Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem K. 628. (September 2019)

K. 628. Four identical rectangular sheets of paper are placed on the table to form a larger rectangle without gaps or overlaps. The area of the resulting rectangle is $\displaystyle 1200~\mathrm{cm}^{2}$. Given that it is not possible to transfer any rectangle to any other only by translation, find the perimeter of the large rectangle.

(6 pont)

Deadline expired on October 10, 2019.

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

Megoldás. A négy téglalap a feltételeknek megfelelően lényegében háromféleképpen adhat ki egy nagyobbat, de a harmadik elrendezést az utolsó feltétel nem engedi meg.

Az első elrendezés esetén $\displaystyle b = 3a$, $\displaystyle T = 3a(a + b) = 3a(4a) = 1200$, ahonnan $\displaystyle a^{2} = 100$, $\displaystyle a = 10$ és $\displaystyle b = 30$. A kerülete ekkor $\displaystyle 2(10 + 30 + 3\cdot10) = 140$ cm.

A második elrendezésben $\displaystyle d = 2c$, $\displaystyle T = 2c(2c+d) = 2c(4c) = 1200$, ahonnan $\displaystyle 8c^{2} = 1200$, $\displaystyle c^{2} = 150$, $\displaystyle c = 5 \sqrt 6$ és $\displaystyle d = 10 \sqrt 6$. A kerülete ekkor $\displaystyle 2\big(2\cdot 5 \sqrt 6 + 10 \sqrt 6 + 10 \sqrt 6 \big) = 60 \sqrt 6$ cm.

### Statistics:

 161 students sent a solution. 6 points: 56 students. 5 points: 12 students. 4 points: 47 students. 3 points: 7 students. 2 points: 2 students. 1 point: 3 students. 0 point: 19 students. Unfair, not evaluated: 4 solutionss. Not shown because of missing birth date or parental permission: 11 solutions.

Problems in Mathematics of KöMaL, September 2019