Mathematical and Physical Journal
for High Schools
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Problem C. 1409. (March 2017)

C. 1409. The trunk of an old plane tree is cylindrical, with a perimeter of 2 metres. On one side of the tree trunk, a snail is crawling up the tree, in the plane of the axis of the cylinder. There are only 3 cm remaining from a height of 2 metres. In that time instant, another snail is just about to start its journey up the tree on the opposite side when they both notice each other. The two snails immediately start to crawl towards each other along the shortest possible path. What distance will they cover until they meet if their speeds are equal?

(5 pont)

Deadline expired on April 10, 2017.

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

Megoldás. Terítsük ki a fa henger alakú törzsét síkba 2 méteres magasságig. Mivel a kerület is 2 m, így az \(\displaystyle ABCD\) négyzetet kapjuk.

Az egyik csiga legyen az \(\displaystyle AB\) alap \(\displaystyle F\) felezőpontjában, ekkor a másik csiga a \(\displaystyle BC\) oldalon \(\displaystyle C\)-től 3 cm-re, \(\displaystyle B\)-től pedig 197 cm-re lévő \(\displaystyle G\) pontjában van. A köztük lévő legrövidebb távolság az \(\displaystyle FBG\) derékszögű háromszög \(\displaystyle FG\) átfogójának hossza. \(\displaystyle FB=100\) cm, \(\displaystyle BG=197\) cm, így

\(\displaystyle FG=\sqrt{FB^2+BG^2}=\sqrt{100^2+197^2}≈220,93 \mathrm{~cm}.\)

Mivel a csigák egyenlő sebességgel haladnak, így mindkettő \(\displaystyle s=\frac{FG}{2}≈110,46\) cm-t tesz a találkozásig (a csigák méretét elhanyagoljuk).


189 students sent a solution.
5 points:143 students.
4 points:27 students.
3 points:6 students.
2 points:2 students.
1 point:8 students.
0 point:1 student.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, March 2017