Problem 12 · The grey area of the three squares
The equal marks force sides $3c$, $2c$, $c$, and two lines do the rest.
Integer answer, at most 4 digitsThe figure shows three squares. The segments marked with double ticks have the same length. What is the grey area?
Reasoned solution
Key idea: the three squares hang aligned along the top. The three equal marks (of length $t$) are: the whole right side of the small square, the lower stretch of the middle square's right side, and the lower stretch of the large square's right side.
Sizes. The small square has side $c = t$. The middle one reaches depth $2t$ (its right side has the marked stretch of length $t$ below level $t$): side $2t$. The large one, likewise: side $3t$. Since $3t + 2t + t = 24$, we get $t = 4$: sides $12$, $8$ and $4$.
The two lines. From the top-left corner: one line goes to the bottom-right corner of the large square $(12, 12)$, the other to the bottom-right corner of the small one $(24, 4)$ (slope $\tfrac{1}{6}$). The grey region lies between the two lines inside the large square, and between the shallow line and the bases inside the middle and small squares.
Areas. Triangle in the large square, with vertical base at $x = 12$ from $y = 2$ (shallow line) to $y = 12$:
Trapezoid in the middle square, between the shallow line and its base $y = 8$ (vertical sides $6$ and $\tfrac{14}{3}$, width $8$): $A_2 = \tfrac{1}{2}\left(6 + \tfrac{14}{3}\right) \cdot 8 = \tfrac{128}{3}$. Triangle in the small one, between the line and its base $y = 4$: $A_3 = \tfrac{1}{2} \cdot \tfrac{2}{3} \cdot 4 = \tfrac{4}{3}$.