Problem 3

Let

ABCD

be a parallelogram. Let

W

X

Y

, and

Z

be points on sides

AB

BC

CD

, and

DA

, respectively, such that the incenters of triangles

AWZ

BXW

CYX

, and

DZY

form a parallelogram. Prove that

WXYZ

is a parallelogram.

Step 1 of 3: Equal opposite inradii from parallel projections

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r_2-r_1=I_1I_2\sin\theta=I_3I_4\sin\theta=r_4-r_3\ \Longrightarrow\ r_1+r_4=r_2+r_3;\quad\text{similarly }r_1+r_2=r_3+r_4\ \Longrightarrow\ r_1=r_3,\ r_2=r_4

Detailed analysis

Let $I_1,I_2,I_3,I_4$ and $r_1,r_2,r_3,r_4$ be the incenters and inradii of $\triangle AWZ,\triangle BXW,\triangle CYX,\triangle DZY$ . Since $I_1,I_2$ are at distances $r_1,r_2$ from $AB$ and $I_3,I_4$ are at distances $r_3,r_4$ from the parallel side $CD$ , and $I_1I_2$ is parallel and equal to $I_4I_3$ in parallelogram $I_1I_2I_3I_4$ , projecting onto the normal to $AB$ (at acute angle $\theta$ with $I_1I_2$ ) gives $r_2-r_1=r_4-r_3$ , i.e. $r_1+r_4=r_2+r_3$ . Projecting onto the normal to $BC$ similarly gives $r_1+r_2=r_3+r_4$ . Adding and subtracting the two relations gives $r_1=r_3$ and $r_2=r_4$ .