Divide a Square into Acute Triangles
Hareendra Yalamanchili
Is it possible to divide a square into triangles which have only acute
angles?
Hareendra Yalamanchili
hyal...@mit.edu
Matthew Daly
Yes. It is an entertaining puzzle to figure out how though.
Hint: Assume it's possible, and consider all of the vertices. Some of
those vertices are internal to the square, some lie on the edge, and
the four corners must be included to keep right angles away. Consider
how many edges have to come from each vertex.
-Matthew
--
Matthew Daly I don't buy everything I read ... I haven't
da...@ppd.kodak.com even read everything I've bought.
My opinions are not necessarily those of my employer, of course.
Ed Murphy
Hareendra Yalamanchili <hyal...@mit.edu> wrote:
> Is it possible to divide a square into triangles which have only acute
> angles?
Yes.
The diagonals divide the square into four right isosceles triangles. For each
of these, draw lines from the midpoints of the sides to points on the hypotenuse
so that the triangle is divided into two acute triangles and an irregular
pentagon. Then draw lines from a point near the middle of the pentagon to each
of its vertices so that it is divided into five acute triangles.
Obtuse triangles can be similarly divided into seven acute triangles. Thus, all
planar closed curves with straight edges can be divided into acute triangles.
--
Ed Murphy <zvy...@ix.netcom.com> "A person who is 'of sound mind'
http://www.bayside.net/users/ford/ is one who keeps the inner madman
#include <std/disclaimer.h> under lock and key." -Paul Valery
Mariza Cabral
In article <52710p$5...@sjx-ixn6.ix.netcom.com>,
Ed Murphy <zvy...@ix.netcom.com> wrote:
>Hareendra Yalamanchili <hyal...@mit.edu> wrote:
>> Is it possible to divide a square into triangles which have only acute
>> angles?
>
>Yes.
>
>The diagonals divide the square into four right isosceles triangles. For each
>of these, draw lines from the midpoints of the sides to points on the
>hypotenuse so that the triangle is divided into two acute triangles and an
>irregular pentagon. Then draw lines from a point near the middle of the
>pentagon to each of its vertices so that it is divided into five acute
>triangles.
>
>Obtuse triangles can be similarly divided into seven acute triangles. Thus,
>all planar closed curves with straight edges can be divided into acute
>triangles.
That's cool! Ah, but that makes 28 triangles, whereas *my* construction
(he he he...) has only 16 triangles:
+-----+-----+ the reason this does not look
|\ /|\ /| like a square is that the keyboard
| \ / | \ / | only allows representation of one
| /|\ | /|\ | acute angle (via symbols "/" and "\")
|/ | \|/ | \|
|\b| /|\ | /| But if angle a is 60 deg. and angle b
| \|/ | \|/ | is 82.3 deg., then this is indeed a
| / \ | / \ | square. There are other number
|/a \|/ \| combinations. And yes, you're right,
+-----+-----+ I have nothing better to do 8^).
My first idea was a construction with an infinite number of triangles,
found recursively (there's many such possibilities).
Susan Hoover
mar...@u.washington.edu (Mariza Cabral) wrote:
>In article <52710p$5...@sjx-ixn6.ix.netcom.com>,
>Ed Murphy <zvy...@ix.netcom.com> wrote:
>>Hareendra Yalamanchili <hyal...@mit.edu> wrote:
>>> Is it possible to divide a square into triangles which have only acute
>>> angles?
>>
>>Yes.
>>
>>The diagonals divide the square into four right isosceles triangles. For each
>>of these, draw lines from the midpoints of the sides to points on the
>>hypotenuse so that the triangle is divided into two acute triangles and an
>>irregular pentagon. Then draw lines from a point near the middle of the
>>pentagon to each of its vertices so that it is divided into five acute
>>triangles.
