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?
foldworks
in reply to Colin the Mathmo • • •Sensitive content
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)
Colin the Mathmo
in reply to foldworks • • •Sensitive content
@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.
foldworks
in reply to Colin the Mathmo • • •Sensitive content
@GerardWestendorp I thought a simpler problem might help, i.e. en.wikipedia.org/wiki/Varignon…
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
theorem that the midpoints of the sides of an arbitrary quadrilateral form a parallelogram
Contributors to Wikimedia projects (Wikimedia Foundation, Inc.)Colin the Mathmo
in reply to foldworks • • •Sensitive content
@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".
CC: @GerardWestendorp
foldworks
in reply to Colin the Mathmo • • •Sensitive content
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
Colin the Mathmo
in reply to foldworks • • •Sensitive content
@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!
CC: @GerardWestendorp
foldworks
in reply to Colin the Mathmo • • •Sensitive content
Colin Beveridge
in reply to foldworks • • •Sensitive content
@foldworks @GerardWestendorp I think I've got this right, but (as always) am happy to be corrected: 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.
GeoGebra Classic - GeoGebra
www.geogebra.orgColin the Mathmo
in reply to Colin Beveridge • • •Sensitive content
@icecolbeveridge Brilliant ... thank you. I'll pass this on to Donald.
Cheers!
CC: @foldworks @GerardWestendorp
Colin Beveridge
in reply to Colin the Mathmo • • •Sensitive content
Gerard Westendorp
in reply to Colin the Mathmo • • •Gerard Westendorp
in reply to Colin the Mathmo • • •Sensitive content
Andrew Stacey (he/him)
in reply to Colin the Mathmo • • •Sensitive content
This was a nice nerd-snipe!
Areas behave nicely under affine transformations, and in particular the ratio of the areas of two regions is preserved, so we may as well assume that three of the vertices of the quadrilateral are at nice coordinates. I chose \((0,0)\), \((2,0)\), and \((0,2)\) (the choice of \(2\) was so that the midpoints were still integers). So then the fourth point is at, say \((a,b)\).
The quadrilateral then has area \(a + b\) (which took me by surprise when I saw it!)
Playing with this in Geogebra shows that the ratio of \(5\), to within the display precision, is surprisingly persistent. But it doesn't take much experimenting to get into the region where it is obviously false.
As a counter-example, with the fourth vertex at \((20,2)\) then I get the ratio to be \(5.33\). But even by \((100,2)\) then it only gets up to about \(5.7\).
I have formulae for the intersection points. They're okay, but not the nicest formulae I've ever seen. So in theory I could give you a formula for the area of the smaller quadrilateral - I just haven't quite got the energy to put it all together just yet! If I knew how to use a symbolic algebra system then it would be a bit easier.