Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer Italy

Problem:

The canvas of a rectangular painting is surrounded by a passe-partout (that is, a border) $10~\mathrm{cm}$ wide. Around the latter there is then a frame, also $10~\mathrm{cm}$ wide (in the figure, the white rectangle represents the canvas, the hatched surface the passe-partout, the black surface the frame).
It is known that the area of the entire picture (frame included) is equal to twice the sum of the areas of the passe-partout and of the canvas.

One can then conclude that:

Pick one

Solution

Solution:

The answer is (D). Indeed, let $a$ and $b$ be the dimensions of the canvas.
Those of the passe-partout are then $(a+20)$ and $(b+20)$ and those of the frame are $(a+40)$ and $(b+40)$ . The hypothesis says that $(a+40)(b+40)=2(a+20)(b+20)$ holds, from which $a b=800$ , a condition obviously equivalent to the hypothesis itself.
On the other hand, the area of the passe-partout is $(a+20)(b+20)-a b=20(a+b)+400$ and therefore it depends on the perimeter of the canvas.
The same conclusion holds for the area of the frame, which measures
$(a+40)(b+40)-(a+20)(b+20)=20(a+b)+1200$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.