Problem B. 5439. (February 2025)

B. 5439. Rectangle \(\displaystyle ABCD\) satisfies \(\displaystyle AD<AB<2AD\). Let point \(\displaystyle O\) be chosen on side \(\displaystyle AB\) satisfying \(\displaystyle OB=AD\). The circle with center \(\displaystyle O\) and radius \(\displaystyle OB\) intersects side \(\displaystyle AD\) at point \(\displaystyle E\). Prove that the area of rectangle \(\displaystyle ABCD\) equals \(\displaystyle BE^2/2\).

Proposed by: Viktor Vígh, Sándorfalva

(3 pont)

Deadline expired on March 10, 2025.

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

Megoldás. Jelölje a téglalap oldalhosszúságait \(\displaystyle AB = CD = a\) és \(\displaystyle BC = DA = b\). Ekkor a feltételek szerint \(\displaystyle OB = OE = b\) és \(\displaystyle AO = a-b\).

A Pitagorasz-tétel szerint az \(\displaystyle AOE\) derékszögű háromszögben:

\(\displaystyle AE^2 = OE^2 - AO^2 = b^2 - (a-b)^2 = 2ab - a^2; \)

míg az \(\displaystyle ABE\) derékszögű háromszögben:

\(\displaystyle BE^2 = AB^2 + AE^2 = a^2 + (2ab - a^2) = 2ab, \)

azaz valóban az \(\displaystyle ABCD\) téglalap területének kétszerese.

Statistics:

111 students sent a solution.
3 points:	98 students.
2 points:	7 students.
1 point:	3 students.
Not shown because of missing birth date or parental permission:	1 solutions.

Problems in Mathematics of KöMaL, February 2025