Problem 2

Suppose

ABCD

is a square of side length

a

. Two parallel lines

\ell_1

and

\ell_2

in the plane are

a

units apart. The square is placed so that

AB

and

AD

meet

\ell_1

E

and

F

, while

CB

and

CD

meet

\ell_2

G

and

H

. If the perimeters of

\triangle AEF

and

\triangle CGH

are

m_1

and

m_2

, prove that

m_1+m_2

is constant, regardless of the placement.

Step 4 of 4: Add the two perimeters

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m_1+m_2=(OM+OQ)+(ON+OP)=MP+NQ=2a

Detailed analysis

Using the collinear orders on the two perpendicular lines, $(OM+OP)=MP$ and $(ON+OQ)=NQ$ . Hence $m_1+m_2=(OM+OQ)+(ON+OP)=MP+NQ=2a$ , which is independent of the placement.