On a math exam, there was a question that asked for the largest angle of the triangle with sidelengths \(21\), \(41\), and \(50\). A student obtained the correct answer as follows:

Let \(x\) be the largest angle. Then,
\[
\sin x = \frac{50}{41} = 1 + \frac{9}{41}.
\]
Since \( \sin 90^{\circ} = 1 \) and \( \sin 12^{\circ} 40′ 49” = 9/41 \), the angle \( x = 90^{\circ} + 12^{\circ} 40′ 49” = 102^{\circ} 40′ 49”\).

Find the triangle with the smallest area with integer sidelengths and possessing this property (that the wrong argument as above gives the correct answer).

The best solution was submitted by Han, Junho (한준호, 수리과학과 2015학번). Congratulations!

Here is his solution of problem 2018-11.

Alternative solutions were submitted by 이종원 (수리과학과 2014학번, +3),
채지석 (수리과학과 2016학번, +3), 고성훈 (2018학번, +2), 이본우 (수리과학과 2017학번, +2). One incorrect solution was submitted.