So here's a nice puzzle from Donald Bell.
It's a lovely question to show that the shaded area in the first image here has area one fifth of the original square.
Using the same construction, joining vertices to midpoints, does the same thing hold true for a general quadrilateral?
How would you prove it? Do you have a counter-example?
@ColinTheMathmo Nice, I expected shearing to preserve the fraction, e.g. rhombus, but not work for the second quadrilateral. Works nicely #geogebra and other dynamic geometry software.
The closest to a counter-example I found were concave quadrilaterals and quadrilaterals with three points colinear (a triangle so a bit of a cheat)
@foldworks The question is ... what's the easiest proof that the middle area in the general quadrilateral is one fifth of the area as a whole?
I'm tempted to brute force it to verify that it's true, but a clean and elegant proof of the general case would be nice to see.
@ColinTheMathmo @GerardWestendorp I thought a simpler problem might help, i.e. https://en.wikipedia.org/wiki/Varignon%27s_theorem
However, I then found some counter-examples showing a small divergence from the desired area. I need to be sure that these aren't rounding errors
#geometry #iTeachMath #puzzle #quadrilateral #midpoint #geogebra
@foldworks So you are suggesting that the proposed theorem is false?
Interesting.
I'd be interested in reproducing the calculation by hand, just to be sure it's not rounding errors. If the result is false it should be possible to find coordinates that give "nice numbers".
@ColinTheMathmo @GerardWestendorp
OK, here’s a counterexample with mostly nice numbers.
It wasn’t easy to find because when all numbers are ‘nice’ then the proposed theorem usually holds true 😆 e.g. rhombus, parallelogram, etc.
#geometry #iTeachMath #puzzle #quadrilateral #midpoint #geogebra
#geometry #iTeachMath #puzzle #quadrilateral #midpoint #geogebra
@foldworks That's very impressive ... thank you.
I'll pass that on to the person who posed the question, who I'm sure will be equally impressed. Maybe this will persuade him finally to join Mastodon.
Cheers!
@foldworks @ColinTheMathmo @GerardWestendorp I think I've got this right, but (as always) am happy to be corrected: https://www.geogebra.org/classic/sebwvtsm
As you move the slider to zero, the shape approaches a triangle and the ratio approaches 1/6 rather than 1/5.
@icecolbeveridge Brilliant ... thank you. I'll pass this on to Donald.
Cheers!
@GerardWestendorp Forgive me for being a bit dense (I'm on holiday, which only makes it worse) -- I don't understand what's going on there.
I can see you've decomposed the quad into its nine constituent parts, and rearranged them to tessellate, but beyond that I'm a bit lost.
Colin the Mathmo
foldworks
Colin Beveridge
Gerard Westendorp