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Problem 3

Let ABCDABCD be a parallelogram. Let WW, XX, YY, and ZZ be points on sides ABAB, BCBC, CDCD, and DADA, respectively, such that the incenters of triangles AWZAWZ, BXWBXW, CYXCYX, and DZYDZY form a parallelogram. Prove that WXYZWXYZ is a parallelogram.
Step 2 of 3: Monotonicity of one leg in terms of the other for fixed angle and inradius
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w+z−w2+z2−2wzcos⁡α=2rcot⁡α2=K ⟹ z=K(w−K/2)(1+cos⁡α)w−Kw+z-\sqrt{w^2+z^2-2wz\cos\alpha}=2r\cot\frac{\alpha}{2}=K\ \Longrightarrow\ z=\frac{K(w-K/2)}{(1+\cos\alpha)w-K}
Detailed analysis

In any triangle with apex angle α\alpha, adjacent sides w,zw,z, opposite side c=w2+z2−2wzcos⁡αc=\sqrt{w^2+z^2-2wz\cos\alpha}, and inradius rr, the tangent segments from the apex have length (w+z−c)/2=rcot⁡(α/2)(w+z-c)/2=r\cot(\alpha/2). Squaring w+z−K=cw+z-K=c (where K=2rcot⁡(α/2)K=2r\cot(\alpha/2)) and solving for zz expresses zz as the strictly decreasing function of ww above (since (1+cos⁡α)>0(1+\cos\alpha)>0 and K>0K>0).