Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1169. (April 2013)

C. 1169. Consider a sphere of diameter d, a cylinder whose diameter and height are both d, and a cone of base diameter d. What may be the height of the cone if the volumes of the three solids form an arithmetic progression in some order?

(5 pont)

Deadline expired on May 10, 2013.

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

Megoldás. A gömb, a henger és a kúp térfogata rendre $\displaystyle V_{g}=\frac{\pi}{6}d^3$, $\displaystyle V_h=2\pi\cdot\left(\frac d2\right)^3=\frac{\pi}{4}d^3$ és $\displaystyle V_k=\frac{\pi}{3}\cdot\left(\frac d2\right)^2\cdot m=\frac{\pi}{12}d^2m$.

Mivel $\displaystyle \frac{\pi}{6}<\frac{\pi}{4}$, ezért $\displaystyle V_g<V_h$. Így három esetet kell megvizsgálnunk. 1. eset: $\displaystyle V_k<V_g<V_h$; 2. eset: $\displaystyle V_g<V_h<V_k$; 3. eset: $\displaystyle V_g<V_k<V_h$.

1. eset:

$\displaystyle \frac{\pi}{4}d^3-\frac{\pi}{6}d^3=\frac{\pi}{6}d^3-\frac{\pi}{12}d^2m.$

Mindkét oldalt $\displaystyle \frac{12}{d^2\pi}$-vel szorozva (nyilván $\displaystyle d\neq0$) kapjuk, hogy $\displaystyle 3d-2d=2d-m$, amiből $\displaystyle m_1=d$ következik.

2. eset:

$\displaystyle \frac{\pi}{12}d^2m-\frac{\pi}{4}d^3=\frac{\pi}{4}d^3-\frac{\pi}{6}d^3.$

Mindkét oldalt $\displaystyle \frac{12}{d^2\pi}$-vel szorozva kapjuk, hogy $\displaystyle m-3d=3d-2d$, amiből $\displaystyle m_2=4d$ következik.

3. eset: $\displaystyle 2V_k=V_g+V_h$, vagyis $\displaystyle \frac{2\pi}{12}d^2m=\frac{\pi}{6}d^3+\frac{\pi}{4}d^3=\frac{5\pi}{12}d^3$. Ebből pedig $\displaystyle \frac{6}{d^2\pi}$-vel szorozva $\displaystyle m_3=2,5d$ következik.

### Statistics:

 135 students sent a solution. 5 points: 86 students. 4 points: 15 students. 2 points: 14 students. 1 point: 9 students. 0 point: 7 students. Unfair, not evaluated: 4 solutionss.

Problems in Mathematics of KöMaL, April 2013