[snip]
>That's cool! Ah, but that makes 28 triangles, whereas *my* construction
>(he he he...) has only 16 triangles:
[snip]
What a coincidence! I was just this afternoon re-reading chapter 3 of
Martin Gardner's _New Mathematical Diversions from Sci Am_ (1966, originally
appeared 1959-1961), in which he poses, and answers, this very problem!
No, really -- I was reading this book at work during compiles and this
is the chapter that I had gotten to today...
Anyway, Gardner gives a solution in 8 triangles. ASCII graphics can't
do it justice, but I'll try to describe it.
A E B
J K
P Q
D H F G C
Call the square ABCD, AB being the top edge and CD the bottom edge.
Bisect AB and call the midpoint E. Bisect CD and call the midpoint F.
Bisect CF and call the midpoint G. Bisect FD and call the midpoint H.
With the compass point at G and H draw semicircles that run from a
corner (C or D) to F. Bisect AD and call the midpoint J.
Bisect BC and call the midpoint K. With the compass point at J and K
draw semicircles that run from A to D and from B to C.
Within the area bounded by the four semicircles, pick two points P and Q
such that the bilateral symmetry of the construction is maintained.
Draw AP, EP, EQ, BQ, CQ, FQ, FP, DP.
The 8 acute triangles are APD, AEP, EQP, EBQ, BCQ, CFQ, FPQ, and DPF.
--
Susan
Jianghua
I think it is impossible to do so. We can consider a equivalent question:
Is it possible to divide a obtuse or right angle triangle into triangles
which have only acute angles? When an obtuse or right angle divided
into acute angles by lines, a new obtuse or right angle will begotten.
Qiang Li
i-p...@microsoft.com
Hareendra Yalamanchili <hyal...@mit.edu> 被写入到文章
<3246D7...@mit.edu> 中...
> Is it possible to divide a square into triangles which have only acute
> angles?
>
> Hareendra Yalamanchili
> hyal...@mit.edu
>
Hareendra Yalamanchili
> >> Is it possible to divide a square into triangles which have only acute
> >> angles?
> That's cool! Ah, but that makes 28 triangles, whereas *my* construction
> (he he he...) has only 16 triangles:
Now that it is known to be possible, what is the fewest number of
triangles needed? It is less than 16.
Hareendra Yalamanchili
hyal...@mit.edu
Danh Dang
Matthew Daly <da...@PPD.Kodak.COM> wrote in article
<526i3t$m...@kodak.rdcs.Kodak.COM>...
> Hareendra Yalamanchili <hyal...@mit.edu> writes:
> >Is it possible to divide a square into triangles which have only acute
> >angles?
Wait a sec... lemme remember some high school geometry. The sum of all
angles in a triangle = 180. If all three angles of a triangle are acute,
then that condition is not sastisfied; therefore it's impossilbe to have
only acute angles.
James R Yingst
In article <527l64$1...@nntp5.u.washington.edu>,
Mariza Cabral <mar...@u.washington.edu> wrote:
>In article <52710p$5...@sjx-ixn6.ix.netcom.com>,
>Ed Murphy <zvy...@ix.netcom.com> wrote:
>>Hareendra Yalamanchili <hyal...@mit.edu> wrote:
>>> angles?
>>
>>
>>The diagonals divide the square into four right isosceles triangles. For each
>>of these, draw lines from the midpoints of the sides to points on the
>>hypotenuse so that the triangle is divided into two acute triangles and an
>>irregular pentagon. Then draw lines from a point near the middle of the
>>pentagon to each of its vertices so that it is divided into five acute
>>triangles.
>>
>>all planar closed curves with straight edges can be divided into acute
>>triangles.
>
>(he he he...) has only 16 triangles:
[deleted]
Of course, it's easy to modify Ed's method to give 14 triangles - just use
*one* diagonal to separate the square into two right isosceles triangles,
then use his method to divide each of these into seven acute triangles.
