Cut a square into the fewest number of acute triangles.
geometry/dissections/acute.triangles New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Cut a square into the fewest number of acute triangles. By "cut", I mean draw line segments on a square and form acute triangles out of the entire square (the acute triangles fill the entire square). I saw this posted a long time ago, and the poster hinted that the solution was 8. I was only able to find 9. Can anyone show me the answer for 8? -- Jim Hsu <jimhsu@athena.mit.edu>
Show solution
Label the corners of your square ABCD. Call X the midpoint of AB. Call Y
the midpoint of CD. Draw the circles that contain AX, XB, AD, BC, CY and
YD as diameters. Pick two points U and V interior to the square but
exterior to all of the circles such that UV is parallel to AB and XY
bisects UV. The eight triangles AUX, XVB, UVX, AUD, CVY, UVY, YUD and
VBC are all acute.
The conditions on possible UVs are more general than I gave but it's
easier to demonstrate their acuteness with the above construction.
-- Robert Galejs <galejs@ll.mit.edu>
Draw a coordinate system around the square so that the
coordinates of the vertices are (0,0) (0,100) (100,0) and (100,100).
Call them A, B, C and D, respectively.
Mark the midpoints of the bottom and top sides, (50,0) and (50,100).
Call these E and F, respectively.
Mark two points inside the square, for example at (49,20) and (51,20).
Call these G and H.
Draw vectors AG, GB, EG, GF, EH, HF, CH, HD and GH.
The square is now divided into 8 acute triangles.
Triangles EGH and FGH are isosceles. It looks like triangles
AGB, AGE, CHD and CHE can be made isosceles by "adjusting" the
exact location of points G and H (I haven't done the math yet).
BFG and DFH can not be isosceles.
B---------------F---------------D
|\ | /|
| \ | / |
| \ | / |
| \ | | / |
| \ | | / |
| \ | | / |
| \ | | / |
| ___---G-H---___ |
| ___--- \ / ---___ |
A---------------E---------------C
-- Tapani Lindgren <Tapani.Lindgren@HUT.Fi>
[ Editor: further material on this puzzle, edited separately from other collected postings (batch 19); it may repeat some of the above. ]
[ Editor: problem as stated there: ]
Cut a square into the fewest number of acute triangles.
In article <CI37Ct.FKA@fc.hp.com> cfb@fc.hp.com (Charlie Brett) writes:
>I came up with:
> ***********************
> ** * * **
> * * * * * *
> * * * * * *
> * * * * * *
> * ** ** *
> * *********** *
> * ** ** *
> * * * * * *
> * * * * * *
> * * * * * *
> ** * * **
> ***********************
It looks to me like the triangles on the left and right edges of your
solution MUST have angles greater than 90 degrees (no matter how short
you make the center horizontal line). Unless you meant to cheat and use a
rectangle.
I'll follow up on my own post by apologizing (before anyone gets mad).
The above solution will work if Charlie had meant for the horizontal
line in the center to appear higher or lower:
*****************************
*** * ***
* *** 1 * * 3 *** *
* *** *2* *** *
* **** **** *
* ***** *
* * * *
* 4 ** 5 ** 6 *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * 7 * * 8 * *
* * * * * *
* * * * * *
* * * * *
** * **
*****************************
-- Jay Cotton <jay@calc.vet.uga.edu>
[ Editor: the puzzle was posted by Jim Hsu <jimhsu@athena.mit.edu>. ]
[ Editor: Jim Hsu's explanation of why eight is the minimum, sent to
the rec.puzzles oracle in August 1994 as "a submission for the
archive". ]
I recall an early article about this puzzle where the poster asked for
a proof that eight is minimal. I tried to prove this, but ended up
with only a good explanation of the answer.
"Cut a square into the fewest number of acute triangles"
(The square is composed entirely of acute triangles with no left-over
space. Acute triangles, as you know, are triangles that have angles
less than ninety degrees.)
Here are three "theorems" to help my explanation.
T1. At least one line must emerge from each corner.
| , angles a,b
|a/ a + b = 90
|/b so a,b are both acute
`----
T2. A line cannot intersect an edge by itself.
____________
a/b a + b = 180
/ but if both are a and b are acute, a + b < 180
/ thus, either a is obtuse or b is obtuse
'
T3. Any point of intersection inside the square must either be a 180
degree straight line cut by at least three acute angles or be at least
five separate acute angles. I call these "interior points".
____________ . |
a/\b \. b| c
/c \ a \,L____ all angles can easily be acute
/ \ / \ e
/ \ / d \
' ` ' `
By T1, T2, and T3, we know a single point of intersection inside the
square is impossible. Suppose it were possible. Four lines would
originate from a corner. A fifth line would be necessary to satisfy
T3. But that fifth line cannot intersect a corner, it must intersect
an edge. By T2, we know this cannot work.
,------ ------.
|\ /|
| \ / |
| ` ' | No interior point `a' can satisfy
a all of our criterions
| , . |
| / \ |
|/ \|
`------ ------'
Since a single point of intersection is out, we resort to using two
points of intersection. It can be shown that these two points must be
connected by a line. (Otherwise the two points are vertices of
different triangles with a common edge. Such a configuration will
either result in an obtuse angle, or involve more than two points of
intersection within the square.)
a a
| / \
| / \
| <----->
| \ /
| \ /
b b
this works this doesn't work
Now, we will show that both of these interior points cannot be a line
with three acute angles on one side. Suppose it were true. Then, the
interior point must be on a line connecting diagonals. (Not too hard
to show this must be true and that any interior point with a line
connecting diagonal corners and three acute angles on one side won't
give us a working solution)
,-----------------.
| /|
| ,-' |
| / | Such a configuration results in
| a,-' | an eventual violation of our
| .-' | criterions
| / |
| .-' |
|/ |
`-----------------'
So, this leaves two interior points, both of which have five or more
acute angles. Supposing it were five (we're trying to minimize the
number of triangles). This means there are five lines that intersect
at each interior point. Obviously, all four corners cannot connect to
one point. And lines from three corners to a point will cause the
problem of the previous paragraph. So, lines from two, non-diagonal
corners must connect to any interior point. With two interior points,
all four corners are accounted for.
,-----------------.
|\ /|
| `\ /' |
| `\ /' | there are two interior points `a' and `b'
| a\_____/b | that connect to two non-diagonal corners
| / \ | and are connected by a line
| /' `\ |
| /' `\ |
|/ \|
`-----------------'
We are left with two unaccounted lines on each interior point. (We
know there are a total of five. Two emerge from corners, and the third
goes to the other interior point.) Interior points and corners have
already been accounted for, so the two remaining lines can only
intersect the edges.
Because we want acute triangles, we place two points on opposing edges
and connect each edge-point to the two interior points. This gives us
a working solution. (after we shift the interior points towards one of
the edge-points to eliminate the remaining obtuse angles) All three
theorems are satisfied.
,--------+--------.
|\ | | /|
| \ | | / |
| \ | | / | there are two interior points `a' and `b'
| \ | | / | that connect to two non-diagonal corners,
| \| |/ | two opposing edges, and are connected by
| a*_____*b | a line
| .-'\ /`-. |
|.-' \ / `-.|
`--------+--------'
I tried to be brief with my answer, and had to sacrifice quite a bit of
mathematical rigor in the process. It is hoped that the reader will be
forgiving.
-- Jim Hsu <jimhsu@mit.edu> (August 1994)