How to slice it

Geometry Level 1

Six circles, each of area 1, are placed in the following manner:

Does there exist a straight line which divides the figure into 2 parts, such that the red, blue and green areas on each side are all equal to 1?

Assumption: The black border around the circles is of zero width, and the circles abut right up against each other.

Inspiration

Yes No

1 solution

Geoff Pilling
Feb 21, 2017

Consider this figure:

By symmetry, any line with an equal area of blue on each side would need to pass through point A.

Or to be a bit more clear...

Take a line passing through A with equal areas of blue on each side (like the example given in the solution). The line can be rotated without losing the property, which would maintain a symmetrical figure, but can't be translated, because symmetry would be lost.

Similarly, the line would have to pass through point B in order to have an equal red area on each side.

However, as you can see, this line cuts off a bit of the lower green circle, having unequal areas of green on each side. Therefore, $\boxed{\text{no}}$ you can't draw a single line that has equal areas of each color on either side.

Moderator note:

Not every symmetric figure has such a point A, so this isn't completely rigorous. Can you find a figure that is centrally symmetric about P, but there is a line which splits the area into 2 but doesn’t pass through P?

The crux of the solution is to show that "A line divides the blue area into equal parts if and only if it passes through point A”, with an outline something like:

Any line through $A$ divides the blue area into 2 equal halves.
Suppose that there is a line $l$ that divides the blue area into 2 equal halves. Consider line $m$ that is parallel to $l$ and passes through $A.$ Then, there is no blue area between line $l$ and $m.$
Convince yourself that this is not possible for any "strip" that contains/touches $A$ to have no blue area. Hence we have a contradiction.

Could you explain why "any line with an equal area of blue on each side would need to pass through point A"?

I understand that any line passing through A will have equal blue on each side of it (because of rotational symmetry about A). Why is the converse true?

Pranshu Gaba - 4 years, 3 months ago

If any line that passes through A has an equal area of blue on each side, then any line parallel to one of those lines (which is all possible lines that don't pass through A) would necessarily have an unbalanced area... Does that make sense?

Otherwise, lemme think about a cleaner way to explain it...

Geoff Pilling - 4 years, 3 months ago

Yup, this makes sense. Displacing the line will reduce the amount of blue on one side and increase on the other side. So if the line does not pass through A, then it cannot divide the blue in two parts with equal areas.

Pranshu Gaba - 4 years, 3 months ago

@Pranshu Gaba – Here's how I would express it (which is similar to what you presented):

Clearly, any line through $A$ divides the blue area into 2.
Suppose that there is a line $l$ that divides the blue area into 2. Consider line $l_A$ that is parallel to $l$ and passes through $A$ . Then, there is no blue area between line $l$ and $l_A$ .
Convince yourself that this is not possible for any "strip" that contains/touches $A$ to have no blue area. Hence we have a contradiction.

Calvin Lin Staff - 4 years, 3 months ago

Stupid of me to overlook that fact.

Peter van der Linden - 4 years, 3 months ago

Good point... I've added the clarification to my solution.

Geoff Pilling - 4 years, 3 months ago

I believe the answer depends on how thick the black border around each circle is. We were not told to assume it had zero width.

Anthony Cutler - 4 years, 3 months ago

Good point... There I've clarified the question.

Geoff Pilling - 4 years, 3 months ago

Why would that matter? Even with non-zero width, there's still symmetry. And there's still a bit of green on the wrong side of AB. Unless you imply that the black border can reasonably be larger than 10% of the whole disc?

C . - 2 years, 9 months ago

Obviously, but this is kind of dumb..

Matthew Kieffer - 4 years, 3 months ago

Not really. The beauty of the solution is that the obvious fact can be used to prove a non-obvious idea.

Agnishom Chattopadhyay - 4 years, 3 months ago

How to slice it

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