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From: hanrahan398 on 8 May 2010 04:15
This is a problem from Engel's book 'Problem Solving Strategies', chap
4, prob 22.

Show that in any convex hexagon, there is a diagonal which cuts off a
triangle with area <= (1/6)* area of hexagon.

I've done the first part. Engel says the second part is easier, but
I'm stuck on it!

OK take the case where the three corner-to-opposite-corner diagonals
are concurrent. Draw these diagonals, giving 6 triangles. One of these
will have area <= (1/6)* area of hexagon. It will also have area no
smaller than one of the "cut-off" triangles with which it shares a
base. This cut-off triangle therefore has area as required.

Now consider the case where the three corner-to-opposite-corner
diagonals are not concurrent. Where do we go from here?

Thanks!

Michael
From: William Elliot on 8 May 2010 06:27
On Sat, 8 May 2010 hanrahan398(a)yahoo.co.uk wrote:

> This is a problem from Engel's book 'Problem Solving Strategies', chap
> 4, prob 22.
>
> Show that in any convex hexagon, there is a diagonal which cuts off a
> triangle with area <= (1/6)* area of hexagon.
>
> I've done the first part.

What first part?

> Engel says the second part is easier, but I'm stuck on it!
>
What second part?

> OK take the case where the three corner-to-opposite-corner diagonals are
> concurrent. Draw these diagonals, giving 6 triangles.

Yes, each corner has three concurrent diagonals to the opposite corners.

What six triangles? I see four and only two of
them are a triangle cut off by a single diagonal.

> One of these will have area <= (1/6)* area of hexagon.

One of those will have an area <= 1/4 * area of hexagon.

> It will also have area no smaller than one of the "cut-off" triangles
> with which it shares a base.

I doubt it.

> This cut-off triangle therefore has area as required.
>
It does?

> Now consider the case where the three corner-to-opposite-corner
> diagonals are not concurrent. Where do we go from here?

Nowhere.
From: hanrahan398 on 8 May 2010 09:14
On May 8, 11:27 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> On Sat, 8 May 2010 hanrahan...(a)yahoo.co.uk wrote:
>
> >This is a problem from Engel's book 'Problem Solving Strategies',
> >chap 4, prob 22.
>
> > Show that in any convex hexagon, there is a diagonal which cuts off
> > a triangle with area <= (1/6)* area of hexagon.
>
> > I've done the first part.
>
> What first part?

The first case, where the corner-to-opposite-corner diagonals are
concurrent.

> > Engel says the second part is easier, but I'm stuck on it!
>
> What second part?

The other case, where they aren't.

> > OK take the case where the three corner-to-opposite-corner
> >diagonals are concurrent. Draw these diagonals, giving 6 triangles.
>
> Yes, each corner has three concurrent diagonals to the opposite
> corners.

I said the corner-to-opposite-corner diagonals. A hexagon has three of
these, and they may or may not be concurrent. First case is where they
are; second case is where they aren't.

> What six triangles?  I see four and only two of
> them are a triangle cut off by a single diagonal.

If the three corner-to-opposite-corner diagonals are concurrent, then
drawing them gives six triangles.

A cut-off triangle, as I defined it, is a triangle cut off by a
diagonal which goes from one vertex to the next-but-one vertex. That's
the only kind of diagonal that cuts off a triangle inside a convex
hexagon.

The corner-to-opposite-corner diagonals are drawn to prove the
statement about the existence of a cut-off triangle.

> > One of these will have area <= (1/6)* area of hexagon.
>
> One of those will have an area <= 1/4 * area of hexagon.
>
> > It will also have area no smaller than one of the "cut-off" triangles
> > with which it shares a base.
>
> I doubt it.
>
> > This cut-off triangle therefore has area as required.
>
> It does?
>
> > Now consider the case where the three corner-to-opposite-corner
> > diagonals are not concurrent. Where do we go from here?
>
> Nowhere.

Please either make a little effort to understand my post, or don't
bother posting to this thread.
From: William Elliot on 9 May 2010 03:19
On Sat, 8 May 2010 hanrahan398(a)yahoo.co.uk wrote:

> On May 8, 11:27�am, William Elliot <ma...(a)rdrop.remove.com> wrote:
>> On Sat, 8 May 2010 hanrahan...(a)yahoo.co.uk wrote:
>>
>>> This is a problem from Engel's book 'Problem Solving Strategies',
>>> chap 4, prob 22.
>>
>>> Show that in any convex hexagon, there is a diagonal which cuts off
>>> a triangle with area <= (1/6)* area of hexagon.
>>
>>> I've done the first part.
>> What first part?
>
> The first case, where the corner-to-opposite-corner diagonals are
> concurrent.

>>> Engel says the second part is easier, but I'm stuck on it!
>>
>> What second part?
> The other case, where they aren't.
>
>>> OK take the case where the three corner-to-opposite-corner
>>> diagonals are concurrent. Draw these diagonals, giving 6 triangles.
>>
>> Yes, each corner has three concurrent diagonals to the opposite
>> corners.
>
> I said the corner-to-opposite-corner diagonals. A hexagon has three of
> these, and they may or may not be concurrent. First case is where they
> are; second case is where they aren't.
>
> If the three corner-to-opposite-corner diagonals are concurrent, then
> drawing them gives six triangles.
>
> A cut-off triangle, as I defined it, is a triangle cut off by a
> diagonal which goes from one vertex to the next-but-one vertex. That's
> the only kind of diagonal that cuts off a triangle inside a convex
> hexagon.
>
> The corner-to-opposite-corner diagonals are drawn to prove the
> statement about the existence of a cut-off triangle.
>
>>> 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? 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?

>>> This cut-off triangle therefore has area as required.
>> It does?
>>
>>> Now consider the case where the three corner-to-opposite-corner
>>> diagonals are not concurrent. Where do we go from here?
>
> Please either make a little effort to understand my post, or don't
> bother posting to this thread.
>
If abcdef are the consecutive points of a convex hexagon, then
a is a corner and the opposite corner diagonals are ac, ad, ae.

As I got chewed out using this definition of an opposite corner diagonal,
I'm put at the disadvantage of deducing another definition of opposite
corner diagonal.

If abcdef are the consecutive points of a convex hexagon, then
then a is a corner and the opposite corner diagonal is ad.

Without a little effort to assure the readers know the definitions
and needed details of the problem and hints, you can expect that
you may be misunderstood nor would you be correct were you to
think everybody in this group has a nearby copy of Engles.

From: hanrahan398 on 9 May 2010 05:59
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:

> > On May 8, 11:27 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> >> On Sat, 8 May 2010 hanrahan...(a)yahoo.co.uk wrote:

> >>> 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.
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
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