# Problem 7.

B-1 The height $B D$ of an acute-angled triangle $A B C$ intersects with its other heights at point $H$. Point $K$ lies on segment $A C$ such that the measure of angle $B K H$ is maximized. Find $D K$, if $A D=2, D C=3$.
Answer: $\sqrt{6}$

Solution. The larger the acute angle, the larger its tangent. Therefore, the condition of the maximum angle $B K H$ can be replaced by the condition of the maximum of its tangent. By the tangent of the difference formula, we have

$$
\begin{aligned}
\operatorname{tg} \angle B K H & =\operatorname{tg}(\angle B K D-\angle D K H)=\frac{\operatorname{tg} \angle B K D-\operatorname{tg} \angle D K H}{1+\operatorname{tg} \angle B K D \cdot \operatorname{tg} \angle D K H}=\frac{\frac{B D}{D K}-\frac{D H}{D K}}{1+\frac{B D}{D K} \cdot \frac{D H}{D K}} \\
& =\frac{(B D-D H) \cdot D K}{D K^{2}+B D \cdot D H}
\end{aligned}
$$

The maximum of this expression is achieved at the same value of $D K$ as the minimum of the expression $y=\frac{x^{2}+B D \cdot D H}{x}=x+\frac{B D \cdot D H}{x}$, where $x=D K$. The derivative $y^{\prime}$ is equal to $1-\frac{B D \cdot D H}{x^{2}}$ and is zero at $x=\sqrt{B D \cdot D H}$ (we are only interested in positive values of $x$).

Notice that $D H: A D=\operatorname{tg} \angle H A D=\operatorname{ctg} \angle B C D=D C: B D$, hence $A D \cdot D C=D H \cdot B D$. Thus, $x=\sqrt{A D \cdot D C}$.

V-2 The height $A M$ of an acute-angled triangle $A B C$ intersects with its other heights at point $O$. Point $K$ lies on segment $B C$ such that the measure of angle $A K O$ is maximized. Find $M K$, if $B M=5, M C=3$.

Answer: $\sqrt{15}$

V-3 The height $C L$ of an acute-angled triangle $A B C$ intersects with its other heights at point $Q$. Point $S$ lies on segment $A B$ such that the measure of angle $C S Q$ is maximized. Find $L S$, if $A L=2, L B=5$.

Answer: $\sqrt{10}$

V-4 The height $B F$ of an acute-angled triangle $A B C$ intersects with its other heights at point $P$. Point $N$ lies on segment $A C$ such that the measure of angle $B N P$ is maximized. Find $F N$, if $A F=7, F C=2$.

Answer: $\sqrt{14}$