Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

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 solutions.

Problems in Mathematics of KöMaL, December 2007