha}2 \sqrt{1-\sin^2\frac{\beta-\alpha}2} =
AB\cdot \sqrt{1-\frac{AB^2}4}.

Tehát


t_{AA_XB_XB} + t_{AA_YB_YB} = AB\cdot \sqrt{1-\frac{AB^2}4}.


Statistics:

72 students sent a solution.
4 points:60 students.
3 points:7 students.
2 points:2 students.
1 point:3 students.

Problems in Mathematics of KöMaL, February 2013