Problems of Ukrainian Authors

Let $A_2$ be a projection of the point $H$ onto the edge $BC$. Then $\angle AOS=180^{\circ }-\angle H{A}_{2}S=180^{\circ }-\alpha$, and so $\angle AOH=\alpha$. This implies that $△AOH\sim △S{A}_{2}H\Rightarrow \frac{AH}{SH}=\frac{OH}{H{A}_2}\Rightarrow AH\cdot H{A}_2=OH\cdot SH$. (fig. 43).

Analogously, $BH\cdot H{B}_2=CH\cdot H{C}_2=OH\cdot SH$. We will next prove that $H$ is the orthocenter of $△ABC$. Since $\frac{AH}{H{C}_2}=\frac{CH}{H{A}_2}$ we have that $\sin \angle HA{C}_2=\sin \angle HC{A}_2$. In view of the fact that $\angle HA{C}_2+\angle HC{A}_2<180^{\circ }$ this implies that $\angle HAB=\angle HCB$, from which $\angle AH{C}_2=\angle CH{A}_2$. Similarly we obtain that $\angle AH{B}_2=\angle BH{A}_2$ and $\angle BH{C}_2=\angle CH{B}_2$. From this it easily follows that $\angle HAB=\angle HCB=90^{\circ }-\angle B$, which implies that $H$ belongs to the altitude of $△ABC$ starting at $A$. Analogously, $H$ belongs to the other two altitudes of $△ABC$. 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 $A{A}_2,B{B}_2$ and $C{C}_2$ are the altitudes of the base (fig. 44).

Consider the $△AS{A}_2$ (see fig. 45). Since $△AOH\sim △S{A}_{2}H$, we have that $A{A}_1\perp S{A}_2$.

Fig. 44

So the lines $S{A}_2$ and $A{A}_2$ are both perpendicular to the line $BC$, which means that the whole plane $AS{A}_2$ is perpendicular to $BC$. This implies that $AO\perp BC$. Similarly, in the plane $SBC$ we have two non-parallel lines $BC$ and $S{A}_2$ that are perpendicular to $AO$, and so $AO\perp SBC$. Analogously, $BO\perp SAC,CO\perp SBA$.

Now, since $ABC\parallel A_1{B}_1{C}_1$, we have that $AB\parallel A_1{B}_1$, and so the triangles $AOB$ and $A_{1}O{B}_1$ are similar. It follows that $\frac{O{A}_1}{AO}=\frac{O{B}_1}{BO}=t$. Also we have that $O{A}_1=SO\sin \angle OS{A}_1$ and $OH=AO\sin \angle OAH=AO\sin \angle OS{A}_1$. This implies that $O{A}_1\cdot OA=SO\cdot OH$. Analogously we can obtain that $SO\cdot OH=OA\cdot O{A}_1=OB\cdot O{B}_1=OC\cdot O{C}_1\Rightarrow SO\cdot OH=t\cdot O{A}^2=t\cdot O{B}^2=t\cdot O{C}^2$.

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