Mathematical and Physical Journal
for High Schools
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# Problem B. 4289. (September 2010)

B. 4289. The diagonals of a trapezium A1A2A3A4 are A1A3=e and A2A4=f. Let ri denote the radius of the circumscribed circle of triangle AjAkAl, 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