Problem K/C. 758. (February 2023)

K/C. 758. A sail is shaped like a right angled triangle and it has the red symbol of the boat class at a height such that \(\displaystyle MA + AC = CB + BM\) (see the figure). If \(\displaystyle BM = 7\) m and \(\displaystyle CB = 5\) m, what is the height of the peak of the sail above the red mark?

(5 pont)

Deadline expired on March 10, 2023.

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

Megoldás. Az \(\displaystyle ABC\) derékszögű háromszögre felírva a Pitagorasz-tételt: \(\displaystyle AB^2 + CB^2 = AC^2\), azaz \(\displaystyle (MA + 7)^2 + 5^2 = AC^2\), ugyanakkor \(\displaystyle AC^2 = (CB + BM) - MA = 12 - MA\), amit az első egyenletbe helyettesítve:

\(\displaystyle (MA + 7)^2 + 5^2 = (12 – MA)^2.\)

Ebből \(\displaystyle MA^2 + 14MA + 49 + 25 = 144 - 24MA + MA^2\), egyszerűsítve \(\displaystyle 14MA + 74 = 144 - 24MA\), ahonnan \(\displaystyle 38MA = 70\), azaz \(\displaystyle MA \approx 1,84\) m távolságra van a vitorla felső csúcsától a jelzés.

Statistics:

208 students sent a solution.
5 points:	168 students.
4 points:	15 students.
3 points:	5 students.
2 points:	1 student.
0 point:	1 student.
Unfair, not evaluated:	10 solutionss.
Not shown because of missing birth date or parental permission:	5 solutions.

Problems in Mathematics of KöMaL, February 2023