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From: William Elliot on 9 May 2010 08:16
On Sun, 9 May 2010 hanrahan398(a)yahoo.co.uk wrote:
> On May 9, 8:19�am, William Elliot <ma...(a)rdrop.remove.com> wrote:
>> On Sat, 8 May 2010 hanrahan...(a)yahoo.co.uk wrote:
>
Yuck, the problem and previous work have been removed.
Don't do that because, now as the details fade from memory,
my replies will be ever more missing the point,

>>>>> One of these will have area <= (1/6)* area of hexagon.
>>
>> Ok.
>>
>>>>> It will also have area no smaller than one of the "cut-off" triangles
>>>>> with which it shares a base.
>>
>> It does? �Why?
>
> I thought that was clear, but I'll explain below.
>
>> The cut off triangle is not contained in the
>> smallest of six triangles formed by the three central diagonals.
>> In fact, the height of the cut off triangle may be longer than the
>> height of the smallest of the six triangles. �How is it shown that
>> won't happen?
>
> The key to this first part is that if one cut-off triangle has a
> greater height than the smallest of the six triangles, another will
> have a smaller height.
>
> In more detail: the corner-to-opposite-corner diagonals (AD,BE,CF) of
> convex hexagon ABCDEF are concurrent at O. Draw these diagonals and
> let the smallest triangle be ABO. If cut-off triangle ABC has a
> greater height than ABO, then cut-off triangle ABF will have a smaller
> height. The corner-to-opposite-corner diagonal CF is either parallel
> to AB, in which case both cut-off triangles with AB as their base have
> the same area as ABO, so we're done. Or it's not parallel to AB, in
> which case the line containing CF meets the line containing AB. Say
> without loss of generality that it meets on the other side of A from
> B. Then ABF has a smaller height than ABO, and therefore a smaller
> area.
>
> Now we come to the case where the three corner-to-opposite-corner
> diagonals are not concurrent. I've now solved this.
>
> Again, draw AD, BE, CF. Once again we will take six triangles. They
> won't cover the entire hexagon, but they won't overlap either, so
> that's no problem - in fact it allows our inequality to be strict.

Three non-concurring lines form a triangle. Three of the triangles will
miss the inner triangle, the other three triangles will include the
inner triangle. Three triangles overlap, three others don't.

> Once we've drawn AD, BE, and CF, label points of the middle triangle
> P,Q,R so that no lines cross triangles ABR, CDP and EFQ. Now all we
> need do is go round the hexagon either clockwise or counterclockwise,
> slicing each of the quadrilaterals BCPR, DEQP, FARQ into two
> triangles. For example, we slice FARQ into triangles FRQ and FRA. Then
> we slice the other two quadrilaterals by drawing DQ and BP.
>
> Now look at the six triangles ARF, FRE, DQC, DQC, CPB, and BPA. One of
> them must have an area strictly smaller than (1/6)*area of hexagon.
> Say it's BPA. Now we've got an corner-to-opposite-corner diagonal CF
> going through the apex of that triangle, namely P. We can proceed as
> before, to show that either both cut-off triangles ABC and ABF have
> the same height as BPA and therefore the same area, or one has smaller
> height.
>
> Michael
>
From: rgr on 9 May 2010 09:49
Yes, that is the solution I came up with too after thinking about that
for some time. Nice problem!
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