Подготовка

[rsoldo]

Let be $\alpha,\beta,\gamma$ angles of a triangle $ABC$ and point $D$ on side $AB$.
i) Determine the maximum of function:
$f(\varphi)=\sin\varphi\cdot\sin(\gamma-\varphi)\quad \left(0\leq\varphi\leq\frac{\gamma}{2}\right).$
ii) If it's $0<\sin\alpha\sin\beta\leq max(f(\varphi)),$ prove that there is a point $D$ on side $AB$ so that $CD$ geometric mean proportional of lines $AD$ and $BD$ ( e.g. $|CD|^{2}=|AD|\cdot|BD|$ ) and converse.

[rsoldo]

a) Note that $f(x)=f(\gamma-x)$ so we can also look at $x\in [0,\gamma]$. We have $f(0)=f(\gamma)=0$ and $2f(x)=2\sin x \sin (\gamma-x)=\cos(2x-\gamma)-\cos \gamma=2\cos^2 (\gamma/2 -x)-1-\cos \gamma$.
We have $\gamma/2-x\in [0,\gamma/2]$ and the form is obviously maximal if $\gamma/2-x=0 \implies x=\gamma/2$. The maximal value is $\sin^2(\gamma/2)$.

P.S. I hope this part is ok!?

[StaticZero]

Consider the triangle (angles $\alpha = A, \ldots, \gamma = C$)

$$ \begin{tikzpicture} \coordinate [left] (A) at (0, 0) \coordinate [above] (B) at (2, 2) \coordinate [right] (C) at (3, 0) \coordinate [left] (D) at (0.667, 0.667) \draw (A) node[left] {$A$} -- (B) node[above] {$B$} -- (C) node[right] {$C$} -- cycle; \path (A) -- (D) node [pos=0.5, left] {$p$} \path (D) -- (B) node [pos=0.5, left] {$q$} \node[left] at (D) {$D$} \draw (C)--(D) node [pos=0.5, above] {$d$} \end{tikzpicture} $$

Let $\delta$ be the angle $ADC$. Using the sine theorem

$$ \frac{d}{\sin \alpha} = \frac{AC}{\sin \delta}, \qquad \frac{d}{\cos \alpha} = \frac{BC}{\sin \delta} $$

and multiplying both

$$ \frac{d^2}{\sin \alpha \sin \beta} = \frac{AC \cdot BC}{\sin^2 \delta} $$

On the other hand, let $\gamma_1$ be the angle $ACD$ so $\gamma_2$ is defined using $\gamma_1 + \gamma_2 = \gamma$. We use the same trick again

$$ \frac{p}{\sin \gamma_1} = \frac{AC}{\sin \delta}, \qquad \frac{q}{\sin \gamma_2} = \frac{BC}{\sin \delta} $$

now multiplying both and inserting earlier equations yields

$$ \frac{d^2}{\sin \alpha \sin \beta} = \frac{AC \cdot BC}{\sin^2 \delta} = \frac{pq}{\sin \gamma_1 \sin \gamma_2} $$

So if $\sin \alpha \sin \beta$ exceeds the maximal value of $\sin \gamma_1 \sin \gamma_2$, we are in trouble :) And what is that maximal value is a completely separate question.

We can make an observation. The value of $d$ is between $\min$ and $\max$ of the pair $AC, BC$. So, if $d \geqslant \min(AC, BC) > (p + q)/2 = AB/2 \geqslant \sqrt{pq}$, there are no $D$ points such that $d = \sqrt{pq}$. Consider $AC = \min(AC, BC)$ for certainty. We have

$$ \frac{AC}{\sin \beta} = \frac{BC}{\sin \alpha} = \frac{AB}{\sin \gamma}, $$

and if $AC/AB > 1/2$, we have $\sin \beta / \sin \gamma > 1/2$. Generally speaking,

$$ \min(\sin \alpha, \sin \beta) \leqslant \frac{\sin \gamma}{2} $$

is an additional constraint. Is it automatically fulfilled? We have

$$ \min(\sin \alpha, \sin \beta) \leqslant \sin \frac{\gamma}{2} \cos \frac{\gamma}{2} $$

and $\sin (\gamma/2) \geqslant \sqrt{\sin \alpha \sin \beta} \geqslant \min(\sin \alpha, \sin \beta)$,
but here we have a $\cos$ in the right side, so it occurs to be a stronger constraint than presented!

[kotenok gav]

rsoldo в сообщении #1492799 писал(а):

a) Note that $f(x)=f(\gamma-x)$ so we can also look at $x\in [0,\gamma]$. We have $f(0)=f(\gamma)=0$ and $2f(x)=2\sin x \sin (\gamma-x)=\cos(2x-\gamma)-\cos \gamma=2\cos^2 (\gamma/2 -x)-1-\cos \gamma$.
We have $\gamma/2-x\in [0,\gamma/2]$ and the form is obviously maximal if $\gamma/2-x=0 \implies x=\gamma/2$. The maximal value is $\sin^2(\gamma/2)$.

P.S. I hope this part is ok!?

It is much easier. $f(\varphi)=\dfrac{\cos(\gamma-2\varphi)-\cos\gamma}2$ and maximum of $\cos(\gamma-2\varphi)$ is at $\varphi=\dfrac\gamma2$.