Problem 4

Let

ABC

be a triangle with

AB = AC

. The angle bisectors of

\angle CAB

and

\angle ABC

meet the sides

BC

and

CA

D

and

E

, respectively. Let

K

be the incenter of triangle

ADC

. Suppose that

\angle BEK = 45^\circ

. Find all possible values of

\angle CAB

Step 1 of 5: Set up coordinates on the axis of symmetry

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In plain words

Because AB=AC, the bisector AD is also the perpendicular bisector of BC, giving a natural right-angled coordinate frame at D.

A=(0,1),\ C=(\tan2x,0),\ B=(-\tan2x,0),\quad \angle CAB=4x,\ 0<x<45^\circ

Detailed analysis

Since $AB=AC$ , the bisector $AD$ of $\angle CAB$ is perpendicular to $BC$ and $D$ is the midpoint of $BC$ ; in particular $\angle ADC=90^\circ$ . Let $\angle DAC=2x$ (so $\angle CAB=\angle BAD+\angle DAC=4x$ , with $0<x<45^\circ$ ). Place $D$ at the origin with $AD$ along the positive $y$ -axis and normalize $AD=1$ , so $A=(0,1)$ . In right triangle $ADC$ , $\tan(\angle DAC)=DC/AD$ , so $DC=\tan2x$ and $C=(\tan2x,0)$ ; by the reflection symmetry across $AD$ , $B=(-\tan2x,0)$ . Also $AC=AD\sec2x=\sec2x$ .