Problems of Ukrainian Authors

Let $A_2$ be a projection of the point $H$ onto the edge $BC$. Then $\angle AOS = 180^\circ - \angle HA_2S = 180^\circ - \alpha$, and so $\angle AOH = \alpha$. This implies that $\triangle AOH \sim \triangle SA_2H \Rightarrow \frac{AH}{SH} = \frac{OH}{HA_2} \Rightarrow AH \cdot HA_2 = OH \cdot SH$. (fig. 43).

Analogously, $BH \cdot HB_2 = CH \cdot HC_2 = OH \cdot SH$. We will next prove that $H$ is the orthocenter of $\triangle ABC$. Since $\frac{AH}{HC_2} = \frac{CH}{HA_2}$ we have that $\sin \angle HAC_2 = \sin \angle HCA_2$. In view of the fact that $\angle HAC_2 + \angle HCA_2 < 180^\circ$ this implies that $\angle HAB = \angle HCB$, from which $\angle AHC_2 = \angle CHA_2$. Similarly we obtain that $\angle AHB_2 = \angle BHA_2$ and $\angle BHC_2 = \angle CHB_2$. From this it easily follows that $\angle HAB = \angle HCB = 90^\circ - \angle B$, which implies that $H$ belongs to the altitude of $\triangle ABC$ starting at $A$. Analogously, $H$ belongs to the other two altitudes of $\triangle ABC$. So, we see that the projection $H$ of the point $S$ onto the base of the tetrahedron is the orthocenter of the base, and the line segments $AA_2, BB_2$ and $CC_2$ are the altitudes of the base (fig. 44).

Consider the $\triangle ASA_2$ (see fig. 45). Since $\triangle AOH \sim \triangle SA_2H$, we have that $AA_1 \perp SA_2$.

Fig. 44

So the lines $SA_2$ and $AA_2$ are both perpendicular to the line $BC$, which means that the whole plane $ASA_2$ is perpendicular to $BC$. This implies that $AO \perp BC$. Similarly, in the plane $SBC$ we have two non-parallel lines $BC$ and $SA_2$ that are perpendicular to $AO$, and so $AO \perp SBC$. Analogously, $BO \perp SAC, CO \perp SBA$.

Now, since $ABC \parallel A_1B_1C_1$, we have that $AB \parallel A_1B_1$, and so the triangles $AOB$ and $A_1OB_1$ are similar. It follows that $\frac{OA_1}{AO} = \frac{OB_1}{BO} = t$. Also we have that $OA_1 = SO \sin \angle OSA_1$ and $OH = AO \sin \angle OAH = AO \sin \angle OSA_1$. This implies that $OA_1 \cdot OA = SO \cdot OH$. Analogously we can obtain that $SO \cdot OH = OA \cdot OA_1 = OB \cdot OB_1 = OC \cdot OC_1 \Rightarrow SO \cdot OH = t \cdot OA^2 = t \cdot OB^2 = t \cdot OC^2$.

It follows that $OA = OB = OC \Rightarrow AH = BH = CH$ and finally that $SA = SB = SC$.