10. The lateral edges of the triangular pyramid $TABC$ form right angles with each other. What is the minimum area that the section of the pyramid by a plane passing through vertex $C$ and the midpoint of side $AB$ of the base can have, if the base edge $AC=5$ and the lateral edges $TA=4, TB=12$? Which of the lateral edges does the plane intersect in this case, and into what parts does it divide it?
# Solution:

If the secant plane intersects the lateral edge $T B$, the area of the section will be the smallest if $K N$ is the common perpendicular to the lines $T B$ and $C S$.

If the secant plane intersects the lateral edge $T A$, the area of the section will be the smallest if $M E$ is the common perpendicular to the lines $T A$ and $C S$.

By constructing the common perpendiculars $K N=H T$, where $H T \perp C G$, and $M E=T D$, where $T D \perp C F$. In all variants, $T A<T B$, i.e., we need to determine the length of $T H$ and the position of point $N$ on the edge $T B$.

If we denote $T A=a, T B=b, T C=c$, then $C F=\sqrt{(b / 2)^{2}+c^{2}}$;

$C S=\sqrt{T F^{2}+F S^{2}}=\sqrt{(a / 2)^{2}+c^{2}+(b / 2)^{2}}, h=H T=\frac{T C \cdot T G}{T G}=\frac{a c}{2 \sqrt{(a / 2)^{2}+c^{2}}}$.

$S_{\triangle C S N}=\frac{1}{2} K N \cdot C S=\frac{a c \sqrt{(a / 2)^{2}+(b / 2)^{2}+c^{2}}}{2 \cdot 2 \sqrt{(a / 2)^{2}+c^{2}}}$.

Point $N$ divides the segment $T B$ in the ratio $\frac{B N}{T N}=1+\frac{1}{2} \frac{a^{2}}{c^{2}}$.

| $T A$ | $T B$ | $T C$ | $C S$ | $S_{\triangle N C S}$ | $B N: T N$ | $B N$ | $T N$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 4 | 12 | 3 | 7 | $21 / \sqrt{13}$ | $17: 9$ | $108 / 13$ | $54 / 13$ |

(*) $\frac{H K}{G S}=\frac{C H}{C G} ; \frac{H K}{G S}=\frac{\sqrt{T C^{2}-H T^{2}}}{C G}$;

$\frac{H K}{b / 2}=\frac{\sqrt{c^{2}-h^{2}}}{\sqrt{c^{2}+(a / 2)^{2}}}=\frac{\sqrt{c^{2}-a^{2} c^{2} /\left(a^{2}+4 c^{2}\right)}}{\sqrt{c^{2}+(a / 2)^{2}}}=\frac{4 c^{2}}{a^{2}+4 c^{2}}$.
$T N=H K=\frac{b}{2} \cdot \frac{4 c^{2}}{a^{2}+4 c^{2}}=\frac{2 b c^{2}}{a^{2}+4 c^{2}} \cdot B N=b-\frac{2 b c^{2}}{a^{2}+4 c^{2}}=b \frac{a^{2}+2 c^{2}}{a^{2}+4 c^{2}}$.

$\frac{B N}{T N}=\frac{b\left(a^{2}+2 c^{2}\right)\left(a^{2}+4 c^{2}\right)}{\left(a^{2}+4 c^{2}\right) \cdot 2 b c^{2}}=\frac{a^{2}+2 c^{2}}{2 c^{2}}=1+\frac{1}{2} \frac{a^{2}}{c^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_50f0bbe00f908a027b9ag-08.jpg?height=1102&width=1171&top_left_y=1565&top_left_x=565)

## Solution for Variant 28