Problem 5

Construct a triangle

ABC

AC=b

AB=c

, and

\angle AMB=\omega

with

\omega<90^\circ

, where

M

is the midpoint of

BC

. Prove that the construction has a solution if and only if

b\tan(\omega/2)\le c<b

. In what case does equality hold?

Step 5 of 6: Derive the necessary and sufficient inequality

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\frac{AB}{AX}=\tan\frac{\omega}{2}\quad\Longrightarrow\quad b\tan\frac{\omega}{2}\le c<b

Detailed analysis

The right triangle formed by $A$ , $X$ , and the circle geometry gives $AB/AX=\tan(\omega/2)$ . Thus $AX\ge b$ is equivalent to $c\ge b\tan(\omega/2)$ , while $b>c$ is the second inequality. Therefore the construction exists exactly when $b\tan(\omega/2)\le c<b$ .