We note first that there are in fact two cases: Case 1: The area formed is a triangle Then let the indicated length be $x$. We are varying this length. The base of the area overlapping can be determined as $1-x$ (the base length is 1 unit). Due to the congruency between the two triangles, it can be easily determined that the height of the area overlapping is $\dfrac {1-x}2$. The area of the triangle is then $A=\dfrac 12 bh= \dfrac14 (1-x)^2$ Then diffrentiating gets $x=1$, but verification by the second diffrential shows this is in fact the min (and this should be obvious anyway) Case 2: The area formed is a pentagon. Let the indicated length be $x$. Now, the pentagon can be split up into two trapeziums, such that the Area to be computed is now $A= \dfrac 12 h (a+b) *2=h(a+b)$. Since the two triangles are congruent, it does not take much to get $h= \dfrac{1-x}2$. By similarity, one can also show that $a=x$ and $b=\dfrac {1+x}2$. Putting this together $A= \dfrac14 (1-x)(3x+1)$ and diffrentiating yields the solution x=1/3. Substitution then shows that $A=\dfrac13$ and the second diffrential shows that this in fact the max.

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Two congruent right-angled isosceles triangles (with baselength $1$) slide on a line as on the picture. What is the maximal area of overlap?