Friday Puzzler -- Cake and Box

Re: Friday Puzzler -- Cake and Box

SamForce

I would start at the base of the triangle at say angle A, make a 90 degree cut at that angle. Then measure the length of the cut and measure that length on the base and make another 90 degree cut. the 2 cut offs should now be placed at the opposing cuts to fit into the box. I hope that makes sense.

Re: Friday Puzzler -- Cake and Box

Ed in Leaside

I would think that all three cuts need to be the same length.

Re: Friday Puzzler -- Cake and Box

Alex Y

Not following your explanation, but parts of it make me think you might be onto it. Can you draw an example and take a phone picture of your drawing and submit it here?

Same length cuts

Alex Y

Correct, Ed, and well on the way to the solution!

Re: Friday Puzzler -- Cake and Box

SamForce

I have revised my thinking, the 1st cut would be a 90 degree cut at the angle opposite the long leg. the second cut would be a 45 degree angle at the newly formed angle. I have a diagram but not sure how to post it.

Re: Same length cuts

Ed in Leaside

Yes, the solution is the point isn't it?

Yes :-)

Alex Y

And you will have to reveal it if no one gets the point of your solution.

Well done!

Re: Yes :-)

Ed in Leaside

Imagine a circle of which the circumference intersects the 3 points of the triangle. Make cuts from the circle's center to the 3 points. Reassemble as required.

I, with many years in the print world, would wonder if the box could be turned inside out.

Double Credit!

Alex Y

Yes, cutting from the incenter (a new term to me -- the center of a circle inscribed in a triangle, found where the bisections of each of the triangles angles intersect) to each edge at 90* yields three interchangeable pieces, and swapping two of them gives a mirror image of the original.

And Bravo! for the idea to turn the box inside out. I ruled out the cake flipping idea by specifying that it is iced, but never thought of this idea!!

Re: Double Credit!

Ed in Leaside

I don't think the incenter is it.

When you posed the question I could visualize the geometric construction, but didn't know the terminology to describe it. So, I searched for "Center of Triangle" ... and found there is more than 1. What is needed to make the cuts is the circumcenter of the circumcircle.

Re: Double Credit!

Alex Y

Try it with an obtuse triangle. The circumcenter will be outside the triangle. An extreme one my help to illustrate the problem, e.g. a 10-120-50 triangle.

Maybe the circumcenter works, but I can't see it.

Re: Double Credit!

Ed in Leaside

All three cuts need to be the same length. When the circumcenter falls outside of the triangle it is not possible to make those cuts. Put another way: the cuts have to be made on the radius of the circumcircle.

Edit: Oops,I looked at incenter again. There are two methods that can be followed to fix the cake. From the right angle at side to the incenter works in all cases. From the points to the circumcenter fails as noted above. Points to Samforce I think.