Mathematical and Physical Journal
for High Schools
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Problem C. 959. (October 2008)

C. 959. The diameter of one base of a truncated cone is 100 mm. If the diameter is increased by 21%, leaving the height and the size of the other base unchanged, the volume of the truncated cone will also increase by 21%. What is the diameter of the other base?

(5 pont)

Deadline expired on November 17, 2008.

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

Megoldás. Az adott alapkör sugara 50 mm. Jelölje a másik alapkör sugarát r, a csonkakúp magasságát pedig m. Ekkor a csonkakúp térfogata \frac{\pi\cdot m}{3}\cdot(r^2+50r+50^2). A feltétel szerint

1,21\cdot\frac{\pi\cdot m}{3}\cdot(r^2+50r+50^2)=
\frac{\pi\cdot m}{3}\cdot(r^2+50\cdot1,21r+(1,21\cdot50)^2),

az egyenlet mindkét oldalát osztva \frac{\pi\cdot m}{3}-mal, majd rendezve kapjuk, hogy:


amiből r=55 mm.

A másik alapkör átmérője 2r=110 mm.


421 students sent a solution.
5 points:290 students.
4 points:88 students.
3 points:12 students.
2 points:7 students.
1 point:8 students.
0 point:8 students.
Unfair, not evaluated:8 solutions.

Problems in Mathematics of KöMaL, October 2008