Puzzle - Various Visual Thinking - week 4

Areas in the Big Square by Peter Grabarchuk

The shape consists of overlapped color circles. Which two colors have their total visible areas equal?

It is a "Gestalt Principle: Symmetry" image. So I saw it as whole of a figure. 

In my way, I draw the black line to restore the circular original. (restore the individual part). I follow the “finding” that shape out each circular figure and make the circular figure more basic.

So I can easy to see the area which is hiding by other circular figure.

For red area- it has six areas that are explicit while other three areas are hide by yellow area and two blue areas.   (+6 and -3)

For green area – there are seven explicit areas and three areas are hide by two red and one blue. (+7 and -3)

For blue area –there are seven explicit areas and three areas are hide by two yellow and one green. (+7 and -3)

For yellow area – there are eight explicit areas and two areas are hide by one blue and one red. (+8 and -2)

So I count the blue area and green area has same visible areas.

My friend’s (Kent) solution: The picture shows the fragments of circular figures to viewer. But in our brain, we still can see the whole circular figures. It is the visual memory in our brain.

So he thought each circular are perfect circulars. And he counts the parts that are hided within another circular figures.

For green area – the fourth row far left, it is only one part that was hiding by another green.  The fourth row middle and right one; it is also same as for the fourth row left one. Finally, the last one hides in the third row far left. There are three part hiding on the green and red.  So it is 1 1 1 3 = 6.

And so on,

For red area – 2 2 2 1, there are seven areas covered

For blue area – 0 2 3 1, there are six areas covered

For yellow area- 0 1 2 2, there are five areas covered.

Accounting to the data, we can know the green area and blue area have same visible areas.

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Three Bracelets by Peter Grabarchuk

In my solution, I compare the right bracelet with the bottom bracelet. The one triangle of the right bracelet is point toward a different direction than the one in the bottom bracelet. I compare the two bracelet by using the method of “matching,” but it is not the same one. I keep comparing the left bracelet and bottom bracelet. I discover the two bracelets are the same. The only different is the left bracelet is flipped. It is “visual induction II”.

So I find the left bracelet is same with the bottom bracelet.

Kent’s  solution.

He looked for the common parts within the three bracelets. First he keeps remembering the set of two triangle that point toward each other and the set of the two triangle that point toward the same direction. “Matching” another two bracelets. It can find the right bracelet is different with another two bracelets. The left bracelet and the bottom bracelet are the same one.