# A rectangle, a circle, and a triangle are drawn on a plane

A rectangle, a circle and a triangle are drawn on a plane. What is the maximum possible number of points of intersection? The sides of the triangle are not collinear with any of the sides of the rectangle.

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• Welcome to Puzzling! Is this a question you came up with yourself, or is it from some other source? – Deusovi Oct 16 at 19:42
• Answers should probably separately count and sum the C-R, C-T, R-T points of intersection. They're totally separable. – smci Oct 17 at 12:29
• If only there was a lateral-thinking tag so I could draw a rectangle, a circle and a triangle on an outline of an aeroplane - even more points of intersection there methinks ;-) – Stiv Oct 17 at 17:54

Here is a proof that the answer by @Avi is the largest possible.

We have the following lemma, which is intuitively clear and also can be proved rigorously:

Lemma: Let $$l$$ be a straight line on the plane $$\Bbb R^2$$ and $$X$$ be a closed convex subset of $$\Bbb R^2$$. Then the intersection of $$l$$ and the boundary of $$X$$ contains either at most two points or infinitely many points.

Proof: Suppose $$A, B, C$$ are three different points in the intersection, such that $$B$$ lies between $$A$$ and $$C$$.
Since $$B$$ is on the boundary of $$X$$, there exists a nonzero affine function $$f:\Bbb R^2 \rightarrow \Bbb R$$ such that $$f(B) = 0$$ and $$f(x) \geq 0$$ for all $$x \in X$$.
In particular, $$f(A)$$ and $$f(C)$$ are nonnegative. But $$B$$ lies between $$A$$ and $$C$$, thus $$f(B) = 0$$ implies that $$f(A) = f(C) = 0$$.
Hence we have $$f(x) = 0$$ for all $$x$$ belonging to the line segment $$AC$$, and therefore the whole segment must be in the boundary of $$X$$, as any neighborhood of any point in the segment contains a point $$y$$ with $$f(y) < 0$$.

With the lemma, we now see that

the intersection of a triangle with a circle has at most $$6$$ points (apply the lemma with $$l =$$ one side of the triangle and $$X =$$ the circle together with its interior);

the intersection of a triangle with a square has at most $$6$$ points, unless there are two sides that coinside, in which case we get infinitely many intersection points (apply the lemma with $$l =$$ one side of the triangle and $$X =$$ the square together with its interior);

the intersection of a square with a circle has at most $$8$$ points (apply the lemma with $$l =$$ one side of the square and $$X =$$ the circle together with its interior).

and adding them together gives the maximum number.

I've seen 20, according to the following image: The idea is to first maximize the rectangle's intersections with the circle, and then maximize the triangle's intersections with the resulting figure.

• "Not collinear" does not mean "not parallel." – Daniel Mathias Oct 16 at 21:16
• True, but if it's not parallel, it can never be collinear – Avi Oct 16 at 21:19
• The answer part about "not parallel" is totally irrelevant and should be omitted. The question only specifies that the R and T can't have a collinear edge. As long as that's satisfied, it doesn't matter (for the R-T) what angle they're at. – smci Oct 17 at 12:37

Just a brute force exercise: When maximizing interesections, no intersection would be of all three objects, given that moving any of the edges by epsilon will create (at least) one more intersection. Therefore the maximum involves only intersections of two objects. This allows us to examine each pair of objects alone. Play around with a triangle and circle, and it's easy to get two intersections per triangle side, but seems impossible to get more. That'd be a total of 6. Doing the same with the triangle and rectangle, and it's again easy to get two intersections per triangle side and impossible to get more. That'd be a total of 6. Finally, rectangle and circle? Again two intersections per side is easy. That'd be a total of 8. That leads to an upper limit of 20, but it's not clear that this is achievable. A moment's trial and error shows that it is.

To prove that a circle can only intercept a line at most places, I'd say: say there is a third such intersection. It must therefore be either between the two points we already have, or to their outside. Then draw some triangles and show that any point to the inside would be closer than the radius distance to the circle center and thus couldn't be on the circle, and likewise points outside the two points we already have likewise would be too far and thus not on the circle.

I don't have a quick proof that a triangle can't intercept a square on more than two points though.

I'm getting 18 possible intersections, not sure how to add an image.

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Draw a rectangle. Use the point of the intersection of the two diagonals as centre and and draw a circle with radius of length less than the distance from the intersection of the diagonals to one of the vertices, as shown. Then draw a triangle as shown. The number of points of intersection of the rectangle, the circle and the triangle is twenty. • This doesn't add anything new to the previous answers. – bobble Oct 17 at 18:19