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 6 of 6: Characterize equality

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c=b\tan\frac{\omega}{2}\quad\Longleftrightarrow\quad AC=AX

Detailed analysis

Equality in the lower bound occurs exactly when $AX=AC=b$ . The two circles are then tangent at the unique point $C=X$ , so the construction has one (coincident) solution rather than two. The strict upper bound $c<b$ cannot be an equality for an intersection on the required major arc.