Problem B. 4289. (September 2010)

B. 4289. The diagonals of a trapezium A₁A₂A₃A₄ are A₁A₃=e and A₂A₄=f. Let r_i denote the radius of the circumscribed circle of triangle A_jA_kA_l, where {1,2,3,4}={i,j,k,l}. Show that .

(4 pont)

Deadline expired on October 11, 2010.

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

Megoldás. A trapéz \(\displaystyle A_i\) csúcsánál levő szöget jelölje \(\displaystyle \alpha_i\). A szinusz-tétel szerint

\(\displaystyle e=2r_2\sin\alpha_4=2r_4\sin\alpha_2,\quad f=2r_1\sin\alpha_3=2r_3\sin\alpha_1.\)

Ha a trapéz alapjai \(\displaystyle A_1A_2\) és \(\displaystyle A_3A_4\), akkor \(\displaystyle \alpha_1+\alpha_4=\alpha_2+\alpha_3=\pi\), vagyis \(\displaystyle \sin\alpha_1=\sin\alpha_4\) és \(\displaystyle \sin\alpha_2=\sin\alpha_3\). Ezáltal

\(\displaystyle \frac{r_2+r_4}{e}= \frac{1}{2}\left(\frac{1}{\sin\alpha_4}+\frac{1}{\sin\alpha_2}\right)= \frac{1}{2}\left(\frac{1}{\sin\alpha_1}+\frac{1}{\sin\alpha_3}\right)= \frac{r_3+r_1}{f}.\)

Statistics:

74 students sent a solution.
4 points:	63 students.
3 points:	2 students.
2 points:	3 students.
1 point:	2 students.
0 point:	3 students.
Unfair, not evaluated:	1 solutions.

Problems in Mathematics of KöMaL, September 2010