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 2 of 5: Locate K as the incenter of the right triangle ADC

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K=(\rho,\rho),\qquad \rho=\tfrac12\left(1+\tan2x-\sec2x\right)

Detailed analysis

Triangle $ADC$ is right-angled at $D$ with legs $AD=1$ , $DC=\tan2x$ along the two coordinate axes, and hypotenuse $AC=\sec2x$ . For a right triangle with the right angle at the origin and legs along the axes, the incenter lies at distance $\rho=\tfrac12(\text{leg}_1+\text{leg}_2-\text{hypotenuse})$ from each leg, i.e. at $(\rho,\rho)$ . Here $\rho=\tfrac12\left(1+\tan2x-\sec2x\right)$ , so $K=(\rho,\rho)$ .