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In a pyramid with a square as a base the side edges form angle $\alpha$ with the base. There is also a half-ball inside it, such that its radius $R$ is tangent to the sides and the great cicle is contained in the base of the pyramid. Find the volume of the pyramid.

Could someone point out where to start?

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    Well, first of all we can say: $$\mathcal{V}_\text{piramid}>\mathcal{V}_\text{half-ball}\space\Longleftrightarrow\space\frac{\text{b}^2\times\text{h}}{3}>\frac{1}{2}\times\frac{4\pi\times\text{R}^3}{3}\space\Longleftrightarrow\space\text{h}>2\pi\times\frac{\text{R}^3}{\text{b}^2}\tag1$$ And, we also can say: $$\text{R}=\frac{\text{b}}{2}\tag2$$ So, we get: $$\frac{\text{b}^2\times\text{h}}{3}>\frac{1}{2}\times\frac{4\pi\times\left(\frac{\text{b}}{2}\right)^3}{3}=\frac{\pi}{12}\times\text{b}^3\space\Longleftrightarrow\space\text{h}>\frac{\pi}{4}\times\text{b}\tag3$$ – 2017-02-27
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    Do you have any picture of your problem? – 2017-02-27

1 Answers 1

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Let the pyramid be $ABCDE$ where $ABCD$ is the base square and $E$ is the apex. Let $H$ be the orthogonal projection on $E$ onto the base plane defined by the square $ABCD$. Then, by assumption, $$\angle \, EAH = \angle \, EBH = \angle \, ECH= \angle \, EDH = \alpha$$ Consequently, all four right triangles $AHE, \, BHE, \, CHE$ and $DHE$ are congruent yielding $$AE = BE = CE = DE \,\,\text{ and } \,\, AH = BH = CH = DH$$ The latter equalities imply that $H$ is the center of the circle circumscribed around the square $ABCD$ so $H$ is in fact the center of the square, where the diagonals intersect. In particular, $H$ lies on $AC$ and on $BD$ so $EH$ lies in each of the two planes $ACE$ and $BDE$, meaning that $EH$ is the intersection line of the planes $ACE$ and $BDE$.

Now let $s$ be a sphere with center $O$ tangent to $AE, \, BE, \, CE$ and $DE$ and let $T_{AE}, \, T_{BE}, \, T_{CE}$ and $T_{DE}$ be the corresponding points of tangency. Then $$ET_{AE} = ET_{BE} = ET_{CE} = ET_{DE} \,\, \text{ and } \,\, OT_{AE} = OT_{BE} = OT_{CE} = OT_{DE}$$ Therefore the segments $T_{AE}T_{BE}, \, T_{BE}T_{CE},\, T_{CE}T_{DE},\, T_{DE}T_{AE}$ are parallel to $AB, \, BC, \, CD, \, DA$ respectively and are all equal to each other, so $T_{AE}T_{BE}T_{CE}T_{DE}$ is also a square parallel to square $ABCD$ so the plane of the former square is parallel to the plane of the latter one. Therefore, the line $EH$ which is orthogonal to $ABCD$ is also orthogonal to $T_{AE}T_{BE}T_{CE}T_{DE}$ so it passes through the center $H'$ of the square $T_{AE}T_{BE}T_{CE}T_{DE}$, by an argument analogous to the one above concerning $EH$ and $ABCD$. The circumcircle of $T_{AE}T_{BE}T_{CE}T_{DE}$ is centered at $H'$ and it coincides with the circle optained by intersection the sphere $s$ with the plane $T_{AE}T_{BE}T_{CE}T_{DE}$. Since $O$ lies on the line passing through the center $H'$ and perpendicular to the plane $T_{AE}T_{BE}T_{CE}T_{DE}$, the lines $OH'$ and $EH'$ coincide, which means that $O$ lies on $EH$. Finally, one concludes that when the center of the sphere $s$ lies on the plane $ABCD$, point $O$ must coincide with point $H$.

Finally, if the radius of $s$ is $R$ then look at one triangle say $AHE$. In it, $HT_{AE}$ is the altitude from the right angle vertex $H$. Therefore, triangle $AHT_{AE}$ is right so $$AH = \frac{HT_{AE}}{\sin{\alpha}} = \frac{R}{\sin{\alpha}}$$ Analogously, triangle $HET_{AE}$ is right and $\angle \, EHT_{AE} = \alpha$ so $$EH = \frac{HT_{AE}}{\cos{\alpha}} = \frac{R}{\cos{\alpha}}$$ Finally the area of square $ABCD$ is $$S_{ABCD} = \frac{1}{2} \, AC \cdot BD = \frac{1}{2} \cdot \, 2 \, AH \cdot 2 BH = 2 AH^2 = \frac{2\, R^2}{\sin^2{\alpha}}$$ Thus the volume of the pyramid is $$V = \frac{1}{3} \cdot \, S_{ABCD} \, \cdot \, EH = \frac{2 \, R^3}{3 \,\sin^2{(\alpha)} \,\cos{(\alpha)}}$$