Problem B. 4289.

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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 $\frac{r_2+r_4}{e}=\frac{r_1+r_3}{f}$ .

(4 points)

Deadline expired.

Sorry, the solution is published in Hungarian only.

A trapéz A_i csúcsánál levő szöget jelölje $\alpha$ _i. A szinusz-tétel szerint

e=2r₂sin $\alpha$ ₄=2r₄sin $\alpha$ ₂, f=2r₁sin $\alpha$ ₃=2r₃sin $\alpha$ ₁.

Ha a trapéz alapjai A₁A₂ és A₃A₄, akkor $\alpha$ ₁+ $\alpha$ ₄= $\alpha$ ₂+ $\alpha$ ₃= $\pi$ , vagyis sin $\alpha$ ₁=sin $\alpha$ ₄ és sin $\alpha$ ₂=sin $\alpha$ ₃. Ezáltal

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

Statistics on problem B. 4289.

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 solution.

Problems in Mathematics of KöMaL, September 2010

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