I've subsequently come up with a solution using ten acute triangles, but I
won't bother posting it as Martin Gardener's solution (thanks Susan!) uses
eight (though I haven't read it yet.) Good luck!
Hareendra Yalamanchili
> >>> Is it possible to divide a square into triangles which have only acute
> >>> angles?
SPOILER
The minimum number of triangles needed is 8.
Using the previous method of 7 triangles for an isosceles right triangle
or in fact any obtuse or right triangle, it can be done in 10
triangles. Just draw the two diagonals and then move the intersection
point so that you get three acute and one obtuse triangle. You then can
dothe obtuse with seven for a total of 10.
Here is a way of doing 8.
Let the square have coordinates A(0,0) B(1,0) C(0,1) D(1,1)
Then place two points at E(.5,0) and F(.5,1)
Then place two points at G(.401,.2) and H(.599,.2)
Connect AG CG EG FG GH BH DH EH FH
This yields all acute angles.
Note that if G and H are at .4,.2 and .6,.2
Then four right triangles are formed. Moving slightly towards .5,.2
makes these acute.
Proof that 8 is minimal
Let i be the number of interior vertices
Let e be the number of exterior vertices,other than the corners
Let n be the number of triangles.
Note that anytime line segments intersect in the interior, at least 5
must do so since the total angle measure is 360. Similarly there must
be 3 line segments at each exterior point.
Then the total angle measure in the triangles is 180n
Also the total angle measure is 360*i+180*e+4*90
So n=2i+e+2
The total number of angles in the triangles is 3n
This must be at least 5*i +3*e +4*2
So 3n=6i+3e+6>=5i+3e+8
i>=2
If i>=2 then n>=6+e
If n=6 then i=2,e=0
Then each interior point must be connected to all four corners. Connect
the first point in this way. Then the square is divided into four
regions. The second point must be in one of these regions and cannot be
connected to the corners outside that region. Note that if the second
point was right on the edge of a region, then the first point is not
connected to all four corners.
So n=6 is impossible
If n=7 Then i=2 and e=1
If an interior point connected to all four corners than the second point
would be limited to 4 connections.
So both interior points connect to 3 corners, each other and the
exterior point. Connect the first interior to three corners and
theexterior point. This divides the square into four regions. None of
these contain the exterior point and three corners, so the second
interior cannot be connected.
So n=7 is impossible
Thus n=8 isthe minimum solution.
Jon Orwant
Hareendra Yalamanchili <hyal...@mit.edu> writes:
> >>> Is it possible to divide a square into triangles which have only acute
> >>> angles?
Pictures of various solutions can be found at
http://table-saw.media.mit.edu/people/walter/squareans.html
-Jon
-----------------------------
Jon Orwant
The Perl Journal
http://work.media.mit.edu/tpj
Hareendra Yalamanchili
> How do you move the intersection point? This is a square, remember - if
> we move a diagonal, it will no longer intersect two vertices.
I am not moving a diagonal; I am moving an intersection point. Select
the new point and connect it to the four corners.
Ed Murphy
"Danh Dang" <dd...@fas.harvard.edu> wrote:
> Hareendra Yalamanchili <hyal...@mit.edu> writes:
> >Is it possible to divide a square into triangles which have only acute
> >angles?
> angles in a triangle = 180. If all three angles of a triangle are acute,
> then that condition is not sastisfied; therefore it's impossilbe to have
> only acute angles.
How do you explain a triangle with angles of 50, 60 and 70 degrees?
Ed Murphy
Hareendra Yalamanchili <hyal...@mit.edu> wrote:
> Using the previous method of 7 triangles for an isosceles right triangle
> or in fact any obtuse or right triangle, it can be done in 10
> triangles. Just draw the two diagonals and then move the intersection
> point so that you get three acute and one obtuse triangle. You then can
> dothe obtuse with seven for a total of 10.
How do you move the intersection point? This is a square, remember - if
we move a diagonal, it will no longer intersect two vertices.
--