Mathematical and Physical Journal
for High Schools
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Problem B. 4814. (October 2016)

B. 4814. Let \(\displaystyle P\) be a given point in the interior of a sphere. Consider the pairwise perpendicular planes \(\displaystyle S_1\), \(\displaystyle S_2\) and \(\displaystyle S_3\) that intersect at \(\displaystyle P\). Show that the sum of the areas of the circles cut out of the planes by the sphere is independent of the choice of the planes \(\displaystyle S_1\), \(\displaystyle S_2\) and \(\displaystyle S_3\).

(Italian problem)

(4 pont)

Deadline expired on November 10, 2016.


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

Megoldás. Legyen az adott \(\displaystyle \mathcal G\) gömb középpontja \(\displaystyle O\), sugara \(\displaystyle r\), az \(\displaystyle O\) pont merőleges vetülete az \(\displaystyle S_1, S_2\) és \(\displaystyle S_3\) síkokra pedig rendre \(\displaystyle X_1, X_2\) és \(\displaystyle X_3\). Ekkor a térbeli Pitagorasz-tétel szerint

\(\displaystyle \overline{OX_1}^2+\overline{OX_2}^2+\overline{OX_3}^2=\overline{OP}^2.\)

Vegyük észre, hogy \(\displaystyle \mathcal G \cap S_1\) egy \(\displaystyle X_1\) középpontú kör, aminek sugara \(\displaystyle r_1=\sqrt{r^2-\overline{OX_1}^2}\), így területe \(\displaystyle t_1=r_1^2\pi=(r^2-\overline{OX_1}^2)\pi\).

Innen

\(\displaystyle t_1+t_2+t_3=\pi(3r^2-(\overline{OX_1}^2+\overline{OX_2}^2+\overline{OX_3}^2))=\pi(3r^2-\overline{OP}^2),\)

ami valóban csak \(\displaystyle P\)-től és \(\displaystyle \mathcal G\)-től függ, de független a síkok konkrét választásától.


Statistics:

89 students sent a solution.
4 points:87 students.
3 points:1 student.
0 point:1 student.

Problems in Mathematics of KöMaL, October 2016