In triangle $ABC$ , $CH$ is a height and we have: $AB\leq 2CH$.Find maximum value of $\angle C$.

Question

In triangle $ABC$ , $CH$ is a height and we have: $AB\leq 2CH$.Find maximum value of $\angle C$.

From the given relation I derived the following:
$$tan(A)+tan(B)\leq 2$$
But can't go on this way.Also I think the rule of Cosines may help...

Answer 1

$$AB\leq 2CH$$ gives $$c\leq\frac{4S}{c}$$ or $$2ab\sin\gamma\geq c^2$$ or $$2ab\sin\gamma\geq a^2+b^2-2ab\cos\gamma$$ or by AM-GM $$\sin(45^{\circ}+\gamma)\geq\frac{a^2+b^2}{2\sqrt2 ab}\geq\frac{1}{\sqrt{2}}.$$ Thus, $$\gamma+45^{\circ}\leq135^{\circ}$$ or $$\gamma\leq90^{\circ}.$$ The equality occurs for $a=b$ and $CH=\frac{c}{2}$, which says that $90^{\circ}$ is a maximal value.

Answer 2

Here is my idea. First, try to break $∠C$ into two angles, namely $\angle_1$ and $\angle_2$. Note that $\tan( \angle_1) = \frac{HB}{CH}$ and $\tan(\angle_2) = \frac{HA}{CH}$. Hence by using tangent formula, we get $\tan(\angle C) = \frac{\tan(\angle_1)+ \tan(\angle_2)}{ 1-\tan(\angle_1)\tan(\angle_2)} $.

Notice that we have the following inequality $$\frac{a+b}2 \geq \sqrt{ab} $$ In the end, we shall use the fact from the question, that is $AB\leq 2CH$.

Hopefully, this can help. First time answering question!

Answer 3

Geometry:

Let $AB$ be given. Draw a parallel at a distance $h(=CH =(1/2) AB)$ from $AB.$

1)Consider any parallel to $AB$ with distance $h' \gt h$.

Look at triangles $ABC',$ with $C'$ on parallel $h'$.

$AC'$ intersects parallel $h$ at $C$.

Compare $\angle C$ with angle $C'.$

$\angle C' \lt \angle C$, since

$\angle C$ is an exterior angle of $\triangle ABC'$, the sum of the two non adjacent angles.

Left to do:

2) Show that on parallel $h$ $\angle C$ is maximal when $\triangle ABC$ is isosceles.

Compare $\triangle ABC''$ with $C''$ on $h$:

Consider a circle through $A,B,C$, $\triangle ABC$ is isosceles with height h$.

For any $\triangle ABC''$ with $C''$ on parallel $h$:

$AC"$ intersects the above circle in $D$.

$\angle D = \angle C$, same circle, same chord $AB$.

$\angle D$ is exterior angle to $\triangle DC"B$.

$\rightarrow:$

$\angle C" \lt \angle C$.

Finally:

In $\triangle ABC$, isosceles with height $h =(1/2) AB$:

$\tan (\angle C/2) = 1$.

$\rightarrow:$

$C=90°$, maximal value.