MathLabs

Problem 4

Let ABCABC be a triangle with AB=ACAB = AC. The angle bisectors of ∠CAB\angle CAB and ∠ABC\angle ABC meet the sides BCBC and CACA at DD and EE, respectively. Let KK be the incenter of triangle ADCADC. Suppose that ∠BEK=45∘\angle BEK = 45^\circ. Find all possible values of ∠CAB\angle CAB.
Step 4 of 5: Impose the 45° condition and factor in t = tan x
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t=tan⁡x:8t2(t2−4t+1)(t2+2t−1)=0t=\tan x:\qquad 8t^2\left(t^2-4t+1\right)\left(t^2+2t-1\right)=0
Detailed analysis

The condition ∠BEK=45∘\angle BEK=45^\circ means the vectors EB⃗\vec{EB} and EK⃗\vec{EK} make a 45∘45^\circ angle, equivalently EB⃗×EK⃗=EB⃗⋅EK⃗\vec{EB}\times\vec{EK}=\vec{EB}\cdot\vec{EK} (cross product equals dot product, both computed from the coordinates of Steps 1–3). Substituting t=tan⁡xt=\tan x and writing sin⁡2x=2t1+t2\sin2x=\tfrac{2t}{1+t^2}, cos⁡2x=1−t21+t2\cos2x=\tfrac{1-t^2}{1+t^2} turns every coordinate into a rational function of tt. Clearing denominators, the equation EB⃗×EK⃗−EB⃗⋅EK⃗=0\vec{EB}\times\vec{EK}-\vec{EB}\cdot\vec{EK}=0 factors exactly as 8t2(t2−4t+1)(t2+2t−1)=08t^2\left(t^2-4t+1\right)\left(t^2+2t-1\right)=0.