Problem C. 924. (December 2007)

C. 924. The areas of the rectangles obtained by intersecting a certain cuboid with a plane passing through two parallel edges may be t₁=60, $t_2=4\sqrt{153}$ , or $t_3=12\sqrt{10}$ . Calculate the volume and surface area of the cuboid.

(5 pont)

Deadline expired on January 15, 2008.

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

Megoldás. Tudjuk, hogy $t_1=a\sqrt{b^2+c^2}=60$ , $t_2=b\sqrt{c^2+a^2}=4\sqrt{153}$ , $t_3=c\sqrt{a^2+b^2}=12\sqrt{10}$ . Vagyis a következő egyenletrendszert írhatjuk fel:

a²b²+a²c²=3600,

b²c²+a²b²=2448,

a²c²+b²c²=1440.

A három egyenlet összegének a fele: a²b²+a²c²+b²c²=3744. Ebből az egyenletből az egyenletrendszer egy-egy egyenletét kivonva kapjuk, hogy b²c²=144, a²c²=1296, a²b²=2304. Azaz: bc=12, ac=36, ab=48.

A=2(ab+ac+bc)=2(48+36+12)=192.

$V=\sqrt{ab\cdot ac \cdot bc}=\sqrt{48\cdot36\cdot12}=144.$

Statistics:

262 students sent a solution.

5 points: 202 students.

4 points: 22 students.

3 points: 8 students.

2 points: 11 students.

1 point: 9 students.

0 point: 7 students.

Unfair, not evaluated: 3 solutionss.

Problems in Mathematics of KöMaL, December 2007

262 students sent a solution.
5 points:	202 students.
4 points:	22 students.
3 points:	8 students.
2 points:	11 students.
1 point:	9 students.
0 point:	7 students.
Unfair, not evaluated:	3 solutionss.

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